[Solution Manual] Shigley's Mechanical Engineering Design 7th com

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Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5

Impending motion to left E

1

1

f

f

A

B

D

G Fcr ␪cr

C

F

␪

Facc

Consider force F at G, reactions at B and D. Extend lines of action for fully-developed friction D E and B E to find the point of concurrency at E for impending motion to the left. The critical angle is θcr . Resolve force F into components Facc and Fcr . Facc is related to mass and acceleration. Pin accelerates to left for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin. Impending motion to right E⬘

1 f

⭈E

1 f B

A

G

d

⬘ Fcr D

C

⬘ ␪cr

F⬘ ␪⬘ F⬘acc

Consider force F at G, reactions at A and C. Extend lines of action for fully-developed friction AE and C E to find the point of concurrency at E for impending motion to the left. The and F . F is related to mass critical angle is θcr . Resolve force F into components Facc acc cr and acceleration. Pin accelerates to right for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin. The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches. The graphic approach accomplishes this quickly. It is important to point out that this understanding enables a mathematical model to be constructed, and that there are two of them. This is the simplest problem in mechanical engineering. Using it is a good way to begin a course. What is the role of pin diameter d? Yes, changing the sense of F changes the response.

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2

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

1-6

(a)

y

F

T r

␪ N

Fy = −F − f N cos θ + N sin θ = 0

(1)

T =0 r Ans.

Fx = f N sin θ + N cos θ − F = N (sin θ − f cos θ) T = Nr( f sin θ + cos θ)

x

fN

Combining T = Fr

1 + f tan θ = KFr tan θ − f

(b) If T → ∞ detent self-locking tan θ − f = 0 (Friction is fully developed.) Check: If

F = 10 lbf, N=

Ans.

∴ θcr = tan−1 f

f = 0.20, θ = 45◦ ,

(2) Ans.

r = 2 in

10 = 17.68 lbf −0.20 cos 45◦ + sin 45◦

T = 17.28(0.20 sin 45◦ + cos 45◦ ) = 15 lbf r f N = 0.20(17.28) = 3.54 lbf θcr = tan−1 f = tan−1 (0.20) = 11.31◦ 11.31° < θ < 90° 1-7 (a) F = F0 + k(0) = F0 T1 = F0r Ans. (b) When teeth are about to clear F = F0 + kx2 From Prob. 1-6 T2 = Fr T2 = r

f tan θ + 1 tan θ − f

( F0 + kx2 )( f tan θ + 1) tan θ − f

Ans.

1-8 Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦ , f = 0.25, xi = 0, x f = 0.2 Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.

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3

Chapter 1

From Eq. (1) of Prob. 1-6 N=

F − f cos θ + sin θ 10 = 13.49 lbf Ni = −0.25 cos 60◦ + sin 60◦

Ans.

10.5 13.49 = 14.17 lbf Ans. 10

Nf = From Eq. (2) of Prob. 1-6 K =

1 + f tan θ 1 + 0.25 tan 60◦ = = 0.967 Ans. tan θ − f tan 60◦ − 0.25

Ti = 0.967(10)(2) = 19.33 lbf · in Tf = 0.967(10.5)(2) = 20.31 lbf · in 1-9 (a) Point vehicles v x

Q=

v 42.1v − v 2 cars = = hour x 0.324

Seek stationary point maximum 42.1 − 2v dQ =0= ∴ v* = 21.05 mph dv 0.324 Q* =

42.1(21.05) − 21.052 = 1367.6 cars/h Ans. 0.324

(b)

v l 2

v Q= = x +l

x

l 2

l 0.324 + v(42.1) − v 2 v

−1

Maximize Q with l = 10/5280 mi v 22.18 22.19 22.20 22.21 22.22 % loss of throughput

Q 1221.431 1221.433 1221.435 ← 1221.435 1221.434 1368 − 1221 = 12% 1221

Ans.

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4


22.2 − 21.05 = 5.5% 21.05 Modest change in optimal speed Ans.

(c) % increase in speed

1-10

This and the following problem may be the student’s first experience with a figure of merit. • Formulate fom to reflect larger figure of merit for larger merit. • Use a maximization optimization algorithm. When one gets into computer implementation and answers are not known, minimizing instead of maximizing is the largest error one can make. FV = F1 sin θ − W = 0 FH = −F1 cos θ − F2 = 0 From which F1 = W/sin θ F2 = −W cos θ/sin θ fom = −S = −¢γ (volume) . = −¢γ(l1 A1 + l2 A2 ) W F1 l1 = , l2 = A1 = S S sin θ cos θ F2 W cos θ A2 = = S S sin θ l2 W cos θ l2 W + fom = −¢γ cos θ S sin θ S sin θ −¢γ W l2 1 + cos2 θ = S cos θ sin θ Set leading constant to unity θ◦

fom

0 20 30 40 45 50 54.736 60

−∞ −5.86 −4.04 −3.22 −3.00 −2.87 −2.828 −2.886

θ* = 54.736◦ fom* = −2.828

Ans.

Alternative: d 1 + cos2 θ =0 dθ cos θ sin θ And solve resulting transcendental for θ*.

Check second derivative to see if a maximum, minimum, or point of inflection has been found. Or, evaluate fom on either side of θ*.

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Chapter 1

1-11 (a) x1 + x2 = X 1 + e1 + X 2 + e2 error = e = (x1 + x2 ) − ( X 1 + X 2 ) = e1 + e2 Ans. (b) x1 − x2 = X 1 + e1 − ( X 2 + e2 ) e = (x1 − x2 ) − ( X 1 − X 2 ) = e1 − e2 Ans. (c) x1 x2 = ( X 1 + e1 )( X 2 + e2 ) e = x1 x2 − X 1 X 2 = X 1 e2 + X 2 e1 + e1 e2 e2 e1 . Ans. + = X 1 e2 + X 2 e1 = X 1 X 2 X1 X2 x1 X 1 + e1 X 1 1 + e1 / X 1 = = (d) x2 X 2 + e2 X 2 1 + e2 / X 2 −1 e2 e2 e1 e2 . e1 e2 . 1+ =1− and 1+ 1− =1+ − X2 X2 X1 X2 X1 X2 x1 X 1 . X 1 e1 e2 e= − − = Ans. x2 X2 X2 X1 X2 1-12 (a)

x1 =

√ 5 = 2.236 067 977 5

X 1 = 2.23 3-correct digits √ x2 = 6 = 2.449 487 742 78 X 2 = 2.44 3-correct digits √ √ x1 + x2 = 5 + 6 = 4.685 557 720 28 √ e1 = x1 − X 1 = 5 − 2.23 = 0.006 067 977 5 √ e2 = x2 − X 2 = 6 − 2.44 = 0.009 489 742 78 √ √ e = e1 + e2 = 5 − 2.23 + 6 − 2.44 = 0.015 557 720 28 Sum = x1 + x2 = X 1 + X 2 + e = 2.23 + 2.44 + 0.015 557 720 28 = 4.685 557 720 28 (Checks) Ans. (b) X 1 = 2.24, X 2 = 2.45 √ e1 = 5 − 2.24 = −0.003 932 022 50 √ e2 = 6 − 2.45 = −0.000 510 257 22 e = e1 + e2 = −0.004 442 279 72 Sum = X 1 + X 2 + e = 2.24 + 2.45 + (−0.004 442 279 72) = 4.685 557 720 28 Ans.

5

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6


1-13 (a) (b) (c) (d) (e) (f) (g) (h) (i)

σ = 20(6.89) = 137.8 MPa F = 350(4.45) = 1558 N = 1.558 kN M = 1200 lbf · in (0.113) = 135.6 N · m A = 2.4(645) = 1548 mm2 I = 17.4 in4 (2.54) 4 = 724.2 cm4 A = 3.6(1.610) 2 = 9.332 km2 E = 21(1000)(6.89) = 144.69(103 ) MPa = 144.7 GPa v = 45 mi/h (1.61) = 72.45 km/h V = 60 in3 (2.54) 3 = 983.2 cm3 = 0.983 liter

(a) (b) (c) (d) (e) (f) (g) (h) (i)

l = 1.5/0.305 = 4.918 ft = 59.02 in σ = 600/6.89 = 86.96 kpsi p = 160/6.89 = 23.22 psi Z = 1.84(105 )/(25.4) 3 = 11.23 in3 w = 38.1/175 = 0.218 lbf/in δ = 0.05/25.4 = 0.00197 in v = 6.12/0.0051 = 1200 ft/min = 0.0021 in/in V = 30/(0.254) 3 = 1831 in3

1-14

1-15 200 = 13.1 MPa 15.3 42(103 ) = 70(106 ) N/m2 = 70 MPa (b) σ = 6(10−2 ) 2

(a) σ =

(c) y =

1200(800) 3 (10−3 ) 3 = 1.546(10−2 ) m = 15.5 mm 3(207)(6.4)(109 )(10−2 ) 4

(d) θ =

1100(250)(10−3 ) = 9.043(10−2 ) rad = 5.18◦ 4 9 − 3 4 79.3(π/32)(25) (10 )(10 )

1-16 600 = 5 MPa 20(6) 1 (b) I = 8(24) 3 = 9216 mm4 12 π (c) I = 324 (10−1 ) 4 = 5.147 cm4 64 16(16) = 5.215(106 ) N/m2 = 5.215 MPa (d) τ = π(253 )(10−3 ) 3 (a) σ =

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Chapter 1

1-17 (a) τ =

120(103 ) = 382 MPa (π/4)(202 )

32(800)(800)(10−3 ) = 198.9(106 ) N/m2 = 198.9 MPa π(32) 3 (10−3 ) 3 π (364 − 264 ) = 3334 mm3 (c) Z = 32(36) (b) σ =

(d) k =

(1.6) 4 (79.3)(10−3 ) 4 (109 ) = 286.8 N/m 8(19.2) 3 (32)(10−3 ) 3

7

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Chapter 2 2-1 (a)

12 10 8 6 4 2 0

60

70

80

90 100 110 120 130 140 150 160 170 180 190 200 210

(b) f/(Nx) = f/(69 · 10) = f/690

Eq. (2-9) Eq. (2-10)

x

f

fx

f x2

f/(Nx)

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69

120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 8480

7200 4900 19 200 40 500 80 000 145 200 86 400 169 000 156 800 112 500 51 200 86 700 64 800 36 100 0 44 100 1 104 600

0.0029 0.0015 0.0043 0.0072 0.0116 0.0174 0.0087 0.0145 0.0116 0.0174 0.0029 0.0043 0.0029 0.0015 0 0.0015

8480 = 122.9 kcycles 69 1/2 1 104 600 − 84802 /69 sx = 69 − 1 x¯ =

= 30.3 kcycles

Ans.

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9

Chapter 2

2-2 Data represents a 7-class histogram with N = 197. x

f

fx

f x2

174 182 190 198 206 214 220

6 9 44 67 53 12 6 197

1044 1638 8360 13 266 10 918 2568 1320 39 114

181 656 298 116 1 588 400 2 626 688 2 249 108 549 552 290 400 7 789 900

39 114 = 198.55 kpsi Ans. 197 1/2 7 783 900 − 39 1142 /197 sx = 197 − 1 x¯ =

= 9.55 kpsi Ans. 2-3 Form a table: x

f

fx

f x2

64 68 72 76 80 84 88 92

2 6 6 9 19 10 4 2 58

128 408 432 684 1520 840 352 184 4548

8192 27 744 31 104 51 984 121 600 70 560 30 976 16 928 359 088

4548 = 78.4 kpsi 58 1/2 359 088 − 45482 /58 sx = = 6.57 kpsi 58 − 1 x¯ =

From Eq. (2-14)

1 x − 78.4 2 1 f (x) = √ exp − 2 6.57 6.57 2π

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2-4 (a) y

f

fy

f y2

y

f/(Nw)

f (y)

g(y)

5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125

1 0 0 3 3 6 14 15 10 2 1 55

5.625 0 0 19.125 19.875 41.25 99.75 110.625 76.25 15.75 8.125 396.375

31.640 63 0 0 121.9219 131.6719 283.5938 710.7188 815.8594 581.4063 124.0313 66.015 63 2866.859

5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125

0.072 727 0 0 0.218 182 0.218 182 0.436 364 1.018 182 1.090 909 0.727 273 0.145 455 0.072 727

0.001 262 0.008 586 0.042 038 0.148 106 0.375 493 0.685 057 0.899 389 0.849 697 0.577 665 0.282 608 0.099 492

0.000 295 0.004 088 0.031 194 0.140 262 0.393 667 0.725 002 0.915 128 0.822 462 0.544 251 0.273 138 0.106 72

For a normal distribution,

2866.859 − (396.3752 /55) y¯ = 396.375/55 = 7.207, sy = 55 − 1 1 x − 7.207 2 1 f ( y) = √ exp − 2 0.4358 0.4358 2π

1/2 = 0.4358

For a lognormal distribution, √ √ x¯ = ln 7.206 818 − ln 1 + 0.060 4742 = 1.9732, sx = ln 1 + 0.060 4742 = 0.0604 1 ln x − 1.9732 2 1 exp − g( y) = √ 2 0.0604 x(0.0604)( 2π) (b) Histogram f 1.2

Data N LN

1 0.8 0.6 0.4 0.2 0 5.63

5.88

6.13

6.38

6.63

6.88 7.13 log N

7.38

7.63

7.88

8.13

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Chapter 2

11

2-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000, b = 0.5008 in. (a) Eq. (2-22)

µx =

0.5000 + 0.5008 a+b = = 0.5004 2 2

Eq. (2-23)

σx =

0.5008 − 0.5000 b−a = 0.000 231 √ = √ 2 3 2 3

(b) PDF from Eq. (2-20) f (x) =

1250 0.5000 ≤ x ≤ 0.5008 in 0 otherwise

(c) CDF from Eq. (2-21)  x < 0.5000 0 F(x) = (x − 0.5)/0.0008 0.5000 ≤ x ≤ 0.5008  1 x > 0.5008 If all smaller diameters are removed by inspection, a = 0.5002, b = 0.5008 µx =

0.5002 + 0.5008 = 0.5005 in 2

0.5008 − 0.5002 = 0.000 173 in √ 2 3 1666.7 0.5002 ≤ x ≤ 0.5008 f (x) = 0 otherwise σˆ x =

 x < 0.5002 0 F(x) = 1666.7(x − 0.5002) 0.5002 ≤ x ≤ 0.5008  1 x > 0.5008 2-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution is uniform. From Eqs. (2-22) and (2-23), √ √ a = µx − 3s = 0.6241 − 3(0.000 581) = 0.6231 in √ √ b = µx + 3s = 0.6241 + 3(0.000 581) = 0.6251 in We suspect the dimension was

0.623 in 0.625

Ans.

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2-7 F(x) = 0.555x − 33 mm (a) Since F(x) is linear, the distribution is uniform at x = a F(a) = 0 = 0.555(a) − 33 ∴ a = 59.46 mm. Therefore, at x = b F(b) = 1 = 0.555b − 33 ∴ b = 61.26 mm. Therefore,  x < 59.46 mm 0 F(x) = 0.555x − 33 59.46 ≤ x ≤ 61.26 mm  1 x > 61.26 mm The PDF is d F/dx , thus the range numbers are: 0.555 59.46 ≤ x ≤ 61.26 mm f (x) = 0 otherwise

Ans.

From the range numbers, µx =

59.46 + 61.26 = 60.36 mm Ans. 2

61.26 − 59.46 = 0.520 mm Ans. σˆ x = √ 2 3

1

(b) σ is an uncorrelated quotient F¯ = 3600 lbf, A¯ = 0.112 in2 C F = 300/3600 = 0.083 33,

C A = 0.001/0.112 = 0.008 929

From Table 2-6, for σ µF 3600 = 32 143 psi Ans. = µA 0.112 1/2 (0.083332 + 0.0089292 ) σˆ σ = 32 143 = 2694 psi Ans. (1 + 0.0089292 ) σ¯ =

Cσ = 2694/32 143 = 0.0838 Ans. Since F and A are lognormal, division is closed and σ is lognormal too. σ = LN(32 143, 2694) psi Ans.

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13

Chapter 2

2-8 Cramer’s rule

y x 2 x y x 3 yx 3 − x yx 2 = a1 = Ans. x x 2 xx 3 − (x 2 ) 2 x 2 x 3 y x 2 xx y − yx 2 x x y = a2 = Ans. x x 2 xx 3 − (x 2 ) 2 x 2 x 3 x

y

x2

x3

xy

0 0.2 0.4 0.6 0.8 1.0 3.0

0.01 0.15 0.25 0.25 0.17 −0.01 0.82

0 0.04 0.16 0.36 0.64 1.00 2.20

0 0.008 0.064 0.216 0.512 1.000 1.800

0 0.030 0.100 0.150 0.136 −0.010 0.406

a1 = 1.040 714

a2 = −1.046 43

Ans.

x

Data y

Regression y

0 0.2 0.4 0.6 0.8 1.0

0.01 0.15 0.25 0.25 0.17 −0.01

0 0.166 286 0.248 857 0.247 714 0.162 857 −0.005 71

y 0.3

Data Regression

0.25 0.2 0.15 0.1 0.05 0 0.05

x 0

0.2

0.4

0.6

0.8

1

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2-9 Su 0 60 64 65 82 101 119 120 130 134 145 180 195 205 207 210 213 225 225 227 230 238 242 265 280 295 325 325 355 5462

Data Se 30 48 29.5 45 51 50 48 67 60 64 84 78 96 87 87 75 99 87 116 105 109 106 105 96 99 114 117 122 2274.5

m = 0.312 067 Se 140

Regression Se 20.356 75 39.080 78 40.329 05 40.641 12 45.946 26 51.875 54 57.492 75 57.804 81 60.925 48 62.173 75 65.606 49 76.528 84 81.209 85 84.330 52 84.954 66 85.890 86 86.827 06 90.571 87 90.571 87 91.196 92.132 2 94.628 74 95.877 01 103.054 6 107.735 6 112.416 6 121.778 6 121.778 6 131.140 6

Su2

Su Se

3 600 4 096 4 225 6 724 10 201 14 161 14 400 16 900 17 956 21 025 32 400 38 025 42 025 42 849 44 100 45 369 50 625 50 625 51 529 52 900 56 644 58 564 70 225 78 400 87 025 105 625 105 625 126 025 1 251 868

1 800 3 072 1 917.5 3 690 5 151 5 950 5 760 8 710 8 040 9 280 15 120 15 210 19 680 18 009 18 270 15 975 22 275 19 575 26 332 24 150 25 942 25 652 27 825 26 880 29 205 37 050 38 025 43 310 501 855.5

b = 20.356 75

Ans.

Data Regression

120 100 80 60 40 20 0

0

100

200

300

400

Su

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15

Chapter 2

2-10

E

=

y − a0 − a2 x 2

2

∂E = −2 y − a0 − a2 x 2 = 0 ∂a0 y = na0 + a2 x2 = 0 ⇒ x2 y − na0 − a2 ∂E =2 y − a0 − a2 x 2 (2x) = 0 ⇒ x + a2 x3 x y = a0 ∂a2 Cramer’s rule

y x y a0 = n x n x a2 = n x

x 2 x 3 x 3 y − x 2 x y = x 2 nx 3 − xx 2 3 x y nx y − xy x y 2 = x nx 3 − xx 2 x 3

x

Data y

Regression y

x2

20 40 60 80 200

19 17 13 7 56

19.2 16.8 12.8 7.2

400 1600 3600 6400 12 000

x3

xy

8 000 380 64 000 680 216 000 780 512 000 560 800 000 2400

a0 =

800 000(56) − 12 000(2400) = 20 4(800 000) − 200(12 000)

a2 =

4(2400) − 200(56) = −0.002 4(800 000) − 200(12 000)

y 25

Data Regression

20

15

10

5

0

x 0

20

40

60

80

100

Ans.

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2-11

x

Data y

Regression y

0.2 0.4 0.6 0.8 1 2 5

7.1 10.3 12.1 13.8 16.2 25.2 84.7

7.931 803 9.884 918 11.838 032 13.791 147 15.744 262 25.509 836

x2

y2

xy

x − x¯

(x − x) ¯ 2

0.04 0.16 0.36 0.64 1.00 4.00 6.2

50.41 106.09 146.41 190.44 262.44 635.04 1390.83

1.42 4.12 7.26 11.04 16.20 50.40 90.44

−0.633 333 −0.433 333 −0.233 333 −0.033 333 0.166 666 1.166 666 0

0.401 111 111 0.187 777 778 0.054 444 444 0.001 111 111 0.027 777 778 1.361 111 111 2.033 333 333

mˆ = k =

6(90.44) − 5(84.7) = 9.7656 6(6.2) − (5) 2

84.7 − 9.7656(5) = 5.9787 bˆ = Fi = 6 F 30

Data Regression

25 20 15 10 5 0

x 0

0.5

5 x¯ = ; 6

(a)

1

y¯ =

1.5

2

2.5

84.7 = 14.117 6

Eq. (2-37) s yx =

1390.83 − 5.9787(84.7) − 9.7656(90.44) 6−2

= 0.556 Eq. (2-36) sbˆ = 0.556

1 (5/6) 2 + = 0.3964 lbf 6 2.0333

Fi = (5.9787, 0.3964) lbf

Ans.

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17

Chapter 2

(b) Eq. (2-35) 0.556 = 0.3899 lbf/in smˆ = √ 2.0333 k = (9.7656, 0.3899) lbf/in 2-12

Ans.

The expression = δ/l is of the form x/y. Now δ = (0.0015, 0.000 092) in, unspecified distribution; l = (2.000, 0.0081) in, unspecified distribution; C x = 0.000 092/0.0015 = 0.0613 C y = 0.0081/2.000 = 0.000 75 From Table 2-6, ¯ = 0.0015/2.000 = 0.000 75 1/2 0.06132 + 0.004 052 σˆ = 0.000 75 1 + 0.004 052 = 4.607(10−5 ) = 0.000 046 We can predict ¯ and σˆ but not the distribution of .

2-13

σ = E = (0.0005, 0.000 034) distribution unspecified; E = (29.5, 0.885) Mpsi, distribution unspecified; C x = 0.000 034/0.0005 = 0.068, C y = 0.0885/29.5 = 0.030 σ is of the form x, y Table 2-6 σ¯ = ¯ E¯ = 0.0005(29.5)106 = 14 750 psi σˆ σ = 14 750(0.0682 + 0.0302 + 0.0682 + 0.0302 ) 1/2 = 1096.7 psi Cσ = 1096.7/14 750 = 0.074 35

2-14 δ=

Fl AE

F = (14.7, 1.3) kip, A = (0.226, 0.003) in2 , l = (1.5, 0.004) in, E = (29.5, 0.885) Mpsi distributions unspecified. C F = 1.3/14.7 = 0.0884 ;

C A = 0.003/0.226 = 0.0133 ;

C E = 0.885/29.5 = 0.03 Mean of δ:

Fl 1 1 = Fl δ= AE A E

Cl = 0.004/1.5 = 0.00267 ;

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From Table 2-6, ¯ A)(1/ ¯ ¯ δ¯ = F¯ l(1/ E) 1 1 δ¯ = 14 700(1.5) 0.226 29.5(106 ) = 0.003 31 in

Ans.

For the standard deviation, using the first-order terms in Table 2-6,

1/2 1/2 . F¯ l¯ 2 = δ¯ C F2 + Cl2 + C 2A + C 2E σˆ δ = C F + Cl2 + C 2A + C E2 A¯ E¯ σˆ δ = 0.003 31(0.08842 + 0.002672 + 0.01332 + 0.032 ) 1/2 = 0.000 313 in

Ans.

COV Cδ = 0.000 313/0.003 31 = 0.0945

Ans.

Force COV dominates. There is no distributional information on δ. 2-15

M = (15 000, 1350) lbf · in, distribution unspecified; d = (2.00, 0.005) in distribution unspecified. σ=

32M , πd3

CM =

1350 = 0.09 , 15 000

Cd =

0.005 = 0.0025 2.00

σ is of the form x/y, Table 2-6. Mean: σ¯ =

32 M¯ . 32 M¯ 32(15 000) = = π(23 ) π d¯3 π d3

= 19 099 psi

Ans.

Standard Deviation:

From Table 2-6,

1/2 2 σˆ σ = σ¯ C M + Cd23 1 + Cd23 . Cd 3 = 3Cd = 3(0.0025) = 0.0075 1/2 2 + (3Cd ) 2 (1 + (3Cd )) 2 σˆ σ = σ¯ C M = 19 099[(0.092 + 0.00752 )/(1 + 0.00752 )]1/2 = 1725 psi

Ans.

COV: Cσ =

1725 = 0.0903 19 099

Ans.

Stress COV dominates. No information of distribution of σ.

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19

Chapter 2

2-16 f (x)

␣

␤ x1

x2

x

Fraction discarded is α + β. The area under the PDF was unity. Having discarded α + β fraction, the ordinates to the truncated PDF are multiplied by a. a=

1 1 − (α + β)

New PDF, g(x) , is given by

f (x)/[1 − (α + β)] x1 ≤ x ≤ x2 0 otherwise

g(x) =

More formal proof: g(x) has the property

x2

1=

x1

f (x) dx

x1

1=a

x2

g(x) dx = a ∞

−∞

f (x) dx −

x1

f (x) dx −

0

∞

f (x) dx

x2

1 = a {1 − F(x1 ) − [1 − F(x2 )]} a=

1 1 1 = = F(x2 ) − F(x1 ) (1 − β) − α 1 − (α + β)

2-17 (a) d = U[0.748, 0.751] µd =

0.751 + 0.748 = 0.7495 in 2

σˆ d =

0.751 − 0.748 = 0.000 866 in √ 2 3

f (x) =

1 1 = = 333.3 in−1 b−a 0.751 − 0.748

F(x) =

x − 0.748 = 333.3(x − 0.748) 0.751 − 0.748

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F(x1 ) = F(0.748) = 0

(b)

F(x2 ) = (0.750 − 0.748)333.3 = 0.6667 If g(x) is truncated, PDF becomes g(x) =

g(x) 500 f (x) 333.3

0.748

0.749

0.750

0.751

x

333.3 f (x) = = 500 in−1 F(x2 ) − F(x1 ) 0.6667 − 0

µx =

0.748 + 0.750 a + b = = 0.749 in 2 2

σˆ x =

0.750 − 0.748 b − a = 0.000 577 in √ = √ 2 3 2 3

2-18 From Table A-10, 8.1% corresponds to z 1 = −1.4 and 5.5% corresponds to z 2 = +1.6. k1 = µ + z 1 σˆ k2 = µ + z 2 σˆ From which 1.6(9) − (−1.4)11 z 2 k1 − z 1 k2 = z2 − z1 1.6 − (−1.4)

µ=

= 9.933 σˆ = =

k2 − k1 z2 − z1 11 − 9 = 0.6667 1.6 − (−1.4)

The original density function is

1 f (k) = √ exp − 2 0.6667 2π 1

k − 9.933 0.6667

2-19 From Prob. 2-1, µ = 122.9 kcycles and σˆ = 30.3 kcycles. z 10 =

x10 − 122.9 x10 − µ = σˆ 30.3

x10 = 122.9 + 30.3z 10 From Table A-10, for 10 percent failure, z 10 = −1.282 x10 = 122.9 + 30.3(−1.282) = 84.1 kcycles Ans.

2 Ans.

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21

Chapter 2

2-20 x

f

fx

f x2

x

f/(Nw)

f (x)

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69

120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 8480

7200 4900 19 200 40 500 80 000 145 200 86 400 169 000 156 800 112 500 51 200 86 700 64 800 36 100 0 44 100

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

0.002 899 0.001 449 0.004 348 0.007 246 0.011 594 0.017 391 0.008 696 0.014 493 0.011 594 0.007 246 0.002 899 0.004 348 0.002 899 0.001 449 0 0.001 449

0.000 399 0.001 206 0.003 009 0.006 204 0.010 567 0.014 871 0.017 292 0.016 612 0.013 185 0.008 647 0.004 685 0.002 097 0.000 776 0.000 237 5.98E-05 1.25E-05

x¯ = 122.8986

sx = 22.887 19

x

f/(Nw)

f (x)

x

f/(Nw)

f (x)

55 55 65 65 75 75 85 85 95 95 105 105 115 115 125 125 135 135

0 0.002 899 0.002 899 0.001 449 0.001 449 0.004 348 0.004 348 0.007 246 0.007 246 0.011 594 0.011 594 0.017 391 0.017 391 0.008 696 0.008 696 0.014 493 0.014 493 0.011 594

0.000 214 0.000 214 0.000 711 0.000 711 0.001 951 0.001 951 0.004 425 0.004 425 0.008 292 0.008 292 0.012 839 0.012 839 0.016 423 0.016 423 0.017 357 0.017 357 0.015 157 0.015 157

145 145 155 155 165 165 175 175 185 185 195 195 205 205 215 215

0.011 594 0.007 246 0.007 246 0.002 899 0.002 899 0.004 348 0.004 348 0.002 899 0.002 899 0.001 449 0.001 449 0 0 0.001 499 0.001 499 0

0.010 935 0.010 935 0.006 518 0.006 518 0.003 21 0.003 21 0.001 306 0.001 306 0.000 439 0.000 439 0.000 122 0.000 122 2.8E-05 2.8E-05 5.31E-06 5.31E-06

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f 0.02

Histogram PDF

0.018 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0

0

50

100

150

200

250

x

2-21 x

f

fx

f x2

f/(Nw)

f (x)

174 182 190 198 206 214 222 1386

6 9 44 67 53 12 6 197

1044 1638 8360 13 266 10 918 2568 1332 39 126

181 656 298 116 1 588 400 2 626 668 2 249 108 549 552 295 704 7 789 204

0.003 807 0.005 711 0.027 919 0.042 513 0.033 629 0.007 614 0.003 807

0.001 642 0.009 485 0.027 742 0.041 068 0.030 773 0.011 671 0.002 241

x¯ = 198.6091 x 170 170 178 178 186 186 194 194 202 202 210 210 218 218 226 226

f/(Nw) 0 0.003 807 0.003 807 0.005 711 0.005 711 0.027 919 0.027 919 0.042 513 0.042 513 0.033 629 0.033 629 0.007 614 0.007 614 0.003 807 0.003 807 0

sx = 9.695 071

f (x) 0.000 529 0.000 529 0.004 297 0.004 297 0.017 663 0.017 663 0.036 752 0.036 752 0.038 708 0.038 708 0.020 635 0.020 635 0.005 568 0.005 568 0.000 76 0.000 76

f 0.045 0.04

Data PDF

0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 150

x 170

190

210

230

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23

Chapter 2

2-22 f

fx

f x2

f/(Nw)

2 6 6 9 19 10 4 2 58

128 408 432 684 1520 840 352 184 4548

8192 27 744 31 104 51 984 121 600 70 560 30 976 16 928 359 088

0.008 621 0.025 862 0.025 862 0.038 793 0.081 897 0.043 103 0.017 241 0.008 621

x 64 68 72 76 80 84 88 92 624

x¯ = 78.413 79

f (x) 0.005 48 0.017 299 0.037 705 0.056 742 0.058 959 0.042 298 0.020 952 0.007 165

sx = 6.572 229

x

f/(Nw)

f(x)

x

f/(Nw)

f(x)

62 62 66 66 70 70 74 74 78 78

0 0.008 621 0.008 621 0.025 862 0.025 862 0.025 862 0.025 862 0.038 793 0.038 793 0.081 897

0.002 684 0.002 684 0.010 197 0.010 197 0.026 749 0.026 749 0.048 446 0.048 446 0.060 581 0.060 581

82 82 86 86 90 90 94 94

0.081 897 0.043 103 0.043 103 0.017 241 0.017 241 0.008 621 0.008 621 0

0.052 305 0.052 305 0.031 18 0.031 18 0.012 833 0.012 833 0.003 647 0.003 647

f 0.09

Data PDF

0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01

x

0 60

70

80

90

2-23 σ¯ =

4(40) 4 P¯ = 50.93 kpsi = 2 πd π(12 )

σˆ σ =

4 σˆ P 4(8.5) = 10.82 kpsi = 2 πd π(12 )

σˆ s y = 5.9 kpsi

100

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For no yield, m = S y − σ ≥ 0 m − µm 0 − µm µm = =− σˆ m σˆ m σˆ m ¯ µm = S y − σ¯ = 27.47 kpsi, 1/2 2 2 σˆ m = σˆ σ + σˆ Sy = 12.32 kpsi z=

z=

−27.47 = −2.230 12.32

m 0

From Table A-10, p f = 0.0129 R = 1 − p f = 1 − 0.0129 = 0.987 2-24

Ans.

For a lognormal distribution,

µ y = ln µx − ln 1 + C x2 σˆ y = ln 1 + C x2

Eq. (2-18) Eq. (2-19) From Prob. (2-23)

µm = S¯ y − σ¯ = µx 2 2 µ y = ln S¯ y − ln 1 + C Sy − ln σ¯ − ln 1 + Cσ

S¯ y = ln σ¯

1 + Cσ2 1 + C S2y

1/2 σˆ y = ln 1 + C S2y + ln 1 + Cσ2 2 2 = ln 1 + C Sy 1 + Cσ S¯ y 1 + Cσ2 ln σ¯ 1 + C S2 µ y z = − = −

σˆ ln 1 + C S2y 1 + Cσ2 σ¯ =

4(30) 4 P¯ = 38.197 kpsi = 2 πd π(12 )

4 σˆ P 4(5.1) = 6.494 kpsi = 2 πd π(12 ) 6.494 = 0.1700 Cσ = 38.197 3.81 C Sy = = 0.076 81 49.6 σˆ σ =

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25

Chapter 2

 ln 

49.6 38.197



1 + 0.170  1 + 0.076 812 2

z = − = −1.470 ln (1 + 0.076 812 )(1 + 0.1702 ) From Table A-10 p f = 0.0708 R = 1 − p f = 0.929

Ans.

2-25 (a) a = 1.000 ± 0.001 in b = 2.000 ± 0.003 in c = 3.000 ± 0.005 in d = 6.020 ± 0.006 in w¯ = d − a − b − c = 6.020 − 1 − 2 − 3 = 0.020 in ! tw = tall = 0.001 + 0.003 + 0.005 + 0.006 = 0.015 in w = 0.020 ± 0.015 in

Ans.

(b) w¯ = 0.020 0.001 2 0.003 2 0.005 2 0.006 2 2 + √ + √ + √ σˆ w = σˆ all = √ 3 3 3 3 = 0.004 86 → 0.005 in (uniform) w = 0.020 ± 0.005 in Ans. 2-26 V + V = (a + a)(b + b)(c + c) V + V = abc + bca + acb + abc + small higher order terms V . a b c = + + Ans. a b c V¯ V¯ = a¯ b¯ c¯ = 1.25(1.875)(2.75) = 6.4453 in3 V 0.001 0.002 0.003 + + = 0.00296 = 1.250 1.875 2.750 V¯ V ¯ V = V = 0.00296(6.4453) = 0.0191 in3 V¯ Lower range number: V¯ − V = 6.4453 − 0.0191 = 6.4262 in3

Ans.

Upper range number: V¯ + V = 6.4453 + 0.0191 = 6.4644 in3

Ans.

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2-27 a

(a)

c

b

w

wmax = 0.014 in, wmin = 0.004 in w¯ = (0.014 + 0.004)/2 = 0.009 in w = 0.009 ± 0.005 in w¯ = x¯ − y¯ = a¯ − b¯ − c¯ 0.009 = a¯ − 0.042 − 1.000 a¯ = 1.051 in tw = tall 0.005 = ta + 0.002 + 0.002 ta = 0.005 − 0.002 − 0.002 = 0.001 in a = 1.051 ± 0.001 in Ans. 2 = σˆ a2 + σˆ b2 + σˆ c2 σˆ w = σˆ all

(b)

σˆ a2 = σˆ w2 − σˆ b2 − σˆ c2 0.005 2 0.002 2 0.002 2 = − √ − √ √ 3 3 3 2 −6 σˆ a = 5.667(10 ) " σˆ a = 5.667(10−6 ) = 0.00238 in a¯ = 1.051 in, 2-28

σˆ a = 0.00238 in Ans.

Choose 15 mm as basic size, D, d. Table 2-8: fit is designated as 15H7/h6. From Table A-11, the tolerance grades are D = 0.018 mm and d = 0.011 mm. Hole: Eq. (2-38) Dmax = D + D = 15 + 0.018 = 15.018 mm Dmin = D = 15.000 mm Ans.

Ans.

Shaft: From Table A-12, fundamental deviation δ F = 0. From Eq. (2-39) dmax = d + δ F = 15.000 + 0 = 15.000 mm Ans. dmin = d + δ R − d = 15.000 + 0 − 0.011 = 14.989 mm 2-29

Ans.

Choose 45 mm as basic size. Table 2-8 designates fit as 45H7/s6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm Hole: Eq. (2-38) Dmax = D + D = 45.000 + 0.025 = 45.025 mm Dmin = D = 45.000 mm Ans.

Ans.

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27

Chapter 2

Shaft: From Table A-12, fundamental deviation δ F = +0.043 mm. From Eq. (2-40) dmin = d + δ F = 45.000 + 0.043 = 45.043 mm Ans. dmax = d + δ F + d = 45.000 + 0.043 + 0.016 = 45.059 mm 2-30

Ans.

Choose 50 mm as basic size. From Table 2-8 fit is 50H7/g6. From Table A-11, the tolerance grades are D = 0.025 mm and d = 0.016 mm. Hole: Dmax = D + D = 50 + 0.025 = 50.025 mm Dmin = D = 50.000 mm Ans.

Ans.

Shaft: From Table A-12 fundamental deviation = −0.009 mm dmax = d + δ F = 50.000 + (−0.009) = 49.991 mm dmin = d + δ F − d = 50.000 + (−0.009) − 0.016 = 49.975 mm 2-31

Ans.

Choose the basic size as 1.000 in. From Table 2-8, for 1.0 in, the fit is H8/f7. From Table A-13, the tolerance grades are D = 0.0013 in and d = 0.0008 in. Hole: Dmax = D + (D) hole = 1.000 + 0.0013 = 1.0013 in Dmin = D = 1.0000 in Ans.

Ans.

Shaft: From Table A-14: Fundamental deviation = −0.0008 in dmax = d + δ F = 1.0000 + (−0.0008) = 0.9992 in Ans. dmin = d + δ F − d = 1.0000 + (−0.0008) − 0.0008 = 0.9984 in Ans. Alternatively, dmin = dmax − d = 0.9992 − 0.0008 = 0.9984 in. Ans. 2-32 W

Di

W

Do

Do = W + Di + W D¯ o = W¯ + D¯ i + W¯ = 0.139 + 3.734 + 0.139 = 4.012 in tall = 0.004 + 0.028 + 0.004 t Do = = 0.036 in Do = 4.012 ± 0.036 in Ans.

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2-33 Do = Di + 2W D¯ o = D¯ i + 2W¯ = 208.92 + 2(5.33) = 219.58 mm t = t Di + 2tw t Do = all

= 1.30 + 2(0.13) = 1.56 mm Do = 219.58 ± 1.56 mm Ans. 2-34 Do = Di + 2W D¯ o = D¯ i + 2W¯ = 3.734 + 2(0.139) = 4.012 mm 2 1/2 t Do = t 2 = tD + (2 tw ) 2 o all

= [0.0282 + (2) 2 (0.004) 2 ]1/2 = 0.029 in Do = 4.012 ± 0.029 in Ans. 2-35 Do = Di + 2W D¯ o = D¯ i + 2W¯ = 208.92 + 2(5.33) = 219.58 mm t Do = t 2 = [1.302 + (2) 2 (0.13) 2 ]1/2 all

= 1.33 mm Do = 219.58 ± 1.33 mm Ans. 2-36 (a)

w F

W

w= F−W w¯ = F¯ − W¯ = 0.106 − 0.139 = −0.033 in tw = t = 0.003 + 0.004 all

tw = 0.007 in wmax = w¯ + tw = −0.033 + 0.007 = −0.026 in wmin = w¯ − tw = −0.033 − 0.007 = −0.040 in The minimum “squeeze” is 0.026 in.

Ans.

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Chapter 2

(b)

Ymax = D¯ o = 4.012 in Ymin = max[0.99 D¯ o , D¯ o − 0.06] = max[3.9719, 3.952] = 3.972 in

Y w

Do

29

Y = 3.992 ± 0.020 in Do + w − Y = 0 w = Y − D¯ o w¯ = Y¯ − D¯ o = 3.992 − 4.012 = −0.020 in tw = t = tY + t Do = 0.020 + 0.036 = 0.056 in all

wmax = 0.036 in wmin = −0.076 in

w = −0.020 ± 0.056 in O-ring is more likely compressed than free prior to assembly of the end plate.

2-37 (a) Figure defines w as gap. w=F−W w¯ = F¯ − W¯ = 4.32 − 5.33 = −1.01 mm t = t F + tW = 0.13 + 0.13 = 0.26 mm tw =

w F

W

all

wmax = w¯ + tw = −1.01 + 0.26 = −0.75 mm wmin = w¯ − tw = −1.01 − 0.26 = −1.27 mm The O-ring is “squeezed” at least 0.75 mm. (b)

Ymax = D¯ o = 219.58 mm

Y w

Do

Ymin = max[0.99 D¯ o , D¯ o − 1.52] = max[0.99(219.58, 219.58 − 1.52)] = 217.38 mm Y = 218.48 ± 1.10 mm

From the figure, the stochastic equation is: or,

Do + w = Y w = Y − Do w¯ = Y¯ − D¯ o = 218.48 − 219.58 = −1.10 mm t = tY + t Do = 1.10 + 0.34 = 1.44 mm tw = all

wmax = w¯ + tw = −1.10 + 1.44 = 0.34 mm wmin = w¯ − tw = −1.10 − 1.44 = −2.54 mm The O-ring is more likely to be circumferentially compressed than free prior to assembly of the end plate.

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2-38

wmax = −0.020 in, wmin = −0.040 in 1 w¯ = (−0.020 + (−0.040)) = −0.030 in 2 1 tw = (−0.020 − (−0.040)) = 0.010 in 2 b = 0.750 ± 0.001 in c = 0.120 ± 0.005 in d = 0.875 ± 0.001 in w¯ = a¯ − b¯ − c¯ − d¯ −0.030 = a¯ − 0.875 − 0.120 − 0.750 a¯ = 0.875 + 0.120 + 0.750 − 0.030 a¯ = 1.715 in

w b

c

d

a

Absolute: tw =

t = 0.010 = ta + 0.001 + 0.005 + 0.001

all

ta = 0.010 − 0.001 − 0.005 − 0.001 = 0.003 in a = 1.715 ± 0.003 in Ans. Statistical: For a normal distribution of dimensions tw2 = t 2 = ta2 + tb2 + tc2 + td2 all

1/2 ta = tw2 − tb2 − tc2 − td2 = (0.0102 − 0.0012 − 0.0052 − 0.0012 ) 1/2 = 0.0085 a = 1.715 ± 0.0085 in Ans. 2-39 x

n

nx

nx 2

93 95 97 99 101 103 105 107 109 111

19 25 38 17 12 10 5 4 4 2 136

1767 2375 3685 1683 1212 1030 525 428 436 222 13 364

164 311 225 625 357 542 166 617 122 412 106 090 55 125 45 796 47 524 24 624 1315 704

x¯ = 13 364/136 = 98.26 kpsi 1/2 1 315 704 − 13 3642 /136 sx = = 4.30 kpsi 135

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31

Chapter 2

Under normal hypothesis, z 0.01 = (x0.01 − 98.26)/4.30 x0.01 = 98.26 + 4.30z 0.01 = 98.26 + 4.30(−2.3267) . = 88.26 = 88.3 kpsi Ans. 2-40

From Prob. 2-39, µx = 98.26 kpsi, and σˆ x = 4.30 kpsi. C x = σˆ x /µx = 4.30/98.26 = 0.043 76 From Eqs. (2-18) and (2-19), µ y = ln(98.26) − 0.043 762 /2 = 4.587 " σˆ y = ln(1 + 0.043 762 ) = 0.043 74 For a yield strength exceeded by 99% of the population, z 0.01 = (ln x0.01 − µ y )/σˆ y

⇒

ln x0.01 = µ y + σˆ y z 0.01

From Table A-10, for 1% failure, z 0.01 = −2.326. Thus, ln x0.01 = 4.587 + 0.043 74(−2.326) = 4.485 x0.01 = 88.7 kpsi Ans. The normal PDF is given by Eq. (2-14) as

1 x − 98.26 2 1 f (x) = √ exp − 2 4.30 4.30 2π

For the lognormal distribution, from Eq. (2-17), defining g(x), 1 1 ln x − 4.587 2 g(x) = √ exp − 2 0.043 74 x(0.043 74) 2π x (kpsi)

f/(Nw)

f (x)

g (x)

x (kpsi)

f/(Nw)

f (x)

g (x)

92 92 94 94 96 96 98 98 100 100 102

0.000 00 0.069 85 0.069 85 0.091 91 0.091 91 0.139 71 0.139 71 0.062 50 0.062 50 0.044 12 0.044 12

0.032 15 0.032 15 0.056 80 0.056 80 0.080 81 0.080 81 0.092 61 0.092 61 0.085 48 0.085 48 0.063 56

0.032 63 0.032 63 0.058 90 0.058 90 0.083 08 0.083 08 0.092 97 0.092 97 0.083 67 0.083 67 0.061 34

102 104 104 106 106 108 108 110 110 112 112

0.036 76 0.036 76 0.018 38 0.018 38 0.014 71 0.014 71 0.014 71 0.014 71 0.007 35 0.007 35 0.000 00

0.063 56 0.038 06 0.038 06 0.018 36 0.018 36 0.007 13 0.007 13 0.002 23 0.002 23 0.000 56 0.000 56

0.061 34 0.037 08 0.037 08 0.018 69 0.018 69 0.007 93 0.007 93 0.002 86 0.002 86 0.000 89 0.000 89

Note: rows are repeated to draw histogram

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Histogram f (x) g(x)

0.16

Probability density

0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 90

92

94

96

98

100 102 x (kpsi)

104

106

108

110

112

The normal and lognormal are almost the same. However the data is quite skewed and perhaps a Weibull distribution should be explored. For a method of establishing the Weibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design, McGraw-Hill, 5th ed., 1989, Sec. 4-12. 2-41

Let x = (S fe ) 104 x0 = 79 kpsi, θ = 86.2 kpsi, b = 2.6 Eq. (2-28) x¯ = x0 + (θ − x0 )(1 + 1/b) x¯ = 79 + (86.2 − 79)(1 + 1/2.6) = 79 + 7.2 (1.38) From Table A-34, (1.38) = 0.888 54 x¯ = 79 + 7.2(0.888 54) = 85.4 kpsi Eq. (2-29)

Ans.

σˆ x = (θ − x0 )[(1 + 2/b) − 2 (1 + 1/b)]1/2 = (86.2 − 79)[(1 + 2/2.6) − 2 (1 + 1/2.6)]1/2 = 7.2[0.923 76 − 0.888 542 ]1/2 = 2.64 kpsi Ans. 2.64 σˆ x = = 0.031 Ans. Cx = x¯ 85.4

2-42 x = Sut x0 = 27.7, θ = 46.2, b = 4.38 µx = 27.7 + (46.2 − 27.7)(1 + 1/4.38) = 27.7 + 18.5 (1.23) = 27.7 + 18.5(0.910 75) = 44.55 kpsi Ans.

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33

Chapter 2

σˆ x = (46.2 − 27.7)[(1 + 2/4.38) − 2 (1 + 1/4.38)]1/2 = 18.5[(1.46) − 2 (1.23)]1/2 = 18.5[0.8856 − 0.910 752 ]1/2 = 4.38 kpsi Ans. 4.38 = 0.098 Ans. Cx = 44.55 From the Weibull survival equation x − x0 b R = exp − =1− p θ − x0 x40 − x0 b = 1 − p40 R40 = exp − θ − x0 40 − 27.7 4.38 = exp − = 0.846 46.2 − 27.7 p40 = 1 − R40 = 1 − 0.846 = 0.154 = 15.4%

Ans.

2-43 x = Sut

x0 = 151.9, θ = 193.6, b = 8 µx = 151.9 + (193.6 − 151.9)(1 + 1/8) = 151.9 + 41.7 (1.125) = 151.9 + 41.7(0.941 76) = 191.2 kpsi Ans. σˆ x = (193.6 − 151.9)[(1 + 2/8) − 2 (1 + 1/8)]1/2 = 41.7[(1.25) − 2 (1.125)]1/2 = 41.7[0.906 40 − 0.941 762 ]1/2 = 5.82 kpsi Ans. 5.82 = 0.030 Cx = 191.2

2-44 x = Sut

x0 = 47.6, θ = 125.6, b = 11.84 x¯ = 47.6 + (125.6 − 47.6)(1 + 1/11.84) x¯ = 47.6 + 78 (1.08) = 47.6 + 78(0.959 73) = 122.5 kpsi σˆ x = (125.6 − 47.6)[(1 + 2/11.84) − 2 (1 + 1/11.84)]1/2 = 78[(1.08) − 2 (1.17)]1/2 = 78(0.959 73 − 0.936 702 ) 1/2 = 22.4 kpsi

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From Prob. 2-42

x − x0 b p = 1 − exp − θ − θ0 100 − 47.6 11.84 = 1 − exp − 125.6 − 47.6 = 0.0090 Ans.

y = Sy y0 = 64.1, θ = 81.0, b = 3.77 y¯ = 64.1 + (81.0 − 64.1)(1 + 1/3.77) = 64.1 + 16.9 (1.27) = 64.1 + 16.9(0.902 50) = 79.35 kpsi σ y = (81 − 64.1)[(1 + 2/3.77) − (1 + 1/3.77)]1/2 σ y = 16.9[(0.887 57) − 0.902 502 ]1/2 = 4.57 kpsi y − y0 3.77 p = 1 − exp − θ − y0 70 − 64.1 3.77 p = 1 − exp − = 0.019 Ans. 81 − 64.1 2-45 x = Sut = W[122.3, 134.6, 3.64] kpsi, p(x > 120) = 1 = 100% since x0 > 120 kpsi 133 − 122.3 3.64 p(x > 133) = exp − 134.6 − 122.3 = 0.548 = 54.8% Ans. 2-46

Using Eqs. (2-28) and (2-29) and Table A-34, µn = n 0 + (θ − n 0 )(1 + 1/b) = 36.9 + (133.6 − 36.9)(1 + 1/2.66) = 122.85 kcycles σˆ n = (θ − n 0 )[(1 + 2/b) − 2 (1 + 1/b)] = 34.79 kcycles For the Weibull density function, Eq. (2-27), 2.66−1 2.66 n − 36.9 2.66 n − 36.9 exp − f W (n) = 133.6 − 36.9 133.6 − 36.9 133.6 − 36.9 For the lognormal distribution, Eqs. (2-18) and (2-19) give, µ y = ln(122.85) − (34.79/122.85) 2 /2 = 4.771 " σˆ y = [1 + (34.79/122.85) 2 ] = 0.2778

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35

Chapter 2

From Eq. (2-17), the lognormal PDF is

1 f L N (n) = √ exp − 2 0.2778 n 2π 1

ln n − 4.771 0.2778

2

We form a table of densities f W (n) and f L N (n) and plot. n (kcycles)

f W (n)

f L N (n)

40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220

9.1E-05 0.000 991 0.002 498 0.004 380 0.006 401 0.008 301 0.009 822 0.010 750 0.010 965 0.010 459 0.009 346 0.007 827 0.006 139 0.004 507 0.003 092 0.001 979 0.001 180 0.000 654 0.000 336

1.82E-05 0.000 241 0.001 233 0.003 501 0.006 739 0.009 913 0.012 022 0.012 644 0.011 947 0.010 399 0.008 492 0.006 597 0.004 926 0.003 564 0.002 515 0.001 739 0.001 184 0.000 795 0.000 529

f (n) 0.014

LN W

0.012 0.010 0.008 0.006 0.004 0.002 0

0

50

100

150

200

250

n, kcycles

The Weibull L10 life comes from Eq. (2-26) with a reliability of R = 0.90. Thus, n 0.10 = 36.9 + (133 − 36.9)[ln(1/0.90)]1/2.66 = 78.1 kcycles Ans.

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The lognormal L10 life comes from the definition of the z variable. That is, ln n 0 = µ y + σˆ y z

or n 0 = exp(µ y + σˆ y z)

From Table A-10, for R = 0.90, z = −1.282. Thus, n 0 = exp[4.771 + 0.2778(−1.282)] = 82.7 kcycles Ans. 2-47

Form a table

i

x L(10−5)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

3.05 3.55 4.05 4.55 5.05 5.55 6.05 6.55 7.05 7.55 8.05 8.55 9.05 9.55 10.05

−5

fi

f i x(10 )

3 7 11 16 21 13 13 6 2 0 4 3 0 0 1 100

9.15 24.85 44.55 72.80 106.05 72.15 78.65 39.30 14.10 0 32.20 25.65 0 0 10.05 529.50

2

−10

f i x (10

)

27.9075 88.2175 180.4275 331.24 535.5525 400.4325 475.8325 257.415 99.405 0 259.21 219.3075 0 0 101.0025 2975.95

g(x) (105 ) 0.0557 0.1474 0.2514 0.3168 0.3216 0.2789 0.2151 0.1517 0.1000 0.0625 0.0375 0.0218 0.0124 0.0069 0.0038

x¯ = 529.5(105 )/100 = 5.295(105 ) cycles Ans. 1/2 2975.95(1010 ) − [529.5(105 )]2 /100 sx = 100 − 1 = 1.319(105 ) cycles Ans. C x = s/x¯ = 1.319/5.295 = 0.249 µ y = ln 5.295(105 ) − 0.2492 /2 = 13.149 " σˆ y = ln(1 + 0.2492 ) = 0.245 1 ln x − µ y 2 1 g(x) = √ exp − 2 σˆ y x σˆ y 2π 1 ln x − 13.149 2 1.628 exp − g(x) = x 2 0.245

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37

Chapter 2

105 g(x) 0.5

Superposed histogram and PDF

0.4

0.3

0.2

0.1

0

3.05(105)

10.05(105) x, cycles

2-48 Eq. (2-28)

Eq. (2-29)

x = Su = W[70.3, 84.4, 2.01] µx = 70.3 + (84.4 − 70.3)(1 + 1/2.01) = 70.3 + (84.4 − 70.3)(1.498) = 70.3 + (84.4 − 70.3)0.886 17 = 82.8 kpsi Ans. σˆ x = (84.4 − 70.3)[(1 + 2/2.01) − 2 (1 + 1/2.01)]1/2 σˆ x = 14.1[0.997 91 − 0.886 172 ]1/2 = 6.502 kpsi Cx =

2-49

6.502 = 0.079 Ans. 82.8

Take the Weibull equation for the standard deviation σˆ x = (θ − x0 )[(1 + 2/b) − 2 (1 + 1/b)]1/2 and the mean equation solved for x¯ − x0 x¯ − x0 = (θ − x0 )(1 + 1/b) Dividing the first by the second, [(1 + 2/b) − 2 (1 + 1/b)]1/2 σˆ x = x¯ − x0 (1 + 1/b) √ 4.2 (1 + 2/b) = − 1 = R = 0.2763 49 − 33.8 2 (1 + 1/b)

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Make a table and solve for b iteratively b 3 4 4.1

1 + 2/b 1 + 1/b (1 + 2/b) (1 + 1/b) 1.67 1.5 1.49

1.33 1.25 1.24

0.903 30 0.886 23 0.885 95

0.893 38 0.906 40 0.908 52

0.363 0.280 0.271

. b = 4.068 Using MathCad Ans. 49 − 33.8 x¯ − x0 = 33.8 + θ = x0 + (1 + 1/b) (1 + 1/4.068) = 49.8 kpsi Ans. 2-50

x = S y = W[34.7, 39, 2.93] kpsi x¯ = 34.7 + (39 − 34.7)(1 + 1/2.93) = 34.7 + 4.3(1.34) = 34.7 + 4.3(0.892 22) = 38.5 kpsi σˆ x = (39 − 34.7)[(1 + 2/2.93) − 2 (1 + 1/2.93)]1/2 = 4.3[(1.68) − 2 (1.34)]1/2 = 4.3[0.905 00 − 0.892 222 ]1/2 = 1.42 kpsi Ans. C x = 1.42/38.5 = 0.037 Ans.

2-51 x (Mrev) 1 2 3 4 5 6 7 8 9 10 11 12 Sum 78

f

fx

f x2

11 22 38 57 31 19 15 12 11 9 7 5 237

11 44 114 228 155 114 105 96 99 90 77 60 1193

11 88 342 912 775 684 735 768 891 900 847 720 7673

µx = 1193(106 )/237 = 5.034(106 ) cycles 7673(1012 ) − [1193(106 )]2 /237 = 2.658(106 ) cycles σˆ x = 237 − 1 C x = 2.658/5.034 = 0.528

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39

Chapter 2

From Eqs. (2-18) and (2-19), µ y = ln[5.034(106 )] − 0.5282 /2 = 15.292 " σˆ y = ln(1 + 0.5282 ) = 0.496 From Eq. (2-17), defining g(x),

1 g (x) = √ exp − 2 x(0.496) 2π 1

x (Mrev)

f/(Nw)

g(x) · (106 )

0.5 0.5 1.5 1.5 2.5 2.5 3.5 3.5 4.5 4.5 5.5 5.5 6.5 6.5 7.5 7.5 8.5 8.5 9.5 9.5 10.5 10.5 11.5 11.5 12.5 12.5

0.000 00 0.046 41 0.046 41 0.092 83 0.092 83 0.160 34 0.160 34 0.240 51 0.240 51 0.130 80 0.130 80 0.080 17 0.080 17 0.063 29 0.063 29 0.050 63 0.050 63 0.046 41 0.046 41 0.037 97 0.037 97 0.029 54 0.029 54 0.021 10 0.021 10 0.000 00

0.000 11 0.000 11 0.052 04 0.052 04 0.169 92 0.169 92 0.207 54 0.207 54 0.178 48 0.178 48 0.131 58 0.131 58 0.090 11 0.090 11 0.059 53 0.059 53 0.038 69 0.038 69 0.025 01 0.025 01 0.016 18 0.016 18 0.010 51 0.010 51 0.006 87 0.006 87

z=

ln x − µ y σˆ y

⇒

ln x − 15.292 0.496

2

Histogram PDF

0.25

0.2

g(x)(106)

shi20396_ch02.qxd

0.15

0.1

0.05

0

0

2

4

6 x, Mrev

8

ln x = µ y + σˆ y z = 15.292 + 0.496z

L10 life, where 10% of bearings fail, from Table A-10, z = −1.282. Thus, ln x = 15.292 + 0.496(−1.282) = 14.66 ∴ x = 2.32 × 106 rev Ans.

10

12

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Page 40

Chapter 3 3-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 kpsi)

Ans.

3-2 From Table A-20 Sut = 620 MPa (90 kpsi), S y = 340 MPa (49.5 kpsi) Ans. 3-3 Comparison of yield strengths: Sut of G10 500 HR is S yt of SAE1020 CD is

620 = 1.32 times larger than SAE1020 CD 470

Ans.

390 = 1.15 times larger than G10500 HR Ans. 340

From Table A-20, the ductilities (reduction in areas) show, SAE1020 CD is

40 = 1.14 times larger than G10500 35

Ans.

The stiffness values of these materials are identical Ans. Sut MPa (kpsi)

Sy MPa (kpsi)

Table A-20 Ductility R%

SAE1020 CD 470(68) UNS10500 HR 620(90)

390 (57) 340(495)

40 35

Table A-5 Stiffness GPa (Mpsi) 207(30) 207(30)

3-4 From Table A-21 1040 Q&T S¯ y = 593 (86) MPa (kpsi)

at 205◦C (400◦F)

Ans.

3-5 From Table A-21 1040 Q&T

R = 65%

at 650◦C (1200◦F)

Ans.

3-6 Using Table A-5, the specific strengths are: Sy 39.5(103 ) = = 1.40(105 ) in Ans. UNS G10350 HR steel: W 0.282 Sy 43(103 ) = = 4.39(105 ) in Ans. 2024 T4 aluminum: W 0.098 Ti-6Al-4V titanium:

Sy 140(103 ) = = 8.75(105 ) in W 0.16

ASTM 30 gray cast iron has no yield strength.

Ans.

Ans.

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41

Chapter 3

3-7 The specific moduli are: 30(106 ) E = = 1.06(108 ) in W 0.282

UNS G10350 HR steel:

Ans.

2024 T4 aluminum:

10.3(106 ) E = = 1.05(108 ) in W 0.098

Ans.

Ti-6Al-4V titanium:

16.5(106 ) E = = 1.03(108 ) in W 0.16

Ans.

Gray cast iron:

14.5(106 ) E = = 5.58(107 ) in W 0.26

Ans.

2G(1 + ν) = E

3-8

⇒ ν=

E − 2G 2G

From Table A-5 Steel: ν = Aluminum: ν = Beryllium copper: ν = Gray cast iron: ν =

30 − 2(11.5) = 0.304 2(11.5)

Ans.

10.4 − 2(3.90) = 0.333 2(3.90) 18 − 2(7) = 0.286 2(7) 14.5 − 2(6) = 0.208 2(6)

Ans.

Ans. Ans.

3-9 E U 80 70 60 Stress P兾A0 kpsi

shi20396_ch03.qxd

50 Y

40

Su ⫽ 85.5 kpsi Ans. Sy ⫽ 45.5 kpsi Ans.

30

E ⫽ 90兾0.003 ⫽ 30 000 kpsi Ans.

20

R⫽

A 0 ⫺ AF 0.1987 ⫺ 0.1077 (100) ⫽ 45.8% Ans. ⫽ 0.1987 A0

10

⫽

A0 l ⫺ l0 ⌬l l ⫽ ⫽ ⫺ 1⫽ ⫺1 A l0 l0 l0

0

0

0.002 0.1

0.004 0.2

0.006 0.3

0.008 0.4 Strain,

0.010 0.5

0.012 0.6

0.014 0.7

0.016 (Lower curve) 0.8 (Upper curve)

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3-10

To plot σtrue vs. ε, the following equations are applied to the data. A0 = Eq. (3-4)

π(0.503) 2 = 0.1987 in2 4

ε = ln

l l0

ε = ln

A0 A

σtrue =

for 0 ≤ L ≤ 0.0028 in for L > 0.0028 in

P A

The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log σ vs log ε m = 0.2306

The curve fit gives

log σ0 = 5.1852 ⇒ σ0 = 153.2 kpsi

Ans.

For 20% cold work, Eq. (3-10) and Eq. (3-13) give, A = A0 (1 − W ) = 0.1987(1 − 0.2) = 0.1590 in2 ε = ln

0.1987 A0 = ln = 0.2231 A 0.1590

Eq. (3-14): S y = σ0 εm = 153.2(0.2231) 0.2306 = 108.4 kpsi

Ans.

Eq. (3-15), with Su = 85.5 kpsi from Prob. 3-9, Su = P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

85.5 Su = = 106.9 kpsi 1−W 1 − 0.2

Ans.

L

A

ε

σtrue

log ε

log σtrue

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 4 0.197 8 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7

0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 149 0.001 399 0.001 575 0.004 604 0.012 216 0.032 284 0.058 082 0.240 083 0.418 956 0.612 511

0 5032.388 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 354.84 46 511.63 46 357.62 68 607.07 81 066.67 108 765.2 125 478.2 137 418.8

−3.699 01 −3.522 94 −3.301 14 −3.187 23 −2.939 55 −2.854 18 −2.802 61 −2.336 85 −1.913 05 −1.491 01 −1.235 96 −0.619 64 −0.377 83 −0.212 89

3.701 774 4.002 804 4.178 895 4.303 834 4.546 872 4.626 053 4.646 941 4.667 562 4.666 121 4.836 369 4.908 842 5.036 49 5.098 568 5.138 046

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43

Chapter 3

160000 140000 120000 true (psi)

100000 80000 60000 40000 20000 0

0

0.1

0.2

0.3

0.4 true

0.5

0.6

0.7

5.2

y ⫽ 0.2306x ⫹ 5.1852 5

log

5.1

4.9

⫺1.6

3-11

⫺1.4

Tangent modulus at σ = 0 is E0 = At σ = 20 kpsi

⫺1

⫺0.8 log

⫺0.6

⫺0.4

⫺0.2

0

4.8

5000 − 0 σ . = = 25(106 ) psi ε 0.2(10−3 ) − 0

. (26 − 19)(103 ) E 20 = = 14.0(106 ) psi Ans. − 3 (1.5 − 1)(10 )

ε(10−3 )

σ (kpsi)

60

0 5 10 16 19 26 32 40 46 49 54

50

0 0.20 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 3-12

⫺1.2

40 (kpsi)

shi20396_ch03.qxd

(Sy)0.001 ⫽ ˙ 35 kpsi Ans.

30 20 10 0

0

1

2

3 (10⫺3)

From Prob. 2-8, for y = a1 x + a2 x 2 a1 =

yx 3 − x yx 2 xx 3 − (x 2 ) 2

a2 =

xx y − yx 2 xx 3 − (x 2 ) 2

4

5

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Let x represent ε(10−3 ) and y represent σ (kpsi), x 0 0.2 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 = 21.14

y

x2

x3

xy

0 5 10 16 19 26 32 40 46 49 54 297

0 0.04 0.1936 0.64 1.00 2.25 4.00 7.84 11.56 16.00 25.00 68.5236

0 0.008 0.085 184 0.512 1.000 3.375 8.000 21.952 39.304 64.000 125.000 263.2362

0 1.0 4.4 12.8 19.0 39.0 64.0 112.0 156.4 196.0 270.0 874.6

Substituting, 297(263.2362) − 874.6(68.5236) = 20.993 67 21.14(263.2362) − (68.5236) 2 21.14(874.6) − 297(68.5236) a2 = = −2.142 42 21.14(263.2362) − (68.5236) 2 The tangent modulus is a1 =

dy dσ = = 20.993 67 − 2(2.142 42)x = 20.993 67 − 4.284 83x dx dε At σ = 0, E 0 = 20.99 Mpsi Ans. At σ = 20 kpsi 20 = 20.993 67x − 2.142 42x 2 ⇒ x = 1.069, 8.73 Taking the first root, ε = 1.069 and the tangent modulus is E 20 = 20.993 67 − 4.284 83(1.069) = 16.41 Mpsi Ans. Determine the equation for the 0.1 percent offset line y = 20.99x + b at y = 0, x = 1 ∴ b = −20.99 y = 20.99x − 20.99 = 20.993 67x − 2.142 42x 2 2.142 42x 2 − 20.99 = 0 ⇒ x = 3.130 (S y ) 0.001 = 20.99(3.13) − 2.142(3.13) 2 = 44.7 kpsi 3-13

Since |εo | = |εi |

R+h R R + N ln = ln = −ln R+ N R+ N R R+N R+h = R+N R ( R + N ) 2 = R( R + h)

From which,

N 2 + 2R N − Rh = 0

Ans.

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45

Chapter 3

h 1/2 N = R −1 ± 1 + R

The roots are:

The + sign being significant,

h 1/2 N=R 1+ −1 R

Ans.

Substitute for N in εo = ln  Gives

  ε0 = ln  

R+h R+N

R+h

h 1/2 R+ R 1+ −R R

  h 1/2   = ln 1 +  R

Ans.

These constitute a useful pair of equations in cold-forming situations, allowing the surface strains to be found so that cold-working strength enhancement can be estimated. 3-14 16T 16T 10−6 = = 2.6076T MPa πd 3 π(12.5) 3 (10−3 ) 3 π π ◦ ◦ θ r (12.5) θ 180 180 = = 6.2333(10−4 )θ ◦ γ = L 350 τ=

For G, take the first 10 data points for the linear part of the curve.

T

θ (deg.) γ (10−3 )

0 0 7.7 0.38 15.3 0.80 23.0 1.24 30.7 1.64 38.3 2.01 46.0 2.40 53.7 2.85 61.4 3.25 69.0 3.80 76.7 4.50 80.0 5.10 85.0 6.48 90.0 8.01 95.0 9.58 100.0 11.18

0 0.236 865 0.498 664 0.772 929 1.022 261 1.252 893 1.495 992 1.776 491 2.025 823 2.368 654 2.804 985 3.178 983 4.039 178 4.992 873 5.971 501 6.968 829

τ (MPa) 0 20.078 52 39.896 28 59.974 8 80.053 32 99.871 08 119.949 6 140.028 1 160.106 6 179.924 4 200.002 9 = 208.608 221.646 234.684 247.722 260.76

γ (10−3 ) x

τ (MPa) y

x2

0 0.236 865 0.498 664 0.772 929 1.022 261 1.252 893 1.495 992 1.776 491 2.025 823 2.368 654 11.450 57

0 20.078 52 39.896 28 59.974 8 80.053 32 99.871 08 119.949 6 140.028 1 160.106 6 179.924 4 899.882 8

0 0.056 105 0.248 666 0.597 420 1.045 018 1.569 742 2.237 992 3.155 918 4.103 957 5.610 522 18.625 34

xy 0 4.7559 19.8948 46.3563 81.8354 125.1278 179.4436 248.7586 324.3476 426.1786 1456.6986

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Page 46


300 250

(MPa)

200 150 100 50 0

0

1

2

3

4

5

6

7

(10⫺3)

y = mx + b,

τ = y,

γ = x where m is the shear modulus G,

m=

MPa N x y − xy = 77.3 −3 = 77.3 GPa 2 2 N x − (x) 10

b=

y − mx = 1.462 MPa N

Ans.

. From curve S ys = 200 MPa Ans. Note since τ is not uniform, the offset yield does not apply, so we are using the elastic limit as an approximation. 3-15 x 38.5 39.5 40.5 41.5 42.5 43.5 44.5 45.5 46.5 47.5 48.5 49.5 = 528.0

f

fx

f x2

2 9 30 65 101 112 90 54 25 9 2 1 500

77.0 355.5 1215.0 2697.5 4292.5 4872.0 4005.0 2457.0 1162.5 427.5 97.0 49.5 21 708.0

2 964.50 14 042.25 49 207.50 111 946.30 182 431.30 211 932.00 178 222.50 111793.50 54 056.25 20 306.25 4 704.50 2 450.25 944 057.00

x¯ = 21 708/500 = 43.416, σˆ x =

944 057 − (21 7082 /500) = 1.7808 500 − 1

C x = 1.7808/43.416 = 0.041 02, y¯ = ln 43.416 − ln(1 + 0.041 022 ) = 3.7691

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47

Chapter 3

ln(1 + 0.041 022 ) = 0.0410, 1 ln x − 3.7691 2 1 g(x) = √ exp − 2 0.0410 x(0.0410) 2π σˆ y =

x

f /( N w)

g(x)

x

f /( N w)

g(x)

38 38 39 39 40 40 41 41 42 42 43 43 44 44

0 0.004 0.004 0.018 0.018 0.060 0.060 0.130 0.130 0.202 0.202 0.224 0.224 0.180

0.001 488 0.001 488 0.009 057 0.009 057 0.035 793 0.035 793 0.094 704 0.094 704 0.172 538 0.172 538 0.222 074 0.222 074 0.206 748 0.206 748

45 45 46 46 47 47 48 48 49 49 50 50

0.180 0.108 0.108 0.050 0.050 0.018 0.018 0.004 0.004 0.002 0.002 0

0.142 268 0.142 268 0.073 814 0.073 814 0.029 410 0.029 410 0.009 152 0.009 152 0.002 259 0.002 259 0.000 449 0.000 449

f (x) 0.25

Histogram PDF

0.2

0.15

0.1

0.05

0 35

40

45

S y = LN(43.42, 1.781) kpsi 3-16

50

x

Ans.

From Table A-22 AISI 1212

S y = 28.0 kpsi, σ f = 106 kpsi, Sut = 61.5 kpsi σ0 = 110 kpsi,

From Eq. (3-12)

m = 0.24,

ε f = 0.85

εu = m = 0.24

Eq. (3-10)

1 1 A0 = = 1.25 = Ai 1−W 1 − 0.2

Eq. (3-13)

εi = ln 1.25 = 0.2231 ⇒ εi < εu

shi20396_ch03.qxd

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8/18/03

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S y = σ0 εim = 110(0.2231) 0.24 = 76.7 kpsi Ans.

Eq. (3-14)

Su =

Eq. (3-15) 3-17

Page 48

61.5 Su = = 76.9 kpsi Ans. 1−W 1 − 0.2

For H B = 250, Su = 0.495 (250) = 124 kpsi = 3.41 (250) = 853 MPa

Eq. (3-17)

3-18

For the data given, H B = 2530

Ans.

H B2 = 640 226 640 226 − (2530) 2 /10 2530 = 3.887 = 253 σˆ H B = H¯ B = 9 10

Eq. (3-17)

S¯u = 0.495(253) = 125.2 kpsi

Ans.

σ¯ su = 0.495(3.887) = 1.92 kpsi Ans. 3-19

From Prob. 3-18,

H¯ B = 253

and σˆ HB = 3.887

Eq. (3-18) S¯u = 0.23(253) − 12.5 = 45.7 kpsi σˆ su = 0.23(3.887) = 0.894 kpsi 3-20

. 45.52 = 34.5 in · lbf/in3 uR = 2(30)

(a)

Ans.

Ans.

Ans.

(b) P

L

0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

A

0.1963 0.1924 0.1875 0.1563 0.1307 0.1077

A0 /A − 1

ε

0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845059

0 0.0002 0.0003 0.0005 0.000 65 0.001 15 0.0014 0.0018 0.004 45 0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845 059

σ = P/A0 0 5 032.39 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 285.02 46 297.97 45 794.73 66 427.53 76 492.30 85 550.60 82 531.17 74 479.35

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49

Chapter 3 90000 80000 70000 60000 50000 40000 30000 20000 10000 0

0

0.2

0.4 All data points

0.6

0.8

50000 45000 40000 35000 30000 25000 20000 A1

15000

A2

10000 5000 0

0

0.001

0.002 0.003 First 9 data points

0.004

0.005

90000 80000 70000 60000 50000 40000

A4

A5

30000 20000 A3 10000 0

0

0.2

0.4 0.6 Last 6 data points

0.8

5 1 . uT = Ai = (43 000)(0.001 5) + 45 000(0.004 45 − 0.001 5) 2 i=1 1 + (45 000 + 76 500)(0.059 8 − 0.004 45) 2 +81 000(0.4 − 0.059 8) + 80 000(0.845 − 0.4)

. = 66.7(103 )in · lbf/in3

Ans.

MECHANICAL ENGINEERING DESIGN TUTORIAL 4 –15: PRESSURE VESSEL DESIGN PRESSURE VESSEL DESIGN MODELS FOR CYLINDERS: 1. Thick-walled Cylinders 2. Thin-walled Cylinders THICK-WALL THEORY •

Thick-wall theory is developed from the Theory of Elasticity which yields the state of stress as a continuous function of radius over the pressure vessel wall. The state of stress is defined relative to a convenient cylindrical coordinate system: 1. σ t — Tangential Stress 2. σ r — Radial Stress 3. σ l — Longitudinal Stress

• •

Stresses in a cylindrical pressure vessel depend upon the ratio of the inner radius to the outer radius ( ro / ri ) rather than the size of the cylinder. Principal Stresses ( σ 1 , σ 2 , σ 3 ) 1. Determined without computation of Mohr’s Circle; 2. Equivalent to cylindrical stresses ( σ t , σ r , σ l )

•

Applicable for any wall thickness-to-radius ratio.

Cylinder under Pressure Consider a cylinder, with capped ends, subjected to an internal pressure, pi, and an external pressure, po, σr

ri

σl

σt σl σ r

pi

σt

ro po

FIGURE T4-15-1 Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations and figures with the prefix T refer to the present tutorial. †

The cylinder geometry is defined by the inside radius, ri , the outside radius, ro , and the cylinder length, l. In general, the stresses in the cylindrical pressure vessel ( σ t , σ r , σ l ) can be computed at any radial coordinate value, r, within the wall thickness bounded by ri and ro , and will be characterized by the ratio of radii, ζ = ro / ri . These cylindrical stresses represent the principal stresses and can be computed directly using Eq. 4-50 and 4-52. Thus we do not need to use Mohr’s circle to assess the principal stresses. Tangential Stress:

σt =

pi ri2 − po ro2 − ri2 ro2 ( po − pi ) / r 2 ro2 − ri2

for ri ≤ r ≤ ro

(Text Eq. 4-50)

for ri ≤ r ≤ ro

(Text Eq. 4-50)

Radial Stress:

σr =

pi ri2 − po ro2 + ri2 ro2 ( po − pi ) / r 2 ro2 − ri2

Longitudinal Stress:

• •

Applicable to cases where the cylinder carries the longitudinal load, such as capped ends. Only valid far away from end caps where bending, nonlinearities and stress concentrations are not significant. pi ri2 − po ro2

σl =

for ri ≤ r ≤ ro

ro2 − ri2

(Modified Text Eq. 4-52)

Two Mechanical Design Cases

1. Internal Pressure Only ( p o = 0 ) 2. External Pressure Only ( p i = 0 ) Design Case 1: Internal Pressure Only

•

Only one case to consider — the critical section which exists at r = ri .

•

Substituting p o = 0 into Eqs. (4-50) and incorporating ζ = ro / ri , the largest value of each stress component is found at the inner surface:

σ t (r = ri ) = σ t ,max

Shigley, Mischke & Budynas

ro2 + ri2 ζ 2 +1 = pi 2 2 = pi 2 = pi Cti ζ −1 ro − ri

Machine Design Tutorial 4–15: Pressure Vessel Design

(T-1)

2/10

ζ 2 + 1 ro2 + ri 2 = is a function of cylinder geometry only. where Cti = 2 ζ − 1 ro2 − ri 2 σ r (r = ri ) = σ r ,max = − pi •

Natural Boundary Condition

(T-2)

Longitudinal stress depends upon end conditions:

pi Cli

Capped Ends

(T-3a)

Uncapped Ends

(T-3b)

σl = 0

where Cli =

1 . ζ −1 2

Design Case 2: External Pressure Only

•

The critical section is identified by considering the state of stress at two points on the cylinder: r = ri and r = ro. Substituting pi = 0 into Text Eqs. (4-50) for each case: r = ri

σ r (r = ri ) = 0


σ t (r = ri ) = σ t ,max = − po where, Cto = r = ro

2ro2 2ζ 2 = − = − poCto p o ζ 2 −1 ro2 − ri2

(T-4a)

(T-4b)

2ro2 2ζ 2 = . ζ 2 − 1 ro2 − ri 2

σ r (r = ro ) = σ r ,max = − po Natural Boundary Condition (T-5a)

σ t (r = ro ) = − po

ro2 + ri2 ζ 2 +1 = − p = − poCti o ζ 2 −1 ro2 − ri2

(T-5b)

• Longitudinal stress for a closed cylinder now depends upon external pressure and radius while that of an open-ended cylinder remains zero: − poClo

Capped Ends

(T-6a)

Uncapped Ends

(T-6b)

σl = 0



3/10

where Clo =

ζ2 . ζ 2 −1

Example T4.15.1: Thick-wall Cylinder Analysis Problem Statement: Consider a cylinder subjected to an external pressure of 150 MPa and an internal pressure of zero. The cylinder has a 25 mm ID and a 50 mm OD, respectively. Assume the cylinder is capped. Find:

1. the state of stress ( σ r , σ t , σ l ) at the inner and outer cylinder surfaces; 2. the Mohr’s Circle plot for the inside and outside cylinder surfaces; 3. the critical section based upon the estimate of τ max .

Solution Methodology:

Since we have an external pressure case, we need to compute the state of stress ( σ r , σ t , σ l ) at both the inside and outside radius in order to determine the critical section. 1. As the cylinder is closed and exposed to external pressure only, Eq. (T-6a) may be applied to calculate the longitudinal stress developed. This result represents the average stress across the wall of the pressure vessel and thus may be used for both the inner and outer radii analyses. 2. Assess the radial and tangential stresses using Eqs. (T-4) and (T-5) for the inner and outer radii, respectively. 3. Assess the principal stresses for the inner and outer radii based upon the magnitudes of ( σ r , σ t , σ l ) at each radius. 4. Use the principal stresses to calculate the maximum shear stress at each radius. 5. Draw Mohr’s Circle for both states of stress and determine which provides the critical section. Solution:

1. Longitudinal Stress Calculation: ro =

OD 50 mm = = 25 mm ; 2 2

ri =

ID 25 mm = = 12.5 mm 2 2

Compute the radius ratio, ζ

ζ =

ro 25 mm = = 2.0 ri 12.5 mm



4/10

Then,

ζ2 (2) 2 = = 1.3333 mm 2 2 2 ζ − 1 (2) − 1 ζ2 = − poClo = (−150MPa)(1.3333 mm 2 ) σ l (r = ri ) = σ l (r = ro ) = − po 2 ζ −1 σ l = −200 MPa Clo =

2. Radial & Tangential Stress Calculations: Inner Radius (r = ri) 2ζ 2 2(2) 2 Cto = 2 = = 2.6667 ζ − 1 (2) 2 − 1 2ro2 σ t (r = ri ) = σ t ,max = − po 2 2 = − poCto = (−150 MPa)(2.6667) ro − ri σt (r = ri ) = −400 MPa Compressive σ r (r = ri ) = 0 Natural Boundary Condition for pi = 0

Outer Radius (r = ro) Cti =

ζ 2 + 1 (2)2 + 1 = = 1.6667 ζ 2 − 1 (2) 2 − 1

σ t (r = ro ) = σ t ,min = − po

ro2 + ri 2 = − poCti = (−150 MPa)(1.6667) ro2 − ri 2

σ t (r = ro ) = −250 MPa Compressive σ r (r = ri ) = − po = −150 MPa


3. Define Principal Stresses: Inner Radius (r = ri )

Outer Radius (r = ro )

σ 1 = σ r = 0 MPa σ 2 = σ l = −200 MPa σ 3 = σ t = −400 MPa

σ 1 = σ r = −150 MPa σ 2 = σ l = −200 MPa σ 3 = σ t = −250 MPa

4. Maximum Shear Stress Calculations: Inner Radius (r = ri ) τ max (r = ri ) =


σ 1 − σ 3 0 − (−400) = = 200 MPa 2 2


5/10

Outer Radius (r = ro ) τ max (r = ro ) =

σ 1 − σ 3 (−150) − (−250) = = 50 MPa 2 2

5. Mohr’s Circles: Inner Radius (r = ri )

τ FIGURE T4-15-2 τ max = 200 MPa

σ 1 = 0 MPa

σ 3 = -400 MPa

σ

σ 2 = -200 MPa

Outer Radius (r = ro )

τ FIGURE T4-15-3

σ 3 = -250 MPa

σ 1 = −150 MPa

•

τmax = 50 MPa

σ σ 2 = -200 MPa

Critical Section

τ max ( r = ri ) = 200 MPa ⇐ Critical Section is at Inside Radius!



6/10

THIN-WALL THEORY

• •

•

Thin-wall theory is developed from a Strength of Materials solution which yields the state of stress as an average over the pressure vessel wall. Use restricted by wall thickness-to-radius ratio: t 1 According to theory, Thin-wall Theory is justified for ≤ r 20 t 1 In practice, typically use a less conservative rule, ≤ r 10 State of Stress Definition: 1. Hoop Stress, σ t , assumed to be uniform across wall thickness. 2. Radial Stress is insignificant compared to tangential stress, thus, σ r 0. 3. Longitudinal Stress, σ l S Exists for cylinders with capped ends; S Assumed to be uniformly distributed across wall thickness; S This approximation for the longitudinal stress is only valid far away from the end-caps.

4. These cylindrical stresses (σ t , σ r , σ l ) are principal stresses (σ t , σ r , σ l ) which can be determined without computation of Mohr’s circle plot. •

Analysis of Cylinder Section

FIGURE T4-15-4 di t

FV

1

FHoop

FHoop Pressure Acting over Projected Vertical Area



7/10

The internal pressure exerts a vertical force, FV, on the cylinder wall which is balanced by the tangential hoop stress, FHoop. FV = pA proj = p{( d i )(1)} = pd i FHoop = σ t Astressed = σ t {(t )(1)} = σ t t

¦ Fy = 0 = FV − 2 FHoop = pd i − 2σ t t Solving for the tangential stress,

σt = •

pd i 2t

(Text Eq. 4-53)

Hoop Stress

Comparison of state of stress for cylinder under internal pressure verses external pressure: Internal Pressure Only pdi 2t σr = 0

σt =

σl =

pdi σ t = 4t 2

Hoop Stress By Definition Capped Case

(Text Eq.4-55)

External Pressure Only pd o 2t σr = 0

σt =

σl =

pd o σ t = 4t 2

Hoop Stress By Definition Capped Case

Example T4.15.2: Thin-wall Theory Applied to Cylinder Analysis

Problem Statement: Repeat Example T1.1 using the Thin-wall Theory (po = 150 MPa, pi = 0, ID = 25 mm, OD = 50 mm). Find: The percent difference of the maximum shear stress estimates found using the Thick-wall and Thin-wall Theories.



8/10


1. 2. 3. 4. 5.

Check t/r ratio to determine if Thin-wall Theory is applicable. Use the Thin-wall Theory to compute the state of stress Identify the principal stresses based upon the stress magnitudes. Use the principal stresses to assess the maximum shear stress. Calculate the percent difference between the maximum shear stresses derived using the Thick-wall and Thin-wall Theories.

Solution:

1. Check t/r Ratio:

t 12.5 mm 1 = = ² r 25 mm 2

1 20

or

1 10

The application of Thin-wall Theory to estimate the stress state of this cylinder is thus not justified. 2. Compute stresses using the Thin-wall Theory to compare with Thickwall theory estimates. a. Hoop Stress (average stress, uniform across wall) − po d o − (150 MPa )(50 mm) σt = = = −300 MPa 2t 2(12.5 mm) b. Radial Stress σ r = 0 by definition c. Longitudinal Stress (average stress, uniform across wall) −p d σ σ l = o o = t = −150 MPa 4t 2 3. Identify Principal Stresses in terms of “Average” Stresses:

σ 1 = σ r = 0 MPa σ 2 = σ l = −150 MPa σ 3 = σ t = −300 MPa 4. Maximum Shear Stress Calculation:

τ max =

σ 1 − σ 3 0 − (−300 MPa ) = = +150 MPa 2 2

5. Percent Difference between Thin- and Thick-wall Estimates for the Critical Section: Shigley, Mischke & Budynas


9/10

% Difference = =

τ max,Thin − τ max,Thick ∗100% τ max,Thick (+150) − (+200) ∗ (100%) = −25% (+200)

Thin -wall estimate is 25% low! ⇐



10/10

MECHANICAL ENGINEERING DESIGN TUTORIAL 4-17: PRESS AND SHRINK FITS APPLICATION OF THICK-WALL CYLINDRICAL PRESSURE VESSEL THEORY RELATING TO STRESSES DEVELOPED FROM INTERFERENCE FITS: 1. Design application which uses the cylindrical pressure vessel Thick-Wall Theory. 2. Stresses develop between cylinders due to the contact pressure generated by an interference fit. The interference fit is achieved by pressing a larger inside member into the smaller opening of an outside member. In the specific case of a shaft press fit into the hub of a gear, the outside diameter (OD) of the shaft is slightly larger than the inside hole diameter (ID) of the hub. The diametral difference between the shaft OD and the ID of the hub hole is referred to as the interference fit. •

• • •

The radial deformation required by the interference fit causes an interfacial pressure, p, to develop at the nominal radius, at r = R. Consequently, radial and tangential stresses, σ r and σ t , are produced. Assuming uncapped ends (σ l = 0), a biaxial state of stress exists for which two non-zero principal stresses must be considered. From the cylindrical pressure vessel theory, the radial and tangential stresses represent principal stresses. The length of the outer member is assumed to be equal to the length of the inner member. ro

Outer Member

δ R ri

Inner Member

R

ro

ri

(a) End view of inner and outer members, press fit together.

(b) Cross-section of cylinders showing internal outside radius larger than external inside radius by a small amount of δ.

FIGURE T4-17-1 Interference fit of two cylinders of finite length and equal lengths.

†

Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations with the prefix T refer to the present tutorial.

3. Referring to Fig. T4-17-1, the geometric features of the cylindrical parts are defined as: ri = the inside radius of the inner cylinder R = nominal radius of internal outside radius and external inside radius after assembly ro = outside radius of the outer cylinder

δ = radial interference

INSIDE CYLINDER • •

Inner member experiences an external pressure, po = p, resulting in compressive tangential and radial stresses. Thick-Wall Theory may be applied with ro = R: § R 2 + ri2 · = − = − pCit p r =R o¨ 2 2 ¸ © R − ri ¹ (σ r )i r = R = − po = − p (σ t )i

(Text Eq. 4-57)

OUTSIDE CYLINDER • •

Outer member only experiences internal pressure, pi = p, resulting in tensile tangential stress and compressive radial stress. Thick-Wall Theory is, as always, applicable with ri = R: § ro2 + R 2 · (σ t )o r = R = pi ¨ 2 = pCot 2 ¸ © ro − R ¹ (σ r )i r = R = − pi = − p

(Text Eq. 4-58)

DEFINITION OF INTERFACIAL PRESSURE We presently have two equations and three unknowns for both the inside and outside cylinder analyses. A third equation which relates the contact pressure and the interference can be derived by examining the deformation of the members. Deflection Equation The total radial interference may be defined as:

δ total = δ i + δ o where, Shigley, Mischke & Budynas

Machine Design Tutorial 4-17: Press and Shrink Fits

2/11

δ i = decrease in radius of inner cylinder δ o = increase in radius of hole The deformation may also be expressed as:

δ total = pRK i + pRK o


where the outside member constant, Ko, is defined as, 1 Eo

Ko =

ª§ ro2 + R 2 · º 1 + ν [Co +ν o ] «¨ 2 ¸ o» = 2 ¬«© ro − R ¹ ¼» Eo

Using the radius ratio form defined for the cylindrical pressure vessel formulation, ζ o = ro / R , we can define

ζ o2 + 1 Co = 2 ζ o −1 Note that the Co term is a function of geometry only, while the member constant term Ko is a function of both geometry and material parameters. Similarly for the inside member constant Ki, Ki =

1 Ei

ª§ R 2 + ri2 · º 1 «¨ 2 2 ¸ − ν i » = [Ci − ν i ] «¬© R − ri ¹ »¼ Ei

with ζ i = R / ri , Ci is defined as: Ci =

ζ i2 + 1 ζ i2 − 1

For the case of a solid shaft, ri = 0, ζ i = R / ri = ∞ and Ci = 1. We can now solve for the deformation for a given class of interference fits,

δ total = [ K o + K i ] pR Or, rearranging, the contact pressure, p, may be expressed as a function of the interference without assumptions regarding material property values: ª º δ total 1 p=« » ¬ Ko + Ki ¼ R




3/11

Example T4.17.1: Shaft & Hub Shrink Fit Problem Statement: A carbon-steel gear hub having a nominal hole diameter of 1 inch is to be shrink-fitted to a carbon-steel shaft using a class FN4 fit. The hub has a nominal thickness of ½ inch. Find:

1. the maximum tangential and radial stresses in the hub and shaft when the loosest fit is obtained; 2. the same as (a), except using the tightest fit as a condition. Solution Methodology:

1. Using Table A-1-2 of this document, identify the diametral size ranges for the shaft and the hole based on an ANSI US Customary Standard class FN4 fit which is discussed below. 2. Calculate the interferences for the loosest and tightest fits. 3. Compute the interfacial pressure for the loosest and tightest fits. 4. For each fit, calculate the radial and tangential stresses for both the hub and the shaft. Schematic: Shaft

Hub

Solution:

1. Shaft and Hub Size Ranges: This problem specifies a class FN4 fit. The class FN4 fit is a specification defined by the American National Standards Institute (ANSI), American Standard Limits for Cylindrical Parts ANSI B4.1-1978. The FN4 fit is a “Force fit suitable for parts which can be highly stressed or for shrink fits where the heavy pressing forces required are impractical.” Excerpts from the ANSI B4.1-1978 standard are provided in Table A-1 of this tutorial. From Table A-1-2, for a nominal shaft/hole diameter of 1 in., the appropriate size range is 0.95–1.19 in. The allowable tolerances, in thousandths of an inch, are:

†

Text Eq. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations with the prefix T refer to the present tutorial.

Largest Tolerance (10-3 in.)

Smallest Tolerance (10-3 in.)

+0.8 +2.3

–0.0 +1.8

Hub hole Shaft

The shaft and hub size ranges for a nominal 1 in. diameter are: Largest Diameter (in.) Hub hole Shaft

Smallest Diameter (in.)

1.0008 1.0023

1.0000 1.0018

2. Diametral and Radial Interference Calculations:

Hub hole Shaft

Loosest Fit

Tightest Fit

1.0008 1.0018 –0.0010

1.0000 1.0023 –0.0023

The radial interferences are therefore δ = –0.0005 in. (loosest fit) and δ = –0.001 15 in. (tightest fit). The interference is taken as a positive number by convention. Consequently, the radial interferences for the two cases are: δ = 0.0005 in. (loosest fit) δ = 0.001 15 in. (tightest fit)

3. Interfacial Pressure The interface pressure can be computed from Modified Eq. (4-60): ª º δ total 1 p=« » ¬ Ko + Ki ¼ R For this problem, the geometric features have been defined as: ro =

d 1 in. +t = + 0.5 in. = 1.0 in. 2 2

ri = 0.0

R=

d = 0.5 in. 2

Thus, for the hub (outside member),



5/11

ζo =

ro (1 in.) = =2 R (0.5 in.)

ζ o2 + 1 (2) 2 + 1 5 = = = 1.6667 ζ o2 − 1 (2) 2 − 1 3 1 1 Ko = [Co +ν o ] = [1.667 + 0.292] = 6.5289 ×10−8 (1/psi) 6 Eo ( 30 ×10 psi )

Co =

Similarly for the shaft (inside member)

ζi =

R =∞ ri

ζ i2 + 1 = 1.0 ζ i2 − 1 1 1 K i = [Ci − ν i ] = [1.0 − 0.292] = 2.360 ×10−8 (1/psi) 6 Ei ( 30 ×10 psi )

Ci =

For the loosest fit case:

ª 1 º δ loosest fit ª 1 º § 0.0005 · =« ploosest fit = « » ¸ −8 −8 » ¨ ¬ 6.5289 ×10 + 2.360 ×10 ¼ © 0.5 ¹ ¬ Ko + Ki ¼ R = 11 250 psi (loosest fit) For the tightest fit case: ª 1 º δ tightest fit ª 1 º § 0.00115 · =« ptightest fit = « » ¸ −8 −8 » ¨ ¬ 6.5289 × 10 + 2.360 × 10 ¼ © 0.5 ¹ ¬ K o + Ki ¼ R = 25 875 psi (tightest fit) 4. Radial & Tangential Stress Calculations for Shaft and Hub: Loosest Fit (p = 11 250 psi)

Hub:

σ r (r = R) = − p = −11 250 psi

ro2 + R 2 = pCo = (11 250 psi)(1.6667) = 18 750 psi ro2 − R 2 Shaft: σ r (r = R) = − p = −11 250 psi

σ t (r = R) = p

R 2 + ri 2 σ t (r = R) = − p 2 2 = − pCi = (−11 250 psi)(1) = −11 250 psi R − ri Tightest Fit (p = 25 875 psi) Shigley, Mischke & Budynas


6/11

Hub:

σ r (r = R) = − p = −25 875 psi

ro2 + R 2 = pCo = (25 875 psi)(1.6667) = 43 125 psi ro2 − R 2 Shaft: σ r (r = R) = − p = −25 875 psi

σ t (r = R) = p

σ t (r = R) = − p

R 2 + ri 2 = − pCi = (−25 875 psi)(1) = −25 875 psi R 2 − ri 2

Thus the shaft has equal radial and tangential stress for each tightness condition, whereas, the hub’s tangential stress is consistently higher than its radial stress for both the loosest and the tightest condition. Note: Since the shaft length is greater than the hub length, which is typical in practice, this design case violates one of the assumptions of the interfacial pressure development. In this case, there would be an increase in the interfacial pressure at each end of the hub. This condition of increased interfacial pressure would typically be accounted for by applying a stress concentration factor, Kt, for stresses calculated at points at the end of the hub such as,

σ t actual = K t , tangentialσ t σ r actual = K t , radialσ r



7/11

Table A-1 LIMITS AND FITS FOR CYLINDRICAL PARTS† The limits shown in the accompanying tabulations are in thousandths of an inch. The size ranges include all sizes over the smallest size in the range, up to and including the largest size in the range. The letter symbols are defined as follows: Running and sliding fits are intended to provide a similar running performance, with suitable lubrication allowance, throughout the range of sizes. The clearance for the first two classes, used chiefly as slide fits, increases more slowly with diameter than the other classes, so that accurate location is maintained even at the expense of free relative motion. Close sliding fits are intended for the accurate location of parts which must assemble without perceptible RC1 play. Sliding fits are intended for accurate location but with greater maximum clearance than class RC1. Parts RC2 made to this fit move and turn easily, but are not intended to run freely, and in the larger sizes may seize with small temperature changes. Precision running fits are about the closest fits which can be expected to run freely and are intended for RC3 precision work at slow speeds and light journal pressures, but are not suitable where appreciable temperature differences are likely to be encountered. Close running fits are intended chiefly for running fits on accurate machinery with moderate surface speeds RC4 and journal pressures, where accurate location and minimum play is desired. RC5–RC6 Medium running fits are intended for higher running speeds, heavy journal pressures, or both. Free running fits are intended for use where accuracy is not essential or where large temperature variations RC7 are likely to be encountered, or under both of these conditions. RC8–RC9 Loose running fits are intended for use where wide commercial tolerances may be necessary, together with an allowance, on the external member. Locational fits are fits intended to determine only the location of the mating parts; they may provide rigid or L accurate location, as with interference fits, or provide some freedom of location, as with clearance fits. Accordingly, they are divided into three groups: clearance fits, transition fits, and interference fits. Locational clearance fits are intended for parts which are normally stationary but which can be freely LC assembled or disassembled. They run from snug fits for parts requiring accuracy of location, through the medium clearance fits for parts such as ball, race, and housing, to the looser fastener fits where freedom of assembly is of prime importance. Locational transition fits are a compromise between clearance and interference fits, for application where LT accuracy of location is important, but either a small amount of clearance or interference is permissible. Locational interference fits are used where accuracy of location is of prime importance and for parts LN requiring rigidity and alignment with no special requirements for bore pressure. Such fits are not intended for parts designed to transmit frictional loads form one part to another by virtue of the tightness of fit, since these conditions are covered by force fits. Force and shrink fits constitute a special type of interference fit, normally characterized by maintenance of FN constant bore pressure throughout the range of sizes. The interference therefore varies almost directly with diameter, and the difference between its minimum and maximum value is small so as to maintain the resulting pressures within reasonable limits. Light drive fits are those requiring light assembly pressures and producing more or less permanent FN1 assemblies. They are suitable for thin sections or long fits or in cast-iron external members. Medium drive fits are suitable for ordinary steel parts or for shrink fits on light sections. They are about the FN2 tightest fits that can be used with high-grade cast-iron external members. Heavy drive fits are suitable for heavier steel parts or for shrink fits in medium sections. FN3 FN4–FN5 Force fits are suitable for parts which can be highly stressed or for shrink fits where the heavy pressing forces required are impractical. RC

†

Extracted from American Standard Limits for Cylindrical Parts ANSI B4.1-1978, with the permission of the th publishers, The American Society of Mechanical Engineers, United Engineering Center, 345 East 47 Street, New York 10017. Limit dimensions are tabulated in this standard for nominal sizes up to and including 200 in. An SI version is also available.



8/11

Table A-1-1 RUNNING AND SLIDING FITS Diameter Size Range (in.) Class RC1 RC2 RC3 RC4 RC5 RC6 RC7 RC8 RC9

0.00 - 0.12 Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft

+0.20 +0.10 +0.25 -0.10 +0.40 -0.30 +0.60 -0.30 +0.60 -0.60 +1.00 -0.60 +1.00 -1.00 +1.60 -2.50 +2.50 -4.00

-0.00 -0.25 -0.00 -0.30 -0.00 -0.55 -0.00 -0.70 -0.00 -1.00 -0.00 -1.20 -0.00 -1.60 -0.00 -3.50 -0.00 -5.60

0.12 - 0.24 +0.20 -0.15 +0.30 -0.15 +0.50 -0.40 +0.70 -0.40 +0.70 -0.80 +1.20 -0.80 +1.20 -1.20 +1.80 -2.80 +3.00 -4.50

-0.00 -0.30 -0.00 -0.35 -0.00 -0.70 -0.00 -0.90 -0.00 -1.30 -0.00 -1.50 -0.00 -1.90 -0.00 -4.00 -0.00 -6.00

0.24 - 0.40 +0.25 -0.20 +0.40 -0.20 +0.60 -0.50 +0.90 -0.50 +0.90 -1.00 +1.40 -1.00 +1.40 -1.60 +2.20 -3.00 +3.50 -5.00

-0.00 -0.35 -0.00 -0.45 -0.00 -0.90 -0.00 -1.10 -0.00 -1.60 -0.00 -1.90 -0.00 -2.50 -0.00 -4.40 -0.00 -7.20

0.40 - 0.71 +0.30 -0.25 +0.40 -0.25 +0.70 -0.60 +1.00 -0.60 +1.00 -1.20 +1.60 -1.20 +1.60 -2.00 +2.80 -3.50 +4.00 -6.00

-0.00 -0.45 -0.00 -0.55 -0.00 -1.00 -0.00 -1.30 -0.00 -1.90 -0.00 -2.20 -0.00 -3.00 -0.00 -5.10 -0.00 -8.80

Diameter Size Range (in.) Class RC1 RC2 RC3 RC4 RC5 RC6 RC7 RC8 RC9

0.71 - 1.19 Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft

+0.40 -0.30 +0.50 -0.30 +0.80 -0.80 +1.20 -0.80 +1.20 -1.60 +2.00 -1.60 +2.00 -2.50 +3.50 -4.50 +5.00 -7.00


-0.00 -0.55 -0.00 -0.70 -0.00 -1.30 -0.00 -1.60 -0.00 -2.40 -0.00 -2.80 -0.00 -3.70 -0.00 -6.50 -0.00 -10.50

1.19 - 1.97 +0.40 -0.40 +0.60 -0.40 +1.00 -1.00 +1.60 -1.00 +1.60 -2.00 +2.50 -2.00 +2.50 -3.00 +4.00 -5.00 +6.00 -8.00

-0.00 -0.70 -0.00 -0.80 -0.00 -1.60 -0.00 -2.00 -0.00 -3.00 -0.00 -3.60 -0.00 -4.60 -0.00 -7.50 -0.00 -12.00

1.97 - 3 .15 +0.50 -0.40 +0.70 -0.40 +1.20 -1.20 +1.80 -1.20 +1.80 -2.50 +3.00 -2.50 +3.00 -4.00 +4.50 -6.00 +7.00 -9.00

-0.00 -0.70 -0.00 -0.90 -0.00 -1.90 -0.00 -2.40 -0.00 -3.70 -0.00 -4.30 -0.00 -5.80 -0.00 -9.00 -0.00 -13.50


3.15 - 4.73 +0.60 -0.50 +0.90 -0.50 +1.40 -1.40 +2.20 -1.40 +2.20 -3.00 +3.50 -3.00 +3.50 -5.00 +5.00 -7.00 +9.00 -10.00

-0.00 -0.90 -0.00 -1.10 -0.00 -2.30 -0.00 -2.80 -0.00 -4.40 -0.00 -5.20 -0.00 -7.20 -0.00 -10.50 -0.00 -15.00

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Table A-1-2 FORCE AND SHRINK FITS Diameter Size Range (in.) Class FN1 FN2 FN3 FN4 FN5

0.00 - 0.12 Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft

0.12 - 0.24

0.24 - 0.40

0.40 - 0.56

+0.25 +0.50 +0.40 +0.85

-0.00 +0.30 -0.00 +0.60

+0.30 +0.60 +0.50 +1.00

-0.00 +0.40 -0.00 +0.70

+0.40 +0.75 +0.60 +1.40

-0.00 +0.50 -0.00 +1.00

+0.40 +0.80 +0.70 +1.60

-0.00 +0.50 -0.00 +1.20

+0.40 +0.95 +0.60 +1.30

-0.00 +0.70 -0.00 +0.90

+0.50 +1.20 +0.70 +1.70

-0.00 +0.90 -0.00 +1.20

+0.60 +1.60 +0.90 +2.00

-0.00 +1.20 -0.00 +1.40

+0.70 +1.80 +1.00 +2.30

-0.00 +1.40 -0.00 +1.60

Diameter Size Range (in.) Class FN1 FN2 FN3 FN4 FN5


0.71 - 0.95

+0.40 +0.90 +0.70 +1.60

-0.00 +0.60 -0.00 +1.20

+0.50 +1.10 +0.80 +1.90

-0.00 +0.70 -0.00 +1.40

+0.70 +1.80 +1.00 +2.50

-0.00 +1.40 -0.00 +1.80

+0.80 +2.10 +1.20 +3.00

-0.00 +1.60 -0.00 +2.20

0.95 - 1.19 +0.50 +1.20 +0.80 +1.90 +0.80 +2.10 +0.80 +2.30 +1.20 +3.30

-0.00 +0.80 -0.00 +1.40 -0.00 +1.60 -0.00 +1.80 -0.00 +2.50

1.19 - 1.58 +0.60 +1.30 +1.00 +2.40 +1.00 +2.60 +1.00 +3.10 +1.60 +4.00

-0.00 +0.90 -0.00 +1.80 -0.00 +2.00 -0.00 +2.50 -0.00 +3.00

Diameter Size Range (in.) Class FN1 FN2 FN3 FN4 FN5


+0.60 +1.40 +1.00 +2.40 +1.00 +2.80 +1.00 +3.40 +1.60 +5.00


-0.00 +1.00 -0.00 +1.80 -0.00 +2.20 -0.00 +2.80 -0.00 +4.00

1.97 - 2.56 +0.70 +1.80 +1.20 +2.70 +1.20 +3.20 +1.20 +4.20 +1.80 +6.20

-0.00 +1.30 -0.00 +2.00 -0.00 +2.50 -0.00 +3.50 -0.00 +5.00

2.56 - 3 .15 +0.70 +1.90 +1.20 +2.90 +1.20 +3.70 +1.20 +4.70 +1.80 +7.20


-0.00 +1.40 -0.00 +2.20 -0.00 +3.00 -0.00 +4.00 -0.00 +6.00

3.15 - 3.94 +0.90 +2.40 +1.40 +3.70 +1.40 +4.40 +1.40 +5.90 +2.20 +8.40

10/11

-0.00 +1.80 -0.00 +2.80 -0.00 +3.50 -0.00 +5.00 -0.00 +7.00

Table A-1-2 FORCE AND SHRINK FITS (CONTINUED) Diameter Size Range (in.) Class FN1 FN2 FN3 FN4 FN5


+0.90 +2.60 +1.40 +3.90 +1.40 +4.90 +1.40 +6.90 +2.20 +9.40


-0.00 +2.00 -0.00 +3.00 -0.00 +4.00 -0.00 +6.00 -0.00 +8.00

4.73 - 5.52 +1.00 +2.90 +1.60 +4.50 +1.60 +6.00 +1.60 +8.00 +2.50 +11.60

-0.00 +2.20 -0.00 +3.50 -0.00 +5.00 -0.00 +7.00 -0.00 +10.00

5.52 - 6.30 +1.00 +3.20 +1.60 +5.00 +1.60 +6.00 +1.60 +8.00 +2.50 +13.60

-0.00 +2.50 -0.00 +4.00 -0.00 +5.00 -0.00 +7.00 -0.00 +12.00


6.30 - 7.09 +1.00 +3.50 +1.60 +5.50 +1.60 +7.00 +1.60 +9.00 +2.50 +13.60

11/11

-0.00 +2.80 -0.00 +4.50 -0.00 +6.00 -0.00 +8.00 -0.00 +12.00

MECHANICAL ENGINEERING DESIGN TUTORIAL 4-20: HERTZ CONTACT STRESSES CHARACTERISTICS OF CONTACT STRESSES 1. Represent compressive stresses developed from surface pressures between two curved bodies pressed together; 2. Possess an area of contact. The initial point contact (spheres) or line contact (cylinders) become area contacts, as a result of the force pressing the bodies against each other; 3. Constitute the principal stresses of a triaxial (three dimensional) state of stress; 4. Cause the development of a critical section below the surface of the body; 5. Failure typically results in flaking or pitting on the bodies’ surfaces. TWO DESIGN CASES Two design cases will be considered, 1. Sphere – Sphere Contact (Point Contact Circular Contact Area) 2. Cylinder – Cylinder Contact (Line Contact Rectangular Contact Area) SPHERE – SPHERE CONTACT F

F

x

x

d1 y

y

d2 2a

F

F z

z (a) Two spheres held in contact by force F.

(b) Contact stress has an elliptical distribution across contact over zone of diameter 2a.

TEXT FIGURE 4-42 Two Spheres in Contact

†

Text. refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke, and Richard G. Budynas; equations and figures with the prefix T refer to the present tutorial.

Consider two solid elastic spheres held in contact by a force F such that their point of contact expands into a circular area of radius a, given as: a = Ka 3 F

(Modified Text Eq. 4-72) 1/ 3

ª 3 (1 − ν12 ) / E1 + (1 − ν 22 ) / E2 º where K a = « » (1/ d1 ) + (1/ d 2 ) ¬8 ¼ F = applied force

ν1 ,ν 2 = Poisson's ratios for spheres 1 and 2 E1 , E2 = elastic modulii for spheres 1 and 2 d1 , d 2 = diameters of spheres 1 and 2 This general expression for the contact radius can be applied to two additional common cases: 1. Sphere in contact with a plane (d 2 = ∞); 2. Sphere in contact with an internal spherical surface or ‘cup’ (d 2 = − d ). Returning to the sphere-sphere case, the maximum contact pressure, pmax , occurs at the center point of the contact area. pmax =

3F 2π a 2

(Text Eq. 4-73)

State of Stress •

The state of stress is computed based on the following mechanics: 1. Two planes of symmetry in loading and geometry dictates that σ x = σ y ; 2. The dominant stress occurs on the axis of loading: σ max = σ z ; 3. The principal stresses are σ 1 = σ 2 = σ x = σ y and σ 3 = σ z given σ 1 , σ 2 ≥ σ 3 ; 4. Compressive loading leads to σ x , σ y , and σ z being compressive stresses.

•

Calculation of Principal Stresses ªª § 1 « σ x = − pmax «1 − ζ a tan −1 ¨¨ « ¬« © ζa ¬ = σ y = σ1 = σ 2


·º 1 ¸¸ » (1 + ν ) − 2 1 + ζ a2 ¹ ¼»

(

)

º » » ¼


Machine Design Tutorial 4-20: Hertz Contact Stresses

2/10

σ3 = σ z = where

•

− pmax 1 + ζ a2


ζ a = z / a = nondimensional depth below the surface ν = Poisson's ratio for the sphere examined (1 or 2)

Mohr’s Circle Plotting the principal stresses on a Mohr’s circle plot results in: one circle, defined by σ 1 = σ 2 , shrinking to a point; and two circles, defined by σ 1 , σ 3 and σ 2 , σ 3 , plotted on top of each other. The maximum shear stress, τ max , for the plot is calculated as:

τ max =

σ1 − σ 3 σ x − σ z σ y − σ z = = 2 2 2


If the maximum shear stress, τ max , and principal stresses, σ 1 , σ 2 , and σ 3 , are plotted as a function of maximum pressure, p max , below the surface contact point, the plot of Fig. 4-43 is generated. This plot, based on a Poisson’s ratio of ν = 0.3 , reveals that a critical section exists on the load axis, approximately 0.48a below the sphere surface. Many authorities theorize that this maximum shear stress is responsible for the surface fatigue failure of such contacting elements; a crack, originating at the point of maximum shear, progresses to the surface where lubricant pressure wedges a chip loose and thus creates surface pitting.

TEXT FIGURE 4-43: Magnitude of the stress components below the surface as a function of maximum pressure of contacting spheres. Shigley, Mischke & Budynas


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CYLINDER–CYLINDER CONTACT

Consider two solid elastic cylinders held in contact by forces F uniformly distributed along the cylinder length l.

x

F

F

x

d1 l

y

y

d2 2b

F

F

z

z (a) Two right circular cylinders held in contact by forces F uniformly distributed along cylinder length l.

(b) Contact stress has an elliptical distribution across contact zone of width 2b.

TEXT FIGURE 4-44 Two Cylinders in Contact

The resulting pressure causes the line of contact to become a rectangular contact zone of halfwidth b given as: b = Kb F


ª 2 (1 − ν12 ) / E1 + (1 − ν 22 ) / E2 º where Kb = « » (1/ d1 ) + (1/ d 2 ) ¬π l ¼ F = applied force ν1 ,ν 2 = Poisson's ratios for cylinders 1 and 2 E1 , E2 = elastic modulii for cylinders 1 and 2 d1 , d 2 = diameters of spheres 1 and 2 l = length of cylinders 1 and 2 (l1 = l2 assumed)



4/10

This expression for the contact half-width, b, is general and can be used for two additional cases which are frequently encountered: 1. Cylinder in contact with a plane, e.g. a rail (d 2 = ∞); 2. Cylinder in contact with an internal cylindrical surface, for example the race of a roller bearing (d 2 = − d ). The maximum contact pressure between the cylinders acts along a longitudinal line at the center of the rectangular contact area, and is computed as: pmax =

2F π bl

(Text Eq. 4-78)

State of Stress

•

The state of stress is computed based on the following mechanics: 1. One plane of symmetry in loading and geometry dictates that σ x ≠ σ y ; 2. The dominant stress occurs along the axis of loading: σ max = σ z ; 3. The principal stresses are equal to σ x , σ y , and σ z with σ 3 = σ z ; 4. Compressive loading leads to σ x , σ y , and σ z being compressive stresses.

•

Calculation of Principal Stresses and Maximum Shear Stress

σ 3 = σ z = − pmax

1


1 + ζ b2

°σ x for 0 ≤ ζ b ≤ 0.436 σ1 = ® °¯σ y for 0.436 ≤ ζ b where,

σ x = −2ν pmax ª« 1 + ζ b2 − ζ b º» ¬ ¼ ª§ 1 + 2ζ 2 · b ¸ σ y = − pmax «¨ − 2 ζb «¨ 1 + ζ 2 ¸ b ¹ ¬© ζb = z /b


(Modified Text Eq. 4-79) º » » ¼



5/10

The maximum shear stress is thus given as: °τ1/ 3 = (σ z − σ x ) / 2 for 0 ≤ ζ b ≤ 0.436 τ max = ® °¯τ1/ 3 = (σ z − σ y ) / 2 for 0.436 ≤ ζ b

When these equations are plotted as a function of maximum contact pressure up to a distance 3b below the surface contact point, the plot of Fig. 4-45 is generated. Based on a Poisson’s ratio of 0.3, this plot reveals that τ max attains a maxima for ζ b = z / b = 0.786 and 0.3pmax.

TEXT FIGURE 4-45: Magnitude of stress components below the surface as a function of maximum pressure for contacting cylinders.

Example T4.20.1: Problem Statement: A 6-in-diameter cast-iron wheel, 2 in wide, rolls on a flat steel surface carrying a 800 lbf load. Find:


1. The Hertzian stresses σ x , σ y , σ z and τ 1/3 in the cast iron wheel at the critical section; 2. The comparative state of stress and maximum shear stress, arising during a revolution, at point A located 0.015 inch below the wheel rim surface.


6/10


1. Compute the value of the contact half-width, b. 2. Compute the maximum pressure generated by the normal force of the wheel. 3. Use the results of steps (1) and (2) to calculate the contact stresses in the cast iron wheel for the critical section, z/b = 0.786. 4. Evaluate the principal stresses based upon the contact stress calculations. 5. Calculate the maximum shear stress. 6. Compare these results with those obtained by using Fig. 4-45. 7. During a single revolution of the wheel, point A will experience a cycle of stress values varying from zero (when point A lies well outside the contact zone) to a maximum state of stress (when A lies within the contact zone and on the line of action of the 800 lbf force.) We expect point A to “feel” the effects of a semi-elliptical contact pressure distribution as point A moves into and through the contact zone. Thus, we need to calculate the contact stresses for a depth of z = 0.015 inch, which we expect to lie within the contact zone.

Schematic: Ecast iron = 14.5 × 106 psi

ν cast iron = 0.211

800 lbf

6 in

2 in A·

Esteel = 30 × 106 psi

ν steel = 0.292

Solution:

1. Compute contact half-width, b Material Properties: E1 = Ecast iron = 14.5 × 106 psi; ν 1 = ν cast iron = 0.211 E2 = Esteel = 30.0 × 106 psi; ν 2 = ν steel = 0.292



7/10

d1 = 6.0 in; d 2 = ∞; l = 2.0 in

Dimensions: b = Kb F


ª 2 (1 − ν12 ) / E1 + (1 − ν 22 ) / E2 º Kb = « » (1/ d1 ) + (1/ d 2 ) ¬π l ¼

1/ 2

° 2 ª¬1 − (0.211) 2 º¼ /(14.5 ×106 ) + ª¬1 − (0.292) 2 º¼ /(30.0 × 106 ) ½° =® ¾ (1/ 6.0) + (1/ ∞) °¯ π (2.0) °¿ = 4.291× 10−4 in/ lbf b = Kb F = (4.291×10 −4 in/ lbf )(800 lbf )1/2 = 1.214 × 10-2 in

2. Maximum Pressure, pmax pmax =

2F 2(800 lbf) = = 20 980 psi π bl π (1.214 ×10−2 in)(2.0 in)

3. Hertz Contact Stresses in Cast Iron Wheel At the critical section, ζ b = z / b = 0.786 ,

σ x = −2ν1 pmax ª« 1 + ζ b2 − ζ b º» ¬ ¼ = −2(0.211)(20 980 psi) ª 1+(0.786)2 − 0.786 º «¬ »¼ = −4302 psi ª§ 1 + 2ζ 2 b σ y = − pmax «¨ 2 «¨ ¬© 1 + ζ b

· ¸ − 2 ζb ¸ ¹

º » » ¼

ª 1 + 2(0.786) 2 º ½ ° ° « » = (−20 980 psi) ® − 2(0.786) ¾ 2 °¯ «¬ 1 + (0.786) »¼ °¿ = −3895 psi

σ z = − pmax

1 1 + ζ b2

=

−20 980 psi 1+(0.786)2

= −16 490 psi



8/10

Note that the small contact area involved in this type of problem gives rise to very high pressure, relative to the applied force, and thus exceptionally high stresses. 4. Since σ x , σ y , and σ z are all principal stresses, we can conclude:

σ 1 = σ y = − 3895 psi σ 2 = σ x = − 4302 psi σ 3 = σ z = −16 490 psi 5. Maximum Shear Stress

σ 1 − σ 3 σ y − σ z −3895 psi − (−16 490 psi) = = 2 2 2 = 6298 psi

τ max = τ1/ 3 =

6. Comparison with results based on Text Figure 4-45: For z/b ≈ 0.75,

σ x ≈ −0.3 pmax = −6294 psi σ y ≈ −0.2 pmax = −4196 psi σ z ≈ −0.8 pmax = −16 780 psi τ max ≈ 0.3 pmax = 6294 psi Comparing these results with those calculated using a value of ν = 0.211, we find that only σ x is a function of ν; σ y , σ z , and τ max are independent of ν since the graphical estimates of their values are within 3 % of those obtained from the plot which assumes a Poisson’s ratio of 0.3. 7. For a depth of 0.015 in below the cylinder surface,

ζb =

0.015 in = 1.236 1.214 × 10 −2 in

Substituting, Shigley, Mischke & Budynas


9/10

σ x = −2ν pmax ª« 1 + ζ b2 − ζ b º» ¬ ¼ = −2(0.211)(20 980 psi) ª 1+(1.236)2 − 1.236 º «¬ »¼ = −3133 psi ª§ 1 + 2ζ 2 b σ y = − pmax «¨ «¨ 1 + ζ 2 b ¬©

· ¸ − 2 ζb ¸ ¹

º » » ¼

ª 1 + 2(1.236) 2 º ½ ° ° » − 2 1.236 ¾ = (−20 980 psi) ® « 2 » « ¯° ¬ 1 + (1.236) ¼ ¿° = −1652 psi

σ z = − pmax

1 1 + ζ b2

=

−20 980 psi 1+(1.236) 2

= −13 200 psi

τ max =

σ 1 − σ 3 σ y − σ z −1652 − (−13 200) = = = 5774 psi 2 2 2

As expected, at a depth corresponding greater than the critical section ( z/b = 1.236 > 0.786), the magnitudes of all three principal stresses are smaller than those calculated for z/b = 0.786. The difference between the principal stresses is also smaller and consequently, τmax also decreases.



10/10

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Chapter 4 4-1 W

W 2

2 B

A RA

B

A

RB

RA

1

RB 1

1

(a)

(b) 1

1 RD

RC

D

2

3

C

A

RA RB

B

W

(c)

W 1

RC RB

RA RB

2 RA W

(d) (e) A

2 W RBx B RBx 1

RB

RBy

RBy

Scale of corner magnified

(f)

RA

1

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51

Chapter 4

4-2 (a)

2 kN

R A = 2 sin 60 = 1.732 kN

RB

60°

90°

R B = 2 sin 30 = 1 kN

2 2 kN

30°

60° RB

Ans.

Ans.

RA

30°

RA

1

(b)

RA

0.4 m

S = 0.6 m

B

A

α = tan−1

45° 800 N

0.6 = 30.96◦ 0.4 + 0.6

0.6 m

RO

O s

800 RA = sin 135 sin 30.96 800 RO = sin 14.04 sin 30.96

RO RA 135° 30.96°

800 N

45 30.96 14.04

⇒

R A = 1100 N Ans.

⇒

R O = 377 N Ans.

30.96°

(c)

1.2 = 2.078 kN Ans. tan 30 1.2 = 2.4 kN Ans. RA = sin 30

1.2 kN

RO = 30° RA

RO

60°

90°

60° 1.2 kN

RA RO

(d) Step 1: Find R A and R E

4.5 = 7.794 m tan 30 MA = 0 +

h=

4.5 m

C

30° y

400 N

4

2

9R E − 7.794(400 cos 30) − 4.5(400 sin 30) = 0 R E = 400 N Ans.

h B RAx RA

3

D 60°

A RAy

x

E 9m

RE

Fx = 0

R Ax + 400 cos 30 = 0

⇒

R Ax = −346.4 N

Fy = 0

R Ay + 400 − 400 sin 30 = 0 ⇒ R Ay = −200 N R A = 346.42 + 2002 = 400 N Ans.

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Step 2: Find components of RC on link 4 and R D (RCy)4 MC = 0 + (RCx)4

C

400(4.5) − (7.794 − 1.9) R D = 0 ⇒ Fx = 0 ⇒ ( RC x ) 4 = 305.4 N Fy = 0 ⇒ ( RC y ) 4 = −400 N

4 RD

D

R D = 305.4 N

Ans.

E 400 N

Step 3: Find components of RC on link 2 (RCy)2 Fx = 0 (RCx)2

C

( RC x ) 2 + 305.4 − 346.4 = 0 Fy = 0

2

⇒

( RC x ) 2 = 41 N

( RC y ) 2 = 200 N

B

305.4 N

346.4 N A 200 N 400 N

200 N 41 N

200 N C

305.4 N

400 N C

41 N

305.4 N

30° Pin C

400 N

305.4 N

B

D

B

346.4 N

305.4 N

D

E

A 200 N

400 N

400 N

Ans. 4-3 (a)

y

40 lbf 4"

O

60 lbf 4"

6"

A R1

4"

B

C

30 lbf

R2

D

+

−18(60) + 14R2 + 8(30) − 4(40) = 0 R2 = 71.43 lbf Fy = 0: R1 − 40 + 30 + 71.43 − 60 = 0

x

R1 = −1.43 lbf

60 11.43

41.43

M (lbf • in) O

M4 M1 M2

M3

M0 = 0

x

V (lbf) O 1.43

x

M1 M2 M3 M4

= −1.43(4) = −5.72 lbf · in = −5.72 − 41.43(4) = −171.44 lbf · in = −171.44 − 11.43(6) = −240 lbf · in = −240 + 60(4) = 0 checks!

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53

Chapter 4

(b)

y

A

O MO

4 kN/m

2 kN

200 mm

RO

B

C

150 mm

R0 = 2 + 4(0.150) = 2.6 kN M0 = 0

x

150 mm

M0 = 2000(0.2) + 4000(0.150)(0.425) = 655 N · m

V (kN) 2.6 0.6

O

O

x

M1 = −655 + 2600(0.2) = −135 N · m

M (N • m) M3

O

M2 = −135 + 600(0.150) = −45 N · m 1 M3 = −45 + 600(0.150) = 0 checks! 2

x

M2

M1 655

(c)

Fy = 0

y

1000 lbf 6 ft

O

4 ft

B

x

A

M0 = 0: 10R2 − 6(1000) = 0 Fy = 0: R1 − 1000 + 600 = 0

⇒

R2 = 600 lbf

⇒

R2

R1 V (lbf) 400

x

O

600 M (lbf • ft)

M1 M2

O

(d)

y

O

2 ft

+

2000 lbf

1000 lbf 6 ft

2 ft

A

C

B

x

R2

R1 1200

x

M1 = 400(6) = 2400 lbf · ft M2 = 2400 − 600(4) = 0 checks!

200

x

MC = 0

−10R1 + 2(2000) + 8(1000) = 0 R1 = 1200 lbf Fy = 0: 1200 − 1000 − 2000 + R2 = 0 R2 = 1800 lbf

1800 M M1 O

M2 M3

x

M1 = 1200(2) = 2400 lbf · ft M2 = 2400 + 200(6) = 3600 lbf · ft M3 = 3600 − 1800(2) = 0 checks!

R1 = 400 lbf

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(e)

y

400 lbf 4 ft

O

+

800 lbf

3 ft B 3 ft A

R1

R1 = −171.4 lbf Fy = 0: −171.4 − 400 + R2 − 800 = 0

R2

V (lbf)

800

R2 = 1371.4 lbf

x

O 171.4

MB = 0

−7R1 + 3(400) − 3(800) = 0

x

C

571.4

M M3

O

M1 = −171.4(4) = −685.7 lbf · ft M2 = −685.7 − 571.4(3) = −2400 lbf · ft M3 = −2400 + 800(3) = 0 checks!

x

M1 M2

(f) Break at A 40 lbf/in

O

1 R1 = V A = 40(8) = 160 lbf 2

A

8" R1

VA

y

+

320 lbf

160 lbf 5"

A B 2"

5"

D R3

40 lbf/in

320 lbf x

160 lbf

352 lbf

MD = 0

12(160) − 10R2 + 320(5) = 0 R2 = 352 lbf

C

R2

128 lbf

Fy = 0

−160 + 352 − 320 + R3 = 0 R3 = 128 lbf

V (lbf) 192

160

x

O 128

160 M

M4 M1

O

M5 M2 M3

x

1 M1 = 160(4) = 320 lbf · in 2 1 M2 = 320 − 160(4) = 0 checks! (hinge) 2 M3 = 0 − 160(2) = −320 lbf · in M4 = −320 + 192(5) = 640 lbf · in M5 = 640 − 128(5) = 0

checks!

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55

Chapter 4

4-4 (a)

q = R1 x−1 − 40x − 4−1 + 30x − 8−1 + R2 x − 14−1 − 60x − 18−1 V = R1 − 40x − 40 + 30x − 80 + R2 x − 140 − 60x − 180 M = R1 x − 40x − 41 + 30x − 81 + R2 x − 141 − 60x − 181 for x = 18+

V =0

and M = 0

0 = R1 − 40 + 30 + R2 − 60

(1) (2)

Eqs. (1) and (2) give ⇒

R1 + R2 = 70

0 = R1 (18) − 40(14) + 30(10) + 4R2

⇒

(3)

9R1 + 2R2 = 130

(4)

Solve (3) and (4) simultaneously to get R1 = −1.43 lbf, R2 = 71.43 lbf. Ans. From Eqs. (1) and (2), at x = 0+ , V = R1 = −1.43 lbf, M = 0 x = 4+ : x = 8+ :

V V M x = 14+ : V M + x = 18 : V

= −1.43 − 40 = −41.43, M = −1.43x = −1.43 − 40 + 30 = −11.43 = −1.43(8) − 40(8 − 4) 1 = −171.44 = −1.43 − 40 + 30 + 71.43 = 60 = −1.43(14) − 40(14 − 4) + 30(14 − 8) = −240 . = 0, M = 0 See curves of V and M in Prob. 4-3 solution.

(b) q = R0 x−1 − M0 x−2 − 2000x − 0.2−1 − 4000x − 0.350 + 4000x − 0.50 V = R0 − M0 x−1 − 2000x − 0.20 − 4000x − 0.351 + 4000x − 0.51 M = R0 x − M0 − 2000x − 0.21 − 2000x − 0.352 + 2000x − 0.52

(1) (2)

at x = 0.5+ m, V = M = 0, Eqs. (1) and (2) give R0 − 2000 − 4000(0.5 − 0.35) = 0 ⇒ R1 = 2600 N = 2.6 kN R0 (0.5) − M0 − 2000(0.5 − 0.2) − 2000(0.5 − 0.35) 2 = 0

Ans.

with R0 = 2600 N, M0 = 655 N · m Ans. With R0 and M0, Eqs. (1) and (2) give the same V and M curves as Prob. 4-3 (note for V, M0 x−1 has no physical meaning). q = R1 x−1 − 1000x − 6−1 + R2 x − 10−1 V = R1 − 1000x − 60 + R2 x − 100 M = R1 x − 1000x − 61 + R2 x − 101

(c)

at x = 10+ ft, V = M = 0, Eqs. (1) and (2) give R1 − 1000 + R2 = 0 10R1 − 1000(10 − 6) = 0 0 ≤ x ≤ 6: 6 ≤ x ≤ 10:

⇒ ⇒

R1 + R2 = 1000 R1 = 400 lbf , R2 = 1000 − 400 = 600 lbf

V = 400 lbf, M = 400x V = 400 − 1000(x − 6) 0 = 600 lbf M = 400x − 1000(x − 6) = 6000 − 600x

See curves of Prob. 4-3 solution.

(1) (2)

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(d)

q = R1 x−1 − 1000x − 2−1 − 2000x − 8−1 + R2 x − 10−1 V = R1 − 1000x − 20 − 2000x − 80 + R2 x − 100 M = R1 x − 1000x − 21 − 2000x − 81 + R2 x − 101

(1) (2)

At x = 10+ , V = M = 0 from Eqs. (1) and (2) R1 − 1000 − 2000 + R2 = 0

⇒

R1 + R2 = 3000

10R1 − 1000(10 − 2) − 2000(10 − 8) = 0

⇒

R1 = 1200 lbf , R2 = 3000 − 1200 = 1800 lbf

0 ≤ x ≤ 2: 2 ≤ x ≤ 8:

V = 1200 lbf, M = 1200x lbf · ft V = 1200 − 1000 = 200 lbf M = 1200x − 1000(x − 2) = 200x + 2000 lbf · ft

8 ≤ x ≤ 10:

V = 1200 − 1000 − 2000 = −1800 lbf M = 1200x − 1000(x − 2) − 2000(x − 8) = −1800x + 18 000 lbf · ft

Plots are the same as in Prob. 4-3. (e)

q = R1 x−1 − 400x − 4−1 + R2 x − 7−1 − 800x − 10−1 V = R1 − 400x − 40 + R2 x − 70 − 800x − 100 M = R1 x − 400x − 41 + R2 x − 71 − 800x − 101

(1) (2)

at x = 10+ , V = M = 0 R1 − 400 + R2 − 800 = 0 10R1 − 400(6) + R2 (3) = 0

⇒ ⇒

R1 + R2 = 1200 10R1 + 3R2 = 2400

(3) (4)

Solve Eqs. (3) and (4) simultaneously: R1 = −171.4 lbf, R2 = 1371.4 lbf 0 ≤ x ≤ 4: 4 ≤ x ≤ 7: 7 ≤ x ≤ 10:

V = −171.4 lbf, M = −171.4x lbf · ft V = −171.4 − 400 = −571.4 lbf M = −171.4x − 400(x − 4) lbf · ft = −571.4x + 1600 V = −171.4 − 400 + 1371.4 = 800 lbf M = −171.4x − 400(x − 4) + 1371.4(x − 7) = 800x − 8000 lbf · ft

Plots are the same as in Prob. 4-3. (f) q = R1 x−1 − 40x0 + 40x − 80 + R2 x − 10−1 − 320x − 15−1 + R3 x − 20 V = R1 − 40x + 40x − 81 + R2 x − 100 − 320x − 150 + R3 x − 200 (1) 2 2 1 1 1 M = R1 x − 20x + 20x − 8 + R2 x − 10 − 320x − 15 + R3 x − 20 (2) 2 M = 0 at x = 8 in ∴ 8R1 − 20(8) = 0 ⇒ R1 = 160 lbf at x = 20+ ,

V and M = 0

160 − 40(20) + 40(12) + R2 − 320 + R3 = 0 160(20) − 20(20) 2 + 20(12) 2 + 10R2 − 320(5) = 0

⇒ ⇒

R2 + R3 = 480 R2 = 352 lbf R3 = 480 − 352 = 128 lbf 2 0 ≤ x ≤ 8: V = 160 − 40x lbf, M = 160x − 20x lbf · in 8 ≤ x ≤ 10: V = 160 − 40x + 40(x − 8) = −160 lbf , M = 160x − 20x 2 + 20(x − 8) 2 = 1280 − 160x lbf · in

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57

Chapter 4

10 ≤ x ≤ 15:

V = 160 − 40x + 40(x − 8) + 352 = 192 lbf M = 160x − 20x 2 + 20(x − 8) + 352(x − 10) = 192x − 2240

15 ≤ x ≤ 20:

V = 160 − 40x + 40(x − 8) + 352 − 320 = −128 lbf M = 160x − 20x 2 − 20(x − 8) + 352(x − 10) − 320(x − 15) = −128x + 2560

Plots of V and M are the same as in Prob. 4-3. 4-5 Solution depends upon the beam selected. 4-6 (a) Moment at center, xc = (l − 2a)/2 2 w l wl l l Mc = (l − 2a) − −a = 2 2 2 2 4 At reaction, |Mr | = wa 2 /2 a = 2.25, l = 10 in, w = 100 lbf/in 100(10) Mc = 2 Mr =

10 − 2.25 = 125 lbf · in 4

100(2.252 ) = 253.1 lbf · in Ans. 2

(b) Minimum occurs when Mc = |Mr | wa 2 wl l −a = 2 4 2

⇒

a 2 + al − 0.25l 2 = 0

Taking the positive root l √

1 −l + l 2 + 4(0.25l 2 ) = a= 2 − 1 = 0.2071l 2 2 for l = 10 in and w = 100 lbf,

Mmin = (100/2)[(0.2071)(10)]2 = 214.5 lbf · in

4-7 For the ith wire from bottom, from summing forces vertically (a)

Ti xi

Ti = (i + 1)W

a W

Ans.

iW

From summing moments about point a, Ma = W (l − xi ) − i W xi = 0 Giving, xi =

l i +1

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So l l = 1+1 2 l l = x= 2+1 3 l l = y= 3+1 4 l l = z= 4+1 5

W =

(b) With straight rigid wires, the mobile is not stable. Any perturbation can lead to all wires becoming collinear. Consider a wire of length l bent at its string support: Ti il i1

l i1

iW

W

Ma = 0 Ma =

ilW iWl cos α − cos β = 0 i +1 i +1

iWl (cos α − cos β) = 0 i +1 Moment vanishes when α = β for any wire. Consider a ccw rotation angle β, which makes α → α + β and β → α − β iWl [cos(α + β) − cos(α − β)] i +1 2i W l . 2i W lβ = sin α sin β = sin α i +1 i +1

Ma =

There exists a correcting moment of opposite sense to arbitrary rotation β. An equation for an upward bend can be found by changing the sign of W . The moment will no longer be correcting. A curved, convex-upward bend of wire will produce stable equilibrium too, but the equation would change somewhat. 4-8 (a)

1

x

cw 2s 2p 2

C R

(6, 4

ccw

ccw y

) 2

D

1

12 + 6 =9 2 12 − 6 CD = =3 2 R = 32 + 42 = 5 C=

(12, 4cw)

σ1 = 5 + 9 = 14 σ2 = 9 − 5 = 4

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59

Chapter 4

1 −1 4 = 26.6◦ cw φ p = tan 2 3

4

x 26.6 14

φs = 45◦ − 26.6◦ = 18.4◦ ccw

τ1 = R = 5, 3

3 18.4

x

5

3 3

(b)

y

1

(9, 5cw) R D

2

C

1

2p

σ1 = 6.1 + 12.5 = 18.6 1 5 φ p = tan−1 = 27.5◦ ccw 2 3.5 σ2 = 12.5 − 6.1 = 6.4

2s (16, 5ccw)

ccw

9 + 16 = 12.5 2 16 − 9 CD = = 3.5 2 R = 52 + 3.52 = 6.10 C=

cw

2

x

6.4 18.6 27.5 x

τ1 = R = 6.10 , φs = 45◦ − 27.5◦ = 17.5◦ cw 12.5 6.10 x 17.5 12.5

(c)

1 cw

y

C=

(24, 6cw) R D

2

C

(10, 6 ccw

)

2s

x 2

24 − 10 =7 2 R = 72 + 62 = 9.22

CD = 1

2p ccw

24 + 10 = 17 2

σ1 = 17 + 9.22 = 26.22 σ2 = 17 − 9.22 = 7.78

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1 −1 7 90 + tan = 69.7◦ ccw φp = 2 6

26.22 69.7 7.78 x

φs = 69.7◦ − 45◦ = 24.7◦ ccw

τ1 = R = 9.22, 17

9.22 17 24.7 x

(d)

x cw

1

C=

(9, 8cw) 2s

D

1

C R

(19, 8ccw)

ccw

2

19 − 9 =5 2 R = 52 + 82 = 9.434

CD =

2p 2

σ1 = 14 + 9.43 = 23.43 σ2 = 14 − 9.43 = 4.57

y

1 −1 5 90 + tan = 61.0◦ cw φp = 2 8

4.57 x 61 23.43

τ1 = R = 9.434,

φs = 61◦ − 45◦ = 16◦ cw

14

x 16 14 9.434

9 + 19 = 14 2

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61

Chapter 4

4-9 (a)

1

cw

C=

y (12, 7cw) R

2

D

1

2p

σ1 = 4 + 10.63 = 14.63 σ2 = 4 − 10.63 = −6.63

2s

x 2

ccw

12 + 4 =8 2 R = 82 + 72 = 10.63

CD =

C

(4, 7ccw)

12 − 4 =4 2

1 −1 8 90 + tan = 69.4◦ ccw φp = 2 7

14.63 69.4 6.63 x

φs = 69.4◦ − 45◦ = 24.4◦ ccw

τ1 = R = 10.63, 4

10.63 4 24.4 x

(b)

y

cw 1

C=

(5, 8cw)

R

6−5 = 0.5 2

6+5 = 5.5 2 R = 5.52 + 82 = 9.71

CD = C

2

D

1

2p 2s (6, 8ccw) ccw

2

σ1 = 0.5 + 9.71 = 10.21 σ2 = 0.5 − 9.71 = −9.21

x

9.21

φp =

10.21 27.75 x

1 −1 8 tan = 27.75◦ ccw 2 5.5

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φs = 45◦ − 27.75◦ = 17.25◦ cw

τ1 = R = 9.71, 0.5 9.71

x 17.25 0.5

(c)

cw 1

C=

x (8, 6cw)

8+7 = 7.5 2 R = 7.52 + 62 = 9.60

2s

CD =

2p 2

D

−8 + 7 = −0.5 2

1

C

σ1 = 9.60 − 0.5 = 9.10

R (7, 6ccw)

σ2 = −0.5 − 9.6 = −10.1

y 2 ccw

1 −1 7.5 90 + tan = 70.67◦ cw φp = 2 6

10.1 x

70.67 9.1

φs = 70.67◦ − 45◦ = 25.67◦ cw

τ1 = R = 9.60, 0.5

x 25.67 0.5 9.60

(d)

cw

1

C= x

2s

2

C

R

(6, 3ccw)

D 1

9+6 = 7.5 2 R = 7.52 + 32 = 8.078

CD =

(9, 3cw) 2p

9−6 = 1.5 2

σ1 = 1.5 + 8.078 = 9.58

y

ccw

2

σ2 = 1.5 − 8.078 = −6.58

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63

Chapter 4 6.58

φp =

1 −1 3 tan = 10.9◦ cw 2 7.5

x

10.9 9.58

φs = 45◦ − 10.9◦ = 34.1◦ ccw

τ1 = R = 8.078, 1.5

1.5 34.1 x 8.08

4-10 (a)

1

cw

C= x (20, 8cw)

2s 2p 2

D 1

C

R (10, 8ccw)

20 − 10 =5 2

20 + 10 = 15 2 R = 152 + 82 = 17

CD =

σ1 = 5 + 17 = 22

y

σ2 = 5 − 17 = −12

2

ccw 12

φp =

1 −1 8 tan = 14.04◦ cw 2 15

x 14.04 22

τ1 = R = 17,

φs = 45◦ − 14.04◦ = 30.96◦ ccw

5 5 30.96 x 17

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(b)

1

cw

C= y (10, 10cw)

R D

2

C

2p

1

(30, 10ccw) x

2

12.36

φp = 32.36 13.28

τ1 = R = 22.36,

30 + 10 = 20 2 R = 202 + 102 = 22.36

CD =

2s

ccw

30 − 10 = 10 2

σ1 = 10 + 22.36 = 32.36 σ2 = 10 − 22.36 = −12.36

1 −1 10 tan = 13.28◦ ccw 2 20

x

φs = 45◦ − 13.28◦ = 31.72◦ cw

10 22.36 x 31.72 10

(c)

1

cw

2s

(10, 9cw)

2

2p

D

1

C R

(18, 9ccw) y ccw

−10 + 18 =4 2 10 + 18 CD = = 14 2 R = 142 + 92 = 16.64 C=

x

2

σ1 = 4 + 16.64 = 20.64 σ2 = 4 − 16.64 = −12.64

1 −1 14 90 + tan = 73.63◦ cw φp = 2 9

12.64 x

73.63

20.64

τ1 = R = 16.64,

φs = 73.63◦ − 45◦ = 28.63◦ cw

4

x 28.63 4 16.64

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65

Chapter 4

(d)

1

cw x 2s

(12, 12cw)

12 + 22 = 17 2 R = 172 + 122 = 20.81

CD =

2p 2

D

1

C R

σ1 = 5 + 20.81 = 25.81 σ2 = 5 − 20.81 = −15.81

(22, 12ccw) y 2

ccw

−12 + 22 =5 2

C=

1 −1 17 90 + tan = 72.39◦ cw φp = 2 12

15.81 x

72.39 25.81

φs = 72.39◦ − 45◦ = 27.39◦ cw

τ1 = R = 20.81, 5

x 27.39 5 20.81

4-11 (a)

1/3

14 7 2

1/2 5 2/3 2 4 3 y

y 2 0

1/3

(b) y

x 10 x 1

(0, 4cw)

3

R 2

C

D

1

(10, 4ccw) x

τ1/3 = R = 6.40,

0 + 10 =5 2 10 − 0 CD = =5 2 R = 52 + 42 = 6.40 C=

1/2

2/3

τ1/2 =

σ1 = 5 + 6.40 = 11.40 σ2 = 0, σ3 = 5 − 6.40 = −1.40

11.40 = 5.70, 2

τ2/3 =

1.40 = 0.70 2

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(c)

(2, 4cw) x

2 circles

Point is a circle C

3

1 2

D

−2 − 8 = −5 2

C=

8−2 =3 2 R = 32 + 42 = 5

CD =

(8, 4ccw) y

σ1 = −5 + 5 = 0, σ2 = 0 σ3 = −5 − 5 = −10 10 = 5, 2

τ1/3 =

τ1/2 = 0, 1/3

(d)

τ2/3 = 5

10 − 30 = −10 2 10 + 30 CD = = 20 2 R = 202 + 102 = 22.36 C=

2/3 y 1/2

(30, 10cw) R C

D

2

3

1

(10, 10ccw) x

σ1 = −10 + 22.36 = 12.36 σ2 = 0 σ3 = −10 − 22.36 = −32.36

τ1/3 = 22.36,

τ1/2 =

12.36 = 6.18, 2

τ2/3 =

32.36 = 16.18 2

4-12 1/3

(a)

C=

2/3 x (80, 20cw)

1/2

C 3

D

R

2

1

(30, 20ccw) y

−80 − 30 = −55 2

80 − 30 = 25 2 R = 252 + 202 = 32.02

CD =

σ1 = 0 σ2 = −55 + 32.02 = −22.98 = −23.0 σ3 = −55 − 32.0 = −87.0

τ1/2 =

23 = 11.5, 2

τ2/3 = 32.0,

τ1/3 =

87 = 43.5 2

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67

Chapter 4 1/3

(b) 2/3

C=

x (30, 30cw)

R

60 + 30 = 45 2 R = 452 + 302 = 54.1

CD =

1/2 3

C

2

30 − 60 = −15 2

1

D

(60, 30ccw)

σ1 = −15 + 54.1 = 39.1

y

σ2 = 0 σ3 = −15 − 54.1 = −69.1 τ1/3 = (c)

39.1 + 69.1 = 54.1, 2

y

τ1/2 =

1/3

C=

1/2 (0, 20cw)

2/3

τ2/3 =

69.1 = 34.6 2

40 + 0 = 20 2

40 − 0 = 20 2 R = 202 + 202 = 28.3

CD =

R

3

39.1 = 19.6, 2

D

2

1

C

(40, 20ccw)

σ1 = 20 + 28.3 = 48.3

x

σ2 = 20 − 28.3 = −8.3 σ3 = σz = −30 τ1/3 =

48.3 + 30 = 39.1, 2

(d)

1/3

τ1/2 = 28.3,

τ2/3 =

C=

50 = 25 2

x

1/2

(50, 30cw)

2/3 3

2

C

D

(0, 30ccw)

1

30 − 8.3 = 10.9 2

50 = 25 2 R = 252 + 302 = 39.1

CD =

σ1 = 25 + 39.1 = 64.1

y

σ2 = 25 − 39.1 = −14.1 σ3 = σz = −20 τ1/3 =

64.1 + 20 = 42.1, 2

τ1/2 = 39.1,

τ2/3 =

20 − 14.1 = 2.95 2

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4-13 σ =

2000 F = = 10 190 psi = 10.19 kpsi A (π/4)(0.52 )

Ans.

FL L 72 = σ = 10 190 = 0.024 46 in Ans. AE E 30(106 ) 0.024 46 δ = 340(10−6 ) = 340µ Ans. 1 = = L 72 δ=

From Table A-5, ν = 0.292 2 = −ν1 = −0.292(340) = −99.3µ

Ans.

d = 2 d = −99.3(10−6 )(0.5) = −49.6(10−6 ) in 4-14

4-15

4-16

Ans.

From Table A-5, E = 71.7 GPa L 3 = 5.65(10−3 ) m = 5.65 mm δ = σ = 135(106 ) E 71.7(109 )

Ans.

From Table 4-2, biaxial case. From Table A-5, E = 207 GPa and ν = 0.292 σx =

E(x + ν y ) 207(109 )[0.0021 + 0.292(−0.000 67)] −6 = (10 ) = 431 MPa 1 − ν2 1 − 0.2922

σy =

207(109 )[−0.000 67 + 0.292(0.0021)] −6 (10 ) = −12.9 MPa 1 − 0.2922

Ans.

Ans.

The engineer has assumed the stress to be uniform. That is, t F

Ft = −F cos θ + τ A = 0 ⇒ τ =

F cos θ A

When failure occurs in shear F cos θ A The uniform stress assumption is common practice but is not exact. If interested in the details, see p. 570 of 6th edition. Ssu =

4-17

From Eq. (4-15) σ 3 − (−2 + 6 − 4)σ 2 + [−2(6) + (−2)(−4) + 6(−4) − 32 − 22 − (−5) 2 ]σ − [−2(6)(−4) + 2(3)(2)(−5) − (−2)(2) 2 − 6(−5) 2 − (−4)(3) 2 ] = 0 σ 3 − 66σ + 118 = 0 Roots are: 7.012, 1.89, −8.903 kpsi Ans.

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69

Chapter 4

7.012 − 1.89 = 2.56 kpsi 2 8.903 + 1.89 = 5.40 kpsi = 2

1/3 (kpsi)

τ1/2 = τ2/3

τmax = τ1/3 =

2/3 1/2

8.903 + 7.012 = 7.96 kpsi Ans. 2

8.903

1.89

7.012

(kpsi)

Note: For Probs. 4-17 to 4-19, one can also find the eigenvalues of the matrix σx τx y τzx [σ ] = τx y σ y τ yz τzx τ yz σz for the principal stresses 4-18

From Eq. (4-15)

√ 2 σ 3 − (10 + 0 + 10)σ 2 + 10(0) + 10(10) + 0(10) − 202 − −10 2 − 02 σ √ 2 √

− 10(0)(10) + 2(20) −10 2 (0) − 10 −10 2 − 0(0) 2 − 10(20) 2 = 0 σ 3 − 20σ 2 − 500σ + 6000 = 0

Roots are: 30, 10, −20 MPa 30 − 10 = 10 MPa τ1/2 = 2

Ans.

10 + 20 = 15 MPa τ2/3 = 2 30 + 20 = 25 MPa τmax = τ1/3 = 2

4-19

(MPa)

1/3

2/3 1/2

20

10

Ans.

From Eq. (4-15) σ 3 − (1 + 4 + 4)σ 2 + [1(4) + 1(4) + 4(4) − 22 − (−4) 2 − (−2) 2 ]σ −[1(4)(4) + 2(2)(−4)(−2) − 1(−4) 2 − 4(−2) 2 − 4(2) 2 ] = 0 σ 3 − 9σ 2 = 0

30

(MPa)

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Roots are: 9, 0, 0 kpsi

1/2 1/3

(kpsi)

2/3 O 0

τ2/3 = 0, 4-20 (a) R1 =

τ1/2 = τ1/3 = τmax =

c F l

Mmax = R1 a = σ =

(b)

9

(kpsi)

9 = 4.5 kpsi Ans. 2

ac F l

6M 6 ac = 2 F 2 bh bh l

⇒

F=

σ bh 2l 6ac

Ans.

(σm /σ )(bm /b) (h m / h) 2 (lm /l) 1(s)(s) 2 (s) Fm = = = s2 F (am /a) (cm /c) (s)(s)

Ans.

For equal stress, the model load varies by the square of the scale factor. 4-21

wl , R1 = 2 σ =

Mmax |x=l/2

l wl 2 wl l− = = 22 2 8

6M 3W l 6 wl 2 = = 2 2 bh bh 8 4bh 2

⇒

W =

4 σ bh 2 3 l

Wm (σm /σ )(bm /b) (h m / h) 2 1(s)(s) 2 = = = s2 W lm /l s wm lm = s2 wl

⇒

s2 wm = =s w s

Ans.

Ans.

Ans.

For equal stress, the model load w varies linearily with the scale factor. 4-22 (a) Can solve by iteration or derive equations for the general case. W1

W2

W3 . . .

A a23 RA

d3

a13

WT . . .

Find maximum moment under wheel W3

WT = W at centroid of W’s

Wn B RB

x3 l

RA =

l − x3 − d3 WT l

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71

Chapter 4

Under wheel 3 M3 = R A x3 − W1 a13 − W2 a23 = For maximum,

(l − x3 − d3 ) WT x3 − W1 a13 − W2 a23 l

WT d M3 = 0 = (l − d3 − 2x3 ) dx3 l

substitute into M,

M3 =

⇒

⇒

x3 =

l − d3 2

(l − d3 ) 2 WT − W1 a13 − W2 a23 4l

This means the midpoint of d3 intersects the midpoint of the beam For wheel i

i−1 (l − di ) 2 Mi = WT − W j a ji 4l j=1

l − di , xi = 2

Note for wheel 1: W j a ji = 0 WT = 104.4,

W1 = W2 = W3 = W4 =

(1200 − 238) 2 M1 = (104.4) = 20 128 kip · in 4(1200)

476 = 238 in, Wheel 1: d1 = 2 Wheel 2:

104.4 = 26.1 kip 4

d2 = 238 − 84 = 154 in (1200 − 154) 2 M2 = (104.4) − 26.1(84) = 21 605 kip · in = Mmax 4(1200)

Check if all of the wheels are on the rail 84" 77"

84"

315"

xmax 600"

600"

(b) xmax = 600 − 77 = 523 in (c) See above sketch. (d) inner axles 4-23 (a)

D c1 0.833"

a

a

0.75"

Ga

0.083"

C

1

1" 4

y c2 0.667"

Gb

3" 8

b

0.167"

B

0.5"

A 1" 4

1 12 "

1" 4

Aa = Ab = 0.25(1.5) = 0.375 in2 A = 3(0.375) = 1.125 in2

1

1.5"

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y¯ =

2(0.375)(0.75) + 0.375(0.5) = 0.667 in 1.125

0.25(1.5) 3 Ia = = 0.0703 in4 12 Ib =

1.5(0.25) 3 = 0.001 95 in4 12

I1 = 2[0.0703 + 0.375(0.083) 2 ] + [0.001 95 + 0.375(0.167) 2 ] = 0.158 in4 σA =

10 000(0.667) = 42(10) 3 psi Ans. 0.158

σB =

10 000(0.667 − 0.375) = 18.5(10) 3 psi Ans. 0.158

σC =

10 000(0.167 − 0.125) = 2.7(10) 3 psi Ans. 0.158

σD = −

10 000(0.833) = −52.7(10) 3 psi 0.158

(b)

Ans.

Ans.

D

c1 1.155"

C

a

b

1.732" 1

1

c2 0.577"

y

B

0.982" Ga

0.327"

0.577" 0.25" A

0.577" 1.134" A

2"

Here we treat the hole as a negative area. Aa = 1.732 in2 0.982 = 0.557 in2 Ab = 1.134 2 A = 1.732 − 0.557 = 1.175 in2 y¯ =

1.732(0.577) − 0.557(0.577) = 0.577 in 1.175

Ia =

2(1.732) 3 bh 3 = = 0.289 in4 36 36

Ib =

1.134(0.982) 3 = 0.0298 in4 36

I1 = Ia − Ib = 0.289 − 0.0298 = 0.259 in4

Gb

Ans.

Ans.

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73

Chapter 4

because the centroids are coincident. 10 000(0.577) = 22.3(10) 3 psi Ans. 0.259 10 000(0.327) = 12.6(10) 3 psi Ans. σB = 0.259 10 000(0.982 − 0.327) = −25.3(10) 3 psi σC = − 0.259 10 000(1.155) = −44.6(10) 3 psi Ans. σD = − 0.259 (c) Use two negative areas. σA =

Ans.

D C

c1 1.708"

b 1

1 Gb Gc

c2 2.292"

1.5" 2"

c

B a G a

0.25"

A

Ab = 9 in2 , Ac = 16 in2 , A = 16 − 9 − 1 = 6 in2 ; Aa = 1 in2 , y¯a = 0.25 in, y¯b = 2.0 in, y¯c = 2 in 16(2) − 9(2) − 1(0.25) = 2.292 in Ans. y¯ = 6 c1 = 4 − 2.292 = 1.708 in 2(0.5) 3 = 0.020 83 in4 12 3(3) 3 = 6.75 in4 Ib = 12 4(4) 3 = 21.333 in4 Ic = 12 Ia =

I1 = [21.333 + 16(0.292) 2 ] − [6.75 + 9(0.292) 2 ] − [0.020 83 + 1(2.292 − 0.25) 2 ] = 10.99 in4 Ans. 10 000(2.292) = 2086 psi Ans. 10.99 10 000(2.292 − 0.5) σB = = 1631 psi Ans. 10.99 10 000(1.708 − 0.5) = −1099 psi Ans. σC = − 10.99 10 000(1.708) σD = − = −1554 psi Ans. 10.99 σA =

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(d) Use a as a negative area. C c1 1.355"

b

B

1

1 1.490"

a

3.464"

c2 2.645" Ga 1.155" A

Aa = 6.928 in2 , y¯a = 1.155 in,

Ab = 16 in2 , y¯b = 2 in

A = 9.072 in2 ;

2(16) − 1.155(6.928) = 2.645 in 9.072 c1 = 4 − 2.645 = 1.355 in y¯ =

Ans.

4(3.464) 3 bh 3 = = 4.618 in4 36 36 3 4(4) = 21.33 in4 Ib = 12 I1 = [21.33 + 16(0.645) 2 ] − [4.618 + 6.928(1.490) 2 ] = 7.99 in4 Ans. 10 000(2.645) = 3310 psi Ans. σA = 7.99 10 000(3.464 − 2.645) σB = − = −1025 psi Ans. 7.99 10 000(1.355) = −1696 psi Ans. σC = − 7.99 Ia =

(e)

C c1 1.422"

a B

c2 2.828"

3.625(7.5) + 1.5(4.5) = 2.828 in Ans. 12 1 1 = (6)(1.25) 3 + 7.5(3.625 − 2.828) 2 + (1.5)(3) 3 + 4.5(2.828 − 1.5) 2 12 12 = 17.05 in4 Ans. 10 000(2.828) = 1659 psi Ans. = 17.05 10 000(3 − 2.828) = −101 psi Ans. =− 17.05 10 000(1.422) = −834 psi Ans. =− 17.05 b

A

I

σA σB σC

Aa = 6(1.25) = 7.5 in2 Ab = 3(1.5) = 4.5 in2 A = Ac + Ab = 12 in2 y¯ =

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75

Chapter 4

(f)

Let a = total area

D a

c 1.5

A = 1.5(3) − 1(1.25) = 3.25 in2 1 1 I = Ia − 2Ib = (1.5)(3) 3 − (1.25)(1) 3 12 12 = 3.271 in4 Ans.

C b

b

c 1.5

B A

1.5

10 000(1.5) = 4586 psi, 3.271

σ D = −4586 psi

10 000(0.5) = 1529 psi, σB = 3.271

σC = −1529 psi

σA =

Ans.

4-24 (a) The moment is maximum and constant between A and B I =

M = −50(20) = −1000 lbf · in ,

1 (0.5)(2) 3 = 0.3333 in4 12

E I 1.6(106 )(0.3333) = 533.3 in ρ = = M 1000 (x, y) = (30, −533.3) in

Ans.

(b) The moment is maximum and constant between A and B M = 50(5) = 250 lbf · in, ρ=

I = 0.3333 in4

1.6(106 )(0.3333) = 2133 in 250

(x, y) = (20, 2133) in

Ans.

Ans.

4-25 (a)

1000 lbf 12"

O

6"

I = B

A 333 lbf

A = 0.75(1.5) = 1.125 in 667 lbf

Mmax is at A. At the bottom of the section,

V (lbf)

x 667

O

4000(0.75) Mc = = 14 225 psi Ans. I 0.2109 Due to V, τmax constant is between A and B at y = 0

σmax =

333 O

M (lbf • in)

1 (0.75)(1.5) 3 = 0.2109 in4 12

4000

τmax = x

3 667 3V = = 889 psi 2A 2 1.125

Ans.

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(b)

1000 lbf

1000 lbf

8"

A

I =

8"

x

B

O 2000 lbf V (lbf)

1 (1)(2) 3 = 0.6667 in4 12

Mmax is at A at the top of the beam σmax =

1000 x

O

|Vmax | = 1000 lbf τmax =

1000

8000(1) = 12 000 psi 0.6667

Ans.

from O to B at y = 0

3 1000 3V = = 750 psi 2A 2 (2)(1)

Ans.

M (lbf • in) x

O

8000

(c)

120 lbf/in x

O

A

5"

B

15"

1500 lbf

5" C

1500 lbf

V (lbf) 900 600 x

O 600

900

M (lbf • in)

x

O M1

M3

τmax =

100 lbf/in x O

A

6"

B

12"

1350 lbf V (lbf)

450 lbf

750 x

O 7.5" 600 M (lbf • in)

1875(1) = 3750 psi 0.5 At A and B at y = 0

σmax =

M2

(d)

1 (0.75)(2) 3 = 0.5 in4 12 1 M1 = − 600(5) = −1500 lbf · in = M3 2 1 M2 = −1500 + (900)(7.5) = 1875 lbf · in 2 Mmax is at span center. At the bottom of the beam, I =

450

M1

Ans.

1 (1)(2) 3 = 0.6667 in4 12 600 (6) = −1800 lbf · in M1 = − 2 1 M2 = −1800 + 750(7.5) = 1013 lbf · in 2 At A, top of beam

x

O

3 900 = 900 psi 2 (0.75)(2)

I =

σmax =

M2

Ans.

1800(1) = 2700 psi 0.6667

Ans.

At A, y = 0 τmax =

3 750 = 563 psi 2 (2)(1)

Ans.

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77

Chapter 4

4-26 Mmax =

wl 2 8

⇒

σmax =

wl 2 c 8I

⇒

w=

8σ I cl 2

(a) l = 12(12) = 144 in, I = (1/12)(1.5)(9.5) 3 = 107.2 in4 w=

8(1200)(107.2) = 10.4 lbf/in 4.75(1442 )

Ans.

(b) l = 48 in, I = (π/64)(24 − 1.254 ) = 0.6656 in4 w=

8(12)(103 )(0.6656) = 27.7 lbf/in 1(48) 2

Ans.

8(12)(103 )(2.051) = 57.0 lbf/in 1.5(48) 2

Ans.

. l = 48 in, I = (1/12)(2)(33 ) − (1/12)(1.625)(2.6253 ) = 2.051 in4

(c)

w=

(d) l = 72 in; Table A-6, I = 2(1.24) = 2.48 in4 0.842"

cmax = 2.158"

2.158"

8(12)(103 )(2.48) = 21.3 lbf/in 2.158(72) 2 (e) l = 72 in; Table A-7, I = 3.85 in4 w=

2

w=

8(12)(103 )(3.85) = 35.6 lbf/in 2(722 )

Ans.

Ans.

(f) l = 72 in, I = (1/12)(1)(43 ) = 5.333 in4 w= 4-27

8(12)(103 )(5.333) = 49.4 lbf/in (2)(72) 2

(a) Model (c)

π (0.54 ) = 3.068(10−3 ) in4 64 π A = (0.52 ) = 0.1963 in2 4 218.75(0.25) Mc = σ = I 3.068(10−3 ) = 17 825 psi = 17.8 kpsi Ans. I =

500 lbf 500 lbf 0.4375 1.25 in 500 lbf

Ans.

500 lbf

V (lbf) 500

O

τmax =

500 M (lbf ¥ in) O

Mmax 500(0.4375) 218.75 lbf ¥ in

4 500 4V = = 3400 psi 3A 3 0.1963

Ans.

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(b) Model (d) 1333 lbf/in 0.25" 1.25" 500 lbf

1 Mmax = 500(0.25) + (500)(0.375) 2 = 218.75 lbf · in

500 lbf

V (lbf) 500

Vmax = 500 lbf

O

Same M and V 500 M

∴ σ = 17.8 kpsi Ans. τmax = 3400 psi

Mmax

Ans.

O

4-28

If support R B is between F1 and F2 at position x = l, maximum moments occur at x = 3 and l. M B = R A l − 2000(l − 3) + 1100(7.75 − l) = 0 R A = 3100 − 14 525/l Mx=3 = 3R A = 9300 − 43 575/l M B = R A l − 2000(l − 3) = 1100 l − 8525 To minimize the moments, equate Mx=3 to −M B giving 9300 − 43 575/l = −1100l + 8525 Multiply by l and simplify to l 2 + 0.7046l − 39.61 = 0 The positive solution for l is 5.95 in and the magnitude of the moment is M = 9300 − 43 575/5.95 = 1976 lbf · in Placing the bearing to the right of F2 , the bending moment would be minimized by placing it as close as possible to F2 . If the bearing is near point B as in the original figure, then we need to equate the reaction forces. From statics, R B = 14 525/l, and R A = 3100 − R B . For R A = R B , then R A = R B = 1550 lbf, and l = 14 575/1550 = 9.37 in.

4-29 F l

p2 b

p1 a

p1 + p2 x − l1 + terms for x > l + a a p1 + p2 1 x − l2 + terms for x > l + a V = −F + p1 x − l − 2a p1 p 1 + p2 M = −F x + x − l2 − x − l3 + terms for x > l + a 2 6a q = −Fx−1 + p1 x − l0 −

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79

Chapter 4

At x = (l + a) + , V = M = 0, terms for x > l + a = 0 p1 + p2 2 a =0 2a

⇒

p1 − p2 =

p1 + p2 3 p1 a 2 − a =0 −F(l + a) + 2 6a

⇒

2 p1 − p2 =

−F + p1 a −

p1 =

From (1) and (2)

2F (3l + 2a), a2

a b = p2 p1 + p2

From similar triangles

p2 =

⇒

2F a 6F(l + a) a2

(1) (2)

2F (3l + a) a2

(3)

ap2 p1 + p2

(4)

b=

Mmax occurs where V = 0 F a 2b

l

p2 p2

p2 b

p1

xmax = l + a − 2b

b

Mmax = −F(l + a − 2b) +

p1 p1 + p2 (a − 2b) 2 − (a − 2b) 3 2 6a

= −Fl − F(a − 2b) +

p1 p1 + p2 (a − 2b) 2 − (a − 2b) 3 2 6a

Normally Mmax = −Fl The fractional increase in the magnitude is

=

F(a − 2b) − ( p1 /2)(a − 2b) 2 − [( p1 + p2 )/6a](a − 2b) 3 Fl

For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in (3)

(4)

p1 =

2(1500) [3(1.5) + 2(1.2)] = 14 375 lbf/in 1.22

p2 =

2(1500) [3(1.5) + 1.2] = 11 875 lbf/in 1.22

b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in

Substituting into (5) yields

= 0.036 89 4-30

or

Computer program; no solution given here.

3.7% higher than −Fl

(5)

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4-31 y a

c

x

l R2

R1

σ =

c F l c M = Fx l

R1 =

F

6M 6(c/l) Fx = 2 bh bh 2

0≤x ≤a ⇒

6cFx blσmax

h=

0≤x ≤a

4-32 F a l R1

b F l b M = Fx l 32M 32 b σmax = Fx = 3 πd πd 3 l 32 bFx 1/3 d= 0≤x ≤a π lσmax R1 =

b R2

4-33

Ans.

t

t

b

Square:

b

Am = (b − t) 2 Tsq = 2Am tτall = 2(b − t) 2 tτall

Round:

Am = π(b − t) 2 /4 Trd = 2π(b − t) 2 tτall /4

Ratio of torques

Tsq 4 2(b − t) 2 tτall = = 1.27 = 2 Trd π(b − t) tτall /2 π

Twist per unit length square: 2Gθ1 t θsq = tτall Round:

L 4(b − t) L = C = C A m A m (b − t) 2

4(b − t) π(b − t) L =C =C θrd = C 2 A m π(b − t) /4 (b − t) 2

Ratio equals 1, twists are the same.

Ans.

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81

Chapter 4

Note the weight ratio is Wsq b−t ρl(b − t) 2 = = Wrd ρlπ(b − t)(t) πt

thin-walled assumes b ≥ 20t

19 = 6.04 π = 2.86 =

4-34

with b = 20 with b = 10t

l = 40 in, τall = 11 500 psi, G = 11.5(106 ) psi, t = 0.050 in rm = ri + t/2 = ri + 0.025 for ri > 0 =0 for ri = 0 π Am = (1 − 0.05) 2 − 4 rm2 − rm2 = 0.952 − (4 − π)rm2 4 L m = 4(1 − 0.05 − 2rm + 2πrm /4) = 4[0.95 − (2 − π/2)rm ] T = 2Am tτ = 2(0.05)(11 500) Am = 1150Am

Eq. (4-45): Eq. (4-46):

θ(deg) = θ1 l

T L m l 180 T L m (40) 180 180 = = π 4G A2m t π 4(11.5)(106 ) A2m (0.05) π

= 9.9645(10−4 )

T Lm A2m

Equations can then be put into a spreadsheet resulting in: ri 0 0.10 0.20 0.30 0.40 0.45

rm

Am

Lm

ri

T(lbf · in)

ri

θ(deg)

0 0.125 0.225 0.325 0.425 0.475

0.902 5 0.889 087 0.859 043 0.811 831 0.747 450 0.708 822

3.8 3.585 398 3.413 717 3.242 035 3.070 354 2.984 513

0 0.10 0.20 0.30 0.40 0.45

1037.9 1022.5 987.9 933.6 859.6 815.1

0 0.10 0.20 0.30 0.40 0.45

4.825 4.621 4.553 4.576 4.707 4.825

1200 1000 800 T (lbf • in)

shi20396_ch04.qxd

600 400 200 0

0

0.1

0.2

0.3 ri (in)

0.4

0.5

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4.85 4.80

(deg)

4.75 4.70 4.65 4.60 4.55 4.50

0

0.1

0.2

0.3

0.4

0.5

ri (in)

Torque carrying capacity reduces with ri . However, this is based on an assumption of uniform stresses which is not the case for small ri . Also note that weight also goes down with an increase in ri . 4-35

From Eq. (4-47) where θ1 is the same for each leg. 1 1 T1 = Gθ1 L 1 c13 , T2 = Gθ1 L 2 c23 3 3

1 1 T = T1 + T2 = Gθ1 L 1 c13 + L 2 c23 = Gθ1 L i ci3 3 3 τ1 = Gθ1 c1 , τ2 = Gθ1 c2 τmax = Gθ1 cmax Ans.

Ans.

4-36 (a)

τmax = Gθ1 cmax τmax 11 500 = 1.227(105 ) psi · rad = cmax 3/32 1 1 T1/16 = Gθ1 (Lc3 ) 1/16 = (1.227)(105 )(0.5)(1/16) 3 = 4.99 lbf · in Ans. 3 3 1 T3/32 = (1.227)(105 )(0.5)(3/32) 3 = 16.85 lbf · in Ans. 3 τ1/16 = 1.227(105 )1/16 = 7669 psi, τ3/32 = 1.227(105 )3/32 = 11 500 psi Ans. Gθ1 =

(b)

4-37

θ1 =

1.227(105 ) = 1.0667(10−2 ) rad/in = 0.611◦ /in 11.5(106 )

Ans.

Separate strips: For each 1/16 in thick strip, T = ∴ Tmax

Lc2 τ (1)(1/16) 2 (11 500) = = 14.97 lbf · in 3 3

= 2(14.97) = 29.95 lbf · in

Ans.

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83

Chapter 4

For each strip, 3T l 3(14.97)(12) = = 0.192 rad Ans. Lc3 G (1)(1/16) 3 (11.5)(106 )

θ=

kt = T /θ = 29.95/0.192 = 156.0 lbf · in

Ans.

Solid strip: From Example 4-12, Tmax = 59.90 lbf · in

4-38

Ans.

θ = 0.0960 rad

Ans.

kt = 624 lbf · in

Ans.

τall = 8000 psi, 50 hp (a) n = 2000 rpm 63 025(50) 63 025H = = 1575.6 lbf · in n 2000 16T 1/3 16(1575.6) 1/3 ⇒ d= = = 1.00 in Ans. πτmax π(8000)

T =

Eq. (4-40) τmax =

16T πd 3

(b) n = 200 rpm

4-39

∴

T = 15 756 lbf · in 16(15 756) 1/3 d= = 2.157 in π(8000)

Ans.

τall = 110 MPa, θ = 30◦ , d = 15 mm, l = ? τ=

16T πd 3

Tl θ= JG

⇒

180 π

T =

π 3 τd 16

π π dGθ π J Gθ π d 4 Gθ = = l= 3 180 T 180 32 (π/16) τ d 360 τ =

4-40

π (0.015)(79.3)(109 )(30) = 2.83 m Ans. 360 110(106 )

d = 70 mm, replaced by 70 mm hollow with t = 6 mm (a)

Tsolid =

π τ (703 ) 16

Thollow =

π (704 − 584 ) τ 32 35

(π/16)(703 ) − (π/32) [(704 − 584 )/35] (100) = 47.1% % T = (π/16)(703 )

Ans.

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(b) Wsolid = kd 2 = k(702 ) , Whollow = k(702 − 582 ) % W =

4-41

k(702 ) − k(702 − 582 ) (100) = 68.7% k(702 )

Ans.

T = 5400 N · m, τall = 150 MPa 4.023(104 ) 5400(d/2) Tc = ⇒ 150(106 ) = (π/32)[d 4 − (0.75d) 4 ] d3 J 1/3 4.023(104 ) d= = 6.45(10−2 ) m = 64.5 mm 150(106 ) τ=

(a)

From Table A-17, the next preferred size is d = 80 mm; ID = 60 mm π (0.084 − 0.064 ) = 2.749(10−6 ) mm4 32 5400(0.030) = 58.9(106 ) Pa = 58.9 MPa τi = 2.749(10−6 )

Ans.

J=

(b)

Ans.

4-42 63 025(1) 63 025H = = 12 605 lbf · in n 5 16T 16T 1/3 16(12 605) 1/3 τ= ⇒ dC = = = 1.66 in πτ π(14 000) πdC3

T =

(a)

Ans.

From Table A-17, select 1 3/4 in τstart =

16(2)(12 605) = 23.96(103 ) psi = 23.96 kpsi 3 π(1.75 )

(b) design activity 4-43

ω = 2πn/60 = 2π(8)/60 = 0.8378 rad/s 1000 H = = 1194 N · m ω 0.8378 16T 1/3 16(1194) 1/3 dC = = = 4.328(10−2 ) m = 43.3 mm 6 πτ π(75)(10 ) T =

From Table A-17, select 45 mm 4-44

s=

√

A, d =

Ans.

4A/π

Square: Eq. (4-43) with b = c

4.8T c3 4.8T (τmax ) sq = ( A) 3/2 τmax =

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Chapter 4

(τmax ) rd =

Round:

16 T 16T 3.545T = = 3 3 / 2 π d π(4A/π) ( A) 3/2 (τmax ) sq 4.8 = 1.354 = (τmax ) rd 3.545

Square stress is 1.354 times the round stress 4-45

s=

√

A,

d=

Ans.

4A/π

Square: Eq. (4-44) with b = c, β = 0.141 θsq =

Tl Tl = 4 0.141c G 0.141( A) 4/2 G

Round: θrd =

Tl Tl 6.2832T l = = JG ( A) 4/2 G (π/32) (4A/π) 4/2 G

θsq 1/0.141 = 1.129 = θrd 6.2832 Square has greater θ by a factor of 1.13 4-46

Ans.

Text Eq. (4-43) gives τmax =

T T 1 = 2· 2 αbc bc α

From in-text table, p. 139, α is a function of b/c. Arrange equation in the form 1 b2 cτmax 1 = = y = a0 + a1 = a0 + a1 x T α b/c To plot 1/α vs 1/(b/c) , first form a table.

b/c 1 1.5 1.75 2 2.5 3 4 6 8 10 ∞

α

x 1/(b/c)

y 1/α

0.208 0.231 0.239 0.246 0.258 0.267 0.282 0.299 0.307 0.313 0.333

1 0.666 667 0.571 429 0.5 0.4 0.333 333 0.25 0.166 667 0.125 0.1 0

4.807 692 4.329 004 4.184 100 4.065 041 3.875 969 3.745 318 3.546 099 3.344 482 3.257 329 3.194 888 3.003 003

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 1兾 5

4.5

y 1.867x 3.061 4

3.5

3

0

0.2

0.4

0.6

0.8

1

1兾(b兾c)

Plot is a gentle convex-upward curve. Roark uses a polynomial, which in our notation is 1 3T τmax = 1 + 0.6095 + ··· 8(b/2)(c/2) 2 b/c 1 . T τmax = 2 3 + 1.8285 bc b/c Linear regression on table data y = 3.06 + 1.87x 1 1 = 3.06 + 1.87 α b/c 1 T τmax = 2 3.06 + 1.87 bc b/c 1.8 T τmax = 2 3 + bc b/c

Eq. (4-43)

4-47 Gear F 1000 lbf • in

1000 = 400 lbf 2.5 Fn = 400 tan 20 = 145.6 lbf Ft =

2.5R

Shaft ABCD y

Fn

Ft

Torque at C TC = 400(5) = 2000 lbf · in

RAy 666.7 lbf A 3" z

P=

2000 lbf • in B 145.6 lbf

RAz

2000 lbf • in 10"

RDy C C 400 lbf

5" D RDz

x

2000 = 666.7 lbf 3

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87

Chapter 4

( M A)z = 0 ( M A) y = 0

⇒

18R Dy − 145.6(13) − 666.7(3) = 0

⇒

−18R Dz + 400(13) = 0

Fy = 0

⇒

R Ay + 216.3 − 666.7 − 145.6 = 0

Fz = 0

⇒

⇒

⇒

R Dy = 216.3 lbf

R Dz = 288.9 lbf ⇒

R Ay = 596.0 lbf

R Az + 288.9 − 400 = 0 ⇒ R Az = 111.1 lbf M B = 3 5962 + 111.12 = 1819 lbf · in MC = 5 216.32 + 288.92 = 1805 lbf · in

∴ Maximum stresses occur at B. Ans. 32M B 32(1819) = 9486 psi = σB = 3 πd π(1.253 ) 16TB 16(2000) = 5215 psi = πd 3 π(1.253 ) σB 9486 σB 2 9486 2 2 + + = + τB = + 52152 = 11 792 psi 2 2 2 2 σB 2 τmax = + τ B2 = 7049 psi Ans. 2 τB =

σmax

Ans.

4-48 (a) At θ = 90◦ , σr = τrθ = 0, σθ = −σ Ans. θ = 0◦ , σr = τrθ = 0, σθ = 3σ Ans. (b) r

σθ /σ

5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

3.000 2.071 1.646 1.424 1.297 1.219 1.167 1.132 1.107 1.088 1.074 1.063 1.054 1.048 1.042 1.037

兾 3 2.5 2 1.5 1 0.5 0

0

5

10 r (mm)

15

20

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4-49

1.5 = 1.5 1 1/8 r/d = = 0.125 1 . K ts = 1.39 . K t = 1.60

D/d =

Fig. A-15-8: Fig. A-15-9: σ A = Kt

32K t M 32(1.6)(200)(14) Mc = = 45 630 psi = 3 I πd π(13 )

16K ts T 16(1.39)(200)(15) Tc = = 21 240 psi = 3 J πd π(13 ) 2 σA 45.63 σA 45.63 2 2 + + = + τA = + 21.242 2 2 2 2

τ A = K ts σmax

τmax

4-50

= 54.0 kpsi Ans. 45.63 2 = + 21.242 = 31.2 kpsi Ans. 2

As shown in Fig. 4-34, the maximum stresses occur at the inside fiber where r = ri . Therefore, from Eq. (4-51) ri2 pi ro2 1+ 2 σt, max = 2 ro − ri2 ri

σr, max

4-51

ro2 + ri2 Ans. = pi ro2 − ri2 ri2 pi ro2 = 2 1 − 2 = − pi ro − ri2 ri

If pi = 0, Eq. (4-50) becomes − poro2 − ri2ro2 po /r 2 ro2 − ri2 ri2 poro2 =− 2 1+ 2 r ro − ri2

σt =

The maximum tangential stress occurs at r = ri . So σt, max = −

2 poro2 ro2 − ri2

Ans.

Ans.

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89

Chapter 4

For σr , we have − poro2 + ri2ro2 po /r 2 ro2 − ri2 ri2 poro2 = 2 −1 ro − ri2 r 2

σr =

So σr = 0 at r = ri . Thus at r = ro σr, max

por 2 = 2 o2 r o − ri

ri2 − ro2 ro2

= − po

Ans.

4-52 rav t

2 F = p A = πrav p 2 prav F πrav p σ1 = σ2 = = = Awall 2πrav t 2t

p

Ans.

F

4-53

σt > σl > σr τmax = (σt − σr )/2 at r = ri where σl is intermediate in value. From Prob. 4-50 1 τmax = (σt, max − σr, max ) 2 pi ro2 + ri2 +1 τmax = 2 ro2 − ri2 Now solve for pi using ro = 3 in, ri = 2.75 in, and τmax = 4000 psi. This gives pi = 639 psi Ans.

4-54

Given ro = 120 mm, ri = 110 mm and referring to the solution of Prob. 4-53, 2.4 MPa (120) 2 + (110) 2 τmax = +1 2 (120) 2 − (110) 2 = 15.0 MPa Ans.

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From Table A-20, S y = 57 kpsi; also, ro = 0.875 in and ri = 0.625 in From Prob. 4-51 σt, max = −

4-56

Rearranging

2

ro − ri2 (0.8S y ) po = 2ro2

Solving, gives po = 11 200 psi

Ans.

From Table A-20, S y = 57 kpsi; also ro = 1.1875 in, ri = 0.875 in. From Prob. 4-50

σt, max = pi

ro2 + ri2 ro2 − ri2

solving gives 4-57

2 poro2 ro2 − ri2

therefore

pi = 0.8S y

pi = 13 510 psi

ro2 − ri2 ro2 + ri2

Ans.

Since σt and σr are both positive and σt > σr τmax = (σt ) max /2 where σt is max at ri Eq. (4-56) for r = ri = 0.375 in 0.282 2π(7200) 2 3 + 0.292 (σt ) max = 386 60 8 (0.3752 )(52 ) 1 + 3(0.292) 2 2 2 (0.375 ) = 8556 psi × 0.375 + 5 + − 0.3752 3 + 0.292 τmax =

8556 = 4278 psi Ans. 2

Radial stress:

r 2r 2 σr = k ri2 + ro2 − i 2o − r 2 r

ri2ro2 √ dσr = k 2 3 − 2r = 0 ⇒ r = ri ro = 0.375(5) = 1.3693 in Maxima: dr r 0.282 2π(7200) 2 3 + 0.292 0.3752 (52 ) 2 2 2 0.375 + 5 − − 1.3693 (σr ) max = 386 60 8 1.36932 = 3656 psi Ans.

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91

Chapter 4

ω = 2π(2069)/60 = 216.7 rad/s,

4-58

ρ = 3320 kg/m3 , ν = 0.24, ri = 0.0125 m, ro = 0.15 m; use Eq. (4-56) σt = 3320(216.7)

2

3 + 0.24 8

(0.0125) 2 + (0.15) 2 + (0.15) 2 1 + 3(0.24) 2 (0.0125) (10) −6 − 3 + 0.24

= 2.85 MPa Ans. 4-59 ρ=

(6/16) 386(1/16)(π/4)(62 − 12 )

= 5.655(10−4 ) lbf · s2 /in

4

τmax is at bore and equals Eq. (4-56) (σt ) max

σt 2

2π(10 000) = 5.655(10 ) 60 = 4496 psi

τmax =

−4

2

3 + 0.20 8

1 + 3(0.20) 2 2 2 2 0.5 + 3 + 3 − (0.5) 3 + 0.20

4496 = 2248 psi Ans. 2 ω = 2π(3000)/60 = 314.2 rad/s

4-60

m=

0.282(1.25)(12)(0.125) 386

= 1.370(10−3 ) lbf · s2 /in F

F 6"

F = mω2r = 1.370(10−3 )(314.22 )(6) = 811.5 lbf Anom = (1.25 − 0.5)(1/8) = 0.093 75 in2 811.5 = 8656 psi Ans. 0.093 75 . Note: Stress concentration Fig. A-15-1 gives K t = 2.25 which increases σmax and fatigue. σnom =

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4-61 to 4-66 ν = 0.292,

E = 30 Mpsi (207 GPa), ri = 0

R = 0.75 in (20 mm), ro = 1.5 in (40 mm) Eq. (4-60)

30(106 )δ (1.52 − 0.752 )(0.752 − 0) = 1.5(107 )δ ppsi = 2 2 0.75 in 2(0.75 )(1.5 − 0) 207(109 )δ (0.042 − 0.022 )(0.022 − 0) = 3.881(1012 )δ pPa = 2 2 0.020 2(0.02 )(0.04 − 0)

4-61 1 δmax = [40.042 − 40.000] = 0.021 mm 2

Ans.

1 δmin = [40.026 − 40.025] = 0.0005 mm 2

Ans.

From (2) pmax = 81.5 MPa,

pmin = 1.94 MPa

Ans.

4-62 1 δmax = (1.5016 − 1.5000) = 0.0008 in 2 1 δmin = (1.5010 − 1.5010) = 0 Ans. 2 Eq. (1)

pmax = 12 000 psi,

pmin = 0

Ans.

Ans.

4-63 1 δmax = (40.059 − 40.000) = 0.0295 mm Ans. 2 1 δmin = (40.043 − 40.025) = 0.009 mm Ans. 2 Eq. (2)

pmax = 114.5 MPa,

pmin = 34.9 MPa

Ans.

4-64 1 δmax = (1.5023 − 1.5000) = 0.001 15 in Ans. 2 1 δmin = (1.5017 − 1.5010) = 0.000 35 in Ans. 2 Eq. (1)

pmax = 17 250 psi

pmin = 5250 psi Ans.

(1)

(2)

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93

Chapter 4

4-65 1 δmax = (40.076 − 40.000) = 0.038 mm Ans. 2 1 δmin = (40.060 − 40.025) = 0.0175 mm Ans. 2 pmax = 147.5 MPa

Eq. (2)

pmin = 67.9 MPa

Ans.

4-66 1 δmax = (1.5030 − 1.500) = 0.0015 in Ans. 2 1 δmin = (1.5024 − 1.5010) = 0.0007 in Ans. 2 pmax = 22 500 psi

Eq. (1)

pmin = 10 500 psi

Ans.

4-67 1 δ = (1.002 − 1.000) = 0.001 in ri = 0, R = 0.5 in, 2 ν = 0.292, E = 30 Mpsi

ro = 1 in

Eq. (4-60) 30(106 )(0.001) (12 − 0.52 )(0.52 − 0) = 2.25(104 ) psi Ans. p= 2 2 0.5 2(0.5 )(1 − 0) Eq. (4-51) for outer member at ri = 0.5 in 12 0.52 (2.25)(104 ) (σt ) o = 1+ = 37 500 psi Ans. 12 − 0.52 0.52 Inner member, from Prob. 4-51 ri2 0 poro2 2.25(104 )(0.52 ) 1+ = −22 500 psi Ans. (σt ) i = − 2 1+ 2 =− ro 0.52 − 0 0.52 ro − ri2 Eqs. (d) and (e) above Eq. (4-59) 2 2.25(104 ) 1 + 0.52 δo = 0.5 2 + 0.292 = 0.000 735 in Ans. 30(106 ) 1 − 0.52 2.25(104 )(0.5) 0.52 + 0 δi = − − 0.292 = −0.000 265 in Ans. 30(106 ) 0.52 − 0

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4-68 νi = 0.292,

E i = 30(106 ) psi,

νo = 0.211,

1 δ = (1.002 − 1.000) = 0.001 in, 2

ri = 0,

E o = 14.5(106 ) psi R = 0.5,

ro = 1

Eq. (4-59)

0.5 0.001 = 14.5(106 )

2 12 + 0.52 0.5 + 0 0.5 − 0.292 p + 0.211 + 12 − 0.52 30(106 ) 0.52 − 0

p = 13 064 psi Ans. Eq. (4-51) for outer member at ri = 0.5 in 12 0.52 (13 064) 1+ = 21 770 psi (σt ) o = 12 − 0.52 0.52

Ans.

Inner member, from Prob. 4-51

0 13 064(0.52 ) 1+ = −13 064 psi Ans. (σt )i = − 0.52 − 0 0.52

Eqs. (d ) and (e) above Eq. (4-59) 13 064(0.5) 12 + 0.52 δo = + 0.211 = 0.000 846 in Ans. 14.5(106 ) 12 − 0.52 13 064(0.5) 0.52 + 0 − 0.292 = −0.000 154 in Ans. δi = − 30(106 ) 0.52 − 0 4-69 1 δmax = (1.003 − 1.000) = 0.0015 in ri = 0, R = 0.5 in, ro = 1 in 2 1 δmin = (1.002 − 1.001) = 0.0005 in 2 Eq. (4-60) pmax

30(106 )(0.0015) (12 − 0.52 )(0.52 − 0) = 33 750 psi Ans. = 0.5 2(0.52 )(12 − 0)

Eq. (4-51) for outer member at r = 0.5 in 12 0.52 (33 750) (σt ) o = 1+ = 56 250 psi Ans. 12 − 0.52 0.52 For inner member, from Prob. 4-51, with r = 0.5 in (σt ) i = −33 750 psi Ans.

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95

Chapter 4

Eqs. (d) and (e) just above Eq. (4-59) 33 750(0.5) 12 + 0.52 + 0.292 = 0.001 10 in Ans. δo = 30(106 ) 12 − 0.52 33 750(0.5) 0.52 + 0 − 0.292 = −0.000 398 in Ans. δi = − 30(106 ) 0.52 − 0 For δmin all answers are 0.0005/0.0015 = 1/3 of above answers

Ans.

4-70 νi = 0.292,

E i = 30 Mpsi,

νo = 0.334,

E o = 10.4 Mpsi

1 δmax = (2.005 − 2.000) = 0.0025 in 2 1 δmin = (2.003 − 2.002) = 0.0005 in 2 2 2 1.0 2 + 12 1 +0 1.0 − 0.292 pmax 0.0025 = + 0.334 + 10.4(106 ) 22 − 12 30(106 ) 12 − 0 pmax = 11 576 psi Ans. Eq. (4-51) for outer member at r = 1 in 22 12 (11 576) 1 + 2 = 19 293 psi Ans. (σt ) o = 22 − 12 1 Inner member from Prob. 4-51 with r = 1 in (σt ) i = −11 576 psi Ans. Eqs. (d) and (e) just above Eq. (4-59) 11 576(1) 22 + 12 δo = + 0.334 = 0.002 23 in Ans. 10.4(106 ) 22 − 12 11 576(1) 12 + 0 δi = − − 0.292 = −0.000 273 in Ans. 30(106 ) 12 − 0 For δmin all above answers are 0.0005/0.0025 = 1/5 Ans. 4-71 (a) Axial resistance Normal force at fit interface N = p A = p(2π Rl) = 2π p Rl Fully-developed friction force Fax = f N = 2π f p Rl

Ans.

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(b) Torsional resistance at fully developed friction is T = f R N = 2π f p R 2l 4-72

Ans.

d = 1 in, ri = 1.5 in, ro = 2.5 in. From Table 4-5, for R = 0.5 in, rc = 1.5 + 0.5 = 2 in rn =

0.52 √

= 1.968 245 8 in 2 2 − 22 − 0.52

e = rc − rn = 2.0 − 1.968 245 8 = 0.031 754 in ci = rn − ri = 1.9682 − 1.5 = 0.4682 in co = ro − rn = 2.5 − 1.9682 = 0.5318 in A = πd 2 /4 = π(1) 2 /4 = 0.7854 in2 M = Frc = 1000(2) = 2000 lbf · in Using Eq. (4-66) Mci 2000(0.4682) F 1000 + + = 26 300 psi Ans. = A Aeri 0.7854 0.7854(0.031 754)(1.5) Mco 2000(0.5318) F 1000 σo = − − = −15 800 psi Ans. = A Aero 0.7854 0.7854(0.031 754)(2.5) σi =

4-73

Section AA: D = 0.75 in, ri = 0.75/2 = 0.375 in, ro = 0.75/2 + 0.25 = 0.625 in From Table 4-5, for R = 0.125 in, rc = (0.75 + 0.25)/2 = 0.500 in 0.1252 rn = √

= 0.492 061 5 in 2 0.5 − 0.52 − 0.1252 e = 0.5 − rn = 0.007 939 in co = ro − rn = 0.625 − 0.492 06 = 0.132 94 in ci = rn − ri = 0.492 06 − 0.375 = 0.117 06 in A = π(0.25) 2 /4 = 0.049 087 M = Frc = 100(0.5) = 50 lbf · in 50(0.117 06) 100 + = 42 100 ksi Ans. 0.049 09 0.049 09(0.007 939)(0.375) 50(0.132 94) 100 σo = − = −25 250 psi Ans. 0.049 09 0.049 09(0.007 939)(0.625) σi =

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97

Chapter 4

Section BB: Abscissa angle θ of line of radius centers is r2 + d/2 −1 θ = cos r2 + d + D/2 0.375 + 0.25/2 −1 = 60◦ = cos 0.375 + 0.25 + 0.75/2 D+d cos θ = 100(0.5) cos 60◦ = 25 lbf · in 2 ri = r2 = 0.375 in

M=F

ro = r2 + d = 0.375 + 0.25 = 0.625 in e = 0.007 939 in σi = = σo =

(as before)

Mci Fcos θ − A Aeri 25(0.117 06) 100 cos 60◦ − = −19 000 psi 0.049 09 0.049 09(0.007 939)0.375 25(0.132 94) 100 cos 60◦ + = 14 700 psi 0.049 09 0.049 09(0.007 939)0.625

Ans. Ans.

On section BB, the shear stress due to the shear force is zero at the surface. 4-74 ri = 0.125 in, ro = 0.125 + 0.1094 = 0.2344 in From Table 4-5 for h = 0.1094 rc = 0.125 + 0.1094/2 = 0.1797 in rn = 0.1094/ln(0.2344/0.125) = 0.174 006 in e = rc − rn = 0.1797 − 0.174 006 = 0.005 694 in ci = rn − ri = 0.174 006 − 0.125 = 0.049 006 in co = ro − rn = 0.2344 − 0.174 006 = 0.060 394 in A = 0.75(0.1094) = 0.082 050 in2 M = F(4 + h/2) = 3(4 + 0.1094/2) = 12.16 lbf · in 12.16(0.0490) 3 σi = − − = −10 240 psi Ans. 0.082 05 0.082 05(0.005 694)(0.125) σo = −

12.16(0.0604) 3 + = 6670 psi Ans. 0.082 05 0.082 05(0.005 694)(0.2344)

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Find the resultant of F1 and F2 . Fx = F1x + F2x = 250 cos 60◦ + 333 cos 0◦ = 458 lbf Fy = F1 y + F2 y = 250 sin 60◦ + 333 sin 0◦ = 216.5 lbf F = (4582 + 216.52 ) 1/2 = 506.6 lbf This is the pin force on the lever which acts in a direction θ = tan−1

Fy 216.5 = 25.3◦ = tan−1 Fx 458

On the 25.3◦ surface from F1 2000 lbf • in 25.3 206

142

Ft = 250 cos(60◦ − 25.3◦ ) = 206 lbf Fn = 250 sin(60◦ − 25.3◦ ) = 142 lbf

507

A = 2[0.8125(0.375) + 1.25(0.375)] = 1.546 875 in2 The denomenator of Eq. (3-67), given below, has four additive parts. A rn = (d A/r) For d A/r , add the results of the following equation for each of the four rectangles. ro ro bdr = b ln , b = width r ri ri

1.8125 2.1875 3.6875 4.5 dA = 0.375 ln + 1.25 ln + 1.25 ln + 0.375 ln r 1 1.8125 3.3125 3.6875 = 0.666 810 6 rn =

1.546 875 = 2.3198 in 0.666 810 6

e = rc − rn = 2.75 − 2.3198 = 0.4302 in ci = rn − ri = 2.320 − 1 = 1.320 in co = ro − rn = 4.5 − 2.320 = 2.180 in Shear stress due to 206 lbf force is zero at inner and outer surfaces. σi = −

2000(1.32) 142 + = 3875 psi 1.547 1.547(0.4302)(1)

σo = −

2000(2.18) 142 − = −1548 psi 1.547 1.547(0.4302)(4.5)

Ans. Ans.

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Chapter 4

4-76 A = (6 − 2 − 1)(0.75) = 2.25 in2 6+2 = 4 in rc = 2 Similar to Prob. 4-75, 3.5 6 dA = 0.75 ln + 0.75 ln = 0.635 473 4 in r 2 4.5 rn =

2.25 A = = 3.5407 in 0.635 473 4 (d A/r)

e = 4 − 3.5407 = 0.4593 in σi =

5000 20 000(3.5407 − 2) + = 17 130 psi Ans. 2.25 2.25(0.4593)(2)

σo =

5000 20 000(6 − 3.5407) − = −5710 psi Ans. 2.25 2.25(0.4593)(6)

4-77

(a)

ro

A= ri

b dr = 2

6

2 6 dr = 2 ln r 2

= 2.197 225 in2 6 2r 1 ro 1 dr rc = br dr = A ri 2.197 225 2 r =

2 (6 − 2) = 3.640 957 in 2.197 225

r n = ro ri

=

A 2.197 225 = 6 2 (b/r) dr 2 (2/r ) dr

2.197 225 = 3.295 837 in 2[1/2 − 1/6]

e = R − rn = 3.640 957 − 3.295 837 = 0.345 12 ci = rn − ri = 3.2958 − 2 = 1.2958 in co = ro − rn = 6 − 3.2958 = 2.7042 in σi =

20 000 20 000(3.641)(1.2958) + = 71 330 psi Ans. 2.197 2.197(0.345 12)(2)

σo =

20 000 20 000(3.641)(2.7042) − = −34 180 psi Ans. 2.197 2.197(0.345 12)(6)

99

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(b) For the centroid, Eq. (4-70) gives,

r(2/r) s

s 4 rb s = = = rc = b s (2/r) s

s/r

s/r Let s = 4 × 10−3 in with the following visual basic program, “cen.” Function cen(R) DS = 4 / 1000 R = R + DS / 2 Sum = 0 For I = 1 To 1000 Step 1 Sum = Sum + DS / R R = R + DS Next I cen = 4 / Sum End Function For eccentricity, Eq. (4-71) gives,

[(s/r)/(rc − s)] s [s/(rc − s)](2/r) s = e= (2/r) s/(rc − s) [(1/r)/(rc − s)] s Program the following visual basic program, “ecc.” Function ecc(RC) DS = 4 / 1000 S = −(6 − RC) + DS / 2 R = 6 − DS / 2 SUM1 = 0 SUM2 = 0 For I = 1 To 1000 Step 1 SUM1 = SUM1 + DS * (S / R) / (RC − S) SUM2 = SUM2 + DS * (1 / R) / (RC − S) S = S + DS R = R − DS Next I ecc = SUM1 / SUM2 End Function In the spreadsheet enter the following,

1 2

A

B

C

D

ri 2

rc = cen(A2)

e = ecc(B2)

rn = B2 − C2

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101

Chapter 4

which results in,

1 2

A

B

C

D

ri 2

rc 3.640 957

e 0.345 119

rn 3.295 838

which are basically the same as the analytical results of part (a) and will thus yield the same final stresses. Ans. 4-78

rc = 12

s

s 2 (b/2) 2 2 /4 = + = 1 ⇒ b = 2 1 − s 4 − s2 22 12

√ [(s 4 − s 2 )/(rc − s)] s e= √ [( 4 − s 2 )/(rc − s)] s

1"

2"

A = πab = π(2)(1) = 6.283 in2 Function ecc(rc) DS = 4 / 1000 S = −2 + DS/2 SUM1 = 0 SUM2 = 0 For I = 1 To 1000 Step 1 SUM1 = SUM1 + DS * (S * Sqr(4 − S ^ 2)) / (rc − S) SUM2 = SUM2 + DS * Sqr(4 − S ^ 2) / (rc − S) S = S + DS Next I ecc = SUM1/SUM2 End Function rc

e

rn = rc − e

12

0.083 923

11.91 608

ci = 11.916 08 − 10 = 1.9161 co = 14 − 11.916 08 = 2.0839 M = F(2 + 2) = 20(4) = 80 kip · in σi =

80(1.9161) 20 + = 32.25 kpsi Ans. 6.283 6.283(0.083 923)(10)

σo =

80(2.0839) 20 − = −19.40 kpsi Ans. 6.283 6.283(0.083 923)(14)

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4-79 0.4"

0.4"

0.4"R 1"

For circle,

1"

For rectangle,

A r2 =

, (d A/r) 2 rc − rc2 − r 2

dA = b ln ro /ri r

Ao = πr 2

dA = 2π rc − rc2 − r 2 ∴ r dA 2.6 = 1 ln − 2π 1.8 − 1.82 − 0.42 = 0.672 723 4 r 1 A = 1(1.6) − π(0.42 ) = 1.097 345 2 in2 rn =

1.097 345 2 = 1.6312 in 0.672 723 4

e = 1.8 − rn = 0.1688 in ci = 1.6312 − 1 = 0.6312 in co = 2.6 − 1.6312 = 0.9688 in M = 3000(5.8) = 17 400 lbf · in

4-80

σi =

17.4(0.6312) 3 + = 62.03 kpsi 1.0973 1.0973(0.1688)(1)

σo =

17.4(0.9688) 3 − = −32.27 kpsi Ans. 1.0973 1.0973(0.1688)(2.6)

100 and 1000 elements give virtually the same results as shown below. Visual basic program for 100 elements: Function ecc(RC) DS = 1.6 / 100 S = −0.8 + DS / 2 SUM1 = 0 SUM2 = 0 For I = 1 To 25 Step 1 SUM1 = SUM1 + DS * S / (RC − S) SUM2 = SUM2 + DS / (RC − S) S = S + DS Next I

Ans.

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103

Chapter 4

For I = 1 To 50 Step 1 SUM1 = SUM1 + DS * S * (1 − 2 * Sqr(0.4 ^ 2 − S ^ 2)) / (RC − S) SUM2 = SUM2 + DS * (1 − 2 * Sqr(0.4 ^ 2 − S ^ 2)) / (RC − S) S = S + DS Next I For I = 1 To 25 Step 1 SUM1 = SUM1 + DS * S / (RC − S) SUM2 = SUM2 + DS / (RC − S) S = S + DS Next ecc = SUM1 / SUM2 End Function 100 elements

1000 elements

rc

e

rn = rc − e

rc

e

rn = rc − e

1.8

0.168 811

1.631 189

1.8

0.168 802

1.631 198

e = 0.1688 in, rn = 1.6312 in Yields same results as Prob. 4-79. 4-81

Ans.

From Eq. (4-72)

a = KF

1/3

=F

1/3

3 2[(1 − ν 2 )/E] 8 2(1/d)

1/3

Use ν = 0.292, F in newtons, E in N/mm2 and d in mm, then 1/3 3 [(1 − 0.2922 )/207 000] K = = 0.0346 8 1/25 pmax = =

3F 3F = 2πa 2 2π(K F 1/3 ) 2 3F 1/3 3F 1/3 = 2π K 2 2π(0.0346) 2

= 399F 1/3 MPa = |σmax | τmax = 0.3 pmax = 120F 1/3 MPa 4-82

From Prob. 4-81,

3 2[(1 − 0.2922 )/207 000] K = 8 1/25 + 0 pmax =

1/3 = 0.0436

3F 1/3 3F 1/3 = = 251F 1/3 2π K 2 2π(0.0436) 2

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and so,

σz = −251F 1/3 MPa τmax = 0.3(251) F

1/3

Ans.

= 75.3F 1/3 MPa 1/3

z = 0.48a = 0.48(0.0436)18 4-83

Ans.

= 0.055 mm

Ans.

ν1 = 0.334, E 1 = 10.4 Mpsi, l = 2 in, d1 = 1 in, ν2 = 0.211, E 2 = 14.5 Mpsi, d2 = −8 in. With b = K c F 1/2 1/2 2 (1 − 0.3342 )/[10.4(106 )] + (1 − 0.2112 )/[14.5(106 )] Kc = π(2) 1 − 0.125 = 0.000 234 6 Be sure to check σx for both ν1 and ν2 . Shear stress is maximum in the aluminum roller. So, τmax = 0.3 pmax 4000 = 13 300 psi 0.3

pmax = Since pmax = 2F/(πbl) we have pmax = So,

F= =

2F 2F 1/2 = πl K c F 1/2 πl K c

πl K c pmax 2

π(2)(0.000 234 6)(13 300) 2

= 96.1 lbf 4-84

Good class problem

4-85

From Table A-5, ν = 0.211

2 2

Ans.

1 σx 1 = (1 + ν) − = (1 + 0.211) − = 0.711 pmax 2 2 σy = 0.711 pmax σz =1 pmax These are principal stresses 1 1 τmax = (σ1 − σ3 ) = (1 − 0.711) = 0.1445 pmax 2 2

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Chapter 4

4-86

105

From Table A-5: ν1 = 0.211, ν2 = 0.292, E 1 = 14.5(106 ) psi, E 2 = 30(106 ) psi, d1 = 6 in, d2 = ∞, l = 2 in 2(800) (1 − 0.2112 )/14.5(106 ) + (1 − 0.2922 )/[30(106 )] b= (a) π(2) 1/6 + 1/∞ = 0.012 135 in pmax =

2(800) = 20 984 psi π(0.012 135)(2)

For z = 0 in, σx 1 = −2ν1 pmax = −2(0.211)20 984 = −8855 psi in wheel σx 2 = −2(0.292)20 984 = −12 254 psi In plate σ y = − pmax = −20 984 psi σz = −20 984 psi These are principal stresses. (b) For z = 0.010 in, σx1 = −4177 psi in wheel σx2 = −5781 psi in plate σ y = −3604 psi σz = −16 194 psi

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Chapter 5 5-1 (a)

F k1

k2

k3

so

y

F F F + + k1 k2 k3

k=

F ; y

k=

1 (1/k1 ) + (1/k2 ) + (1/k3 )

y=

Ans.

(b) k1

F

k2

y

F = k1 y + k2 y + k3 y k = F/y = k1 + k2 + k3

Ans.

k3

(c) k2 k1

1 1 1 = + k k1 k2 + k3

k=

1 1 + k1 k2 + k3

−1

k3

5-2 For a torsion bar, k T = T /θ = Fl/θ, and so θ = Fl/k T . For a cantilever, kC = F/δ, δ = F/kC . For the assembly, k = F/y, y = F/k = lθ + δ So

y=

Fl 2 F F = + k kT kC

Or

k=

1 (l 2 /k T ) + (1/kC )

Ans.

5-3 For a torsion bar, k = T/θ = GJ/l where J = πd 4 /32. So k = πd 4 G/(32l) = K d 4 /l . The springs, 1 and 2, are in parallel so d4 d4 +K l1 l2 1 4 1 + = Kd x l−x

k = k1 + k2 = K

And

Then

θ=

T = k

K d4

T 1 1 + x l−x

K d 4θ K d4 θ+ T = kθ = x l−x

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107

Chapter 5

K d4 K d 4θ T1 = θ; T2 = Thus x l−x If x = l/2, then T1 = T2 . If x < l/2, then T1 > T2 Using τ = 16T /πd 3 and θ = 32T l/(Gπd 4 ) gives πd 3 τ T = 16 and so θall =

2lτall 32l πd 3 τ = · 4 Gπd 16 Gd

Thus, if x < l/2, the allowable twist is 2xτall θall = Ans. Gd 1 4 1 k = Kd + Since x l−x π Gd 4 1 1 + = 32 x l−x Then the maximum torque is found to be Tmax

πd 3 xτall = 16

Ans.

1 1 + x l−x

Ans.

5-4 Both legs have the same twist angle. From Prob. 5-3, for equal shear, d is linear in x. Thus, d1 = 0.2d2 Ans. d24 π G (0.2d2 )4 πG + 1.258d24 k= = Ans. 32 0.2l 0.8l 32l 2(0.8l)τall θall = Ans. Gd2 Tmax = kθall = 0.198d23 τall

Ans.

5-5 A = πr 2 = π(r1 + x tan α)2

F r1

Fdx Fdx = AE Eπ(r1 + x tan α) 2 l dx F δ= π E 0 (r1 + x tan α) 2 l 1 F − = πE tan α(r1 + x tan α) 0

dδ = x ␣

l dx

= F

F 1 π E r1 (r1 + l tan α)

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Then

π Er1 (r1 + l tan α) F = δ l 2l E A1 1 + tan α = Ans. l d1

k=

5-6

dT = −w dx

Enlarged free body of length dx

x

T l dx

F = (T + dT ) + w dx − T = 0

Solution is T = −wx + c T |x=0 = P + wl = c

w dx

T = −wx + P + wl T dT

T = P + w(l − x)

w is cable’s weight per foot P

The infinitesmal stretch of the free body of original length dx is T dx AE P + w(l − x) dx = AE

dδ =

Integrating,

δ= 0

δ=

l

[P + w(l − x)] dx AE

wl 2 Pl + AE 2AE

Ans.

5-7 M = wlx − EI

wl 2 wx 2 − 2 2

dy wlx 2 wl 2 wx 3 = − x− + C1 , dx 2 2 6

EIy = y=

wlx 3 wl 2 x 2 wx 4 − − + C2 , 6 4 24 wx 2 (4lx − 6l 2 − x 2 ) 24E I

Ans.

dy = 0 at x = 0, dx

C1 = 0

y = 0 at x = 0,

C2 = 0

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109

Chapter 5

5-8 M = M1 = M B EI

dy = M B x + C1 , dx

EIy = y= 5-9 y ds dy dx

␭

MB x 2 + C2 , 2 MB x 2 2E I

dy = 0 at x = 0, dx

C1 = 0

y = 0 at x = 0,

C2 = 0

Ans.

2 dy ds = dx 2 + dy 2 = dx 1 + dx

Expand right-hand term by Binomial theorem 2 1/2 1 dy 2 dy =1+ + ··· 1+ dx 2 dx Since dy/dx is small compared to 1, use only the first two terms, dλ = ds − dx

1 dy 2 = dx 1 + − dx 2 dx 1 dy 2 dx = 2 dx 1 l dy 2 dx Ans. λ= 2 0 dx This contraction becomes important in a nonlinear, non-breaking extension spring. 5-10

4a 2 4ax 4ax − 2x y = − 2 (l − x) = − l l l dy 4a 8ax =− − 2 dx l l 2 16a 2 64a 2 x 64a 2 x 2 dy = 2 − + dx l l3 l4 1 l dy 2 8 a2 Ans. λ= dx = 2 0 dx 3 l

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5-11 y = a sin

πx l

aπ πx dy = cos dx l l 2 dy a2π 2 πx = 2 cos2 dx l l 1 l dy 2 λ= dx 2 0 dx λ=

a2 π 2 a2 = 2.467 4 l l

Ans.

Compare result with that of Prob. 5-10. See Charles R. Mischke, Elements of Mechanical Analysis, Addison-Wesley, Reading, Mass., 1963, pp. 244–249, for application to a nonlinear extension spring. 5-12 I = 2(5.56) = 11.12 in4 ymax = y1 + y2 = −

Fa 2 wl 4 + (a − 3l) 8E I 6E I

Here w = 50/12 = 4.167 lbf/in, and a = 7(12) = 84 in, and l = 10(12) = 120 in.

So

y1 = −

4.167(120) 4 = −0.324 in 8(30)(106 )(11.12)

y2 = −

600(84) 2 [3(120) − 84] = −0.584 in 6(30)(106 )(11.12)

ymax = −0.324 − 0.584 = −0.908 in

Ans.

M0 = −Fa − (wl 2 /2) = −600(84) − [4.167(120) 2 /2] = −80 400 lbf · in c = 4 − 1.18 = 2.82 in (−80 400)(−2.82) −M y =− (10−3 ) I 11.12 = −20.4 kpsi Ans.

σmax =

σmax is at the bottom of the section.

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111

Chapter 5

5-13

7 (800) + 10 3 RC = (800) + 10

800 lbf 600 lbf 3 ft

O

2 ft

RO = 5 ft

A

C

B

RO

RC

V (lbf)

5 (600) = 860 lbf 10 5 (600) = 540 lbf 10

860

O

60

M1 = 860(3)(12) = 30.96(103 ) lbf · in

540 M (lbf • in)

M1

M2 = 30.96(103 ) + 60(2)(12)

M2

= 32.40(103 ) lbf · in 32.40 Mmax ⇒ 6= Z = 5.4 in3 Z Z 2 F2l 3 F1 a[l − (l/2)] l l 2 − = + a − 2l 6E I l 2 2 48E I

σmax = y|x=5ft −

800(36)(60) 600(1203 ) 1 2 2 2 = [60 + 36 − 120 ] − 16 6(30)(106 ) I (120) 48(30)(106 ) I I = 23.69 in4

⇒

I /2 = 11.84 in4

Select two 6 in-8.2 lbf/ft channels; from Table A-7, I = 2(13.1) = 26.2 in4 , Z = 2(4.38) in3 1 23.69 − = −0.0565 in ymax = 26.2 16 32.40 = 3.70 kpsi σmax = 2(4.38) 5-14 I =

π (1.54 ) = 0.2485 in4 64

Superpose beams A-9-6 and A-9-7, yA =

300(24)(16) (162 + 242 − 402 ) 6(30)(106 )(0.2485)(40) +

12(16) [2(40)(162 ) − 163 − 403 ] 24(30)(106 )(0.2485)

y A = −0.1006 in Ans. y|x=20 =

300(16)(20) [202 + 162 − 2(40)(20)] 6(30)(106 )(0.2485)(40) −

% difference =

5(12)(404 ) = −0.1043 in Ans. 384(30)(106 )(0.2485)

0.1043 − 0.1006 (100) = 3.79% Ans. 0.1006

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5-15 1 I = 12

3 (1.53 ) = 0.105 47 in4 8

From Table A-9-10 F

␪A A

E

B

a

D

C a

Fa 2 (l + a) 3E I dy AB Fa 2 = (l − 3x 2 ) dx 6E I l yC = −

Thus, Fal Fal 2 = 6E I l 6E I Fa 2l y D = −θ A a = − 6E I With both loads, θA =

yD = − =−

Fa 2 Fa 2l − (l + a) 6E I 3E I Fa 2 120(102 ) (3l + 2a) = − [3(20) + 2(10)] 6E I 6(30)(106 )(0.105 47)

= −0.050 57 in Ans. 2 2Fa(l/2) 2 3 Fal 2 l yE = = l − 6E I l 2 24 E I = 5-16

120(10)(202 ) 3 = 0.018 96 in Ans. 24 (30)(106 )(0.105 47)

a = 36 in, l = 72 in, I = 13 in4, E = 30 Mpsi y= =

F2l 3 F1 a 2 (a − 3l) − 6E I 3E I 400(72) 3 400(36) 2 (36 − 216) − 6(30)(106 )(13) 3(30)(106 )(13)

= −0.1675 in Ans. I = 2(1.85) = 3.7 in4

5-17

Adding the weight of the channels, 2(5)/12 = 0.833 lbf/in, yA = −

Fl 3 10.833(484 ) 220(483 ) wl 4 − =− − 8E I 3E I 8(30)(106 )(3.7) 3(30)(106 )(3.7)

= −0.1378 in Ans.

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113

Chapter 5

5-18 I = πd 4 /64 = π(2)4 /64 = 0.7854 in4 Tables A-9-5 and A-9-9 y =− =−

F1 a F2l 3 + (4a 2 − 3l 2 ) 48E I 24E I 120(40) 3 80(10)(400 − 4800) + = −0.0130 in Ans. 6 48(30)(10 )(0.7854) 24(30)(106 )(0.7854)

5-19 (a) Useful relations k=

F 48E I = 3 y l

I =

2400(48) 3 kl 3 = = 0.1843 in4 48E 48(30)106

From I = bh 3 /12

12(0.1843) b Form a table. First, Table A-17 gives likely available fractional sizes for b: h=

3

8 12 , 9, 9 12 , 10 in

For h:

1 9 5 11 3 , , , , 2 16 8 16 4 For available b what is necessary h for required I?

3

b 8.5 9.0 9.5 10.0

12(0.1843) b 0.638 5" 0.626 ← choose 9"× 8 0.615 0.605

Ans.

(b) I = 9(0.625)3 /12 = 0.1831 in4 48E I 48(30)(106 )(0.1831) = = 2384 lbf/in l3 483 4(90 000)(0.1831) 4σ I = = 4394 lbf F= cl (0.625/2)(48) F 4394 y= = = 1.84 in Ans. k 2384 k=

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5-20 z R1 353 lbf 12"

502 lbf 21"

A

15"

O

C

x

B

R2 175 lbf

680 lbf

Torque = (600 − 80)(9/2) = 2340 lbf · in (T2 − T1 )

12 = T2 (1 − 0.125)(6) = 2340 2 2340 = 446 lbf, T1 = 0.125(446) = 56 lbf T2 = 6(0.875) M0 = 12(680) − 33(502) + 48R2 = 0 R2 =

33(502) − 12(680) = 175 lbf 48

R1 = 680 − 502 + 175 = 353 lbf We will treat this as two separate problems and then sum the results. First, consider the 680 lbf load as acting alone. z R1 510 lbf O

12"

A

R2 170 lbf 21"

B

15"

C

x

680 lbf

Fbx 2 (x + b2 − l 2 ); here b = 36 ", 6E I l x = 12", l = 48", F = 680 lbf

zO A = −

Also, πd 4 π(1.5) 4 = = 0.2485 in4 64 64 680(36)(12)(144 + 1296 − 2304) zA = − 6(30)(106 )(0.2485)(48) = +0.1182 in I =

z AC = −

Fa(l − x) 2 (x + a 2 − 2lx) 6E I l

where a = 12" and x = 21 + 12 = 33" 680(12)(15)(1089 + 144 − 3168) 6(30)(106 )(0.2485)(48) = +0.1103 in

zB = −

Next, consider the 502 lbf load as acting alone.

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115

Chapter 5

z 502 lbf 12"

O

A

21"

15"

B

C

R1

x

R2

Fbx 2 (x + b2 − l 2 ), where b = 15 ", 6E I l x = 12", l = 48", I = 0.2485 in4

zO B =

502(15)(12)(144 + 225 − 2304) = −0.081 44 in 6(30)(106 )(0.2485)(48) For z B use x = 33" Then,

zA =

502(15)(33)(1089 + 225 − 2304) 6(30)(106 )(0.2485)(48) = −0.1146 in

zB =

Therefore, by superposition z A = +0.1182 − 0.0814 = +0.0368 in Ans. z B = +0.1103 − 0.1146 = −0.0043 in Ans. 5-21 (a) Calculate torques and moment of inertia T = (400 − 50)(16/2) = 2800 lbf · in (8T2 − T2 )(10/2) = 2800 ⇒ T2 = 80 lbf, T1 = 8(80) = 640 lbf π I = (1.254 ) = 0.1198 in4 64 y RO O

720 lbf 9"

11"

450 lbf 12"

B

A

C RB

Due to 720 lbf, flip beam A-9-6 such that y AB → b = 9, x = 0, l = 20, F = −720 lbf dy Fb θB = (3x 2 + b2 − l 2 ) =− dx x=0 6E I l =−

−720(9) (0 + 81 − 400) = −4.793(10−3 ) rad 6(30)(106 )(0.1198)(20)

yC = −12θ B = −0.057 52 in Due to 450 lbf, use beam A-9-10, 450(144)(32) Fa 2 (l + a) = − = −0.1923 in yC = − 3E I 3(30)(106 )(0.1198)

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Adding the two deflections, yC = −0.057 52 − 0.1923 = −0.2498 in Ans. (b) At O: Due to 450 lbf:

θO = −

Fa 2 Fal dy 2 (l = − 3x ) = dx x=0 6E I l 6E I x=0

450(12)(20) 720(11)(0 + 112 − 400) + = 0.010 13 rad = 0.5805◦ 6 6(30)(10 )(0.1198)(20) 6(30)(106 )(0.1198)

At B: 450(12) [202 − 3(202 )] 6(30)(106 )(0.1198)(20) = −0.014 81 rad = 0.8485◦ 0.8485◦ = 1.694 in4 I = 0.1198 0.06◦ 64I 1/4 64(1.694) 1/4 d= = = 2.424 in π π

θ B = −4.793(10−3 ) +

Use d = 2.5 in Ans. π (2.54 ) = 1.917 in4 64 0.1198 = −0.015 61 in Ans. yC = −0.2498 1.917 I =

5-22 (a) l = 36(12) = 432 in y

x l

ymax = −

5(5000/12)(432) 4 5wl 4 =− 384E I 384(30)(106 )(5450) = −1.16 in

The frame is bowed up 1.16 in with respect to the bolsters. It is fabricated upside down and then inverted. Ans. (b) The equation in xy-coordinates is for the center sill neutral surface y=

wx (2lx 2 − x 3 − l 3 ) 24E I

Ans.

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117

Chapter 5

Differentiating this equation and solving for the slope at the left bolster gives w dy = (6lx 2 − 4x 3 − l 3 ) dx 24E I

(5000/12)(432) 3 wl 3 dy = − = − dx x=0 24E I 24(30)(106 )(5450)

Thus,

= −0.008 57 The slope at the right bolster is 0.008 57, so equation at left end is y = −0.008 57x and at the right end is y = 0.008 57(x − l). Ans. 5-23

From Table A-9-6, yL =

Fbx 2 (x + b2 − l 2 ) 6E I l

yL =

Fb 3 (x + b2 x − l 2 x) 6E I l

Fb dy L = (3x 2 + b2 − l 2 ) dx 6E I l Fb(b2 − l 2 ) dy L = dx x=0 6E I l Fb(b2 − l 2 ) ξ = 6E I l

Let And set

I =

And solve for d L

πd L4 64 32Fb(b2 − l 2 ) 1/4 dL = 3π Elξ

Ans.

For the other end view, observe the figure of Table A-9-6 from the back of the page, noting that a and b interchange as do x and −x 32Fa(l 2 − a 2 ) 1/4 Ans. dR = 3π Elξ For a uniform diameter shaft the necessary diameter is the larger of d L and d R .

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5-24

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Incorporating a design factor into the solution for d L of Prob. 5-23,

1/4 32n 2 2 Fb(l − b ) d= 3π Elξ 3 3 −9 ) 1/4 kN mm (10 10 − 3 = (mm 10 ) GPa mm 109 (10−3 )

32(1.28)(3.5)(150)|(2502 − 1502 )| −12 10 d=4 3π(207)(250)(0.001)

3.5 kN 100

150

d 250

= 36.4 mm Ans. 5-25

The maximum occurs in the right section. Flip beam A-9-6 and use y=

Fbx 2 (x + b2 − l 2 ) 6E I l

where b = 100 mm

Fb dy = (3x 2 + b2 − l 2 ) = 0 dx 6E I l Solving for x,

x= y=

l 2 − b2 = 3

2502 − 1002 = 132.29 mm from right 3

3.5(103 )(0.1)(0.132 29) [0.132 292 + 0.12 − 0.252 ](103 ) 6(207)(109 )(π/64)(0.03644 )(0.25)

= −0.0606 mm Ans. 5-26 The slope at x = 0 due to F1 in the xy plane is F1 b1 b12 − l 2 θx y = 6E I l

y a1

F1

b1 z

a2 F2

b2 x

and in the xz plane due to F2 is F2 b2 b22 − l 2 θx z = 6E I l

For small angles, the slopes add as vectors. Thus 1/2 θ L = θx2y + θx2z  2 2 2 2 1/2 2 2 F2 b2 b2 − l F1 b1 b1 − l  = + 6E I l 6E I l

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Chapter 5

119

Designating the slope constraint as ξ, we then have 2 1/2 1 ξ = |θ L | = Fi bi bi2 − l 2 6E I l Setting I = πd 4 /64 and solving for d 32 2 2 1/2 1/4 2 Fi bi bi − l d = 3π Elξ For the LH bearing, E = 30 Mpsi, ξ = 0.001, b1 = 12, b2 = 6, and l = 16. The result is d L = 1.31 in. Using a similar flip beam procedure, we get d R = 1.36 in for the RH bearing. So use d = 1 3/8 in Ans. 5-27

For the xy plane, use yBC of Table A-9-6 y=

100(4)(16 − 8) 2 [8 + 42 − 2(16)8] = −1.956(10−4 ) in 6(30)(106 )(16)

For the xz plane use yAB z=

300(6)(8) [82 + 62 − 162 ] = −7.8(10−4 ) in 6(30)(106 )(16)

δ = (−1.956j − 7.8k)(10−4 ) in |δ| = 8.04(10−4 ) in Ans. 5-28 32n 2 2 1/2 1/4 2 Fi bi bi − l d L = 3π Elξ 1/4 32(1.5) 2 2 2 2 2 2 1/2 [800(6)(6 − 10 )] + [600(3)(3 − 10 )] = 3π(29.8)(106 )(10)(0.001) = 1.56 in 1/2 1/4 32(1.5) 2 2 2 2 2 2 d R = [800(4)(10 − 4 )] + [600(7)(10 − 7 )] 3π(29.8)(106 )(10)(0.001) = 1.56 in choose d ≥ 1.56 in Ans. 5-29

From Table A-9-8 we have yL =

MB x 2 (x + 3a 2 − 6al + 2l 2 ) 6E I l

MB dy L = (3x 2 + 3a 2 − 6al + 2l 2 ) dx 6E I l

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At x = 0, the LH slope is θL =

MB dy L = (3a 2 − 6al + 2l 2 ) dx 6E I l

from which ξ = |θ L | =

MB 2 (l − 3b2 ) 6E I l

Setting I = πd 4 /64 and solving for d 32M B (l 2 − 3b2 ) 1/4 d= 3π Elξ For a multiplicity of moments, the slopes add vectorially and 32 2 1/2 1/4 2 2 Mi l − 3bi dL = 3π Elξ 1/2 1/4 32 2 2 Mi 3ai − l 2 d R = 3π Elξ The greatest slope is at the LH bearing. So 32(1200)[92 − 3(42 )] 1/4 = 0.706 in d = 3π(30)(106 )(9)(0.002) So use d = 3/4 in Ans. 5-30 RO

FAC

80 lbf

6

12 B

6FAC = 18(80) FAC = 240 lbf R O = 160 lbf I =

1 (0.25)(23 ) = 0.1667 in4 12

Initially, ignore the stretch of AC. From Table A-9-10 80(122 ) Fa 2 (l + a) = − (6 + 12) = −0.041 47 in 3E I 3(10)(106 )(0.1667) FL 240(12) δ= = = 1.4668(10−3 ) in 2 6 AE AC (π/4)(1/2) (10)(10 )

yB1 = −

Stretch of AC:

Due to stretch of AC By superposition,

y B 2 = −3δ = −4.400(10−3 ) in y B = −0.041 47 − 0.0044 = −0.045 87 in Ans.

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121

Chapter 5

5-31 θ=

(0.1F)(1.5) TL = = 9.292(10−4 ) F JG (π/32)(0.0124 )(79.3)(109 )

Due to twist δ B 1 = 0.1(θ) = 9.292(10−5 ) F Due to bending δB2 =

F(0.13 ) F L3 = = 1.582(10−6 ) F 3E I 3(207)(109 )(π/64)(0.0124 )

δ B = 1.582(10−6 ) F + 9.292(10−5 ) F = 9.450(10−5 ) F k=

1 = 10.58(103 ) N/m = 10.58 kN/m Ans. 9.450(10−5 )

5-32 F A

a

B

b

C

l

Fb l

R2 =

Fa l

δ1 =

R1 k1

δ2 =

R2 k2

R2

R1 ␦1

␦2

Spring deflection

Fb Fa Fb x =− + x − k1l k1l 2 k2l 2 Fb Fbx 2 Fx b a 2 2 (x + b − l ) + 2 Ans. − = − 6E I l l k1 k2 k1l Fb Fa(l − x) 2 Fx b a 2 − (x + a − 2lx) + 2 Ans. = − 6E I l l k1 k2 k1l

δ1 − δ2 yS = −δ1 + l y AB

y BC

5-33

R1 =

See Prob. 5-32 for deflection due to springs. Replace Fb/l and Fa/l with wl/2 wl wx 1 wl wl 1 wl − x= − + + yS = − 2k1 2k1l 2k2l 2 k1 k2 2k1 wl wx wx 1 1 2 3 3 (2lx − x − l ) + − y= + Ans. 24E I 2 k1 k2 2k1

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5-34

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Let the load be at x > l/2. The maximum deflection will be in Section AB (Table A-9-10) Fbx 2 (x + b2 − l 2 ) y AB = 6E I l Fb dy AB = (3x 2 + b2 − l 2 ) = 0 ⇒ 3x 2 + b2 − l 2 = 0 dx 6E I l

2 2 l −b l2 = 0.577l Ans. x= xmax = , 3 3 xmin = l − 0.577l = 0.423l

For x < l/2 5-35

50 lbf/ft O

M O = 50(10)(60) + 600(84)

600 lbf

= 80 400 lbf · in

B

A

7'

MO

Ans.

10'

R O = 50(10) + 600 = 1100 lbf

RO

I = 11.12 in4 from Prob. 5-12 M = −80 400 + 1100x − EI

4.167x 2 − 600x − 841 2

dy = −80 400x + 550x 2 − 0.6944x 3 − 300x − 842 + C1 dx

dy = 0 at x = 0 C1 = 0 dx E I y = −402 00x 2 + 183.33x 3 − 0.1736x 4 − 100x − 843 + C2 y = 0 at x = 0

C2 = 0

yB =

1 30(106 )(11.12)

[−40 200(1202 ) + 183.33(1203 )

− 0.1736(1204 ) − 100(120 − 84) 3 ] = −0.9075 in Ans. 5-36

See Prob. 5-13 for reactions: R O = 860 lbf, RC = 540 lbf M = 860x − 800x − 361 − 600x − 601 dy = 430x 2 − 400x − 362 − 300x − 602 + C1 EI dx E I y = 143.33x 3 − 133.33x − 363 − 100x − 603 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 120 in ⇒ C1 = −1.2254(106 ) lbf · in2 Substituting C1 and C2 and evaluating at x = 60, 1 6 = 143.33(603 ) − 133.33(60 − 36) 3 − 1.2254(106 )(60) E I y = 30(10 ) I − 16 I = 23.68 in4 Agrees with Prob. 5-13. The rest of the solution is the same.

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Chapter 5

5-37 I = 0.2485 in4 24 R O = 12(20) + (300) = 420 lbf 40 12 M = 420x − x 2 − 300x − 161 2 dy = 210x 2 − 2x 3 − 150x − 162 + C1 EI dx E I y = 70x 3 − 0.5x 4 − 50x − 163 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 40 in ⇒ C1 = −6.272(104 ) lbf · in2 Substituting for C1 and C2 and evaluating at x = 16, 1 [70(163 ) − 0.5(164 ) − 6.272(104 )(16)] yA = 6 30(10 )(0.2485) = −0.1006 in Ans. y|x=20 =

1 [70(203 ) − 0.5(204 ) − 50(20 − 16) 3 − 6.272(104 )(20)] 6 30(10 )(0.2485)

= 0.1043 in Ans. 3.7% difference Ans. 5-38 w[(l + a)/2][(l − a)/2)] l w a la = (l 2 − a 2 ) 2 4l w w w R2 = (l + a) − (l 2 − a 2 ) = (l + a)2 2 4l 4l 2 w w wx + (l + a) 2 x − l1 M = (l 2 − a 2 )x − 4l 2 4l w dy w w = (l 2 − a 2 )x 2 − x 3 + (l + a)2 x − l2 + C1 EI dx 8l 6 8l w 2 w w (l − a 2 )x 3 − x 4 + (l + a) 2 x − l3 + C1 x + C2 EIy = 24l 24 24l

冢l 2 a冣 w

R1 =

y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = l w 2 w 4 wa 2l 2 3 0= (l − a )l − l + C1l ⇒ C1 = 24l 24 24 w [(l 2 − a 2 )x 3 − lx 4 + (l + a) 2 x − l3 + a 2l 2 x] Ans. y= 24E I l

123

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5-39

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From Prob. 5-15, R A = R B = 120 lbf, and I = 0.105 47 in4 First half of beam, M = −120x + 120x − 101 dy = −60x 2 + 60x − 102 + C1 EI dx dy/dx = 0 at x = 20 in ⇒ 0 = −60(202 ) + 60(20 − 10) 2 + C1 ⇒ C1 = 1.8(104 ) lbf · in2 E I y = −20x 3 + 20x − 103 + 1.8(104 )x + C2 y = 0 at x = 10 in ⇒ C2 = −1.6(105 ) lbf · in3 y|x=0 =

1 30(106 )(0.105 47)

(−1.6)(105 )

= −0.050 57 in Ans. y|x=20 =

1 30(106 )(0.105 47)

[−20(203 ) + 20(20 − 10) 3 + 1.8(104 )(20) − 1.6(105 )]

= 0.018 96 in Ans. 5-40

From Prob. 5-30, R O = 160 lbf ↓, FAC = 240 lbf

I = 0.1667 in4

M = −160x + 240x − 61 dy = −80x 2 + 120x − 62 + C1 EI dx E I y = −26.67x 3 + 40x − 63 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 240(12) FL = −1.4668(10−3 ) in =− yA = − AE AC (π/4)(1/2) 2 (10)(106 ) at x = 6 10(106 )(0.1667)(−1.4668)(10−3 ) = −26.67(63 ) + C1 (6) C1 = 552.58 lbf · in2 yB =

1 [−26.67(183 ) + 40(18 − 6) 3 + 552.58(18)] 6 10(10 )(0.1667)

= −0.045 87 in 5-41 I1 = R1 =

Ans.

π (1.54 ) = 0.2485 in4 64 200 (12) = 1200 lbf 2

For 0 ≤ x ≤ 16 in,

M = 1200x −

I2 =

π 4 (2 ) = 0.7854 in4 64 MI

200 x − 42 2

x

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Chapter 5

1200x M = − 4800 I I1

125

1 1 100 1 1 0 x − 4 − 1200 x − 41 − − − x − 42 I1 I2 I1 I2 I2

= 4829x − 13 204x − 40 − 3301.1x − 41 − 127.32x − 42 E

dy = 2414.5x 2 − 13 204x − 41 − 1651x − 42 − 42.44x − 43 + C1 dx dy = 0 at x = 10 in dx 0 = 2414.5(102 ) − 13 204(10 − 4) 1 − 1651(10 − 4) 2 − 42.44(10 − 4) 3 + C1 C1 = −9.362(104 )

Boundary Condition:

E y = 804.83x 3 − 6602x − 42 − 550.3x − 43 − 10.61x − 44 − 9.362(104 )x + C2 y=0

at x = 0 ⇒ C2 = 0

For 0 ≤ x ≤ 16 in y=

1 [804.83x 3 − 6602x − 42 − 550.3x − 43 6 30(10 ) − 10.61x − 44 − 9.362(104 )x] Ans.

at x = 10 in y|x=10 =

1 [804.83(103 ) − 6602(10 − 4) 2 − 550.3(10 − 4) 3 6 30(10 ) − 10.61(10 − 4) 4 − 9.362(104 )(10)]

= −0.016 72 in Ans. 5-42

Define δi j as the deflection in the direction of the load at station i due to a unit load at station j. If U is the potential energy of strain for a body obeying Hooke’s law, apply P1 first. Then U=

1 P1 ( P1 δ11 ) 2

When the second load is added, U becomes U=

1 1 P1 ( P1 δ11 ) + P2 ( P2 δ22 ) + P1 ( P2 δ12 ) 2 2

For loading in the reverse order U =

1 1 P2 ( P2 δ22 ) + P1 ( P1 δ11 ) + P2 ( P1 δ21 ) 2 2

Since the order of loading is immaterial U = U and P1 P2 δ12 = P2 P1 δ21

when P1 = P2 , δ12 = δ21

which states that the deflection at station 1 due to a unit load at station 2 is the same as the deflection at station 2 due to a unit load at 1. δ is sometimes called an influence coefficient.

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5-43 (a) From Table A-9-10 y AB δ12

y

Fcx(l 2 − x 2 ) = 6E I l y ca(l 2 − a 2 ) = = F x=a 6E I l

y2 = Fδ21 = Fδ12 = Substituting I =

400 lbf b

a

A

c

B 1

2 7"

9"

Fca(l 2 − a 2 ) 6E I l

x

23"

πd 4 64 y2 =

400(7)(9)(232 − 92 )(64) = 0.00347 in Ans. 6(30)(106 )(π)(2) 4 (23)

(b) The slope of the shaft at left bearing at x = 0 is θ=

Fb(b2 − l 2 ) 6E I l

Viewing the illustration in Section 6 of Table A-9 from the back of the page provides the correct view of this problem. Noting that a is to be interchanged with b and −x with x leads to θ=

Fa(l 2 − a 2 )(64) Fa(l 2 − a 2 ) = 6E I l 6Eπd 4l

θ=

400(9)(232 − 92 )(64) = 0.000 496 in/in 6(30)(106 )(π)(2) 4 (23)

So y2 = 7θ = 7(0.000 496) = 0.00347 in Ans. 5-44

Place a dummy load Q at the center. Then, wx Qx (l − x) + 2 2 l/2 2 M dx ∂U U =2 , ymax = 2E I ∂ Q Q=0 0 l/2 2M ∂ M dx ymax = 2 2E I ∂ Q 0 Q=0 l/2 wx Qx x 2 (l − x) + dx ymax = EI 2 2 2 0 Q=0 M=

Set Q = 0 and integrate

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127

Chapter 5

ymax

w = 2E I

ymax =

lx 3 x 4 − 3 4

l/2 0

4

5wl 384E I

Ans.

5-45 I = 2(1.85) = 3.7 in4 Adding weight of channels of 0.833 lbf · in, ∂M 10.833 2 x = −F x − 5.417x 2 = −x M = −F x − 2 ∂F 48 48 1 1 ∂M δB = dx = M ( F x + 5.417x 2 )(x) dx EI 0 ∂F EI 0 =

(220/3)(483 ) + (5.417/4)(484 ) = 0.1378 in 30(106 )(3.7)

in direction of 220 lbf

y B = −0.1378 in Ans. 5-46 IO B

1 = (0.25)(23 ) = 0.1667 in4 , 12

O

FAC 3F A 6"

F

right

left

B

M = −F x¯

M = −2F x

12" x¯

x

1 U= 2E I

l

2 1 = 0.196 35 in2 2

∂ FAC =3 ∂F

FAC = 3F, 2F

A AC

π = 4

M 2 dx +

0

∂M = −x¯ ∂F

∂M = −2x ∂F

2 L AC FAC 2A AC E

l 1 FAC (∂ FAC /∂ F)L AC ∂U ∂M = dx + M δB = ∂F EI 0 ∂F A AC E 12 6 3F(3)(12) 1 −F x(− ¯ x) ¯ d x¯ + (−2F x)(−2x) dx + = EI 0 A AC E 0 3 1 F 108F 6 = (123 ) + 4F + EI 3 3 A AC E =

108F 864F + EI A AC E

=

108(80) 864(80) + = 0.045 86 in Ans. 6 10(10 )(0.1667) 0.196 35(10)(106 )

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5-47 Torsion

T = 0.1F

∂T = 0.1 ∂F

Bending

M = −F x¯

∂M = −x¯ ∂F

1 U= 2E I

M 2 dx +

F x

T 2L 2J G

1 ∂M T (∂ T /∂ F)L ∂U = M dx + δB = ∂F EI ∂F JG 0.1 0.1F(0.1)(1.5) 1 −F x(− ¯ x) ¯ d x¯ + = EI 0 JG =

0.015F F (0.13 ) + 3E I JG

Where π (0.012) 4 = 1.0179(10−9 ) m4 64 J = 2I = 2.0358(10−9 ) m4 0.015 0.001 + = 9.45(10−5 ) F δB = F 9 − 9 − 9 9 3(207)(10 )(1.0179)(10 ) 2.0358(10 )(79.3)(10 ) I =

k= 5-48

1 = 10.58(103 ) N/m = 10.58 kN/m Ans. 9.45(10−5 )

From Prob. 5-41, I1 = 0.2485 in4 , I2 = 0.7854 in4 For a dummy load ↑ Q at the center 200 −x Q ∂M x − 42 , = M = 1200x − x − 0 ≤ x ≤ 10 in 2 2 ∂Q 2 ∂U y|x=10 = ∂ Q Q=0 x x 2 1 4 1 10 2 = (1200x) − dx + [1200x − 100(x − 4) ] − dx E I1 0 2 I2 4 2 200(43 ) 1.566(105 ) 2 − − = E I1 I2 2 1.28(104 ) 1.566(105 ) =− + 30(106 ) 0.2485 0.7854 = −0.016 73 in Ans.

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129

Chapter 5

5-49

O x 3 Fa 5

l A O

4 Fa 5 l

4 F 5

A A

3 F 5

a

a

3 F 5

B

B 3 F 5

x

4 F 5

4 F 5

AB M = Fx OA

1 ∂u δB = = ∂F EI

I =

π 4 d , 64

N=

3 F 5

∂N 3 = ∂F 5

T =

4 Fa 5

∂T 4 = a ∂F 5

M1 =

4 F x¯ 5

∂ M1 4 = x¯ ∂F 5

M2 =

3 Fa 5

∂ M2 3 = a ∂F 5

(3/5) F(3/5)l (4/5) Fa(4a/5)l + AE JG

F x(x) dx +

0

1 + EI 9 Fa 3 + = 3E I 25

a

l

0

Fl AE

J = 2I,

4 F x¯ 5

∂M =x ∂F

16 + 25

A=

l 3 4 1 3 x¯ d x¯ + Fa a d x¯ 5 EI 0 5 5

Fa 2l JG

16 + 75

Fl 3 EI

9 + 25

Fa 2l EI

π 2 d 4

4Fl 16 32Fa 2l 16 64Fl 3 9 64Fa 2l 64Fa 3 9 + + + δB = + 3Eπd 4 25 πd 2 E 25 πd 4 G 75 Eπd 4 25 Eπd 4 4F 3 2 2 E 3 2 400a + 27ld + 384a l + 256l + 432a l Ans. = 75π Ed 4 G

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5-50

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The force applied to the copper and steel wire assembly is Fc + Fs = 250 lbf Since δc = δs Fs L Fc L = 2 6 3(π/4)(0.0801) (17.2)(10 ) (π/4)(0.0625) 2 (30)(106 ) Fc = 2.825Fs ∴ 3.825Fs

= 250

⇒

Fs = 65.36 lbf,

Fc = 2.825Fs = 184.64 lbf

σc =

184.64 = 12 200 psi = 12.2 kpsi Ans. 3(π/4)(0.0801) 2

σs =

65.36 = 21 300 psi = 21.3 kpsi Ans. (π/4)(0.06252 )

5-51 σb = 0.9(85) = 76.5 kpsi Ans. π (0.3752 ) = 50.69 kips Fb = 6(76.5) Bolt force 4 Fb 50.69 = −15.19 kpsi Ans. σc = − =− Cylinder stress Ac (π/4)(4.52 − 42 ) (b) Force from pressure

(a) Bolt stress

P=

π D2 π(42 ) p= (600) = 7540 lbf = 7.54 kip 4 4

50.69 Pc 50.69 Pb

6 bolts

x

Fx = 0

Pb + Pc = 7.54 (1)

P 7.54 kip

Pb L Pc L = 2 2 (π/4)(4.5 − 4 ) E 6(π/4)(0.3752 ) E

Since δc = δb ,

Pc = 5.037Pb

(2)

Substituting into Eq. (1) 6.037Pb = 7.54

⇒

Pb = 1.249 kip;

and from Eq (2),

Pc = 6.291 kip

Using the results of (a) above, the total bolt and cylinder stresses are σb = 76.5 +

1.249 = 78.4 kpsi Ans. 6(π/4)(0.3752 )

σc = −15.19 +

6.291 = −13.3 kpsi Ans. (π/4)(4.52 − 42 )

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131

Chapter 5

5-52 T = Tc + Ts

and θc = θs

Ts L Tc L = (G J ) c (G J ) s

Also,

Tc =

(G J ) c Ts (G J ) s

Substituting into equation for T ,

(G J ) c Ts T = 1+ (G J ) s

%Ts =

(G J ) s Ts = T (G J ) s + (G J ) c

Ans.

5-53 RB

RO + RB = W δ O A = δ AB

B

30"

20" O RO

(2)

30R B 20R O = , AE AE

W 800 lbf

A

RO =

3 RB 2

3 R B + R B = 800 2 1600 RB = = 320 lbf Ans. 5 R O = 800 − 320 = 480 lbf Ans. 480 = −480 psi Ans. σO = − 1 σB = +

5-54

(1)

320 = 320 psi Ans. 1

Since θ O A = θ AB T AB (6) TO A (4) = , GJ GJ

3 TO A = T AB 2

Also TO A + T AB = T T AB

3 + 1 = T, 2

T AB =

3 3T TO A = T AB = 2 5

T 2.5

Ans.

Ans.

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5-55 T2 T1 T2 T1 = = ⇒ r1 r2 1.25 3 3 T2 = T1 1.25 3 4π ∴ θ1 + θ2 = rad 1.25 180 3 4π T1 (48) (3/1.25)T1 (48) + = 4 6 4 6 (π/32)(7/8) (11.5)(10 ) 1.25 (π/32)(1.25) (11.5)(10 ) 180 F1 = F2

⇒

T1 = 403.9 lbf · in 3 T1 = 969.4 lbf · in T2 = 1.25 16T1 16(403.9) = = 3071 psi Ans. τ1 = 3 πd π(7/8)3 τ2 =

16(969.4) = 2528 psi Ans. π(1.25)3

5-56 10 kip

5 kip

FA

RO

FB

x

RC

(1) Arbitrarily, choose RC as redundant reaction Fx = 0, 10(103 ) − 5(103 ) − R O − RC = 0 (2) R O + RC = 5(103 ) lbf (3)

δC =

RC (15) [10(103 ) − 5(103 ) − RC ]20 [5(103 ) + RC ] − (10) − =0 AE AE AE −45RC + 5(104 ) = 0

⇒

RC = 1111 lbf Ans.

R O = 5000 − 1111 = 3889 lbf Ans. 5-57 w A

MC

B a

C RC

RB l x

(1) Choose R B as redundant reaction (2)

R B + RC = wl

(a)

R B (l − a) −

wl 2 + MC = 0 (b) 2

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133

Chapter 5

R B (l − a) 3 w(l − a) 2 yB = + [4l(l − a) − (l − a) 2 − 6l 2 ] = 0 3E I 24E I w [6l 2 − 4l(l − a) + (l − a)2 ] RB = 8(l − a) w (3l 2 + 2al + a 2 ) Ans. = 8(l − a)

(3)

Substituting, w (5l 2 − 10al − a 2 ) Ans. 8(l − a) wl 2 w − R B (l − a) = (l 2 − 2al − a 2 ) Ans. MC = 2 8 RC = wl − R B =

Eq. (a) Eq. (b)

5-58 w MC A

a x

B

C

RB

RC

∂M wx 2 + R B x − a1 , = x − a1 M =− 2 ∂ RB l ∂U 1 ∂M = M dx ∂ RB EI 0 ∂ RB a l −wx 2 1 −wx 2 1 (0) dx + + R B (x − a) (x − a) dx = 0 = EI 0 2 EI a 2 a 3 RB w 1 4 4 3 (l − a ) − (l − a ) + (l − a) 3 − (a − a) 3 = 0 − 2 4 3 3 RB =

w w (3l 2 + 2al + a 2 ) Ans. [3(L 4 − a 4 ) − 4a(l 3 − a 3 )] = 3 (l − a) 8(l − a)

RC = wl − R B = MC =

w (5l 2 − 10al − a 2 ) Ans. 8(l − a)

wl 2 w − R B (l − a) = (l 2 − 2al − a 2 ) Ans. 2 8

5-59 FBE 500

A

FDF 500

C

500

B RA

(1)

D

A=

π (0.0122 ) = 1.131(10−4 ) m2 4

20 kN

R A + FB E + FD F = 20 kN M A = 3FD F − 2(20) + FB E = 0

(a)

FB E + 3FD F = 40 kN

(b)

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M = R A x + FB E x − 0.51 − 20(103 )x − 11

(2) EI

FB E dy x2 = RA + x − 0.52 − 10(103 )x − 12 + C1 dx 2 2

E I y = RA

FB E 10 x3 + x − 0.53 − (103 )x − 13 + C1 x + C2 6 6 3

(3) y = 0 at x = 0 C2 = 0 FB E (1) Fl = −4.2305(10−8 ) FB E =− yB = − AE B E 1.131(10−4 )209(109 ) Substituting and evaluating at x = 0.5 m 0.53 + C1 (0.5) E I y B = 209(109 )(8)(10−7 )(−4.2305)(10−8 ) FB E = R A 6 2.0833(10−2 ) R A + 7.0734(10−3 ) FB E + 0.5C1 = 0 FD F (1) Fl yD = − = −4.2305(10−8 ) FD F =− − 4 9 AE D F 1.131(10 )(209)(10 )

(c)

Substituting and evaluating at x = 1.5 m FB E 10 1.53 + (1.5 − 0.5) 3 − (103 )(1.5 − 1) 3 +1.5C1 E I y D = −7.0734(10−3 ) FD F = R A 6 6 3 0.5625R A + 0.166 67FB E + 7.0734(10−3 ) FD F + 1.5C1 = 416.67 (d )      1 1 1 0  20 000  RA             0 1 3 0 40 000 F B E   =  2.0833(10−2 ) 7.0734(10−3 ) 0 0.5   0  FD F          0.5625 0.166 67 7.0734(10−3 ) 1.5 416.67 C1 Solve simultaneously or use software R A = −3885 N, FB E = 15 830 N, σB E =

15 830 = 140 MPa Ans., (π/4)(122 )

FD F = 8058 N, σD F =

C1 = −62.045 N · m2

8058 = 71.2 MPa Ans. (π/4)(122 )

E I = 209(109 )(8)(10−7 ) = 167.2(103 ) N · m2 3885 3 15 830 1 10 3 3 3 − x + x − 0.5 − (10 )x − 1 − 62.045x y= 167.2(103 ) 6 6 3 B: x = 0.5 m, C: x = 1 m,

D: x = 1.5,

y B = −6.70(10−4 ) m = −0.670 mm Ans. 3885 3 1 15 830 3 − (1 ) + (1 − 0.5) − 62.045(1) yC = 167.2(103 ) 6 6 = −2.27(10−3 ) m = −2.27 mm Ans. 3885 1 15 830 − (1.53 ) + (1.5 − 0.5) 3 yD = 3 167.2(10 ) 6 6 10 3 3 − (10 )(1.5 − 1) − 62.045(1.5) 3 = −3.39(10−4 ) m = −0.339 mm Ans.

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135

Chapter 5

5-60 y FBE

500 lbf 3" A

3"

C

3"

D

E I = 30(106 )(0.050) = 1.5(106 ) lbf · in2

x

B FFD

RC

RC + FB E − FF D = 500 3RC + 6FB E = 9(500) = 4500 M = −500x + FB E x − 31 + RC x − 61

(1) (2)

(a) (b)

dy FB E RC = −250x 2 + x − 32 + x − 62 + C1 dx 2 2 250 3 FB E RC x + x − 33 + x − 63 + C1 x + C2 EIy = − 3 6 6 FB E (2) Fl = −8.692(10−7 ) FB E yB = =− AE B E (π/4)(5/16) 2 (30)(106 )

EI

Substituting and evaluating at x = 3 in E I y B = 1.5(106 )[−8.692(10−7 ) FB E ] = −

250 3 (3 ) + 3C1 + C2 3

1.3038FB E + 3C1 + C2 = 2250

(c)

250 3 FB E (6 ) + (6 − 3) 3 + 6C1 + C2 3 6 4.5FB E + 6C1 + C2 = 1.8(104 )

(d)

Since y = 0 at x = 6 in E I y|=0 = − yD =

Fl AE

= DF

FD F (2.5) = 1.0865(10−6 ) FD F 2 6 (π/4)(5/16) (30)(10 )

Substituting and evaluating at x = 9 in 250 3 FB E (9 ) + (9 − 3) 3 3 6 RC (9 − 6) 3 + 9C1 + C2 + 6

E I y D = 1.5(106 )[1.0865(10−6 ) FD F ] = −

4.5RC + 36FB E − 1.6297FD F  1 1 −1 0  3 6 0 0   0 1.3038 0 3   0 4.5 0 6 4.5 36 −1.6297 9 RC = −590.4 lbf,

+ 9C1 + C2 = 6.075(104 )     500 RC  0                0  FB E   4500   2250 = 1 F  D F     1.8(104 )      1  C1          4 1 C2 6.075(10 )

FB E = 1045.2 lbf,

C1 = 4136.4 lbf · in2 ,

FD F = −45.2 lbf

C2 = −11 522 lbf · in3

(e)

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σB E =

1045.2 = 13 627 psi = 13.6 kpsi Ans. (π/4)(5/16) 2

σD F = −

45.2 = −589 psi Ans. (π/4)(5/16) 2

1 (−11 522) = −0.007 68 in Ans. 1.5(106 ) 250 3 1 yB = − (3 ) + 4136.4(3) − 11 522 = −0.000 909 in Ans. 1.5(106 ) 3 250 3 1 1045.2 −590.4 3 3 − (9 ) + (9 − 3) + (9 − 6) + 4136.4(9) − 11 522 yD = 1.5(106 ) 3 6 6 yA =

= −4.93(10−5 ) in Ans. 5-61

␪

F Q (dummy load)

M = −P R sin θ + Q R(1 − cos θ)

∂M = R(1 − cos θ) ∂Q

π ∂U 1 P R3 δQ = = (−P R sin θ) R(1 − cos θ) R dθ = −2 ∂ Q Q=0 EI 0 EI Deflection is upward and equals 2( P R 3 /E I ) 5-62

Equation (4-34) becomes

U =2 0

π

M 2 R dθ 2E I

Ans.

R/ h > 10

∂M = R(1 − cos θ) ∂F π ∂U 2 ∂M δ= = R dθ M ∂F EI 0 ∂F π 2 F R 3 (1 − cos θ) 2 dθ = EI 0

where M = F R(1 − cos θ) and

=

3π F R 3 EI

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137

Chapter 5

Since I = bh 3 /12 = 4(6) 3 /12 = 72 mm4 and R = 81/2 = 40.5 mm, we have δ=

3π(40.5) 3 F = 66.4F mm Ans. 131(72)

where F is in kN. 5-63

P

R ␪ l x

M = −P x,

∂M = −x ∂P

0≤x ≤l

∂M = l + R(1 − cos θ) 0 ≤ θ ≤ l M = Pl + P R(1 − cos θ), ∂P l π/2 1 2 δP = −P x(−x) dx + P[l + R(1 − cos θ)] R dθ EI 0 0 = 5-64

P {4l 3 + 3R[2πl 2 + 4(π − 2)l R + (3π − 8) R 2 ]} Ans. 12E I

A: Dummy load Q is applied at A. Bending in AB due only to Q which is zero. Q

A

C P

B

␪

∂M π M = P R sin θ + Q R(1 + sin θ), = R(1 + sin θ) 0 ≤ θ ≤ ∂Q 2 π/2 ∂U 1 = ( P R sin θ)[R(1 + sin θ)]R dθ (δ A ) V = ∂ Q Q=0 EI 0 θ sin 2θ π/2 π P R3 P R3 −cos θ + − 1+ = = EI 2 4 EI 4 0 =

π + 4 P R3 4 EI

Ans.

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M = P R sin θ,

B:

1 ∂U = (δ B ) V = ∂P EI =

π P R3 4 EI

∂M = R sin θ ∂P π/2

( P R sin θ)( R sin θ) R dθ

0

Ans.

5-65

y T

M x 200 N

␪

z

R A 200 N

M = P R sin θ,

∂M = R sin θ ∂P

0<θ <

π 2

∂T = R(1 − cos θ) ∂P π/2 π/2 1 ∂U 1 2 2 (δ A ) y = − =− P( R sin θ) R dθ + P[R(1 − cos θ)] R dθ ∂P EI 0 GJ 0 T = P R(1 − cos θ),

Integrating and substituting J = 2I and G = E/[2(1 + ν)] 3π P R3 P R3 π + (1 + ν) − 2 = −[4π − 8 + (3π − 8)ν] (δ A ) y = − EI 4 4 4E I (200)(100) 3 = −[4π − 8 + (3π − 8)(0.29)] = −40.6 mm 4(200)(103 )(π/64)(5) 4 5-66

Consider the horizontal reaction, to be applied at B, subject to the constraint (δ B ) H = 0.

P R A

(a) (δ B ) H =

␪ B

∂U =0 ∂H

H

P 2

Due to symmetry, consider half of the structure. P does not deflect horizontally.

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139

Chapter 5

∂M PR π (1 − cos θ) − H R sin θ, = −R sin θ, 0 < θ < 2 ∂H 2 π/2 1 ∂U PR = (1 − cos θ) − H R sin θ (−R sin θ) R dθ = 0 ∂H EI 0 2

M=

−

P P π + +H =0 2 4 4

⇒

H=

P π

Ans.

Reaction at A is the same where H goes to the left (b) For 0 < θ <

π , 2

PR PR (1 − cos θ) − sin θ 2 π PR M= [π(1 − cos θ) − 2 sin θ] Ans. 2π

M=

Due to symmetry, the solution for the left side is identical. R ∂M = [π(1 − cos θ) − 2 sin θ] ∂P 2π π/2 P R2 2 ∂U = δP = [π(1 − cos θ) − 2 sin θ]2 R dθ ∂P EI 0 4π 2 π/2 P R3 = (π 2 + π 2 cos2 θ + 4 sin2 θ − 2π 2 cos θ 2 2π E I 0 − 4π sin θ + 4π sin θ cos θ) dθ π P R3 2 π 2 π 2 π +π +4 − 2π − 4π + 2π = 2π 2 E I 2 4 4

(c)

= 5-67

(3π 2 − 8π − 4) P R 3 8π EI

Ans.

Must use Eq. (5-34) A = 80(60) − 40(60) = 2400 mm2 R=

(25 + 40)(80)(60) − (25 + 20 + 30)(40)(60) = 55 mm 2400

Section is equivalent to the “T” section of Table 4-5 rn =

60(20) + 20(60) = 45.9654 mm 60 ln[(25 + 20)/25] + 20 ln[(80 + 25)/(25 + 20)] e = R − rn = 9.035 mm

Straight section y z

Iz = 30 mm 50 mm

1 (60)(203 ) + 60(20)(30 − 10) 2 12 1 3 2 (10)(60 ) + 10(60)(50 − 30) +2 12

= 1.36(106 ) mm4

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Fr F

F␪

␪ 100 mm x

M

For 0 ≤ x ≤ 100 mm ∂M = −x; M = −F x, ∂F

V = F,

∂V =1 ∂F

For θ ≤ π/2

∂ Fr ∂ Fθ = cos θ; Fθ = F sin θ, = sin θ ∂F ∂F ∂M M = F(100 + 55 sin θ), = (100 + 55 sin θ) ∂F Use Eq. (5-34), integrate from 0 to π/2, double the results and add straight part π/2 100 2 1 100 (100 + 55 sin θ) 2 (1) F(1) dx 2 δ= + dθ F x dx + F E I 0 2400(G/E) 2400(9.035) 0 0 π/2 π/2 F sin2 θ(55) F(100 + 55 sin θ) + dθ − sin θdθ 2400 2400 0 0 Fr = F cos θ,

− 0

π/2

F sin θ(100 + 55 sin θ) dθ + 2400

π/2

(1) F cos2 θ(55) dθ 2400(G/E)

0

Substitute I = 1.36(103 ) mm2 , F = 30(103 ) N, E = 207(103 ) N/mm2 , G = 79(103 ) N/mm2 207 100 2.908(104 ) 55 2 π 1003 3 30(10 ) + + + δ= 3 6 207(10 ) 3(1.36)(10 ) 79 2400 2400(9.035) 2400 4 207 2 55 π (143.197) + = 0.476 mm Ans. − 2400 79 2400 4 5-68

Fr

M

R F␪

␪

F Q

M = F R sin θ − Q R(1 − cos θ), Fθ = Q cos θ + F sin θ,

∂M = −R(1 − cos θ) ∂Q

∂ Fθ = cos θ ∂Q

∂ ( M Fθ ) = [F R sin θ − Q R(1 − cos θ)] cos θ ∂Q + [−R(1 − cos θ)][Q cos θ + F sin θ] Fr = F cos θ − Q sin θ,

∂ Fr = −sin θ ∂Q

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141

Chapter 5

From Eq. (5-34) π π ∂U 1 R δ= = ( F R sin θ)[−R(1 − cos θ)] dθ + F sin θ cos θdθ ∂ Q Q=0 AeE 0 AE 0 π 1 [F R sin θ cos θ − F R sin θ(1 − cos θ)] dθ − AE 0 CR π −F cos θ sin θdθ + AG 0 2F R 2F R 2 R 2F R +0+ +0=− −1 Ans. =− AeE AE e AE 5-69

The cross section at A does not rotate, thus for a single quadrant we have ∂U =0 ∂ MA The bending moment at an angle θ to the x axis is M = MA − because x = R cos θ. Next,

F FR ( R − x) = M A − (1 − cos θ) 2 2

M2 ds = 2E I

U= since ds = R dθ. Then

∂U R = ∂ MA EI But ∂ M/∂ M A = 1. Therefore π/2 M dθ = 0

Since this term is zero, we have

Substituting into Eq. (1)

π/2

0

0

π/2

M 0

π/2

M2 R dθ 2E I

∂M dθ = 0 ∂ MA

FR (1 − cos θ) dθ = 0 MA − 2

2 FR 1− MA = 2 π FR 2 M= cos θ − 2 π

The maximum occurs at B where θ = π/2. It is MB = −

FR π

Ans.

(1)

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For one quadrant

2 ∂M R 2 FR cos θ − ; = cos θ − M= 2 π ∂F 2 π π/2 ∂U M ∂M =4 R dθ δ= ∂F EI ∂F 0 2 2 F R 3 π/2 cos θ − = dθ EI 0 π 2 F R3 π − = Ans. EI 4 π

5-71 Cπ 2 E I l2 π π D4 (1 − K 4 ) I = ( D4 − d 4) = 64 64 2 4 Cπ E π D 4 Pcr = (1 − K ) l2 64 1/4 64Pcr l 2 D= Ans. π 3 C E(1 − K 4 )

Pcr =

5-72 A=

π 2 D (1 − K 2 ), 4

I =

π 4 π 4 D (1 − K 4 ) = D (1 − K 2 )(1 + K 2 ), 64 64

k2 =

D2 I = (1 + K 2 ) A 16

From Eq. (5-50) S y2 l 2 S y2 l 2 Pcr = Sy − = Sy − (π/4) D 2 (1 − K 2 ) 4π 2 k 2 C E 4π 2 ( D 2 /16)(1 + K 2 )C E 4Pcr = π D 2 (1 − K 2 )S y −

4S y2 l 2 π D 2 (1 − K 2 ) π 2 D 2 (1 + K 2 )C E

π D 2 (1 − K 2 )S y = 4Pcr +

4S y2 l 2 (1 − K 2 ) π(1 + K 2 )C E

4S y2 l 2 (1 − K 2 ) 4Pcr D= + π S y (1 − K 2 ) π(1 + K 2 )C Eπ(1 − K 2 )S y

Sy l 2 Pcr + =2 π S y (1 − K 2 ) π 2 C E(1 + K 2 )

1/2

1/2 Ans.

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143

Chapter 5

5-73

(a) + M A = 0,

3 2.5(180) − √ FB O (1.75) = 0 32 + 1.752

⇒

FB O = 297.7 lbf

Using n d = 5, design for Fcr = n d FB O = 5(297.7) = 1488 lbf, l = 3.473 ft, S y = 24 kpsi In plane: Try 1" × 1/2" section

32 + 1.752 =

k = 0.2887h = 0.2887", C = 1.0

3.473(12) l = = 144.4 k 0.2887 1/2 2 l 2π (1)(30)(106 ) = = 157.1 k 1 24(103 ) Since (l/k) 1 > (l/k) use Johnson formula 2 3 1 1 24(10 ) 144.4 Pcr = (1) 24(103 ) − = 6930 lbf 2 2π 1(30)(106 ) Pcr = 3465 lbf

Try 1" × 1/4":

k = 0.2887(0.5) = 0.1444 in,

Out of plane:

C = 1.2

3.473(12) l = = 289 k 0.1444 Since (l/k) 1 < (l/k) use Euler equation 1.2(π 2 )(30)(106 ) Pcr = 1(0.5) = 2127 lbf 2892 2 l 1/4" increases l/k by 2, by 4, and A by 1/2 k k = 0.2887(0.375) = 0.1083 in

Try 1" × 3/8": l = 385, k Use 1" × 1/2" (b) σb = − 5-74 5-75

Pcr = 1(0.375)

1.2(π 2 )(30)(106 ) = 899 lbf 3852

(too low)

Ans.

298 P =− = −379 psi πdl π(0.5)(0.5)

No, bearing stress is not significant.

This is a design problem with no one distinct solution. π F = 800 (32 ) = 5655 lbf, S y = 37.5 kpsi 4 Pcr = n d F = 3(5655) = 17 000 lbf

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(a) Assume Euler with C = 1 Pcr l 2 π I = d4 = 64 Cπ 2 E

64Pcr l 2 d= π 3C E

⇒

1/4

64(17)(103 )(602 ) = π 3 (1)(30)(106 )

1/4 = 1.433 in

Use d = 1.5 in; k = d/4 = 0.375 60 l = = 160 k 0.375 1/2 2 l 2π (1)(30)(106 ) = = 126 k 1 37.5(103 ) Pcr =

∴

use Euler

π 2 (30)(106 )(π/64)(1.54 ) = 20 440 lbf 602

d = 1.5 in

is satisfactory. Ans. 1/4 64(17)(103 )(182 ) d= = 0.785 in, π 3 (1)(30)(106 )

(b)

k=

so use 0.875 in

0.875 = 0.2188 in 4

18 = 82.3 try Johnson 0.2188 2 π 37.5(103 ) 1 2 3 Pcr = (0.875 ) 37.5(10 ) − 82.3 = 17 714 lbf 4 2π 1(30)(106 )

l/k =

Use d = 0.875 in Ans. 20 440 = 3.61 Ans. 5655 17 714 = 3.13 Ans. = 5655

n (a) =

(c)

n (b) 5-76 W 9.8(400) 3920 N

4F sin θ = 3920

2 bars 2F

2F

␪

␪

F=

3920 4 sin θ

In range of operation, F is maximum when θ = 15◦ 3920 = 3786 N per bar Fmax = 4 sin 15 Pcr = n d Fmax = 2.5(3786) = 9465 N l = 300 mm, h = 25 mm

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Chapter 5

√ Try b = 5 mm: out of plane k = (5/ 12) = 1.443 mm 300 l = = 207.8 k 1.443 1/2 l (2π 2 )(1.4)(207)(109 ) = = 123 k 1 380(106 ) Pcr = (25)(5)

∴

use Euler

(1.4π 2 )(207)(103 ) = 8280 N (207.8) 2

√ Try: 5.5 mm: k = 5.5/ 12 = 1.588 mm

300 l = = 189 k 1.588 (1.4π 2 )(207)(103 ) Pcr = 25(5.5) = 11 010 N 1892 Use 25 × 5.5 mm bars Ans. The factor of safety is thus n=

5-77

2000

A

F = 0 = 2000 + 10 000 − P ⇒ P = 12 000 lbf Ans. 5.68 M A = 12 000 − 10 000(5.68) + M = 0 2

10,000

C

P M

11 010 = 2.91 Ans. 3786

M = 22 720 lbf · in M 22 720 e= = = 1.893 in Ans. P 12 000

From Table A-8, A = 4.271 in2 , I = 7.090 in4 7.090 I = = 1.66 in2 A 4.271 1.893(2) 12 000 1+ = −9218 psi σc = − 4.271 1.66 1.893(2) 12 000 σt = − 1− = 3598 psi 4.271 1.66 k2 =

3598

␴

9218

5-78

This is a design problem so the solutions will differ.

Ans.

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For free fall with y ≤ h Fy − m y¨ = 0 mg − m y¨ = 0,

y

so y¨ = g

mg

Using y = a + bt + ct 2 , we have at t = 0, y = 0, and y˙ = 0, and so a = 0, b = 0, and c = g/2. Thus 1 y = gt 2 and y˙ = gt for y ≤ h 2 At impact, y = h, t = (2h/g) 1/2 , and v0 = (2gh) 1/2 After contact, the differential equatioin (D.E.) is

y k(y h)

mg − k( y − h) − m y¨ = 0

for y > h mg

Now let x = y − h; then x˙ = y˙ and x¨ = y¨ . So the D.E. is x¨ + (k/m)x = g with solution ω = (k/m) 1/2 and mg x = A cos ωt + B sin ωt + k

At contact, t = 0, x = 0, and x˙ = v0 . Evaluating A and B then yields v0 mg mg cos ωt + sin ωt + x =− k ω k or W v0 W y = − cos ωt + sin ωt + +h k ω k and Wω y˙ = sin ωt + v0 cos ωt

k To find ymax set y˙ = 0. Solving gives v0 k tan ωt = − Wω v0 k

−1 (ωt )* = tan − or Wω The first value of (ωt )* is a minimum and negative. So add π radians to it to find the maximum. Numerical example: h = 1 in, W = 30 lbf, k = 100 lbf/in. Then ω = (k/m) 1/2 = [100(386)/30]1/2 = 35.87 rad/s W/k = 30/100 = 0.3 v0 = (2gh) 1/2 = [2(386)(1)]1/2 = 27.78 in/s Then y = −0.3 cos 35.87t +

27.78 sin 35.87t + 0.3 + 1 35.87

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Chapter 5

For ymax

27.78(100) v0 k =− = −2.58 Wω 30(35.87) (ωt )* = −1.20 rad (minimum)

tan ωt = −

(ωt )* = −1.20 + π = 1.940 (maximum) Then t * = 1.940/35.87 = 0.0541 s. This means that the spring bottoms out at t * seconds. Then (ωt )* = 35.87(0.0541) = 1.94 rad 27.78 ymax = −0.3 cos 1.94 + sin 1.94 + 0.3 + 1 = 2.130 in Ans. So 35.87 The maximum spring force is Fmax = k( ymax − h) = 100(2.130 − 1) = 113 lbf Ans. The action is illustrated by the graph below. Applications: Impact, such as a dropped package or a pogo stick with a passive rider. The idea has also been used for a one-legged robotic walking machine. y Free fall

0

Speeds agree 0.05 Time of release

0.01 Inflection point of trig curve (The maximum speed about this point is 29.8 in/s.)

0.01

1

2

0.05

Time t Equilibrium, rest deflection

During contact ymax

5-80

Choose t = 0 at the instant of impact. At this instant, v1 = (2gh) 1/2 . Using momentum, m 1 v1 = m 2 v2 . Thus W1 + W2 W1 (2gh) 1/2 = v2 g g

Therefore at t = 0, y = 0, and y˙ = v2 y

W1 W2 ky

W1 (2gh) 1/2 v2 = W1 + W2

Let W = W1 + W2 W1

Because the spring force at y = 0 includes a reaction to W2 , the D.E. is W y¨ = −ky + W1 g With ω = (kg/W ) 1/2 the solution is y = A cos ωt + B sin ωt + W1 /k y˙ = −Aω sin ωt + Bω cos ωt

At t = 0, y = 0 ⇒ A = −W1 /k At t = 0, y˙ = v2 ⇒ v2 = Bω

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Then B=

W1 (2gh) 1/2 v2 = ω (W1 + W2 )[kg/(W1 + W2 )]1/2

We now have

1/2 W1 2h W1

y=− cos ωt + W1 sin ωt + k k(W1 + W2 ) k Transforming gives 1/2 2hk W1 W1 +1 cos(ωt − φ) + y= k W1 + W2 k where φ is a phase angle. The maximum deflection of W2 and the maximum spring force are thus 1/2 2hk W1 W1 Ans. +1 + ymax = k W1 + W2 k 1/2 2hk Fmax = kymax + W2 = W1 +1 + W1 + W2 Ans. W1 + W2 5-81

Assume x > y to get a free-body diagram. y k1(x y)

W

k2 y

x

Then W y¨ = k1 (x − y) − k2 y g A particular solution for x = a is k1 a y= k1 + k2 Then the complementary plus the particular solution is y = A cos ωt + B sin ωt +

where

(k1 + k2 )g ω= W

1/2

k1 a k1 + k2

At t = 0, y = 0, and y˙ = 0. Therefore B = 0 and k1 a A=− k1 + k2 Substituting, k1 a (1 − cos ωt) y= k1 + k2 Since y is maximum when the cosine is −1 2k1 a ymax = k1 + k2

Ans.

MECHANICAL ENGINEERING DESIGN TUTORIAL 6, PART A: INTRODUCTION TO STATIC FAILURE THEORIES APPROACH Theories have been developed for the static failure of metals based upon the two classes of material failure; ductile metals yield while brittle metals fracture. Thus separate failure theories exist for ductile and brittle metals: Failure Theories for Ductile Materials 1. Maximum Shear Stress (MSS) 2. Distortion Energy (DE) 3. Ductile Coulomb-Mohr (DCM) Failure Theories for Brittle Materials 1. Maximum Normal Stress (MNS) 2. Brittle Coulomb-Mohr (BCM) These theories have grown out of hypotheses and experimental data in the following manner. 1. 2. 3. 4.

Experimental failure data is first collected through tensile tests. The state of stress is correlated to the experimental data using Mohr’s circle plots. A failure theory is developed from a concept of the responsible failure mechanism. A design envelope is established based upon the theoretical and empirical design equations.

In light of the extensive dependence of failure theories on experimental data, we will first review the acquisition and correlation of tensile test data to failure theory. Subsequently, the criteria and application of specific failure theories will be discussed. TENSILE TEST d0

l0 TEXT FIGURE 3-1: A typical tension-test specimen. Some of the standard dimensions used for d0 are 2.5, 6.25, and 12.5 mm and 0.505 in, but other sections and sizes are in use. Common gauge lengths l0 used are 10, 25, and 50 mm and 1 and 2 in.

†

Text refers to Mechanical Engineering Design, 7th edition text by Joseph Edward Shigley, Charles R. Mischke and Richard G. Budynas; equations and examples with the prefix T refer to the present tutorial.

The tensile test is a standardized test (ASTM Standard E8 or E8m) and thus allows for the sharing of experimental data amongst researchers, typically in the form of stress-strain curves. Standard dimensions for the test-specimen are provided in Text Figure 3-1 while a comparison of characteristic stress-strain curves for ductile and brittle materials are shown in Text Figure 3-2. These engineering stress-strain diagrams graphically demonstrate the difference in the failure behavior of ductile and brittle metals, and the need for separate failure criteria. However, the curves do not represent true values of stress and strain; rather, they are calculated based upon the original specimen cross-sectional area, prior to loading. Engineering Stress

σ=

P 4P = A0 π d 02

(Text Eq. 3-1)

Engineering Strain

ε=

l − l0 l0

(Text Eq. 3-2)

u

Su Sf Sy

f

Stress σ = P/A0 O

y

el pl

a

εy

εu

εf

Sut Sy

u, f y

a

Strain ε

Strain ε

(a)

(b)

TEXT FIGURE 3-2: Stress-strain diagram obtained from the standard tensile test (a) Ductile material; (b) brittle material. pl marks the proportional limit; el, the elastic limit; y, the offset yield strength as defined by offset strain Oa; u, the maximum or ultimate strength; and f, the fracture strength.

Referring to Text Fig. 3-2 (a), point el, the elastic limit, defines the onset of permanent set while point a represents 0.2 percent permanent set with respect to the original gauge length (ε = 0.002). A measure of the “true” stress and strain can be obtained by taking simultaneous measurements of the load and cross-sectional area during the tensile test experiment. Text


Machine Design Tutorial 6: Static Failure Theories

2/5

Figure 3-4 shows a typical true stress-strain diagram for a ductile material. The curve between points u and f corresponds to a reduction in stress as the specimen necks down.

σf

f

σu True stress

u

εf

εu True strain

TEXT FIGURE 3-4: True stressstrain diagram plotted in Cartesian coordinates.

TEXT FIGURE 3-3: Tension specimen after necking.

CORRELATION OF STATE OF STRESS WITH TEST DATA For design, we need to relate the expected state of stress in a part to the actual state of stress and thus, the material strength, as determined through the tensile test. We accomplish this by applying principal stresses since they characterize a state of stress independent of the original coordinate system. y

σy =0 τ yx = τ xy = 0

σ x = P / A0

σ x = P / A0

x

σ1

τ yx = τ xy = 0

σ1 σ2 =σ3 = 0

σy =0 (a) State of stress for simple tension.

(b) Principal stresses for simple tension.

FIGURE 6A-1: Correlation of state of stress with principal stresses for simple tension.



3/5

Since the tensile test generates a uniaxial state of stress, the principal stresses can be defined as,

σ 1 = σ axial =

P A0

and σ 2 = σ 3 = 0

When plotted on a Mohr’s circle diagram, these stress values translate into what looks like a single circle passing through the origin where σ 2 is coincident with σ 3 . Actually, there are still three circles on the Mohr’s circle diagram. Two circles, defined by principal stresses (σ 1,σ 2 ) and (σ 1,σ 3 ) , are drawn on top of each other. The third circle degenerates to a point defined by principal stresses (σ 2,σ 3 ) . τ

σ3 σ2

σ1

σ

FIGURE 6A-2: Mohr’s circle for simple tension.

DEVELOPMENT OF STATIC FAILURE THEORIES Design for static loading dictates that all loading variables remain constant: 1. Magnitude of load is constant; 2. Direction of load is constant; 3. Point of application of the load is fixed. These conditions, in conjunction with criteria specific to ductile and brittle materials, have been used in the development of the static failure theories outlined earlier. Characteristics of Ductile Materials 1. The strain at failure is, ε f ≥ 0.05 , or percent elongation greater than five percent. 2. Ductile materials typically have a well defined yield point. The value of the stress at the yield point defines the yield strength, Sy. 3. For typical ductile materials, the yield strength has approximately the same value for tensile and compressive loading ( S yt ≈ S yc = S y ). Shigley, Mischke & Budynas


4/5

4. A single tensile test is sufficient to characterize the material behavior of a ductile material, Sy and Sut. Characteristics of Brittle Materials 1. The strain at failure is, ε f ≤ 0.05 or percent elongation less than five percent. 2. Brittle materials do not exhibit an identifiable yield point; rather, they fail by brittle fracture. The value of the largest stress in tension and compression defines the ultimate strength, Sut and Suc respectively. 3. The compressive strength of a typical brittle material is significantly higher than its tensile strength, ( Suc Sut ). 4. Two material tests, a tensile test and a compressive test, are required to characterize the material behavior of a brittle material, Sut and Suc. SUMMARY This tutorial has attempted to provide a focused introduction to the development of static failure theory by summarizing the theories associated with specific material classifications. In addition, the experimental and analytical models, which have been employed historically to relate the experimental data to strength quantities used for static design, are presented. Subsequent tutorials, Static Failure of Ductile Materials and Static Failure of Brittle Materials, will respectively provide detailed reviews and examples, respectively, of the failure theories associated with ductile and brittle materials.



5/5

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Chapter 6 6-1 σ1 − σ3 = S y /n

MSS:

⇒

Sy σ1 − σ3

n=

Sy σ 1/2 2 1/2 σ = σ A2 − σ A σ B + σ B2 = σx − σx σ y + σ y2 + 3τx2y n=

DE:

(a) MSS:

σ1 = 12, σ2 = 6, σ3 = 0 kpsi 50 n= = 4.17 Ans. 12

σ = (122 − 6(12) + 62 ) 1/2 = 10.39 kpsi, 12 2 12 ± + (−8) 2 = 16, −4 kpsi (b) σ A , σ B = 2 2 DE:

n=

50 = 4.81 10.39

Ans.

σ1 = 16, σ2 = 0, σ3 = −4 kpsi MSS:

n=

50 = 2.5 16 − (−4)

Ans.

50 = 2.73 Ans. σ = (122 + 3(−82 )) 1/2 = 18.33 kpsi, n = 18.33 −6 + 10 2 −6 − 10 ± + (−5) 2 = −2.615, −13.385 kpsi (c) σ A , σ B = 2 2 DE:

σ1 = 0, σ2 = −2.615, σ3 = −13.385 kpsi MSS:

50 = 3.74 n= 0 − (−13.385)

␴B

Ans.

σ = [(−6) 2 − (−6)(−10) + (−10) 2 + 3(−5) 2 ]1/2 = 12.29 kpsi 50 n= = 4.07 Ans. 12.29 12 − 4 2 12 + 4 ± + 12 = 12.123, 3.877 kpsi (d) σ A , σ B = 2 2 DE:

σ1 = 12.123, σ2 = 3.877, σ3 = 0 kpsi MSS: DE:

n=

50 = 4.12 12.123 − 0

Ans.

σ = [122 − 12(4) + 42 + 3(12 )]1/2 = 10.72 kpsi n=

50 = 4.66 10.72

Ans.

␴A

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6-2 S y = 50 kpsi Sy σ1 − σ3

MSS:

σ1 − σ3 = S y /n

DE:

1/2 2 = S y /n σ A − σ A σ B + σ B2

n=

DE:

[122

n=

[122

1/2 n = S y / σ A2 − σ A σ B + σ B2

⇒

50 = 4.17 Ans. 12 − 0

50 = 4.17 − (12)(12) + 122 ]1/2

Ans.

50 = 4.81 − (12)(6) + 62 ]1/2

Ans.

σ1 = 12 kpsi, σ3 = −12 kpsi, n =

(c) MSS:

n=

DE:

[122

n=

50 = 2.08 Ans. 12 − (−12)

50 = 2.41 − (12)(−12) + (−12) 2 ]1/3

σ1 = 0, σ3 = −12 kpsi, n =

DE:

Ans.

50 = 4.17 12

σ1 = 12 kpsi, σ3 = 0, n =

DE:

(d) MSS:

n=

σ1 = 12 kpsi, σ3 = 0, n =

(a) MSS:

(b) MSS:

⇒

[(−6) 2

Ans.

50 = 4.17 Ans. −(−12)

50 = 4.81 − (−6)(−12) + (−12) 2 ]1/2

6-3 S y = 390 MPa Sy σ1 − σ3

MSS:

σ1 − σ3 = S y /n

DE:

1/2 2 = S y /n σ A − σ A σ B + σ B2

(a) MSS:

⇒

n=

σ1 = 180 MPa, σ3 = 0, n =

⇒

1/2 n = S y / σ A2 − σ A σ B + σ B2

390 = 2.17 Ans. 180

390 = 2.50 Ans. [1802 − 180(100) + 1002 ]1/2 180 180 2 ± + 1002 = 224.5, −44.5 MPa = σ1 , σ3 (b) σ A , σ B = 2 2 DE:

n=

MSS:

n=

DE:

n=

390 = 1.45 Ans. 224.5 − (−44.5) [1802

390 = 1.56 + 3(1002 )]1/2

Ans.

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Chapter 6

160 2 160 ± − + 1002 = 48.06, −208.06 MPa = σ1 , σ3 (c) σ A , σ B = − 2 2 390 = 1.52 Ans. 48.06 − (−208.06) 390 n= = 1.65 Ans. 2 [−160 + 3(1002 )]1/2 n=

MSS: DE:

(d) σ A , σ B = 150, −150 MPa = σ1 , σ3 380 = 1.27 Ans. n= MSS: 150 − (−150) 390 n= = 1.50 Ans. DE: [3(150) 2 ]1/2 6-4 S y = 220 MPa (a) σ1 = 100, σ2 = 80, σ3 = 0 MPa MSS: DET:

220 = 2.20 Ans. 100 − 0 σ = [1002 − 100(80) + 802 ]1/2 = 91.65 MPa 220 = 2.40 Ans. n= 91.65 n=

(b) σ1 = 100, σ2 = 10, σ3 = 0 MPa MSS: DET:

n=

220 = 2.20 Ans. 100

σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa n=

220 = 2.31 Ans. 95.39

(c) σ1 = 100, σ2 = 0, σ3 = −80 MPa MSS: DE:

n=

220 = 1.22 Ans. 100 − (−80)

σ = [1002 − 100(−80) + (−80) 2 ]1/2 = 156.2 MPa 220 = 1.41 Ans. n= 156.2

(d) σ1 = 0, σ2 = −80, σ3 = −100 MPa 220 n= = 2.20 Ans. MSS: 0 − (−100) σ = [(−80) 2 − (−80)(−100) + (−100) 2 ] = 91.65 MPa DE: n=

220 = 2.40 Ans. 91.65

151

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6-5 (a) MSS: DE: (b) MSS: DE:

n=

2.23 OB = = 2.1 OA 1.08

n=

OC 2.56 = = 2.4 OA 1.08

n=

OE 1.65 = = 1.5 OD 1.10

n=

OF 1.8 = = 1.6 OD 1.1 ␴B

(a) C B

A Scale 1" ⫽ 200 MPa ␴A

O D E F

J K

G

L (d)

H I (c)

(c) MSS: DE: (d) MSS: DE:

n=

OH 1.68 = = 1.6 OG 1.05

n=

OI 1.85 = = 1.8 OG 1.05

n=

OK 1.38 = = 1.3 OJ 1.05

n=

OL 1.62 = = 1.5 OJ 1.05

(b)

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153

Chapter 6

6-6 S y = 220 MPa OB 2.82 = = 2.2 OA 1.3 OC 3.1 n= = = 2.4 OA 1.3 OE 2.2 n= = = 2.2 OD 1

n=

(a) MSS: DE: (b) MSS:

n=

DE:

OF 2.33 = = 2.3 OD 1 ␴B

(a) C B

A

1" ⫽ 100 MPa

E D

F

O

␴A

G J

H I (c)

K

L (d)

(c) MSS: DE: (d) MSS: DE:

n=

OH 1.55 = = 1.2 OG 1.3

n=

OI 1.8 = = 1.4 OG 1.3

OK 2.82 = = 2.2 OJ 1.3 OL 3.1 n= = = 2.4 OJ 1.3

n=

(b)

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6-7 Sut = 30 kpsi, Suc = 100 kpsi; σ A = 20 kpsi, σ B = 6 kpsi n=

(a) MNS: Eq. (6-30a)

Sut 30 = 1.5 Ans. = σx 20

30 = 1.5 Ans. 20 30 n= = 1.5 Ans. 20 30 n= = 1.5 Ans. 20 n=

BCM: Eq. (6-31a) M1M: Eq. (6-32a) M2M: Eq. (6-33a) (b) σx = 12 kpsi,τx y = −8 kpsi 12 ± σ A, σB = 2

12 2

2 + (−8) 2 = 16, −4 kpsi

30 = 1.88 Ans. 16 16 (−4) 1 = − ⇒ n 30 100 30 n= = 1.88 Ans. 16 30 n= = 1.88 Ans. 16 n=

MNS: Eq. (6-30a) BCM: Eq. (6-31b) M1M: Eq. (6-32a) M2M: Eq. (6-33a)

n = 1.74 Ans.

(c) σx = −6 kpsi, σ y = −10 kpsi,τx y = −5 kpsi −6 − 10 −6 + 10 2 ± + (−5) 2 = −2.61, −13.39 kpsi σ A, σB = 2 2 MNS: Eq. (6-30b)

n=−

100 = 7.47 Ans. −13.39

BCM: Eq. (6-31c)

n=−

100 = 7.47 Ans. −13.39

M1M: Eq. (6-32c)

n=−

100 = 7.47 Ans. −13.39

M2M: Eq. (6-33c)

n=−

100 = 7.47 Ans. −13.39

(d) σx = −12 kpsi,τx y = 8 kpsi 12 σ A, σB = − ± 2 MNS: Eq. (6-30b)

n=

12 2 − + 82 = 4, −16 kpsi 2

−100 = 6.25 Ans. −16

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155

Chapter 6

4 (−16) 1 = − n 30 100

BCM: Eq. (6-31b)

n = 3.41 Ans.

(100 − 30)4 −16 1 = − ⇒ n = 3.95 Ans. n 100(30) 100 4 n(−16) + 30 2 n + =1 30 30 − 100

M1M: Eq. (6-32b) M2M: Eq. (6-33b) Reduces to

⇒

n 2 − 1.1979n − 15.625 = 0 1.1979 + 1.19792 + 4(15.625) = 4.60 n= 2

Ans.

␴B

B

1" ⫽ 20 kpsi

(a)

A O

␴A C

E D

K F

G

H

I

L (c)

J (d)

(b)

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6-8 See Prob. 6-7 for plot. (a) For all methods:

n=

1.55 OB = = 1.5 OA 1.03

(b) BCM:

n=

OD 1.4 = = 1.75 OC 0.8

n=

OE 1.55 = = 1.9 OC 0.8

(c) For all methods:

n=

OL 5.2 = = 7.6 OK 0.68

(d) MNS:

n=

OJ 5.12 = = 6.2 OF 0.82

BCM:

n=

OG 2.85 = = 3.5 OF 0.82

M1M:

n=

OH 3.3 = = 4.0 OF 0.82

M2M:

n=

3.82 OI = = 4.7 OF 0.82

All other methods:

6-9 Given: S y = 42 kpsi, Sut = 66.2 kpsi, ε f = 0.90. Since ε f > 0.05, the material is ductile and thus we may follow convention by setting S yc = S yt . Use DE theory for analytical solution. For σ , use Eq. (6-13) or (6-15) for plane stress and Eq. (6-12) or (6-14) for general 3-D. (a) σ = [92 − 9(−5) + (−5) 2 ]1/2 = 12.29 kpsi n=

42 = 3.42 Ans. 12.29

(b) σ = [122 + 3(32 )]1/2 = 13.08 kpsi n=

42 = 3.21 Ans. 13.08

(c) σ = [(−4) 2 − (−4)(−9) + (−9) 2 + 3(52 )]1/2 = 11.66 kpsi n=

42 = 3.60 Ans. 11.66

(d) σ = [112 − (11)(4) + 42 + 3(12 )]1/2 = 9.798 n=

42 = 4.29 Ans. 9.798

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157

Chapter 6 ␴B

(d) H

1 cm ⫽ 10 kpsi G O C

D

(b)

␴A

A E

B (a)

F (c)

For graphical solution, plot load lines on DE envelope as shown. σ A = 9, σ B = −5 kpsi OB 3.5 n= = = 3.5 Ans. OA 1 12 2 12 ± + 32 = 12.7, −0.708 kpsi (b) σ A , σ B = 2 2

(a)

4.2 OD = = 3.23 OC 1.3 4−9 2 −4 − 9 ± + 52 = −0.910, −12.09 kpsi (c) σ A , σ B = 2 2 n=

4.5 OF = = 3.6 Ans. OE 1.25 11 − 4 2 11 + 4 ± + 12 = 11.14, 3.86 kpsi (d) σ A , σ B = 2 2 n=

n= 6-10

5.0 OH = = 4.35 Ans. OG 1.15

This heat-treated steel exhibits S yt = 235 kpsi, S yc = 275 kpsi and ε f = 0.06. The steel is ductile (ε f > 0.05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis (DCM) of Fig. 6-27 applies — confine its use to first and fourth quadrants.

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(a) σx = 90 kpsi, σ y = −50 kpsi, σz = 0 σ A = 90 kpsi and σ B = −50 kpsi. For the fourth quadrant, from Eq. (6-13) 1 1 = = 1.77 Ans. n= (σ A /S yt ) − (σ B /Suc ) (90/235) − (−50/275) (b) σx = 120 kpsi, τx y = −30 kpsi ccw. σ A , σ B = 127.1, −7.08 kpsi. For the fourth quadrant 1 = 1.76 Ans. n= (127.1/235) − (−7.08/275) (c) σx = −40 kpsi, σ y = −90 kpsi, τx y = 50 kpsi . σ A , σ B = −9.10, −120.9 kpsi. Although no solution exists for the third quadrant, use S yc 275 = 2.27 Ans. n=− =− σy −120.9 (d) σx = 110 kpsi, σ y = 40 kpsi, τx y = 10 kpsi cw. σ A , σ B = 111.4, 38.6 kpsi. For the first quadrant S yt 235 = 2.11 Ans. = n= σA 111.4 Graphical Solution: ␴B

OB 1.82 = = 1.78 (a) n = OA 1.02 (b) n =

OD 2.24 = = 1.75 OC 1.28

(c) n =

OF 2.75 = = 2.22 OE 1.24

(d) n =

OH 2.46 = = 2.08 OG 1.18

(d)

H 1 in ⫽ 100 kpsi

G

␴A

O C

D

A

B (a) E

F (c)

(b)

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159

Chapter 6

6-11

The material is brittle and exhibits unequal tensile and compressive strengths. Decision: Use the Modified II-Mohr theory as shown in Fig. 6-28 which is limited to first and fourth quadrants. Sut = 22 kpsi, Suc = 83 kpsi Parabolic failure segment:

S B + 22 2 S A = 22 1 − 22 − 83 SB

SA

SB

SA

−22 −30 −40 −50

22.0 21.6 20.1 17.4

−60 −70 −80 −83

13.5 8.4 2.3 0

(a) σx = 9 kpsi, σ y = −5 kpsi. σ A , σ B = 9, −5 kpsi. For the fourth quadrant, use Eq. (6-33a) Sut 22 = 2.44 Ans. n= = σA 9 (b) σx = 12 kpsi, τx y = −3 kpsi ccw. σ A , σ B = 12.7, 0.708 kpsi. For the first quadrant, n=

Sut 22 = 1.73 Ans. = σA 12.7

(c) σx = −4 kpsi, σ y = −9 kpsi, τx y = 5 kpsi . σ A , σ B = −0.910, −12.09 kpsi. For the third quadrant, no solution exists; however, use Eq. (6-33c) −83 = 6.87 Ans. n= −12.09 (d) σx = 11 kpsi,σ y = 4 kpsi,τx y = 1 kpsi. σ A , σ B = 11.14, 3.86 kpsi. For the first quadrant n=

S yt SA 22 = 1.97 Ans. = = σA σA 11.14 ␴B 30 Sut ⫽ 22

30

–50

Sut ⫽ 83 –90

␴A

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. Since ε f < 0.05, the material is brittle. Thus, Sut = Suc and we may use M2M which is basically the same as MNS. (a) σ A , σ B = 9, −5 kpsi 35 = 3.89 Ans. 9 (b) σ A , σ B = 12.7, −0.708 kpsi n=

35 = 2.76 Ans. 12.7 (c) σ A , σ B = −0.910, −12.09 kpsi (3rd quadrant) n=

n=

36 = 2.98 Ans. 12.09

␴B

(d) σ A , σ B = 11.14, 3.86 kpsi n=

35 = 3.14 Ans. 11.14

1 cm ⫽ 10 kpsi

Graphical Solution: O

OD 3.45 = = 2.70 Ans. OC 1.28

(c) n =

OF 3.7 = = 2.85 Ans. (3rd quadrant) OE 1.3

(d) n =

OH 3.6 = = 3.13 Ans. OG 1.15

D

(b)

B

(a)

E

F (c)

Sut = 30 kpsi, Suc = 109 kpsi Use M2M: (a) σ A , σ B = 20, 20 kpsi 30 n= = 1.5 Ans. Eq. (6-33a): 20 (b) σ A , σ B = ± (15) 2 = 15, −15 kpsi n=

30 = 2 Ans. 15

(c) σ A , σ B = −80, −80 kpsi For the 3rd quadrant, there is no solution but use Eq. (6-33c). Eq. (6-33c):

C A

(b) n =

Eq. (6-33a)

(d)

G

OB 4 = = 4.0 Ans. (a) n = OA 1

6-13

H

n=−

109 = 1.36 Ans. −80

␴A

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161

Chapter 6

(d) σ A , σ B = 15, −25 kpsi n(15) −25n + 30 2 + =1 30 30 − 109

Eq. (6-33b):

n = 1.90 Ans. OB 4.25 = = 1.50 OA 2.83 OD 4.24 = = 2.00 (b) n = OC 2.12 (a) n =

(c) n =

OF 15.5 = = 1.37 (3rd quadrant) OE 11.3

(d) n =

5.3 OH = = 1.83 OG 2.9

␴B

B (a)

A

␴A

O

C 1 cm ⫽ 10 kpsi G D (b)

H (d)

E

F (c)

6-14

Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the DE theory to stress elements A and B with S y = 280 MPa A:

σx =

4(8)(103 ) 32Fl 4P 32(0.55)(103 )(0.1) + + = πd 3 πd 2 π(0.0203 ) π(0.0202 )

= 95.49(106 ) Pa = 95.49 MPa

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16T 16(30) = 19.10(106 ) Pa = 19.10 MPa = 3 πd π(0.0203 ) 1/2 σ = σx2 + 3τx2y = [95.492 + 3(19.1) 2 ]1/2 = 101.1 MPa

τx y =

n= B:

Sy 280 = 2.77 = σ 101.1

Ans.

4P 4(8)(103 ) = 25.47(106 ) Pa = 25.47 MPa = σx = 3 2 πd π(0.020 ) 16(30) 4 0.55(103 ) 16T 4V = + + τx y = πd 3 3 A π(0.0203 ) 3 (π/4)(0.0202 ) = 21.43(106 ) Pa = 21.43 MPa σ = [25.472 + 3(21.432 )]1/2 = 45.02 MPa n=

6-15

280 = 6.22 45.02

Ans.

Design decisions required: • • • •

Material and condition Design factor Failure model Diameter of pin

Using F = 416 lbf from Ex. 6-3 32M πd 3 32M 1/3 d= πσmax

σmax =

Decision 1: Select the same material and condition of Ex. 6-3 (AISI 1035 steel, S y = 81 000). Decision 2: Since we prefer the pin to yield, set n d a little larger than 1. Further explanation will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set n d = 1 Sy Sy = 81 000 psi = nd 1 32(416)(15) 1/3 d= = 0.922 in π(81 000)

σmax =

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163

Chapter 6

Choose preferred size of d = 1.000 in F=

π(1) 3 (81 000) = 530 lbf 32(15)

n=

530 = 1.274 416

Set design factor to n d = 1.274 Adequacy Assessment: Sy 81 000 = 63 580 psi = nd 1.274 32(416)(15) 1/3 d= = 1.000 in (OK ) π(63 580)

σmax =

6-16

F=

π(1) 3 (81 000) = 530 lbf 32(15)

n=

530 = 1.274 (OK) 416

For a thin walled cylinder made of AISI 1018 steel, S y = 54 kpsi, Sut = 64 kpsi. The state of stress is p(8) pd pd = = 40 p, σl = = 20 p, 4t 4(0.05) 8t These three are all principal stresses. Therefore, σt =

σr = − p

1 σ = √ [(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 ]1/2 2 1 = √ [(40 p − 20 p) 2 + (20 p + p) 2 + (− p − 40 p) 2 ] 2 = 35.51 p = 54 ⇒ p = 1.52 kpsi (for yield) Ans. . . For rupture, 35.51 p = 64 ⇒ p = 1.80 kpsi Ans. 6-17

For hot-forged AISI steel w = 0.282 lbf/in3 , S y = 30 kpsi and ν = 0.292. Then ρ = w/g = 0.282/386 lbf · s2 /in; ri = 3 in; ro = 5 in; ri2 = 9; ro2 = 25; 3 + ν = 3.292; 1 + 3ν = 1.876. Eq. (4-56) for r = ri becomes σt = ρω

2

3+ν 8

1 + 3ν 2 2 2ro + ri 1 − 3+ν

Rearranging and substituting the above values: Sy 1.876 0.282 3.292 50 + 9 1 − = ω2 386 8 3.292 = 0.016 19

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Setting the tangential stress equal to the yield stress, 30 000 1/2 = 1361 rad/s ω= 0.016 19 or

n = 60ω/2π = 60(1361)/(2π) = 13 000 rev/min

Now check the stresses at r = (rori ) 1/2 , or r = [5(3)]1/2 = 3.873 in 2 3+ν (ro − ri ) 2 σr = ρω 8 0.282ω2 3.292 (5 − 3) 2 = 386 8 = 0.001 203ω2 Applying Eq. (4-56) for σt 9(25) 1.876(15) 3.292 2 0.282 9 + 25 + − σt = ω 386 8 15 3.292 = 0.012 16ω2 Using the Distortion-Energy theory 1/2 = 0.011 61ω2 σ = σt2 − σr σt + σr2 Solving

ω=

30 000 0.011 61

1/2

= 1607 rad/s

So the inner radius governs and n = 13 000 rev/min 6-18

Ans.

For a thin-walled pressure vessel, di = 3.5 − 2(0.065) = 3.37 in σt =

p(di + t) 2t

σt =

500(3.37 + 0.065) = 13 212 psi 2(0.065)

σl =

500(3.37) pdi = = 6481 psi 4t 4(0.065)

σr = − pi = −500 psi

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165

Chapter 6

These are all principal stresses, thus, 1 σ = √ {(13 212 − 6481) 2 + [6481 − (−500)]2 + (−500 − 13 212) 2 }1/2 2 σ = 11 876 psi n=

Sy 46 000 46 000 = = σ σ 11 876

= 3.87 Ans. 6-19

Table A-20 gives S y as 320 MPa. The maximum significant stress condition occurs at ri where σ1 = σr = 0, σ2 = 0, and σ3 = σt . From Eq. (4-50) for r = ri σt = −

2ro2 po 2(1502 ) po = − = −3.6 po 1502 − 1002 ro2 − ri2

σ = 3.6 po = S y = 320 po = 6-20

320 = 88.9 MPa Ans. 3.6

Sut = 30 kpsi, w = 0.260 lbf/in3 , ν = 0.211, 3 + ν = 3.211, 1 + 3ν = 1.633. At the inner radius, from Prob. 6-17 1 + 3ν 2 σt 3+ν 2 2 2ro + ri − r =ρ ω2 8 3+ν i Here ro2 = 25, ri2 = 9, and so σt 0.260 = ω2 386

3.211 8

1.633(9) 50 + 9 − = 0.0147 3.211

Since σr is of the same sign, we use M2M failure criteria in the first quadrant. From Table A-24, Sut = 31 kpsi, thus, 31 000 1/2 ω= = 1452 rad/s 0.0147 rpm = 60ω/(2π) = 60(1452)/(2π) = 13 866 rev/min Using the grade number of 30 for Sut = 30 000 kpsi gives a bursting speed of 13640 rev/min. 6-21

TC = (360 − 27)(3) = 1000 lbf · in , TB = (300 − 50)(4) = 1000 lbf · in y 223 lbf A

127 lbf B

8"

C 8" 350 lbf xy plane

6"

D

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In x y plane, M B = 223(8) = 1784 lbf · in and MC = 127(6) = 762 lbf · in. 387 lbf 8"

A

8" B

6" C

106 lbf

D 281 lbf

xz plane

In the x z plane, M B = 848 lbf · in and MC = 1686 lbf · in. The resultants are M B = [(1784) 2 + (848) 2 ]1/2 = 1975 lbf · in MC = [(1686) 2 + (762) 2 ]1/2 = 1850 lbf · in So point B governs and the stresses are 16T 16(1000) 5093 = = 3 psi 3 3 πd πd d 32M B 32(1975) 20 120 = = psi σx = πd 3 πd 3 d3 2 1/2 σx σx ± + τx2y σ A, σB = 2 2  1/2  2   20.12 1 20.12 ± σ A, σB = 3 + (5.09) 2  d  2 2 τx y =

Then

= Then

(10.06 ± 11.27) kpsi · in3 d3

σA =

10.06 + 11.27 21.33 = kpsi d3 d3

σB =

10.06 − 11.27 1.21 = − 3 kpsi 3 d d

and

For this state of stress, use the Brittle-Coulomb-Mohr theory for illustration. Here we use Sut (min) = 25 kpsi, Suc (min) = 97 kpsi, and Eq. (6-31b) to arrive at 1 21.33 −1.21 − = 25d 3 97d 3 2.8 Solving gives d = 1.34 in. So use d = 1 3/8 in

Ans.

Note that this has been solved as a statics problem. Fatigue will be considered in the next chapter. 6-22

As in Prob. 6-21, we will assume this to be statics problem. Since the proportions are unchanged, the bearing reactions will be the same as in Prob. 6-21. Thus x y plane: x z plane:

M B = 223(4) = 892 lbf · in M B = 106(4) = 424 lbf · in

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167

Chapter 6

So Mmax = [(892) 2 + (424) 2 ]1/2 = 988 lbf · in σx =

32M B 32(988) 10 060 = = psi πd 3 πd 3 d3

Since the torsional stress is unchanged, τx z = 5.09/d 3 kpsi 

1/2  2   10.06 10.06 1 σ A, σB = 3 ± + (5.09) 2  d  2 2 σ A = 12.19/d 3

and σ B = −2.13/d 3

Using the Brittle-Coulomb-Mohr, as was used in Prob. 6-21, gives 1 12.19 −2.13 − = 3 3 25d 97d 2.8 Solving gives d = 1 1/8 in. Now compare to Modified II-Mohr theory 6-23

Ans.

( FA ) t = 300 cos 20 = 281.9 lbf, ( FA )r = 300 sin 20 = 102.6 lbf 3383 = 676.6 lbf 5

T = 281.9(12) = 3383 lbf · in, ( FC ) t = ( FC )r = 676.6 tan 20 = 246.3 lbf y ROy = 193.7 lbf O

ROz = 233.5 lbf

246.3 lbf

A

C

B

20"

16" 281.9 lbf

10"

x

RBy = 158.1 lbf

xy plane

O

676.6 lbf

A

B

20"

16"

z 102.6 lbf

C 10" RBz = 807.5 lbf

xz plane

M A = 20 193.72 + 233.52 = 6068 lbf · in M B = 10 246.32 + 676.62 = 7200 lbf · in (maximum) σx =

32(7200) 73 340 = 3 πd d3

16(3383) 17 230 = 3 πd d3 1/2 Sy σ = σx2 + 3τx2y = n 2 1/2 79 180 60 000 73 340 17 230 2 +3 = = 3 3 3 d d d 3.5

τx y =

d = 1.665 in

so use a standard diameter size of 1.75 in Ans.

x

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6-24

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From Prob. 6-23, τmax =

73 340 2d 3

2

σx 2

1/2

2 + τx y

=

2

17 230 + d3

2 1/2 =

Sy 2n

40 516 60 000 = d3 2(3.5)

d = 1.678 in so use 1.75 in Ans. 6-25

T = (270 − 50)(0.150) = 33 N · m , S y = 370 MPa (T1 − 0.15T1 )(0.125) = 33

⇒

T1 = 310.6 N,

T2 = 0.15(310.6) = 46.6 N

(T1 + T2 ) cos 45 = 252.6 N 107.0 N

y 163.4 N O

300

252.6 N A

89.2 N

400

150

300

400

z

C

B

252.6 N 150

320 N

174.4 N

xz plane

xy plane

M A = 0.3 163.42 + 1072 = 58.59 N · m M B = 0.15 89.22 + 174.42 = 29.38 N · m

(maximum)

32(58.59) 596.8 = 3 πd d3 16(33) 168.1 τx y = = 3 πd d3 2 2 1/2 596.8 664.0 370(106 ) 168.1 2 2 1/2 σ = σx + 3τx y = +3 = = d3 d3 d3 3.0 σx =

d = 17.5(10−3 ) m = 17.5 mm, 6-26

so use 18 mm Ans.

From Prob. 6-25,

τmax =

596.8 2d 3

2

d = 17.7(10−3 ) m = 17.7 mm,

σx 2

168.1 + d3

1/2

2 + τx2y 2 1/2

=

=

Sy 2n

342.5 370(106 ) = d3 2(3.0)

so use 18 mm Ans.

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169

Chapter 6

6-27

For the loading scheme shown in Figure (c), 4.4 F a b + = (6 + 4.5) Mmax = 2 2 4 2

V

M

= 23.1 N · m y

For a stress element at A:

B C

32M 32(23.1)(103 ) σx = = = 136.2 MPa πd 3 π(12) 3

x

A

The shear at C is 4(4.4/2)(103 ) 4( F/2) = = 25.94 MPa 3πd 2 /4 3π(12) 2 /4 1/2 136.2 2 = = 68.1 MPa 2

τx y = τmax

Since S y = 220 MPa, Ssy = 220/2 = 110 MPa, and n=

Ssy 110 = 1.62 Ans. = τmax 68.1

For the loading scheme depicted in Figure (d) 2 F 1 F a+b F a b b Mmax = − + = 2 2 2 2 2 2 2 4 This result is the same as that obtained for Figure (c). At point B, we also have a surface compression of −F −4.4(103 ) −F = − = −20.4 MPa σy = A bd 18(12) With σx = −136.2 MPa. From a Mohrs circle diagram, τmax = 136.2/2 = 68.1 MPa. n= 6-28

110 = 1.62 MPa 68.1

Ans.

Based on Figure (c) and using Eq. (6-15) 1/2 σ = σx2 = (136.22 ) 1/2 = 136.2 MPa n=

Sy 220 = 1.62 Ans. = σ 136.2

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Based on Figure (d) and using Eq. (6-15) and the solution of Prob. 6-27, 1/2 σ = σx2 − σx σ y + σ y2 = [(−136.2) 2 − (−136.2)(−20.4) + (−20.4) 2 ]1/2 = 127.2 MPa n=

Sy 220 = 1.73 Ans. = σ 127.2

6-29 w dF

r

When the ring is set, the hoop tension in the ring is equal to the screw tension. ri2 pi ro2 1+ 2 σt = 2 r ro − ri2

We have the hoop tension at any radius. The differential hoop tension d F is d F = wσt dr ro ro wri2 pi ro2 F= 1 + 2 dr = wri pi wσt dr = 2 r ro − ri2 ri ri

(1)

The screw equation is Fi =

T 0.2d

(2)

From Eqs. (1) and (2) F T = wri 0.2dwri d Fx = f pi ri dθ 2π f pi wri dθ = Fx =

pi ri d␪

pi =

dFx

o

=

2π f T 0.2d

f Tw ri 0.2dwri

2π

dθ o

Ans.

6-30 (a) From Prob. 6-29,

(b) From Prob. 6-29,

T = 0.2Fi d 190 T Fi = = = 3800 lbf 0.2d 0.2(0.25)

Ans.

F = wri pi

pi =

F Fi 3800 = 15 200 psi = = wri wri 0.5(0.5)

Ans.

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171

Chapter 6

pi ri2 + ro2 ri2 pi ro2 1+ = σt = 2 r r=ri ro − ri2 ro2 − ri2

(c)

15 200(0.52 + 12 ) = = 25 333 psi Ans. 12 − 0.52 σr = − pi = −15 200 psi σt − σr σ1 − σ3 τmax = = 2 2 25 333 − (−15 200) = 20 267 psi Ans. = 2 1/2 σ = σ A2 + σ B2 − σ A σ B

(d)

= [25 3332 + (−15 200) 2 − 25 333(−15 200)]1/2 = 35 466 psi Ans. (e) Maximum Shear hypothesis n=

0.5S y Ssy 0.5(63) = 1.55 Ans. = = τmax τmax 20.267

Distortion Energy theory n=

Sy 63 = 1.78 Ans. = σ 35 466

6-31 1"R

re

1" R 2

␴t r

The moment about the center caused by force F is Fre where re is the effective radius. This is balanced by the moment about the center caused by the tangential (hoop) stress. ro rσt w dr Fre = ri

ro2 r+ dr r ri ro2 − ri2 wpi ri2 ro 2 re = 2 + ro ln 2 ri F ro − ri2 wpi r 2 = 2 i2 r o − ri

ro

From Prob. 6-29, F = wri pi . Therefore, ro2 − ri2 ri r o + ro2 ln re = 2 2 ri ro − ri2 For the conditions of Prob. 6-29, ri = 0.5 and ro = 1 in 2 1 − 0.52 0.5 1 2 + 1 ln = 0.712 in re = 2 1 − 0.52 2 0.5

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δnom = 0.0005 in (a) From Eq. (4-60) 30(106 )(0.0005) (1.52 − 12 )(12 − 0.52 ) = 3516 psi Ans. p= 1 2(12 )(1.52 − 0.52 ) Inner member: Eq. (4-57)

Eq. (6-13)

2 R 2 + ri2 1 + 0.52 = −5860 psi (σt ) i = − p 2 = −3516 2 1 − 0.52 R − ri2

(σr ) i = − p = −3516 psi 1/2 σi = σ A2 − σ A σ B + σ B2 = [(−5860) 2 − (−5860)(−3516) + (−3516) 2 ]1/2 = 5110 psi Ans.

Outer member: Eq. (4-58)

1.52 + 12 (σt ) o = 3516 1.52 − 12

= 9142 psi

(σr ) o = − p = −3516 psi Eq. (6-13)

σo = [91422 − 9142(−3516) + (−3516) 2 ]1/2 = 11 320 psi Ans.

(b) For a solid inner tube,

30(106 )(0.0005) (1.52 − 12 )(12 ) = 4167 psi Ans. p= 1 2(12 )(1.52 )

(σt ) i = − p = −4167 psi,

(σr ) i = −4167 psi

σi = [(−4167) 2 − (−4167)(−4167) + (−4167) 2 ]1/2 = 4167 psi Ans. 2 1.5 + 12 = 10 830 psi, (σr ) o = −4167 psi (σt ) o = 4167 1.52 − 12 σo = [10 8302 − 10 830(−4167) + (−4167) 2 ]1/2 = 13 410 psi Ans. 6-33

Using Eq. (4-60) with diametral values, 207(103 )(0.02) (752 − 502 )(502 − 252 ) = 19.41 MPa Ans. p= 50 2(502 )(752 − 252 ) 2 50 + 252 (σt ) i = −19.41 = −32.35 MPa Eq. (4-57) 502 − 252 (σr ) i = −19.41 MPa Eq. (6-13)

σi = [(−32.35) 2 − (−32.35)(−19.41) + (−19.41) 2 ]1/2 = 28.20 MPa Ans.

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Chapter 6

Eq. (4-58)

752 + 502 (σt ) o = 19.41 752 − 502

173

= 50.47 MPa,

(σr ) o = −19.41 MPa σo = [50.472 − 50.47(−19.41) + (−19.41) 2 ]1/2 = 62.48 MPa Ans. 6-34 δ=

1.9998 1.999 − = 0.0004 in 2 2

Eq. (4-59)

2 2 p(1) p(1) 2 + 12 1 +0 0.0004 = − 0.292 + 0.211 + 14.5(106 ) 22 − 12 30(106 ) 12 − 0 p = 2613 psi

Applying Eq. (4-58) at R,

Eq. (6-33b)

22 + 12 = 4355 psi (σt ) o = 2613 2 2 − 12 (σr ) o = −2613 psi, Sut = 20 kpsi, Suc = 83 kpsi n(4355) −2613n + 20 000 2 + =1 20 000 20 000 − 83 000

n = 4.52 Ans. 6-35

E = 30(106 ) psi, ν = 0.292, I = (π/64)(24 − 1.54 ) = 0.5369 in4 Eq. (4-60) can be written in terms of diameters,

2 Eδd do2 − D 2 D 2 − di2 30(106 ) (2 − 1.752 )(1.752 − 1.52 ) (0.002 46) p= = D 1.75 2(1.752 )(22 − 1.52 ) 2D 2 do2 − di2 = 2997 psi = 2.997 kpsi Outer member: Outer radius: Inner radius:

1.752 (2.997) (2) = 19.58 kpsi, (σr ) o = 0 22 − 1.752 22 1.752 (2.997) 1+ = 22.58 kpsi, (σr ) i = −2.997 kpsi (σt ) i = 2 2 − 1.752 1.752

(σt ) o =

Bending: ro :

(σx ) o =

6.000(2/2) = 11.18 kpsi 0.5369

ri :

(σx ) i =

6.000(1.75/2) = 9.78 kpsi 0.5369

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J = 2I = 1.0738 in4

Torsion: ro :

(τx y ) o =

8.000(2/2) = 7.45 kpsi 1.0738

ri :

(τx y ) i =

8.000(1.75/2) = 6.52 kpsi 1.0738

Outer radius is plane stress σx = 11.18 kpsi,

σ y = 19.58 kpsi,

τx y = 7.45 kpsi

σ = [11.182 − (11.18)(19.58) + 19.582 + 3(7.452 )]1/2 =

Eq. (6-15)

21.35 =

60 no

⇒

Inner radius, 3D state of stress

Sy 60 = no no

n o = 2.81 Ans. z –2.997 kpsi 9.78 kpsi

22.58 kpsi x

6.52 kpsi

y

From Eq. (6-14) with τ yz = τzx = 0 1 60 σ = √ [(9.78 − 22.58) 2 + (22.58 + 2.997) 2 + (−2.997 − 9.78) 2 + 6(6.52) 2 ]1/2 = ni 2 60 24.86 = ⇒ n i = 2.41 Ans. ni 6-36

From Prob. 6-35: p = 2.997 kpsi, I = 0.5369 in4 , J = 1.0738 in4 Inner member: Outer radius:

(0.8752 + 0.752 ) = −19.60 kpsi (σt ) o = −2.997 (0.8752 − 0.752 )

(σr ) o = −2.997 kpsi Inner radius:

2(2.997)(0.8752 ) = −22.59 kpsi (σt ) i = − 0.8752 − 0.752 (σr ) i = 0

Bending: ro :

(σx ) o =

6(0.875) = 9.78 kpsi 0.5369

ri :

(σx ) i =

6(0.75) = 8.38 kpsi 0.5369

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175

Chapter 6

Torsion: ro :

(τx y ) o =

8(0.875) = 6.52 kpsi 1.0738

ri :

(τx y ) i =

8(0.75) = 5.59 kpsi 1.0738

The inner radius is in plane stress: σx = 8.38 kpsi, σ y = −22.59 kpsi, τx y = 5.59 kpsi σi = [8.382 − (8.38)(−22.59) + (−22.59) 2 + 3(5.592 )]1/2 = 29.4 kpsi ni =

Sy 60 = 2.04 Ans. = σi 29.4

Outer radius experiences a radial stress, σr 1/2 1 σo = √ (−19.60 + 2.997) 2 + (−2.997 − 9.78) 2 + (9.78 + 19.60) 2 + 6(6.52) 2 2 = 27.9 kpsi 60 = 2.15 Ans. no = 27.9 6-37

KI KI θ θ θ 3θ 2 1 2√ cos ± sin cos sin σp = √ 2 2 2 2 2 2πr 2πr 1/2 θ 3θ 2 θ KI sin cos cos + √ 2 2 2 2πr 1/2

KI θ 2 θ 2 θ 2 3θ 2 θ 2 θ 2 3θ + sin cos cos =√ cos ± sin cos sin 2 2 2 2 2 2 2 2πr θ θ θ KI θ θ KI cos ± cos sin =√ cos 1 ± sin =√ 2 2 2 2 2 2πr 2πr Plane stress: The third principal stress is zero and θ θ θ θ KI KI cos 1 + sin , σ2 = √ cos 1 − sin , σ1 = √ 2 2 2 2 2πr 2πr

σ3 = 0 Ans.

Plane strain: σ1 and σ2 equations still valid however, θ KI cos σ3 = ν(σx + σ y ) = 2ν √ 2 2πr 6-38

Ans.

For θ = 0 and plane strain, the principal stress equations of Prob. 6-37 give KI , σ1 = σ2 = √ 2πr

KI σ3 = 2ν √ = 2νσ1 2πr

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1 √ [(σ1 − σ1 ) 2 + (σ1 − 2νσ1 ) 2 + (2νσ1 − σ1 ) 2 ]1/2 = S y 2 σ1 − 2νσ1 = S y 1 1 1−2 σ1 = S y ⇒ σ1 = 3S y Ans. For ν = , 3 3

(a) DE:

σ1 − σ3 = S y

(b) MSS:

ν=

⇒

σ1 − 2νσ1 = S y

σ1 = 3S y

Ans.

2 σ3 = σ1 3

␶

2␴ 3 1

6-39

1 3

⇒

␴1, ␴2

␴

Radius of largest circle σ1 1 2 σ1 − σ1 = R= 2 3 6

(a) Ignoring stress concentration F = S y A = 160(4)(0.5) = 320 kips Ans. (b) From Fig. 6-36: h/b = 1, a/b = 0.625/4 = 0.1563, β = 1.3 70 = 1.3

Eq. (6-51)

F π(0.625) 4(0.5)

F = 76.9 kips Ans. 6-40

Given: a = 12.5 mm, K I c = 80 MPa · ro = a/(ro − ri ) = ri /ro = Fig. 6-40: Eq. (6-51):

√ m, S y = 1200 MPa, Sut = 1350 MPa

350 = 175 mm, 2 12.5 = 0.5 175 − 150 150 = 0.857 175

. β = 2.5

√ K I c = βσ πa 80 = 2.5σ π(0.0125) σ = 161.5 MPa

ri =

350 − 50 = 150 mm 2

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177

Chapter 6

Eq. (4-51) at r = ro : ri2 pi σ = 2 (2) ro − ri2 161.5 =

1502 pi (2) 1752 − 1502

pi = 29.2 MPa Ans. 6-41 (a) First convert the data to radial dimensions to agree with the formulations of Fig. 4-25. Thus ro = 0.5625 ± 0.001in ri = 0.1875 ± 0.001 in Ro = 0.375 ± 0.0002 in Ri = 0.376 ± 0.0002 in The stochastic nature of the dimensions affects the δ = |Ri | − |Ro | relation in Eq. (4-60) but not the others. Set R = (1/2)( Ri + Ro ) = 0.3755. From Eq. (4-60)

Eδ ro2 − R 2 R 2 − ri2 p= R 2R 2 ro2 − ri2 Substituting and solving with E = 30 Mpsi gives p = 18.70(106 ) δ

Since δ = Ri − Ro

δ¯ = R¯ i − R¯ o = 0.376 − 0.375 = 0.001 in and

σˆ δ =

0.0002 4

2

0.0002 + 4

2 1/2

= 0.000 070 7 in Then Cδ =

σˆ δ 0.000 070 7 = 0.0707 = 0.001 δ¯

The tangential inner-cylinder stress at the shrink-fit surface is given by R¯ 2 + r¯i2 σit = −p 2 R¯ − r¯ 2 i

0.37552 + 0.18752 = −18.70(10 ) δ 0.37552 − 0.18752 6

= −31.1(106 ) δ σ¯ it = −31.1(106 ) δ¯ = −31.1(106 )(0.001) = −31.1(103 ) psi

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Also σˆ σit = |Cδ σ¯ it | = 0.0707(−31.1)103 = 2899 psi σit = N(−31 100, 2899) psi Ans. (b) The tangential stress for the outer cylinder at the shrink-fit surface is given by 2 r¯o + R¯ 2 σot = p 2 r¯o − R¯ 2 0.56252 + 0.37552 6 = 18.70(10 ) δ 0.56252 − 0.37552 = 48.76(106 ) δ psi σ¯ ot = 48.76(106 )(0.001) = 48.76(103 ) psi σˆ σot = Cδ σ¯ ot = 0.0707(48.76)(103 ) = 34.45 psi σot = N(48 760, 3445) psi Ans. 6-42

From Prob. 6-41, at the fit surface σot = N(48.8, 3.45) kpsi. The radial stress is the fit pressure which was found to be p = 18.70(106 ) δ p¯ = 18.70(106 )(0.001) = 18.7(103 ) psi σˆ p = Cδ p¯ = 0.0707(18.70)(103 ) = 1322 psi and so p = N(18.7, 1.32) kpsi and σor = −N(18.7, 1.32) kpsi These represent the principal stresses. The von Mises stress is next assessed. σ¯ A = 48.8 kpsi,

σ¯ B = −18.7 kpsi

k = σ¯ B /σ¯ A = −18.7/48.8 = −0.383 σ¯ = σ¯ A (1 − k + k 2 ) 1/2 = 48.8[1 − (−0.383) + (−0.383) 2 ]1/2 = 60.4 kpsi σˆ σ = C p σ¯ = 0.0707(60.4) = 4.27 kpsi

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Chapter 6

179

Using the interference equation S¯ − σ¯ z = − 1/2 σˆ S2 + σˆ σ2 =−

95.5 − 60.4 = −4.5 [(6.59) 2 + (4.27) 2 ]1/2

p f = α = 0.000 003 40, or about 3 chances in a million.

Ans.

6-43 σt =

6000N(1, 0.083 33)(0.75) pd = 2t 2(0.125)

= 18N(1, 0.083 33) kpsi σl =

6000N(1, 0.083 33)(0.75) pd = 4t 4(0.125)

= 9N(1, 0.083 33) kpsi σr = −p = −6000N(1, 0.083 33) kpsi These three stresses are principal stresses whose variability is due to the loading. From Eq. (6-12), we find the von Mises stress to be 1/2 (18 − 9) 2 + [9 − (−6)]2 + (−6 − 18) 2 σ = 2 = 21.0 kpsi σˆ σ = C p σ¯ = 0.083 33(21.0) = 1.75 kpsi S¯ − σ¯ z = − 1/2 σˆ S2 + σˆ σ2 =

50 − 21.0 = −6.5 (4.12 + 1.752 ) 1/2

The reliability is very high . R = 1 − (6.5) = 1 − 4.02(10−11 ) = 1 Ans.

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Chapter 7 7-1 H B = 490 Eq. (3-17):

Sut = 0.495(490) = 242.6 kpsi > 212 kpsi Se = 107 kpsi

Eq. (7-8):

a = 1.34, b = −0.085

Table 7-4:

ka = 1.34(242.6) −0.085 = 0.840 3/16 −0.107 kb = = 1.05 0.3

Eq. (7-18): Eq. (7-19):

Se = ka kb Se = 0.840(1.05)(107) = 94.4 kpsi

Eq. (7-17):

Ans.

7-2 (a) Sut = 68 kpsi, Se = 0.495(68) = 33.7 kpsi

Ans.

(b) Sut = 112 kpsi, Se = 0.495(112) = 55.4 kpsi (c) 2024T3 has no endurance limit (d) Eq. (3-17): 7-3

Se

= 107 kpsi

Ans.

Ans.

Ans.

σ F = σ0 εm = 115(0.90) 0.22 = 112.4 kpsi Eq. (7-8):

Se = 0.504(66.2) = 33.4 kpsi log(112.4/33.4) = −0.083 64 log(2 · 106 )

Eq. (7-11):

b=−

Eq. (7-9):

f =

112.4 (2 · 103 ) −0.083 64 = 0.8991 66.2

Eq. (7-13):

a=

[0.8991(66.2)]2 = 106.1 kpsi 33.4

Eq. (7-12): Eq. (7-15):

S f = a N b = 106.1(12 500) −0.083 64 = 48.2 kpsi Ans. σ 1/b 36 −1/0.083 64 a N= = = 409 530 cycles Ans. a 106.1

7-4 From S f = a N b log S f = log a + b log N Substituting (1, Sut ) log Sut = log a + b log (1) From which

a = Sut

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181

Chapter 7

Substituting (103 , f Sut ) and a = Sut log f Sut = log Sut + b log 103 From which b= ∴

1 log f 3

S f = Sut N (log f )/3

1 ≤ N ≤ 103

For 500 cycles as in Prob. 7-3 500S f ≥ 66.2(500) (log 0.8991)/3 = 60.2 kpsi Ans. 7-5 Read from graph: (103, 90) and (106, 50). From S = a N b log S1 = log a + b log N1 log S2 = log a + b log N2 From which log a =

log S1 log N2 − log S2 log N1 log N2 /N1

log 90 log 106 − log 50 log 103 log 106 /103 = 2.2095

=

a = 10log a = 102.2095 = 162.0 b=

log 50/90 = −0.085 09 3

(S f ) ax = 162−0.085 09 Check:

103 ≤ N ≤ 106 in kpsi Ans.

103 (S f ) ax = 162(103 ) −0.085 09 = 90 kpsi 106 (S f ) ax = 162(106 ) −0.085 09 = 50 kpsi

The end points agree. 7-6 Eq. (7-8): Table 7-4: Eq. (7-18): Eq. (7-19): Eq. (7-17):

Se = 0.504(710) = 357.8 MPa a = 4.51,

b = −0.265

ka = 4.51(710) −0.265 = 0.792 d −0.107 32 −0.107 kb = = = 0.858 7.62 7.62 Se = ka kb Se = 0.792(0.858)(357.8) = 243 MPa Ans.

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7-7 For AISI 4340 as forged steel, Se = 107 kpsi Eq. (7-8): a = 39.9,

Table 7-4:

b = −0.995

ka = 39.9(260) −0.995 = 0.158 0.75 −0.107 kb = = 0.907 Eq. (7-19): 0.30 Each of the other Marin factors is unity. Eq. (7-18):

Se = 0.158(0.907)(107) = 15.3 kpsi For AISI 1040: Se = 0.504(113) = 57.0 kpsi ka = 39.9(113) −0.995 = 0.362 kb = 0.907 (same as 4340) Each of the other Marin factors is unity. Se = 0.362(0.907)(57.2) = 18.7 kpsi Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see why? 7-8 (a) For an AISI 1018 CD-machined steel, the strengths are

2.5 mm

20 mm

Eq. (3-17):

25 mm

Sut = 440 MPa

⇒

HB =

S y = 370 MPa Ssu = 0.67(440) = 295 MPa Fig. A-15-15: Fig. 7-21: Eq. (7-31):

2.5 D 25 r = = 0.125, = = 1.25, d 20 d 20 qs = 0.94 K f s = 1 + 0.94(1.4 − 1) = 1.376

K ts = 1.4

For a purely reversing torque of 200 N · m K f s 16T 1.376(16)(200 × 103 N · mm) = τmax = πd 3 π(20 mm) 3 τmax = 175.2 MPa = τa Se = 0.504(440) = 222 MPa The Marin factors are

Eq. (7-17):

ka = 4.51(440) −0.265 = 0.899 20 −0.107 = 0.902 kb = 7.62 kc = 0.59, kd = 1, ke = 1 Se = 0.899(0.902)(0.59)(222) = 106.2 MPa

440 = 129 3.41

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183

Chapter 7

[0.9(295)]2 a= = 664 106.2

Eq. (7-13):

1 0.9(295) b = − log = −0.132 65 3 106.2 175.2 1/−0.132 65 N= 664

Eq. (7-14): Eq. (7-15):

N = 23 000 cycles Ans. (b) For an operating temperature of 450°C, the temperature modification factor, from Table 7-6, is kd = 0.843 Se = 0.899(0.902)(0.59)(0.843)(222) = 89.5 MPa Thus [0.9(295)]2 = 788 a= 89.5 1 0.9(295) b = − log = −0.157 41 3 89.5 175.2 1/−0.157 41 N= 788 N = 14 100 cycles

Ans.

7-9 F 1 kN 800 mm b b

f = 0.9 n = 1.5 N = 104 cycles

For AISI 1045 HR steel, Sut = 570 MPa and S y = 310 MPa Se = 0.504(570 MPa) = 287.3 MPa Find an initial guess based on yielding: σa = σmax =

M(b/2) 6M Mc = = 3 3 I b(b )/12 b

Mmax = (1 kN)(800 mm) = 800 N · m σmax

Sy = n

⇒

6(800 × 103 N · mm) 310 N/mm2 = b3 1.5

b = 28.5 mm Eq. (7-24): Eq. (7-19):

de = 0.808b 0.808b −0.107 kb = = 1.2714b−0.107 7.62 kb = 0.888

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The remaining Marin factors are ka = 57.7(570) −0.718 = 0.606 kc = kd = ke = k f = 1 Eq. (7-17): Eq. (7-13): Eq. (7-14): Eq. (7-12):

Se = 0.606(0.888)(287.3 MPa) = 154.6 MPa a=

[0.9(570)]2 = 1702 154.6

1 0.9(570) b = − log = −0.173 64 3 154.6 S f = a N b = 1702[(104 ) −0.173 64 ] = 343.9 MPa n=

Sf σa

Sf n

or σa =

343.9 6(800 × 103 ) ⇒ b = 27.6 mm = 3 b 1.5 Check values for kb , Se , etc. kb = 1.2714(27.6) −0.107 = 0.891 Se = 0.606(0.891)(287.3) = 155.1 MPa a=

[0.9(570)]2 = 1697 155.1

1 0.9(570) b = − log = −0.173 17 3 155.1 S f = 1697[(104 ) −0.173 17 ] = 344.4 MPa 344.4 6(800 × 103 ) = b3 1.5 b = 27.5 mm Ans. 7-10

10 12

Fa

Fa

60 1018

Table A-20:

Sut = 440 MPa, Se

Table 7-4:

S y = 370 MPa

= 0.504(440) = 221.8 MPa

ka = 4.51(440)−0.265 = 0.899 kb = 1 (axial loading)

Eq. (7-25):

kc = 0.85 Se = 0.899(1)(0.85)(221.8) = 169.5 MPa

Table A-15-1:

d/w = 12/60 = 0.2,

K t = 2.5

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Chapter 7

185

From Eq. (7-35) and Table 7-8 Kf =

Kt 2.5 √ = 2.09 √ = √ 1 + 2/ r [(K t − 1)/K t ] a 1 + 2/ 6 [(2.5 − 1)/2.5](174/440)

σa = K f

Fa A

Se 169.5 2.09Fa = = nf 10(60 − 12) 1.8

⇒

Fa = 21 630 N = 21.6 kN Ans. Sy Fa = A ny

⇒

370 Fa = 10(60 − 12) 1.8

Fa = 98 667 N = 98.7 kN Ans. Largest force amplitude is 21.6 kN. 7-11

Ans.

A priori design decisions: The design decision will be: d Material and condition: 1095 HR and from Table A-20 Sut = 120, S y = 66 kpsi. Design factor: n f = 1.6 per problem statement. Life: (1150)(3) = 3450 cycles Function: carry 10 000 lbf load Preliminaries to iterative solution: Se = 0.504(120) = 60.5 kpsi ka = 2.70(120) −0.265 = 0.759 πd 3 I = = 0.098 17d 3 c 32 6 (10 000)(12) = 30 000 lbf · in M(crit.) = 24 The critical location is in the middle of the shaft at the shoulder. From Fig. A-15-9: D/d = 1.5, r/d = 0.10, and K t = 1.68. With no direct information concerning f, use f = 0.9. For an initial trial, set d = 2.00 in 2.00 −0.107 = 0.816 kb = 0.30 Se = 0.759(0.816)(60.5) = 37.5 kpsi [0.9(120)]2 = 311.0 37.5 1 0.9(120) b = − log = −0.1531 3 37.5

a=

S f = 311.0(3450) −0.1531 = 89.3

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σ0 =

30 M 305.6 = = 3 I /c 0.098 17d d3

305.6 = 38.2 kpsi 23 2 d = = 0.2 r= 10 10 1.68 = 1.584 Kf = √ 1 + 2/ 0.2 [(1.68 − 1)/1.68](4/120) =

Eq. (7-37): (K f ) 103 = 1 − (1.584 − 1)[0.18 − 0.43(10−2 )120 + 0.45(10−5 )1202 ] = 1.158 Eq. (7-38):

1.1582 (3450) −(1/3) log(1.158/1.584) = 1.584

(K f ) N = K 3450

= 1.225 305.6 σ0 = = 38.2 kpsi 23 σa = (K f ) N σ0 = 1.225(38.2) = 46.8 kpsi nf =

(S f )3450 89.3 = 1.91 = σa 46.8

The design is satisfactory. Reducing the diameter will reduce n, but the resulting preferred size will be d = 2.00 in. 7-12 σa = 172 MPa, Yield:

172 + 178.4 =

σm =

√ √ 3τm = 3(103) = 178.4 MPa

Sy 413 = ny ny

⇒

n y = 1.18 Ans.

(a) Modified Goodman, Table 7-9 nf =

1 = 1.06 Ans. (172/276) + (178.4/551)

(b) Gerber, Table 7-10  

2  2  1 551 172 2(178.4)(276) = 1.31 Ans. −1 + 1 + nf =  2 178.4 276  551(172) (c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.32 Ans. (172/276) 2 + (178.4/413) 2

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187

Chapter 7

7-13 σa = 69 MPa, 69 + 239 =

Yield:

σm =

413 ny

√ 3(138) = 239 MPa

⇒

n y = 1.34 Ans.

(a) Modified Goodman, Table 7-9 nf = (b) Gerber, Table 7-10 nf =

1 2

551 239

2

1 = 1.46 Ans. (69/276) + (239/551)

 

 2  69 2(239)(276) = 1.73 Ans. −1 + 1 +  276  551(69)

(c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.59 Ans. (69/276) 2 + (239/413) 2 7-14 σa

= σa2 + 3τa2 = 832 + 3(692 ) = 145.5 MPa, 145.5 + 178.4 =

Yield:

413 ny

⇒

σm =

√ 3(103) = 178.4 MPa

n y = 1.28 Ans.

(a) Modified Goodman, Table 7-9 nf = (b) Gerber, Table 7-10 nf =

1 2

551 178.4

2

1 = 1.18 Ans. (145.5/276) + (178.4/551)  

 2  145.5 2(178.4)(276) −1 + 1 + = 1.47 Ans.  276  551(145.5)

(c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.47 Ans. (145.5/276) 2 + (178.4/413) 2 7-15 σa = Yield:

√ 3(207) = 358.5 MPa,

358.5 =

413 ny

⇒

σm = 0

n y = 1.15 Ans.

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1 = 0.77 Ans. (358.5/276)

(b) Gerber criterion of Table 7-10 does not work; therefore use Eq. (7-50). nf

σa =1 Se

(c) ASME-Elliptic, Table 7-11

⇒

nf =

nf =

Se 276 = 0.77 Ans. = σa 358.5

1 358.5/276

2 = 0.77 Ans.

Let f = 0.9 to assess the cycles to failure by fatigue Eq. (7-13): Eq. (7-14): Eq. (7-15):

[0.9(551)]2 = 891.0 MPa 276

a=

1 0.9(551) b = − log = −0.084 828 3 276 358.5 −1/0.084 828 N= = 45 800 cycles Ans. 891.0

7-16 σa = Yield:

√ 3(103) = 178.4 MPa,

178.4 + 103 =

413 ny

⇒

σm = 103 MPa

n y = 1.47 Ans.


1 = 1.20 Ans. (178.4/276) + (103/551)

(b) Gerber, Table 7-10  

2  2  178.4 1 551 2(103)(276) nf = −1 + 1 + = 1.44 Ans. 2 103 276  551(178.4)  (c) ASME-Elliptic, Table 7-11 1/2 1 nf = = 1.44 Ans. (178.4/276) 2 + (103/413) 2 7-17

Table A-20:

Sut = 64 kpsi, S y = 54 kpsi A = 0.375(1 − 0.25) = 0.2813 in2 σmax =

3000 Fmax = (10−3 ) = 10.67 kpsi A 0.2813

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189

Chapter 7

ny =

54 = 5.06 Ans. 10.67

Se = 0.504(64) = 32.3 kpsi ka = 2.70(64) −0.265 = 0.897 kb = 1,

kc = 0.85

Se = 0.897(1)(0.85)(32.3) = 24.6 kpsi Table A-15-1: w = 1 in, d = 1/4 in, d/w = 0.25 K t = 2.45. From Eq. (7-35) and Table 7-8 2.45 = 1.94 √ 1 + 2/ 0.125 [(2.45 − 1)/2.45](5/64) Fmax − Fmin σa = K f 2A 3.000 − 0.800 = 7.59 kpsi = 1.94 2(0.2813)

Kf =

Fmax + Fmin 2A 3.000 + 0.800 = 13.1 kpsi = 1.94 2(0.2813)

σm = K f

r=

σa 7.59 = 0.579 = σm 13.1

(a) DE-Gerber, Table 7-10 Sa =

 2

2

0.579 (64 )  −1 + 2(24.6)

1+

2(24.6) 0.579(64)

2

  = 18.5 kpsi

18.5 Sa = = 32.0 kpsi r 0.579  

2 2 1 64 7.59  2(13.1)(24.6)  nf = −1 + 1 + 2 13.1 24.6 7.59(64)

Sm =

= 2.44 Ans. (b) DE-Elliptic, Table 7-11 Sa = Sm =

(0.5792 )(24.62 )(542 ) = 19.33 kpsi 24.62 + (0.5792 )(542 ) 19.33 Sa = = 33.40 kpsi r 0.579

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Table 7-16

1 = 2.55 Ans. (7.59/24.6) 2 + (13.1/54) 2

nf =

7-18

Referring to the solution of Prob. 7-17, for load fluctuations of −800 to 3000 lbf 3.000 − (−0.800) = 13.1 kpsi σa = 1.94 2(0.2813) 3.000 + (−0.800) = 7.59 kpsi σm = 1.94 2(0.2813) r= (a) Table 7-10, DE-Gerber nf =

1 2

64 7.59

2

σa 13.13 = 1.728 = σm 7.60  

2 13.1  2(7.59)(24.6)  −1 + 1 + = 1.79 Ans. 24.6 64(13.1)

(b) Table 7-11, DE-Elliptic

nf =

7-19

1 = 1.82 Ans. (13.1/24.6) 2 + (7.59/54) 2

Referring to the solution of Prob. 7-17, for load fluctuations of 800 to −3000 lbf 0.800 − (−3.000) = 13.1 kpsi σa = 1.94 2(0.2813) 0.800 + (−3.000) = −7.59 kpsi σm = 1.94 2(0.2813) r=

σa 13.1 = −1.726 = σm −7.59

(a) We have a compressive midrange stress for which the failure locus is horizontal at the Se level. nf =

Se 24.6 = 1.88 Ans. = σa 13.1

nf =

Se 24.6 = 1.88 Ans. = σa 13.1

(b) Same as (a)

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Chapter 7

191

7-20 Sut = 0.495(380) = 188.1 kpsi Se = 0.504(188.1) = 94.8 kpsi ka = 14.4(188.1) −0.718 = 0.335 For a non-rotating round bar in bending, Eq. (7-23) gives: de = 0.370d = 0.370(3/8) = 0.1388 in 0.1388 −0.107 kb = = 1.086 0.3 Se = 0.335(1.086)(94.8) = 34.49 kpsi 30 − 15 30 + 15 = 7.5 lbf, Fm = = 22.5 lbf Fa = 2 2 32Mm 32(22.5)(16) (10−3 ) = 69.54 kpsi = σm = 3 πd π(0.3753 ) σa = r=

32(7.5)(16) (10−3 ) = 23.18 kpsi 3 π(0.375 ) 23.18 = 0.333 69.54

0


1 = 0.960 (23.18/34.49) + (69.54/188.1)

Since finite failure is predicted, proceed to calculate N Eq. (7-10):

σ F = 188.1 + 50 = 238.1 kpsi log(238.1/34.49) = −0.133 13 log(2 · 106 ) 238.1 f = (2 · 103 ) −0.133 13 = 0.4601 188.1

b=−

Eq. (7-11): Eq. (7-9): Eq. (7-13): σa σm + =1 Sf Sut

[0.4601(188.1)]2 = 217.16 kpsi 34.49 23.18 σa = = 36.78 kpsi Sf = 1 − (σm /Sut ) 1 − (69.54/188.1)

a= ⇒

Eq. (7-15) with σa = S f 36.78 1/−0.133 13 N= = 620 000 cycles Ans. 217.16

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(b) Gerber, Table 7-10 nf =

1 2

188.1 69.54

= 1.20

2

 

 2  23.18 2(69.54)(34.49) −1 + 1 +  34.49  188.1(23.18)

Thus, infinite life is predicted ( N ≥ 106 cycles). Ans.

7-21 (a)

I =

1 (18)(33 ) = 40.5 mm4 12

y=

Fl 3 3E I

⇒

F=

3E I y l3

3(207)(109 )(40.5)(10−12 )(2)(10−3 ) = 50.3 N Ans. (1003 )(10−9 ) 6 = (50.3) = 150.9 N Ans. 2

Fmin = Fmax (b)

F

101.5 mm

M = 0.1015F N · m A = 3(18) = 54 mm2 M

F

Curved beam: rc = 4.5 mm, σi = − σo =

rn =

3 h = = 4.3281 mm ln(ro /ri ) ln(6/3)

e = rc − rn = 4.5 − 4.3281 = 0.1719 mm

(0.1015F)(1.5 − 0.1719) F Mci F − = −4.859F MPa − =− − 3 Aeri A 54(0.1719)(3)(10 ) 54

(0.1015F)(1.5 + 0.1719) F Mco F − = 3.028F MPa − = − 3 Aero A 54(0.1719)(6)(10 ) 54 (σi ) min = −4.859(150.9) = −733.2 MPa (σi ) max = −4.859(50.3) = −244.4 MPa (σo ) max = 3.028(150.9) = 456.9 MPa (σo ) min = 3.028(50.3) = 152.3 MPa

Eq. (3-17)

Sut = 3.41(490) = 1671 MPa

Per the problem statement, estimate the yield as S y = 0.9Sut = 0.9(1671) = 1504 MPa. Then from Eq. (7-8), Se = 740 MPa; Eq. (7-18), ka = 1.58(1671) −0.085 = 0.841; Eq. (7-24) de = 0.808[18(3)]1/2 = 5.938 mm; and Eq. (7-19), kb = (5.938/7.62) −0.107 = 1.027.

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193

Chapter 7

Se = 0.841(1.027)(740) = 639 MPa −733.2 + 244.4 = 244.4 MPa (σi ) a = 2

At Inner Radius

(σi ) m =

−733.2 − 244.4 = −488.8 MPa 2 ␴a 1504 MPa

639

244.4

1504

Load line: Langer (yield) line: Intersection:

488.4

␴m

σm = −244.4 − σa σm = σa − 1504 = −244.4 − σa σa = 629.8 MPa, σm = −874.2 MPa (Note that σa is less than 639 MPa)

Yield:

ny =

629.8 = 2.58 244.4

Fatigue:

nf =

639 = 2.61 244.4

Thus, the spring is not likely to fail in fatigue at the inner radius. Ans.

At Outer Radius

Yield load line: Langer line: Intersection:

Fatigue line:

(σo ) a =

456.9 − 152.3 = 152.3 MPa 2

(σo ) m =

456.9 + 152.3 = 304.6 MPa 2

σm = 152.3 + σa σm = 1504 − σa = 152.3 + σa σa = 675.9 MPa, σm = 828.2 MPa 675.9 = 4.44 ny = 152.3 σa = [1 − (σm /Sut ) 2 ]Se = σm − 152.3 2 σm 639 1 − = σm − 152.3 1671 σm2 + 4369.7σm − 3.4577(106 ) = 0

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σm =

−4369.7 +

4369.72 + 4(3.4577)(106 ) = 684.2 MPa 2

σa = 684.2 − 152.3 = 531.9 MPa nf =

531.9 = 3.49 152.3

Thus, the spring is not likely to fail in fatigue at the outer radius. Ans. 7-22

The solution at the inner radius is the same as in Prob. 7-21. At the outer radius, the yield solution is the same. σm Se = σm − 152.3 σa = 1 − Fatigue line: Sut σm 639 1 − = σm − 152.3 1671 1.382σm = 791.3

⇒

σm = 572.4 MPa

σa = 572.4 − 152.3 = 420 MPa nf = 7-23

420 = 2.76 Ans. 152.3

Preliminaries: Table A-20:

Sut = 64 kpsi,

Sy = 54 kpsi

Se = 0.504(64) = 32.3 kpsi ka = 2.70(64) −0.265 = 0.897 kb = 1 kc = 0.85 Se = 0.897(1)(0.85)(32.3) = 24.6 kpsi Fillet: Fig. A-15-5: D = 3.75 in, d = 2.5 in, D/d = 3.75/2.5 = 1.5, and r/d = 0.25/2.5 = 0.10 ∴ K t = 2.1 Kf = σmax =

2.1 = 1.86 √ 1 + 2/ 0.25 [(2.1 − 1)/2.1](4/64) 4 = 3.2 kpsi 2.5(0.5)

−16 = −12.8 kpsi 2.5(0.5) 3.2 − (−12.8) = 14.88 kpsi σa = 1.86 2

σmin =

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195

Chapter 7

3.2 + (−12.8) = −8.93 kpsi σm = 1.86 2 S y 54 = 4.22 = ny = σmin −12.8

Since the midrange stress is negative, Sa = Se = 24.6 kpsi nf =

Sa 24.6 = 1.65 = σa 14.88

Hole: Fig. A-15-1: d/w = 0.75/3.75 = 0.20, K t = 2.5 Kf = σmax =

2.5 √ = 2.17 1 + 2/ 0.75/2 [(2.5 − 1)/2.5](5/64) 4 = 2.67 kpsi 0.5(3.75 − 0.75)

−16 = −10.67 kpsi 0.5(3.75 − 0.75) 2.67 − (−10.67) = 14.47 kpsi σa = 2.17 2

σmin =

σm = 2.17

2.67 + (−10.67) = −8.68 kpsi 2

Since the midrange stress is negative, S y 54 = = 5.06 n y = σmin −10.67 Sa = Se = 24.6 kpsi Sa 24.6 = 1.70 = σa 14.47 Thus the design is controlled by the threat of fatigue at the fillet; the minimum factor of safety is n f = 1.65. Ans. nf =

7-24 (a)

T

T

M = −T , h = 5 mm, A = 25 mm2 rc = 20 mm, ro = 22.5 mm, ri = 17.5 mm h 5 = 19.8954 mm rn = = ln ro /ri ln (22.5/17.5) e = rc − rn = 20 − 19.8954 = 0.1046 mm co = 2.605 mm,

ci = 2.395 mm

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σi =

Mci −T (0.002 395) = −52.34(106 )T = − 6 Aeri 25(10 )(0.1046)(10−3 )(17.5)(10−3 )

−Mco T (2.605)(10−3 ) = 44.27(106 )T = σo = − 6 − 3 − 3 Aero 25(10 )(0.1046)(10 )(22.5)(10 ) For fatigue, σo is most severe as it represents a tensile stress. 1 σm = σa = (44.27)(106 )T = 22.14(106 )T 2 Se = 0.504Sut = 0.504(770) = 388.1 MPa ka = 4.51(770) −0.265 = 0.775 de = 0.808[5(5)]1/2 = 4.04 mm 4.04 −0.107 = 1.070 kb = 7.62 Se = 0.775(1.07)(388.1) = 321.8 MPa Modified Goodman, Table 7-9 σm 1 σa + = Se Sut nf

⇒

22.14T 1 22.14T + = 321.8 770 3

T = 3.42 N · m Ans. (b) Gerber, Eq. (7-50)

nσa nσm 2 + =1 Se Sut 3(22.14)T 3(22.14)T 2 + =1 321.8 770

T 2 + 27.74T − 134.40 = 0 1 T = −27.74 + 27.742 + 4(134.40) = 4.21 N · m Ans. 2 (c) To guard against yield, use T of part (b) and the inner stress. 420 = 1.91 Ans. ny = 52.34(4.21) 7-25

From Prob. 7-24, Se = 321.8 MPa, S y = 420 MPa, and Sut = 770 MPa (a) Assuming the beam is straight, 6M 6T = 48(106 )T σmax = 2 = 3 bh 5 [(10−3 ) 3 ] Goodman:

24T 1 24T + = 321.8 770 3

⇒

T = 3.15 N · m Ans.

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197

Chapter 7

3(24)T 2 3(24)T + =1 321.8 770

(b) Gerber:

T 2 + 25.59T − 114.37 = 1 1 T = −25.59 + 25.592 + 4(114.37) = 3.88 N · m Ans. 2 (c) Using σmax = 52.34(106 )T from Prob. 7-24, ny =

420 = 2.07 Ans. 52.34(3.88)

7-26 (a)

16K f s Tmax πd 3 . Fig. 7-21 for H B > 200, r = 3 mm, qs = 1 τmax =

K f s = 1 + qs (K ts − 1) K f s = 1 + 1(1.6 − 1) = 1.6 Tmax = 2000(0.05) = 100 N · m, τmax

Tmin =

500 (100) = 25 N · m 2000

16(1.6)(100)(10−6 ) = = 101.9 MPa π(0.02) 3

τmin =

500 (101.9) = 25.46 MPa 2000

1 τm = (101.9 + 25.46) = 63.68 MPa 2 1 τa = (101.9 − 25.46) = 38.22 MPa 2 Ssu = 0.67Sut = 0.67(320) = 214.4 MPa Ssy = 0.577S y = 0.577(180) = 103.9 MPa Se = 0.504(320) = 161.3 MPa ka = 57.7(320) −0.718 = 0.917 de = 0.370(20) = 7.4 mm 7.4 −0.107 kb = = 1.003 7.62 kc = 0.59 Se = 0.917(1.003)(0.59)(161.3) = 87.5 MPa

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Modified Goodman, Table 7-9, nf =

1 1 = = 1.36 Ans. (τa /Se ) + (τm /Ssu ) (38.22/87.5) + (63.68/214.4)

(b) Gerber, Table 7-10 nf =

=

7-27

1 2 1 2

Ssu τm

2

214.4 63.68



τa  −1 + Se

2

1+

2τm Se Ssu τa

2

 

 

2   38.22 2(63.68)(87.5) = 1.70 −1 + 1 + 87.5  214.4(38.22) 

Ans.

S y = 800 MPa, Sut = 1000 MPa

. (a) From Fig. 7-20, for a notch radius of 3 mm and Sut = 1 GPa, q = 0.92. K f = 1 + q(K t − 1) = 1 + 0.92(3 − 1) = 2.84 2.84(4) P 4P σmax = −K f =− = −4018P 2 πd π(0.030) 2 1 σm = σa = (−4018P) = −2009P 2 D+d T = fP 4 0.150 + 0.03 = 0.0135P Tmax = 0.3P 4 . From Fig. 7-21, qs = 0.95. Also, K ts is given as 1.8. Thus, K f s = 1 + qs (K ts − 1) = 1 + 0.95(1.8 − 1) = 1.76 16K f s T 16(1.76)(0.0135P) = = 4482P πd 3 π(0.03) 3 1 τa = τm = (4482P) = 2241P 2 1/2 2 σm = σm + 3τm2 = [(−2009P) 2 + 3(2241P) 2 ]1/2 = 4366P

τmax =

σa = σm = 4366P Se = 0.504(1000) = 504 MPa ka = 4.51(1000) −0.265 = 0.723 30 −0.107 = 0.864 kb = 7.62 kc = 0.85 (Note that torsion is accounted for in the von Mises stress.) Se = 0.723(0.864)(0.85)(504) = 267.6 MPa

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199

Chapter 7

σ 1 σa + m = Se Sut n

Modified Goodman:

4366P 1 4366P + = 6 6 267.6(10 ) 1000(10 ) 3

⇒

P = 16.1(103 ) N = 16.1 kN Ans.

1 σ + σm = a ny Sy

Yield:

ny =

800(106 ) = 5.69 Ans. 2(4366)(16.1)(103 )

(b) If the shaft is not rotating, τm = τa = 0. σm = σa = −2009P kb = 1 (axial) kc = 0.85 (Since there is no tension, kc = 1 might be more appropriate.) Se = 0.723(1)(0.85)(504) = 309.7 MPa ␴a 800

309.7

␴m

800

nf =

⇒

P=

309.7(106 ) = 51.4(103 ) N 3(2009) = 51.4 kN Ans.

800(106 ) = 3.87 Ans. ny = 2(2009)(51.4)(103 )

Yield: 7-28

309.7(106 ) 2009P

From Prob. 7-27, K f = 2.84, K f s = 1.76, Se = 267.6 MPa 4Pmax (4)(80)(10−3 ) = −321.4 MPa = −2.84 σmax = −K f πd 2 π(0.030) 2 20 (−321.4) = −80.4 MPa 80 D+d 0.150 + 0.03 3 = 0.3(80)(10 ) = 1080 N · m = f Pmax 4 4

σmin = Tmax

Tmin =

20 (1080) = 270 N · m 80

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τmax τmin

16Tmax 16(1080) −6 = Kfs = 1.76 (10 ) = 358.5 MPa πd 3 π(0.030) 3 20 = (358.5) = 89.6 MPa 80 321.4 − 80.4 = 120.5 MPa 2 −321.4 − 80.4 = −200.9 MPa = 2 358.5 − 89.6 = 134.5 MPa = 2 358.5 + 89.6 = 224.1 MPa = 2 1/2 = σa2 + 3τa2 = [120.52 + 3(134.5) 2 ]1/2 = 262.3 MPa

σa = σm τa τm σa

σm = [(−200.9) 2 + 3(224.1) 2 ]1/2 = 437.1 MPa Goodman:

σa 262.3 = 466.0 MPa (σa ) e = = 1 − σm /Sut 1 − 437.1/1000

Let f = 0.9 [0.9(1000)]2 = 2928 MPa a= 276.6 1 0.9(1000) b = − log = −0.1708 3 276.6 (σa ) e 1/b 466.0 1/−0.1708 N= = = 47 130 cycles Ans. a 2928 7-29

S y = 490 MPa, Sut = 590 MPa, Se = 200 MPa σm =

420 + 140 420 − 140 = 280 MPa , σa = = 140 MPa 2 2

Goodman: (σa ) e =

σa 140 = 266.5 MPa > Se = 1 − σm /Sut 1 − (280/590)

[0.9(590)]2 = 1409.8 MPa 200 1 0.9(590) b = − log = −0.141 355 3 200

a=

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201

Chapter 7

N=

266.5 1409.8

−1/0.143 55

= 131 200 cycles

Nremaining = 131 200 − 50 000 = 81 200 cycles 350 + (−200) = 75 MPa 2 350 − (−200) (σa ) 2 = = 275 MPa 2 275 (σa ) e2 = = 315.0 MPa 1 − (75/590) (σm ) 2 =

Second loading:

(a) Miner’s method

N2 =

n2 n1 + =1 N1 N2

315 1409.8

−1/0.141 355

⇒

= 40 200 cycles

n2 50 000 + =1 131 200 40 200

n 2 = 24 880 cycles Ans. (b) Manson’s method Two data points:

0.9(590 MPa), 103 cycles 266.5 MPa, 81 200 cycles

a2 (103 ) b2 0.9(590) = 266.5 a2 (81 200) b2 1.9925 = (0.012 315) b2 log 1.9925 = −0.156 789 b2 = log 0.012 315 266.5 a2 = = 1568.4 MPa (81 200) −0.156 789 315 1/−0.156 789 n2 = = 27 950 cycles Ans. 1568.4 7-30

(a) Miner’s method [0.9(76)]2 = 155.95 kpsi a= 30 1 0.9(76) b = − log = −0.119 31 3 30 1/−0.119 31 48 σ1 = 48 kpsi, N1 = = 19 460 cycles 155.95 1/−0.119 31 38 σ2 = 38 kpsi, N2 = = 137 880 cycles 155.95

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1/−0.119 31 32 σ3 = 32 kpsi, N3 = = 582 150 cycles 155.95 n1 n2 n3 + + =1 N1 N2 N3 60 000 n3 4000 + + = 1 ⇒ n 3 = 209 160 cycles Ans. 19 460 137 880 582 150 (b) Manson’s method

The life remaining after the first cycle is N R1 = 19 460 − 4000 = 15 460 cycles . The two data points required to define Se, 1 are [0.9(76), 103 ] and (48, 15 460) . 0.9(76) a2 (103 ) b2 = 48 a2 (15 460)

⇒

1.425 = (0.064 683) b2

log(1.425) = −0.129 342 log(0.064 683) 48 a2 = = 167.14 kpsi (15 460) −0.129 342 −1/0.129 342 38 = 94 110 cycles N2 = 167.14 b2 =

N R2 = 94 110 − 60 000 = 34 110 cycles 0.9(76) a3 (103 ) b3 = 38 a3 (34 110) b3 b3 =

7-31

⇒

1.8 = (0.029 317) b3

log 1.8 38 = −0.166 531, a3 = = 216.10 kpsi log(0.029 317) (34 110) −0.166 531 32 −1/0.166 531 N3 = = 95 740 cycles Ans. 216.1

Using Miner’s method a=

[0.9(100)]2 = 162 kpsi 50

0.9(100) 1 = −0.085 091 b = − log 3 50 70 1/−0.085 091 σ1 = 70 kpsi, N1 = = 19 170 cycles 162 55 1/−0.085 091 = 326 250 cycles σ2 = 55 kpsi, N2 = 162 σ3 = 40 kpsi,

N3 → ∞ 0.5N 0.3N 0.2N + + =1 19 170 326 250 ∞ N = 83 570 cycles Ans.

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Chapter 7

7-32

Given H B = 495LN(1, 0.03) Eq. (3-20)

Sut = 0.495 [LN(1, 0.041)]H B = 0.495 [LN(1, 0.041)][495 LN(1, 0.03)] ¯Sut = 0.495(495) = 245 kpsi

Table 2-6 for the COV of a product. . C x y = C x2 + C y2 = (0.0412 + 0.032 ) 1/2 = 0.0508 Sut = 245LN(1, 0.0508) kpsi From Table 7-13:

a = 1.34, b = −0.086, C = 0.12 −0.086 LN(1, 0.120) ka = 1.34 S¯ut

= 1.34(245) −0.086 LN(1, 0.12) = 0.835LN(1, 0.12) kb = 1.05 (as in Prob. 7-1) Se = 0.835LN(1, 0.12)(1.05)[107LN(1, 0.139)] S¯e = 0.835(1.05)(107) = 93.8 kpsi

Now

7-33

. C Se = (0.122 + 0.1392 ) 1/2 = 0.184 Se = 93.8LN(1, 0.184) kpsi Ans.

A Priori Decisions: • Material and condition: 1018 CD, Sut = 440LN(1, 0.03), and S y = 370LN(1, 0.061) MPa • Reliability goal: R = 0.999 (z = −3.09) • Function: Critical location—hole • Variabilities: Cka = 0.058 Ckc = 0.125 Cφ = 0.138 2 1/2 2 C Se = Cka + Ckc + Cφ2 = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195 Ckc = 0.10 C Fa = 0.20 Cσ a = (0.102 + 0.202 ) 1/2 = 0.234

2 C Se + Cσ2 a 0.1952 + 0.2342 = = 0.297 Cn = 1 + Cσ2 a 1 + 0.2342

203

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Resulting in a design factor n f of, Eq. (6-59): n f = exp[−(−3.09) ln(1 + 0.2972 ) + ln 1 + 0.2972 ] = 2.56 • Decision: Set n f = 2.56 Now proceed deterministically using the mean values: k¯a = 0.887, kb = 1, k¯c = 0.890, and from Prob. 7-10, K f = 2.09 F¯a F¯a S¯e = K¯ f = σ¯ a = K¯ f A t (60 − 12) n¯ f n¯ f K¯ f F¯a 2.56(2.09)(15.103 ) = 9.5 mm = ∴ t= (60 − 12)(175.7) (60 − 12) S¯e Decision: If 10 mm 1018 CD is available, t = 10 mm 7-34

Ans.

1.25"

1.00"

M

M

Rotation is presumed. M and Sut are given as deterministic, but notice that σ is not; therefore, a reliability estimation can be made. From Eq. (7-70): Se = 0.506(110)LN(1, 0.138) = 55.7LN(1, 0.138) kpsi Table 7-13: ka = 2.67(110) −0.265 LN(1, 0.058) = 0.768LN(1, 0.058) Based on d = 1 in, Eq. (7-19) gives kb =

1 0.30

−0.107

= 0.879

Conservatism is not necessary Se = 0.768[LN(1, 0.058)](0.879)(55.7)[LN(1, 0.138)] S¯e = 37.6 kpsi C Se = (0.0582 + 0.1382 ) 1/2 = 0.150 Se = 37.6LN(1, 0.150)

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205

Chapter 7

Fig. A-15-14: D/d = 1.25, r/d = 0.125. Thus K t = 1.70 and Eqs. (7-35), (7-78) and Table 7-8 give 1.70LN(1, 0.15) Kf = √ 1 + 2/ 0.125 [(1.70 − 1)/(1.70)](3/110) = 1.598LN(1, 0.15)

32M 32(1400) = 1.598[LN(1 − 0.15)] σ = Kf πd 3 π(1) 3 = 22.8LN(1, 0.15) kpsi From Eq. (6-57):

ln (37.6/22.8) (1 + 0.152 )/(1 + 0.152 ) z=− = −2.37 ln[(1 + 0.152 )(1 + 0.152 )]

From Table A-10, p f = 0.008 89 ∴ R = 1 − 0.008 89 = 0.991 Ans. Note: The correlation method uses only the mean of Sut ; its variability is already included in the 0.138. When a deterministic load, in this case M, is used in a reliability estimate, engineers state, “For a Design Load of M, the reliability is 0.991.” They are in fact referring to a Deterministic Design Load. 7-35

For completely reversed torsion, ka and kb of Prob. 7-34 apply, but kc must also be considered. kc = 0.328(110) 0.125 LN(1, 0.125) Eq. 7-74: = 0.590LN(1, 0.125) Note 0.590 is close to 0.577. S Se = ka kb kc Se = 0.768[LN(1, 0.058)](0.878)[0.590LN(1, 0.125)][55.7LN(1, 0.138)] S¯ Se = 0.768(0.878)(0.590)(55.7) = 22.2 kpsi C Se = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195 S Se = 22.2LN(1, 0.195) kpsi Fig. A-15-15: D/d = 1.25, r/d = 0.125, then K ts = 1.40. From Eqs. (7-35), (7-78) and Table 7-8 1.40LN(1, 0.15) = 1.34LN(1, 0.15) Kts = √ 1 + 2/ 0.125 [(1.4 − 1)/1.4](3/110) τ = Kts

16T πd 3

16(1.4) τ = 1.34[LN(1, 0.15)] π(1) 3 = 9.55LN(1, 0.15) kpsi

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From Eq. (6-57):

ln (22.2/9.55) (1 + 0.152 )/(1 + 0.1952 ) = −3.43 z=− ln [(1 + 0.1952 )(1 + 0.152 )]

From Table A-10, p f = 0.0003 R = 1 − p f = 1 − 0.0003 = 0.9997 Ans. For a design with completely-reversed torsion of 1400 lbf · in, the reliability is 0.9997. The improvement comes from a smaller stress-concentration factor in torsion. See the note at the end of the solution of Prob. 7-34 for the reason for the phraseology. 7-36

1" 1 D 4

1" D 8

Non-rotating

M

M

Sut = 58 kpsi Se = 0.506(58)LN(1, 0.138) = 29.3LN(1, 0.138) kpsi Table 7-13:

ka = 14.5(58) −0.719 LN(1, 0.11) = 0.782LN(1, 0.11)

Eq. (7-23): de = 0.37(1.25) = 0.463 in 0.463 −0.107 kb = = 0.955 0.30 Se = 0.782[LN(1, 0.11)](0.955)[29.3LN(1, 0.138)] S¯e = 0.782(0.955)(29.3) = 21.9 kpsi C Se = (0.112 + 0.1382 ) 1/2 = 0.150 Table A-16: d/D = 0, a/D = 0.1, A = 0.83 ∴ K t = 2.27. From Eqs. (7-35) and (7-78) and Table 7-8 2.27LN(1, 0.10) = 1.783LN(1, 0.10) Kf = √ 1 + 2/ 0.125 [(2.27 − 1)/2.27](5/58) Table A-16:

π AD 3 π(0.83)(1.253 ) = 0.159 in3 Z= = 2 3 32 M 1.6 = 1.783LN(1, 0.10) σ = Kf Z 0.159 = 17.95LN(1, 0.10) kpsi

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Chapter 7

σ¯ = 17.95 kpsi Cσ = 0.10 2 2 ln (21.9/17.95) (1 + 0.10 )/(1 + 0.15 ) = −1.07 z=− ln[(1 + 0.152 )(1 + 0.102 )]

Eq. (6-57):

p f = 0.1423

Table A-10:

R = 1 − p f = 1 − 0.1423 = 0.858 Ans. For a completely-reversed design load Ma of 1400 lbf · in, the reliability estimate is 0.858. 7-37

For a non-rotating bar subjected to completely reversed torsion of Ta = 2400 lbf · in From Prob. 7-36: Se = 29.3LN(1, 0.138) kpsi ka = 0.782LN(1, 0.11) kb = 0.955 For kc use Eq. (7-74): kc = 0.328(58) 0.125 LN(1, 0.125) = 0.545LN(1, 0.125) S Se = 0.782[LN(1, 0.11)](0.955)[0.545LN(1, 0.125)][29.3LN(1, 0.138)] S¯ Se = 0.782(0.955)(0.545)(29.3) = 11.9 kpsi C Se = (0.112 + 0.1252 + 0.1382 ) 1/2 = 0.216 Table A-16: d/D = 0, a/D = 0.1, A = 0.92, K ts = 1.68 From Eqs. (7-35), (7-78), Table 7-8 Kf s =

1.68LN(1, 0.10) √ 1 + 2/ 0.125 [(1.68 − 1)/1.68](5/58)

= 1.403LN(1, 0.10) Table A-16: Jnet =

π(0.92)(1.254 ) π AD 4 = = 0.2201 32 32

τa = K f s

Ta c Jnet

2.4(1.25/2) = 1.403[LN(1, 0.10)] 0.2201 = 9.56LN(1, 0.10) kpsi

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From Eq. (6-57):

ln (11.9/9.56) (1 + 0.102 )/(1 + 0.2162 ) = −0.85 z=− ln[(1 + 0.102 )(1 + 0.2162 )]

Table A-10, p f = 0.1977 R = 1 − p f = 1 − 0.1977 = 0.80 Ans. 7-38

This is a very important task for the student to attempt before starting Part 3. It illustrates the drawback of the deterministic factor of safety method. It also identifies the a priori decisions and their consequences. The range of force fluctuation in Prob. 7-23 is −16 to +4 kip, or 20 kip. Repeatedlyapplied Fa is 10 kip. The stochastic properties of this heat of AISI 1018 CD are given. Function

Consequences

Axial Fatigue load

Fa = 10 kip C Fa = 0 Ckc = 0.125 z = −3.09 C K f = 0.11

Overall reliability R ≥ 0.998; with twin fillets √ R ≥ 0.998 ≥ 0.999 Cold rolled or machined surfaces Ambient temperature Use correlation method Stress amplitude Significant strength Se

Cka = 0.058 Ckd Cφ CK f Cσ a C Se

=0 = 0.138 = 0.11 = 0.11 = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195

Choose the mean design factor which will meet the reliability goal

0.1952 + 0.112 = 0.223 Cn = 1 + 0.112 n¯ = exp −(−3.09) ln(1 + 0.2232 ) + ln 1 + 0.2232 n¯ = 2.02 Review the number and quantitative consequences of the designer’s a priori decisions to accomplish this. The operative equation is the definition of the design factor Se σa = n K¯ f Fa S¯e S¯e ⇒ = σ¯ a = n¯ w2 h n¯

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209

Chapter 7

Solve for thickness h. To do so we need −0.265 = 2.67(64) −0.265 = 0.887 k¯a = 2.67 S¯ut kb = 1 −0.078 k¯c = 1.23 S¯ut = 1.23(64) −0.078 = 0.889 k¯d = k¯e = 1 S¯e = 0.887(1)(0.889)(1)(1)(0.506)(64) = 25.5 kpsi

Fig. A-15-5: D = 3.75 in, d = 2.5 in, D/d = 3.75/2.5 = 1.5, r/d = 0.25/2.5 = 0.10 ∴ K t = 2.1 K¯ f = h=

2.1 = 1.857 √ 1 + 2/ 0.25 [(2.1 − 1)/(2.1)](4/64) K¯ f n¯ Fa 1.857(2.02)(10) = 0.667 Ans. = ¯ 2.5(25.5) w2 Se

This thickness separates S¯e and σ¯ a so as to realize the reliability goal of 0.999 at each shoulder. The design decision is to make t the next available thickness of 1018 CD steel strap from the same heat. This eliminates machining to the desired thickness and the extra cost of thicker work stock will be less than machining the fares. Ask your steel supplier what is available in this heat. 7-39

3" 4

1" 1 2

1200 lbf

1" 4

Fa = 1200 lbf Sut = 80 kpsi (a) Strength ka = 2.67(80) −0.265 LN(1, 0.058) = 0.836LN(1, 0.058) kb = 1 kc = 1.23(80) −0.078 LN(1, 0.125) = 0.874LN(1, 0.125) Sa = 0.506(80)LN(1, 0.138) = 40.5LN(1, 0.138) kpsi Se = 0.836[LN(1, 0.058)](1)[0.874LN(1, 0.125)][40.5LN(1, 0.138)] S¯e = 0.836(1)(0.874)(40.5) = 29.6 kpsi C Se = (0.0582 + 0.1252 + 0.1382 ) 1/2 = 0.195

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Stress: Fig. A-15-1; d/w = 0.75/1.5 = 0.5, K t = 2.17. From Eqs. (7-35), (7-78) and Table 7-8 2.17LN(1, 0.10) Kf = √ 1 + 2/ 0.375 [(2.17 − 1)/2.17](5/80) = 1.95LN(1, 0.10) K f Fa σa = , Cσ = 0.10 (w − d)t K¯ f Fa 1.95(1.2) = = 12.48 kpsi σ¯ a = (w − d)t (1.5 − 0.75)(0.25) S¯a = S¯e = 29.6 kpsi ln ( S¯a /σ¯ a ) 1 + Cσ2 1 + C S2 z=− ln 1 + Cσ2 1 + C S2 ln (29.6/12.48) (1 + 0.102 )/(1 + 0.1952 ) =− = −3.9 ln (1 + 0.102 )(1 + 0.1952 ) From Table A-20 p f = 4.481(10−5 ) R = 1 − 4.481(10−5 ) = 0.999 955 Ans. (b) All computer programs will differ in detail. 7-40

Each computer program will differ in detail. When the programs are working, the experience should reinforce that the decision regarding n¯ f is independent of mean values of strength, stress or associated geometry. The reliability goal can be realized by noting the impact of all those a priori decisions.

7-41

Such subprograms allow a simple call when the information is needed. The calling program is often named an executive routine (executives tend to delegate chores to others and only want the answers).

7-42

This task is similar to Prob. 7-41.

7-43

Again, a similar task.

7-44

The results of Probs. 7-41 to 7-44 will be the basis of a class computer aid for fatigue problems. The codes should be made available to the class through the library of the computer network or main frame available to your students.

7-45

Peterson’s notch sensitivity q has very little statistical basis. This subroutine can be used to show the variation in q, which is not apparent to those who embrace a deterministic q .

7-46

An additional program which is useful.

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Chapter 8 8-1 (a)

Thread depth = 2.5 mm

5 mm

Width = 2.5 mm Ans. dm = 25 − 1.25 − 1.25 = 22.5 mm dr = 25 − 5 = 20 mm l = p = 5 mm Ans.

2.5 2.5 mm 25 mm

(b)

Ans.

Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm Ans.

5 mm 2.5 5 mm

8-2 From Table 8-1, dr = d − 1.226 869 p dm = d − 0.649 519 p d − 1.226 869 p + d − 0.649 519 p = d − 0.938 194 p d¯ = 2 π π d¯2 = (d − 0.938 194 p) 2 Ans. At = 4 4 8-3 From Eq. (c) of Sec. 8-2, P=F T = e=

tan λ + f 1 − f tan λ

Pdm Fdm tan λ + f = 2 2 1 − f tan λ

T0 Fl/(2π) 1 − f tan λ 1 − f tan λ = = tan λ T Fdm /2 tan λ + f tan λ + f

Using f = 0.08, form a table and plot the efficiency curve. λ, deg. 0 10 20 30 40 45

1

e 0 0.678 0.796 0.838 0.8517 0.8519

e

0

50 ␭, deg.

Ans.

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8-4 Given F = 6 kN, l = 5 mm, and dm = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) 6(0.05)(40) 6(22.5) 5 + π(0.08)(22.5) + TR = 2 π(22.5) − 0.08(5) 2 = 10.23 + 6 = 16.23 N · m Ans. The torque required to lower the load, from Eqs. (8-2) and (8-6) is 6(0.05)(40) 6(22.5) π(0.08)22.5 − 5 + TL = 2 π(22.5) + 0.08(5) 2 = 0.622 + 6 = 6.622 N · m Ans. Since TL is positive, the thread is self-locking. The efficiency is e=

Eq. (8-4):

6(5) = 0.294 Ans. 2π(16.23)

8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom segment of the screws must be in compression. Where as tension specimens and their grips must be in tension. Both screws must be of the same-hand threads. 8-6 Screws rotate at an angular rate of n=

1720 = 22.9 rev/min 75

(a) The lead is 0.5 in, so the linear speed of the press head is V = 22.9(0.5) = 11.5 in/min Ans. (b) F = 2500 lbf/screw dm = 3 − 0.25 = 2.75 in sec α = 1/cos(29/2) = 1.033 Eq. (8-5): 2500(2.75) TR = 2

0.5 + π(0.05)(2.75)(1.033) π(2.75) − 0.5(0.05)(1.033)

= 377.6 lbf · in

Eq. (8-6): Tc = 2500(0.06)(5/2) = 375 lbf · in Ttotal = 377.6 + 375 = 753 lbf · in/screw Tmotor = H=

753(2) = 21.1 lbf · in 75(0.95) Tn 21.1(1720) = = 0.58 hp Ans. 63 025 63 025

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213

Chapter 8

8-7 The force F is perpendicular to the paper. 1" 4 7" 16

3" D. 16 2.406"

3"

7 1 1 − − = 2.406 in 8 4 32 T = 2.406F 7 7 M= L− F = 2.406 − F = 2.188F 32 32 L =3−

S y = 41 kpsi σ = Sy =

32M 32(2.188) F = = 41 000 3 πd π(0.1875) 3

F = 12.13 lbf T = 2.406(12.13) = 29.2 lbf · in Ans. (b) Eq. (8-5), 2α = 60◦ , l = 1/14 = 0.0714 in, f = 0.075, sec α = 1.155, p = 1/14 in 7 1 − 0.649 519 = 0.3911 in dm = 16 14 Fclamp (0.3911) Num TR = 2 Den Num = 0.0714 + π(0.075)(0.3911)(1.155) Den = π(0.3911) − 0.075(0.0714)(1.155) T = 0.028 45Fclamp 29.2 T = = 1030 lbf Ans. Fclamp = 0.028 45 0.028 45 (c) The column has one end fixed and the other end pivoted. Base decision on the mean diameter column. Input: C = 1.2, D = 0.391 in, S y = 41 kpsi, E = 30(106 ) psi, L = 4.1875 in, k = D/4 = 0.097 75 in, L/k = 42.8. For this J. B. Johnson column, the critical load represents the limiting clamping force for bucking. Thus, Fclamp = Pcr = 4663 lbf. (d) This is a subject for class discussion. 8-8

T = 6(2.75) = 16.5 lbf · in 1 5 − = 0.5417 in 8 12 1 29◦ l = = 0.1667 in, α = = 14.5◦ , 6 2

dm =

sec 14.5◦ = 1.033

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0.1667 + π(0.15)(0.5417)(1.033) = 0.0696F T = 0.5417( F/2) π(0.5417) − 0.15(0.1667)(1.033)

Eq. (8-5): Eq. (8-6):

Tc = 0.15(7/16)( F/2) = 0.032 81F Ttotal = (0.0696 + 0.0328) F = 0.1024F F=

16.5 = 161 lbf Ans. 0.1024

8-9 dm = 40 − 3 = 37 mm, l = 2(6) = 12 mm From Eq. (8-1) and Eq. (8-6)

10(0.15)(60) 10(37) 12 + π(0.10)(37) + TR = 2 π(37) − 0.10(12) 2 = 38.0 + 45 = 83.0 N · m

Since n = V /l = 48/12 = 4 rev/s ω = 2πn = 2π(4) = 8π rad/s so the power is H = T ω = 83.0(8π) = 2086 W

Ans.

8-10 (a) dm = 36 − 3 = 33 mm, l = p = 6 mm From Eqs. (8-1) and (8-6) 0.09(90) F 33F 6 + π(0.14)(33) + T = 2 π(33) − 0.14(6) 2 = (3.292 + 4.050) F = 7.34F N · m ω = 2πn = 2π(1) = 2π rad/s H = Tω 3000 H = = 477 N · m T = ω 2π 477 F= = 65.0 kN Ans. 7.34 Fl 65.0(6) = = 0.130 Ans. (b) e = 2π T 2π(477) 8-11 1 = 2(0.5) + 0.25 = 1.25 in Ans. 4 (b) From Table A-32 the washer thickness is 0.109 in. Thus, (a) L T = 2D +

L G = 0.5 + 0.5 + 0.109 = 1.109 in (c) From Table A-31, H =

7 = 0.4375 in 16

Ans.

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215

Chapter 8

(d) L G + H = 1.109 + 0.4375 = 1.5465 in This would be rounded to 1.75 in per Table A-17. The bolt is long enough.

Ans.

(e) ld = L − L T = 1.75 − 1.25 = 0.500 in Ans. lt = L G − ld = 1.109 − 0.500 = 0.609 in Ans. These lengths are needed to estimate bolt spring rate kb . Note: In an analysis problem, you need not know the fastener’s length at the outset, although you can certainly check, if appropriate. 8-12 (a) L T = 2D + 6 = 2(14) + 6 = 34 mm

Ans.

(b) From Table A-33, the maximum washer thickness is 3.5 mm. Thus, the grip is, L G = 14 + 14 + 3.5 = 31.5 mm Ans. (c) From Table A-31, H = 12.8 mm (d) L G + H = 31.5 + 12.8 = 44.3 mm This would be rounded to L = 50 mm. The bolt is long enough. (e) ld = L − L T = 50 − 34 = 16 mm Ans.

Ans.

lt = L G − ld = 31.5 − 16 = 15.5 mm Ans. These lengths are needed to estimate the bolt spring rate kb . 8-13 (a) L T = 2D + (b) L G > h +

1 = 2(0.5) + 0.25 = 1.25 in 4

Ans.

0.5 d d = t1 + = 0.875 + = 1.125 in 2 2 2

Ans.

(c) L > h + 1.5d = t1 + 1.5d = 0.875 + 1.5(0.5) = 1.625 in From Table A-17, this rounds to 1.75 in. The cap screw is long enough. (d) ld = L − L T = 1.75 − 1.25 = 0.500 in Ans. lt = L G − ld = 1.125 − 0.5 = 0.625 in Ans. 8-14 (a) L T = 2(12) + 6 = 30 mm (b) L G = h +

Ans.

12 d d = t1 + = 20 + = 26 mm 2 2 2

Ans.

(c) L > h + 1.5d = t1 + 1.5d = 20 + 1.5(12) = 38 mm This rounds to 40 mm (Table A-17). The fastener is long enough. (d) ld = L − L T = 40 − 30 = 10 mm Ans. l T = L G − ld = 26 − 10 = 16 mm Ans.

Ans.

Ans.

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8-15 Ad = 0.7854(0.75) 2 = 0.442 in2 Atube = 0.7854(1.1252 − 0.752 ) = 0.552 in2

(a)

kb =

0.442(30)(106 ) Ad E = = 1.02(106 ) lbf/in grip 13

km =

0.552(30)(106 ) Atube E = = 1.27(106 ) lbf/in 13 13

C=

(b)

1.02 = 0.445 1.02 + 1.27

Ans. Ans.

Ans.

1 1 1 · = = 0.020 83 in 16 3 48 (13 − 0.020 83) |P|l |δb | = = |P| = 9.79(10−7 )|P| in AE 0.442(30)(106 ) δ=

Grip Original bolt A ␦

Nut advance ␦b

␦m

|δm | =

Equilibrium A

|P|(13) |P|l = = 7.85(10−7 )|P| in AE 0.552(30)(106 )

|δb | + |δm | = δ = 0.020 83 9.79(10−7 )|P| + 7.85(10−7 )|P| = 0.020 83 Fi = |P| =

0.020 83 = 11 810 lbf Ans. + 7.85(10−7 )

9.79(10−7 )

(c) At opening load P0 9.79(10−7 ) P0 = 0.020 83 P0 =

0.020 83 = 21 280 lbf Ans. 9.79(10−7 )

As a check use Fi = (1 − C) P0 P0 = 8-16

11 810 Fi = = 21 280 lbf 1−C 1 − 0.445

The movement is known at one location when the nut is free to turn δ = pt = t/N Letting Nt represent the turn of the nut from snug tight, Nt = θ/360◦ and δ = Nt /N. The elongation of the bolt δb is δb =

Fi kb

The advance of the nut along the bolt is the algebraic sum of |δb | and |δm |

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217

Chapter 8

Nt N Fi Fi Nt + = kb km N 1 1 kb + km Nt = N Fi = Fi N , + kb km kb km |δb | + |δm | =

θ 360◦

Ans.

As a check invert Prob. 8-15. What Turn-of-Nut will induce Fi = 11 808 lbf? 1 1 + Nt = 16(11 808) 1.02(106 ) 1.27(106 ) . = 0.334 turns = 1/3 turn (checks) The relationship between the Turn-of-Nut method and the Torque Wrench method is as follows. kb + km Fi N (Turn-of-Nut) Nt = kb km T = K Fi d Eliminate Fi

Nt =

(Torque Wrench)

kb + km kb km

θ NT = Kd 360◦

Ans.

8-17 (a) From Ex. 8-4, Fi = 14.4 kip, kb = 5.21(106 ) lbf/in, km = 8.95(106 ) lbf/in Eq. (8-27):

T = k Fi d = 0.2(14.4)(103 )(5/8) = 1800 lbf · in Ans.

From Prob. 8-16,

t = N Fi

1 1 + kb km

1 1 = 16(14.4)(10 ) + 6 5.21(10 ) 8.95(106 )

= 0.132 turns = 47.5◦

3

Ans.

Bolt group is (1.5)/(5/8) = 2.4 diameters. Answer is lower than RB&W recommendations. (b) From Ex. 8-5, Fi = 14.4 kip, kb = 6.78 Mlbf/in, and km = 17.4 Mlbf/in T = 0.2(14.4)(103 )(5/8) = 1800 lbf · in Ans. 1 1 3 t = 11(14.4)(10 ) + 6.78(106 ) 17.4(106 ) = 0.0325 = 11.7◦ 8-18

Ans.

Again lower than RB&W.

From Eq. (8-22) for the conical frusta, with d/l = 0.5 0.577π km = 1.11 = Ed (d/l)=0.5 2 ln {5[0.577 + 0.5(0.5)]/[0.577 + 2.5(0.5)]}

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Eq. (8-23), from the Wileman et al. finite element study, km = 0.787 15 exp[0.628 75(0.5)] = 1.08 Ed (d/l)=0.5 8-19

For cast iron, from Table 8-8 For gray iron: A = 0.778 71, B = 0.616 16

0.625 km = 12(10 )(0.625)(0.778 71) exp 0.616 16 = 7.55(106 ) lbf/in 1.5 6

This member’s spring rate applies to both members. We need km for the upper member which represents half of the joint. kci = 2km = 2[7.55(106 )] = 15.1(106 ) lbf/in For steel from Table 8-8: A = 0.787 15, B = 0.628 73 0.625 6 = 19.18(106 ) lbf/in km = 30(10 )(0.625)(0.787 15) exp 0.628 73 1.5 ksteel = 2km = 2(19.18)(106 ) = 38.36(106 ) lbf/in For springs in series 1 1 1 1 1 + = + = 6 km kci ksteel 15.1(10 ) 38.36(106 ) km = 10.83(106 ) lbf/in Ans. 8-20

The external tensile load per bolt is 1 π (150) 2 (6)(10−3 ) = 10.6 kN P= 10 4 Also, L G = 45 mm and from Table A-31, for d = 12 mm, H = 10.8 mm. No washer is specified. L T = 2D + 6 = 2(12) + 6 = 30 mm L G + H = 45 + 10.8 = 55.8 mm L = 60 mm ld = 60 − 30 = 30 mm lt = 45 − 30 = 15 mm

Table A-17:

Ad =

π(12) 2 = 113 mm2 4

At = 84.3 mm2

Table 8-1: Eq. (8-17): kb =

113(84.3)(207) = 466.8 MN/m 113(15) + 84.3(30)

Steel: Using Eq. (8-23) for A = 0.787 15, B = 0.628 73 and E = 207 GPa

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Chapter 8

Eq. (8-23):

km = 207(12)(0.787 15) exp[(0.628 73)(12/40)] = 2361 MN/m ks = 2km = 4722 MN/m

Cast iron: A = 0.778 71, B = 0.616 16, E = 100 GPa km = 100(12)(0.778 71) exp[(0.616 16)(12/40)] = 1124 MN/m kci = 2km = 2248 MN/m 1 1 1 = + km ks kci C=

⇒

km = 1523 MN/m

466.8 = 0.2346 466.8 + 1523

Table 8-1: At = 84.3 mm2 , Table 8-11, Sp = 600 MPa Fi = 0.75(84.3)(600)(10−3 ) = 37.9 kN Eqs. (8-30) and (8-31): Eq. (8-28): Sp At − Fi 600(10−3 )(84.3) − 37.9 = = 5.1 Ans. n= CP 0.2346(10.6) 8-21

Computer programs will vary.

8-22

D3 = 150 mm, A = 100 mm, B = 200 mm, C = 300 mm, D = 20 mm, E = 25 mm. ISO 8.8 bolts: d = 12 mm, p = 1.75 mm, coarse pitch of p = 6 MPa. 1 π (1502 )(6)(10−3 ) = 10.6 kN/bolt P= 10 4 L G = D + E = 20 + 25 = 45 mm L T = 2D + 6 = 2(12) + 6 = 30 mm Table A-31: H = 10.8 mm L G + H = 45 + 10.8 = 55.8 mm Table A-17: L = 60 mm D1 dw

20 2.5 45

22.5 25

ld = 60 − 30 = 30 mm, Table 8-1: At = 84.3 mm2

lt = 45 − 30 = 15 mm,

Ad = π(122 /4) = 113 mm2

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Eq. (8-17): kb =

113(84.3)(207) = 466.8 MN/m 113(15) + 84.3(30)

There are three frusta: dm = 1.5(12) = 18 mm D1 = (20 tan 30◦ )2 + dw = (20 tan 30◦ )2 + 18 = 41.09 mm Upper Frustum: t = 20 mm, E = 207 GPa, D = 1.5(12) = 18 mm k1 = 4470 MN/m

Eq. (8-20):

Central Frustum: t = 2.5 mm, D = 41.09 mm, E = 100 GPa (Table A-5) ⇒ k2 = 52 230 MN/m Lower Frustum: t = 22.5 mm, E = 100 GPa, D = 18 mm ⇒ k3 = 2074 MN/m km = [(1/4470) + (1/52 230) + (1/2074)]−1 = 1379 MN/m 466.8 C= = 0.253 Eq. (e), p. 421: 466.8 + 1379 Eqs. (8-30) and (8-31): From Eq. (8-18):

Fi = K Fp = K At Sp = 0.75(84.3)(600)(10−3 ) = 37.9 kN Eq. (8-28):

8-23

n=

Sp At − Fi 600(10−3 )(84.3) − 37.9 = = 4.73 Ans. Cm P 0.253(10.6)

1 π P= (1202 )(6)(10−3 ) = 8.48 kN 8 4 From Fig. 8-21, t1 = h = 20 mm and t2 = 25 mm l = 20 + 12/2 = 26 mm t = 0 (no washer), L T = 2(12) + 6 = 30 mm L > h + 1.5d = 20 + 1.5(12) = 38 mm Use 40 mm cap screws. ld = 40 − 30 = 10 mm lt = l − ld = 26 − 10 = 16 mm Ad = 113 mm , 2

12 13

At = 84.3 mm

2

h 20 l 26

Eq. (8-17):

7

13 6

113(84.3)(207) kb = 113(16) + 84.3(10)

t2 25

= 744 MN/m Ans. dw = 1.5(12) = 18 mm D = 18 + 2(6)(tan 30) = 24.9 mm

D

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Chapter 8

From Eq. (8-20): Top frustum: D = 18, t = 13, E = 207 GPa

⇒ k1 = 5316 MN/m

Mid-frustum: t = 7, E = 207 GPa, D = 24.9 mm

⇒ k2 = 15 620 MN/m

D = 18, t = 6, E = 100 GPa ⇒ k3 = 3887 MN/m 1 = 2158 MN/m Ans. km = (1/5316) + (1/55 620) + (1/3887)

Bottom frustum:

C=

744 = 0.256 Ans. 744 + 2158

From Prob. 8-22, Fi = 37.9 kN Sp At − Fi 600(0.0843) − 37.9 = = 5.84 Ans. n= CP 0.256(8.48) 8-24

Calculation of bolt stiffness: H = 7/16 in L T = 2(1/2) + 1/4 = 1 1/4 in L G = 1/2 + 5/8 + 0.095 = 1.22 in L > 1.125 + 7/16 + 0.095 = 1.66 in Use L = 1.75 in 1.454 1.327 3 4 1 2

0.61 1.22

5 8

3 4

0.095

0.860

ld = L − L T = 1.75 − 1.25 = 0.500 in lt = 1.125 + 0.095 − 0.500 = 0.72 in Ad = π(0.502 )/4 = 0.1963 in2 At = 0.1419 in2 (UNC) At E 0.1419(30) = 5.9125 Mlbf/in = lt 0.72 Ad E 0.1963(30) kd = = 11.778 Mlbf/in = ld 0.500 1 kb = = 3.936 Mlbf/in Ans. (1/5.9125) + (1/11.778) kt =

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Member stiffness for four frusta and joint constant C using Eqs. (8-20) and (e). ⇒ k1 = 33.30 Mlbf/in

Top frustum:

D = 0.75, t = 0.5, d = 0.5, E = 30

2nd frustum:

D = 1.327, t = 0.11, d = 0.5, E = 14.5

3rd frustum:

D = 0.860, t = 0.515, E = 14.5

Fourth frustum:

⇒ k2 = 173.8 Mlbf/in

⇒ k3 = 21.47 Mlbf/in

D = 0.75, t = 0.095, d = 0.5, E = 30 ⇒ k4 = 97.27 Mlbf/in −1 4 1/ki = 10.79 Mlbf/in Ans. km = i=1

C = 3.94/(3.94 + 10.79) = 0.267 Ans. 8-25 0.095"

0.4225" 0.845"

1.238" 0.5"

0.1725"

1.018"

0.595" Steel

0.25" 0.75"

kb =

0.625"

Cast iron

0.1419(30) At E = = 5.04 Mlbf/in Ans. l 0.845

From Fig. 8-21, h=

1 + 0.095 = 0.595 in 2

l=h+

0.5 d = 0.595 + = 0.845 2 2

D1 = 0.75 + 0.845 tan 30◦ = 1.238 in l/2 = 0.845/2 = 0.4225 in From Eq. (8-20): Frustum 1: D = 0.75, t = 0.4225 in, d = 0.5 in, E = 30 Mpsi ⇒ k1 = 36.14 Mlbf/in Frustum 2: Frustum 3:

D = 1.018 in, t = 0.1725 in, E = 70 Mpsi, d = 0.5 in ⇒ k2 = 134.6 Mlbf/in D = 0.75, t = 0.25 in, d = 0.5 in, E = 14.5 Mpsi ⇒ k3 = 23.49 Mlbf/in 1 km = = 12.87 Mlbf/in Ans. (1/36.14) + (1/134.6) + (1/23.49) C=

5.04 = 0.281 Ans. 5.04 + 12.87

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Chapter 8

8-26

Refer to Prob. 8-24 and its solution. Additional information: A = 3.5 in, Ds = 4.25 in, static pressure 1500 psi, Db = 6 in, C (joint constant) = 0.267, ten SAE grade 5 bolts. P=

1 π(4.252 ) (1500) = 2128 lbf 10 4

From Tables 8-2 and 8-9, At = 0.1419 in2 Sp = 85 000 psi Fi = 0.75(0.1419)(85) = 9.046 kip From Eq. (8-28), n= 8-27

Sp At − Fi 85(0.1419) − 9.046 = = 5.31 Ans. CP 0.267(2.128)

From Fig. 8-21, t1 = 0.25 in h l D1 D2

= 0.25 + 0.065 = 0.315 in = h + (d/2) = 0.315 + (3/16) = 0.5025 in = 1.5(0.375) + 0.577(0.5025) = 0.8524 in = 1.5(0.375) = 0.5625 in

l/2 = 0.5025/2 = 0.251 25 in Frustum 1: Washer 0.5625"

0.6375"

0.065"

E = 30 Mpsi, t = 0.065 in, D = 0.5625 in k = 78.57 Mlbf/in (by computer) Frustum 2: Cap portion 0.6375" 0.18625" 0.8524"

E = 14 Mpsi, t = 0.186 25 in D = 0.5625 + 2(0.065)(0.577) = 0.6375 in k = 23.46 Mlbf/in (by computer) Frustum 3: Frame and Cap 0.8524" 0.25125" 0.5625"

E = 14 Mpsi, t = 0.251 25 in, D = 0.5625 in k = 14.31 Mlbf/in (by computer) 1 = 7.99 Mlbf/in Ans. km = (1/78.57) + (1/23.46) + (1/14.31)

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For the bolt, L T = 2(3/8) + (1/4) = 1 in. So the bolt is threaded all the way. Since At = 0.0775 in2 kb =

0.0775(30) = 4.63 Mlbf/in Ans. 0.5025

8-28 (a) Fb = R Fb, max sin θ Half of the external moment is contributed by the line load in the interval 0 ≤ θ ≤ π. π π M 2 = Fb R sin θ dθ = Fb, max R 2 sin2 θ dθ 2 0 0 π M = Fb, max R 2 2 2 from which Fb, max = Fmax =

φ2

φ1

M π R2

Fb R sin θ

M dθ = π R2

φ2

φ1

R sin θ dθ =

M (cos φ1 − cos φ2 ) πR

Noting φ1 = 75◦ , φ2 = 105◦ 12 000 (cos 75◦ − cos 105◦ ) = 494 lbf Ans. π(8/2) 2M M 2π = ( R) Fmax = Fb, max R φ = 2 πR N RN 2(12 000) = 500 lbf Ans. Fmax = (8/2)(12)

Fmax = (b)

(c) F = Fmax sin θ M = 2Fmax R[(1) sin2 90◦ + 2 sin2 60◦ + 2 sin2 30◦ + (1) sin2 (0)] = 6Fmax R from which Fmax =

12 000 M = = 500 lbf Ans. 6R 6(8/2)

The simple general equation resulted from part (b) Fmax = 8-29

(a) Table 8-11: Eq. (8-30): Table (8-15): Eq. (8-27)

2M RN

Sp = 600 MPa Fi = 0.9At Sp = 0.9(245)(600)(10−3 ) = 132.3 kN K = 0.18 T = 0.18(132.3)(20) = 476 N · m Ans.

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225

Chapter 8

(b) Washers: t = 3.4 mm, d = 20 mm, D = 30 mm, E = 207 GPa ⇒ k1 = 42 175 MN/m Cast iron: t = 20 mm, d = 20 mm, D = 30 + 2(3.4) tan 30◦ = 33.93 mm, E = 135 GPa

⇒

k2 = 7885 MN/m

Steel: t = 20 mm, d = 20 mm, D = 33.93 mm, E = 207 GPa ⇒ k3 = 12 090 MN/m km = (2/42 175 + 1/7885 + 1/12 090) −1 = 3892 MN/m Bolt: L G = 46.8 mm. Nut: H = 18 mm. L > 46.8 + 18 = 64.8 mm. Use L = 80 mm. L T = 2(20) + 6 = 46 mm, ld = 80 − 46 = 34 mm, lt = 46.8 − 34 = 12.8 mm, At = 245 mm2 , kb =

Ad = π202 /4 = 314.2 mm2

Ad At E 314.2(245)(207) = 1290 MN/m = Ad lt + At ld 314.2(12.8) + 245(34)

C = 1290/(1290 + 3892) = 0.2489, n=

Sp = 600 MPa,

Fi = 132.3 kN

Sp At − Fi 600(0.245) − 132.3 = = 15.7 Ans. C( P/N ) 0.2489(15/4)

Bolts are a bit oversized for the load. 8-30

(a) ISO M 20 × 2.5 grade 8.8 coarse pitch bolts, lubricated. Table 8-2

At = 245 mm2

Table 8-11

Sp = 600 MPa Ad = π(20) 2 /4 = 314.2 mm2 Fp = 245(0.600) = 147 kN Fi = 0.90Fp = 0.90(147) = 132.3 kN T = 0.18(132.3)(20) = 476 N · m Ans.

(b) L ≥ L G + H = 48 + 18 = 66 mm . Therefore, set L = 80 mm per Table A-17. L T = 2D + 6 = 2(20) + 6 = 46 mm ld = L − L T = 80 − 46 = 34 mm lt = L G − ld = 48 − 34 = 14 mm 80 48 grip Not to scale 14 34

46

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kb =

Ad At E 314.2(245)(207) = 1251.9 MN/m = Ad lt + At ld 314.2(14) + 245(34) 30

24

24

30

Use Wileman et al. Eq. (8-23) A = 0.787 15, B = 0.628 73 km 20 Bd = 0.787 15 exp 0.628 73 = A exp = 1.0229 Ed LG 48 km = 1.0229(207)(20) = 4235 MN/m 1251.9 = 0.228 1251.9 + 4235 Bolts carry 0.228 of the external load; members carry 0.772 of the external load. Ans. Thus, the actual loads are C=

Fb = C P + Fi = 0.228(20) + 132.3 = 136.9 kN Fm = (1 − C) P − Fi = (1 − 0.228)20 − 132.3 = −116.9 kN 8-31

Given pmax = 6 MPa, pmin = 0 and from Prob. 8-20 solution, C = 0.2346, Fi = 37.9 kN, At = 84.3 mm2 . For 6 MPa, P = 10.6 kN per bolt σi =

Fi 37.9(103 ) = 450 MPa = At 84.3

Eq. (8-35): σa =

CP 0.2346(10.6)(103 ) = 14.75 MPa = 2At 2(84.3)

σm = σa + σi = 14.75 + 450 = 464.8 MPa (a) Goodman Eq. (8-40) for 8.8 bolts with Se = 129 MPa, Sut = 830 MPa Sa =

Se (Sut − σi ) 129(830 − 450) = 51.12 MPa = Sut + Se 830 + 129 nf =

Sa 51.12 = 3.47 Ans. = σa 14.75

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227

Chapter 8

(b) Gerber Eq. (8-42) 1

2 2 Sa = Sut Sut + 4Se (Se + σi ) − Sut − 2σi Se 2Se 1 = 830 8302 + 4(129)(129 + 450) − 8302 − 2(450)(129) 2(129) = 76.99 MPa nf =

76.99 = 5.22 Ans. 14.75

(c) ASME-elliptic Eq. (8-43) with Sp = 600 MPa Se 2 2 − σ2 − σ S Sa = 2 S + S S p i e p e i Sp + Se2

129 600 6002 + 1292 − 4502 − 450(129) = 65.87 MPa = 2 2 600 + 129 65.87 = 4.47 Ans. nf = 14.75 8-32 P= Table 8-11:

pA π D2 p π(0.92 )(550) = = = 9.72 kN/bolt N 4N 4(36)

Sp = 830 MPa, Sut = 1040 MPa, S y = 940 MPa At = 58 mm2

Table 8-1:

Ad = π(102 )/4 = 78.5 mm2 L G = D + E = 20 + 25 = 45 mm L T = 2(10) + 6 = 26 mm H = 8.4 mm

Table A-31:

L ≥ L G + H = 45 + 8.4 = 53.4 mm Choose L = 60 mm from Table A-17 ld = L − L T = 60 − 26 = 34 mm lt = L G − ld = 45 − 34 = 11 mm kb =

Ad At E 78.5(58)(207) = 332.4 MN/m = Ad lt + At ld 78.5(11) + 58(34) 15

20

22.5

2.5 25

22.5

10

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Frustum 1: Top, E = 207, t = 20 mm, d = 10 mm, D = 15 mm

k1 = ln

0.5774π(207)(10) 15 + 10 1.155(20) + 15 − 10 1.155(20) + 15 + 10 15 − 10

= 3503 MN/m Frustum 2: Middle, E = 96 GPa, D = 38.09 mm, t = 2.5 mm, d = 10 mm 0.5774π(96)(10) k2 = 38.09 + 10 1.155(2.5) + 38.09 − 10 ln 1.155(2.5) + 38.09 + 10 38.09 − 10 = 44 044 MN/m could be neglected due to its small influence on km . Frustum 3: Bottom, E = 96 GPa, t = 22.5 mm, d = 10 mm, D = 15 mm

k3 = ln

0.5774π(96)(10) 15 + 10 1.155(22.5) + 15 − 10 1.155(22.5) + 15 + 10 15 − 10

= 1567 MN/m 1 = 1057 MN/m (1/3503) + (1/44 044) + (1/1567) 332.4 C= = 0.239 332.4 + 1057

km =

Fi = 0.75At Sp = 0.75(58)(830)(10−3 ) = 36.1 kN Table 8-17: Se = 162 MPa σi =

Fi 36.1(103 ) = 622 MPa = At 58

(a) Goodman Eq. (8-40) Sa =

Se (Sut − σi ) 162(1040 − 622) = 56.34 MPa = Sut + Se 1040 + 162 nf =

56.34 = 2.82 Ans. 20

(b) Gerber Eq. (8-42) 1

2 2 Sa = + 4Se (Se + σi ) − Sut − 2σi Se Sut Sut 2Se 1 = 1040 10402 + 4(162)(162 + 622) − 10402 − 2(622)(162) 2(162) = 86.8 MPa

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Chapter 8

CP 0.239(9.72)(103 ) = 20 MPa σa = = 2At 2(58) nf =

Sa 86.8 = 4.34 Ans. = σa 20

(c) ASME elliptic Se 2 2 − σ2 − σ S S + S S p i e p e i Sp2 + Se2

162 2 + 1622 − 6222 − 622(162) = 84.90 MPa 830 830 = 8302 + 1622

Sa =

nf = 8-33

84.90 = 4.24 Ans. 20

Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi. Referring to the Figure of Prob. 4-73, the following notation will be used for the radii of Section AA. ri = 1 in,

ro = 2 in,

rc = 1.5 in

From Table 4-5, with R = 0.5 in 0.52 rn = √ = 1.457 107 in 2 1.5 − 1.52 − 0.52 e = rc − rn = 1.5 − 1.457 107 = 0.042 893 in co = ro − rn = 2 − 1.457 109 = 0.542 893 in ci = rn − ri = 1.457 107 − 1 = 0.457 107 in A = π(12 )/4 = 0.7854 in2 If P is the maximum load M = Prc = 1.5P rc ci 1.5(0.457) P P 1+ 1+ = 21.62P = σi = A eri 0.7854 0.0429(1) σa = σm =

21.62P σi = = 10.81P 2 2

(a) Eye: Section AA ka = 14.4(93.7) −0.718 = 0.553 de = 0.37d = 0.37(1) = 0.37 in 0.37 −0.107 kb = = 0.978 0.30 kc = 0.85 Se = 0.504(93.7) = 47.2 kpsi Se = 0.553(0.978)(0.85)(47.2) = 21.7 kpsi

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Since no stress concentration exists, use a load line slope of 1. From Table 7-10 for Gerber   2 2 93.7  2(21.7)  −1 + 1 + = 20.65 kpsi Sa = 2(21.7) 93.7 Note the mere 5 percent degrading of Se in Sa 1910 Sa 20.65(103 ) = nf = = σa 10.81P P Thread: Die cut. Table 8-17 gives 18.6 kpsi for rolled threads. Use Table 8-16 to find Se for die cut threads Se = 18.6(3.0/3.8) = 14.7 kpsi Table 8-2: At = 0.663 in2 σ = P/At = P/0.663 = 1.51P σa = σm = σ/2 = 1.51P/2 = 0.755P From Table 7-10, Gerber Sa =



2

120  −1 + 2(14.7)

1+

2(14.7) 120

2

  = 14.5 kpsi

19 200 Sa 14 500 = = σa 0.755P P Comparing 1910/P with 19 200/P, we conclude that the eye is weaker in fatigue. Ans. nf =

(b) Strengthening steps can include heat treatment, cold forming, cross section change (a round is a poor cross section for a curved bar in bending because the bulk of the material is located where the stress is small). Ans. (c) For n f = 2 P=

8-34

1910 = 955 lbf, max. load Ans. 2

41 = 2.41 in . Use L = 2 12 in Ans. 64 (b) Four frusta: Two washers and two members

(a) L ≥ 1.5 + 2(0.134) +

1.125" 0.134" D1

1.280"

0.75"

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231

Chapter 8

Washer: E = 30 Mpsi, t = 0.134 in, D = 1.125 in, d = 0.75 in k1 = 153.3 Mlbf/in Eq. (8-20): Member: E = 16 Mpsi, t = 0.75 in, D = 1.280 in, d = 0.75 in k2 = 35.5 Mlbf/in Eq. (8-20): 1 = 14.41 Mlbf/in Ans. km = (2/153.3) + (2/35.5) Bolt: L T = 2(3/4) + 1/4 = 13/4 in L G = 2(0.134) + 2(0.75) = 1.768 in ld = L − L T = 2.50 − 1.75 = 0.75 in lt = L G − ld = 1.768 − 0.75 = 1.018 in At = 0.373 in2 (Table 8-2) Ad = π(0.75) 2 /4 = 0.442 in2 Ad At E 0.442(0.373)(30) = 6.78 Mlbf/in Ans. kb = = Ad lt + At ld 0.442(1.018) + 0.373(0.75) 6.78 C= = 0.320 Ans. 6.78 + 14.41 (c) From Eq. (8-40), Goodman with Se = 18.6 kpsi, Sut = 120 kpsi 18.6[120 − (25/0.373)] = 7.11 kpsi 120 + 18.6

Sa = The stress components are σa =

CP 0.320(6) = 2.574 kpsi = 2At 2(0.373)

σm = σa + nf =

Fi 25 = 69.6 kpsi = 2.574 + At 0.373 Sa 7.11 = 2.76 = σa 2.574

Ans.

(d) Eq. (8-42) for Gerber 1 25 25 Sa = − 1202 − 2 18.6 120 1202 + 4(18.6) 18.6 + 2(18.6) 0.373 0.373 = 10.78 kpsi nf = (e) n proof =

10.78 = 4.19 2.574

85 = 1.17 Ans. 2.654 + 69.8

Ans.

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8-35 At = 0.1419 in2 Sp = 85 kpsi, Sut = 120 kpsi Se = 18.6 kpsi

(a) Table 8-2: Table 8-9: Table 8-17:

Fi = 0.75At Sp = 0.75(0.1419)(85) = 9.046 kip c= σa =

4.94 = 0.236 4.94 + 15.97 CP 0.236P = 0.832P kpsi = 2At 2(0.1419)

Eq. (8-40) for Goodman criterion 18.6(120 − 9.046/0.1419) Sa = = 7.55 kpsi 120 + 18.6 nf =

Sa 7.55 =2 = σa 0.832P

⇒

P = 4.54 kip Ans.

(b) Eq. (8-42) for Gerber criterion 1 9.046 9.046 − 1202 − 2 18.6 Sa = 120 1202 + 4(18.6) 18.6 + 2(18.6) 0.1419 0.1419 = 11.32 kpsi nf =

Sa 11.32 =2 = σa 0.832P

From which P=

11.32 = 6.80 kip Ans. 2(0.832)

(c) σa = 0.832P = 0.832(6.80) = 5.66 kpsi σm = Sa + σa = 11.32 + 63.75 = 75.07 kpsi Load factor, Eq. (8-28) n=

Sp At − Fi 85(0.1419) − 9.046 = = 1.88 Ans. CP 0.236(6.80)

Separation load factor, Eq. (8-29) n= 8-36

Table 8-2: Table 8-9: Table 8-17:

9.046 Fi = = 1.74 Ans. (1 − C) P 6.80(1 − 0.236) At = 0.969 in2 (coarse) At = 1.073 in2 (fine) Sp = 74 kpsi, Sut = 105 kpsi Se = 16.3 kpsi

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Chapter 8

233

Coarse thread, UNC Fi = 0.75(0.969)(74) = 53.78 kip

Eq. (8-42):

σi =

Fi 53.78 = 55.5 kpsi = At 0.969

σa =

CP 0.30P = 0.155P kpsi = 2At 2(0.969)

Sa =

1 105 1052 + 4(16.3)(16.3 + 55.5) − 1052 − 2(55.5)(16.3) = 9.96 kpsi 2(16.3)

nf =

Sa 9.96 =2 = σa 0.155P

From which P=

9.96 = 32.13 kip Ans. 0.155(2)

Fine thread, UNF Fi = 0.75(1.073)(74) = 59.55 kip 59.55 σi = = 55.5 kpsi 1.073 0.32P σa = = 0.149P kpsi 2(1.073) Sa = 9.96 (as before) Sa 9.96 nf = =2 = σa 0.149P From which P=

9.96 = 33.42 kip Ans. 0.149(2)

Percent improvement 33.42 − 32.13 . (100) = 4% Ans. 32.13 8-37

For a M 30 × 3.5 ISO 8.8 bolt with P = 80 kN/bolt and C = 0.33 Table 8-1:

At = 561 mm2

Table 8-11:

Sp = 600 MPa Sut = 830 MPa

Table 8-17:

Se = 129 MPa

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Fi = 0.75(561)(10−3 )(600) = 252.45 kN σi =

252.45(10−3 ) = 450 MPa 561

σa =

CP 0.33(80)(103 ) = 23.53 MPa = 2At 2(561)

Eq. (8-42): Sa =

1 830 8302 + 4(129)(129 + 450) − 8302 − 2(450)(129) = 77.0 MPa 2(129)

Fatigue factor of safety nf =

Sa 77.0 = 3.27 Ans. = σa 23.53

Load factor from Eq. (8-28), n=

Sp At − Fi 600(10−3 )(561) − 252.45 = = 3.19 Ans. CP 0.33(80)

Separation load factor from Eq. (8-29), n=

252.45 Fi = = 4.71 Ans. (1 − C) P (1 − 0.33)(80)

8-38 (a) Table 8-2: Table 8-9: Table 8-17: Unthreaded grip

At = 0.0775 in2 Sp = 85 kpsi, Sut = 120 kpsi Se = 18.6 kpsi

π(0.375) 2 (30) Ad E kb = = = 0.245 Mlbf/in per bolt Ans. l 4(13.5) π π [( D + 2t) 2 − D 2 ] = (4.752 − 42 ) = 5.154 in2 4 4 5.154(30) 1 Am E = = 2.148 Mlbf/in/bolt. Ans. km = l 12 6

Am =

(b)

Fi = 0.75(0.0775)(85) = 4.94 kip σi = 0.75(85) = 63.75 kpsi 2000 π 2 P = pA = (4) = 4189 lbf/bolt 6 4 C=

0.245 = 0.102 0.245 + 2.148

σa =

CP 0.102(4.189) = 2.77 kpsi = 2At 2(0.0775)

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Chapter 8

235

Eq. (8-40) for Goodman Sa =

18.6(120 − 63.75) = 7.55 kpsi 120 + 18.6

nf =

Sa 7.55 = 2.73 Ans. = σa 2.77

(c) From Eq. (8-42) for Gerber fatigue criterion,

1 Sa = 120 1202 + 4(18.6)(18.6 + 63.75) − 1202 − 2(63.75)(18.6) 2(18.6) = 11.32 kpsi nf =

Sa 11.32 = 4.09 Ans. = σa 2.77

(d) Pressure causing joint separation from Eq. (8-29) Fi =1 (1 − C) P Fi 4.94 P= = = 5.50 kip 1−C 1 − 0.102 5500 P = 6 = 2626 psi Ans. p= A π(42 )/4 n=

8-39

This analysis is important should the initial bolt tension fail. Members: S y = 71 kpsi, Ssy = 0.577(71) = 41.0 kpsi. Bolts: SAE grade 8, S y = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi Shear in bolts

π(0.3752 ) = 0.221 in2 As = 2 4 As Ssy 0.221(75.01) = = 5.53 kip Fs = n 3

Bearing on bolts Ab = 2(0.375)(0.25) = 0.188 in2 Ab S yc 0.188(130) = = 12.2 kip Fb = n 2 Bearing on member Fb =

0.188(71) = 5.34 kip 2.5

Tension of members At = (1.25 − 0.375)(0.25) = 0.219 in2 0.219(71) = 5.18 kip 3 F = min(5.53, 12.2, 5.34, 5.18) = 5.18 kip Ans.

Ft =

The tension in the members controls the design.

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8-40

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Members: S y = 32 kpsi Bolts: S y = 92 kpsi, Ssy = (0.577)92 = 53.08 kpsi Shear of bolts

π(0.375) 2 = 0.221 in2 As = 2 4

τ=

Fs 4 = 18.1 kpsi = As 0.221

n=

Ssy 53.08 = = 2.93 Ans. τ 18.1

Bearing on bolts Ab = 2(0.25)(0.375) = 0.188 in2 −4 = −21.3 kpsi σb = 0.188 n=

Sy 92 = = 4.32 Ans. |σb | |−21.3|

n=

S yc 32 = = 1.50 Ans. |σb | |−21.3|

Bearing on members

Tension of members At = (2.375 − 0.75)(1/4) = 0.406 in2 4 = 9.85 kpsi σt = 0.406 Sy 32 = 3.25 Ans. n= = At 9.85 8-41

Members: S y = 71 kpsi Bolts: S y = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts F = Ssy A/n Fs =

53.08(2)(π/4)(7/8) 2 = 35.46 kip 1.8

Bearing on bolts Fb =

2(7/8)(3/4)(92) = 54.89 kip 2.2

Fb =

2(7/8)(3/4)(71) = 38.83 kip 2.4

Bearing on members

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237

Chapter 8

Tension in members Ft =

(3 − 0.875)(3/4)(71) = 43.52 kip 2.6

F = min(35.46, 54.89, 38.83, 43.52) = 35.46 kip 8-42

Ans.

Members: S y = 47 kpsi Bolts: S y = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts π(0.75) 2 = 0.442 in2 4 20 = 15.08 kpsi τs = 3(0.442)

Ad =

n=

Ssy 53.08 = 3.52 Ans. = τs 15.08

Bearing on bolt 20 = −14.22 kpsi 3[(3/4) · (5/8)] Sy 92 = 6.47 Ans. n=− =− σb −14.22

σb = −

Bearing on members F 20 = −14.22 kpsi =− Ab 3[(3/4) · (5/8)] Sy 47 = 3.31 Ans. n=− =− σb 14.22

σb = −

Tension on members 20 F = = 6.10 kpsi A (5/8)[7.5 − 3(3/4)] Sy 47 = 7.71 Ans. n= = σt 6.10

σt =

8-43

Members: S y = 57 kpsi Bolts: S y = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear of bolts

π(3/8) 2 As = 3 = 0.3313 in2 4 5.4 F = = 16.3 kpsi τs = A 0.3313 Ssy 53.08 n= = 3.26 Ans. = τs 16.3

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Bearing on bolt

5 3 = 0.3516 in2 Ab = 3 8 16 F 5.4 = −15.36 kpsi σb = − =− Ab 0.3516 Sy 92 = 5.99 Ans. n=− =− σb −15.36

Bearing on members Ab = 0.3516 in2 (From bearing on bolt calculations) σb = −15.36 kpsi (From bearing on bolt calculations) Sy 57 = 3.71 Ans. n=− =− σb −15.36 Tension in members Failure across two bolts

5 3 3 A= 2 −2 = 0.5078 in2 16 8 8 5.4 F = = 10.63 kpsi A 0.5078 Sy 57 = 5.36 Ans. n= = σt 10.63

σ =

8-44

2.8 kN

350

By symmetry, R1 = R2 = 1.4 kN

350

C

R2

R1

RA 200 1.4 kN

B

A 50

RB

M B = 0 1.4(250) − 50R A = 0 MA = 0

200(1.4) − 50R B = 0

Members: S y = 370 MPa Bolts: S y = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Bolt shear:

π (102 ) = 78.54 mm2 4 7(103 ) = 89.13 MPa τ= 78.54

As =

n=

Ssy 242.3 = = 2.72 τ 89.13

⇒ ⇒

R A = 7 kN R B = 5.6 kN

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239

Chapter 8

Ab = td = 10(10) = 100 mm2

Bearing on member:

−7(103 ) = −70 MPa 100 Sy −370 = 5.29 n=− = σ −70

σb =

Strength of member M = 1.4(200) = 280 N · m At A, 1 I A = [10(503 ) − 10(103 )] = 103.3(103 ) mm4 12 Mc 280(25) (103 ) = 67.76 MPa σA = = 3 IA 103.3(10 ) Sy 370 = 5.46 n= = σA 67.76 At C, M = 1.4(350) = 490 N · m IC =

1 (10)(503 ) = 104.2(103 ) mm4 12

σC =

490(25) (103 ) = 117.56 MPa 3 104.2(10 )

n=

Sy 370 = 3.15 < 5.46 = σC 117.56

C more critical

n = min(2.72, 5.29, 3.15) = 2.72 Ans. 8-45

Fs P

Fs = 3000 lbf 3000(3) = 1286 lbf P= 7

3" 3000 lbf

7"

3"

Pivot about this point O

H= 1" 2

1 1 + + 0.095 = 1.095 in 2 2 L ≥ L G + H = 1.095 + (7/16) = 1.532 in l = LG =

l 1" 2

7 in 16

1

3" 4

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Use 1

3" bolts 4

1 = 2(0.5) + 0.25 = 1.25 in 4 = 1.75 − 1.25 = 0.5 = 1.095 − 0.5 = 0.595 π(0.5) 2 = 0.1963 in2 = 4 = 0.1419 in

L T = 2D + ld lt Ad At

Ad At E Ad lt + At ld 0.1963(0.1419)(30) = 0.1963(0.595) + 0.1419(0.5) = 4.451 Mlbf/in

kb =

0.75" t 0.5475"

0.5"

Two identical frusta A = 0.787 15, B = 0.628 73 d km = Ed A exp 0.628 73 L G 0.5 = 30(0.5)(0.787 15) exp 0.628 73 1.095 km = 15.733 Mlbf/in C=

4.451 = 0.2205 4.451 + 15.733

Sp = 85 kpsi Fi = 0.75(0.1419)(85) = 9.046 kip σi = 0.75(85) = 63.75 kpsi σb =

C P + Fi 0.2205(1.286) + 9.046 = 65.75 kpsi = At 0.1419

τs =

Fs 3 = 15.28 kpsi = As 0.1963

von Mises stress 1/2 = [65.742 + 3(15.282 )]1/2 = 70.87 kpsi σ = σb2 + 3τs2 Stress margin m = Sp − σ = 85 − 70.87 = 14.1 kpsi Ans.

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241

Chapter 8

8-46

2P(200) = 12(50)

2Fs 12 kN

2P 50

12(50) = 1.5 kN per bolt 2(200) = 6 kN/bolt = 380 MPa π = 245 mm2 , Ad = (202 ) = 314.2 mm2 4 = 0.75(245)(380)(10−3 ) = 69.83 kN

P=

200

Fs Sp

O

At Fi

69.83(103 ) σi = = 285 MPa 245 C P + Fi 0.30(1.5) + 69.83 σb = (103 ) = 287 MPa = At 245 Fs 6(103 ) = 19.1 MPa τ= = Ad 314.2 σ = [2872 + 3(19.12 )]1/2 = 289 MPa m = Sp − σ = 380 − 289 = 91 MPa Thus the bolt will not exceed the proof stress. 8-47

Ans.

Using the result of Prob. 6-29 for lubricated assembly 2π f T Fx = 0.18d n With a design factor of d gives 0.18(3)(1000)d 0.18n d Fx d = = 716d T = 2π f 2π(0.12) or T /d = 716. Also T = K (0.75Sp At ) d = 0.18(0.75)(85 000) At = 11 475At Form a table Size 1 4

28

5 16 3 8

24

24

At

T /d = 11 475At

n

0.0364

417.7

1.75

0.058

665.55

2.8

0.0878

1007.5

4.23

The factor of safety in the last column of the table comes from 2π f (T /d) 2π(0.12)(T /d) n= = 0.0042(T /d) = 0.18Fx 0.18(1000)

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Select a 38 " − 24 UNF capscrew. The setting is given by T = (11 475At )d = 1007.5(0.375) = 378 lbf · in Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in. Check the factor of safety 2π(0.12)(400) 2π f T = = 4.47 n= 0.18Fx d 0.18(1000)(0.375) 8-48

26

F'A

F'C

F'B 50

A

50 C

B M

F"A

F"C

152

Bolts: Sp = 380 MPa, S y = 420 MPa Channel: t = 6.4 mm, S y = 170 MPa Cantilever: S y = 190 MPa Nut: H = 10.8 mm

L T = 2(12) + 6 = 30 mm L > 12 + 6.4 + 10.8 = 29.2 mm Therefore, use L = 30 mm All threads, so At = 84.3 mm2 FA + FB + FC = F/3 M = (50 + 26 + 125) F = 201F 201F = 2.01F 2(50) 1 + 2.01 F = 2.343F FC = FC + FC = 3 FA = FC =

Bolts: The shear bolt area is As = At = 84.3 mm2 Ssy = 0.577(420) = 242.3 MPa Ssy 242.3(84.3)(10−3 ) As = = 3.11 kN F= n 2.343 2.8(2.343) Bearing on bolt: For a 12-mm bolt, dm = 12 − 0.649 519(1.75) = 10.86 mm Ab = tdm = (6.4)(10.86) = 69.5 mm2 Sy Ab 420 69.5(10−3 ) = = 4.45 kN F= n 2.343 2.8 2.343

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243

Chapter 8

Bearing on member: Ab = 12(10.86) = 130.3 mm2 170 (130.3)(10−3 ) = 3.38 kN F= 2.8 2.343 Strength of cantilever: I =

1 (12)(503 − 123 ) = 1.233(105 ) mm4 12

I 1.233(105 ) = = 4932 c 25 4932(190) M = = 2.22 kN F= 151 2.8(151)(103 ) So F = 2.22 kN based on bending of cantilever 8-49

F = 4 kN; M = 12(200) = 2400 N · m 2400 = 37.5 kN FA = FB = 64 FA = FB = (4) 2 + (37.5) 2 = 37.7 kN FO = 4 kN

Ans. F'A 4 kN A

F"A 37.5 kN

Ans.

32

F'O 4 kN

Ans. O

Bolt shear: As =

π(12) 2 = 113 mm2 4

37.7(10) 3 = 334 MPa Ans. τ= 113

32

B

F'B 4 kN

F"B 37.5 kN

Bearing on member: Ab = 12(8) = 96 mm2 σ =−

37.7(10) 3 = −393 MPa Ans. 96

Bending stress in plate:

b

3 bh 3 bd 3 bd 2 − −2 + a bd I = 12 12 12 8(12) 3 8(136) 3 8(12) 3 2 − −2 + (32) (8)(12) = 12 12 12 = 1.48(10) 6 mm4

Ans.

M = 12(200) = 2400 N · m σ =

Mc 2400(68) = (10) 3 = 110 MPa Ans. I 1.48(10) 6

a h a

d

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8-50

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M 16.5(300) 4950 lbf • in

F'A 150 lbf

300 lbf

16 1" 2

y

1 1" 2

A

F'B 150 lbf 1 1" 2

O V 300 lbf

F"A 1650 lbf

x

B

F"B 1650 lbf

4950 = 1650 lbf 3 FA = 1500 lbf, FB = 1800 lbf FA = FB =

Shear of bolt: π (0.5) 2 = 0.1963 in2 4 1800 F = = 9170 psi τ= A 0.1963

As =

Ssy = 0.577(92) = 53.08 kpsi

Bearing on bolt: 1 3 = 0.1875 in2 Ab = 2 8 σ =−

53.08 n= = 5.79 Ans. 9.17

n=

Bearing on members: S y = 54 kpsi, n =

1800 F =− = −9600 psi A 0.1875

92 = 9.58 Ans. 9.6

54 = 5.63 Ans. 9.6

Strength of members: Considering the right-hand bolt M = 300(15) = 4500 lbf · in I =

0.375(2) 3 0.375(0.5) 3 − = 0.246 in4 12 12

4500(1) Mc = = 18 300 psi I 0.246 54(10) 3 n= = 2.95 Ans. 18 300

1" 2

2"

σ =

8-51

3" 8

The direct shear load per bolt is F = 2500/6 = 417 lbf. The moment is taken only by the four outside bolts. This moment is M = 2500(5) = 12 500 lbf · in. Thus F =

12 500 = 1250 lbf and the resultant bolt load is 2(5) F = (417) 2 + (1250) 2 = 1318 lbf

Bolt strength, S y = 57 kpsi; Channel strength, S y = 46 kpsi; Plate strength, S y = 45.5 kpsi As = π(0.625) 2 /4 = 0.3068 in2

Shear of bolt: n=

Ssy (0.577)(57 000) = = 7.66 Ans. τ 1318/0.3068

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245

Chapter 8

Bearing on bolt: Channel thickness is t = 3/16 in; Ab = (0.625)(3/16) = 0.117 in2 ; n = Bearing on channel: Bearing on plate:

n=

57 000 = 5.07 Ans. 1318/0.117

46 000 = 4.08 Ans. 1318/0.117

Ab = 0.625(1/4) = 0.1563 in2 n=

45 500 = 5.40 Ans. 1318/0.1563

1" 4

Strength of plate: 0.25(7.5) 3 0.25(0.625) 3 − I = 12 12 1 5 0.25(0.625) 3 2 + (2.5) = 6.821 in4 −2 12 4 8

5 "D 8

5"

7

1" 2

M = 6250 lbf · in per plate 6250(3.75) Mc = = 3436 psi I 6.821 45 500 = 13.2 Ans. n= 3436

σ =

8-52

Specifying bolts, screws, dowels and rivets is the way a student learns about such components. However, choosing an array a priori is based on experience. Here is a chance for students to build some experience.

8-53

Now that the student can put an a priori decision of an array together with the specification of fasteners.

8-54

A computer program will vary with computer language or software application.

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Chapter 9 9-1 Eq. (9-3): F = 0.707hlτ = 0.707(5/16)(4)(20) = 17.7 kip Ans. 9-2 Table 9-6: τall = 21.0 kpsi f = 14.85h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64(4) = 18.56 kip Ans. 9-3 Table A-20: 1018 HR: Sut = 58 kpsi, S y = 32 kpsi 1018 CR: Sut = 64 kpsi, S y = 54 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τall = min(0.30Sut , 0.40S y ) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi for both materials. Eq. (9-3):

F = 0.707hlτall F = 0.707(5/16)(4)(12.8) = 11.3 kip Ans.

9-4 Eq. (9-3)

√ √ 2F 2(32) = = 18.1 kpsi τ= hl (5/16)(4)(2)

Ans.

9-5 b = d = 2 in F 1.414 7"

(a) Primary shear

Table 9-1 τ y =

F V = = 1.13F kpsi A 1.414(5/16)(2)

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247

Chapter 9

Secondary shear

Table 9-1 Ju =

2[(3)(22 ) + 22 ] d(3b2 + d 2 ) = = 5.333 in3 6 6

J = 0.707h Ju = 0.707(5/16)(5.333) = 1.18 in4 τx = τ y =

Mr y 7F(1) = = 5.93F kpsi J 1.18

Maximum shear τmax = τx2 + (τ y + τ y ) 2 = F 5.932 + (1.13 + 5.93) 2 = 9.22F kpsi F=

20 τall = = 2.17 kip Ans. 9.22 9.22

(1)

(b) For E7010 from Table 9-6, τall = 21 kpsi Table A-20: HR 1020 Bar:

Sut = 55 kpsi, S y = 30 kpsi

HR 1015 Support:

Sut = 50 kpsi, S y = 27.5 kpsi

Table 9-5, E7010 Electrode: Sut = 70 kpsi, S y = 57 kpsi Therefore, the bar controls the design. Table 9-4: τall = min[0.30(50), 0.40(27.5)] = min[15, 11] = 11 kpsi The allowable load from Eq. (1) is F=

11 τall = = 1.19 kip Ans. 9.22 9.22

9-6 b = d = 2 in F

7"

Primary shear τ y =

F V = = 0.566F A 4(0.707)(5/16)(2)

Secondary shear Table 9-1:

Ju =

(2 + 2) 3 (b + d) 3 = = 10.67 in3 6 6

J = 0.707h Ju = 0.707(5/16)(10.67) = 2.36 in4 τx = τ y =

Mr y (7F)(1) = = 2.97F J 2.36

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Maximum shear τmax

= τx2 + (τ y + τ y ) 2 = F 2.972 + (0.556 + 2.97) 2 = 4.61F kpsi

F=

τall 4.61

Ans.

which is twice τmax /9.22 of Prob. 9-5. 9-7 Weldment, subjected to alternating fatigue, has throat area of A = 0.707(6)(60 + 50 + 60) = 721 mm2 Members’ endurance limit: AISI 1010 steel Sut = 320 MPa,

Se = 0.504(320) = 161.3 MPa

ka = 272(320) −0.995 = 0.875 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 kf =

1 1 = 0.370 = Kfs 2.7

Sse = 0.875(1)(0.59)(0.37)(161.3) = 30.81 MPa Electrode’s endurance: 6010 Sut = 62(6.89) = 427 MPa Se = 0.504(427) = 215 MPa ka = 272(427) −0.995 = 0.657 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 k f = 1/K f s = 1/2.7 = 0.370 . Sse = 0.657(1)(0.59)(0.37)(215) = 30.84 MPa = 30.81 Thus, the members and the electrode are of equal strength. For a factor of safety of 1, Fa = τa A = 30.8(721)(10−3 ) = 22.2 kN Ans.

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Chapter 9

τ = 0

9-8 Primary shear

249

(why?)

Secondary shear Ju = 2πr 3 = 2π(4) 3 = 402 cm3

Table 9-1:

J = 0.707h Ju = 0.707(0.5)(402) = 142 cm4 M = 200F N · m ( F in kN) (200F)(4) Mr = = 2.82F (2 welds) 2J 2(142) τall 140 F = = = 49.2 kN Ans. τ 2.82

τ =

9-9

2 a 3 /12 a2 a Ju = = = 0.0833 fom = lh ah 12h h 2 a2 a(3a 2 + a 2 ) a = = 0.3333 fom = 6(2a)h 3h h 2 5a 2 a (2a) 4 − 6a 2 a 2 = = 0.2083 fom = 12(a + a)2ah 24h h 2 a4 11 a 2 a 1 8a 3 + 6a 3 + a 3 − = = 0.3056 fom = 3ah 12 2a + a 36 h h 2 8a 3 a2 a (2a) 3 1 = = = 0.3333 fom = 6h 4a 24ah 3h h 2 a3 a2 a 2π(a/2) 3 = = = 0.25 fom = πah 4ah 4h h

Rank 5

1

4

2

1

3

These rankings apply to fillet weld patterns in torsion that have a square area a × a in which to place weld metal. The object is to place as much metal as possible to the border. If your area is rectangular, your goal is the same but the rankings may change. Students will be surprised that the circular weld bead does not rank first. 9-10 1 Iu = fom = lh a

fom =

1 Iu = lh 2ah

fom =

1 Iu = lh 2ah

2 1 1 a2 a = = 0.0833 h 12 h h a3 a2 = 0.0833 6 h 2 2 2 1 a a a = = 0.25 2 4 h h

a3 12

5 5 1

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1 Iu = fom = lh [2(2a)]h

2 1 a2 a2 a (3a + a) = = 0.1667 6 6 h h

d2 a2 a = = b + 2d 3a 3 2 2a 3 2a 3 a3 2d 3 a2 a 2 a Iu = − 2d + (b + 2d) = − + 3a = 3 3 9 3 3 9 3 a 3 /3 1 a2 a2 Iu = = = 0.1111 fom = lh 3ah 9 h h 3 πa Iu = πr 3 = 8 2 πa 3 /8 a2 a Iu = = = 0.125 fom = lh πah 8h h x¯ =

b a = , 2 2

2

y¯ =

4

3

The CEE-section pattern was not ranked because the deflection of the beam is out-of-plane. If you have a square area in which to place a fillet weldment pattern under bending, your objective is to place as much material as possible away from the x-axis. If your area is rectangular, your goal is the same, but the rankings may change. 9-11

Materials: Attachment (1018 HR) S y = 32 kpsi, Sut = 58 kpsi S y = 36 kpsi, Sut ranges from 58 to 80 kpsi, use 58. Member (A36) The member and attachment are weak compared to the E60XX electrode. Decision Specify E6010 electrode Controlling property:

τall = min[0.3(58), 0.4(32)] = min(16.6, 12.8) = 12.8 kpsi

For a static load the parallel and transverse fillets are the same. If n is the number of beads, F = τall n(0.707)hl F 25 = 0.921 nh = = 0.707lτall 0.707(3)(12.8) τ=

Make a table. Number of beads n

Leg size h

1 2 3 4

0.921 0.460 → 1/2" 0.307 → 5/16" 0.230 → 1/4"

Decision: Specify 1/4" leg size Decision: Weld all-around

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Chapter 9

Weldment Specifications: Pattern: All-around square Electrode: E6010 Type: Two parallel fillets Two transverse fillets Length of bead: 12 in Leg: 1/4 in

251

Ans.

For a figure of merit of, in terms of weldbead volume, is this design optimal? 9-12

Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-9) and have thus reduced a synthesis problem to an analysis problem: A = 1.414hd = 1.414(h)(3) = 4.24h in3

Table 9-1: Primary shear

τ y =

3000 707 V = = A 4.24h h

Secondary shear Table 9-1:

Ju =

3[3(32 ) + 32 ] d(3b2 + d 2 ) = = 18 in3 6 6

J = 0.707(h)(18) = 12.7h in4 Mr y 3000(7.5)(1.5) 2657 = = = τ y J 12.7h h 1 4287 = τx2 + (τ y + τ y ) 2 = 26572 + (707 + 2657) 2 = h h

τx = τmax

Attachment (1018 HR): S y = 32 kpsi, Sut = 58 kpsi Member (A36): S y = 36 kpsi The attachment is weaker Decision: Use E60XX electrode τall = min[0.3(58), 0.4(32)] = 12.8 kpsi τmax = τall = h=

4287 = 12 800 psi h

4287 = 0.335 in 12 800

Decision: Specify 3/8" leg size Weldment Specifications: Pattern: Parallel fillet welds Electrode: E6010 Type: Fillet Length of bead: 6 in Leg size: 3/8 in

Ans.

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An optimal square space (3" × 3" ) weldment pattern is or or . In Prob. 9-12, there was roundup of leg size to 3/8 in. Consider the member material to be structural A36 steel. Decision: Use a parallel horizontal weld bead pattern for welding optimization and convenience. Materials: Attachment (1018 HR): S y = 32 kpsi, Sut = 58 kpsi Member (A36): S y = 36 kpsi, Sut 58–80 kpsi; use 58 kpsi From Table 9-4 AISC welding code, τall = min[0.3(58), 0.4(32)] = min(16.6, 12.8) = 12.8 kpsi Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Throat area and other properties: A = 1.414hd = 1.414(h)(3) = 4.24h in2 x¯ = b/2 = 3/2 = 1.5 in y¯ = d/2 = 3/2 = 1.5 in 3[3(32 ) + 32 ] d(3b2 + d 2 ) = = 18 in3 6 6 J = 0.707h Ju = 0.707(h)(18) = 12.73h in4

Ju =

Primary shear: τx =

3000 707.5 V = = A 4.24h h ␶x ␶y rx

r

ry

␶x

␶ y

x

Secondary shear: τ =

Mr J

Mr Mr x cos 45◦ = J J 2651 3000(6 + 1.5)(1.5) τx = = 12.73h h 2651 τ y = τx = h

τx = τ cos 45◦ =

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τmax

= (τx + τx ) 2 + τ y2 1 (2651 + 707.5) 2 + 26512 h 4279 psi = h =

Relate stress and strength: τmax = τall 4279 = 12 800 h 4279 h= = 0.334 in → 3/8 in 12 800 Weldment Specifications: Pattern: Horizontal parallel weld tracks Electrode: E6010 Type of weld: Two parallel fillet welds Length of bead: 6 in Leg size: 3/8 in Additional thoughts: Since the round-up in leg size was substantial, why not investigate a backward C weld pattern. One might then expect shorter horizontal weld beads which will have the advantage of allowing a shorter member (assuming the member has not yet been designed). This will show the inter-relationship between attachment design and supporting members. 9-14

Materials: S y = 36 kpsi, Sut = 58 to 80 kpsi; use Sut = 58 kpsi Member (A36): Attachment (1018 HR): S y = 32 kpsi, Sut = 58 kpsi τall = min[0.3(58), 0.4(32)] = 12.8 kpsi Decision: Use E6010 electrode. From Table 9-3: S y = 50 kpsi, Sut = 62 kpsi, τall = min[0.3(62), 0.4(50)] = 20 kpsi Decision: Since A36 and 1018 HR are weld metals to an unknown extent, use τall = 12.8 kpsi Decision: Use the most efficient weld pattern – square, weld-all-around. Choose6" × 6" size. Attachment length: l1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties: A = 1.414h(b + d) = 1.414(h)(6 + 6) = 17.0h x¯ =

6 b = = 3 in, 2 2

y¯ =

d 6 = = 3 in 2 2

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Primary shear τ y =

F 20 000 1176 V = = = psi A A 17h h

Secondary shear Ju =

(6 + 6) 3 (b + d) 3 = = 288 in3 6 6

J = 0.707h(288) = 203.6h in4 Mr y 20 000(6.25 + 3)(3) 2726 = = psi J 203.6h h 1 4760 psi = τx2 + (τ y + τ y ) 2 = 27262 + (2726 + 1176) 2 = h h

τx = τ y = τmax

Relate stress to strength τmax = τall 4760 = 12 800 h 4760 h= = 0.372 in 12 800 Decision: Specify 3/8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: 3/8 in Attachment length: 12.25 in 9-15

This is a good analysis task to test the students’ understanding (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d. (3) Find h and round and output all parameters on a single screen. Allow return to Step 1 or Step 2 (4) When the iteration is complete, the final display can be the bulk of your adequacy assessment. Such a program can teach too.

9-16

The objective of this design task is to have the students teach themselves that the weld patterns of Table 9-3 can be added or subtracted to obtain the properties of a comtemplated weld pattern. The instructor can control the level of complication. I have left the

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presentation of the drawing to you. Here is one possibility. Study the problem’s opportunities, then present this (or your sketch) with the problem assignment. Section AA b1

A 1" 2 8"

d

1018 HR A

A36 a

10000 lbf

8"

b

Body welds not shown

Attachment weld pattern considered

Use b1 as the design variable. Express properties as a function of b1 . From Table 9-3, category 3: A = 1.414h(b − b1 ) x¯ = b/2,

y¯ = d/2

(b − b1 )d 2 bd 2 b1 d 2 − = 2 2 2 I = 0.707h Iu F V = τ = A 1.414h(b − b1 ) Iu =

Fa(d/2) Mc = I 0.707h Iu = τ 2 + τ 2

τ = τmax Parametric study

Let a = 10 in, b = 8 in, d = 8 in, b1 = 2 in, τall = 12.8 kpsi, l = 2(8 − 2) = 12 in A = 1.414h(8 − 2) = 8.48h in2 Iu = (8 − 2)(82 /2) = 192 in3 I = 0.707(h)(192) = 135.7h in4 τ =

1179 10 000 = psi 8.48h h

2948 10 000(10)(8/2) = psi 135.7h h 1 3175 = 12 800 = 11792 + 29482 = h h

τ = τmax

from which h = 0.248 in. Do not round off the leg size – something to learn. fom =

192 Iu = = 64.5 hl 0.248(12)

A = 8.48(0.248) = 2.10 in2 I = 135.7(0.248) = 33.65 in4

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vol = I = vol τ = τ = τmax =

0.2482 h2 l= 12 = 0.369 in3 2 2 33.65 = 91.2 = eff 0.369 1179 = 4754 psi 0.248 2948 = 11 887 psi 0.248 4127 . = 12 800 psi 0.248

Now consider the case of uninterrupted welds, b1 = 0 A = 1.414(h)(8 − 0) = 11.31h Iu = (8 − 0)(82 /2) = 256 in3 I = 0.707(256)h = 181h in4 884 10 000 = 11.31h h 2210 10 000(10)(8/2) τ = = 181h h 1 2380 τmax = = τall 8842 + 22102 = h h τmax 2380 = 0.186 in h= = τall 12 800 τ =

Do not round off h. A = 11.31(0.186) = 2.10 in2 I = 181(0.186) = 33.67 0.1862 884 = 4753 psi, vol = 16 = 0.277 in3 τ = 0.186 2 2210 = 11 882 psi τ = 0.186 256 Iu fom = = = 86.0 hl 0.186(16) I 33.67 eff = 2 = = 121.7 (h /2)l (0.1862 /2)16 Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ . To meet the shortened bead length, h is increased proportionately. However, volume of bead laid down increases as h 2 . The uninterrupted bead is superior. In this example, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the next larger standard weld fillet size will decrease the merit.

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Had the weld bead gone around the corners, the situation would change. Here is a followup task analyzing an alternative weld pattern. b1

d1

d

b

9-17

From Table 9-2 A = 1.414h(b + d)

For the box

Subtracting b1 from b and d1 from d A = 1.414 h(b − b1 + d − d1 ) d 3 b1 d 2 d2 (3b + d) − 1 − 6 6 2 1 1 = (b − b1 )d 2 + d 3 − d13 2 6

Iu =

length of bead

l = 2(b − b1 + d − d1 ) fom = Iu / hl

9-18

Below is a Fortran interactive program listing which, if imitated in any computer language of convenience, will reduce to drudgery. Furthermore, the program allows synthesis by interaction or learning without fatigue. C Weld2.f for rect. fillet beads resisting bending C. Mischke Oct 98 1 print*,’weld2.f rectangular fillet weld-beads in bending,’ print*,’gaps allowed - C. Mischke Oct. 98’ print*,’ ’ print*,’Enter largest permissible shear stress tauall’ read*,tauall print*,’Enter force F and clearance a’ read*,F,a 2 print*,’Enter width b and depth d of rectangular pattern’ read*,b,d xbar=b/2. ybar=d/2. 3 print*,’Enter width of gap b1, and depth of gap d1’ print*,’both gaps central in their respective sides’ read*,b1,d1 xIu=(b-b1)*d**2/2.+(d**3-d1**3)/6. xl=2.*(d-d1)+2.*(b-b1)

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C Following calculations based on unit leg h = 1 AA=1.414*(b-b1+d-d1) xI=0.707*xIu tau2=F*a*d/2./xI tau1=F/AA taumax=sqrt(tau2**2+tau1**2) h=taumax/tauall C Adjust parameters for now-known h AA=AA*h xI=xI*h tau2=tau2/h tau1=tau1/h taumax=taumax/h fom=xIu/h/xl print*,’F=’,F,’ a=’,a,’ bead length = ’,xl print*,’b=’,b,’ b1=’,b1,’ d=’,d,’ d1=’,d1 print*,’xbar=’,xbar,’ ybar=’,ybar,’ Iu=’,xIu print*,’I=’,xI,’ tau2=’,tau2,’ tau1=’,tau1 print*,’taumax=’,taumax,’ h=’,h,’ A=’,AA print*,’fom=’,fom, ’I/(weld volume)=’,2.*xI/h**2/xl print*,’ ’ print*,’To change b1,d1: 1; b,d: 2; New: 3; Quit: 4’ read*,index go to (3,2,1,4), index 4 call exit end

9-19

τall = 12 800 psi. Use Fig. 9-17(a) for general geometry, but employ beads. Horizontal parallel weld bead pattern 6"

8"

b = 6 in d = 8 in

From Table 9-2, category 3 A = 1.414 hb = 1.414(h)(6) = 8.48 h in2 x¯ = b/2 = 6/2 = 3 in, y¯ = d/2 = 8/2 = 4 in 6(8) 2 bd 2 = = 192 in3 Iu = 2 2 I = 0.707h Iu = 0.707(h)(192) = 135.7h in4 τ =

1179 10 000 = psi 8.48h h

beads and then

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Chapter 9

10 000(10)(8/2) 2948 Mc = = psi I 135.7h h 1 3175 psi = τ 2 + τ 2 = (11792 + 29482 ) 1/2 = h h

τ = τmax

Equate the maximum and allowable shear stresses. τmax = τall =

3175 = 12 800 h

from which h = 0.248 in. It follows that I = 135.7(0.248) = 33.65 in4 The volume of the weld metal is 0.2482 (6 + 6) h 2l = = 0.369 in3 vol = 2 2 The effectiveness, (eff) H , is 33.65 I = = 91.2 in vol 0.369 192 Iu = = 64.5 in (fom ) H = hl 0.248(6 + 6) Vertical parallel weld beads (eff) H =

6" 8"

b = 6 in d = 8 in

From Table 9-2, category 2 A = 1.414hd = 1.414(h)(8) = 11.31h in2 x¯ = b/2 = 6/2 = 3 in,

y¯ = d/2 = 8/2 = 4 in

83 d3 = = 85.33 in3 Iu = 6 6 I = 0.707h Iu = 0.707(h)(85.33) = 60.3h τ =

10 000(10)(8/2) 6633 Mc = = psi I 60.3 h h 1 = τ 2 + τ 2 = (8842 + 66332 ) 1/2 h 6692 psi = h

τ = τmax

884 10 000 = psi 11.31h h

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Equating τmax to τall gives h = 0.523 in. It follows that I = 60.3(0.523) = 31.5 in4 0.5232 h 2l = (8 + 8) = 2.19 in3 2 2 31.6 I = = 14.4 in (eff) V = vol 2.19 85.33 Iu (fom ) V = = = 10.2 in hl 0.523(8 + 8) vol =

The ratio of (eff) V /(eff) H is 14.4/91.2 = 0.158. The ratio (fom ) V /(fom ) H is 10.2/64.5 = 0.158. This is not surprising since eff =

I I 0.707 h Iu Iu = 2 = = 1.414 = 1.414 fom 2 vol (h /2)l (h /2)l hl

The ratios (eff) V /(eff) H and (fom ) V /(fom ) H give the same information. 9-20

Because the loading is pure torsion, there is no primary shear. From Table 9-1, category 6: Ju = 2πr 3 = 2π(1) 3 = 6.28 in3 J = 0.707 h Ju = 0.707(0.25)(6.28) = 1.11 in4 τ=

20(1) Tr = = 18.0 kpsi Ans. J 1.11

h = 0.375 in,

9-21

d = 8 in,

b = 1 in

From Table 9-2, category 2: A = 1.414(0.375)(8) = 4.24 in2 Iu =

83 d3 = = 85.3 in3 6 6

I = 0.707h Iu = 0.707(0.375)(85.3) = 22.6 in4 τ =

5 F = = 1.18 kpsi A 4.24

M = 5(6) = 30 kip · in c = (1 + 8 + 1 − 2)/2 = 4 in 30(4) Mc = = 5.31 kpsi I 22.6 = τ 2 + τ 2 = 1.182 + 5.312

τ = τmax

= 5.44 kpsi Ans.

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h = 0.6 cm,

9-22

b = 6 cm,

d = 12 cm.

Table 9-3, category 5: A = 0.707h(b + 2d)

6 B

= 0.707(0.6)[6 + 2(12)] = 12.7 cm2

4.8 G

y¯ =

7.2

Iu =

A

122 d2 = = 4.8 cm b + 2d 6 + 2(12) 2d 3 − 2d 2 y¯ + (b + 2d) y¯ 2 3

2(12)3 − 2(122 )(4.8) + [6 + 2(12)]4.82 = 3 = 461 cm3 I = 0.707h Iu = 0.707(0.6)(461) = 196 cm4 τ =

7.5(103 ) F = = 5.91 MPa A 12.7(102 )

M = 7.5(120) = 900 N · m c A = 7.2 cm,

c B = 4.8 cm

The critical location is at A. 900(7.2) Mc A = = 33.1 MPa I 196 = τ 2 + τ 2 = (5.912 + 33.12 ) 1/2 = 33.6 MPa

τ A = τmax

n=

9-23

120 τall = 3.57 Ans. = τmax 33.6

The largest possible weld size is 1/16 in. This is a small weld and thus difficult to accomplish. The bracket’s load-carrying capability is not known. There are geometry problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern – Table 9-2, category 6: A = 1.414 h(b + d) = 1.414(1/16)(1 + 7.5) = 0.751 in2 7.5"

x¯ = b/2 = 0.5 in y¯ = 1"

7.5 d = = 3.75 in 2 2

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7.52 d2 [3(1) + 7.5] = 98.4 in3 Iu = (3b + d) = 6 6 I = 0.707h Iu = 0.707(1/16)(98.4) = 4.35 in4 M = (3.75 + 0.5)W = 4.25W W V = = 1.332W A 0.751 4.25W (7.5/2) Mc = = 3.664W τ = I 4.35 τmax = τ 2 + τ 2 = W 1.3322 + 3.6642 = 3.90W τ =

Material properties: The allowable stress given is low. Let’s demonstrate that. For the A36 structural steel member, S y = 36 kpsi and Sut = 58 kpsi. For the 1020 CD attachment, use HR properties of S y = 30 kpsi and Sut = 55. The E6010 electrode has strengths of S y = 50 and Sut = 62 kpsi. Allowable stresses: A36:

τall = min[0.3(58), 0.4(36)] = min(17.4, 14.4) = 14.4 kpsi

1020:

τall = min[0.3(55), 0.4(30)] τall = min(16.5, 12) = 12 kpsi

E6010:

τall = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi

Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is τall = min(14.4, 12, 18.0) = 12 kpsi However, the allowable stress in the problem statement is 0.9 kpsi which is low from the weldment perspective. The load associated with this strength is τmax = τall = 3.90W = 900 W =

900 = 231 lbf 3.90

If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is 12 000 psi and the load W = 3047 lbf. Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carrying capability of the top bend is unknown. These uncertainties may require the use of a different weld pattern. Our solution provides the best weldment and thus insight for comparing a welded joint to one which employs screw fasteners.

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Chapter 9

9-24 y x

F

FB

x

RA

60

A RyA B

F = 100 lbf,

τall = 3 kpsi

FB = 100(16/3) = 533.3 lbf FBx = −533.3 cos 60◦ = −266.7 lbf FB = −533.3 cos 30◦ = −462 lbf y

y

It follows that R A = 562 lbf and R xA = 266.7 lbf, R A = 622 lbf M = 100(16) = 1600 lbf · in 100 462 266.7 16

3

266.7

562

The OD of the tubes is 1 in. From Table 9-1, category 6: A = 1.414(π hr)(2) = 2(1.414)(πh)(1/2) = 4.44h in2 Ju = 2πr 3 = 2π(1/2)3 = 0.785 in3 J = 2(0.707)h Ju = 1.414(0.785)h = 1.11h in4 622 140 V = = A 4.44h h Mc 1600(0.5) 720.7 Tc = = = τ = J J 1.11h h τ =

The shear stresses, τ and τ , are additive algebraically 861 1 (140 + 720.7) = psi h h 861 τmax = τall = = 3000 h 861 h= = 0.287 → 5/16" 3000 τmax =

Decision: Use 5/16 in fillet welds Ans.

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9-25 y 1" 4 1" 4B

g

g G

g

g 9"

3" 8 3" 8

x

7"

For the pattern in bending shown, find the centroid G of the weld group. 6(0.707)(1/4)(3) + 6(0.707)(3/8)(13) 6(0.707)(1/4) + 6(0.707)(3/8) = 9 in = 2 IG + A2x¯ 0.707(1/4)(63 ) 2 + 0.707(1/4)(6)(6 ) =2 12

x¯ =

I1/4

= 82.7 in4 0.707(3/8)(63 ) 2 + 0.707(3/8)(6)(4 ) =2 12

I3/8

= 60.4 in4 I = I1/4 + I3/8 = 82.7 + 60.4 = 143.1 in4 The critical location is at B. From Eq. (9-3), F = 0.189F 2[6(0.707)(3/8 + 1/4)] (8F)(9) Mc τ = = = 0.503F I 143.1 τmax = τ 2 + τ 2 = F 0.1892 + 0.5032 = 0.537F τ =

Materials: A36 Member: S y = 36 kpsi 1015 HR Attachment: S y = 27.5 kpsi E6010 Electrode: S y = 50 kpsi τall = 0.577 min(36, 27.5, 50) = 15.9 kpsi F= 9-26

15.9/2 τall /n = = 14.8 kip Ans. 0.537 0.537

Figure P9-26b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. M = 1200(0.366) = 439 lbf · in Ans. (a) Fy = 1200 sin 30◦ = 600 lbf Ans. (b) Fx = 1200 cos 30◦ = 1039 lbf Ans. (c)

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Chapter 9

(d) From Table 9-2, category 6: A = 1.414(0.25)(0.25 + 2.5) = 0.972 in2 2.52 d2 (3b + d) = [3(0.25) + 2.5] = 3.39 in3 6 6 The second area moment about an axis through G and parallel to z is Iu =

I = 0.707h Iu = 0.707(0.25)(3.39) = 0.599 in4

Ans.

(e) Refer to Fig. P.9-26b. The shear stress due to Fy is τ1 =

Fy 600 = = 617 psi A 0.972

The shear stress along the throat due to Fx is τ2 =

1039 Fx = = 1069 psi A 0.972

The resultant of τ1 and τ2 is in the throat plane 1/2 τ = τ12 + τ22 = (6172 + 10692 ) 1/2 = 1234 psi The bending of the throat gives τ =

439(1.25) Mc = = 916 psi I 0.599

The maximum shear stress is τmax = (τ 2 + τ 2 ) 1/2 = (12342 + 9162 ) 1/2 = 1537 psi Ans. (f) Materials: 1018 HR Member: S y = 32 kpsi, Sut = 58 kpsi (Table A-20) S y = 50 kpsi (Table 9-3) E6010 Electrode: n=

Ssy 0.577S y 0.577(32) = 12.0 Ans. = = τmax τmax 1.537

(g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. . A1 = bh = 0.25(2.5) = 0.625 in2 τx y =

Fx 1039 = 1662 psi = A1 0.625

bd 2 0.25(2.5) 2 I = = = 0.260 in3 c 6 6 At location A σy =

Fy M + A1 I /c

σy =

439 600 + = 2648 psi 0.625 0.260

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The von Mises stress σ is 1/2 σ = σ y2 + 3τx2y = [26482 + 3(1662) 2 ]1/2 = 3912 psi Thus, the factor of safety is, n=

Sy 32 = 8.18 Ans. = σ 3.912

The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole −1200 F = = −9600 psi σ = td 0.25(0.50) n=−

Sy 32(103 ) = 3.33 Ans. = − σ −9600

Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 with Sut = 58 kpsi Se = 0.504Sut = 0.504(58) = 29.2 kpsi Table 7-4:

ka = 14.4(58) −0.718 = 0.780

For the size factor estimate, we first employ Eq. (7-24) for the equivalent diameter. √ de = 0.808 0.707hb = 0.808 0.707(2.5)(0.25) = 0.537 in Eq. (7-19) is used next to find kb de −0.107 0.537 −0.107 kb = = = 0.940 0.30 0.30 The load factor for shear kc , is kc = 0.59 The endurance strength in shear is Sse = 0.780(0.940)(0.59)(29.2) = 12.6 kpsi From Table 9-5, the shear stress-concentration factor is K f s = 2.7. The loading is repeatedly-applied. τa = τm = k f

1.537 τmax = 2.7 = 2.07 kpsi 2 2

Table 7-10: Gerber factor of safety n f , adjusted for shear, with Ssu = 0.67Sut   2  2  1 0.67(58) 2.07 2(2.07)(12.6) −1 + 1 + = 5.55 Ans. nf = 2 2.07 12.6  0.67(58)(2.07)  Attachment metal should be checked for bending fatigue.

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9-27

267

Use b = d = 4 in. Since h = 5/8 in, the primary shear is F = 0.283F 1.414(5/8)(4) The secondary shear calculations, for a moment arm of 14 in give τ =

Ju =

4[3(42 ) + 42 ] = 42.67 in3 6

J = 0.707h Ju = 0.707(5/8)42.67 = 18.9 in4 Mr y 14F(2) = = 1.48F J 18.9

τx = τ y =

Thus, the maximum shear and allowable load are: τmax = F 1.482 + (0.283 + 1.48) 2 = 2.30F F=

20 τall = = 8.70 kip Ans. 2.30 2.30

From Prob. 9-5b, τall = 11 kpsi 11 τall = = 4.78 kip 2.30 2.30 The allowable load has thus increased by a factor of 1.8 Ans. Fall =

9-28

Purchase the hook having the design shown in Fig. P9-28b. Referring to text Fig. 9-32a, this design reduces peel stresses.

9-29

(a) 1 τ¯ = l

l/2

−l/2

= A1

Pω cosh(ωx) dx 4b sinh(ωl/2)

l/2

−l/2

cosh(ωx) dx

=

l/2 A1 sinh(ωx) −l/2 ω

=

A1 [sinh(ωl/2) − sinh(−ωl/2)] ω

=

A1 [sinh(ωl/2) − (−sinh(ωl/2))] ω

=

2A1 sinh(ωl/2) ω

=

Pω [2 sinh(ωl/2)] 4bl sinh(ωl/2) P Ans. τ¯ = 2bl

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(b) (c)

τ (l/2) =

Pω Pω cosh(ωl/2) = 4b sinh(ωl/2) 4b tanh(ωl/2)

Pω τ (l/2) = K = τ¯ 4b sinh(ωl/2) K =

ωl/2 tanh(ωl/2)

2bl P

Ans.

Ans.

For computer programming, it can be useful to express the hyperbolic tangent in terms of exponentials: K = 9-30

ωl exp(ωl/2) − exp(−ωl/2) 2 exp(ωl/2) + exp(−ωl/2)

Ans.

This is a computer programming exercise. All programs will vary.

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Chapter 10 10-1

1" 2

1" 4"

10-2

1" 2

1" 4"

A = Sd m dim( Auscu ) = dim(S) dim(d m ) = kpsi · inm dim( ASI ) = dim(S1 ) dim d1m = MPa · mmm MPa mmm . · m Auscu = 6.894 757(25.40) m Auscu = 6.895(25.4) m Auscu ASI = kpsi in For music wire, from Table 10-4: Auscu = 201, m = 0.145; what is ASI ? ASI = 6.89(25.4) 0.145 (201) = 2214 MPa · mmm

10-3

Ans.

Given: Music wire, d = 0.105 in, OD = 1.225 in, plain ground ends, Nt = 12 coils. Na = Nt − 1 = 12 − 1 = 11 L s = d Nt = 0.105(12) = 1.26 in

Table 10-1:

A = 201, m = 0.145 201 Sut = = 278.7 kpsi (0.105) 0.145

Table 10-4: (a) Eq. (10-14):

Ssy = 0.45(278.7) = 125.4 kpsi D = 1.225 − 0.105 = 1.120 in D 1.120 C= = = 10.67 d 0.105

Table 10-6:

Eq. (10-6): Eq. (10-3): Eq. (10-9):

KB = F| Ssy

4(10.67) + 2 = 1.126 4(10.67) − 3

πd 3 Ssy π(0.105) 3 (125.4)(103 ) = = 45.2 lbf = 8K B D 8(1.126)(1.120)

(0.105) 4 (11.75)(106 ) d4G = 11.55 lbf/in = 8D 3 Na 8(1.120) 3 (11) F| Ssy 45.2 + Ls = + 1.26 = 5.17 in Ans. L0 = k 11.55 k=

Ans.

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270 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

(b) F| Ssy = 45.2 lbf Ans. (c) k = 11.55 lbf/in Ans. 2.63(1.120) 2.63D = = 5.89 in (d) (L 0 ) cr = α 0.5 Many designers provide (L 0 ) cr /L 0 ≥ 5 or more; therefore, plain ground ends are not often used in machinery due to buckling uncertainty. 10-4

Referring to Prob. 10-3 solution, C = 10.67, Na = 11, Ssy = 125.4 kpsi, (L 0 ) cr = 5.89 in and F = 45.2 lbf (at yield). 4 ≤ C ≤ 12 C = 10.67 O.K. Eq. (10-18): Na = 11 O.K. 3 ≤ Na ≤ 15 Eq. (10-19): y1

F1

ys

L0

Fs L1 Ls

L 0 = 5.17 in, y1 =

L s = 1.26 in

30 F1 = = 2.60 in k 11.55

L 1 = L 0 − y1 = 5.17 − 2.60 = 2.57 in ξ=

5.17 − 1.26 ys − 1 = 0.50 −1= y1 2.60

ξ ≥ 0.15, ξ = 0.50 O.K. Eq. (10-20): From Eq. (10-3) for static service 8F1 D 8(30)(1.120) τ1 = K B = 1.126 = 83 224 psi πd 3 π(0.105) 3 Ssy 125.4(103 ) ns = = 1.51 = τ1 83 224 n s ≥ 1.2, n s = 1.51 O.K. Eq. (10-21): 45.2 45.2 = 83 224 = 125 391 psi τs = τ1 30 30 . Ssy /τs = 125.4(103 )/125 391 = 1 Ssy /τs ≥ (n s ) d : Not solid-safe. Not O.K. L 0 ≤ (L 0 ) cr :

5.17 ≤ 5.89

Margin could be higher,

Design is unsatisfactory. Operate over a rod?

Ans.

Not O.K.

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Chapter 10

10-5

Static service spring with: HD steel wire, d = 2 mm, OD = 22 mm, Nt = 8.5 turns plain and ground ends. Preliminaries A = 1783 MPa · mmm , m = 0.190 Table 10-5: 1783 = 1563 MPa Sut = Eq. (10-14): (2) 0.190 Ssy = 0.45(1563) = 703.4 MPa Table 10-6: Then, D = OD − d = 22 − 2 = 20 mm C = 20/2 = 10 KB =

4(10) + 2 4C + 2 = = 1.135 4C − 3 4(10) − 3

Na = 8.5 − 1 = 7.5 turns L s = 2(8.5) = 17 mm Eq. (10-21): Use (n s ) d = 1.2 for solid-safe property. πd 3 Ssy /n d π(2) 3 (703.4/1.2) (10−3 ) 3 (106 ) = = 81.12 N Fs = 8K B D 8(1.135)(20) 10−3 (2) 4 (79.3) (10−3 ) 4 (109 ) d4G = 0.002 643(106 ) = 2643 N/m = k= 8D 3 Na 8(20) 3 (7.5) (10−3 ) 3 ys =

81.12 Fs = = 30.69 mm k 2643(10−3 )

(a) L 0 = y + L s = 30.69 + 17 = 47.7 mm Ans. L0 47.7 = 5.61 mm Ans. = (b) Table 10-1: p = Nt 8.5 (c) Fs = 81.12 N (from above) Ans. (d) k = 2643 N/m (from above) Ans. (e) Table 10-2 and Eq. (10-13): 2.63(20) 2.63D = = 105.2 mm α 0.5 (L 0 ) cr /L 0 = 105.2/47.7 = 2.21 (L 0 ) cr =

This is less than 5. Operate over a rod? Plain and ground ends have a poor eccentric footprint. 10-6

Ans.

Referring to Prob. 10-5 solution: C = 10, Na = 7.5, k = 2643 N/m, d = 2 mm, D = 20 mm, Fs = 81.12 N and Nt = 8.5 turns. 4 ≤ C ≤ 12, C = 10 O.K. Eq. (10-18):

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3 ≤ Na ≤ 15,

Eq. (10-19):

Na = 7.5 O.K.

75 F1 = = 28.4 mm k 2643(10−3 ) 81.12(1.2) ( y) for yield = = 36.8 mm 2643(10−3 ) 81.12 = 30.69 mm ys = 2643(10−3 ) ( y) for yield 36.8 ξ= − 1 = 0.296 −1= y1 28.4 ξ ≥ 0.15, ξ = 0.296 O.K. Ssy = 0.45Sut O.K. y1 =

Eq. (10-20): Table 10-6: As-wound τs = K B

8Fs D πd 3

8(81.12)(20) = 1.135 π(2) 3

10−3 = 586 MPa (10−3 ) 3 (106 )

Ssy 703.4 = 1.2 O.K. (Basis for Prob. 10-5 solution) = τs 586 L s = Nt d = 8.5(2) = 17 mm Table 10-1: Fs 81.12 + Ls = + 17 = 47.7 mm L0 = k 2.643 2.63(20) 2.63D = = 105.2 mm α 0.5 (L 0 ) cr 105.2 = 2.21 = L0 47.7 which is less than 5. Operate over a rod? Not O.K. Eq. (10-21):

Plain and ground ends have a poor eccentric footprint. 10-7

Ans.

Given: A228 (music wire), SQ&GRD ends, d = 0.006 in, OD = 0.036 in, L 0 = 0.63 in, Nt = 40 turns. Table 10-4:

Table 10-1:

A = 201 kpsi · inm , m = 0.145 D = OD − d = 0.036 − 0.006 = 0.030 in C = D/d = 0.030/0.006 = 5 4(5) + 2 = 1.294 KB = 4(5) − 3 Na = Nt − 2 = 40 − 2 = 38 turns Sut =

201 = 422.1 kpsi (0.006) 0.145

Ssy = 0.45(422.1) = 189.9 kpsi Gd 4 12(106 )(0.006) 4 = 1.895 lbf/in k= = 8D 3 Na 8(0.030) 3 (38)

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Chapter 10

L s = Nt d = 40(0.006) = 0.240 in

Table 10-1:

Now Fs = kys where ys = L 0 − L s = 0.390 in. Thus, 8(kys ) D 8(1.895)(0.39)(0.030) = 1.294 (10−3 ) = 338.2 kpsi τs = K B πd 3 π(0.006) 3

(1)

τs > Ssy , that is, 338.2 > 189.9 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(189 900/1.2)(π)(0.006) 3 (τs /n)(πd 3 ) = = 0.182 in 8K B k D 8(1.294)(1.895)(0.030)

Using a design factor of 1.2, L 0 = L s + ys = 0.240 + 0.182 = 0.422 in The spring should be wound to a free length of 0.422 in. 10-8

Ans.

Given: B159 (phosphor bronze), SQ&GRD ends, d = 0.012 in, OD = 0.120 in, L 0 = 0.81 in, Nt = 15.1 turns. m=0

Table 10-4:

A = 145 kpsi · inm ,

Table 10-5:

G = 6 Mpsi D = OD − d = 0.120 − 0.012 = 0.108 in C = D/d = 0.108/0.012 = 9 KB =

4(9) + 2 = 1.152 4(9) − 3

Na = Nt − 2 = 15.1 − 2 = 13.1 turns

Table 10-1:

Sut =

145 = 145 kpsi 0.0120

Ssy = 0.35(145) = 50.8 kpsi

Table 10-6:

k=

6(106 )(0.012) 4 Gd 4 = 0.942 lbf/in = 8D 3 Na 8(0.108) 3 (13.1)

L s = d Nt = 0.012(15.1) = 0.181 in

Table 10-1:

Now Fs = kys , ys = L 0 − L s = 0.81 − 0.181 = 0.629 in 8(kys ) D 8(0.942)(0.6)(0.108) = 1.152 (10−3 ) = 108.6 kpsi τs = K B πd 3 π(0.012) 3

(1)

τs > Ssy , that is, 108.6 > 50.8 kpsi; the spring is not solid safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)πd 3 (50.8/1.2)(π)(0.012) 3 (103 ) = = 0.245 in 8K B k D 8(1.152)(0.942)(0.108)

L 0 = L s + ys = 0.181 + 0.245 = 0.426 in Wind the spring to a free length of 0.426 in.

Ans.

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10-9

Given: A313 (stainless steel), SQ&GRD ends, d = 0.040 in, OD = 0.240 in, L 0 = 0.75 in, Nt = 10.4 turns. Table 10-4:


Table 10-5:

G = 10(106 ) psi

m = 0.146

D = OD − d = 0.240 − 0.040 = 0.200 in C = D/d = 0.200/0.040 = 5 KB =

4(5) + 2 = 1.294 4(5) − 3

Table 10-6:

Na = Nt − 2 = 10.4 − 2 = 8.4 turns 169 Sut = = 270.4 kpsi (0.040) 0.146

Table 10-13:

Ssy = 0.35(270.4) = 94.6 kpsi k=

10(106 )(0.040) 4 Gd 4 = 47.62 lbf/in = 8D 3 Na 8(0.2) 3 (8.4)

L s = d Nt = 0.040(10.4) = 0.416 in

Table 10-6:

Now Fs = kys , ys = L 0 − L s = 0.75 − 0.416 = 0.334 in 8(kys ) D 8(47.62)(0.334)(0.2) τs = K B = 1.294 (10−3 ) = 163.8 kpsi 3 3 πd π(0.040)

(1)

τs > Ssy , that is, 163.8 > 94.6 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (94 600/1.2)(π)(0.040) 3 = = 0.161 in 8K B k D 8(1.294)(47.62)(0.2)

L 0 = L s + ys = 0.416 + 0.161 = 0.577 in Wind the spring to a free length 0.577 in. 10-10

Ans.

Given: A227 (hard drawn steel), d = 0.135 in, OD = 2.0 in, L 0 = 2.94 in, Nt = 5.25 turns. Table 10-4:


Table 10-5:

G = 11.4(106 ) psi

m = 0.190

D = OD − d = 2 − 0.135 = 1.865 in C = D/d = 1.865/0.135 = 13.81 KB =

4(13.81) + 2 = 1.096 4(13.81) − 3

Na = Nt − 2 = 5.25 − 2 = 3.25 turns Sut =

140 = 204.8 kpsi (0.135) 0.190

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Chapter 10

Table 10-6:

Ssy = 0.45(204.8) = 92.2 kpsi k=

Gd 4 11.4(106 )(0.135) 4 = 22.45 lbf/in = 8D 3 Na 8(1.865) 3 (3.25)

L s = d Nt = 0.135(5.25) = 0.709 in

Table 10-1:

Now Fs = kys , ys = L 0 − L s = 2.94 − 0.709 = 2.231 in 8(kys ) D 8(22.45)(2.231)(1.865) = 1.096 (10−3 ) = 106.0 kpsi (1) τs = K B πd 3 π(0.135) 3 τs > Ssy , that is, 106 > 92.2 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (92 200/1.2)(π)(0.135) 3 = = 1.612 in 8K B k D 8(1.096)(22.45)(1.865)

L 0 = L s + ys = 0.709 + 1.612 = 2.321 in Wind the spring to a free length of 2.32 in. 10-11

Ans.

Given: A229 (OQ&T steel), SQ&GRD ends, d = 0.144 in, OD = 1.0 in, L 0 = 3.75 in, Nt = 13 turns. A = 147 kpsi · inm , m = 0.187 Table 10-4: Table 10-5:

G = 11.4(106 ) psi D = OD − d = 1.0 − 0.144 = 0.856 in C = D/d = 0.856/0.144 = 5.944 4(5.944) + 2 KB = = 1.241 4(5.944) − 3

Table 10-1:

Na = Nt − 2 = 13 − 2 = 11 turns 147 Sut = = 211.2 kpsi (0.144) 0.187

Table 10-6:

Ssy = 0.50(211.2) = 105.6 kpsi Gd 4 11.4(106 )(0.144) 4 k= = 88.8 lbf/in = 8D 3 Na 8(0.856) 3 (11)

Table 10-1:

L s = d Nt = 0.144(13) = 1.872 in

Now Fs = kys , ys = L 0 − L s = 3.75 − 1.872 = 1.878 in 8(kys ) D 8(88.8)(1.878)(0.856) = 1.241 (10−3 ) = 151.1 kpsi τs = K B 3 3 πd π(0.144)

(1)

τs > Ssy , that is,151.1 > 105.6 kpsi; the spring is not solid-safe. Solving Eq. (1) for ys gives (Ssy /n)(πd 3 ) (105 600/1.2)(π)(0.144) 3 = = 1.094 in ys = 8K B k D 8(1.241)(88.8)(0.856) L 0 = L s + ys = 1.878 + 1.094 = 2.972 in Wind the spring to a free length 2.972 in.

Ans.

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10-12

Given: A232 (Cr-V steel), SQ&GRD ends, d = 0.192 in, OD = 3 in, L 0 = 9 in, Nt = 8 turns. m = 0.168

Table 10-4:


Table 10-5:

G = 11.2(106 ) psi D = OD − d = 3 − 0.192 = 2.808 in C = D/d = 2.808/0.192 = 14.625 KB =

4(14.625) + 2 = 1.090 4(14.625) − 3

Na = Nt − 2 = 8 − 2 = 6 turns

Table 10-1:

Sut =

169 = 223.0 kpsi (0.192) 0.168

Ssy = 0.50(223.0) = 111.5 kpsi

Table 10-6:

k=

11.2(106 )(0.192) 4 Gd 4 = 14.32 lbf/in = 8D 3 Na 8(2.808) 3 (6)

L s = d Nt = 0.192(8) = 1.536 in

Table 10-1:

Now Fs = kys , ys = L 0 − L s = 9 − 1.536 = 7.464 in 8(kys ) D 8(14.32)(7.464)(2.808) = 1.090 (10−3 ) = 117.7 kpsi τs = K B πd 3 π(0.192) 3

(1)

τs > Ssy , that is,117.7 > 111.5 kpsi; the spring is not solid safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (111 500/1.2)(π)(0.192) 3 = = 5.892 in 8K B k D 8(1.090)(14.32)(2.808)

L 0 = L s + ys = 1.536 + 5.892 = 7.428 in Wind the spring to a free length of 7.428 in. 10-13

Ans.

Given: A313 (stainless steel) SQ&GRD ends, d = 0.2 mm, OD = 0.91 mm, L 0 = 15.9 mm, Nt = 40 turns. Table 10-4:

A = 1867 MPa · mmm , m = 0.146

Table 10-5:

G = 69.0 GPa D = OD − d = 0.91 − 0.2 = 0.71 mm C = D/d = 0.71/0.2 = 3.55 KB =

4(3.55) + 2 = 1.446 4(3.55) − 3

Na = Nt − 2 = 40 − 2 = 38 turns Sut =

1867 = 2361.5 MPa (0.2) 0.146

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Chapter 10

277

Table 10-6: Ssy = 0.35(2361.5) = 826.5 MPa (0.2) 4 (69.0) (10−3 ) 4 (109 ) d4G = k= 8D 3 Na 8(0.71) 3 (38) (10−3 ) 3 = 1.0147(10−3 )(106 ) = 1014.7 N/m or 1.0147 N/mm L s = d Nt = 0.2(40) = 8 mm Fs = kys ys = L 0 − L s = 15.9 − 8 = 7.9 8(kys ) D 8(1.0147)(7.9)(0.71) 10−3 (10−3 )(10−3 ) = 1.446 τs = K B πd 3 π(0.2) 3 (10−3 ) 3 = 2620(1) = 2620 MPa

(1)

τs > Ssy , that is, 2620 > 826.5 MPa; the spring is not solid safe. Solve Eq. (1) for ys giving ys =

(Ssy /n)(πd 3 ) (826.5/1.2)(π)(0.2) 3 = = 2.08 mm 8K B k D 8(1.446)(1.0147)(0.71)

L 0 = L s + ys = 8.0 + 2.08 = 10.08 mm Wind the spring to a free length of 10.08 mm. This only addresses the solid-safe criteria. There are additional problems. Ans. 10-14

Given: A228 (music wire), SQ&GRD ends, d = 1 mm, OD = 6.10 mm, L 0 = 19.1 mm, Nt = 10.4 turns. Table 10-4:

A = 2211 MPa · mmm , m = 0.145

Table 10-5: G = 81.7 GPa D = OD − d = 6.10 − 1 = 5.1 mm C = D/d = 5.1/1 = 5.1 Na = Nt − 2 = 10.4 − 2 = 8.4 turns KB =

4(5.1) + 2 = 1.287 4(5.1) − 3

Sut =

2211 = 2211 MPa (1) 0.145

Table 10-6: Ssy = 0.45(2211) = 995 MPa

(1) 4 (81.7) (10−3 ) 4 (109 ) d4G = 0.009 165(106 ) = k= 3 3 − 3 3 8D Na 8(5.1) (8.4) (10 ) = 9165 N/m or 9.165 N/mm

L s = d Nt = 1(10.4) = 10.4 mm Fs = kys

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ys = L 0 − L s = 19.1 − 10.4 = 8.7 mm 8(kys ) D 8(9.165)(8.7)(5.1) = 1.287 = 1333 MPa τs = K B πd 3 π(1) 3

(1)

τs > Ssy , that is, 1333 > 995 MPa; the spring is not solid safe. Solve Eq. (1) for ys giving ys =

(Ssy /n)(πd 3 ) (995/1.2)(π)(1) 3 = = 5.43 mm 8K B k D 8(1.287)(9.165)(5.1)

L 0 = L s + ys = 10.4 + 5.43 = 15.83 mm Wind the spring to a free length of 15.83 mm. 10-15

Ans.

Given: A229 (OQ&T spring steel), SQ&GRD ends, d = 3.4 mm, OD = 50.8 mm, L 0 = 74.6 mm, Nt = 5.25. Table 10-4:

A = 1855 MPa · mmm ,

Table 10-5:

G = 77.2 GPa

m = 0.187

D = OD − d = 50.8 − 3.4 = 47.4 mm C = D/d = 47.4/3.4 = 13.94 Na = Nt − 2 = 5.25 − 2 = 3.25 turns

Table 10-6:

KB =

4(13.94) + 2 = 1.095 4(13.94) − 3

Sut =

1855 = 1476 MPa (3.4) 0.187

Ssy = 0.50(1476) = 737.8 MPa

(3.4) 4 (77.2) (10−3 ) 4 (109 ) d4G = 0.003 75(106 ) = k= 8D 3 Na 8(47.4) 3 (3.25) (10−3 ) 3 = 3750 N/m or 3.750 N/mm

L s = d Nt = 3.4(5.25) = 17.85 Fs = kys ys = L 0 − L s = 74.6 − 17.85 = 56.75 mm 8(kys ) D τs = K B πd 3 8(3.750)(56.75)(47.4) = 720.2 MPa = 1.095 π(3.4) 3 τs < Ssy , that is, 720.2 < 737.8 MPa

(1)

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Chapter 10

∴ The spring is solid safe. With n s = 1.2, ys =

(Ssy /n)(πd 3 ) (737.8/1.2)(π)(3.4) 3 = = 48.76 mm 8K B k D 8(1.095)(3.75)(47.4)

L 0 = L s + ys = 17.85 + 48.76 = 66.61 mm Wind the spring to a free length of 66.61 mm. 10-16

Ans.

Given: B159 (phosphor bronze), SQ&GRD ends, d = 3.7 mm, OD = 25.4 mm, L 0 = 95.3 mm, Nt = 13 turns. Table 10-4:

A = 932 MPa · mmm , m = 0.064

Table 10-5:

G = 41.4 GPa D = OD − d = 25.4 − 3.7 = 21.7 mm C = D/d = 21.7/3.7 = 5.865 KB =

4(5.865) + 2 = 1.244 4(5.865) − 3

Na = Nt − 2 = 13 − 2 = 11 turns Sut =

932 = 857.1 MPa (3.7) 0.064

Table 10-6: Ssy = 0.35(857.1) = 300 MPa d4G (3.7) 4 (41.4) (10−3 ) 4 (109 ) = 0.008 629(106 ) k= = 3 3 − 3 3 8D Na 8(21.7) (11) (10 ) = 8629 N/m or 8.629 N/mm L s = d Nt = 3.7(13) = 48.1 mm Fs = kys ys = L 0 − L s = 95.3 − 48.1 = 47.2 mm 8(kys ) D τs = K B πd 3 8(8.629)(47.2)(21.7) = 553 MPa = 1.244 π(3.7) 3

(1)

τs > Ssy , that is, 553 > 300 MPa; the spring is not solid-safe. Solving Eq. (1) for ys gives ys =

(Ssy /n)(πd 3 ) (300/1.2)(π)(3.7) 3 = = 21.35 mm 8K B k D 8(1.244)(8.629)(21.7)

L 0 = L s + ys = 48.1 + 21.35 = 69.45 mm Wind the spring to a free length of 69.45 mm.

Ans.

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10-17

Given: A232 (Cr-V steel), SQ&GRD ends, d = 4.3 mm, OD = 76.2 mm, L 0 = 228.6 mm, Nt = 8 turns. Table 10-4:

A = 2005 MPa · mmm ,

Table 10-5:

G = 77.2 GPa

m = 0.168

D = OD − d = 76.2 − 4.3 = 71.9 mm C = D/d = 71.9/4.3 = 16.72 KB =

4(16.72) + 2 = 1.078 4(16.72) − 3

Na = Nt − 2 = 8 − 2 = 6 turns Sut =

2005 = 1569 MPa (4.3) 0.168

Table 10-6: Ssy = 0.50(1569) = 784.5 MPa d4G (4.3) 4 (77.2) (10−3 ) 4 (109 ) k= = 0.001 479(106 ) = 3 3 − 3 3 8D Na 8(71.9) (6) (10 ) = 1479 N/m or 1.479 N/mm L s = d Nt = 4.3(8) = 34.4 mm Fs = kys ys = L 0 − L s = 228.6 − 34.4 = 194.2 mm 8(kys ) D 8(1.479)(194.2)(71.9) = 1.078 = 713.0 MPa τs = K B πd 3 π(4.3) 3

(1)

τs < Ssy , that is, 713.0 < 784.5; the spring is solid safe. With n s = 1.2 Eq. (1) becomes ys =

(Ssy /n)(πd 3 ) (784.5/1.2)(π)(4.3) 3 = = 178.1 mm 8K B k D 8(1.078)(1.479)(71.9)

L 0 = L s + ys = 34.4 + 178.1 = 212.5 mm Wind the spring to a free length of L 0 = 212.5 mm. 10-18

Ans.

For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na , k = 20/2 = 10 lbf/in.

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Chapter 10

(a) Spring over a Rod Source

Eq. (10-2) Eq. (10-9) Table 10-1 Table 10-1 1.15y + L s Eq. (10-13) Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-6) Eq. (10-3) Eq. (10-22)

(b) Spring in a Hole

Parameter Values d D ID OD C Na Nt Ls L0 (L 0 ) cr A m Sut Ssy KB n fom

0.075 0.875 0.800 0.950 11.667 6.937 8.937 0.670 2.970 4.603 201.000 0.145 292.626 131.681 1.115 0.973 −0.282

0.08 0.88 0.800 0.960 11.000 8.828 10.828 0.866 3.166 4.629 201.000 0.145 289.900 130.455 1.122 1.155 −0.391

Source 0.085 0.885 0.800 0.970 10.412 11.061 13.061 1.110 3.410 4.655 201.000 0.145 287.363 129.313 1.129 1.357 −0.536

Eq. (10-2) Eq. (10-9) Table 10-1 Table 10-1 1.15y + L s Eq. (10-13) Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-6) Eq. (10-3) Eq. (10-22)

Parameter Values d D ID OD C Na Nt Ls L0 (L 0 ) cr A m Sut Ssy KB n fom

0.075 0.875 0.800 0.950 11.667 6.937 8.937 0.670 2.970 4.603 201.000 0.145 292.626 131.681 1.115 0.973 −0.282

0.08 0.870 0.790 0.950 10.875 9.136 11.136 0.891 3.191 4.576 201.000 0.145 289.900 130.455 1.123 1.167 −0.398

0.085 0.865 0.780 0.950 10.176 11.846 13.846 1.177 3.477 4.550 201.000 0.145 287.363 129.313 1.133 1.384 −0.555

For n s ≥ 1.2, the optimal size is d = 0.085 in for both cases. 10-19

From the figure: L 0 = 120 mm, OD = 50 mm, and d = 3.4 mm. Thus D = OD − d = 50 − 3.4 = 46.6 mm (a) By counting, Nt = 12.5 turns. Since the ends are squared along 1/4 turn on each end, Na = 12.5 − 0.5 = 12 turns Ans. p = 120/12 = 10 mm Ans. The solid stack is 13 diameters across the top and 12 across the bottom. L s = 13(3.4) = 44.2 mm Ans. (b) d = 3.4/25.4 = 0.1339 in and from Table 10-5, G = 78.6 GPa k=

d4G (3.4) 4 (78.6)(109 ) (10−3 ) = 1080 N/m Ans. = 3 3 8D Na 8(46.6) (12)

(c) Fs = k(L 0 − L s ) = 1080(120 − 44.2)(10−3 ) = 81.9 N Ans. (d) C = D/d = 46.6/3.4 = 13.71 KB = τs = 10-20

4(13.71) + 2 = 1.096 4(13.71) − 3 8K B Fs D 8(1.096)(81.9)(46.6) = = 271 MPa Ans. 3 πd π(3.4) 3

One approach is to select A227-47 HD steel for its low cost. Then, for y1 ≤ 3/8 at F1 = 10 lbf, k ≥10/ 0.375 = 26.67 lbf/in . Try d = 0.080 in #14 gauge

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For a clearance of 0.05 in: ID = (7/16) + 0.05 = 0.4875 in; OD = 0.4875 + 0.16 = 0.6475 in D = 0.4875 + 0.080 = 0.5675 in C = 0.5675/0.08 = 7.094 G = 11.5 Mpsi d4G (0.08) 4 (11.5)(106 ) = = 12.0 turns Na = 8k D 3 8(26.67)(0.5675) 3 Nt = 12 + 2 = 14 turns, L s = d Nt = 0.08(14) = 1.12 in O.K. L 0 = 1.875 in, ys = 1.875 − 1.12 = 0.755 in Fs = kys = 26.67(0.755) = 20.14 lbf 4(7.094) + 2 KB = = 1.197 4(7.094) − 3 8Fs D 8(20.14)(0.5675) τs = K B = 1.197 = 68 046 psi πd 3 π(0.08) 3 A = 140 kpsi · inm , m = 0.190 140 Ssy = 0.45 = 101.8 kpsi (0.080) 0.190 101.8 = 1.50 > 1.2 O.K. n= 68.05 F1 10 (68.05) = 33.79 kpsi, τ1 = τs = Fs 20.14 101.8 n1 = = 3.01 > 1.5 O.K. 33.79 There is much latitude for reducing the amount of material. Iterate on y1 using a spread sheet. The final results are: y1 = 0.32 in, k = 31.25 lbf/in, Na = 10.3 turns, Nt = 12.3 turns, L s = 0.985 in, L 0 = 1.820 in, ys = 0.835 in, Fs = 26.1 lbf, K B = 1.197, τs = 88 190 kpsi, n s = 1.15, and n 1 = 3.01. ID = 0.4875 in, OD = 0.6475 in, d = 0.080 in Try other sizes and/or materials. Table 10-4:

10-21

A stock spring catalog may have over two hundred pages of compression springs with up to 80 springs per page listed. • • • •

Students should be aware that such catalogs exist. Many springs are selected from catalogs rather than designed. The wire size you want may not be listed. Catalogs may also be available on disk or the web through search routines. For example, disks are available from Century Spring at 1 − (800) − 237 − 5225 www.centuryspring.com • It is better to familiarize yourself with vendor resources rather than invent them yourself. • Sample catalog pages can be given to students for study.

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Chapter 10

10-22

For a coil radius given by: R2 − R1 θ R = R1 + 2π N The torsion of a section is T = P R where d L = R dθ 2π N 1 ∂T 1 ∂U δp = = T dL = P R 3 dθ ∂P GJ ∂P GJ 0 2π N R2 − R1 3 P θ dθ R1 + = GJ 0 2π N 2π N 2π N R2 − R1 4

P 1 θ R1 + =

GJ 4 R2 − R1 2π N 0 πPN 4 πPN 4 ( R1 + R2 ) R12 + R22 = R2 − R1 = 2G J ( R2 − R1 ) 2G J 2 π 16P N 2 J = d 4 ∴ δp = ( R + R ) R + R 1 2 1 2 32 Gd 4 k=

10-23

P d4G = δp 16N ( R1 + R2 ) R12 + R22

Ans.

For a food service machinery application select A313 Stainless wire. G = 10(106 ) psi 0.013 ≤ d ≤ 0.10 in A = 169, m = 0.146 Note that for 0.10 < d ≤ 0.20 in A = 128, m = 0.263 18 − 4 18 + 4 = 7 lbf, Fm = = 11 lbf, r = 7/11 Fa = 2 2 18 − 4 = 9.333 lbf/in k = F/y = 2.5 − 1 169 d = 0.080 in, Sut = = 244.4 kpsi Try (0.08) 0.146 Ssu = 0.67Sut = 163.7 kpsi, Ssy = 0.35Sut = 85.5 kpsi Try unpeened using Zimmerli’s endurance data: Ssa = 35 kpsi, Ssm = 55 kpsi Gerber:

Ssa 35 = = 39.5 kpsi 2 1 − (Ssm /Ssu ) 1 − (55/163.7) 2   2  2 2  (7/11) (163.7) 2(39.5) −1 + 1 + = 35.0 kpsi =  2(39.5) (7/11)(163.7) 

Sse =

Ssa

α = Ssa /n f = 35.0/1.5 = 23.3 kpsi 8Fa 8(7) −3 β= (10−3 ) = 2.785 kpsi (10 ) = 2 2 πd π(0.08 ) 3(23.3) 2(23.3) − 2.785 2(23.3) − 2.785 2 + = 6.97 − C= 4(2.785) 4(2.785) 4(2.785) D = Cd = 6.97(0.08) = 0.558 in

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4(6.97) + 2 = 1.201 4(6.97) − 3 8Fa D 8(7)(0.558) −3 τa = K B = 1.201 (10 ) = 23.3 kpsi πd 3 π(0.083 )

KB =

n f = 35/23.3 = 1.50 checks Na =

Gd 4 10(106 )(0.08) 4 = = 31.58 turns 8k D 3 8(9.333)(0.558) 3

Nt = 31.58 + 2 = 33.58 turns, L s = d Nt = 0.08(33.58) = 2.686 in ys = (1 + ξ ) ymax = (1 + 0.15)(2.5) = 2.875 in L 0 = 2.686 + 2.875 = 5.561 in 2.63(0.558) D = = 2.935 in α 0.5 τs = 1.15(18/7)τa = 1.15(18/7)(23.3) = 68.9 kpsi

(L 0 ) cr = 2.63

n s = Ssy /τs = 85.5/68.9 = 1.24 kg 9.333(386) = = 107 Hz f = π 2 d 2 D Na γ π 2 (0.082 )(0.558)(31.58)(0.283) These steps are easily implemented on a spreadsheet, as shown below, for different diameters.

d m A Sut Ssu Ssy Sse Ssa α β C D KB τa nf Na Nt Ls ymax L0 (L 0 ) cr τs ns f (Hz)

d1

d2

d3

d4

0.080 0.146 169.000 244.363 163.723 85.527 39.452 35.000 23.333 2.785 6.977 0.558 1.201 23.333 1.500 31.547 33.547 2.684 2.875 5.559 2.936 69.000 1.240 106.985

0.0915 0.146 169.000 239.618 160.544 83.866 39.654 35.000 23.333 2.129 9.603 0.879 1.141 23.333 1.500 13.836 15.836 1.449 2.875 4.324 4.622 69.000 1.215 112.568

0.1055 0.263 128.000 231.257 154.942 80.940 40.046 35.000 23.333 1.602 13.244 1.397 1.100 23.333 1.500 6.082 8.082 0.853 2.875 3.728 7.350 69.000 1.173 116.778

0.1205 0.263 128.000 223.311 149.618 78.159 40.469 35.000 23.333 1.228 17.702 2.133 1.074 23.333 1.500 2.910 4.910 0.592 2.875 3.467 11.220 69.000 1.133 119.639

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Chapter 10

The shaded areas depict conditions outside the recommended design conditions. Thus, one spring is satisfactory–A313, as wound, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, Nt = 15.84 turns 10-24

The steps are the same as in Prob. 10-23 except that the Gerber-Zimmerli criterion is replaced with Goodman-Zimmerli: Ssa Sse = 1 − (Ssm /Ssu ) The problem then proceeds as in Prob. 10-23. The results for the wire sizes are shown below (see solution to Prob. 10-23 for additional details). d1

d m A Sut Ssu Ssy Sse Ssa α β C D

d2

d3

Iteration of d for the first trial d4 d1

0.080 0.0915 0.1055 0.1205 0.146 0.146 0.263 0.263 169.000 169.000 128.000 128.000 244.363 239.618 231.257 223.311 163.723 160.544 154.942 149.618 85.527 83.866 80.940 78.159 52.706 53.239 54.261 55.345 43.513 43.560 43.634 43.691 29.008 29.040 29.090 29.127 2.785 2.129 1.602 1.228 9.052 12.309 16.856 22.433 0.724 1.126 1.778 2.703

d2

d3

d 0.080 0.0915 0.1055 KB 1.151 1.108 1.078 τa 29.008 29.040 29.090 nf 1.500 1.500 1.500 Na 14.444 6.572 2.951 Nt 16.444 8.572 4.951 Ls 1.316 0.784 0.522 ymax 2.875 2.875 2.875 L0 4.191 3.659 3.397 (L 0 ) cr 3.809 5.924 9.354 τs 85.782 85.876 86.022 ns 0.997 0.977 0.941 f (Hz) 138.806 144.277 148.617

d4 0.1205 1.058 29.127 1.500 1.429 3.429 0.413 2.875 3.288 14.219 86.133 0.907 151.618

Without checking all of the design conditions, it is obvious that none of the wire sizes satisfy n s ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting n f = 1.5 for Goodman makes it impossible to reach the yield line (n s < 1) . The table below uses n f = 2. d1 d m A Sut Ssu Ssy Sse Ssa α β C D

d2

Iteration of d for the second trial d3 d4 d1

0.080 0.0915 0.1055 0.1205 0.146 0.146 0.263 0.263 169.000 169.000 128.000 128.000 244.363 239.618 231.257 223.311 163.723 160.544 154.942 149.618 85.527 83.866 80.940 78.159 52.706 53.239 54.261 55.345 43.513 43.560 43.634 43.691 21.756 21.780 21.817 21.845 2.785 2.129 1.602 1.228 6.395 8.864 12.292 16.485 0.512 0.811 1.297 1.986

d KB τa nf Na Nt Ls ymax L0 (L 0 ) cr τs ns f (Hz)

d2

d3

0.080 0.0915 0.1055 1.221 1.154 1.108 21.756 21.780 21.817 2.000 2.000 2.000 40.962 17.594 7.609 42.962 19.594 9.609 3.437 1.793 1.014 2.875 2.875 2.875 6.312 4.668 3.889 2.691 4.266 6.821 64.336 64.407 64.517 1.329 1.302 1.255 98.065 103.903 108.376

d4 0.1205 1.079 21.845 2.000 3.602 5.602 0.675 2.875 3.550 10.449 64.600 1.210 111.418

The satisfactory spring has design specifications of: A313, as wound, unpeened, squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in , Nt = 19.6 turns.

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10-25

This is the same as Prob. 10-23 since Sse = Ssa = 35 kpsi. Therefore, design the spring using: A313, as wound, un-peened, squared and ground, d = 0.915 in, OD = 0.971 in, Nt = 15.84 turns.

10-26

For the Gerber fatigue-failure criterion, Ssu = 0.67S ut , Sse =

Ssa , 1 − (Ssm /Ssu ) 2

Ssa =

2 2

r Ssu  −1 + 2Sse

1+

2Sse r Ssu

2

 

The equation for Ssa is the basic difference. The last 2 columns of diameters of Ex. 10-5 are presented below with additional calculations. d = 0.105

d = 0.112

278.691 186.723 38.325 125.411 34.658 23.105 1.732 12.004 1.260 1.155 1.365

276.096 184.984 38.394 124.243 34.652 23.101 1.523 13.851 1.551 1.439 1.663

Sut Ssu Sse Ssy Ssa α β C D ID OD

Na Ls L0 (L 0 ) cr KB τa nf τs ns fn fom

d = 0.105

d = 0.112

8.915 1.146 3.446 6.630 1.111 23.105 1.500 70.855 1.770 105.433 −0.973

6.190 0.917 3.217 8.160 1.095 23.101 1.500 70.844 1.754 106.922 −1.022

There are only slight changes in the results. 10-27

As in Prob. 10-26, the basic change is Ssa . Sse =

For Goodman,

Ssa 1 − (Ssm /Ssu )

Recalculate Ssa with

r Sse Ssu r Ssu + Sse Calculations for the last 2 diameters of Ex. 10-5 are given below. Ssa =

Sut Ssu Sse Ssy Ssa α β C D ID OD

d = 0.105

d = 0.112

278.691 186.723 49.614 125.411 34.386 22.924 1.732 11.899 1.249 1.144 1.354

276.096 184.984 49.810 124.243 34.380 22.920 1.523 13.732 1.538 1.426 1.650

There are only slight differences in the results.

Na Ls L0 (L 0 ) cr KB τa nf τs ns fn fom

d = 0.105

d = 0.112

9.153 1.171 3.471 6.572 1.112 22.924 1.500 70.301 1.784 104.509 −0.986

6.353 0.936 3.236 8.090 1.096 22.920 1.500 70.289 1.768 106.000 −1.034

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Chapter 10

10-28

287

Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1. d = 0.067 in,

Try

Sut =

140 = 234.0 kpsi (0.067) 0.190

Table 10-6:

Ssy = 0.45Sut = 105.3 kpsi

Table 10-7:

S y = 0.75Sut = 175.5 kpsi

Eq. (10-34) with D/d = C and C1 = C Sy Fmax σA = [(K ) A (16C) + 4] = 2 πd ny πd 2 S y 4C 2 − C − 1 (16C) + 4 = 4C(C − 1) n y Fmax πd 2 S y 2 4C − C − 1 = (C − 1) −1 4n y Fmax πd 2 S y 1 1 πd 2 S y 2 1+ −1 C + −2 =0 C − 4 4n y Fmax 4 4n y Fmax   2 2 2 2 πd S y πd S y πd S y 1 ± − + 2  take positive root C=  2 16n y Fmax 16n y Fmax 4n y Fmax 1 π(0.0672 )(175.5)(103 ) = 2 16(1.5)(18)  2  2 3 2 3 π(0.067) (175.5)(10 ) π(0.067) (175.5)(10 ) + 2 = 4.590 + −  16(1.5)(18) 4(1.5)(18) D = Cd = 0.3075 in πd 3 C −3 πd 3 τi 33 500 = ± 1000 4 − Fi = 8D 8D exp(0.105C) 6.5 Use the lowest Fi in the preferred range. This results in the best fom. 33 500 4.590 − 3 π(0.067) 3 − 1000 4 − = 6.505 lbf Fi = 8(0.3075) exp[0.105(4.590)] 6.5 For simplicity, we will round up to the next integer or half integer; therefore, use Fi = 7 lbf 18 − 7 = 22 lbf/in 0.5 d4G (0.067) 4 (11.5)(106 ) Na = = = 45.28 turns 8k D 3 8(22)(0.3075) 3 11.5 G = 44.88 turns Nb = Na − = 45.28 − E 28.6 L 0 = (2C − 1 + Nb )d = [2(4.590) − 1 + 44.88](0.067) = 3.555 in k=

L 18 lbf = 3.555 + 0.5 = 4.055 in

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Body: K B = τmax = (n y ) body =

4(4.590) + 2 4C + 2 = = 1.326 4C − 3 4(4.590) − 3 8K B Fmax D 8(1.326)(18)(0.3075) = (10−3 ) = 62.1 kpsi 3 3 πd π(0.067) Ssy 105.3 = 1.70 = τmax 62.1

r2 = 2d = 2(0.067) = 0.134 in, C2 =

2(0.134) 2r2 = =4 d 0.067

4(4) − 1 4C2 − 1 = = 1.25 4C2 − 4 4(4) − 4 8Fmax D 8(18)(0.3075) τ B = (K ) B = 1.25 (10−3 ) = 58.58 kpsi πd 3 π(0.067) 3

(K ) B =

(n y ) B =

Ssy 105.3 = 1.80 = τB 58.58

fom = −(1)

π 2 (0.067) 2 (44.88 + 2)(0.3075) π 2 d 2 ( Nb + 2) D =− = −0.160 4 4

Several diameters, evaluated using a spreadsheet, are shown below. d:

0.067

0.072

0.076

0.081

0.085

Sut 233.977 105.290 Ssy Sy 175.483 C 4.589 D 0.307 Fi (calc) 6.505 Fi (rd) 7.0 k 22.000 Na 45.29 Nb 44.89 L0 3.556 L 18 lbf 4.056 KB 1.326 τmax 62.118 (n y ) body 1.695 τB 58.576 (n y ) B 1.797 (n y ) A 1.500 −0.160 fom

230.799 103.860 173.100 5.412 0.390 5.773 6.0 24.000 27.20 26.80 2.637 3.137 1.268 60.686 1.711 59.820 1.736 1.500 −0.144

228.441 102.798 171.331 6.099 0.463 5.257 5.5 25.000 19.27 18.86 2.285 2.785 1.234 59.707 1.722 60.495 1.699 1.500 −0.138

225.692 101.561 169.269 6.993 0.566 4.675 5.0 26.000 13.10 12.69 2.080 2.580 1.200 58.636 1.732 61.067 1.663 1.500 −0.135

223.634 100.635 167.726 7.738 0.658 4.251 4.5 27.000 9.77 9.36 2.026 2.526 1.179 57.875 1.739 61.367 1.640 1.500 −0.133

0.09 221.219 99.548 165.914 8.708 0.784 3.764 4.0 28.000 7.00 6.59 2.071 2.571 1.157 57.019 1.746 61.598 1.616 1.500 −0.135

0.095

0.104

218.958 98.531 164.218 9.721 0.923 3.320 3.5 29.000 5.13 4.72 2.201 2.701 1.139 56.249 1.752 61.712 1.597 1.500 −0.138

215.224 96.851 161.418 11.650 1.212 2.621 3.0 30.000 3.15 2.75 2.605 3.105 1.115 55.031 1.760 61.712 1.569 1.500 −0.154

Except for the 0.067 in wire, all springs satisfy the requirements of length and number of coils. The 0.085 in wire has the highest fom.

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Chapter 10

10-29

Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in. D = 1.5 − 0.162 =1.338 in (a) Eq. (10-39): L 0 = 2( D − d) + ( Nb + 1)d = 2(1.338 − 0.162) + (84 + 1)(0.162) = 16.12 in Ans. 2d + L 0 = 2(0.162) + 16.12 = 16.45 in overall.

or

1.338 D = = 8.26 d 0.162 4(8.26) + 2 KB = = 1.166 4(8.26) − 3 8Fi D 8(16)(1.338) τi = K B = 1.166 = 14 950 psi Ans. πd 3 π(0.162) 3 C=

(b)

(c) From Table 10-5 use: G = 11.4(106 ) psi and E = 28.5(106 ) psi 11.4 G Na = Nb + = 84 + = 84.4 turns E 28.5 d4G (0.162) 4 (11.4)(106 ) = 4.855 lbf/in Ans. k= = 8D 3 Na 8(1.338) 3 (84.4) (d) Table 10-4:

A = 147 psi · inm , m = 0.187 147 = 207.1 kpsi Sut = (0.162) 0.187 S y = 0.75(207.1) = 155.3 kpsi Ssy = 0.50(207.1) = 103.5 kpsi

Body πd 3 Ssy π KB D π(0.162) 3 (103.5)(103 ) = 110.8 lbf = 8(1.166)(1.338) Torsional stress on hook point B F=

2(0.25 + 0.162/2) 2r2 = = 4.086 d 0.162 4(4.086) − 1 4C2 − 1 (K ) B = = = 1.243 4C2 − 4 4(4.086) − 4 π(0.162) 3 (103.5)(103 ) = 103.9 lbf F= 8(1.243)(1.338) Normal stress on hook point A C2 =

1.338 2r1 = = 8.26 d 0.162 4C12 − C1 − 1 4(8.26) 2 − 8.26 − 1 (K ) A = = = 1.099 4C1 (C1 − 1) 4(8.26)(8.26 − 1) C1 =

289

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4 16(K ) A D S yt = σ = F + πd 3 πd 2

155.3(103 ) F= = 85.8 lbf [16(1.099)(1.338)]/[π(0.162) 3 ] + {4/[π(0.162) 2 ]} = min(110.8, 103.9, 85.8) = 85.8 lbf Ans. (e) Eq. (10-48): 85.8 − 16 F − Fi = = 14.4 in Ans. k 4.855

y= 10-30

Fmin = 9 lbf,

Fmax = 18 lbf Fa =

A313 stainless:

18 − 9 = 4.5 lbf, 2

0.013 ≤ d ≤ 0.1 0.1 ≤ d ≤ 0.2 E = 28 Mpsi,

Fm =

18 + 9 = 13.5 lbf 2

A = 169 kpsi · inm , m = 0.146 A = 128 kpsi · inm , m = 0.263 G = 10 Gpsi

Try d = 0.081 in and refer to the discussion following Ex. 10-7 Sut =

169 = 243.9 kpsi (0.081) 0.146

Ssu = 0.67Sut = 163.4 kpsi Ssy = 0.35Sut = 85.4 kpsi S y = 0.55Sut = 134.2 kpsi Table 10-8:

Sr = 0.45Sut = 109.8 kpsi Se =

Sr /2 109.8/2 = = 57.8 kpsi 1 − [Sr /(2Sut )]2 1 − [(109.8/2)/243.9]2

r = Fa /Fm = 4.5/13.5 = 0.333   2 2 2 r Sut  2Se  −1 + 1 + Table 7-10: Sa = 2Se r Sut   2 2 2 (0.333) (243.9 )  2(57.8)  = 42.2 kpsi Sa = −1 + 1 + 2(57.8) 0.333(243.9) Hook bending

Sa 4 Sa 16C = (σa ) A = Fa (K ) A 2 + = πd πd 2 (n f ) A 2 Sa 4.5 (4C 2 − C − 1)16C +4 = 2 πd 4C(C − 1) 2

This equation reduces to a quadratic in C—see Prob. 10-28

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Chapter 10

291

The useable root for C is   2 2 2 2 πd Sa πd Sa πd Sa + + 2 − C = 0.5  144 144 36   2  π(0.081) 2 (42.2)(103 )  2 3 2 3 π(0.081) (42.2)(10 ) π(0.081) (42.2)(10 ) = 0.5 + +2 −   144 144 36 = 4.91

D = Cd = 0.398 in

πd 3 C −3 πd 3 τi 33 500 = ± 1000 4 − Fi = 8D 8D exp(0.105C) 6.5 Use the lowest Fi in the preferred range. 4.91 − 3 π(0.081) 3 33 500 − 1000 4 − Fi = 8(0.398) exp[0.105(4.91)] 6.5 = 8.55 lbf For simplicity we will round up to next 1/4 integer. Fi = 8.75 lbf k=

18 − 9 = 36 lbf/in 0.25

d4G (0.081) 4 (10)(106 ) = = 23.7 turns 8k D 3 8(36)(0.398) 3 10 G = 23.3 turns Nb = Na − = 23.7 − E 28 L 0 = (2C − 1 + Nb )d = [2(4.91) − 1 + 23.3](0.081) = 2.602 in L max = L 0 + ( Fmax − Fi )/k = 2.602 + (18 − 8.75)/36 = 2.859 in 4.5(4) 4C 2 − C − 1 +1 (σa ) A = πd 2 C −1 18(10−3 ) 4(4.912 ) − 4.91 − 1 + 1 = 21.1 kpsi = π(0.0812 ) 4.91 − 1 Na =

(n f ) A = Body:

Sa 42.2 = 2 checks = (σa ) A 21.1 KB =

4(4.91) + 2 4C + 2 = = 1.300 4C − 3 4(4.91) − 3

τa =

8(1.300)(4.5)(0.398) (10−3 ) = 11.16 kpsi π(0.081) 3

τm =

Fm 13.5 (11.16) = 33.47 kpsi τa = Fa 4.5

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The repeating allowable stress from Table 7-8 is Ssr = 0.30Sut = 0.30(243.9) = 73.17 kpsi The Gerber intercept is Sse = From Table 7-10, (n f ) body =

1 2

163.4 33.47

73.17/2 = 38.5 kpsi 1 − [(73.17/2)/163.4]2

2

   2  11.16 2(33.47)(38.5) = 2.53 −1 + 1 + 38.5  163.4(11.16) 

Let r2 = 2d = 2(0.081) = 0.162 2r2 = 4, C2 = d

(K ) B =

4(4) − 1 = 1.25 4(4) − 4

(τa ) B =

(K ) B 1.25 (11.16) = 10.73 kpsi τa = KB 1.30

(τm ) B =

(K ) B 1.25 (33.47) = 32.18 kpsi τm = KB 1.30

Table 10-8: (Ssr ) B = 0.28Sut = 0.28(243.9) = 68.3 kpsi 68.3/2 = 35.7 kpsi 1 − [(68.3/2)/163.4]2   2  2  10.73 1 163.4 2(32.18)(35.7) (n f ) B = −1 + 1 + = 2.51 2 32.18 35.7  163.4(10.73) 

(Sse ) B =

Yield Bending:

4Fmax (4C 2 − C − 1) +1 (σ A ) max = πd 2 C −1 4(18) 4(4.91) 2 − 4.91 − 1 + 1 (10−3 ) = 84.4 kpsi = 2 π(0.081 ) 4.91 − 1 (n y ) A =

Body:

134.2 = 1.59 84.4

τi = ( Fi /Fa )τa = (8.75/4.5)(11.16) = 21.7 kpsi r = τa /(τm − τi ) = 11.16/(33.47 − 21.7) = 0.948 r 0.948 (Ssa ) y = (Ssy − τi ) = (85.4 − 21.7) = 31.0 kpsi r +1 0.948 + 1 (Ssa ) y 31.0 = 2.78 = (n y ) body = τa 11.16

Hook shear: Ssy = 0.3Sut = 0.3(243.9) = 73.2 kpsi τmax = (τa ) B + (τm ) B = 10.73 + 32.18 = 42.9 kpsi

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73.2 = 1.71 42.9 7.6π 2 d 2 ( Nb + 2) D 7.6π 2 (0.081) 2 (23.3 + 2)(0.398) fom = − =− = −1.239 4 4 A tabulation of several wire sizes follow (n y ) B =

d Sut Ssu Sr Se Sa C D OD Fi (calc) Fi (rd) k Na Nb L0 L 18 lbf (σa ) A (n f ) A KB (τa ) body (τm ) body Ssr Sse (n f ) body (K ) B (τa ) B (τm ) B (Ssr ) B (Sse ) B (n f ) B Sy (σ A ) max (n y ) A τi r (Ssy ) body (Ssa ) y (n y ) body (Ssy ) B (τ B ) max (n y ) B fom

0.081 243.920 163.427 109.764 57.809 42.136 4.903 0.397 0.478 8.572 8.75 36.000 23.86 23.50 2.617 2.874 21.068 2.000 1.301 11.141 33.424 73.176 38.519 2.531 1.250 10.705 32.114 68.298 35.708 2.519 134.156 84.273 1.592 21.663 0.945 85.372 30.958 2.779 73.176 42.819 1.709 −1.246

0.085 242.210 162.281 108.994 57.403 41.841 5.484 0.466 0.551 7.874 9.75 36.000 17.90 17.54 2.338 2.567 20.920 2.000 1.264 10.994 32.982 72.663 38.249 2.547 1.250 10.872 32.615 67.819 35.458 2.463 133.215 83.682 1.592 23.820 1.157 84.773 32.688 2.973 72.663 43.486 1.671 −1.234

0.092 239.427 160.416 107.742 56.744 41.360 6.547 0.602 0.694 6.798 10.75 36.000 11.38 11.02 2.127 2.328 20.680 2.000 1.216 10.775 32.326 71.828 37.809 2.569 1.250 11.080 33.240 67.040 35.050 2.388 131.685 82.720 1.592 25.741 1.444 83.800 34.302 3.183 71.828 44.321 1.621 −1.245 optimal fom

The shaded areas show the conditions not satisfied.

0.098 237.229 158.943 106.753 56.223 40.980 7.510 0.736 0.834 5.987 11.75 36.000 8.03 7.68 2.126 2.300 20.490 2.000 1.185 10.617 31.852 71.169 37.462 2.583 1.250 11.200 33.601 66.424 34.728 2.341 130.476 81.961 1.592 27.723 1.942 83.030 36.507 3.438 71.169 44.801 1.589 −1.283

0.105 234.851 157.350 105.683 55.659 40.570 8.693 0.913 1.018 5.141 12.75 36.000 5.55 5.19 2.266 2.412 20.285 2.000 1.157 10.457 31.372 70.455 37.087 2.596 1.250 11.294 33.883 65.758 34.380 2.298 129.168 81.139 1.592 29.629 2.906 82.198 39.109 3.740 70.455 45.177 1.560 −1.357

0.12 230.317 154.312 103.643 54.585 39.786 11.451 1.374 1.494 3.637 13.75 36.000 2.77 2.42 2.918 3.036 19.893 2.000 1.117 10.177 30.532 69.095 36.371 2.616 1.250 11.391 34.173 64.489 33.717 2.235 126.674 79.573 1.592 31.097 4.703 80.611 40.832 4.012 69.095 45.564 1.516 −1.639

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10-31

For the hook, F

M = F R sin θ, ∂ M/∂ F = R sin θ π/2 1 π P R3 δF = F R 2 sin2 R dθ = EI 0 2 EI

R D兾2 ␪

The total deflection of the body and the two hooks π F R3 8F D 3 Nb π F( D/2) 3 8F D 3 Nb + 2 = + d4G 2 EI d4G E(π/64)(d 4 ) 8F D 3 8F D 3 Na G = 4 Nb + = d G E d4G

δ=

10-32

Na = Nb +

G E

QED

Table 10-4 for A227: A = 140 kpsi · inm ,

m = 0.190

E = 28.5(106 ) psi

Table 10-5:

Sut =

140 = 197.8 kpsi (0.162) 0.190

Eq. (10-57): S y = σall = 0.78(197.8) = 154.3 kpsi D = 1.25 − 0.162 = 1.088 in C = D/d = 1.088/0.162 = 6.72 4(6.72) 2 − 6.72 − 1 4C 2 − C − 1 = = 1.125 Ki = 4C(C − 1) 4(6.72)(6.72 − 1) σ = Ki

From

32M πd 3

Solving for M for the yield condition, My =

πd 3 S y π(0.162) 3 (154 300) = 57.2 lbf · in = 32K i 32(1.125)

Count the turns when M = 0 N = 2.5 −

My d 4 E/(10.8D N )

from which N= =

2.5 1 + [10.8D M y /(d 4 E)] 2.5 = 2.417 turns 1 + {[10.8(1.088)(57.2)]/[(0.162) 4 (28.5)(106 )]}

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Chapter 10

This means (2.5 − 2.417)(360◦ ) or 29.9◦ from closed. Treating the hand force as in the middle of the grip 3.5 = 2.75 in r =1+ 2 My 57.2 = = 20.8 lbf Ans. F= r 2.75 10-33

The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500, (a) D = 0.500 − 0.081 = 0.419 in Using E = 28.6 Mpsi for an estimate (0.081) 4 (28.6)(106 ) d4 E k = = = 24.7 lbf · in/turn 10.8D N 10.8(0.419)(11) for each spring. The moment corresponding to a force of 8 lbf Fr = (8/2)(3.3125) = 13.25 lbf · in/spring The fraction windup turn is n=

13.25 Fr = 0.536 turns = k 24.7

The arm swings through an arc of slightly less than 180◦ , say 165◦ . This uses up 165/360 or 0.458 turns. So n = 0.536 − 0.458 = 0.078 turns are left (or 0.078(360◦ ) = 28.1◦ ). The original configuration of the spring was

Ans. 28.1

(b)

C=

0.419 = 5.17 0.081

Ki =

4(5.17) 2 − 5.17 − 1 = 1.168 4(5.17)(5.17 − 1)

32M πd 3 32(13.25) = 296 623 psi Ans. = 1.168 π(0.081) 3

σ = Ki

To achieve this stress level, the spring had to have set removed. 10-34

Consider half and double results 3FR

Straight section:

L兾2

F

M = 3F R,

∂M = 3R ∂P

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Upper 180◦ section: M = F[R + R(1 − cos φ)] ␾ ␪

R

= F R(2 − cos φ),

∂M = R(2 − cos φ) ∂P

F

Lower section:

M = F R sin θ ∂M = R sin θ ∂P

Considering bending only: L/2 π π/2 2 2 2 2 2 δ= 9F R dx + F R (2 − cos φ) R dφ + F( R sin θ) R dθ EI 0 0 0

π π 2F 9 2

3 3 π R L + R 4π − 4 sin φ + +R = 0 EI 2 2 4 2F R 2 19π 9 F R2 = R+ L = (19π R + 18L) Ans. EI 4 2 2E I 10-35


10-36


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Chapter 11 11-1

For the deep-groove 02-series ball bearing with R = 0.90, the design life x D , in multiples of rating life, is xD =

30 000(300)(60) = 540 106

Ans.

The design radial load FD is FD = 1.2(1.898) = 2.278 kN From Eq. (11-6),

C10

540 = 2.278 0.02 + 4.439[ln(1/0.9)]1/1.483 = 18.59 kN

1/3

Ans.

Table 11-2: Choose a 02-30 mm with C10 = 19.5 kN. Ans. Eq. (11-18):

1.483 540(2.278/19.5) 3 − 0.02 R = exp − 4.439 = 0.919 Ans.

11-2

For the Angular-contact 02-series ball bearing as described, the rating life multiple is xD =

50 000(480)(60) = 1440 106

The design load is radial and equal to FD = 1.4(610) = 854 lbf = 3.80 kN Eq. (11-6):

C10

1440 = 854 0.02 + 4.439[ln(1/0.9)]1/1.483

1/3

= 9665 lbf = 43.0 kN Table 11-2: Select a 02-55 mm with C10 = 46.2 kN. Ans. Using Eq. (11-18),

1.483 1440(3.8/46.2) 3 − 0.02 R = exp − 4.439 = 0.927 Ans.

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For the straight-Roller 03-series bearing selection, x D = 1440 rating lives from Prob. 11-2 solution. FD = 1.4(1650) = 2310 lbf = 10.279 kN 1440 3/10 = 91.1 kN C10 = 10.279 1 Table 11-3: Select a 03-55 mm with C10 = 102 kN. Ans. Using Eq. (11-18),

1.483 1440(10.28/102) 10/3 − 0.02 R = exp − = 0.942 Ans. 4.439

11-4

√ We can choose a reliability goal of 0.90 = 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R1 . Then set the reliability goal of the second as R2 =

0.90 R1

or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. 11-5

Establish a reliability goal of tact ball bearing,

√ 0.90 = 0.95 for each bearing. For a 02-series angular con

C10

1440 = 854 0.02 + 4.439[ln(1/0.95)]1/1.483

1/3

= 11 315 lbf = 50.4 kN Select a 02-60 mm angular-contact bearing with C10 = 55.9 kN. 1.483 1440(3.8/55.9) 3 − 0.02 R A = exp − = 0.969 4.439 For a 03-series straight-roller bearing, 3/10 1440 = 105.2 kN C10 = 10.279 0.02 + 4.439[ln(1/0.95)]1/1.483 Select a 03-60 mm straight-roller bearing with C10 = 123 kN. 1.483 1440(10.28/123) 10/3 − 0.02 R B = exp − = 0.977 4.439

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Chapter 11

Form a table of existing reliabilities Rgoal

RA

RB

0.912

0.90 0.95

0.927 0.969

0.941 0.977

0.872 0.947 0.906

The possible products in the body of the table are displayed to the right of the table. One, 0.872, is predictably less than the overall reliability goal. The remaining three are the choices for a combined reliability goal of 0.90. Choices can be compared for the cost of bearings, outside diameter considerations, bore implications for shaft modifications and housing modifications. The point is that the designer has choices. Discover them before making the selection decision. Did the answer to Prob. 11-4 uncover the possibilities? To reduce the work to fill in the body of the table above, a computer program can be helpful. 11-6

Choose a 02-series ball bearing from manufacturer #2, having a service factor of 1. For Fr = 8 kN and Fa = 4 kN xD = Eq. (11-5):

5000(900)(60) = 270 106

C10

270 =8 0.02 + 4.439[ln(1/0.90)]1/1.483

1/3

= 51.8 kN

Trial #1: From Table (11-2) make a tentative selection of a deep-groove 02-70 mm with C0 = 37.5 kN. 4 Fa = 0.107 = C0 37.5 Table 11-1: Fa /(V Fr ) = 0.5 > e X 2 = 0.56,

Y2 = 1.46

Eq. (11-9): Fe = 0.56(1)(8) + 1.46(4) = 10.32 kN Eq. (11-6):

C10

270 = 10.32 1

1/3

= 66.7 kN > 61.8 kN

Trial #2: From Table 11-2 choose a 02-80 mm having C10 = 70.2 and C0 = 45.0. Check: Fa 4 = 0.089 = C0 45

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Table 11-1: X 2 = 0.56, Y2 = 1.53 Fe = 0.56(8) + 1.53(4) = 10.60 kN Eq. (11-6):

C10

270 = 10.60 1

1/3

= 68.51 kN < 70.2 kN

∴ Selection stands. Decision: Specify a 02-80 mm deep-groove ball bearing. 11-7

Ans.

From Prob. 11-6, x D = 270 and the final value of Fe is 10.60 kN. 1/3 270 = 84.47 kN C10 = 10.6 0.02 + 4.439[ln(1/0.96)]1/1.483 Table 11-2: Choose a deep-groove ball bearing, based upon C10 load ratings. Trial #1: Tentatively select a 02-90 mm. C10 = 95.6,

C0 = 62 kN

4 Fa = 0.0645 = C0 62 From Table 11-1, interpolate for Y2 . Fa /C0

Y2

0.056 0.0645 0.070

1.71 Y2 1.63

0.0645 − 0.056 Y2 − 1.71 = = 0.607 1.63 − 1.71 0.070 − 0.056 Y2 = 1.71 + 0.607(1.63 − 1.71) = 1.661 Fe = 0.56(8) + 1.661(4) = 11.12 kN 1/3 270 C10 = 11.12 0.02 + 4.439[ln(1/0.96)]1/1.483 = 88.61 kN < 95.6 kN Bearing is OK. Decision: Specify a deep-groove 02-90 mm ball bearing.

Ans.

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Chapter 11

11-8

For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.90 and Fr = 12 kN 4000(750)(60) = 180 106 180 3/10 = 12 = 57.0 kN Ans. 1

xD = C10 11-9

y R yO

O

z

T

R zO

Pz R yA 11

1" 2

F

Py R zA

A B

2

3" 4

20⬚

x T

Assume concentrated forces as shown.

Pz = 8(24) = 192 lbf Py = 8(30) = 240 lbf T = 192(2) = 384 lbf · in T x = −384 + 1.5F cos 20◦ = 0 F=

384 = 272 lbf 1.5(0.940)

M Oz = 5.75Py + 11.5R A − 14.25F sin 20◦ = 0; y

y

5.75(240) + 11.5R A − 14.25(272)(0.342) = 0

thus y

thus

R A = −4.73 lbf M O = −5.75Pz − 11.5R zA − 14.25F cos 20◦ = 0; y

−5.75(192) − 11.5R zA − 14.25(272)(0.940) = 0

R zA = −413 lbf;

R A = [(−413) 2 + (−4.73) 2 ]1/2 = 413 lbf

z + Pz + R zA + F cos 20◦ = 0 F z = RO z + 192 − 413 + 272(0.940) = 0 RO z = −34.7 lbf RO

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F y = R O + Py + R A − F sin 20◦ = 0 y

y

y

R O + 240 − 4.73 − 272(0.342) = 0 y

R O = −142 lbf R O = [(−34.6) 2 + (−142) 2 ]1/2 = 146 lbf So the reaction at A governs. √ Reliability Goal: 0.92 = 0.96 FD = 1.2(413) = 496 lbf x D = 30 000(300)(60/106 ) = 540 1/3 540 C10 = 496 0.02 + 4.439[ln(1/0.96)]1/1.483 = 4980 lbf = 22.16 kN A 02-35 bearing will do. Decision: Specify an angular-contact 02-35 mm ball bearing for the locations at A and O. Check combined reliability. Ans. 11-10

For a combined reliability goal of 0.90, use

√ 0.90 = 0.95 for the individual bearings.

y

O z

RO A

FA

20 B

FC

RB

x0 = 20⬚

50 000(480)(60) = 1440 106

16 C

x

10

The resultant of the given forces are R O = 607 lbf and R B = 1646 lbf. At O: Fe = 1.4(607) = 850 lbf C10 = 850 Ball:

1440 0.02 + 4.439[ln(1/0.95)]1/1.483

1/3

= 11 262 lbf or 50.1 kN Select a 02-60 mm angular-contact ball bearing with a basic load rating of 55.9 kN. At B: Fe = 1.4(1646) = 2304 lbf C10 = 2304 Roller:

1440 0.02 + 4.439[ln(1/0.95)]1/1.483

3/10

= 23 576 lbf or 104.9 kN Select a 02-80 mm cylindrical roller or a 03-60 mm cylindrical roller. The 03-series roller has the same bore as the 02-series ball.

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Chapter 11

11-11

The reliability of the individual bearings is R =

√ 0.999 = 0.9995

y R yO

R zO

O z

A 300 F zA

FC F yA 400

R yE

C

E 150

x R zE

From statics, y

R O = −163.4 N, y RE

= −89.1 N,

z RO = 107 N,

R Ez = −174.4 N,

R O = 195 N R E = 196 N

60 000(1200)(60) = 4320 106 1/3 4340 = 0.196 0.02 + 4.439[ln(1/0.9995)]1/1.483

xD = C10

= 8.9 kN A 02-25 mm deep-groove ball bearing has a basic load rating of 14.0 kN which is ample. An extra-light bearing could also be investigated. 11-12

Given: Fr A = 560 lbf or 2.492 kN Fr B = 1095 lbf or 4.873 kN Trial #1: Use K A = K B = 1.5 and from Table 11-6 choose an indirect mounting. 0.47Fr B 0.47Fr A − (−1)(0) KA KB 0.47(4.873) 0.47(2.492) 1.5 1.5 0.781 < 1.527 Therefore use the upper line of Table 11-6. 0.47Fr B Fa A = Fa B = = 1.527 kN KB PA = 0.4Fr A + K A Fa A = 0.4(2.492) + 1.5(1.527) = 3.29 kN PB = Fr B = 4.873 kN

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Fig. 11-16:

f T = 0.8

Fig. 11-17:

f V = 1.07

a3l = f T f V = 0.8(1.07) = 0.856 √ Individual reliability: Ri = 0.9 = 0.95

Thus,

Eq. (11-17):

40 000(400)(60) (C10 ) A = 1.4(3.29) 4.48(0.856)(1 − 0.95) 2/3 (90)(106 ) = 11.40 kN

0.3

40 000(400)(60) (C10 ) B = 1.4(4.873) 4.48(0.856)(1 − 0.95) 2/3 (90)(106 )

0.3

= 16.88 kN From Fig. 11-14, choose cone 32 305 and cup 32 305 which provide Fr = 17.4 kN and K = 1.95. With K = 1.95 for both bearings, a second trial validates the choice of cone 32 305 and cup 32 305. Ans. 11-13 R=

R zO

y

T = 240(12)(cos 20◦ ) = 2706 lbf · in

82.1

O

16"

z R yO

210

T A

√ 0.95 = 0.975

F=

451

14"

226 T

B

R zC

12" C

2706 = 498 lbf 6 cos 25◦

x y

RC

In xy-plane:

y

M O = −82.1(16) − 210(30) + 42RC = 0 y

RC = 181 lbf y

R O = 82 + 210 − 181 = 111 lbf In xz-plane:

M O = 226(16) − 452(30) − 42Rcz = 0 RCz = −237 lbf z RO = 226 − 451 + 237 = 12 lbf R O = (1112 + 122 ) 1/2 = 112 lbf Ans. RC = (1812 + 2372 ) 1/2 = 298 lbf FeO = 1.2(112) = 134.4 lbf FeC = 1.2(298) = 357.6 lbf 40 000(200)(60) = 480 xD = 106

Ans.

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Chapter 11

(C10 ) O

480 = 134.4 0.02 + 4.439[ln(1/0.975)]1/1.483

1/3

= 1438 lbf or 6.398 kN 1/3 480 (C10 ) C = 357.6 0.02 + 4.439[ln(1/0.975)]1/1.483 = 3825 lbf or 17.02 kN Bearing at O: Choose a deep-groove 02-12 mm.

Ans.

Bearing at C: Choose a deep-groove 02-30 mm. Ans. There may be an advantage to the identical 02-30 mm bearings in a gear-reduction unit. 11-14

Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (Roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust is heavily loaded compared to the other bearing. The second bearing is thus oversized and does not contribute √ measurably to the chance of failure. This is predictable. The reliability goal is not 0.99, but 0.99 for the ball bearing. The reliability of the roller is 1. Beginning here saves effort. Bearing at A (Ball) Fr = (362 + 2122 ) 1/2 = 215 lbf = 0.957 kN Fa = 555 lbf = 2.47 kN Trial #1: Tentatively select a 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN . 2.47 Fa = 0.0392 = C0 63.0 xD =

25 000(600)(60) = 900 106

Table 11-1: X 2 = 0.56, Y2 = 1.88 Fe = 0.56(0.957) + 1.88(2.47) = 5.18 kN FD = f A Fe = 1.3(5.18) = 6.73 kN 1/3 900 C10 = 6.73 0.02 + 4.439[ln(1/0.99)]1/1.483 = 107.7 kN > 90.4 kN Trial #2: Tentatively select a 02-95 mm angular-contact ball with C10 = 121 kN and C0 = 85 kN. Fa 2.47 = 0.029 = C0 85

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Table 11-1: Y2 = 1.98 Fe = 0.56(0.957) + 1.98(2.47) = 5.43 kN FD = 1.3(5.43) = 7.05 kN 1/3 900 C10 = 7.05 0.02 + 4.439[ln(1/0.99)]1/1.483 = 113 kN < 121 kN O.K. Select a 02-95 mm angular-contact ball bearing.

Ans.

Bearing at B (Roller): Any bearing will do since R = 1. Let’s prove it. From Eq. (11-18) when a f FD 3 x D < x0 R=1 C10 The smallest 02-series roller has a C10 = 16.8 kN for a basic load rating. 0.427 3 (900) < ? > 0.02 16.8 0.0148 < 0.02 ∴R = 1 √ Spotting this early avoided rework from 0.99 = 0.995. Any 02-series roller bearing will do. Same bore or outside diameter is a common choice. (Why?) Ans. 11-15

Hoover Ball-bearing Division uses the same 2-parameter Weibull model as Timken: b = 1.5, θ = 4.48. We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use line AB. In this case, B is to the right of A. 115(2000)(16) = 13.8 106 This establishes point 1 on the R = 0.90 line. (x) 1 =

For F = 18 kN,

log F

F

2

100

39.6

A

B R ⫽ 0.20 R ⫽ 0.90

1

10

0

1

2

1

18

13.8

72

1

10

100

0

1

2

x log x

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Chapter 11

The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter Weibull distribution, x0 = 0 and points A and B are related by: x A = θ[ln(1/0.90)]1/b

(1)

x B = θ[ln(1/0.20)]1/b and x B /x A is in the same ratio as 600/115. Eliminating θ b=

ln[ln(1/0.20)/ ln(1/0.90)] = 1.65 ln(600/115)

Solving for θ in Eq. (1) θ=

xA 1 = = 3.91 1 / 1 . 65 [ln(1/R A )] [ln(1/0.90)]1/1.65

Therefore, for the data at hand,

x R = exp − 3.91

1.65

Check R at point B: x B = (600/115) = 5.217

5.217 R = exp − 3.91

1.65

= 0.20

Note also, for point 2 on the R = 0.20 line. log(5.217) − log(1) = log(xm ) 2 − log(13.8) (xm ) 2 = 72 11-16

This problem is rich in useful variations. Here is one. Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of (0.99) 1/6 = 0.9983. Shaft a FAr = (2392 + 1112 ) 1/2 = 264 lbf or 1.175 kN FBr = (5022 + 10752 ) 1/2 = 1186 lbf or 5.28 kN Thus the bearing at B controls xD =

10 000(1200)(60) = 720 106

0.02 + 4.439[ln(1/0.9983)]1/1.483 = 0.080 26 0.3 720 C10 = 1.2(5.2) = 97.2 kN 0.080 26 Select either a 02-80 mm with C10 = 106 kN or a 03-55 mm with C10 = 102 kN

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Shaft b FCr = (8742 + 22742 ) 1/2 = 2436 lbf or 10.84 kN FDr = (3932 + 6572 ) 1/2 = 766 lbf or 3.41 kN The bearing at C controls 10 000(240)(60) = 144 106 144 0.3 = 1.2(10.84) = 122 kN 0.0826

xD = C10

Select either a 02-90 mm with C10 = 142 kN or a 03-60 mm with C10 = 123 kN Shaft c FEr = (11132 + 23852 ) 1/2 = 2632 lbf or 11.71 kN FFr = (4172 + 8952 ) 1/2 = 987 lbf or 4.39 kN The bearing at E controls x D = 10 000(80)(60/106 ) = 48 0.3 48 C10 = 1.2(11.71) = 94.8 kN 0.0826 Select a 02-80 mm with C10 = 106 kN or a 03-60 mm with C10 = 123 kN 11-17

The horizontal separation of the R = 0.90 loci in a log F-log x plot such as Fig. 11-5 will be demonstrated. We refer to the solution of Prob. 11-15 to plot point G (F = 18 kN, x G = 13.8). We know that (C10 ) 1 = 39.6 kN, x1 = 1. This establishes the unimproved steel R = 0.90 locus, line AG. For the improved steel 360(2000)(60) = 43.2 106 = 43.2), and draw the R = 0.90 locus Am G parallel

(xm ) 1 = We plot point G ( F = 18 kN, x G to AG log F F 2 100 55.8 39.6

Improved steel Am

Unimproved steel R ⫽ 0.90

A R ⫽ 0.90

G

18

G⬘ 1

1

10

0

1

3

13.8 1 0

10 1

1 3

43.2 x 100 2 log x

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Chapter 11

309

We can calculate (C10 ) m by similar triangles. log 39.6 − log 18 log(C10 ) m − log 18 = log 43.2 − log 1 log 13.8 − log 1 log 43.2 39.6 log + log 18 log(C10 ) m = log 13.8 18 (C10 ) m = 55.8 kN The usefulness of this plot is evident. The improvement is 43.2/13.8 = 3.13 fold in life. This result is also available by (L 10 ) m /(L 10 ) 1 as 360/115 or 3.13 fold, but the plot shows the improvement is for all loading. Thus, the manufacturer’s assertion that there is at least a 3-fold increase in life has been demonstrated by the sample data given. Ans. 11-18

Express Eq. (11-1) as a F1a L 1 = C10 L 10 = K

For a ball bearing, a = 3 and for a 02-30 mm angular contact bearing, C10 = 20.3 kN. K = (20.3) 3 (106 ) = 8.365(109 ) At a load of 18 kN, life L 1 is given by: K 8.365(109 ) L1 = a = = 1.434(106 ) rev 3 F1 18 For a load of 30 kN, life L 2 is: 8.365(109 ) L2 = = 0.310(106 ) rev 303 In this case, Eq. (7-57) – the Palmgren-Miner cycle ratio summation rule – can be expressed as l2 l1 + =1 L1 L2 Substituting, l2 200 000 + =1 6 1.434(10 ) 0.310(106 ) l2 = 0.267(106 ) rev Ans. Check: 0.267(106 ) 200 000 + = 1 O.K. 1.434(106 ) 0.310(106 ) 11-19

Total life in revolutions Let: l = total turns f 1 = fraction of turns at F1 f 2 = fraction of turns at F2

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From the solution of Prob. 11-18, L 1 = 1.434(106 ) rev and L 2 = 0.310(106 ) rev. Palmgren-Miner rule: l2 f 1l f 2l l1 + = + =1 L1 L2 L1 L2 from which 1 l= f 1 /L 1 + f 2 /L 2 l=

1 {0.40/[1.434(106 )]} + {0.60/[0.310(106 )]}

= 451 585 rev Ans. Total life in loading cycles 4 min at 2000 rev/min = 8000 rev 12 000 rev 6 min at 2000 rev/min = 10 min/cycle 20 000 rev/cycle 451 585 rev = 22.58 cycles Ans. 20 000 rev/cycle Total life in hours

11-20

min 22.58 cycles 10 = 3.76 h Ans. cycle 60 min/h

While we made some use of the log F-log x plot in Probs. 11-15 and 11-17, the principal use of Fig. 11-5 is to understand equations (11-6) and (11-7) in the discovery of the catalog basic load rating for a case at hand. Point D FD = 495.6 lbf log FD = log 495.6 = 2.70 xD =

30 000(300)(60) = 540 106

log x D = log 540 = 2.73 K D = FD3 x D = (495.6) 3 (540) = 65.7(109 ) lbf3 · turns log K D = log[65.7(109 )] = 10.82 FD has the following uses: Fdesign , Fdesired , Fe when a thrust load is present. It can include application factor a f , or not. It depends on context.

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Point B

x B = 0.02 + 4.439[ln(1/0.99)]1/1.483 = 0.220 turns log x B = log 0.220 = −0.658 1/3 xD 540 1/3 FB = FD = 495.6 = 6685 lbf xB 0.220

Note: Example 11-3 used Eq. (11-7). Whereas, here we basically used Eq. (11-6). log FB = log(6685) = 3.825 K D = 66853 (0.220) = 65.7(109 ) lbf3 · turns

(as it should)

Point A FA log C10 xA log x A K 10

= FB = C10 = 6685 lbf = log(6685) = 3.825 =1 = log(1) = 0 3 = FA3 x A = C10 (1) = 66853 = 299(109 ) lbf3 · turns

Note that K D /K 10 = 65.7(109 )/[299(109 )] = 0.220 , which is x B . This is worth knowing since KD K 10 = xB log K 10 = log[299(109 )] = 11.48 log F

F

4

104

B

6685

3

A

103 D

495.6

⫺0.658 2

540

102 0.1 ⫺1

1 0

10 1

102 2

x 103 3 log x

Now C10 = 6685 lbf = 29.748 kN, which is required for a reliability goal of 0.99. If we select an angular contact 02-40 mm ball bearing, then C10 = 31.9 kN = 7169 lbf.

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Chapter 12 12-1

Given dmax = 1.000 in and bmin = 1.0015 in, the minimum radial clearance is cmin =

1.0015 − 1.000 bmin − dmax = = 0.000 75 in 2 2

l/d = 1 r= ˙ 1.000/2 = 0.500 r/c = 0.500/0.000 75 = 667 N = 1100/60 = 18.33 rev/s P = W/(ld) = 250/[(1)(1)] = 250 psi 8(10−6 )(18.33) 2 = 0.261 S = (667 ) 250

Also

Eq. (12-7):

h 0 /c = 0.595 Q/(rcNl) = 3.98 f r/c = 5.8 Q s /Q = 0.5 h 0 = 0.595(0.000 75) = 0.000 466 in Ans.

Fig. 12-16: Fig. 12-19: Fig. 12-18: Fig. 12-20:

f =

5.8 5.8 = = 0.0087 r/c 667

The power loss in Btu/s is H=

2π(0.0087)(250)(0.5)(18.33) 2π f W r N = 778(12) 778(12)

= 0.0134 Btu/s Ans. Q = 3.98rcNl = 3.98(0.5)(0.000 75)(18.33)(1) = 0.0274 in3 /s Q s = 0.5(0.0274) = 0.0137 in3 /s Ans. 12-2

1.252 − 1.250 bmin − dmax = = 0.001 in 2 2 . r = 1.25/2 = 0.625 in

cmin =

r/c = 0.625/0.001 = 625 N = 1150/60 = 19.167 rev/s P=

400 = 128 psi 1.25(2.5)

l/d = 2.5/1.25 = 2 S=

(6252 )(10)(10−6 )(19.167) = 0.585 128

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Chapter 12

The interpolation formula of Eq. (12-16) will have to be used. From Figs. 12-16, 12-21, and 12-19 Q = 3.09 l/d = ∞, h o /c = 0.96, P/ pmax = 0.84, For rcNl Q l/d = 1, h o /c = 0.77, P/ pmax = 0.52, = 3.6 rcNl 1 l/d = , 2

h o /c = 0.54,

P/ pmax = 0.42,

Q = 4.4 rcNl

1 l/d = , 4

h o /c = 0.31,

P/ pmax = 0.28,

Q = 5.25 rcNl

Equation (12-16) is easily programmed by code or by using a spreadsheet. The results are:

h o /c P/ pmax Q/rcNl ∴

l/d

y∞

y1

y1/2

y1/4

yl/d

2 2 2

0.96 0.84 3.09

0.77 0.52 3.60

0.54 0.42 4.40

0.31 0.28 5.25

0.88 0.64 3.28

h o = 0.88(0.001) = 0.000 88 in Ans.

pmax =

128 = 200 psi Ans. 0.64

Q = 3.28(0.625)(0.001)(19.167)(2.5) = 0.098 in3 /s Ans. 12-3 cmin =

3.005 − 3.000 bmin − dmax = = 0.0025 in 2 2

. r = 3.000/2 = 1.500 in l/d = 1.5/3 = 0.5 r/c = 1.5/0.0025 = 600 N = 600/60 = 10 rev/s P=

800 = 177.78 psi 1.5(3)

Fig. 12-12: SAE 10, µ = 1.75 µreyn 1.75(10−6 )(10) 2 = 0.0354 S = (600 ) 177.78 Figs. 12-16 and 12-21: h o /c = 0.11,

P/ pmax = 0.21

h o = 0.11(0.0025) = 0.000 275 in Ans. pmax = 177.78/0.21 = 847 psi Ans.

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Fig. 12-12: SAE 40, µ = 4.5 µreyn

4.5 S = 0.0354 1.75 h o /c = 0.19,

= 0.0910

P/ pmax = 0.275

h o = 0.19(0.0025) = 0.000 475 in Ans. pmax = 177.78/0.275 = 646 psi Ans. 12-4 cmin =

3.006 − 3.000 bmin − dmax = = 0.003 2 2

. r = 3.000/2 = 1.5 in l/d = 1 r/c = 1.5/0.003 = 500

N = 750/60 = 12.5 rev/s P=

600 = 66.7 psi 3(3)

Fig. 12-14: SAE 10W, µ = 2.1 µreyn 2.1(10−6 )(12.5) 2 S = (500 ) = 0.0984 66.7 From Figs. 12-16 and 12-21: h o /c = 0.34,

P/ pmax = 0.395

h o = 0.34(0.003) = 0.001 020 in Ans. pmax =

66.7 = 169 psi Ans. 0.395

Fig. 12-14: SAE 20W-40, µ = 5.05 µreyn 5.05(10−6 )(12.5) 2 = 0.237 S = (500 ) 66.7 From Figs. 12-16 and 12-21: h o /c = 0.57,

P/ pmax = 0.47

h o = 0.57(0.003) = 0.001 71 in Ans. pmax =

66.7 = 142 psi Ans. 0.47

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12-5

2.0024 − 2 bmin − dmax = = 0.0012 in 2 2 2 . d r = = = 1 in, l/d = 1/2 = 0.50 2 2 r/c = 1/0.0012 = 833 N = 800/60 = 13.33 rev/s 600 = 300 psi P= 2(1)

cmin =

Fig. 12-12: SAE 20, µ = 3.75 µreyn 3.75(10−6 )(13.3) 2 = 0.115 S = (833 ) 300 From Figs. 12-16, 12-18 and 12-19: h o /c = 0.23, r f /c = 3.8, Q/(rcNl) = 5.3 h o = 0.23(0.0012) = 0.000 276 in Ans. 3.8 = 0.004 56 f = 833 The power loss due to friction is 2π(0.004 56)(600)(1)(13.33) 2π f W r N = H= 778(12) 778(12) = 0.0245 Btu/s Ans. Q = 5.3rcNl = 5.3(1)(0.0012)(13.33)(1) = 0.0848 in3 /s Ans. 12-6

25.04 − 25 bmin − dmax = = 0.02 mm 2 2 r= ˙ d/2 = 25/2 = 12.5 mm, l/d = 1 r/c = 12.5/0.02 = 625 N = 1200/60 = 20 rev/s 1250 P= = 2 MPa 252 50(10−3 )(20) 2 = 0.195 S = (625 ) For µ = 50 MPa · s, 2(106 ) cmin =

From Figs. 12-16, 12-18 and 12-20: h o /c = 0.52,

f r/c = 4.5, Q s /Q = 0.57 h o = 0.52(0.02) = 0.0104 mm Ans. 4.5 = 0.0072 f = 625 T = f W r = 0.0072(1.25)(12.5) = 0.1125 N · m

315

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The power loss due to friction is H = 2π T N = 2π(0.1125)(20) = 14.14 W Ans. Q s = 0.57Q 12-7 cmin = r=

The side flow is 57% of Q

Ans.

30.05 − 30.00 bmin − dmax = = 0.025 mm 2 2 30 d = = 15 mm 2 2

15 r = = 600 c 0.025 N=

1120 = 18.67 rev/s 60

2750 = 1.833 MPa 30(50) 60(10−3 )(18.67) 2 S = (600 ) = 0.22 1.833(106 )

P=

50 l = = 1.67 d 30 This l/d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16). From Fig. 12-16, the h o /c values are: y1/4 = 0.18,

y1/2 = 0.34,

y1 = 0.54,

y∞ = 0.89

ho = 0.659 c

Substituting into Eq. (12-16), From Fig. 12-18, the f r/c values are: y1/4 = 7.4, Substituting into Eq. (12-16),

y1/2 = 6.0,

y1 = 5.0,

y∞ = 4.0

fr = 4.59 c

From Fig. 12-19, the Q/(rcNl) values are: y1/4 = 5.65, Substituting into Eq. (12-16),

y1/2 = 5.05, y1 = 4.05, Q = 3.605 rcN l

y∞ = 2.95

h o = 0.659(0.025) = 0.0165 mm Ans. f = 4.59/600 = 0.007 65 Ans. Q = 3.605(15)(0.025)(18.67)(50) = 1263 mm3 /s Ans.

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Chapter 12

12-8

75.10 − 75 bmin − dmax = = 0.05 mm 2 2 l/d = 36/75 = ˙ 0.5 (close enough)

cmin =

r = d/2 = 75/2 = 37.5 mm r/c = 37.5/0.05 = 750 N = 720/60 = 12 rev/s 2000 P= = 0.741 MPa 75(36) Fig. 12-13: SAE 20, µ = 18.5 MPa · s 18.5(10−3 )(12) 2 = 0.169 S = (750 ) 0.741(106 ) From Figures 12-16, 12-18 and 12-21: h o /c = 0.29,

f r/c = 5.1,

P/ pmax = 0.315

h o = 0.29(0.05) = 0.0145 mm Ans. f = 5.1/750 = 0.0068 T = f W r = 0.0068(2)(37.5) = 0.51 N · m The heat loss rate equals the rate of work on the film Hloss = 2π T N = 2π(0.51)(12) = 38.5 W Ans. pmax = 0.741/0.315 = 2.35 MPa Ans. Fig. 12-13: SAE 40, µ = 37 MPa · s S = 0.169(37)/18.5 = 0.338 From Figures 12-16, 12-18 and 12-21: h o /c = 0.42,

f r/c = 8.5,

P/ pmax = 0.38

h o = 0.42(0.05) = 0.021 mm Ans. f = 8.5/750 = 0.0113 T = f W r = 0.0113(2)(37.5) = 0.85 N · m Hloss = 2π T N = 2π(0.85)(12) = 64 W Ans. pmax = 0.741/0.38 = 1.95 MPa Ans. 12-9

50.05 − 50 bmin − dmax = = 0.025 mm 2 2 r = d/2 = 50/2 = 25 mm

cmin =

r/c = 25/0.025 = 1000 l/d = 25/50 = 0.5, N = 840/60 = 14 rev/s 2000 P= = 1.6 MPa 25(50)

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Fig. 12-13: SAE 30, µ = 34 MPa · s

34(10−3 )(14) = 0.2975 S = (1000 ) 1.6(106 )

2

From Figures 12-16, 12-18, 12-19 and 12-20: h o /c = 0.40,

f r/c = 7.8,

Q s /Q = 0.74,

Q/(rcNl) = 4.9

h o = 0.40(0.025) = 0.010 mm Ans. f = 7.8/1000 = 0.0078 T = f W r = 0.0078(2)(25) = 0.39 N · m H = 2π T N = 2π(0.39)(14) = 34.3 W Ans. Q = 4.9rcNl = 4.9(25)(0.025)(14)(25) = 1072 mm2 /s Q s = 0.74(1072) = 793 mm3 /s Ans. 12-10

Consider the bearings as specified by minimum f:

+0 , d−t d

+tb b− 0

maximum W:

+0 , d−t d

+tb b− 0

and differing only in d and d . Preliminaries: l/d = 1 P = 700/(1.252 ) = 448 psi N = 3600/60 = 60 rev/s Fig. 12-16: S= ˙ 0.08 S= ˙ 0.20

minimum f : maximum W: Fig. 12-12:

µ = 1.38(10−6 ) reyn µN/P = 1.38(10−6 )(60/448) = 0.185(10−6 )

Eq. (12-7):

r = c

For minimum f:

S µN/P

r = c

0.08 = 658 0.185(10−6 )

. c = 0.625/658 = 0.000 950 = 0.001 in

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Chapter 12

319

If this is cmin , b − d = 2(0.001) = 0.002 in The median clearance is c¯ = cmin +

td + tb td + tb = 0.001 + 2 2

and the clearance range for this bearing is c =

td + tb 2

which is a function only of the tolerances. For maximum W:

r = c

0.2 = 1040 0.185(10−6 )

. c = 0.625/1040 = 0.000 600 = 0.0005 in If this is cmin b − d = 2cmin = 2(0.0005) = 0.001 in c¯ = cmin + c =

td + tb td + tb = 0.0005 + 2 2

td + tb 2

The difference (mean) in clearance between the two clearance ranges, crange , is td + tb td + tb − 0.0005 + crange = 0.001 + 2 2 = 0.0005 in For the minimum f bearing b − d = 0.002 in or d = b − 0.002 in For the maximum W bearing d = b − 0.001 in For the same b, tb and td , we need to change the journal diameter by 0.001 in. d − d = b − 0.001 − (b − 0.002) = 0.001 in Increasing d of the minimum friction bearing by 0.001 in, defines d of the maximum load bearing. Thus, the clearance range provides for bearing dimensions which are attainable in manufacturing. Ans.

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320


12-11

Given: SAE 30, N = 8 rev/s, Ts = 60°C, l/d = 1, W = 3000 N cmin =

d = 80 mm,

b = 80.08 mm,

80.08 − 80 bmin − dmax = = 0.04 mm 2 2

r = d/2 = 80/2 = 40 mm 40 r = = 1000 c 0.04 P=

3000 = 0.469 MPa 80(80)

Trial #1: From Figure 12-13 for T = 81°C, µ = 12 MPa · s T = 2(81°C − 60°C) = 42°C 12(10−3 )(8) 2 S = (1000 ) = 0.2047 0.469(106 ) From Fig. 12-24, 0.120T = 0.349 + 6.009(0.2047) + 0.0475(0.2047) 2 = 1.58 P 0.469 = 6.2°C T = 1.58 0.120 Discrepancy = 42°C − 6.2°C = 35.8°C Trial #2: From Figure 12-13 for T = 68°C, µ = 20 MPa · s, T = 2(68°C − 60°C) = 16°C 20 = 0.341 S = 0.2047 12 From Fig. 12-24, 0.120T = 0.349 + 6.009(0.341) + 0.0475(0.341) 2 = 2.4 P 0.469 = 9.4°C T = 2.4 0.120 Discrepancy = 16°C − 9.4°C = 6.6°C Trial #3: µ = 21 MPa · s, T = 65°C T = 2(65°C − 60°C) = 10°C 21 = 0.358 S = 0.2047 12

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Chapter 12

From Fig. 12-24, 0.120T = 0.349 + 6.009(0.358) + 0.0475(0.358) 2 = 2.5 P 0.469 T = 2.5 = 9.8°C 0.120 Discrepancy = 10°C − 9.8°C = 0.2°C

O.K.

Tav = 65°C Ans. T1 = Tav − T /2 = 65°C − (10°C/2) = 60°C T2 = Tav + T /2 = 65°C + (10°C/2) = 70°C S = 0.358 From Figures 12-16, 12-18, 12-19 and 12-20: ho = 0.68, c

fr/c = 7.5,

Q = 3.8, rcN l

Qs = 0.44 Q

h o = 0.68(0.04) = 0.0272 mm Ans. f =

7.5 = 0.0075 1000

T = f W r = 0.0075(3)(40) = 0.9 N · m H = 2π T N = 2π(0.9)(8) = 45.2 W Ans. Q = 3.8(40)(0.04)(8)(80) = 3891 mm3 /s Q s = 0.44(3891) = 1712 mm3 /s Ans. 12-12

Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110°F, N = 1120 rev/min, and l = 2.5 in. For a trial film temperature T f = 150°F Tf

µ

S

T (From Fig. 12-24)

150

2.421

0.0921

18.5

Tav = Ts +

18.5°F T = 110°F + = 119.3°F 2 2

T f − Tav = 150°F − 119.3°F which is not 0.1 or less, therefore try averaging (T f ) new =

150°F + 119.3°F = 134.6°F 2

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Proceed with additional trials Trial Tf

µ

S

T

Tav

New Tf

150.0 134.6 128.1 125.5 124.5 124.1 124.0

2.421 3.453 4.070 4.255 4.471 4.515 4.532

0.0921 0.1310 0.1550 0.1650 0.1700 0.1710 0.1720

18.5 23.1 25.8 27.0 27.5 27.7 27.8

119.3 121.5 122.9 123.5 123.8 123.9 123.7

134.6 128.1 125.5 124.5 124.1 124.0 123.9

Note that the convergence begins rapidly. There are ways to speed this, but at this point they would only add complexity. Depending where you stop, you can enter the analysis. (a) µ = 4.541(10−6 ), From Fig. 12-16:

S = 0.1724

ho = 0.482, c

h o = 0.482(0.002) = 0.000 964 in

From Fig. 12-17: φ = 56° Ans. (b) e = c − h o = 0.002 − 0.000 964 = 0.001 04 in Ans. (c) From Fig. 12-18:

fr = 4.10, c

f = 4.10(0.002/1.25) = 0.006 56 Ans.

(d) T = f W r = 0.006 56(1200)(1.25) = 9.84 lbf · in 2π(9.84)(1120/60) 2π T N = = 0.124 Btu/s Ans. 778(12) 778(12) Q 1120 = 4.16, Q = 4.16(1.25)(0.002) (2.5) (e) From Fig. 12-19: rcNl 60 H=

= 0.485 in3 /s Ans. From Fig. 12-20: (f) From Fig. 12-21:

Qs = 0.6, Q P pmax

= 0.45,

Q s = 0.6(0.485) = 0.291 in3 /s Ans. pmax =

1200 = 427 psi Ans. 2.52 (0.45) φ pmax = 16° Ans.

(g) φ p0 = 82° Ans. (h) T f = 123.9°F Ans. (i) Ts + T = 110°F + 27.8°F = 137.8°F Ans.

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Chapter 12

12-13

Given: d = 1.250 in, td = 0.001in, b = 1.252 in, tb = 0.003in, l = 1.25 in, W = 250 lbf, N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120°F. Below is a partial tabular summary for comparison purposes. cmin 0.001 in Tf T Tmax µ S fr c Q rcN j l Qs Q ho c f Q Qs ho

c 0.002 in

cmax 0.003 in

125.8 11.5 131.5 3.014 0.053 7 0.775 0

124.0 7.96 128.0 3.150 0.024 9 0.873

4.317

1.881

1.243

4.129

4.572

4.691

0.582

0.824

0.903

0.501

0.225

0.127

0.006 9 0.094 1 0.054 8 0.000 501

0.006 0.208 0.172 0.000 495

0.005 9 0.321 0.290 0.000 382

132.2 24.3 144.3 2.587 0.184 0.499

Note the variations on each line. There is not a bearing, but an ensemble of many bearings, due to the random assembly of toleranced bushings and journals. Fortunately the distribution is bounded; the extreme cases, cmin and cmax , coupled with c provide the charactistic description for the designer. All assemblies must be satisfactory. The designer does not specify a journal-bushing bearing, but an ensemble of bearings. 12-14

Computer programs will vary—Fortran based, MATLAB, spreadsheet, etc.

12-15

In a step-by-step fashion, we are building a skill for natural circulation bearings. • Given the average film temperature, establish the bearing properties. • Given a sump temperature, find the average film temperature, then establish the bearing properties. • Now we acknowledge the environmental temperature’s role in establishing the sump temperature. Sec. 12-9 and Ex. 12-5 address this problem. The task is to iteratively find the average film temperature, T f , which makes Hgen and Hloss equal. The steps for determining cmin are provided within Trial #1 through Trial #3 on the following page.

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Trial #1: • Choose a value of T f . • Find the corresponding viscosity. • Find the Sommerfeld number. • Find f r/c , then Hgen

2545 WN c = 1050

fr c

• Find Q/(rcNl) and Q s /Q . From Eq. (12–15) T = Hloss =

0.103P( f r/c) (1 − 0.5Q s /Q)[Q/(rcN j l)] h¯ CR A(T f − T∞ ) 1+α

• Display T f , S, Hgen , Hloss Trial #2: Choose another T f , repeating above drill. Trial #3: Plot the results of the first two trials. H

Hgen Hloss, linear with Tf

(Tf )1

(Tf )3

(Tf )2

Tf

Choose (T f ) 3 from plot. Repeat the drill. Plot the results of Trial #3 on the above graph. If you are not within 0.1°F, iterate again. Otherwise, stop, and find all the properties of the bearing for the first clearance, cmin . See if Trumpler conditions are satisfied, and if so, analyze c¯ and cmax . The bearing ensemble in the current problem statement meets Trumpler’s criteria (for n d = 2). This adequacy assessment protocol can be used as a design tool by giving the students additional possible bushing sizes. b (in)

tb (in)

2.254 2.004 1.753

0.004 0.004 0.003

Otherwise, the design option includes reducing l/d to save on the cost of journal machining and vender-supplied bushings.

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Chapter 12

12-16

Continue to build a skill with pressure-fed bearings, that of finding the average temperature of the fluid film. First examine the case for c = cmin Trial #1: • • • • •

Choose an initial T f . Find the viscosity. Find the Sommerfeld number. Find f r/c, h o /c, and . From Eq. (12-24), find T . Tav = Ts +

• Display T f , S, T, and Tav .

T 2

Trial #2: • Choose another T f . Repeat the drill, and display the second set of values for T f , S, T, and Tav . • Plot Tav vs T f : Tav

2

Tav ⫽ Tf

1 (Tf )2

(Tf )3

(Tf )1

Tf

Trial #3: Pick the third T f from the plot and repeat the procedure. If (T f ) 3 and (Tav ) 3 differ by more than 0.1°F, plot the results for Trials #2 and #3 and try again. If they are within 0.1°F, determine the bearing parameters, check the Trumpler criteria, and compare Hloss with the lubricant’s cooling capacity. Repeat the entire procedure for c = cmax to assess the cooling capacity for the maximum radial clearance. Finally, examine c = c¯ to characterize the ensemble of bearings. 12-17

An adequacy assessment associated with a design task is required. Trumpler’s criteria will do. 0.00 0.010 b = 50.084+ d = 50.00+ −0.05 mm, −0.000 mm SAE 30, N = 2880 rev/min or 48 rev/s, W = 10 kN 50.084 − 50 bmin − dmax = = 0.042 mm cmin = 2 2 r = d/2 = 50/2 = 25 mm r/c = 25/0.042 = 595 1 l = (55 − 5) = 25 mm 2 l /d = 25/50 = 0.5 W 10(106 ) p= = 4000 kPa = 4rl 4(0.25)(0.25) $

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Trial #1: Choose (T f ) 1 = 79°C. From Fig. 12-13, µ = 13 MPa · s. 13(10−3 )(48) 2 = 0.055 S = (595 ) 4000(103 ) fr = 2.3, = 0.85. c 978(106 ) ( f r/c)SW 2 T = 1 + 1.5 2 ps r 4 978(106 ) 2.3(0.055)(102 ) = 1 + 1.5(0.85) 2 200(25) 4

From Figs. 12-18 and 12-16: From Eq. (12-25),

= 76.0°C Tav = Ts + T /2 = 55°C + (76°C/2) = 93°C Trial #2: Choose (T f ) 2 = 100°C. From Fig. 12-13, µ = 7 MPa · s. 7 = 0.0296 S = 0.055 13 fr = 1.6, = 0.90 c 978(106 ) 1.6(0.0296)(102 ) T = = 26.8°C 1 + 1.5(0.9) 2 200(25) 4

From Figs. 12-18 and 12-16:

Tav = 55°C +

26.8°C = 68.4°C 2

Tav 100 90 80

Tav ⫽ Tf

(100⬚C, 100⬚C)

(79⬚C, 93⬚C)

(79⬚C, 79⬚C)

70

(100⬚C, 68.4⬚C) 85⬚C 60

70

80

90

100

Tf

Trial #3: Thus, the plot gives (T f ) 3 = 85°C. From Fig. 12-13, µ = 10.8 MPa · s. 10.8 = 0.0457 S = 0.055 13 fr = 2.2, = 0.875 From Figs. 12-18 and 12-16: c 978(106 ) 2.2(0.0457)(102 ) T = = 58.6°C 1 + 1.5(0.8752 ) 200(25) 4 Tav = 55°C + Result is close. Choose

58.6°C = 84.3°C 2

85°C + 84.3°C = 84.7°C T¯ f = 2

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Chapter 12

µ = 10.8 MPa · s 10.8 = 0.0457 S = 0.055 13 ho fr = 2.23, = 0.874, = 0.13 c c 978(106 ) 2.23(0.0457)(102 ) T = = 59.5°C 1 + 1.5(0.8742 ) 200(254 )

Fig. 12-13:

Tav = 55°C +

59.5°C = 84.7°C 2

O.K.

From Eq. (12-22) π ps rc3 3µl π(200)(0.0423 )(25) 2 = [1 + 1.5(0.874 )] 3(10)(10−6 )(25) = 3334 mm3 /s

Q s = (1 + 1.5 2 )

h o = 0.13(0.042) = 0.005 46 mm or 0.000 215 in Trumpler: h o = 0.0002 + 0.000 04(50/25.4) = 0.000 279 in

Not O.K.

Tmax = Ts + T = 55°C + 63.7°C = 118.7°C or 245.7°F O.K. Pst = 4000 kPa or 581 psi n = 1, as done

Not O.K.

Not O.K.

There is no point in proceeding further. 12-18

So far, we’ve performed elements of the design task. Now let’s do it more completely. First, remember our viewpoint. The values of the unilateral tolerances, tb and td , reflect the routine capabilities of the bushing vendor and the in-house capabilities. While the designer has to live with these, his approach should not depend on them. They can be incorporated later. First we shall find the minimum size of the journal which satisfies Trumpler’s constraint of Pst ≤ 300 psi. W Pst = ≤ 300 2dl W W ≤ 300 ⇒ d ≥ 2d 2 l /d 600(l /d) 900 = 1.73 in dmin = 2(300)(0.5)

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In this problem we will take journal diameter as the nominal value and the bushing bore as a variable. In the next problem, we will take the bushing bore as nominal and the journal diameter as free. To determine where the constraints are, we will set tb = td = 0, and thereby shrink the design window to a point. d = 2.000 in

We set

b = d + 2cmin = d + 2c nd = 2

(This makes Trumpler’s n d ≤ 2 tight)

and construct a table. c

b

d

T¯ f*

Tmax

ho

Pst

Tmax

n

fom

0.0010 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019 0.0020

2.0020 2.0022 2.0024 2.0026 2.0028 2.0030 2.0032 2.0034 2.0036 2.0038 2.0040

2 2 2 2 2 2 2 2 2 2 2

215.50 206.75 198.50 191.40 185.23 179.80 175.00 171.13 166.92 163.50 160.40

312.0 293.0 277.0 262.8 250.4 239.6 230.1 220.3 213.9 206.9 200.6

× × × × × × × ×

×

−5.74 −6.06 −6.37 −6.66 −6.94 −7.20 −7.45 −7.65 −7.91 −8.12 −8.32

*Sample calculation for the first entry of this column. T¯ f = 215.5°F Iteration yields: With T¯ f = 215.5°F, from Table 12-1 µ = 0.0136(10−6 ) exp[1271.6/(215.5 + 95)] = 0.817(10−6 ) reyn 900 N = 3000/60 = 50 rev/s, P = = 225 psi 4 2 1 0.817(10−6 )(50) S= = 0.182 0.001 225 From Figs. 12-16 and 12-18: Eq. (12–24):

= 0.7,

f r/c = 5.5

0.0123(5.5)(0.182)(9002 ) = 191.6°F [1 + 1.5(0.72 )](30)(14 ) 191.6°F . Tav = 120°F + = 215.8°F = 215.5°F 2 For the nominal 2-in bearing, the various clearances show that we have been in contact with the recurving of (h o ) min . The figure of merit (the parasitic friction torque plus the pumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we will place the top of the design window at cmin = 0.002 in, and b = d + 2(0.002) = 2.004 in. At this point, add the b and d unilateral tolerances: TF =

0.000 d = 2.000+ −0.001 in,

0.003 b = 2.004+ −0.000 in

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Chapter 12

¯ and cmax . Of immediate interest is the fom Now we can check the performance at cmin , c, of the median clearance assembly, −9.82, as compared to any other satisfactory bearing ensemble. If a nominal 1.875 in bearing is possible, construct another table with tb = 0 and td = 0. c

b

d

T¯ f

Tmax

ho

Pst

Tmax

fos

fom

0.0020 0.0030 0.0035 0.0040 0.0050 0.0055 0.0060

1.879 1.881 1.882 1.883 1.885 1.886 1.887

1.875 1.875 1.875 1.875 1.875 1.875 1.875

157.2 138.6 133.5 130.0 125.7 124.4 123.4

194.30 157.10 147.10 140.10 131.45 128.80 126.80

× ×

−7.36 −8.64 −9.05 −9.32 −9.59 −9.63 −9.64

The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our design window. 0.000 0.003 b = 1.881+ d = 1.875+ −0.001 in, −0.000 in The ensemble median assembly has fom = −9.31. We just had room to fit in a design window based upon the (h o ) min constraint. Further reduction in nominal diameter will preclude any smaller bearings. A table constructed for a d = 1.750 in journal will prove this. We choose the nominal 1.875-in bearing ensemble because it has the largest figure of merit. Ans. 12-19

This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b and radial clearance c. The approach is similar to that of Prob. 12-18 and the tables will change slightly. In the table for a nominal b = 1.875 in, note that at c = 0.003 the constraints are “loose.” Set b = 1.875 in d = 1.875 − 2(0.003) = 1.869 in For the ensemble 0.003 b = 1.875+ −0.001 ,

0.000 d = 1.869+ −0.001

Analyze at cmin = 0.003, c¯ = 0.004 in and cmax = 0.005 in At cmin = 0.003 in: T¯ f = 138.4, µ = 3.160, S = 0.0297, Hloss = 1035 Btu/h and the Trumpler conditions are met. At c¯ = 0.004 in: T¯ f = 130°F, µ = 3.872, S = 0.0205, Hloss = 1106 Btu/h, fom = −9.246 and the Trumpler conditions are O.K. At cmax = 0.005 in: T¯ f = 125.68°F, µ = 4.325 µreyn, S = 0.014 66, Hloss = 1129 Btu/h and the Trumpler conditions are O.K. The ensemble figure of merit is slightly better; this bearing is slightly smaller. The lubricant cooler has sufficient capacity.

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From Table 12-1, Seireg and Dandage, µ0 = 0.0141(106 ) reyn and b = 1360.0 µ(µreyn) = 0.0141 exp[1360/(T + 95)]

(T in °F)

= 0.0141 exp[1360/(1.8C + 127)]

(C in °C)

µ(MPa · s) = 6.89(0.0141) exp[1360/(1.8C + 127)]

(C in °C)

For SAE 30 at 79°C µ = 6.89(0.0141) exp{1360/[1.8(79) + 127]} = 15.2 MPa · s 12-21

Ans.

Originally 0.000 d = 2.000+ −0.001 in,

Doubled,

0.000 d = 4.000+ −0.002 in,

0.003 b = 2.005+ −0.000 in 0.006 b = 4.010+ −0.000

The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were carried out. Some of the results are: Part (a) (b)

c¯

µ

S

T¯f

f r/c

Qs

h o /c

Hloss

ho

Trumpler ho

f

0.007 3.416 0.0310 135.1 0.1612 6.56 0.1032 0.897 9898 0.000 722 0.000 360 0.005 67 0.0035 3.416 0.0310 135.1 0.1612 0.870 0.1032 0.897 1237 0.000 361 0.000 280 0.005 67

The side flow Q s differs because there is a c3 term and consequently an 8-fold increase. Hloss is related by a 9898/1237 or an 8-fold increase. The existing h o is related by a 2-fold increase. Trumpler’s (h o ) min is related by a 1.286-fold increase fom = −82.37 fom = −10.297 12-22

for double size for original size

}

an 8-fold increase for double-size

From Table 12-8: K = 0.6(10−10 ) in3 · min/(lbf · ft · h). P = 500/[(1)(1)] = 500 psi, V = π D N/12 = π(1)(200)/12 = 52.4 ft/min Tables 12-10 and 12-11: Table 12-12:

f 1 = 1.8,

P Vmax = 46 700 psi · ft/min, Pmax = P=

f2 = 1 Pmax = 3560 psi,

Vmax = 100 ft/min

4(500) 4 F = = 637 psi < 3560 psi O.K. π DL π(1)(1) F = 500 psi DL

V = 52.4 ft/min

P V = 500(52.4) = 26 200 psi · ft/min < 46 700 psi · ft/min O.K.

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Chapter 12

Solving Eq. (12-32) for t t=

π(1)(1)(0.005) π DLw = = 1388 h = 83 270 min 4 f1 f2 K V F 4(1.8)(1)(0.6)(10−10 )(52.4)(500)

Cycles = N t = 200(83 270) = 16.7 rev 12-23

Ans.

Estimate bushing length with f 1 = f 2 = 1, and K = 0.6(10−10 ) in3 · min/(lbf · ft · h) L=

Eq. (12-32):

1(1)(0.6)(10−10 )(2)(100)(400)(1000) = 0.80 in 3(0.002)

From Eq. (12-38), with f s = 0.03 from Table 12-9 applying n d = 2 to F and h¯ CR = 2.7 Btu/(h · ft2 · °F) . 720(0.03)(2)(100)(400) = 3.58 in L= 778(2.7)(300 − 70) 0.80 ≤ L ≤ 3.58 in Trial 1: Let L = 1 in, D = 1 in Pmax =

4(2)(100) = 255 psi < 3560 psi O.K. π(1)(1)

P=

2(100) = 200 psi 1(1)

V =

π(1)(400) = 104.7 ft/min > 100 ft/min 12

Not O.K.

Trial 2: Try D = 7/8 in, L = 1 in Pmax =

4(2)(100) = 291 psi < 3560 psi π(7/8)(1)

P=

2(100) = 229 psi 7/8(1)

V =

π(7/8)(400) = 91.6 ft/min < 100 ft/min 12

P V = 229(91.6) = 20 976 psi · ft/min < 46 700 psi · ft/min V 33 91.6 100

O.K.

O.K.

O.K.

f1 1.3 f1 1.8

⇒

91.6 − 33 f 1 = 1.3 + (1.8 − 1.3) 100 − 33 L = 0.80(1.74) = 1.39 in

= 1.74

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Trial 3: Try D = 7/8 in, L = 1.5 in 4(2)(100) = 194 psi < 3560 psi O.K. π(7/8)(1.5) 2(100) P= = 152 psi, V = 91.6 ft/min 7/8(1.5) P V = 152(91.6) = 13 923 psi · ft/min < 46 700 psi · ft/min D = 7/8 in, L = 1.5 in is acceptable Ans.

Pmax =

Suggestion: Try smaller sizes.

O.K.

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Chapter 13 13-1 d P = 17/8 = 2.125 in dG =

N2 1120 (2.125) = 4.375 in dP = N3 544

NG = PdG = 8(4.375) = 35 teeth Ans. C = (2.125 + 4.375)/2 = 3.25 in Ans. 13-2 n G = 1600(15/60) = 400 rev/min Ans. p = πm = 3π mm Ans. C = [3(15 + 60)]/2 = 112.5 mm Ans. 13-3 NG = 20(2.80) = 56 teeth Ans. dG = NG m = 56(4) = 224 mm Ans. d P = N P m = 20(4) = 80 mm Ans. C = (224 + 80)/2 = 152 mm Ans. 13-4

Mesh:

a = 1/P = 1/3 = 0.3333 in Ans. b = 1.25/P = 1.25/3 = 0.4167 in Ans. c = b − a = 0.0834 in Ans. p = π/P = π/3 = 1.047 in Ans. t = p/2 = 1.047/2 = 0.523 in Ans.

Pinion Base-Circle:

d1 = N1 /P = 21/3 = 7 in d1b = 7 cos 20° = 6.578 in Ans.

Gear Base-Circle:

d2 = N2 /P = 28/3 = 9.333 in d2b = 9.333 cos 20° = 8.770 in Ans.

Base pitch:

pb = pc cos φ = (π/3) cos 20° = 0.984 in Ans.

Contact Ratio:

m c = L ab / pb = 1.53/0.984 = 1.55 Ans.

See the next page for a drawing of the gears and the arc lengths.

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O2

10.5

10

Arc of approach 0.87 in Arc of recess 0.77 in Arc of action 1.64 in Lab 1.53 in

P B

A

14

Ans. Ans. Ans.

12.6

O1

13-5

(a) A O =

2.333 2

2

5.333 + 2

2 1/2 = 2.910 in Ans. (b) γ = tan−1 (14/32) = 23.63° Ans.

AO 2

1" 3

= tan−1 (32/14) = 66.37° Ans.

5

(c) d P = 14/6 = 2.333 in, dG = 32/6 = 5.333 in Ans.

1" 3

(d) From Table 13-3, 0.3A O = 0.873 in and 10/P = 10/6 = 1.67 0.873 < 1.67 ∴ F = 0.873 in Ans. 13-6

30

P

G

(a) pn = π/5 = 0.6283 in pt = pn /cos ψ = 0.6283/cos 30° = 0.7255 in px = pt /tan ψ = 0.7255/tan 30° = 1.25 in

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Chapter 13

(b) pnb = pn cos φn = 0.6283 cos 20° = 0.590 in Ans. (c) Pt = Pn cos ψ = 5 cos 30° = 4.33 teeth/in φt = tan−1 (tan φn /cos ψ) = tan−1 (tan 20°/cos 30◦ ) = 22.8° Ans. (d) Table 13-4: a = 1/5 = 0.200 in Ans. b = 1.25/5 = 0.250 in Ans. 17 dP = = 3.926 in Ans. 5 cos 30° dG =

34 = 7.852 in Ans. 5 cos 30°

13-7 P

G 20

φn = 14.5°,

Pn = 10 teeth/in

(a) pn = π/10 = 0.3142 in Ans. pt =

0.3142 pn = = 0.3343 in Ans. cos ψ cos 20°

px =

0.3343 pt = = 0.9185 in Ans. tan ψ tan 20°

(b) Pt = Pn cos ψ = 10 cos 20° = 9.397 teeth/in −1 tan 14.5° φt = tan = 15.39° Ans. cos 20° (c)

a = 1/10 = 0.100 in Ans. b = 1.25/10 = 0.125 in Ans. 19 dP = = 2.022 in Ans. 10 cos 20° dG =

57 = 6.066 in Ans. 10 cos 20°

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From Ex. 13-1, a 16-tooth spur pinion meshes with a 40-tooth gear, m G = 40/16 = 2.5. Equations (13-10) through (13-13) apply. (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10) 4k 2 1 + 1 + 3 sin φ NP ≥ 6 sin2 φ 4(1) 2 1 + 3 sin 20° ≥ 1 + 6 sin2 20° ≥ 12.32 → 13 teeth Ans. (b) The smallest pinion that will mesh with a gear ratio of m G = 2.5, from Eq. (13-11) is

2(1) 2 2 2.5 + 2.5 + [1 + 2(2.5)] sin 20° NP ≥ [1 + 2(2.5)] sin2 20° ≥ 14.64 → 15 pinion teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-12) 4k 4(1) = NP ≥ 2 sin2 φ 2 sin2 20° ≥ 17.097 → 18 teeth Ans. (d) The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-13) is NG ≤ ≤

N P2 sin2 φ − 4k 2 4k − 2N P sin2 φ 132 sin2 20° − 4(1) 2 4(1) − 2(13) sin2 20°

≤ 16.45 → 16 teeth Ans. 13-9

From Ex. 13-2, a 20° pressure angle, 30° helix angle, pt = 6 teeth/in pinion with 18 full depth teeth, and φt = 21.88°. (a) The smallest tooth count that will mesh with a like gear, from Eq. (13-21), is 4k cos ψ 2 1 + 1 + 3 sin φt NP ≥ 6 sin2 φt 4(1) cos 30° 2 ≥ 1 + 1 + 3 sin 21.88° 6 sin2 21.88° ≥ 9.11 → 10 teeth Ans. (b) The smallest pinion-tooth count that will run with a rack, from Eq. (13-23), is 4k cos ψ 2 sin2 φt 4(1) cos 30◦ ≥ 2 sin2 21.88° ≥ 12.47 → 13 teeth Ans.

NP ≥

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Chapter 13

(c) The largest gear tooth possible, from Eq. (13-24) is NG ≤

N P2 sin2 φt − 4k 2 cos2 ψ 4k cos ψ − 2N P sin2 φt

102 sin2 21.88° − 4(12 ) cos2 30° ≤ 4(1) cos 30° − 2(10) sin2 21.88° ≤ 15.86 → 15 teeth 13-10

Pressure Angle:

−1

φt = tan

tan 20° cos 30°

Ans.

= 22.796°

Program Eq. (13-24) on a computer using a spreadsheet or code and increment N P . The first value of N P that can be doubled is N P = 10 teeth, where NG ≤ 26.01 teeth. So NG = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc. Use 10:20 Ans. 13-11

Refer to Prob. 13-10 solution. The first value of N P that can be multiplied by 6 is N P = 11 teeth where NG ≤ 93.6 teeth. So NG = 66 teeth. Use 11:66 Ans.

13-12

Begin with the more general relation, Eq. (13-24), for full depth teeth. NG =

N P2 sin2 φt − 4 cos2 ψ 4 cos ψ − 2N P sin2 φt

Set the denominator to zero 4 cos ψ − 2N P sin2 φt = 0 From which

2 cos ψ NP 2 cos ψ φt = sin−1 NP

sin φt =

For N P = 9 teeth and cos ψ = 1

φt = sin−1

2(1) = 28.126° Ans. 9

13-13 18T

32T

␺ 25, ␾n 20, m 3 mm

(a) pn = πm n = 3π mm Ans. pt = 3π/cos 25° = 10.4 mm Ans. px = 10.4/tan 25° = 22.3 mm Ans.

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(b) m t = 10.4/π = 3.310 mm Ans. tan 20° = 21.88° Ans. φt = tan−1 cos 25° (c) d P = 3.310(18) = 59.58 mm Ans. dG = 3.310(32) = 105.92 mm Ans. 13-14

(a) The axial force of 2 on shaft a is in the negative direction. The axial force of 3 on shaft b is in the positive direction of z. Ans.

b

3

z

a 2

The axial force of gear 4 on shaft b is in the positive z-direction. The axial force of gear 5 on shaft c is in the negative z-direction. Ans. 5

c

z

b

4

14 16 (900) = +103.7 rev/min ccw Ans. 54 36 = 14/(10 cos 30°) = 1.6166 in = 54/(10 cos 30°) = 6.2354 in

(b) n c = n 5 = (c) d P 2 dG 3

Cab =

1.6166 + 6.2354 = 3.926 in Ans. 2

d P 4 = 16/(6 cos 25°) = 2.9423 in dG 5 = 36/(6 cos 25°) = 6.6203 in Cbc = 4.781 in Ans. 13-15

4 20 8 20 = e= 40 17 60 51 4 n d = (600) = 47.06 rev/min cw Ans. 51

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339

Chapter 13

13-16

6 e= 10 na =

18 38

20 48

3 36

3 304

=

3 (1200) = 11.84 rev/min cw Ans. 304

13-17 (a) n c =

12 1 · (540) = 162 rev/min cw about x. Ans. 40 1

(b) d P = 12/(8 cos 23°) = 1.630 in dG = 40/(8 cos 23°) = 5.432 in d P + dG = 3.531 in Ans. 2 32 = 8 in at the large end of the teeth. Ans. (c) d = 4 13-18

(a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus n A = n 3 = 1200(17/54) = 377.8 rev/min Ans. n F = n 5 = 0,

(b)

n L = n6,

e = −1

n 6 − 377.8 0 − 377.8 377.8 = n 6 − 377.8 −1 =

n 6 = 755.6 rev/min Ans. Alternatively, the velocity of the center of gear 4 is v4c ∝ N6 n 3 . The velocity of the left edge of gear 4 is zero since the left wheel is resting on the ground. Thus, the velocity of the right edge of gear 4 is 2v4 c ∝ 2N6 n 3 . This velocity, divided by the radius of gear 6 ∝ N6 , is angular velocity of gear 6–the speed of wheel 6. ∴

n6 =

2N6 n 3 = 2n 3 = 2(377.8) = 755.6 rev/min Ans. N6

(c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans. 13-19

(a) The motive power is divided equally among four wheels instead of two. (b) Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels, rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50.

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Let gear 2 be first, then n F = n 2 = 0. Let gear 6 be last, then n L = n 6 = −12 rev/min. 16 nL − n A 20 16 = , e= e= 30 34 51 nF − nA (0 − n A )

16 = −12 − n A 51 nA =

−12 = −17.49 rev/min (negative indicates cw) Ans. 35/51

Alternatively, since N ∝ r, let v = N n (crazy units). v = N6 n 6 v vA = N4 N4 − N5

4 5

2

vA

=

⇒

vA =

N4 N6 n 6 N4 − N5

vA N4 N6 n 6 = N2 + N4 ( N2 + N4 )( N4 − N5 )

nA =

v0 v

N6 = 20 + 30 − 16 = 34 teeth

30(34)(12) (20 + 30)(30 − 16)

= 17.49 rev/min cw Ans. 13-21

Let gear 2 be first, then n F = n 2 = 180 rev/min. Let gear 6 be last, then n L = n 6 = 0. 20 e= 30 (180 − n A )

16 34

=

16 , 51

e=

nL − n A nF − nA

16 = (0 − n A ) 51 16 nA = − 180 = −82.29 rev/min 35

The negative sign indicates opposite n 2

∴

n A = 82.29 rev/min cw Ans.

Alternatively, since N ∝ r, let v = N n (crazy units). v N2 n 2 vA = = N5 N4 − N5 N4 − N5

v N2 n2

N2 5 2 v0 vA

4

N5 N2 n 2 N4 − N5 vA N5 N2 n 2 nA = = N2 + N4 ( N2 + N4 )( N4 − N5 ) vA =

=

16(20)(180) = 82.29 rev/min cw Ans. (20 + 30)(30 − 16)

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Chapter 13

13-22

N5 = 12 + 2(16) + 2(12) = 68 teeth Ans. Let gear 2 be first, n F = n 2 = 320 rev/min. Let gear 5 be last, n L = n 5 = 0 3 nL − n A 12 16 12 = , e= e= 16 12 68 17 nF − nA 17 320 − n A = (0 − n A ) 3 3 n A = − (320) = −68.57 rev/min 14 The negative sign indicates opposite of n 2 ∴ n A = 68.57 rev/min cw Ans. Alternatively, v0 nA(N2 2N3 N4) 2nA(N2 2N3 N4) n2N2 2nA(N2 N3)

v n 2 N2

nA(N2 N3)

nA =

2nA(N2 2N3 N4) 2nA(N2 N3) n2N2

320(12) n 2 N2 = 2( N3 + N4 ) 2(16 + 12)

= 68.57 rev/min cw Ans. 13-23

Let n F = n 2 then n L = n 7 = 0. 24 18 8 36 e=− =− 18 30 54 15 n L − n5 8 e= =− n F − n5 15 8 15 0−5 =− ⇒ n 2 = 5 + (5) = 14.375 turns in same direction n2 − 5 15 8

13-24

(a) Let n F = n 2 = 0, then n L = n 5 . 99 101 9999 e= = , 100 100 10 000

e=

nL − n A nL − n A = nF − nA 0 − nA

n L − n A = −en A n L = n A (−e + 1) 1 9999 nL = = 0.0001 Ans. =1−e =1− nA 10 000 10 000 101 N4 = = 10.1 in P 10 100 d5 = = 10 in 10 10 d5 dhousing > 2 d4 + = 2 10.1 + = 30.2 in Ans. 2 2

(b) d4 =

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13-25

n2 = nb = n F ,

n A = na ,

n L = n5 = 0 nL − n A 21 = e=− 444 nF − nA

−

21 (n F − n A ) = 0 − n A 444

With shaft b as input −

na =

21 nb, 465

21 21 444 nF + nA + nA = 0 444 444 444 nA na 21 = = nF nb 465

in the same direction as shaft b, the input. Ans.

Alternatively, n 2 N2 vA = N4 N3 + N4

v0 4 vA 3

vA =

n 2 N2 N4 N3 + N4

vA n 2 N2 N4 = N2 + N3 ( N2 + N3 )( N3 + N4 )

na = n A =

n2 N2 2

= 13-26

n F = n2 = na ,

21 18(21)(n b ) = n b in the same direction as b (18 + 72)(72 + 21) 465

n L = n6 = 0 24 22 11 e=− =− , 18 64 24

e=

nL − n A 0 − nb = nF − nA na − nb

0 − nb 11 = ⇒ 24 na − nb Yes, both shafts rotate in the same direction. Ans. −

nb 11 = na 35

Ans.

Alternatively, v0 5 vA

3

n2 N2 2

vA n 2 N2 N2 N2 N5 = = na , v A = na N5 N3 + N5 N3 + N5 N3 + N5 vA N2 N5 na n A = nb = = N2 + N3 ( N2 + N3 )( N3 + N5 ) 11 24(22) nb = = na (24 + 18)(22 + 18) 35 n b rotates ccw

13-27

n 2 = n F = 0,

n L = n5 = nb,

∴

Yes Ans.

n A = na 25 20 20 = e=+ 24 24 36

Ans.

Ans.

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Chapter 13

nb − na 25 = 36 0 − na 11 nb = Ans. na 36 Same sense, therefore shaft b rotates in the same direction as a.

Ans.

Alternatively, ( N2 + N3 )n a v5 = N3 − N4 N3

4

(N2 N3)na

v5 =

3 v5 v0

nb =

( N2 + N3 )( N3 − N4 )n N3

v5 ( N2 + N3 )( N3 − N4 )n a = N5 N3 N5 (20 + 24)(24 − 20) nb = na 24(24) =

13-28

11 36

same sense Ans.

ω = 2πn/60

(a)

H = T ω = 2π T n/60 (T in N · m, H in W) So

T =

60H (103 ) 2πn

= 9550H/n (H in kW, n in rev/min) Ta =

9550(75) = 398 N · m 1800

r2 =

5(17) m N2 = = 42.5 mm 2 2

So t = F32

Ta 398 = 9.36 kN = r2 42.5 F t32

9.36

a 2

Ta2 398 N • m

F3b = −Fb3 = 2(9.36) = 18.73 kN in the positive x-direction. Ans. See the figure in part (b) on the following page.

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r4 =

(b)

5(51) m N4 = = 127.5 mm 2 2 9.36 F t43

3 O b Fb3

18.73

F t23 9.36

Tc4 = 9.36(127.5) = 1193 N · m ccw ∴ T4c

= 1193 N · m cw

Ans.

4

Note: The solution is independent of the pressure angle.

c Tc4 1193 9.36

13-29

2

4 5

6

N 6 d2 = 4 in, d4 = 4 in, d5 = 6 in, d6 = 24 in 24 24 36 e= = 1/6, n P = n 2 = 1000 rev/min 24 36 144 d=

n L = n6 = 0 e=

nL − n A 0 − nA = nF − nA 1000 − n A

n A = −200 rev/min

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Chapter 13

Input torque:

T2 =

63 025H n

T2 =

63 025(25) = 1576 lbf · in 1000

For 100 percent gear efficiency Tarm =

63 025(25) = 7878 lbf · in 200

Gear 2 F ta2 T2 1576 lbf • in

n2

Wt =

r F32 = 788 tan 20° = 287 lbf

F r42

F ra2

1576 = 788 lbf 2

2 Wt

Gear 4 Wt

Wt

4

Fr

FA4 = 2W t = 2(788) = 1576 lbf

Fr n4 FA4

Gear 5 2W t 1576 lbf

F r 287 lbf

Fr Wt

5

W t 788 lbf

Arm 1576 lbf

Tout

4"

13-30

5"

Tout = 1576(9) − 1576(4) = 7880 lbf · in

Ans.

1576 lbf

Given: P = 2 teeth/in, n P = 1800 rev/min cw, N2 = 18T, N3 = 32T, N4 = 18T, N5 = 48T. Pitch Diameters: d2 = 18/2 = 9 in; d3 = 32/2 = 16 in; d4 = 18/2 = 9 in; d5 = 48/2 = 24 in.

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Gear 2 W r 566 lbf

Ta 2 = 63 025(200)/1800 = 7003 lbf · in

W t 1556 lbf

W t = 7003/4.5 = 1556 lbf

a

W r = 1556 tan 20° = 566 lbf

2 Ta2 7003 lbf • in

Gears 3 and 4 y W t 2766 lbf 3

4 b W r 1007 lbf

x

W t (4.5) = 1556(8),

W t = 2766 lbf

W r = 2766 tan 20◦ = 1007 lbf

W t 1556 lbf

Ans.

W r 566 lbf

13-31

P = 5 teeth/in, N2 = 18T, Given: 1800 rev/min.

N3 = 45T,

φn = 20°,

H = 32 hp,

Gear 2 Tin = W r32

Tin F ta2 2

F ra2

18 = 3.600 in 5 45 dG = = 9.000 in 5 1120 t W32 = 622 lbf = 3.6/2 dP =

W t32

a

63 025(32) = 1120 lbf · in 1800

r W32 = 622 tan 20° = 226 lbf t = 622 lbf, Fat 2 = W32

r Far2 = W32 = 226 lbf

Fa 2 = (6222 + 2262 ) 1/2 = 662 lbf Each bearing on shaft a has the same radial load of R A = R B = 662/2 = 331 lbf.

n2 =

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Chapter 13

Gear 3 Fbr3

3

Tout W t23 r3 2799 lbf • in

t t = W32 = 622 lbf W23 r r W23 = W32 = 226 lbf

Fbt3

b

Fb3 = Fb2 = 662 lbf RC = R D = 662/2 = 331 lbf

W t23 W r23

Each bearing on shaft b has the same radial load which is equal to the radial load of bearings, A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans. 13-32

Given: P = 4 teeth/in, φn = 20◦ ,

N P = 20T,

n 2 = 900 rev/min.

d2 =

20 NP = = 5.000 in P 4

Tin =

63 025(30)(2) = 4202 lbf · in 900

t W32 = Tin /(d2 /2) = 4202/(5/2) = 1681 lbf r W32 = 1681 tan 20◦ = 612 lbf y

y

x

3

z 2

y

Load on 2 due to 3

y

Equivalent

4202 lbf • in r 612 lbf W 32

2

z

2

612 lbf

z W t32 1681 lbf

1681 lbf

The motor mount resists the equivalent forces and torque. The radial force due to torque Fr =

4202 = 150 lbf 14(2)

150 B

150

14" C

A

150

D

150

4202 lbf • in

Forces reverse with rotational sense as torque reverses.

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design t , The compressive loads at A and D are absorbed by the base plate, not the bolts. For W32 the tensions in C and D are

M AB = 0

1681(4.875 + 15.25) − 2F(15.25) = 0 B

F = 1109 lbf

F F1

C

A

F 15.25"

F1

4.875

1681 lbf

D

t If W32 reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces change direction. For A and B,

1681(2.875) − 2F1 (13.25) = 0

⇒

F1 = 182.4 lbf

r For W32 153 lbf 612 153 lbf 4

F2

F2

D a 612 lbf 14

153 lbf

11.25 F2

4.875

F2 C 153 lbf

M = 612(4.875 + 11.25/2) = 6426 lbf · in a = (14/2) 2 + (11.25/2) 2 = 8.98 in F2 =

6426 = 179 lbf 4(8.98)

At C and D, the shear forces are: FS 1 = [153 + 179(5.625/8.98)]2 + [179(7/8.98)]2 = 300 lbf At A and B, the shear forces are: FS 2 = [153 − 179(5.625/8.98)]2 + [179(7/8.98)]2 = 145 lbf

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349

Chapter 13

The shear forces are independent of the rotational sense. The bolt tensions and the shear forces for cw rotation are,

A B C D

Tension (lbf)

Shear (lbf)

0 0 1109 1109

145 145 300 300

Tension (lbf)

Shear (lbf)

182 182 0 0

145 145 300 300

For ccw rotation,

A B C D 13-33

Tin = 63 025H/n = 63 025(2.5)/240 = 656.5 lbf · in W t = T /r = 656.5/2 = 328.3 lbf

y 2

γ = tan−1 (2/4) = 26.565°

Wa a G

Wr

= tan−1 (4/2) = 63.435°

Wt

FBz FBx 2 B

1 2

FAz A z

FAx

x

FAy

a = 2 + (1.5 cos 26.565°)/2 = 2.67 in W r = 328.3 tan 20° cos 26.565° = 106.9 lbf W a = 328.3 tan 20° sin 26.565° = 53.4 lbf W = 106.9i − 53.4j + 328.3k lbf

Tin Not to scale

R AG = −2i + 5.17j, R AB = 2.5j

M4 = R AG × W + R AB × F B + T = 0

Solving gives R AB × F B = 2.5FBz i − 2.5FBx k R AG × W = 1697i + 656.6j − 445.9k So

(1697i + 656.6j − 445.9k) + 2.5FBz i − 2.5FBx k + T j = 0 FBz = −1697/2.5 = −678.8 lbf T = −656.6 lbf · in FBx = −445.9/2.5 = −178.4 lbf

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So FB = [(−678.8) 2 + (−178.4) 2 ]1/2 = 702 lbf Ans. F A = −(F B + W) = −(−178.4i − 678.8k + 106.9i − 53.4j + 328.3k) = 71.5i + 53.4j + 350.5k FA (radial) = (71.52 + 350.52 ) 1/2 = 358 lbf Ans. FA (thrust) = 53.4 lbf Ans. 13-34 d2 = 15/10 = 1.5 in, 0.75

γ = tan−1

1.25

DE = y

Wa

Wt

Wr G

1.25"

= 59.04°

9 + 0.5 cos 59.04° = 0.8197 in 16

W r = 30 tan 20° cos 59.04° = 5.617 lbf W a = 30 tan 20° sin 59.04° = 9.363 lbf

R DC = −0.625j

M D = R DG × W + R DC × FC + T = 0

FDx D 5" 8

x

C FCx

25 = 2.5 in 10

R DG = 0.8197j + 1.25i

0.8197"

FDy

0.75 = 30.96°, 1.25

d3 =

W = −5.617i − 9.363j + 30k

E

FDz

W t = 30 lbf,

R DG × W = 24.591i − 37.5j − 7.099k R DC × FC = −0.625FCz i + 0.625FCx k

FCz

T = 37.5 lbf · in Ans.

z Not to scale

FC = 11.4i + 39.3k lbf Ans. FC = (11.42 + 39.32 ) 1/2 = 40.9 lbf Ans.

F=0

F D = −5.78i + 9.363j − 69.3k lbf

FD (radial) = [(−5.78) 2 + (−69.3) 2 ]1/2 = 69.5 lbf Ans. FD (thrust) = W a = 9.363 lbf Ans.

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351

Chapter 13

13-35

Sketch gear 2 pictorially. z

y

Pt = Pn cos ψ = 4 cos 30° = 3.464 teeth/in

Wa 2 W

Wr

t

φt = tan−1

x

tan φn tan 20° = tan−1 = 22.80° cos ψ cos 30°

T

Sketch gear 3 pictorially, z

y

Wt

Wr

x TG

Wa 3

dP =

18 = 5.196 in 3.464

Pinion (Gear 2) W r = W t tan φt = 800 tan 22.80° = 336 lbf W a = W t tan ψ = 800 tan 30° = 462 lbf W = −336i − 462j + 800k lbf Ans. W = [(−336) 2 + (−462) 2 + 8002 ]1/2 = 983 lbf Ans. Gear 3 W = 336i + 462j − 800k lbf Ans. W = 983 lbf Ans. dG =

32 = 9.238 in 3.464

TG = W t r = 800(9.238) = 7390 lbf · in 13-36

From Prob. 13-35 solution, 3 800

462

800 4

462 336

336

336

336 2

462 800

800

462

Notice that the idler shaft reaction contains a couple tending to turn the shaft end-overend. Also the idler teeth are bent both ways. Idlers are more severely loaded than other gears, belying their name. Thus be cautious.

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13-37 y

Gear 3:

Wr r4

Pt = Pn cos ψ = 7 cos 30° = 6.062 teeth/in tan 20° = 0.4203, φt = 22.8° tan φt = cos 30°

Wt W

a

54 = 8.908 in 6.062 W t = 500 lbf d3 =

W a = 500 tan 30° = 288.7 lbf

r3

W r = 500 tan 22.8° = 210.2 lbf

x Wr

Wa

Wt

W3 = 210.2i + 288.7j − 500k lbf Ans. z

Gear 4: 14 = 2.309 in 6.062 8.908 W t = 500 = 1929 lbf 2.309 W a = 1929 tan 30° = 1114 lbf d4 =

W r = 1929 tan 22.8° = 811 lbf W4 = −811i + 1114j − 1929k lbf Ans. 13-38 Pt = 6 cos 30° = 5.196 teeth/in

y 3

d3 =

C

D

T3

φt = 22.8°

T2

d2 = x

B

2

42 = 8.083 in 5.196

A

16 = 3.079 in 5.196

63 025(25) = 916 lbf · in 1720 916 T = 595 lbf Wt = = r 3.079/2 T2 =

W a = 595 tan 30° = 344 lbf W r = 595 tan 22.8° = 250 lbf W = 344i + 250j + 595k lbf R DC = 6i,

R DG = 3i − 4.04j

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353

Chapter 13 y FCy

x FCx

C 3"

3"

FCz

T

D FDz

z

4.04"

FDy Wt G Wa

Wr

M D = R DC × FC + R DG × W + T = 0

R DG × W = −2404i − 1785j + 2140k y R DC × FC = −6FCz j + 6FC k Substituting and solving Eq. (1) gives T = 2404i lbf · in FCz = −297.5 lbf y

F = −356.7 lbf

C F = F D + FC + W = 0 Substituting and solving gives

FCx = −344 lbf y

FD = 106.7 lbf FDz = −297.5 lbf So FC = −344i − 356.7j − 297.5k lbf Ans. F D = 106.7j − 297.5k lbf Ans. 13-39

Pt = 8 cos 15° = 7.727 teeth/in y x

t F32 r F32

a

a 2 Fa2 a F32

t Fa2

r Fa2

z

(1)

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design y

F tc4

x

d2 = 16/7.727 = 2.07 in d3 = 36/7.727 = 4.66 in d4 = 28/7.727 = 3.62 in

c 4

F rc4

F a34 F ac4

T2 =

F r34

z

63 025(7.5) = 274.8 lbf · in 1720

Wt =

274.8 = 266 lbf 2.07/2

F t34

y

F t43 x

W r = 266 tan 20° = 96.8 lbf W a = 266 tan 15° = 71.3 lbf F2a = −266i − 96.8j − 71.3k lbf Ans. F3b = (266 − 96.8)i − (266 − 96.8)j = 169i − 169j lbf Ans. F4c = 96.8i + 266j + 71.3k lbf Ans.

F r43

b

3 F xb3

F a43

Fb3 F yb3 F a23

z

F r23 F t23

y

13-40

FDy D

FDx 2.6"R

b FDz

2"

F r54 1.55"R

3

F a54 G F t54

1" 2

F a23

3"

FCx

3

4

x

H F r23

F t23

C z

FCy

14 N 36 = = 2.021 in, d3 = = 5.196 in Pn cos ψ 8 cos 30° 8 cos 30° 15 45 d4 = = 3.106 in, d5 = = 9.317 in 5 cos 15° 5 cos 15° d2 =

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355

Chapter 13

For gears 2 and 3: φt = tan−1 (tan φn /cos ψ) = tan−1 (tan 20°/cos 30◦ ) = 22.8°, For gears 4 and 5: φt = tan−1 (tan 20°/cos 15°) = 20.6°, t F23 = T2 /r = 1200/(2.021/2) = 1188 lbf 5.196 t = 1987 lbf = 1188 F54 3.106 r t F23 = F23 tan φt = 1188 tan 22.8° = 499 lbf r = 1986 tan 20.6° = 746 lbf F54 a t = F23 tan ψ = 1188 tan 30° = 686 lbf F23 a = 1986 tan 15° = 532 lbf F54

Next, designate the points of action on gears 4 and 3, respectively, as points G and H, as shown. Position vectors are RC G = 1.553j − 3k RC H = −2.598j − 6.5k RC D = −8.5k Force vectors are F54 = −1986i − 748j + 532k F23 = −1188i + 500j − 686k y

FC = FCx i + FC j y

F D = FDx i + FD j + FDz k Now, a summation of moments about bearing C gives

MC = RC G × F54 + RC H × F23 + RC D × F D = 0 The terms for this equation are found to be RC G × F54 = −1412i + 5961j + 3086k RC H × F23 = 5026i + 7722j − 3086k y RC D × F D = 8.5FD i − 8.5FDx j When these terms are placed back into the moment equation, the k terms, representing the shaft torque, cancel. The i and j terms give 3614 = −425 lbf Ans. 8.5 (13 683) = 1610 lbf Ans. FDx = 8.5 Next, we sum the forces to zero.

F = FC + F54 + F23 + F D = 0 y

FD = −

Substituting, gives x y FC i + FC j + (−1987i − 746j + 532k) + (−1188i + 499j − 686k) + (1610i − 425j + FDz k) = 0

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Solving gives FCx = 1987 + 1188 − 1610 = 1565 lbf y

FC = 746 − 499 + 425 = 672 lbf FDz = −532 + 686 = 154 lbf Ans. 13-41 y

VW = B

WW t =

100

H 2000 = 637 N = VW π

L = px N W = 25(1) = 25 mm L λ = tan−1 πdW

100 50 Wa

G

π(0.100)(600) πdW n W = = π m/s 60 60

x

A z

Wr

Wt

= tan−1

Worm shaft diagram

25 = 4.550° π(100)

W =

WW t cos φn sin λ + f cos λ

VS =

π VW = = 3.152 m/s cos λ cos 4.550°

lead angle

In ft/min: VS = 3.28(3.152) = 10.33 ft/s = 620 ft/min Use f = 0.043 from curve A of Fig. 13-42. Then from the first of Eq. (13-43) W =

637 = 5323 N cos 14.5°(sin 4.55°) + 0.043 cos 4.55°

W y = W sin φn = 5323 sin 14.5° = 1333 N W z = 5323[cos 14.5°(cos 4.55°) − 0.043 sin 4.55°] = 5119 N The force acting against the worm is W = −637i + 1333j + 5119k N Thus A is the thrust bearing. Ans. R AG = −0.05j − 0.10k, R AB = −0.20k

M A = R AG × W + R AB × F B + T = 0 R AG × W = −122.6i + 63.7j − 31.85k y

R AB × F B = 0.2FB i − 0.2FBx j Substituting and solving gives T = 31.85 N · m Ans. FBx = 318.5 N, So

y

FB = 613 N

F B = 318.5i + 613j N Ans.

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357

Chapter 13

Or

Radial

FB = [(613) 2 + (318.5) 2 ]1/2 = 691 N radial

F = F A + W + RB = 0 F A = −(W + F B ) = −(−637i + 1333j + 5119k + 318.5i + 613j) = 318.5i − 1946j − 5119k Ans. FrA = 318.5i − 1946j N, FAr = [(318.5) 2 + (−1946) 2 ]1/2 = 1972 N

Thrust 13-42

FAa = −5119 N

From Prob. 13-41

So

WG = 637i − 1333j − 5119k N pt = px dG =

48(25) N G px = = 382 mm π π

Bearing D to take thrust load

M D = R DG × WG + R DC × FC + T = 0 R DG = −0.0725i + 0.191j R DC = −0.1075i The position vectors are in meters. R DG × WG = −977.7i − 371.1j − 25.02k y R DC × FC = 0.1075 FCz j − 0.1075FC k Putting it together and solving y

WG

G Not to scale 191

FC

C

35

72.5 z D

FD

x

Gives T = 977.7 N · m Ans. FC = −233j + 3450k N, FC = 3460 N

F = FC + WG + F D = 0

Ans.

F D = −(FC + WG ) = −637i + 1566j + 1669k N

Ans.

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Radial

FrD = 1566j + 1669k N

Or

FDr = 2289 N (total radial) FtD = −637i N (thrust)

13-43 y y T z

0.75"

z

x WWt

VW =

G

π(1.5)(900) = 353.4 ft/min 12

W x = WW t = pt = px =

33 000(0.5) = 46.69 lbf 353.4

π = 0.314 16 in 10

L = 0.314 16(2) = 0.628 in 0.628 = 7.59° π(1.5) 46.7 W = = 263 lbf cos 14.5° sin 7.59° + 0.05 cos 7.59° λ = tan−1

W y = 263 sin 14.5◦ = 65.8 lbf W z = 263[cos 14.5◦ (cos 7.59◦ ) − 0.05 sin 7.59◦ ] = 251 lbf So

W = 46.7i + 65.8j + 251k lbf Ans. T = 46.7(0.75) = 35 lbf · in Ans.

13-44 100:101 Mesh dP =

100 = 2.083 33 in 48

dG =

101 = 2.104 17 in 48

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359

Chapter 13

Proper center-to-center distance: 2.083 33 + 2.104 17 d P + dG = = 2.093 75 in 2 2 2.0833 cos 20◦ = 0.9788 in = r cos φ = 2

C= rb P 99:100 Mesh

Proper:

dP =

99 = 2.0625 in 48

dG =

100 = 2.083 33 in 48

C=

99/48 + 100/48 = 2.072 917 in 2

rb P = r cos φ = Improper:

2.0625 cos 20◦ = 0.969 06 in 2

d P + dG d P + (100/99)d P = = 2.093 75 in C = 2 2

d P =

2(2.093 75) = 2.0832 in 1 + (100/99) rb P 0.969 06 = cos−1 = 21.5° d P /2 2.0832/2

φ = cos−1 From Ex. 13-1 last line

−1

φ = cos

−1

= cos

−1

= cos 13-45


rb P d P /2

−1

= cos

( N P /P) cos φ (2C /(1 + m G ))

(d P /2) cos φ d P /2

(1 + m G ) N P cos φ 2PC

Ans.

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Chapter 14 14-1 22 N = = 3.667 in P 6 Y = 0.331 d=

Table 14-2:

V = Eq. (14-4b):

πdn π(3.667)(1200) = = 1152 ft/min 12 12

Kv =

1200 + 1152 = 1.96 1200

Wt =

63 025H 63 025(15) T = = = 429.7 lbf d/2 nd/2 1200(3.667/2)

Eq. (14-7): σ = 14-2

1.96(429.7)(6) Kv W t P = = 7633 psi = 7.63 kpsi Ans. FY 2(0.331)

16 = 1.333 in, Y = 0.296 12 π(1.333)(700) V = = 244.3 ft/min 12 1200 + 244.3 Kv = = 1.204 1200 d=

Eq. (14-4b):

Wt =

63 025(1.5) 63 025H = = 202.6 lbf nd/2 700(1.333/2)

Eq. (14-7): σ =

1.204(202.6)(12) Kv W t P = = 13 185 psi = 13.2 kpsi Ans. FY 0.75(0.296)

14-3 d = m N = 1.25(18) = 22.5 mm, Y = 0.309 π(22.5)(10−3 )(1800) = 2.121 m/s 60 6.1 + 2.121 = 1.348 Kv = 6.1 60(0.5)(103 ) 60H Wt = = = 235.8 N πdn π(22.5)(10−3 )(1800) V =

Eq. (14-6b):

Eq. (14-8):

σ =

1.348(235.8) Kv W t = = 68.6 MPa Ans. FmY 12(1.25)(0.309)

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Chapter 14

14-4 d = 5(15) = 75 mm, Y = 0.290 π(75)(10−3 )(200) = 0.7854 m/s 60 Assume steel and apply Eq. (14-6b): V =

Eq. (14-8):

Kv =

6.1 + 0.7854 = 1.129 6.1

Wt =

60(5)(103 ) 60H = = 6366 N πdn π(75)(10−3 )(200)

σ =

1.129(6366) Kv W t = = 82.6 MPa Ans. FmY 60(5)(0.290)

14-5 d = 1(16) = 16 mm, Y = 0.296 π(16)(10−3 )(400) = 0.335 m/s V = 60 Assume steel and apply Eq. (14-6b): Kv =

6.1 + 0.335 = 1.055 6.1

Wt =

60(0.15)(103 ) 60H = = 447.6 N πdn π(16)(10−3 )(400)

F=

Eq. (14-8):

Kv W t 1.055(447.6) = = 10.6 mm σ mY 150(1)(0.296)

From Table A-17, use F = 11 mm Ans. 14-6 d = 1.5(17) = 25.5 mm, Y = 0.303 V = Eq. (14-6b):

π(25.5)(10−3 )(400) = 0.534 m/s 60

Kv =

6.1 + 0.534 = 1.088 6.1

Wt =

60(0.25)(103 ) 60H = = 468 N πdn π(25.5)(10−3 )(400)

F=

Kv W t 1.088(468) = = 14.9 mm σ mY 75(1.5)(0.303)

Eq. (14-8): Use F = 15 mm Ans.

361

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14-7

Eq. (14-4b):

d=

24 = 4.8 in, 5

V =

π(4.8)(50) = 62.83 ft/min 12

Y = 0.337

Kv =

1200 + 62.83 = 1.052 1200

Wt =

63 025(6) 63 025H = = 3151 lbf nd/2 50(4.8/2)

Eq. (14-7):

F=

Kv W t P 1.052(3151)(5) = = 2.46 in σY 20(103 )(0.337)

d=

16 = 3.2 in, 5

V =

π(3.2)(600) = 502.7 ft/min 12

Use F = 2.5 in Ans. 14-8

Eq. (14-4b):

Y = 0.296

Kv =

1200 + 502.7 = 1.419 1200

Wt =

63 025(15) = 984.8 lbf 600(3.2/2)

F=

Eq. (14-7):

1.419(984.8)(5) Kv W t P = = 2.38 in σY 10(103 )(0.296)

Use F = 2.5 in Ans. 14-9

Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309. V = Eq. (14-4b):

Eq. (14-7):

π(2.25)(600) = 353.4 ft/min 12

Kv =

1200 + 353.4 = 1.295 1200

Wt =

63 025(2.5) = 233.4 lbf 600(2.25/2)

F=

Kv W t P 1.295(233.4)(8) = = 0.783 in σY 10(103 )(0.309)

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363

Chapter 14

Using coarse integer pitches from Table 13-2, the following table is formed. P

d

V

Kv

Wt

F

2 3 4 6 8 10 12 16

9.000 6.000 4.500 3.000 2.250 1.800 1.500 1.125

1413.717 942.478 706.858 471.239 353.429 282.743 235.619 176.715

2.178 1.785 1.589 1.393 1.295 1.236 1.196 1.147

58.356 87.535 116.713 175.069 233.426 291.782 350.139 466.852

0.082 0.152 0.240 0.473 0.782 1.167 1.627 2.773

Other considerations may dictate the selection. Good candidates are P = 8 ( F = 7/8 in) and P = 10 ( F = 1.25 in). Ans. 14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309. V =

π(36)(10−3 )(900) = 1.696 m/s 60

6.1 + 1.696 = 1.278 6.1 60(1.5)(103 ) Wt = = 884 N π(36)(10−3 )(900) Kv =

Eq. (14-6b):

F=

Eq. (14-8):

1.278(884) = 24.4 mm 75(2)(0.309)

Using the preferred module sizes from Table 13-2: m

d

V

Kv

Wt

F

1.00 1.25 1.50 2.00 3.00 4.00 5.00 6.00 8.00 10.00 12.00 16.00 20.00 25.00 32.00 40.00 50.00

18.0 22.5 27.0 36.0 54.0 72.0 90.0 108.0 144.0 180.0 216.0 288.0 360.0 450.0 576.0 720.0 900.0

0.848 1.060 1.272 1.696 2.545 3.393 4.241 5.089 6.786 8.482 10.179 13.572 16.965 21.206 27.143 33.929 42.412

1.139 1.174 1.209 1.278 1.417 1.556 1.695 1.834 2.112 2.391 2.669 3.225 3.781 4.476 5.450 6.562 7.953

1768.388 1414.711 1178.926 884.194 589.463 442.097 353.678 294.731 221.049 176.839 147.366 110.524 88.419 70.736 55.262 44.210 35.368

86.917 57.324 40.987 24.382 12.015 7.422 5.174 3.888 2.519 1.824 1.414 0.961 0.721 0.547 0.406 0.313 0.243

Other design considerations may dictate the size selection. For the present design, m = 2 mm ( F = 25 mm) is a good selection. Ans.

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14-11 dP = V = Eq. (14-4b):

22 = 3.667 in, 6

dG =

60 = 10 in 6

π(3.667)(1200) = 1152 ft/min 12

Kv =

1200 + 1152 = 1.96 1200

Wt =

63 025(15) = 429.7 lbf 1200(3.667/2)

C p = 2100 psi

Table 14-8: Eq. (14-12):

3.667 sin 20° 10 sin 20° = 0.627 in, r2 = = 1.710 in 2 2 1 1/2 1 Kv W t + σC = −C p F cos φ r1 r2 1/2 1 1 1.96(429.7) + = −2100 2 cos 20° 0.627 1.710

r1 =

Eq. (14-14):

= −65.6(103 ) psi = −65.6 kpsi Ans. 14-12 dP = V = Eq. (14-4b):

16 = 1.333 in, 12

dG =

48 = 4 in 12

π(1.333)(700) = 244.3 ft/min 12

Kv =

1200 + 244.3 = 1.204 1200

Wt =

63 025(1.5) = 202.6 lbf 700(1.333/2)

√ C p = 2100 psi

Table 14-8: Eq. (14-12):

r1 =

1.333 sin 20° = 0.228 in, 2

Eq. (14-14):

1.202(202.6) σC = −2100 F cos 20° F=

2100 100(103 )

Use F = 0.75 in Ans.

2

r2 =

4 sin 20° = 0.684 in 2

1 1 + 0.228 0.684

1.202(202.6) cos 20°

1/2

= −100(103 )

1 1 + 0.228 0.684

= 0.668 in

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Chapter 14

14-13

365

24 48 = 4.8 in, dG = = 9.6 in 5 5 π(4.8)(50) = 62.83 ft/min = 12 600 + 62.83 = 1.105 = 600 63 025H = 525.2H = 50(4.8/2) √ = 1960 psi

dP = V Kv

Eq. (14-4a):

Wt Cp

Table 14-8: Eq. (14-12): Eq. (14-14): H = 5.77 hp

4.8 sin 20◦ = 0.821 in, r2 = 2r1 = 1.642 in 2 1/2 1 1 1.105(525.2H ) 3 + −100(10 ) = −1960 2.5 cos 20◦ 0.821 1.642 r1 =

Ans.

14-14 d P = 4(20) = 80 mm,

dG = 4(32) = 128 mm

π(80)(10−3 )(1000) = 4.189 m/s 60 3.05 + 4.189 = 2.373 Kv = 3.05 V =

Eq. (14-6a):

60(10)(103 ) = 2387 N π(80)(10−3 )(1000) √ C p = 163 MPa

Wt = Table 14-8: Eq. (14-12): Eq. (14-14):

80 sin 20° 128 sin 20° = 13.68 mm, r2 = = 21.89 mm 2 2 1/2 1 1 2.373(2387) + σC = −163 = −617 MPa 50 cos 20° 13.68 21.89 r1 =

14-15 The pinion controls the design.

Eq. (7-8):

Y P = 0.303, YG = 0.359 17 30 dP = = 1.417 in, dG = = 2.500 in 12 12 π(1.417)(525) πd P n = = 194.8 ft/min V = 12 12 1200 + 194.8 Kv = = 1.162 1200 Se = 0.504(76) = 38 300 psi

Eq. (7-18):

ka = 2.70(76) −0.265 = 0.857

Bending

Eq. (14-4b):

Ans.

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l= Eq. (14-3):

Eq. (7-24): Eq. (7-19):

2.25 2.25 = 0.1875 in = Pd 12

3Y P 3(0.303) = = 0.0379 in 2P 2(12) t = 4(0.1875)(0.0379) = 0.1686 in de = 0.808 0.875(0.1686) = 0.310 in 0.310 −0.107 = 0.996 kb = 0.30 x=

kc = kd = ke = 1,

k f1 = 1.65

0.300 = 0.025 in 12 rf r 0.025 = = = 0.148 d t 0.1686

rf =

Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, K t = 1.68. Eq. (7-35):

Kf =

1.68 = 1.323 4 1.68 − 1 2 1+ √ 1.68 76 0.025

Miscellaneous-Effects Factor: k f = k f 1k f 2 Eq. (7-17):

1 = 1.65 1.323

= 1.247

Se = 0.857(0.996)(1)(1)(1)(1.247)(38 300) = 40 770 psi σall =

40 770 = 18 120 psi 2.25

Wt =

FY P σall 0.875(0.303)(18 120) = K v Pd 1.162(12)

= 345 lbf H= Wear Eq. (14-14): Eq. (14-12):

345(194.8) = 2.04 hp Ans. 33 000

ν1 = ν2 = 0.292, C p = 2300 psi

E 1 = E 2 = 30(106 ) psi

1.417 dP sin φ = sin 20° = 0.242 in 2 2 2.500 dG sin φ = sin 20° = 0.428 r2 = 2 2 1 1 1 1 + = 6.469 in−1 + = r1 r2 0.242 0.428 r1 =

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Chapter 14

From Eq. (7-68), (SC ) 108 = 0.4H B − 10 kpsi = [0.4(149) − 10](103 ) = 49 600 psi −49 600 = −33 067 psi σC, all = √ 2.25 −33 067 2 0.875 cos 20° t W = = 22.6 lbf 2300 1.162(6.469) 22.6(194.8) = 0.133 hp Ans. 33 000 Rating power (pinion controls): H1 = 2.04 hp H=

H2 = 0.133 hp Hall = (min 2.04, 0.133) = 0.133 hp 14-16 Pinion controls: Y P = 0.322, Bending

Ans.

YG = 0.447

d P = 20/3 = 6.667 in,

dG = 33.333 in

V = πd P n/12 = π(6.667)(870)/12 = 1519 ft/min K v = (1200 + 1519)/1200 = 2.266 Se = 0.504(113) = 56.950 kpsi = 56 950 psi ka = 2.70(113) −0.265 = 0.771 l = 2.25/Pd = 2.25/3 = 0.75 in x = 3(0.322)/[2(3)] = 0.161 in t = 4(0.75)(0.161) = 0.695 in de = 0.808 2.5(0.695) = 1.065 in kb = (1.065/0.30) −0.107 = 0.873 kc = kd = ke = 1 r f = 0.300/3 = 0.100 in rf 0.100 r = = = 0.144 d t 0.695 From Table A-15-6, K t = 1.75; Eq. (7-35): K f = 1.597 k f 2 = 1/1.597,

k f = k f 1 k f 2 = 1.65/1.597 = 1.033

Se = 0.771(0.873)(1)(1)(1)(1.033)(56 950) = 39 600 psi σall = 39 600/1.5 = 26 400 psi Wt =

FY P σall 2.5(0.322)(26 400) = 3126 lbf = K v Pd 2.266(3)

H = W t V /33 000 = 3126(1519)/33 000 = 144 hp Ans.

367

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Wear Table 14-8: Eq. (14-12):

C p = 2300 psi r1 = (6.667/2) sin 20° = 1.140 in r2 = (33.333/2) sin 20° = 5.700 in

Eq. (7-68):

SC = [0.4(262) − 10](103 ) = 94 800 psi √ √ σC, all = −SC / n d = −94.800/ 1.5 = −77 404 psi 1 σC, all 2 F cos φ t W = Cp K v 1/r1 + 1/r2 1 −77 404 2 2.5 cos 20° = 2300 2.266 1/1.140 + 1/5.700 = 1115 lbf H=

1115(1519) WtV = = 51.3 hp Ans. 33 000 33 000

For 108 cycles (revolutions of the pinion), the power based on wear is 51.3 hp. Rating power–pinion controls H1 = 144 hp H2 = 51.3 hp Hrated = min(144, 51.3) = 51.3 hp Ans. 14-17 Given: φ = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, N P = 16 milled teeth, NG = 30T, Sut = 900 MPa, H B = 260, n d = 3, Y P = 0.296, and YG = 0.359. Pinion bending d P = m N P = 6(16) = 96 mm

Eq. (14-6b):

dG = 6(30) = 180 mm π(96)(1145)(10−3 )(12) πd P n = = 5.76 m/s V = 12 (12)(60) 6.1 + 5.76 Kv = = 1.944 6.1 Se = 0.504(900) = 453.6 MPa a = 4.45,

b = −0.265

ka = 4.51(900) −0.265 = 0.744 l = 2.25m = 2.25(6) = 13.5 mm x = 3Y m/2 = 3(0.296)6/2 = 2.664 mm √ t = 4lx = 4(13.5)(2.664) = 12.0 mm de = 0.808 75(12.0) = 24.23 mm

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Chapter 14

kb =

24.23 7.62

−0.107

= 0.884

kc = kd = ke = 1 r f = 0.300m = 0.300(6) = 1.8 mm From Fig. A-15-6 for r/d = r f /t = 1.8/12 = 0.15, K t = 1.68. Kf =

1.68 = 1.537 √ 1 + 2 1.8 [(1.68 − 1)/1.68](139/900)

k f 1 = 1.65

(Gerber failure criterion)

k f 2 = 1/K f = 1/1.537 = 0.651 k f = k f 1 k f 2 = 1.65(0.651) = 1.074 Se = 0.744(0.884)(1)(1)(1)(1.074)(453.6) = 320.4 MPa

Eq. (14-8):

σall =

Se 320.4 = 246.5 MPa = nd 1.3

Wt =

FY mσall 75(0.296)(6)(246.5) = 16 890 N = Kv 1.944

H=

16 890(96/2)(1145) Tn = = 97.2 kW 9.55 9.55(106 )

Ans.

Wear: Pinion and gear r1 = (96/2) sin 20◦ = 16.42 mm Eq. (14-12): r2 = (180/2) sin 20◦ = 30.78 mm Table 14-8:

√ ˙ 191 MPa Cp =

Eq. (7-68):

SC = 6.89[0.4(260) − 10] = 647.7 MPa

Eq. (14-14):

647.7 = −568 MPa σC, all = − √ 1.3 1 σC, all 2 F cos φ t W = Cp K v 1/r1 + 1/r2 2 1 75 cos 20◦ −568 = 191 1.944 1/16.42 + 1/30.78 = 3433 N

3433(96) W t dP = = 164 784 N · mm = 164.8 N · m 2 2 Tn 164.8(1145) H= = = 19 758.7 W = 19.8 kW Ans. 9.55 9.55 Thus, wear controls the gearset power rating; H = 19.8 kW. Ans. T =

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14-18 Preliminaries: N P = 17,

NG = 51 dP =

N 17 = 2.833 in = Pd 6

51 = 8.500 in 6 V = πd P n/12 = π(2.833)(1120)/12 = 830.7 ft/min

dG =

K v = (1200 + 830.7)/1200 = 1.692

Eq. (14-4b):

σall =

Sy 90 000 = 45 000 psi = nd 2

Table 14-2:

Y P = 0.303,

Eq. (14-7):

Wt = H=

YG = 0.410

FY P σall 2(0.303)(45 000) = 2686 lbf = K v Pd 1.692(6) Wt V 2686(830.7) = = 67.6 hp 33 000 33 000

Based on yielding in bending, the power is 67.6 hp. (a) Pinion fatigue Bending

Se = 0.504(232/2) = 58 464 psi a = 2.70,

Table 13-1: Eq. (14-3):

l=

b = −0.265,

ka = 2.70(116) −0.265 = 0.766

1 1.25 2.25 2.25 = 0.375 in + = = Pd Pd Pd 6

3Y P 3(0.303) = 0.0758 = 2Pd 2(6) √ t = 4lx = 4(0.375)(0.0758) = 0.337 in √ de = 0.808 Ft = 0.808 2(0.337) = 0.663 in 0.663 −0.107 kb = = 0.919 0.30 x=

kc = kd = ke = 1. Assess two components contributing to k f . First, based upon one-way bending and the Gerber failure criterion, k f 1 = 1.65. Second, due to stressconcentration, rf = Fig. A-15-6:

0.300 0.300 = 0.050 in = Pd 6

rf 0.05 r = = = 0.148 d t 0.338

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Chapter 14

Estimate D/d = ∞ by setting D/d = 3, K t = 1.68. From Eq. (7-35) and Table 7-8, Kf =

1.68 = 1.494 √ 1 + 2 0.05 [(1.68 − 1)/1.68](4/116)

kf 2 =

1 1 = 0.669 = Kf 1.494

k f = k f 1 k f 2 = 1.65(0.669) = 1.104 Se = 0.766(0.919)(1)(1)(1)(1.104)(58 464) = 45 436 psi σall =

Se 45 436 = 22 718 psi = nd 2

Wt =

FY P σall 2(0.303)(22 718) = 1356 lbf = K v Pd 1.692(6)

H=

WtV 1356(830.7) = = 34.1 hp 33 000 33 000

Ans.

(b) Pinion fatigue Wear From Table A-5 for steel:

ν = 0.292, E = 30(106 ) psi

Eq. (14-13) or Table 14-8:

1/2 1 Cp = = 2285 psi 2π[(1 − 0.2922 )/30(106 )] In preparation for Eq. (14-14): Eq. (14-12):

r1 =

2.833 dP sin φ = sin 20◦ = 0.485 in 2 2

8.500 dG sin φ = sin 20◦ = 1.454 in r2 = 2 2 1 1 1 1 = + = 2.750 in + r1 r2 0.485 1.454 Eq. (7-68):

(SC ) 108 = 0.4H B − 10 kpsi

In terms of gear notation σC = [0.4(232) − 10]103 = 82 800 psi We√will introduce the design factor of n d = 2 and apply it to the load W t by dividing by 2. 82 800 σc = −58 548 psi σC, all = − √ = − √ 2 2

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Solve Eq. (14-14) for W t: W = t

Hall =

−58 548 2285

2

2 cos 20◦ = 265 lbf 1.692(2.750)

265(830.7) = 6.67 hp 33 000

Ans.

For 108 cycles (turns of pinion), the allowable power is 6.67 hp. (c) Gear fatigue due to bending and wear Bending 3(0.4103) 3YG = 0.1026 in = 2Pd 2(6) t = 4(0.375)(0.1026) = 0.392 in de = 0.808 2(0.392) = 0.715 in 0.715 −0.107 = 0.911 kb = 0.30 x=

Eq. (14-3):

kc = kd = ke = 1 rf 0.050 r = = = 0.128 d t 0.392 Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; K t = 1.80. Use K f = 1.583. kf 2 =

1 = 0.632, k f = 1.65(0.632) = 1.043 1.583

Se = 0.766(0.911)(1)(1)(1)(1.043)(58 464) = 42 550 psi σall =

42 550 Se = = 21 275 psi 2 2

Wt =

2(0.4103)(21 275) = 1720 lbf 1.692(6)

Hall =

1720(830.7) = 43.3 hp 33 000

Ans.

The gear is thus stronger than the pinion in bending. Wear Since the material of the pinion and the gear are the same, and the contact stresses are the same, the allowable power transmission of both is the same. Thus, Hall = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish SC for 108 /3 revolutions.

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(d) Pinion bending: H1 = 34.1 hp Pinion wear: H2 = 6.67 hp Gear bending: H3 = 43.3 hp Gear wear: H4 = 6.67 hp Power rating of the gear set is thus Hrated = min(34.1, 6.67, 43.3, 6.67) = 6.67 hp

Ans.

14-19 d P = 16/6 = 2.667 in, dG = 48/6 = 8 in π(2.667)(300) = 209.4 ft/min 12 33 000(5) Wt = = 787.8 lbf 209.4 V =

Assuming uniform loading, K o = 1. From Eq. (14-28), Q v = 6, B = 0.25(12 − 6) 2/3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 Eq. (14-27):

Kv =

From Table 14-2,

0.8255 √ 59.77 + 209.4 = 1.196 59.77

N P = 16T, Y P = 0.296 NG = 48T, YG = 0.4056

From Eq. (a), Sec. 14-10 with F = 2 in √ 0.0535 2 0.296 = 1.088 (K s ) P = 1.192 6 √ 0.0535 2 0.4056 (K s ) G = 1.192 = 1.097 6 From Eq. (14-30) with Cmc = 1 2 − 0.0375 + 0.0125(2) = 0.0625 10(2.667) = 1, Cma = 0.093 (Fig. 14-11), Ce = 1

Cp f = C pm

K m = 1 + 1[0.0625(1) + 0.093(1)] = 1.156 Assuming constant thickness of the gears → K B = 1 m G = NG /N P = 48/16 = 3 With N (pinion) = 108 cycles and N (gear) = 108 /3, Fig. 14-14 provides the relations: (Y N ) P = 1.3558(108 ) −0.0178 = 0.977 (Y N ) G = 1.3558(108 /3) −0.0178 = 0.996

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J P = 0.27,

Fig. 14-6:

JG = ˙ 0.38

From Table 14-10 for R = 0.9, K R = 0.85 KT = C f = 1

cos 20◦ sin 20◦ 3 I = = 0.1205 2 3+1 C p = 2300 psi

Eq. (14-23) with m N = 1 Table 14-8:

Strength: Grade 1 steel with H B P = H BG = 200 Fig. 14-2:

(St ) P = (St ) G = 77.3(200) + 12 800 = 28 260 psi

Fig. 14-5:

(Sc ) P = (Sc ) G = 322(200) + 29 100 = 93 500 psi ( Z N ) P = 1.4488(108 ) −0.023 = 0.948

Fig. 14-15:

( Z N ) G = 1.4488(108 /3) −0.023 = 0.973 H B P /H BG = 1

Sec. 14-12:

∴

CH = 1

Pinion tooth bending Eq. (14-15):

Pd K m K B 6 (1.156)(1) = 787.8(1)(1.196)(1.088) (σ ) P = W K o K v K s F J 2 0.27 t

= 13 167 psi

Ans.

Factor of safety from Eq. (14-41) 28 260(0.977)/[(1)(0.85)] St Y N /(K T K R ) = = 2.47 Ans. (S F ) P = σ 13 167 Gear tooth bending

6 (1.156)(1) = 9433 psi Ans. (σ ) G = 787.8(1)(1.196)(1.097) 2 0.38

(S F ) G = Pinion tooth wear Eq. (14-16):

28 260(0.996)/[(1)(0.85)] = 3.51 Ans. 9433

K m C f 1/2 (σc ) P = C p W K o K v K s dP F I P 1/2 1 1.156 = 2300 787.8(1)(1.196)(1.088) 2.667(2) 0.1205 t

= 98 760 psi Eq. (14-42):

Sc Z N /(K T K R ) (S H ) P = σc

Ans.

93 500(0.948)/[(1)(0.85)] = 1.06 Ans. = 98 760

P

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Chapter 14

Gear tooth wear

(K s ) G (σc ) G = (K s ) P (S H ) G =

1/2

(σc ) P =

1.097 1.088

1/2

(98 760) = 99 170 psi

Ans.

93 500(0.973)(1)/[(1)(0.85)] = 1.08 Ans. 99 170

The hardness of the pinion and the gear should be increased. 14-20 d P = 2.5(20) = 50 mm,

dG = 2.5(36) = 90 mm

π(50)(10−3 )(100) πd P n P = = 0.2618 m/s 60 60 60(120) = 458.4 N Wt = π(50)(10−3 )(100) V =

K o = 1,

Eq. (14-28):

Q v = 6,

B = 0.25(12 − 6) 2/3 = 0.8255

A = 50 + 56(1 − 0.8255) = 59.77 √ 0.8255 59.77 + 200(0.2618) = 1.099 Kv = 59.77

Eq. (14-27):

Y P = 0.322,

Table 14-2:

YG = 0.3775

Similar to Eq. (a) of Sec. 14-10 but for SI units: √ 0.0535 1 = 0.8433 m F Y kb √

0.0535 (K s ) P = 0.8433 2.5(18) 0.322 = 1.003 use 1 √

0.0535 (K s ) G = 0.8433 2.5(18) 0.3775 > 1 use 1 Ks =

18 − 0.025 = 0.011 10(50) = 0.247 + 0.0167(0.709) − 0.765(10−4 )(0.7092 ) = 0.259

Cmc = 1,

F = 18/25.4 = 0.709 in,

C pm = 1,

Cma

Cp f =

Ce = 1 K H = 1 + 1[0.011(1) + 0.259(1)] = 1.27 Eq. (14-40): Fig. 14-14:

K B = 1,

m G = NG /N P = 36/20 = 1.8

(Y N ) P = 1.3558(108 ) −0.0178 = 0.977 (Y N ) G = 1.3558(108 /1.8) −0.0178 = 0.987

Fig. 14-6: Eq. (14-38):

(Y J ) P = 0.33,

(Y J ) G = 0.38

Y Z = 0.658 − 0.0759 ln(1 − 0.95) = 0.885 Yθ = Z R = 1

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Eq. (14-23) with m N = 1: cos 20◦ sin 20◦ ZI = 2 √ Z E = 191 MPa

Table 14-8:

1.8 1.8 + 1

= 0.103

Grade 1 steel, H B P = H BG = 200

Strength Fig. 14-2:

(σ F P ) P = (σ F P ) G = 0.533(200) + 88.3 = 194.9 MPa

Fig. 14-5:

(σ H P ) P = (σ H P ) G = 2.22(200) + 200 = 644 MPa ( Z N ) P = 1.4488(108 ) −0.023 = 0.948

Fig. 14-15:

( Z N ) G = 1.4488(108 /1.8) −0.023 = 0.961 H B P /H BG = 1

∴

ZW = 1

Pinion tooth bending 1 KH KB t (σ ) P = W K o K v K s bm t Y J P 1.27(1) 1 = 43.08 MPa Ans. = 458.4(1)(1.099)(1) 18(2.5) 0.33 194.9 0.977 σF P YN (S F ) P = = 4.99 Ans. = σ Yθ Y Z P 43.08 1(0.885) Gear tooth bending

1.27(1) 1 (σ ) G = 458.4(1)(1.099)(1) = 37.42 MPa Ans. 18(2.5) 0.38 194.9 0.987 (S F ) G = = 5.81 Ans. 37.42 1(0.885) Pinion tooth wear K Z H R (σc ) P = Z E W t K o K v K s dw1 b Z I P 1 1.27 = 501.8 MPa Ans. = 191 458.4(1)(1.099)(1) 50(18) 0.103 644 0.948(1) σH P Z N Z W = 1.37 Ans. (S H ) P = = σc Yθ Y Z P 501.8 1(0.885)

Gear tooth wear

1/2 (K s ) G 1/2 1 (σc ) G = (σc ) P = (501.8) = 501.8 MPa (K s ) P 1 644 0.961(1) (S H ) G = = 1.39 Ans. 501.8 1(0.885)

Ans.

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Chapter 14

14-21 Pt = Pn cos ψ = 6 cos 30° = 5.196 teeth/in 16 48 = 3.079 in, dG = (3.079) = 9.238 in 5.196 16 π(3.079)(300) = 241.8 ft/min V = 12 0.8255 √ 33 000(5) 59.77 + 241.8 = 682.3 lbf, K v = = 1.210 Wt = 241.8 59.77 dP =

From Prob. 14-19: Y P = 0.296,

YG = 0.4056

(K s ) P = 1.088, m G = 3,

(K s ) G = 1.097,

(Y N ) P = 0.977,

(St ) P = (St ) G = 28 260 psi, ( Z N ) P = 0.948, The pressure angle is: −1

φt = tan

(Y N ) G = 0.996,

K R = 0.85

C H = 1,

( Z N ) G = 0.973, tan 20° cos 30°

KB = 1 (Sc ) P = (Sc ) G = 93 500 psi √ C p = 2300 psi

= 22.80°

3.079 cos 22.8° = 1.419 in, 2 a = 1/Pn = 1/6 = 0.167 in

(rb ) P =

(rb ) G = 3(rb ) P = 4.258 in

Eq. (14-26): 1/2 1/2 2 2 3.079 9.238 2 2 Z= + 0.167 − 1.419 + 0.167 − 4.258 + 2 2 3.079 9.238 + sin 22.8° − 2 2 = 0.9479 + 2.1852 − 2.3865 = 0.7466 Conditions O.K. for use π p N = pn cos φn = cos 20° = 0.4920 in 6 Eq. (14-23): Eq. (14-23): Fig. 14-7:

0.492 pN = = 0.6937 0.95Z 0.95(0.7466) sin 22.8° cos 22.8° 3 I = = 0.193 2(0.6937) 3+1

mN =

˙ 0.45, J P =

JG = ˙ 0.54

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Fig. 14-8: Corrections are 0.94 and 0.98 J P = 0.45(0.94) = 0.423, JG = 0.54(0.98) = 0.529 2 − 0.0375 + 0.0125(2) = 0.0525 Cmc = 1, C p f = 10(3.079) C pm = 1, Cma = 0.093, Ce = 1 K m = 1 + (1)[0.0525(1) + 0.093(1)] = 1.146 Pinion tooth bending

5.196 (σ ) P = 682.3(1)(1.21)(1.088) 2 (S F ) P =

1.146(1) = 6323 psi Ans. 0.423

28 260(0.977)/[1(0.85)] = 5.14 Ans. 6323

Gear tooth bending

5.196 (σ ) G = 682.3(1)(1.21)(1.097) 2 (S F ) G =

1.146(1) = 5097 psi Ans. 0.529

28 260(0.996)/[1(0.85)] = 6.50 Ans. 5097

Pinion tooth wear 1/2 1 1.146 = 67 700 psi Ans. (σc ) P = 2300 682.3(1)(1.21)(1.088) 3.078(2) 0.193 (S H ) P =

93 500(0.948)/[(1)(0.85)] = 1.54 Ans. 67 700

Gear tooth wear

1.097 (σc ) G = 1.088 (S H ) G =

1/2

(67 700) = 67 980 psi Ans.

93 500(0.973)/[(1)(0.85)] = 1.57 67 980

Ans.

14-22 Given: R = 0.99 at 108 cycles, H B = 232 through-hardening Grade 1, core and case, both gears. N P = 17T, NG = 51T, Y P = 0.303, YG = 0.4103, J P = 0.292, JG = 0.396, d P = 2.833 in, dG = 8.500 in. Pinion bending From Fig. 14-2: 0.99 (St ) 107

= 77.3H B + 12 800 = 77.3(232) + 12 800 = 30 734 psi

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Chapter 14

Fig. 14-14:

Y N = 1.6831(108 ) −0.0323 = 0.928 V = πd P n/12 = π(2.833)(1120/12) = 830.7 ft/min √ K T = K R = 1, S F = 2, S H = 2 σall =

30 734(0.928) = 14 261 psi 2(1)(1)

Q v = 5,

B = 0.25(12 − 5) 2/3 = 0.9148

A = 50 + 56(1 − 0.9148) = 54.77 0.9148 √ 54.77 + 830.7 = 1.472 Kv = 54.77 √ 0.0535 2 0.303 K s = 1.192 = 1.089 ⇒ 6

use 1

K m = Cm f = 1 + Cmc (C p f C pm + Cma Ce ) Cmc = 1 F − 0.0375 + 0.0125F 10d 2 − 0.0375 + 0.0125(2) = 10(2.833)

Cp f =

= 0.0581 C pm = 1 Cma = 0.127 + 0.0158(2) − 0.093(10−4 )(22 ) = 0.1586 Ce = 1 K m = 1 + 1[0.0581(1) + 0.1586(1)] = 1.2167 Kβ = 1 Eq. (14-15):

Wt = = H=

F J P σall K o K v K s Pd K m K β 2(0.292)(14 261) = 775 lbf 1(1.472)(1)(6)(1.2167)(1) 775(830.7) WtV = = 19.5 hp 33 000 33 000

Pinion wear Fig. 14-15:

Z N = 2.466N −0.056 = 2.466(108 ) −0.056 = 0.879 MG = 51/17 = 3 sin 20◦ cos 20◦ I = 2

3 3+1

= 1.205,

CH = 1

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 0.99 (Sc ) 107

Fig. 14-5:

= 322H B + 29 100 = 322(232) + 29 100 = 103 804 psi

103 804(0.879) = 64 519 psi √ 2(1)(1) Fd P I σc, all 2 t W = Cp Ko Kv Ks Km C f 2 2(2.833)(0.1205) 64 519 = 2300 1(1.472)(1)(1.2167)(1) σc, all =

Eq. (14-16):

= 300 lbf 300(830.7) WtV = = 7.55 hp H= 33 000 33 000 The pinion controls therefore Hrated = 7.55 hp

Ans.

4-23 3Y 2Pd √ 3.674 √ 3Y 2.25 = t = 4lx = 4 Y Pd 2Pd Pd √ √ √ F Y 3.674 de = 0.808 Ft = 0.808 F Y = 1.5487 Pd Pd √ −0.107  √ −0.0535 1.5487 F Y/Pd F Y  kb =  = 0.8389 0.30 Pd l = 2.25/Pd ,

x=

1 Ks = = 1.192 kb 14-24

√ 0.0535 F Y Pd

Ans.

Y P = 0.331, YG = 0.422, J P = 0.345, JG = 0.410, K o = 1.25. The service conditions are adequately described by K o . Set S F = S H = 1. d P = 22/4 = 5.500 in dG = 60/4 = 15.000 in V =

π(5.5)(1145) = 1649 ft/min 12

Pinion bending 0.99 (St ) 107

= 77.3H B + 12 800 = 77.3(250) + 12 800 = 32 125 psi

Y N = 1.6831[3(109 )]−0.0323 = 0.832

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Chapter 14

(σall ) P =

Eq. (14-17):

32 125(0.832) = 26 728 psi 1(1)(1)

B = 0.25(12 − 6)2/3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 0.8255 √ 59.77 + 1649 = 1.534 Kv = 59.77 K s = 1,

Cm = 1

F − 0.0375 + 0.0125F 10d

Cmc =

3.25 − 0.0375 + 0.0125(3.25) = 0.0622 10(5.5)

=

Cma = 0.127 + 0.0158(3.25) − 0.093(10−4 )(3.252 ) = 0.178 Ce = 1 K m = Cm f = 1 + (1)[0.0622(1) + 0.178(1)] = 1.240 K β = 1, K T = 1 W1t =

Eq. (14-15):

26 728(3.25)(0.345) = 3151 lbf 1.25(1.534)(1)(4)(1.240)

3151(1649) = 157.5 hp 33 000 By similar reasoning, W2t = 3861 lbf and H2 = 192.9 hp H1 =

Gear bending Pinion wear

m G = 60/22 = 2.727 cos 20◦ sin 20◦ I = 2 0.99 (Sc ) 107

2.727 1 + 2.727

= 0.1176

= 322(250) + 29 100 = 109 600 psi

( Z N ) P = 2.466[3(109 )]−0.056 = 0.727 ( Z N ) G = 2.466[3(109 )/2.727]−0.056 = 0.769 109 600(0.727) = 79 679 psi 1(1)(1) Fd P I σc, all 2 t W3 = Cp Ko Kv Ks Km C f 2 3.25(5.5)(0.1176) 79 679 = 1061 lbf = 2300 1.25(1.534)(1)(1.24)(1)

(σc, all ) P =

H3 =

1061(1649) = 53.0 hp 33 000

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Gear wear W4t = 1182 lbf,

Similarly, Rating

H4 = 59.0 hp

Hrated = min( H1 , H2 , H3 , H4 ) = min(157.5, 192.9, 53, 59) = 53 hp Ans. Note differing capacities. Can these be equalized? 14-25

From Prob. 14-24: W1t = 3151 lbf,

W2t = 3861 lbf,

W3t = 1061 lbf,

W4t = 1182 lbf

Wt =

33 000(1.25)(40) 33 000K o H = = 1000 lbf V 1649

Pinion bending: The factor of safety, based on load and stress, is (S F ) P =

W1t 3151 = = 3.15 1000 1000

Gear bending based on load and stress W2t 3861 = = 3.86 (S F ) G = 1000 1000 Pinion wear based on load: based on stress:

W3t 1061 = = 1.06 1000 1000 √ (S H ) P = 1.06 = 1.03

n3 =

Gear wear based on load: based on stress:

W4t 1182 = = 1.18 1000 1000 √ (S H ) G = 1.18 = 1.09

n4 =

Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06, 1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors (S F ) P , (S F ) G , (S H ) P , (S H ) G are 3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2 and the threat is again from pinion wear. Depending on the magnitude of the numbers, using S F and S H as defined by AGMA, does not necessarily lead to the same conclusion concerning threat. Therefore be cautious.

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Chapter 14

14-26

Solution summary from Prob. 14-24: n = 1145 rev/min, K o = 1.25, Grade 1 materials, N P = 22T, NG = 60T, m G = 2.727, Y P = 0.331, YG = 0.422, J P = 0.345, JG = 0.410, Pd = 4T /in, F = 3.25 in, Q v = 6, ( Nc ) P = 3(109 ), R = 0.99 Pinion H B : 250 core, 390 case Gear H B : 250 core, 390 case K m = 1.240, K T = 1, K β = 1, d P = 5.500 in, dG = 15.000 in, V = 1649 ft/min, K v = 1.534, (K s ) P = (K s ) G = 1, (Y N ) P = 0.832, (Y N ) G = 0.859, K R = 1 Bending (σall ) P = 26 728 psi

(St ) P = 32 125 psi

(σall ) G = 27 546 psi

(St ) G = 32 125 psi

W1t = 3151 lbf,

H1 = 157.5 hp

W2t = 3861 lbf,

H2 = 192.9 hp

Wear φ = 20◦ , ( Z N ) G = 0.769,

I = 0.1176,

( Z N ) P = 0.727, C P = 2300 psi

(Sc ) P = Sc = 322(390) + 29 100 = 154 680 psi (σc, all ) P =

154 680(0.727) = 112 450 psi 1(1)(1)

154 680(0.769) = 118 950 psi 1(1)(1) 112 450 2 t W3 = (1061) = 2113 lbf, 79 679 2 118 950 t W4 = (1182) = 2354 lbf, 109 600(0.769)

(σc, all ) G =

H3 =

2113(1649) = 105.6 hp 33 000

H4 =

2354(1649) = 117.6 hp 33 000

Rated power Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp Prob. 14-24

Ans.

Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hp

The rated power approximately doubled. 14-27

The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell 285 core and Brinell 580–600 case. Table 14-3: 0.99 (St ) 107

= 55 000 psi

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Modification of St by (Y N ) P = 0.832 produces (σall ) P = 45 657 psi, Similarly for (Y N ) G = 0.859 (σall ) G = 47 161 psi, W1t = 4569 lbf,

and H1 = 228 hp

W2t = 5668 lbf,

H2 = 283 hp From Table 14-8, Cp = 2300 psi. Also, from Table 14-6: 0.99 (Sc ) 107

= 180 000 psi

Modification of Sc by (Y N ) produces (σc, all ) P = 130 525 psi (σc, all ) G = 138 069 psi and W3t = 2489 lbf,

H3 = 124.3 hp

W4t = 2767 lbf,

H4 = 138.2 hp

Rating Hrated = min(228, 283, 124, 138) = 124 hp 14-28

Ans.

Grade 2 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27. Summary: Table 14-3:

0.99 (St ) 107

= 65 000 psi

(σall ) P = 53 959 psi (σall ) G = 55 736 psi and it follows that W1t = 5399.5 lbf,

H1 = 270 hp

H2 = 335 hp W2 = 6699 lbf, From Table 14-8, C p = 2300 psi. Also, from Table 14-6: t

Sc = 225 000 psi (σc, all ) P = 181 285 psi (σc, all ) G = 191 762 psi Consequently, W3t = 4801 lbf,

H3 = 240 hp

W4t = 5337 lbf,

H4 = 267 hp

Rating Hrated = min(270, 335, 240, 267) = 240 hp.

Ans.

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Chapter 14

14-29

n = 1145 rev/min, K o = 1.25, N P = 22T, NG = 60T, dG = 7.5 in, Y P = 0.331, YG = 0.422, J P = 0.335, F = 1.625 in, H B = 250, case and core, both gears. C pm = 1, Cma = 0.152, Ce = 1, C f = 0.0419, K β = 1, K s = 1, V = 824 ft/min, (Y N ) P = 0.8318, I = 0.117 58 0.99 (St ) 107 = 32 125 psi (σall ) P = 26 668 psi (σall ) G = 27 546 psi and it follows that

m G = 2.727, d P = 2.75 in, JG = 0.405, P = 8T /in, Cm = 1, F/d P = 0.0591, K m = 1.1942, K T = 1, (Y N ) G = 0.859, K R = 1,

W1t = 879.3 lbf,

H1 = 21.97 hp

W2 = 1098 lbf,

H2 = 27.4 hp

W3t = 304 lbf,

H3 = 7.59 hp

W4t = 340 lbf,

H4 = 8.50 hp

t

For wear

Rating Hrated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp In Prob. 14-24, Hrated = 53 hp Thus 7.59 1 1 = 0.1432 = , not Ans. 53.0 6.98 8 The transmitted load rating is t Wrated = min(879.3, 1098, 304, 340) = 304 lbf In Prob. 14-24 t Wrated = 1061 lbf

Thus 1 304 = 0.2865 = , 1061 3.49 14-30

S P = S H = 1, Bending Table 14-4:

Pd = 4,

0.99 (St ) 107

J P = 0.345,

1 not , 4

JG = 0.410,

Ans. K o = 1.25

= 13 000 psi

13 000(1) = 13 000 psi 1(1)(1) σall F J P 13 000(3.25)(0.345) W1t = = 1533 lbf = K o K v K s Pd K m K β 1.25(1.534)(1)(4)(1.24)(1)

(σall ) P = (σall ) G =

1533(1649) = 76.6 hp 33 000 W2t = W1t JG /J P = 1533(0.410)/0.345 = 1822 lbf H2 = H1 JG /J P = 76.6(0.410)/0.345 = 91.0 hp H1 =

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Wear Table 14-8: Table 14-7:

C p = 1960 psi = 75 000 psi = (σc, all ) P = (σc, all ) G FdpI (σc, all ) P 2 t W3 = Cp Ko Kv Ks Km C f 2 3.25(5.5)(0.1176) 75 000 = 1295 lbf W3t = 1960 1.25(1.534)(1)(1.24)(1)

0.99 (Sc ) 107

W4t = W3t = 1295 lbf H4 = H3 =

1295(1649) = 64.7 hp 33 000

Rating Hrated = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp

Ans.

Notice that the balance between bending and wear power is improved due to CI’s more favorable Sc /St ratio. Also note that the life is 107 pinion revolutions which is (1/300) of 3(109). Longer life goals require power derating. 14-31

From Table A-24a, E av = 11.8(106 ) For φ = 14.5◦ and H B = 156 1.4(81) = 51 693 psi 2 sin 14.5°/[11.8(106 )]

SC = For φ = 20◦

SC =

1.4(112) = 52 008 psi 2 sin 20°/[11.8(106 )]

SC = 0.32(156) = 49.9 kpsi 14-32

Programs will vary.

14-33 (Y N ) P = 0.977,

(Y N ) G = 0.996

(St ) P = (St ) G = 82.3(250) + 12 150 = 32 725 psi (σall ) P =

32 725(0.977) = 37 615 psi 1(0.85)

W1t =

37 615(1.5)(0.423) = 1558 lbf 1(1.404)(1.043)(8.66)(1.208)(1)

H1 =

1558(925) = 43.7 hp 33 000

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Chapter 14

32 725(0.996) = 38 346 psi 1(0.85) 38 346(1.5)(0.5346) W2t = = 2007 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 2007(925) H2 = = 56.3 hp 33 000 ( Z N ) P = 0.948, ( Z N ) G = 0.973 (σall ) G =

Table 14-6:

0.99 (Sc ) 107

(σc, allow ) P W3t H3 (σc, allow ) G W4t

= 150 000 psi 0.948(1) = 167 294 psi = 150 000 1(0.85) 167 294 2 1.963(1.5)(0.195) = 2074 lbf = 2300 1(1.404)(1.043) 2074(925) = 58.1 hp = 33 000 0.973 (167 294) = 171 706 psi = 0.948 171 706 2 1.963(1.5)(0.195) = 2167 lbf = 2300 1(1.404)(1.052) 2167(925) = 60.7 hp 33 000 = min(43.7, 56.3, 58.1, 60.7) = 43.7 hp

H4 = Hrated

Ans.

Pinion bending controlling 14-34

(Y N ) P = 1.6831(108 ) −0.0323 = 0.928 (Y N )G = 1.6831(108 /3.059)−0.0323 = 0.962 Table 14-3:

St = 55 000 psi (σall ) P =

55 000(0.928) = 60 047 psi 1(0.85)

W1t =

60 047(1.5)(0.423) = 2487 lbf 1(1.404)(1.043)(8.66)(1.208)(1)

H1 =

2487(925) = 69.7 hp 33 000

0.962 (60 047) = 62 247 psi 0.928 62 247 0.5346 t W2 = (2487) = 3258 lbf 60 047 0.423

(σall ) G =

H2 =

3258 (69.7) = 91.3 hp 2487

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Sc = 180 000 psi

Table 14-6:

(Z N ) P = 2.466(108 )−0.056 = 0.8790 (Z N )G = 2.466(108 /3.059)−0.056 = 0.9358 180 000(0.8790) = 186 141 psi 1(0.85) 186 141 2 1.963(1.5)(0.195) t = 2568 lbf W3 = 2300 1(1.404)(1.043)

(σc, all ) P =

2568(925) = 72.0 hp 33 000

H3 =

0.9358 (186 141) = 198 169 psi 0.8790 198 169 2 1.043 t (2568) = 2886 lbf W4 = 186 141 1.052

(σc, all ) G =

H4 =

2886(925) = 80.9 hp 33 000

Hrated = min(69.7, 91.3, 72, 80.9) = 69.7 hp

Ans.

Pinion bending controlling 14-35

(Y N ) P = 0.928, Table 14-3:

(Y N ) G = 0.962

(See Prob. 14-34)

St = 65 000 psi (σall ) P =

65 000(0.928) = 70 965 psi 1(0.85)

W1t =

70 965(1.5)(0.423) = 2939 lbf 1(1.404)(1.043)(8.66)(1.208)

H1 =

2939(925) = 82.4 hp 33 000

65 000(0.962) = 73 565 psi 1(0.85) 73 565 0.5346 t (2939) = 3850 lbf W2 = 70 965 0.423

(σall ) G =

H2 =

3850 (82.4) = 108 hp 2939

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Chapter 14

Table 14-6:

Sc = 225 000 psi ( Z N ) P = 0.8790,

( Z N ) G = 0.9358

225 000(0.879) = 232 676 psi 1(0.85) 232 676 2 1.963(1.5)(0.195) t = 4013 lbf W3 = 2300 1(1.404)(1.043)

(σc, all ) P =

H3 =

4013(925) = 112.5 hp 33 000

0.9358 (232 676) = 247 711 psi 0.8790 247 711 2 1.043 t (4013) = 4509 lbf W4 = 232 676 1.052

(σc, all ) G =

H4 =

4509(925) = 126 hp 33 000

Hrated = min(82.4, 108, 112.5, 126) = 82.4 hp The bending of the pinion is the controlling factor.

Ans.

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Chapter 15 15-1

Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, NC = 109 rev of pinion at R = 0.999, N P = 20 teeth, NG = 60 teeth, Q v = 6, Pd = 6 teeth/in, shaft angle 90°, n p = 900 rev/min, J P = 0.249 and JG = 0.216 (Fig. 15-7), F = 1.25 in, S F = S H = 1, K o = 1. d P = 20/6 = 3.333 in

Mesh

dG = 60/6 = 10.000 in Eq. (15-7):

vt = π(3.333)(900/12) = 785.3 ft/min

Eq. (15-6):

B = 0.25(12 − 6) 2/3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 0.8255 √ 59.77 + 785.3 Kv = = 1.374 59.77

Eq. (15-5): Eq. (15-8):

vt, max = [59.77 + (6 − 3)]2 = 3940 ft/min

Since 785.3 < 3904, K v = 1.374 is valid. The size factor for bending is: K s = 0.4867 + 0.2132/6 = 0.5222

Eq. (15-10):

For one gear straddle-mounted, the load-distribution factor is: K m = 1.10 + 0.0036(1.25) 2 = 1.106

Eq. (15-11): Eq. (15-15):

(K L ) P = 1.6831(109 ) −0.0323 = 0.862 (K L ) G = 1.6831(109 /3) −0.0323 = 0.893

Eq. (15-14):

(C L ) P = 3.4822(109 ) −0.0602 = 1 (C L ) G = 3.4822(109 /3) −0.0602 = 1.069 K R = 0.50 − 0.25 log(1 − 0.999) = 1.25 √ C R = K R = 1.25 = 1.118

Eq. (15-19):

(or Table 15-3)

Bending Fig. 15-13: Eq. (15-4): Eq. (15-3):

0.99 St

= sat = 44(300) + 2100 = 15 300 psi

(σall ) P = swt = W Pt = = H1 =

Eq. (15-4):

(σall ) G =

sat K L 15 300(0.862) = 10 551 psi = SF K T K R 1(1)(1.25)

(σall ) P F K x J P Pd K o K v K s K m 10 551(1.25)(1)(0.249) = 690 lbf 6(1)(1.374)(0.5222)(1.106) 690(785.3) = 16.4 hp 33 000 15 300(0.893) = 10 930 psi 1(1)(1.25)

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Chapter 15

WGt =

10 930(1.25)(1)(0.216) = 620 lbf 6(1)(1.374)(0.5222)(1.106)

620(785.3) = 14.8 hp Ans. 33 000 The gear controls the bending rating. H2 =

15-2

Refer to Prob. 15-1 for the gearset specifications. Wear sac = 341(300) + 23 620 = 125 920 psi Fig. 15-12: For the pinion, C H = 1. From Prob. 15-1, C R = 1.118. Thus, from Eq. (15-2): (σc, all ) P =

sac (C L ) P C H SH K T C R

(σc, all ) P =

125 920(1)(1) = 112 630 psi 1(1)(1.118)

For the gear, from Eq. (15-16), B1 = 0.008 98(300/300) − 0.008 29 = 0.000 69 C H = 1 + 0.000 69(3 − 1) = 1.001 38 And Prob. 15-1, (C L ) G = 1.0685. Equation (15-2) thus gives (σc, all ) G =

sac (C L ) G C H SH K T C R

125 920(1.0685)(1.001 38) = 120 511 psi 1(1)(1.118) = 2290 psi = 0.125(1.25) + 0.4375 = 0.593 75 = 0.083 =2 Fd P I (σc, all ) P 2 = Cp K o K v K m Cs C xc 1.25(3.333)(0.083) 112 630 2 = 2290 1(1.374)(1.106)(0.5937)(2)

(σc, all ) G = For steel: Eq. (15-9): Fig. 15-6: Eq. (15-12):

Cp Cs I C xc

Eq. (15-1):

W Pt

= 464 lbf 464(785.3) = 11.0 hp 33 000 1.25(3.333)(0.083) 120 511 2 t WG = 2290 1(1.374)(1.106)(0.593 75)(2) H3 =

= 531 lbf H4 =

531(785.3) = 12.6 hp 33 000

391

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H = 11.0 hp Ans. The pinion controls wear: The power rating of the mesh, considering the power ratings found in Prob. 15-1, is H = min(16.4, 14.8, 11.0, 12.6) = 11.0 hp Ans. 15-3

AGMA 2003-B97 does not fully address cast iron gears, however, approximate comparisons can be useful. This problem is similar to Prob. 15-1, but not identical. We will organize the method. A follow-up could consist of completing Probs. 15-1 and 15-2 with identical pinions, and cast iron gears. Given: Uncrowned, straight teeth, Pd = 6 teeth/in, N P = 30 teeth, NG = 60 teeth, ASTM ◦ 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25, n P = 900 rev/min, √ φn = 20 , one gear straddle-mounted, K o = 1, J P = 0.268, JG = 0.228, S F = 2, S H = 2. Mesh

d P = 30/6 = 5.000 in dG = 60/6 = 10.000 in vt = π(5)(900/12) = 1178 ft/min

Set N L = 107 cycles for the pinion. For R = 0.99, Table 15-7:

sat = 4500 psi

Table 15-5:

sac = 50 000 psi

Eq. (15-4):

swt =

sat K L 4500(1) = 2250 psi = SF K T K R 2(1)(1)

The velocity factor K v represents stress augmentation due to mislocation of tooth profiles along the pitch surface and the resulting “falling” of teeth into engagement. Equation √ (5-67) shows that the induced bending moment in a cantilever (tooth) varies directly with E of the tooth material. If only the material varies (cast iron vs. steel) in the same geometry, I is the same. From the Lewis equation of Section 14-1, σ =

Kv W t P M = I /c FY

We expect the ratio σCI /σsteel to be (K v ) CI σCI = = σsteel (K v ) steel

E CI E steel

In the case of ASTM class 30, from Table A-24(a) ( E CI ) av = (13 + 16.2)/2 = 14.7 kpsi Then

(K v ) CI =

14.7 (K v ) steel = 0.7(K v ) steel 30

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Chapter 15

Our modeling is rough, but it convinces us that (K v ) CI < (K v ) steel , but we are not sure of the value of (K v ) CI . We will use K v for steel as a basis for a conservative rating. B = 0.25(12 − 6) 2/3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 0.8255 √ 59.77 + 1178 Kv = = 1.454 59.77

Eq. (15-6):

Eq. (15-5): Pinion bending

(σall ) P = swt = 2250 psi

From Prob. 15-1,

K x = 1,

Eq. (15-3):

W Pt = = H1 =

K m = 1.106, K s = 0.5222

(σall ) P F K x J P Pd K o K v K s K m 2250(1.25)(1)(0.268) = 149.6 lbf 6(1)(1.454)(0.5222)(1.106) 149.6(1178) = 5.34 hp 33 000

Gear bending WGt

=

H2 =

JG W Pt JP

0.228 = 149.6 0.268

= 127.3 lbf

127.3(1178) = 4.54 hp 33 000

The gear controls in bending fatigue. H = 4.54 hp Ans. 15-4

Continuing Prob. 15-3, Table 15-5:

Eq. (15-1):

sac = 50 000 psi 50 000 swt = σc, all = √ = 35 355 psi 2 Fd P I σc, all 2 t W = Cp K o K v K m Cs C xc I = 0.86

Fig. 15-6:

From Probs. 15-1 and 15-2: Cs = 0.593 75,

K s = 0.5222, K m = 1.106, C xc = 2 C p = 1960 psi From Table 14-8: 1.25(5.000)(0.086) 35 355 2 t W = = 91.6 lbf Thus, 1960 1(1.454)(1.106)(0.59375)(2) H3 = H4 =

91.6(1178) = 3.27 hp 33 000

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Rating

Based on results of Probs. 15-3 and 15-4, H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp Ans.

The mesh is weakest in wear fatigue. 15-5

Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of pinion at R = 0.999, N p = z 1 = 22 teeth, Na = z 2 =24 teeth, √ Q v = 5, m et = 4 mm, shaft angle 90°, n 1 = 1800 rev/min, S F = 1, S H = S F = 1, J P = Y J 1 √ = 0.23, JG = Y J 2 = 0.205, F = b = 25 mm, K o = K A = K T = K θ = 1 and C p = 190 MPa . Mesh Eq. (15-7): Eq. (15-6):

Eq. (15-5): Eq. (15-10): Eq. (15-11) with

= de1 = mz 1 = 4(22) = 88 mm = m et z 2 = 4(24) = 96 mm = 5.236(10−5 )(88)(1800) = 8.29 m/s = 0.25(12 − 5) 2/3 = 0.9148 A = 50 + 56(1 − 0.9148) = 54.77 √ 0.9148 54.77 + 200(8.29) = 1.663 Kv = 54.77

dP dG vet B

K s = Yx = 0.4867 + 0.008 339(4) = 0.520 K mb = 1

(both straddle-mounted),

K m = K Hβ = 1 + 5.6(10−6 )(252 ) = 1.0035 From Fig. 15-8,

Eq. (15-12): Eq. (15-19): From Fig. 15-10, Eq. (15-9): Wear of Pinion Fig. 15-12: Fig. 15-6: Eq. (15-2):

= ( Z N T ) P = 3.4822(109 ) −0.0602 = 1.00 = ( Z N T ) G = 3.4822[109 (22/24)]−0.0602 = 1.0054 = Z xc = 2 (uncrowned) = Y Z = 0.50 − 0.25 log (1 − 0.999) = 1.25 √ C R = Z Z = Y Z = 1.25 = 1.118

(C L ) P (C L ) G C xc KR

C H = Zw = 1 Z x = 0.004 92(25) + 0.4375 = 0.560 σ H lim = 2.35H B + 162.89 = 2.35(180) + 162.89 = 585.9 MPa I = Z I = 0.066 (σ H lim ) P ( Z N T ) P Z W (σ H ) P = SH K θ Z Z 585.9(1)(1) = 524.1 MPa =√ 1(1)(1.118) 2 bde1 Z I σH t WP = Cp 1000K A K v K Hβ Z x Z xc

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Chapter 15

The constant 1000 expresses W t in kN 25(88)(0.066) 524.1 2 t = 0.591 kN WP = 190 1000(1)(1.663)(1.0035)(0.56)(2) 0.591(88/2)(1800) W t rn 1 H3 = = = 4.90 kW 9.55 9.55(10) 3 σ H lim = 585.9 MPa Wear of Gear 585.9(1.0054) (σ H ) G = √ = 526.9 MPa 1(1)(1.118) 526.9 t t (σ H ) G = 0.594 kN = 0.591 WG = W P (σ H ) P 524.1 0.594(88/2)(1800) W t rn H4 = = = 4.93 kW 9.55 9.55(103 ) Thus in wear, the pinion controls the power rating; H = 4.90 kW Ans. We will rate the gear set after solving Prob. 15-6. 15-6

Refer to Prob. 15-5 for terms not defined below. Bending of Pinion (K L ) P = (Y N T ) P = 1.6831(109 ) −0.0323 = 0.862 (K L ) G = (Y N T ) G = 1.6831[109 (22/24)]−0.0323 = 0.864 Fig. 15-13: Eq. (15-13):

σ F lim = 0.30H B + 14.48 = 0.30(180) + 14.48 = 68.5 MPa K x = Yβ = 1

Y Z = 1.25, vet = 8.29 m/s K v = 1.663, K θ = 1, Yx = 0.56, K Hβ = 1.0035 σ F lim Y N T 68.5(0.862) = 47.2 MPa (σ F ) P = = SF K θ Y Z 1(1)(1.25) (σ F ) P bm et Yβ Y J 1 W pt = 1000K A K v Yx K Hβ 47.2(25)(4)(1)(0.23) = 1.16 kN = 1000(1)(1.663)(0.56)(1.0035) 1.16(88/2)(1800) H1 = = 9.62 kW 9.55(103 ) Bending of Gear σ F lim = 68.5 MPa 68.5(0.864) (σ F ) G = = 47.3 MPa 1(1)(1.25) 47.3(25)(4)(1)(0.205) WGt = = 1.04 kN 1000(1)(1.663)(0.56)(1.0035) 1.04(88/2)(1800) H2 = = 8.62 kW 9.55(103 ) From Prob. 15-5:

395

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Rating of mesh is Hrating = min(9.62, 8.62, 4.90, 4.93) = 4.90 kW

Ans.

with pinion wear controlling. 15-7 (S F ) P =

(a)

σ all

σ

= (S F ) G =

P

(sat K L /K T K R ) P t (W Pd K o K v K s K m /F K x J ) P

=

σ all

σ

G

(sat K L /K T K R ) G t (W Pd K o K v K s K m /F K x J ) G

All terms cancel except for sat , K L , and J, (sat ) P (K L ) P J P = (sat ) G (K L ) G JG From which (sat ) G =

(sat ) P (K L ) P J P JP β = (sat ) P m G (K L ) G JG JG

Where β = −0.0178 or β = −0.0323 as appropriate. This equation is the same as Eq. (14-44). Ans. (b) In bending

W = t

In wear

F Kx J σall S F Pd K o K v K s K m

sac C L CU SH K T C R

=

11

= Cp

22

F Kx J sat K L S F K T K R Pd K o K v K s K m

W t K o K v K m Cs C xc Fd P I

(1) 11

1/2 22

t

Squaring and solving for W gives 2 2 2 sac C L C H Fd P I t W = 2 SH K T2 C 2R C 2P 22 K o K v K m Cs C xc 22

(2)

Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing that C R = K R and Pd d P = N P , we obtain

(sac ) 22 =

Cp (C L ) 22

2 SH (sat ) 11 (K L ) 11 K x J11 K T Cs C xc 2 SF CH N P Ks I

For equal W t in bending and wear 2 SH = SF

So we get

√

SF SF

2 =1

Cp (sac ) G = (C L ) G C H

(sat ) P (K L ) P J P K x K T Cs C xc N P I Ks

Ans.

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397

Chapter 15

(c) (S H ) P = (S H ) G =

σc, all σc

= P

σc, all σc

G

Substituting in the right-hand equality gives [sac C L C H /(C R K T )]G [sac C L /(C R K T )] P = C p W t K o K v K m Cs C xc /( Fd P I ) C p W t K o K v K m Cs C xc /( Fd P I ) P

G

Denominators cancel leaving (sac ) P (C L ) P = (sac ) G (C L ) G C H . Solving for (sac ) P gives, with C H = 1 (C L ) G 1 −0.0602 . C H = (sac ) G (sac ) P = (sac ) G (C L ) P mG . (sac ) P = (sac ) G m 0G.0602 Ans. This equation is the transpose of Eq. (14-45). 15-8 Pinion Gear

Core ( H B ) 11 ( H B ) 21

Case ( H B ) 12 ( H B ) 22

Given ( H B ) 11 = 300 Brinell (sat ) P = 44(300) + 2100 = 15 300 psi J P −0.0323 0.249 −0.0323 = 17 023 psi 3 = 15 300 (sat ) G (sat ) P m G JG 0.216

Eq. (15-23):

( H B ) 21 =

17 023 − 2100 = 339 Brinell 44

Ans.

15 300(0.862)(0.249)(1)(0.593 25)(2) = 141 160 psi 20(0.086)(0.5222)

(sac ) G =

2290 1.0685(1)

( H B ) 22 =

141 160 − 23 600 = 345 Brinell 341 0.0602

(sac ) P = (sac ) G m G ( H B ) 12 =

Ans.

1 . 1 0.0602 = 150 811 psi ) = 141 160(3 CH 1

150 811 − 23 600 = 373 Brinell 341 Pinion Gear

Care 300 399

Ans.

Case 373 345

Ans.

(1)

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Pinion core (sat ) P = 44(300) + 2100 = 15 300 psi 15 300(0.862) (σall ) P = = 10 551 psi 1(1)(1.25) 10 551(1.25)(0.249) Wt = = 689.7 lbf 6(1)(1.374)(0.5222)(1.106) Gear core (sat ) G = 44(352) + 2100 = 17 588 psi 17 588(0.893) (σall ) G = = 12 565 psi 1(1)(1.25) 12 565(1.25)(0.216) Wt = = 712.5 lbf 6(1)(1.374)(0.5222)(1.106) Pinion case (sac ) P = 341(372) + 23 620 = 150 472 psi 150 472(1) = 134 590 psi (σc, all ) P = 1(1)(1.118) 1.25(3.333)(0.086) 134 590 2 t = 685.8 lbf W = 2290 1(1.374)(1.106)(0.593 75)(2) Gear case (sac ) G = 341(344) + 23 620 = 140 924 psi 140 924(1.0685)(1) = 134 685 psi (σc, all ) G = 1(1)(1.118) 1.25(3.333)(0.086) 134 685 2 t = 686.8 lbf W = 2290 1(1.374)(1.106)(0.593 75)(2) The rating load would be t Wrated = min(689.7, 712.5, 685.8, 686.8) = 685.8 lbf

which is slightly less than intended. Pinion core (sat ) P = 15 300 psi (σall ) P = 10 551 W t = 689.7

(as before) (as before) (as before)

Gear core (sat ) G = 44(339) + 2100 = 17 016 psi 17 016(0.893) = 12 156 psi (σall ) G = 1(1)(1.25) 12 156(1.25)(0.216) Wt = = 689.3 lbf 6(1)(1.374)(0.5222)(1.106)

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399

Chapter 15

Pinion case (sac ) P = 341(373) + 23 620 = 150 813 psi 150 813(1) = 134 895 psi 1(1)(1.118) 1.25(3.333)(0.086) 134 895 2 t = 689.0 lbf W = 2290 1(1.374)(1.106)(0.593 75)(2)

(σc, all ) P =

Gear case (sac ) G = 341(345) + 23 620 = 141 265 psi 141 265(1.0685)(1) = 135 010 psi 1(1)(1.118) 1.25(3.333)(0.086) 135 010 2 t W = = 690.1 lbf 2290 1(1.1374)(1.106)(0.593 75)(2)

(σc, all ) G =

The equations developed within Prob. 15-7 are effective. In bevel gears, the gear tooth is weaker than the pinion so (C H ) G = 1. (See p. 784.) Thus the approximations in Prob. 15-7 with C H = 1 are really still exact. 15-10

The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also given: N P = 20 teeth, NG = 40 teeth, φn = 20◦ , F = 0.71 in, J P = 0.241, JG = 0.201, Pd = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and Q v = 5 uncrowned. Mesh d P = 20/10 = 2.000 in, vt =

π(2)(1200) πd P n P = = 628.3 ft/min 12 12

K o = 1, Eq. (15-6):

Eq. (15-5): Eq. (15-10):

dG = 40/10 = 4.000 in

S F = 1,

SH = 1

B = 0.25(12 − 5) 2/3 = 0.9148 A = 50 + 56(1 − 0.9148) = 54.77 0.9148 √ 54.77 + 628.3 = 1.412 Kv = 54.77 K s = 0.4867 + 0.2132/10 = 0.508 K mb = 1.25

Eq. (15-11): Eq. (15-15):

K m = 1.25 + 0.0036(0.71) 2 = 1.252 (K L ) P = 1.6831(109 ) −0.0323 = 0.862 (K L ) G = 1.6831(109 /2) −0.0323 = 0.881

Eq. (15-14):

(C L ) P = 3.4822(109 ) −0.0602 = 1.000 (C L ) G = 3.4822(109 /2) −0.0602 = 1.043

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Analyze for 109 pinion cycles at 0.999 reliability Eq. (15-19):

Bending Pinion: Eq. (15-23):

K R = 0.50 − 0.25 log(1 − 0.999) = 1.25 √ C R = K R = 1.25 = 1.118

(sat ) P = 44(300) + 2100 = 15 300 psi

Eq. (15-4):

(σall ) P =

Eq. (15-3):

Wt = = H1 =

15 300(0.862) = 10 551 psi 1(1)(1.25) (σall ) P F K x J P Pd K o K v K s K m 10 551(0.71)(1)(0.241) = 201 lbf 10(1)(1.412)(0.508)(1.252) 201(628.3) = 3.8 hp 33 000

Gear: (sat ) G = 15 300 psi 15 300(0.881) = 10 783 psi 1(1)(1.25)

Eq. (15-4):

(σall ) G =

Eq. (15-3):

Wt =

10 783(0.71)(1)(0.201) = 171.4 lbf 10(1)(1.412)(0.508)(1.252)

H2 =

171.4(628.3) = 3.3 hp 33 000

Wear Pinion: (C H ) G = 1,

I = 0.078, C p = 2290 psi, C xc = 2

Cs = 0.125(0.71) + 0.4375 = 0.526 25 Eq. (15-22):

(sac ) P = 341(300) + 23 620 = 125 920 psi 125 920(1)(1) = 112 630 psi 1(1)(1.118) Fd P I (σc, all ) P 2 t W = Cp K o K v K m Cs C xc 0.71(2.000)(0.078) 112 630 2 = 2290 1(1.412)(1.252)(0.526 25)(2)

(σc, all ) P = Eq. (15-1):

= 144.0 lbf H3 =

144(628.3) = 2.7 hp 33 000

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Chapter 15

401

Gear: (sac ) G = 125 920 psi 125 920(1.043)(1) = 117 473 psi 1(1)(1.118) 0.71(2.000)(0.078) 117 473 2 t W = = 156.6 lbf 2290 1(1.412)(1.252)(0.526 25)(2)

(σc, all ) =

H4 = Rating:

156.6(628.3) = 3.0 hp 33 000 H = min(3.8, 3.3, 2.7, 3.0) = 2.7 hp

Pinion wear controls the power rating. While the basis of the catalog rating is unknown, it is overly optimistic (by a factor of 1.9). 15-11

From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So ( H B ) 11 and ( H B ) 21 are 180 Brinell and the bending stress numbers are: (sat ) P = 44(180) + 2100 = 10 020 psi (sat ) G = 10 020 psi The contact strength of the gear case, based upon the equation derived in Prob. 15-7, is 2 SH Cp (sat ) P (K L ) P K x J P K T Cs C xc (sac ) G = (C L ) G C H S F N P I Ks Substituting (sat ) P from above and the values of the remaining terms from Ex. 15-1, 1.52 10 020(1)(1)(0.216)(1)(0.575)(2) 2290 = 114 331 psi 1.32(1) 1.5 25(0.065)(0.529) ( H B ) 22 =

114 331 − 23 620 = 266 Brinell 341

The pinion contact strength is found using the relation from Prob. 15-7: (sac ) P = (sac ) G m 0G.0602 C H = 114 331(1) 0.0602 (1) = 114 331 psi 114 331 − 23 600 = 266 Brinell 341 Core Case Pinion 180 266 Gear 180 266 Realization of hardnesses The response of students to this part of the question would be a function of the extent to which heat-treatment procedures were covered in their materials and manufacturing ( H B ) 12 =

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prerequisites, and how quantitative it was. The most important thing is to have the student think about it. The instructor can comment in class when students curiosity is heightened. Options that will surface may include: • Select a through-hardening steel which will meet or exceed core hardness in the hotrolled condition, then heat-treating to gain the additional 86 points of Brinell hardness by bath-quenching, then tempering, then generating the teeth in the blank. • Flame or induction hardening are possibilities. • The hardness goal for the case is sufficiently modest that carburizing and case hardening may be too costly. In this case the material selection will be different. • The initial step in a nitriding process brings the core hardness to 33–38 Rockwell C-scale (about 300–350 Brinell) which is too much. Emphasize that development procedures are necessary in order to tune the “Black Art” to the occasion. Manufacturing personnel know what to do and the direction of adjustments, but how much is obtained by asking the gear (or gear blank). Refer your students to D. W. Dudley, Gear Handbook, library reference section, for descriptions of heat-treating processes. 15-12


15-13

A design program would ask the user to make the a priori decisions, as indicated in Sec. 15-5, p. 794, MED7. The decision set can be organized as follows: A priori decisions • • • • •

Function: H, K o , rpm, mG , temp., N L , R √ Design factor: n d (S F = n d , S H = n d ) Tooth system: Involute, Straight Teeth, Crowning, φn Straddling: K mb Tooth count: N P ( NG = m G N P )

Design decisions • • • •

Pitch and Face: Pd , F Quality number: Q v Pinion hardness: ( H B ) 1 , ( H B ) 3 Gear hardness: ( H B ) 2 , ( H B ) 4

n 11

44( HB ) 11 + 2100 48( HB ) 11 + 5980

SF

σall (sat1 ) P (K L ) P = = σ s11 K T K R

n 21

44( HB ) 21 + 2100 48( HB ) 21 + 5980

(sat1 ) G (K L ) G = s21 K T K R

(sat1 ) G =

( HB ) 21

s21 SF K T K R (K L ) G

( HB ) 11

(sat1 ) P =

Note: SF = n d , S H =

Factor of safety

New tabulated strength

Chosen hardness

Associated hardness

(sat ) G =  (sat ) G − 2100   44 bhn = (s ) − 5980  at G  48

s11 SF K T K R (K L ) P

 (sat ) P − 2100   44 bhn = (s ) − 5980  at P  48

(sat ) P =

W t P Ko Kv Km Ks σ = = s21 F K x JG

W t P Ko Kv Km Ks σ = = s11 F K x JP

1/2

n 12

= s12

2

341( HB ) 12 + 23 620 363.6( HB ) 12 + 29 560 (sac1 ) P (C L ) P (C H ) P = s12 K T C R

(sac1 ) P =

( HB ) 12

 (s ) − 23 620 ac P   341 bhn = (s ) − 29 560  ac P  363.6

s12 S H K T C R (C L ) P (C H ) P

W t K o K v Cs C xc Fd P I

(sac ) P =

σc = C p

Pinion Wear

s22 S H K T C R (C L ) G (C H ) G

n 22

2

341( HB ) 22 + 23 620 363.6( HB ) 22 + 29 560 (sac1 ) G (C L ) G (C H ) G = s22 K T C R

(sac1 ) G =

( HB ) 22

 (s ) − 23 620 ac G   341 bhn = (s ) − 29 560  ac G  363.6

(sac ) G =

s22 = s12

Gear Wear

3:25 PM

Tabulated strength

Load-induced stress (Allowable stress)

Gear Bending

8/28/03

Pinion Bending

First gather all of the equations one needs, then arrange them before coding. Find the required hardnesses, express the consequences of the chosen hardnesses, and allow for revisions as appropriate.

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403

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15-14

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N W = 1, NG = 56, Pt = 8 teeth/in, d P = 1.5 in, Ho = 1hp, φn = 20◦ , ta = 70◦ F, K a = 1.25, n d = 1, Fe = 2 in, A = 850 in2 m G = NG /N W = 56,

(a)

px = π/8 = 0.3927 in,

dG = NG /Pt = 56/8 = 7.0 in C = 1.5 + 7 = 8.5 in

Eq. (15-39):

a = px /π = 0.3927/π = 0.125 in

Eq. (15-40):

h = 0.3683 px = 0.1446 in

Eq. (15-41):

h t = 0.6866 px = 0.2696 in

Eq. (15-42):

do = 1.5 + 2(0.125) = 1.75 in

Eq. (15-43):

dr = 3 − 2(0.1446) = 2.711 in

Eq. (15-44):

Dt = 7 + 2(0.125) = 7.25 in

Eq. (15-45):

Dr = 7 − 2(0.1446) = 6.711 in c = 0.1446 − 0.125 = 0.0196 in 2 7.25 2 7 − 0.125 = 2.646 in ( FW ) max = 2 − 2 2

Eq. (15-46): Eq. (15-47):

VW = π(1.5)(1725/12) = 677.4 ft/min π(7)(1725/56) = 56.45 ft/min 12 −1 0.3927 = 4.764◦ L = px N W = 0.3927 in, λ = tan π(1.5) VG =

Eq. (13-28):

Eq. (15-62):

Pn =

8 Pt = = 8.028 cos λ cos 4.764°

pn =

π = 0.3913 in Pn

π(1.5)(1725) = 679.8 ft/min 12 cos 4.764° f = 0.103 exp −0.110(679.8) 0.450 + 0.012 = 0.0250

Vs =

(b) Eq. (15-38): Eq. (13-46): e=

cos 20° − 0.0250 tan 4.764° cos φn − f tan λ = = 0.7563 Ans. cos φn + f cot λ cos 20° + 0.0250 cot 4.764° 33 000(1)(1)(1.25) 33 000 n d Ho K a = = 966 lbf Ans. VG e 56.45(0.7563) cos φn sin λ + f cos λ t = WG cos φn cos λ − f sin λ cos 20° sin 4.764° + 0.025 cos 4.764° = 106.4 lbf Ans. = 966 cos 20° cos 4.764° − 0.025 sin 4.764°

Eq. (15-58): WGt = t Eq. (15-57): WW

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Chapter 15

Eq. (15-36):

Cs = 1190 − 477 log 7.0 = 787 Cm = 0.0107 −562 + 56(56) + 5145 = 0.767

Eq. (15-37):

Cv = 0.659 exp[−0.0011(679.8)] = 0.312

(c) Eq. (15-33):

Eq. (15-38):

(W t ) all = 787(7) 0.8 (2)(0.767)(0.312) = 1787 lbf

Since WGt < (W t ) all , the mesh will survive at least 25 000 h. Eq. (15-61): W f =

0.025(966) = −29.5 lbf 0.025 sin 4.764° − cos 20° cos 4.764°

Eq. (15-63): H f =

29.5(679.8) = 0.608 hp 33 000

HW =

106.4(677.4) = 2.18 hp 33 000

HG =

966(56.45) = 1.65 hp 33 000

The mesh is sufficient Ans. Pn = Pt /cos λ = 8/cos 4.764◦ = 8.028 pn = π/8.028 = 0.3913 in σG =

966 = 39 500 psi 0.3913(0.5)(0.125)

The stress is high. At the rated horsepower, σG =

1 39 500 = 23 940 psi acceptable 1.65

(d) Eq. (15-52): Amin = 43.2(8.5) 1.7 = 1642 in2 < 1700 in2 Eq. (15-49): Hloss = 33 000(1 − 0.7563)(2.18) = 17 530 ft · lbf/min Assuming a fan exists on the worm shaft, Eq. (15-50): Eq. (15-51):

h¯ C R =

1725 + 0.13 = 0.568 ft · lbf/(min · in2 · ◦ F) 3939

ts = 70 +

17 530 = 88.2◦ F Ans. 0.568(1700)

405

406

38.2 36.2 1.47 3 30 854 0.759 0.236 492 2430 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7290 16.71

38.2 36.2 1.87 3 30 607 0.759 0.236 492 2430 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71

HW HG Hf NW NG KW Cs Cm Cv VG WGt t WW f e ( Pt ) G Pn C-to-C ts L λ σG dG

1.75 3.60 1.68 2000

1.75 3.60 2.40 2000

15-16

px dW FG A

15-15

1000 0.759 0.236 492 2430 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 8565 16.71 492 2430 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7247 16.71

38.2 36.2 1.97 3 30 125

1.75 3.60 1.69 2000

15-18

492 2430 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71

38.2 36.2 1.97 3 30 80

1.75 3.60 2.40 2000

15-19

563 2120 1038 0.0183 0.951 1.571 1.732 11.6 171 6.0 24.98 4158 19.099

38.0 36.1 1.85 3 30 50

1.75 4.10 2.25 2000

15-20

492 2524 1284 0.034A 0.913A 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.7

1.75 3.60 2.4 2500 FAN 41.2 37.7 3.59 3 30 115

15-21

492 2524 1284 0.034A 0.913A 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.71

1.75 3.60 2.4 2600 FAN 41.2 37.7 3.59 3 30 185

15-22

3:25 PM

38.2 36.2 1.97 3 30

1.75 3.60 1.43 2000

15-17

8/28/03

#1 #2 #3 #4

Parameters Selected

15-15 to 15-22 Problem statement values of 25 hp, 1125 rev/min, m G = 10, K a = 1.25, n d = 1.1, φn = 20°, ta = 70°F are not referenced in the table.

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Chapter 16 16-1 (a) θ1 = 0°,

θ2 = 120°,

θa = 90°,

0.28 pa (1.5)(6) Eq. (16-2): M f = 1

sin θa = 1,

120°

a = 5 in

sin θ(6 − 5 cos θ) dθ

0°

= 17.96 pa lbf · in pa (1.5)(6)(5) 120° 2 sin θ dθ = 56.87 pa lbf · in Eq. (16-3): M N = 1 0° c = 2(5 cos 30◦ ) = 8.66 in Eq. (16-4):

F=

56.87 pa − 17.96 pa = 4.49 pa 8.66

pa = F/4.49 = 500/4.49 = 111.4 psi for cw rotation 56.87 pa + 17.96 pa 8.66 pa = 57.9 psi for ccw rotation

Eq. (16-7): 500 =

A maximum pressure of 111.4 psioccurs on the RH shoe for cw rotation.

Ans.

(b) RH shoe: 0.28(111.4)(1.5)(6) 2 (cos 0◦ − cos 120◦ ) = 2530 lbf · in Ans. Eq. (16-6): TR = 1 LH shoe: Eq. (16-6): TL =

0.28(57.9)(1.5)(6) 2 (cos 0◦ − cos 120◦ ) = 1310 lbf · in Ans. 1

Ttotal = 2530 + 1310 = 3840 lbf · in Ans. (c)

Force vectors not to scale

Fx F 30⬚ Fy

Fx F

30⬚ Fy

y y

Primary shoe

Ry

Secondary shoe

Rx R

Rx R

x

Ry x

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Fx = 500 sin 30° = 250 lbf, Fy = 500 cos 30° = 433 lbf 2π/3 rad ◦ 1 1 2 120 θ sin θ − sin 2θ A= = 0.375, B = = 1.264 2 2 4 0 0◦

RH shoe: Eqs. (16-8): Eqs. (16-9):

111.4(1.5)(6) [0.375 − 0.28(1.264)] − 250 = −229 lbf 1 111.4(1.5)(6) Ry = [1.264 + 0.28(0.375)] − 433 = 940 lbf 1

Rx =

R = [(−229) 2 + (940) 2 ]1/2 = 967 lbf Ans. Fx = 250 lbf,

LH shoe:

Fy = 433 lbf

Eqs. (16-10): Rx =

57.9(1.5)(6) [0.375 + 0.28(1.264)] − 250 = 130 lbf 1

Ry =

57.9(1.5)(6) [1.264 − 0.28(0.375)] − 433 = 171 lbf 1

R = [(130) 2 + (171) 2 ]1/2 = 215 lbf Ans. 16-2

θ1 = 15°,

θ2 = 105°,

θa = 90°,

sin θa = 1,

a = 5 in

0.28 pa (1.5)(6) 105° sin θ(6 − 5 cos θ) dθ = 13.06 pa Eq. (16-2): M f = 1 15° pa (1.5)(6)(5) 105° 2 sin θ dθ = 46.59 pa Eq. (16-3): M N = 1 15° c = 2(5 cos 30°) = 8.66 in Eq. (16-4):

F=

46.59 pa − 13.06 pa = 3.872 pa 8.66

RH shoe: pa = 500/3.872 = 129.1 psi Eq. (16-6):

TR =

on RH shoe for cw rotation

Ans.

0.28(129.1)(1.5)(62 )(cos 15° − cos 105°) = 2391 lbf · in 1

LH shoe: 500 =

46.59 pa + 13.06 pa 8.66 TL =

⇒

pa = 72.59 psi on LH shoe for ccw rotation

Ans.

0.28(72.59)(1.5)(62 )(cos 15° − cos 105°) = 1344 lbf · in 1

Ttotal = 2391 + 1344 = 3735 lbf · in Ans. Comparing this result with that of Prob. 16-1, a 2.7% reduction in torque is achieved by using 25% less braking material.

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409

Chapter 16

16-3

Given: θ1 = 0°, θ2 = 120°, θa = 90°, sin θa = 1, a = R = 90 mm, F = 1000 N = 1 kN, r = 280/2 = 140 mm, counter-clockwise rotation.

f = 0.30,

LH shoe: f pa br a 2 Mf = r(1 − cos θ2 ) − sin θ2 sin θa 2 0.090 2 0.30 pa (0.030)(0.140) ◦ 0.140(1 − cos 120 ) − sin 120° = 1 2 = 0.000 222 pa N · m pa bra θ2 1 − sin 2θ2 MN = sin θa 2 4 pa (0.030)(0.140)(0.090) 120° π 1 = − sin 2(120°) 1 2 180 4 = 4.777(10−4 ) pa N · m 180◦ − θ2 c = 2r cos = 2(0.090) cos 30◦ = 0.155 88 m 2 4.777(10−4 ) − 2.22(10−4 ) F = 1 = pa = 1.64(10−3 ) pa 0.155 88 pa = 1/1.64(10−3 ) = 610 kPa TL = =

f pa br 2 (cos θ1 − cos θ2 ) sin θa 0.30(610)(103 )(0.030)(0.1402 ) [1 − (−0.5)] 1

= 161.4 N · m Ans. RH shoe:

M f = 2.22(10−4 ) pa N · m M N = 4.77(10−4 ) pa N · m c = 0.155 88 m 4.77(10−4 ) + 2.22(10−4 ) = 4.49(10−3 ) pa F = 1 = pa 0.155 88 pa =

1 = 222.8 kPa Ans. 4.49(10−3 )

TR = (222.8/610)(161.4) = 59.0 N · m Ans.

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16-4 (a) Given: θ1 = 10°, θ2 = 75°, θa = 75°, pa = 106 Pa, f = 0.24, b = 0.075 m (shoe width), a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m. Some of the terms needed are evaluated as: θ2 θ2 θ2 1 2 θ2 sin θ A= r sin θ dθ − a sin θ cos θ dθ = r −cos θ θ1 − a 2 θ1 θ1 θ1 75° 75° 1 2 sin θ = 200 −cos θ 10° − 150 = 77.5 mm 2 10° 75π/180 rad θ2 1 θ 2 − sin 2θ B= sin θ dθ = = 0.528 2 4 θ1 10π/180 rad θ2 sin θ cos θ dθ = 0.4514 C= θ1

Now converting to pascals and meters, we have from Eq. (16-2), Mf =

f pa br 0.24[(10) 6 ](0.075)(0.200) (0.0775) = 289 N · m A= sin θa sin 75°

From Eq. (16-3), MN =

pa bra [(10) 6 ](0.075)(0.200)(0.150) (0.528) = 1230 N · m B= sin θa sin 75°

Finally, using Eq. (16-4), we have MN − M f 1230 − 289 = = 5.70 kN Ans. c 165 (b) Use Eq. (16-6) for the primary shoe. F=

T = =

f pa br 2 (cos θ1 − cos θ2 ) sin θa 0.24[(10) 6 ](0.075)(0.200) 2 (cos 10° − cos 75°) = 541 N · m sin 75°

For the secondary shoe, we must first find pa . Substituting 1230 289 MN = p and M = pa into Eq. (16-7), a f 106 106 5.70 =

(1230/106 ) pa + (289/106 ) pa , 165

solving gives

pa = 619(10) 3 Pa

Then 0.24[0.619(10) 6 ](0.075)(0.200) 2 (cos 10° − cos 75°) = 335 N · m sin 75° so the braking capacity is Ttotal = 2(541) + 2(335) = 1750 N · m Ans. T =

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Chapter 16

(c) Primary shoes: Rx =

pa br (C − f B) − Fx sin θa

(106 )(0.075)(0.200) [0.4514 − 0.24(0.528)](10) −3 − 5.70 = −0.658 kN sin 75° pa br ( B + f C) − Fy Ry = sin θa =

=

(106 )(0.075)(0.200) [0.528 + 0.24(0.4514)](10) −3 − 0 = 9.88 kN sin 75°

Secondary shoes: Rx =

pa br (C + f B) − Fx sin θa

[0.619(10) 6 ](0.075)(0.200) [0.4514 + 0.24(0.528)](10) −3 − 5.70 sin 75° = −0.143 kN =

Ry =

pa br ( B − f C) − Fy sin θa

[0.619(10) 6 ](0.075)(0.200) [0.528 − 0.24(0.4514)](10) −3 − 0 sin 75° = 4.03 kN =

Note from figure that +y for secondary shoe is opposite to +y for primary shoe.

y R RV

Combining horizontal and vertical components, R H = −0.658 − 0.143 = −0.801 kN RV = 9.88 − 4.03 = 5.85 kN R = (0.801) 2 + (5.85) 2 = 5.90 kN Ans. 16-5

Preliminaries: θ1 = 45° − tan−1 (150/200) = 8.13°, θa = 90°, Eq. (16-8): Let

θ2 = 98.13°

a = [(150) 2 + (200) 2 ]1/2 = 250 mm

1 2 98.13° sin θ 8.13° = 0.480 2 θ2

98.13° sin θ dθ = − cos θ 8.13° = 1.1314 C= A=

θ1

RH

y

x x

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Eq. (16-2): Mf =

f pa br 0.25 pa (0.030)(0.150) [0.15(1.1314) − 0.25(0.48)] (rC − a A) = sin θa sin 90°

= 5.59(10−5 ) pa N · m 98.13π/180 rad 1 θ − sin 2θ B= = 0.925 Eq. (16-8): 2 4 8.13π/180 rad Eq. (16-3):

pa bra pa (0.030)(0.150)(0.250) (0.925) B= sin θa 1 = 1.0406(10−3 ) pa N · m

MN =

Using F = ( M N − M f )/c, we obtain 104.06 − 5.59 pa 0.5(105 )

400 =

or

pa = 203 kPa Ans.

0.25(203)(103 )(0.030)(0.150) 2 f pa br 2 C (1.1314) = sin θa 1 = 38.76 N · m Ans.

T =

16-6

For +3σˆ f :

f = f¯ + 3σˆ f = 0.25 + 3(0.025) = 0.325 0.325 −5 = 7.267(10−5 ) pa M f = 5.59(10 ) pa 0.25

Eq. (16-4): 400 =

104.06 − 7.267 pa 105 (0.500)

pa = 207 kPa 0.325 207 = 51.4 N · m Ans. T = 38.75 203 0.25 Similarly, for −3σˆ f : f = f¯ − 3σˆ f = 0.25 − 3(0.025) = 0.175 M f = 3.913(10−5 ) pa pa = 200 kPa T = 26.7 N · m Ans. 16-7

Preliminaries: θ2 = 180° − 30° − tan−1 (3/12) = 136°, θ1 = 20° − tan−1 (3/12) = 6°, θa = 90◦ , a = [(3) 2 + (12) 2 ]1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in. ◦ 0.30(150)(2)(10) 136 Mf = sin θ(10 − 12.37 cos θ) dθ Eq. (16-2): sin 90° 6° = 12 800 lbf · in

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Chapter 16

Eq. (16-3):

150(2)(10)(12.37) MN = sin 90°

136°

sin2 θ dθ = 53 300 lbf · in

6°

LH shoe: c L = 12 + 12 + 4 = 28 in Now note that M f is cw and M N is ccw. Thus, FL =

53 300 − 12 800 = 1446 lbf 28 FL ⫽ 1446 lbf

FR ⫽

4"

lbf 1491 14⬚

Fact ⫽ 361 lbf

16"

Eq. (16-6):

TL =

0.30(150)(2)(10) 2 (cos 6° − cos 136°) = 15 420 lbf · in sin 90°

RH shoe:

pa = 355.3 pa , M N = 53 300 150 On this shoe, both M N and M f are ccw. Also

pa = 85.3 pa M f = 12 800 150

c R = (24 − 2 tan 14°) cos 14° = 22.8 in Fact = FL sin 14° = 361 lbf Ans. FR = FL / cos 14° = 1491 lbf

Thus

1491 =

Then

TR = Ttotal

355.3 + 85.3 pa 22.8

⇒

pa = 77.2 psi

0.30(77.2)(2)(10) 2 (cos 6° − cos 136°) = 7940 lbf · in sin 90° = 15 420 + 7940 = 23 400 lbf · in Ans.

16-8 Mf = 2

θ2

( f d N )(a cos θ − r)

0

= 2 f pbr

θ2

where d N = pbr dθ

(a cos θ − r) dθ = 0

0

From which a

0

a =

θ2

cos θ dθ = r

θ2

dθ 0

rθ2 r(60°)(π/180) = 1.209r = sin θ2 sin 60°

Eq. (16-15) a=

4r sin 60° = 1.170r 2(60)(π/180) + sin[2(60)]

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16-9 (a) Counter-clockwise rotation, θ2 = π/4 rad, a=

r = 13.5/2 = 6.75 in

4r sin θ2 4(6.75) sin(π/4) = 7.426 in = 2θ2 + sin 2θ2 2π/4 + sin(2π/4)

e = 2(7.426) = 14.85 in Ans. (b)

Fy F

x

3"

Rx

P Actuation lever

6.375"

Ry

0.428P tie rod

2.125P

2.125P ␣

α = tan−1 (3/14.85) = 11.4° M R = 0 = 3F x − 6.375P

0.428P

F x = 2.125P Fx = 0 = −F x + R x R x = F x = 2.125P

0.428P 2.125P

F y = F x tan 11.4◦ = 0.428P

P

2.125P

1.428P

Fy = −P − F y + R y R y = P + 0.428P = 1.428P

Fy F

x

Left shoe lever. M R = 0 = 7.78S x − 15.28F x

15.28"

Sx =

Sx Sy

S y = f S x = 0.30(4.174P)

7.78" Rx Ry

15.28 (2.125P) = 4.174P 7.78

= 1.252P Fy = 0 = R y + S y + F y R y = −F y − S y = −0.428P − 1.252P

= −1.68P Fx = 0 = R x − S x + F x Rx = Sx − F x = 4.174P − 2.125P = 2.049P

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415

Chapter 16 1.428P

0.428P 2.125P 2.125P 1.252P 4.174P

Ans.

4.174P 1.252P 2.049P

2.049P

2.68P

1.68P

Right shoe lever

Left shoe lever

(c) The direction of brake pulley rotation affects the sense of S y , which has no effect on the brake shoe lever moment and hence, no effect on S x or the brake torque. The brake shoe levers carry identical bending moments but the left lever carries a tension while the right carries compression (column loading). The right lever is designed and used as a left lever, producing interchangeable levers (identical levers). But do not infer from these identical loadings. 16-10

r = 13.5/2 = 6.75 in,

b = 7.5 in,

θ2 = 45°

From Table 16-3 for a rigid, molded nonasbestos use a conservative estimate of pa = 100 psi, f = 0.31. In Eq. (16-16): 2θ2 + sin 2θ2 = 2(π/4) + sin 2(45°) = 2.571 From Prob. 16-9 solution, N = S x = 4.174P = P=

pa br (2.571) = 1.285 pa br 2

1.285 (100)(7.5)(6.75) = 1560 lbf 4.174

Ans.

Applying Eq. (16-18) for two shoes, T = 2a f N = 2(7.426)(0.31)(4.174)(1560) = 29 980 lbf · in Ans. 16-11

From Eq. (16-22), P1 =

90(4)(14) pa bD = = 2520 lbf Ans. 2 2

f φ = 0.25(π)(270°/180°) = 1.178 Eq. (16-19): P2 = P1 exp(− f φ) = 2520 exp(−1.178) = 776 lbf Ans. T =

( P1 − P2 ) D (2520 − 776)14 = = 12 200 lbf · in Ans. 2 2

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Given: D = 300 mm,

f = 0.28,

b = 80 mm,

φ = 270°,

P1 = 7600 N.

f φ = 0.28(π)(270◦ /180◦ ) = 1.319 P2 = P1 exp(− f φ) = 7600 exp(−1.319) = 2032 N pa =

2(7600) 2P1 = = 0.6333 N/mm2 bD 80(300)

T = ( P1 − P2 )

633 kPa

Ans.

300 D = (7600 − 2032) 2 2

= 835 200 N · mm 16-13

or

835.2 N · m

or

Ans.

125

␣ 200

P1 F

P2 P1

−1

α = cos

125 200

P2

125

275

= 51.32°

φ = 270° − 51.32° = 218.7° f φ = 0.30(218.7) P2 =

π = 1.145 180°

(125 + 275)400 (125 + 275) F = = 1280 N 125 125

P1 = P2 exp( f φ) = 1280 exp(1.145) = 4022 N T = ( P1 − P2 )

250 D = (4022 − 1280) 2 2

= 342 750 N · mm

or

343 N · m Ans.

16-14 D = 16" ,

(a)

b = 3"

n = 200 rev/min P1

f = 0.20,

P2 P2

Eq. (16-22):

P1

pa = 70 psi

P

P1 =

70(3)(16) pa bD = = 1680 lbf 2 2

f φ = 0.20(3π/2) = 0.942

Ans.

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Chapter 16

417

P2 = P1 exp(− f φ) = 1680 exp(−0.942) = 655 lbf

Eq. (16-14):

T = ( P1 − P2 )

16 D = (1680 − 655) 2 2

= 8200 lbf · in Ans. H=

8200(200) Tn = = 26.0 hp Ans. 63 025 63 025

P=

3(1680) 3P1 = = 504 lbf Ans. 10 10

(b)

Force of belt on the drum: R = (16802 + 6552 ) 1/2 = 1803 lbf 1680 lbf 655 lbf

13,440 lbf • in

655 lbf

1680 lbf 5240 lbf • in

Force of shaft on the drum: 1680 and 655 lbf TP1 = 1680(8) = 13 440 lbf · in TP2 = 655(8) = 5240 lbf · in

1803 lbf

Net torque on drum due to brake band: T = TP1 − TP2 = 13 440 − 5240 = 8200 lbf · in

1803 lbf

1680 lbf 655 lbf 8200 lbf ¥ in

The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with the drum at center span, the bearing radial load is 1803/2 = 901 lbf. (c) Eq. (16-22):

2P bD 2(1680) 2P1 = = 70 psi Ans. = 3(16) 3(16)

p= p|θ=0° As it should be

p|θ=270° = 16-15

2(655) 2P2 = = 27.3 psi Ans. 3(16) 3(16)

Given: φ = 270°, b = 2.125 in, f = 0.20, T = 150 lbf · ft, D = 8.25 in, c2 = 2.25 in Notice that the pivoting rocker is not located on the vertical centerline of the drum. (a) To have the band tighten for ccw rotation, it is necessary to have c1 < c2 . When friction is fully developed, P1 = exp( f φ) = exp[0.2(3π/2)] = 2.566 P2

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If friction is not fully developed P1 /P2 ≤ exp( f φ) To help visualize what is going on let’s add a force W parallel to P1 , at a lever arm of c3 . Now sum moments about the rocker pivot. M = 0 = c3 W + c1 P1 − c2 P2 From which W =

c2 P2 − c1 P1 c3

The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0. It follows from the equation above c2 P1 ≥ P2 c1 When friction is fully developed 2.566 = 2.25/c1 2.25 = 0.877 in c1 = 2.566 When P1 /P2 is less than 2.566, friction is not fully developed. Suppose P1 /P2 = 2.25, then 2.25 = 1 in c1 = 2.25 We don’t want to be at the point of slip, and we need the band to tighten. c2 ≤ c1 ≤ c2 P1 /P2 When the developed friction is very small, P1 /P2 → 1 and c1 → c2 (b) Rocker has c1 = 1 in P1 c2 2.25 = 2.25 = = P2 c1 1 ln 2.25 ln( P1 /P2 ) = = 0.172 φ 3π/2 Friction is not fully developed, no slip. D P1 D T = ( P1 − P2 ) = P2 −1 2 P2 2 Solve for P2 f =

P2 =

2(150)(12) 2T = = 349 lbf [( P1 /P2 ) − 1]D (2.25 − 1)(8.25)

P1 = 2.25P2 = 2.25(349) = 785 lbf p=

2(785) 2P1 = = 89.6 psi Ans. bD 2.125(8.25)

Ans.

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419

Chapter 16

(c) The torque ratio is 150(12)/100 or 18-fold. P2 =

349 = 19.4 lbf 18

P1 = 2.25P2 = 2.25(19.4) = 43.6 lbf p=

89.6 = 4.98 psi Ans. 18

Comment: As the torque opposed by the locked brake increases, P2 and P1 increase (although ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be provided by a shear key. 16-16 F=

(a) From Eq. (16-23), since

π pa d ( D − d) 2

then pa =

2F πd( D − d)

and it follows that pa =

2(5000) π(225)(300 − 225)

= 0.189 N/mm2 T =

or 189 000 N/m2

or 189 kPa Ans.

5000(0.25) Ff ( D + d) = (300 + 225) 4 4

= 164 043 N · mm or 164 N · m Ans. (b) From Eq. (16-26), π pa 2 ( D − d 2) 4 4(5000) 4F pa = = 2 2 π( D − d ) π(3002 − 2252 ) F=

= 0.162 N/mm2 = 162 kPa Ans. From Eq. (16-27), π π T = f pa ( D 3 − d 3 ) = (0.25)(162)(103 )(3003 − 2253 )(10−3 ) 3 12 12 = 166 N · m Ans. 16-17 (a) Eq. (16-23): F=

π pa d π(120)(4) ( D − d) = (6.5 − 4) = 1885 lbf 2 2

Ans.

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Eq. (16-24): π f pa d 2 π(0.24)(120)(4) ( D − d 2) N = (6.52 − 42 )(6) 8 8 = 7125 lbf · in Ans.

T =

T =

(b)

π(0.24)(120d) (6.52 − d 2 )(6) 8 d, in

T, lbf · in

2 3 4 5 6

5191 6769 7125 5853 2545

Ans.

(c) The torque-diameter curve exhibits a stationary point maximum in the range of diameter d. The clutch has nearly optimal proportions. 16-18 (a) T =

π f pa d( D 2 − d 2 ) N = C D 2 d − Cd 3 8

Differentiating with respect to d and equating to zero gives dT = C D 2 − 3Cd 2 = 0 dd D d* = √ 3

Ans.

d2T = −6 Cd dd 2 which is negative for all positive d. We have a stationary point maximum. (b)

6.5 d* = √ = 3.75 in Ans. 3 √

π(0.24)(120) 6.5/ 3 [6.52 − (6.52 /3)](6) = 7173 lbf · in T* = 8

(c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in (d) Consider:

0.45 ≤

d ≤ 0.80 D

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421

Chapter 16

Multiply through by D 0.45D ≤ d ≤ 0.80D 0.45(6.5) ≤ d ≤ 0.80(6.5) 2.925 ≤ d ≤ 5.2 in ∗ 1 d = d ∗ /D = √ = 0.577 D 3 which lies within the common range of clutches. Yes. 16-19

Ans.

Given: d = 0.306 m,

l = 0.060 m,

T = 0.200 kN · m,

12

␣ 60

−1

α = tan

165

153 Not to scale

12 60

D = 0.330 m,

f = 0.26.

= 11.31°

CL

Uniform wear Eq. (16-45): π(0.26)(0.306) pa (0.3302 − 0.3062 ) = 0.002 432 pa 8 sin 11.31° 0.200 = 82.2 kPa Ans. pa = 0.002 432

0.200 =

Eq. (16-44): π(82.2)(0.306) π pa d ( D − d) = (0.330 − 0.306) = 0.949 kN 2 2 Uniform pressure Eq. (16-48): F=

0.200 = pa =

Ans.

π(0.26) pa (0.3303 − 0.3063 ) = 0.002 53 pa 12 sin 11.31° 0.200 = 79.1 kPa 0.002 53

Ans.

Eq. (16-47): F= 16-20

π pa 2 π(79.1) ( D − d 2) = (0.3302 − 0.3062 ) = 0.948 kN 4 4

Uniform wear Eq. (16-34):

1 T = (θ2 − θ1 ) f pa ri ro2 − ri2 2

Ans.

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Eq. (16-33): Thus,

F = (θ2 − θ1 ) pa ri (ro − ri )

(1/2)(θ2 − θ1 ) f pa ri ro2 − ri2 T = f FD f (θ2 − θ1 ) pa ri (ro − ri )( D) D/2 + d/2 1 d r o + ri = = 1+ = 2D 2D 4 D

O.K.

Ans.

Uniform pressure

1 T = (θ2 − θ1 ) f pa ro3 − ri3 3

Eq. (16-38): Eq. (16-37):

1 F = (θ2 − θ1 ) pa ro2 − ri2 2

(1/3)(θ2 − θ1 ) f pa ro3 − ri3 T 2 ( D/2) 3 − (d/2) 3

= = f FD 3 [( D/2) 2 − (d/2) 2 D] (1/2) f (θ2 − θ1 ) pa ro2 − ri2 D 1 1 − (d/D) 3 2( D/2) 3 (1 − (d/D) 3 ) = = 3( D/2) 2 [1 − (d/D) 2 ]D 3 1 − (d/D) 2

O.K. Ans.

ω = 2πn/60 = 2π 500/60 = 52.4 rad/s

16-21

H 2(103 ) T = = = 38.2 N · m ω 52.4 Key: F=

38.2 T = = 3.18 kN r 12

Average shear stress in key is τ=

3.18(103 ) = 13.2 MPa 6(40)

Ans.

Average bearing stress is F 3.18(103 ) = −26.5 MPa σb = − =− Ab 3(40) Let one jaw carry the entire load. rav

1 = 2

F=

26 45 + 2 2

= 17.75 mm

T 38.2 = 2.15 kN = rav 17.75

Ans.

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423

Chapter 16

The bearing and shear stress estimates are σb = τ=

−2.15(103 ) = −22.6 MPa 10(22.5 − 13)

Ans.

2.15(103 ) = 0.869 MPa 10[0.25π(17.75) 2 ]

Ans.

ω1 = 2πn/60 = 2π(1800)/60 = 188.5 rad/s ω2 = 0

16-22 From Eq. (16-51),

T t1 320(8.3) I1 I2 = 14.09 N · m · s2 = = I1 + I2 ω1 − ω2 188.5 − 0 Eq. (16-52):

188.52 E = 14.09 (10−3 ) = 250 kJ 2

Eq. (16-55): T = 16-23 n= Cs =

250(103 ) E = = 27.8◦ C Cpm 500(18)

Ans.

260 + 240 n1 + n2 = = 250 rev/min 2 2 260 − 240 = 0.08 250

Ans.

ω = 2π(250)/60 = 26.18 rad/s I =

5000(12) E2 − E1 = = 1094 lbf · in · s2 2 Cs ω 0.08(26.18) 2

Ix =

W 2

m 2 do + di2 = do + di2 8 8g

W =

8g I 8(386)(1094) = = 502 lbf 602 + 562 do2 + di2

w = 0.260 lbf/in3

for cast iron

502 W = = 1931 in3 w 0.260

πt 2

πt 2 do − di2 = 60 − 562 = 364t in3 V = 4 4 V =

Also,

Equating the expressions for volume and solving for t, 1931 = 5.3 in Ans. t= 364

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(a) The useful work performed in one revolution of the crank shaft is U = 35(2000)(8)(0.15) = 84(103 ) in · lbf Accounting for friction, the total work done in one revolution is U = 84(103 )/(1 − 0.16) = 100(103 ) in · lbf Since 15% of the crank shaft stroke is 7.5% of a crank shaft revolution, the energy fluctuation is E 2 − E 1 = 84(103 ) − 100(103 )(0.075) = 76.5(103 ) in · lbf

Ans.

(b) For the flywheel n = 6(90) = 540 rev/min ω=

2πn 2π(540) = = 56.5 rad/s 60 60

Cs = 0.10

Since

I =

E2 − E1 76.5(103 ) = = 239.6 lbf · in · s2 2 2 Cs ω 0.10(56.5)

Assuming all the mass is concentrated at the effective diameter, d, I = W= 16-25

md 2 4 4g I 4(386)(239.6) = = 161 lbf Ans. 2 d 482

Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine. Cs = 0.30 n = 2400 rev/min or 251 rad/s Tm =

3(3368) = 804 in · lbf 4π

Ans.

E 2 − E 1 = 3(3531) = 10 590 in · lbf I = 16-26

E2 − E1 10 590 = 0.560 in · lbf · s2 = 2 2 Cs ω 0.30(251 )

Ans.

(a) (1)

F21

rG

rP

(T2 ) 1 = −F21r P = −

F12 T1

T2

T2 T2 rP = rG −n

Ans.

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425

Chapter 16

(2)

Equivalent energy

rG

rP

(1/2) I2 ω22 = (1/2)( I2 ) 1 w12 ( I2 ) 1 =

IL

IG = (3) IP

rG rP

2

mG mP

=

rG rP

2

(b) Ie = I M + I P + n 2 I P +

IL n2

(c) Ie = 10 + 1 + 102 (1) +

100 102

Ans.

2 = n4

IG n4 I P = = n2 I P 2 2 n n

( I2 ) 1 =

From (2)

rG rP

ω22 I2 I = 2 2 2 n ω1

Ans.

Ans.

= 10 + 1 + 100 + 1 = 112 reflected load inertia reflected gear inertia pinion inertia armature inertia 16-27

Ans.

(a) Reflect I L , IG 2 to the center shaft n IG1

IP

IP ⫹ m2IP ⫹

IM

IL m2

Reflect the center shaft to the motor shaft IP

IM ⫹ n2IP ⫹

Ie = I M + I P + n 2 I P +

IP ⫹ m2IP ⫹ IL 兾m2 n2

IP m2 IL + IP + 2 2 2 2 n n m n

Ans.

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(b) For R = constant = nm, (c) For R = 10,

Ie = I M

IP R2 I P IL + IP + n IP + 2 + 4 + 2 n n R 2

Ans.

2(1) 4(102 )(1) ∂ Ie = 0 + 0 + 2n(1) − 3 − +0=0 ∂n n n5 n 6 − n 2 − 200 = 0

From which n* = 2.430 m* =

Ans.

10 = 4.115 2.430

Ans.

Notice that n*and m* are independent of I L . 16-28

From Prob. 16-27, Ie = I M

IP R2 I P IL + IP + n IP + 2 + 4 + 2 n n R 2

= 10 + 1 + n 2 (1) + = 10 + 1 + n 2 + n 1.00 1.50 2.00 2.43 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00

1 100(1) 100 + + 2 n2 n4 10

1 100 + +1 n2 n4

Ie 114.00 34.40 22.50 20.90 22.30 28.50 37.20 48.10 61.10 76.00 93.00 112.02

Ie 100

20.9 0 1 2

4

6

8

10 n

2.43

Optimizing the partitioning of a double reduction lowered the gear-train inertia to 20.9/112 = 0.187, or to 19% of that of a single reduction. This includes the two additional gears. 16-29

Figure 16-29 applies, t2 = 10 s,

t1 = 0.5 s

10 − 0.5 t2 − t1 = 19 = t1 0.5

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Chapter 16

The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is 1300(12) = 1560 lbf · in TL = 10 The rated motor torque Tr is Tr =

63 025(3) = 168.07 lbf · in 1125

For Eqs. (16-65): 2π (1125) = 117.81 rad/s 60 2π ωs = (1200) = 125.66 rad/s 60

ωr =

a=

−Tr 168.07 = −21.41 =− ωs − ωr 125.66 − 117.81

b=

Tr ωs 168.07(125.66) = ωs − ωr 125.66 − 117.81

= 2690.4 lbf · in The linear portion of the squirrel-cage motor characteristic can now be expressed as TM = −21.41ω + 2690.4 lbf · in Eq. (16-68):

1560 − 168.07 T2 = 168.07 1560 − T2

19

One root is 168.07 which is for infinite time. The root for 10 s is wanted. Use a successive substitution method T2

New T2

0.00 19.30 24.40 26.00 26.50

19.30 24.40 26.00 26.50 26.67

Continue until convergence. T2 = 26.771 Eq. (16-69): I =

−21.41(10 − 0.5) = 110.72 in · lbf · s/rad ln(26.771/168.07)

ω=

T −b a

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ωmax =

26.771 − 2690.4 T2 − b = = 124.41 rad/s Ans. a −21.41

ωmin = 117.81 rad/s Ans. 124.41 + 117.81 = 121.11 rad/s 2 124.41 − 117.81 ωmax − ωmin Cs = = = 0.0545 Ans. (ωmax + ωmin )/2 (124.41 + 117.81)/2 ω¯ =

E1 =

1 2 1 I ωr = (110.72)(117.81) 2 = 768 352 in · lbf 2 2

E2 =

1 2 1 I ω2 = (110.72)(124.41) 2 = 856 854 in · lbf 2 2

E = E 1 − E 2 = 768 352 − 856 854 = −88 502 in · lbf Eq. (16-64): E = Cs I ω¯ 2 = 0.0545(110.72)(121.11) 2 = 88 508 in · lbf,

close enough

Ans.

During the punch T =

63 025H n

H=

1560(121.11)(60/2π) ¯ TL ω(60/2π) = = 28.6 hp 63 025 63 025

The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on the motor shaft. From Table A-18, I =

W 2

m 2 do + di2 = do + di2 8 8g

W =

8g I 8(386)(110.72) = do2 + di2 do2 + di2

If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in di = 30 − (4/2) = 28 in do = 30 + (4/2) = 32 in W =

8(386)(110.72) = 189.1 lbf 322 + 282

Rim volume V is given by V =

πl πl 2 do − di2 = (322 − 282 ) = 188.5l 4 4

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Chapter 16

429

where l is the rim width as shown in Table A-18. The specific weight of cast iron is γ = 0.260 lbf · in3 , therefore the volume of cast iron is V =

189.1 W = = 727.3 in3 γ 0.260

Thus 188.5 l = 727.3 l=

727.3 = 3.86 in wide 188.5

Proportions can be varied. 16-30

Prob. 16-29 solution has I for the motor shaft flywheel as I = 110.72 in · lbf · s2 /rad A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2) I = 102 (110.72) = 11 072 in · lbf · s2 /rad A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under shock conditions. Stating the problem is most of the solution. Satisfy yourself that on the crankshaft: TL = 1300(12) = 15 600 lbf · in Tr = 10(168.07) = 1680.7 lbf · in ωr = 117.81/10 = 11.781 rad/s ωs = 125.66/10 = 12.566 rad/s a = −21.41(100) = −2141 b = 2690.35(10) = 26903.5 TM = −2141ωc + 26 903.5 lbf · in 15 600 − 1680.5 19 T2 = 1680.6 15 600 − T2 The root is 10(26.67) = 266.7 lbf · in ω¯ = 121.11/10 = 12.111 rad/s Cs = 0.0549 (same) ωmax = 121.11/10 = 12.111 rad/s Ans. ωmin = 117.81/10 = 11.781 rad/s Ans. E 1 , E 2 , E and peak power are the same. From Table A-18 W =

8(386)(11 072) 8g I = 2 do2 + di2 o + di

d2

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Scaling will affect do and di , but the gear ratio changed I. Scale up the flywheel in the Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in. d¯ = 30(2.5) = 75 in do = 75 + (10/2) = 80 in di = 75 − (10/2) = 70 in W =

8(386)(11 072) = 3026 lbf 802 + 702

v=

3026 = 11 638 in3 0.26

V =

π l(802 − 702 ) = 1178 l 4

l=

11 638 = 9.88 in 1178

Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the motor armature has its inertia magnified 100-fold, and during the punch there are deceleration stresses in the train. With no motor armature information, we cannot comment. 16-31

This can be the basis for a class discussion.

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Chapter 17 17-1

Preliminaries to give some checkpoints: VR = Hnom = 2 hp,

1750 rev/min,

1 1 = =2 Angular velocity ratio 0.5 C = 9(12) = 108 in, K s = 1.25,

nd = 1

dmin = 1 in, Fa = 35 lbf/in, γ = 0.035 lbf/in3 , f = 0.50, b = 6 in, d = 2 in, from Table 17-2 for F-1 Polyamide: t = 0.05 in; from Table 17-4, C p = 0.70. w = 12γ bt = 12(0.035)(6)(0.05) = 0.126 lbf/ft θd = 3.123 rad, exp( f θ) = 4.766 (perhaps) π(2)(1750) πdn = = 916.3 ft/min 12 12 2 V w 0.126 916.3 2 = = 0.913 lbf Ans. (a) Eq. (e): Fc = 32.174 60 32.174 60 V =

63 025(2)(1.25)(1) = 90.0 lbf · in 1750 2T 2(90) F = = = 90 lbf d 2 T =

Eq. (17-12): ( F1 ) a = bFa C p Cv = 6(35)(0.70)(1) = 147 lbf Ans. F2 = F1a − F = 147 − 90 = 57 lbf Ans. Do not use Eq. (17-9) because we do not yet know f . Eq. (i)

F1a + F2 147 + 57 − Fc = − 0.913 = 101.1 lbf Ans. 2 2 147 − 0.913 1 1 ( F1 ) a − Fc ln = 0.307 = ln f = θd F2 − Fc 3.123 57 − 0.913

Fi =

The friction is thus undeveloped. (b) The transmitted horsepower is, 90(916.3) (F)V = = 2.5 hp Ans. 33 000 33 000 H 2.5 =1 = = Hnom K s 2(1.25)

H= nfs

L = 225.3 in Ans. 3C 2 w dip = (c) From Eq. (17-13), 2Fi where C is the center-to-center distance in feet. 3(108/12) 2 (0.126) = 0.151 in Ans. dip = 2(101.1) From Eq. (17-2),

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Comment: The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will increase f . The limit of narrowing is bmin = 4.680 in, whence w = 0.0983 lbf/ft Fc T F Fi

( F1 ) a = 114.7 lbf F2 = 24.6 lbf f = f = 0.50 dip = 0.173 in

= 0.712 lbf = 90 lbf · in (same) = ( F1 ) a − F2 = 90 lbf = 68.9 lbf

Longer life can be obtained with a 6-inch wide belt by reducing Fi to attain f = 0.50. Prob. 17-8 develops an equation we can use here Fi =

(F + Fc ) exp( f θ) − Fc exp( f θ) − 1

F2 = F1 − F F1 + F2 − Fc 2 1 F1 − Fc f = ln θd F2 − Fc

Fi =

dip =

3(C D/12) 2 w 2Fi

which in this case gives F1 = 114.9 lbf F2 = 24.8 lbf Fi = 68.9 lbf

Fc = 0.913 lbf f = 0.50 dip = 0.222 in

So, reducing Fi from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50, with a corresponding dip of 0.222 in. Having reduced F1 and F2 , the endurance of the belt is improved. Power, service factor and design factor have remained in tack. 17-2

There are practical limitations on doubling the iconic scale. We can double pulley diameters and the center-to-center distance. With the belt we could: • Use the same A-3 belt and double its width; • Change the belt to A-5 which has a thickness 0.25 in rather than 2(0.13) = 0.26 in, and an increased Fa ; • Double the thickness and double tabulated Fa which is based on table thickness. The object of the problem is to reveal where the non-proportionalities occur and the nature of scaling a flat belt drive. We will utilize the third alternative, choosing an A-3 polyamide belt of double thickness, assuming it is available. We will also remember to double the tabulated Fa from 100 lbf/in to 200 lbf/in.

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433

Chapter 17

In assigning this problem, you could outline (or solicit) the three alternatives just mentioned and assign the one of your choice–alternative 3: Ex. 17-2: b = 10 in, d = 16 in, D = 32 in, Polyamide A-3, t = 0.13 in, γ = 0.042, Fa = 100 lbf/in, C p = 0.94, Cv = 1, f = 0.8 63 025(60)(1.15)(1.05) = 5313 lbf · in 860 w = 12 γ bt = 12(0.042)(10)(0.13) = 0.655 lbf/ft V = πdn/12 = π(16)(860/12) = 3602 ft/min θd = 3.037 rad T =

Doubled: b = 20 in, d = 32 in, D = 72 in, Polyamide A-3, t = 0.26 in, γ = 0.042, Fa = 2(100) = 200 lbf/in, C p = 1, Cv = 1, f = 0.8 T = 4(5313) = 21 252 lbf · in w = 12(0.042)(20)(0.26) = 2.62 lbf/ft V = π(32)(860/12) = 7205 ft/min θ = 3.037 rad For fully-developed friction:

For fully-developed friction:

exp( f θd ) = exp[0.8(3.037)] = 11.35

exp( f θd ) = [0.8(3.037)] = 11.35

0.262(7205/60) 2 wV 2 = = 1174.3 lbf Fc = g 32.174

0.655(3602/60) 2 wV 2 = = 73.4 lbf Fc = g 32.174 ( F1 ) a = F1 = bFa C p Cv = 10(100)(0.94)(1) = 940 lbf F = 2T /D = 2(5313)/(16) = 664 lbf F2 = F1 − F = 940 − 664 = 276 lbf F1 + F2 − Fc 2 940 + 276 − 73.4 = 535 lbf = 2 Transmitted power H (or Ha ) : Fi =

F(V ) 664(3602) = = 72.5 hp 33 000 33 000 1 F1 − Fc f = ln θd F2 − Fc 940 − 73.4 1 ln = 3.037 276 − 73.4 = 0.479 undeveloped

H=

Note, in this as well as in the double-size case, exp( f θd ) is not used. It will show up if we relax Fi (and change other parameters to transmit the required power), in order to bring f up to f = 0.80, and increase belt life. You may wish to suggest to your students that solving comparison problems in this manner assists in the design process.

( F1 ) a = 20(200)(1)(1) = 4000 lbf = F1 F = 2T /D = 2(21 252)/(32) = 1328.3 lbf F2 = F1 − F = 4000 − 1328.3 = 2671.7 lbf Fi =

F1 + F2 − Fc 2

4000 + 2671.7 − 1174.3 = 2161.6 lbf 2 Transmitted power H: =

1328.3(7205) F(V ) = = 290 hp 33 000 33 000 1 F1 − Fc f = ln θd F2 − Fc 4000 − 1174.3 1 ln = 3.037 2671.7 − 1174.3 = 0.209 undeveloped

H=

There was a small change in C p . Parameter V Fc F1 F2

Change

Parameter

Change

2-fold 16-fold 4.26-fold 9.7-fold

F Fi Ht f

2-fold 4-fold 4-fold 0.48-fold

Note the change in Fc !

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17-3 48" 192"

As a design task, the decision set on p. 881 is useful. A priori decisions: • • • • • •

Function: Hnom = 60 hp, n = 380 rev/min, VR = 1, C = 192 in, K s = 1.1 Design factor: n d = 1 Initial tension: Catenary Belt material: Polyamide A-3 Drive geometry: d = D = 48 in Belt thickness: t = 0.13 in

Design variable: Belt width of 6 in Use a method of trials. Initially choose b = 6 in V =

π(48)(380) πdn = = 4775 ft/min 12 12

w = 12γ bt = 12(0.042)(6)(0.13) = 0.393 lbf/ft Fc =

0.393(4775/60) 2 wV 2 = = 77.4 lbf g 32.174

T =

63 025(1.1)(1)(60) = 10 946 lbf · in 380

F =

2T 2(10 946) = = 456.1 lbf d 48

F1 = ( F1 ) a = bFa C p Cv = 6(100)(1)(1) = 600 lbf F2 = F1 − F = 600 − 456.1 = 143.9 lbf Transmitted power H H=

456.1(4775) F(V ) = = 66 hp 33 000 33 000

F1 + F2 600 + 143.9 − Fc = − 77.4 = 294.6 lbf 2 2 1 1 600 − 77.4 F1 − Fc f = = ln = 0.656 ln θd F2 − Fc π 143.9 − 77.4

Fi =

L = 534.8 in,

from Eq. (17-2)

Friction is not fully developed, so bmin is just a little smaller than 6 in (5.7 in). Not having a figure of merit, we choose the most narrow belt available (6 in). We can improve the

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Chapter 17

design by reducing the initial tension, which reduces F1 and F2 , thereby increasing belt life. This will bring f to 0.80 F1 =

(F + Fc ) exp( f θ) − Fc exp( f θ) − 1

exp( f θ) = exp(0.80π) = 12.345 Therefore F1 =

(456.1 + 77.4)(12.345) − 77.4 = 573.7 lbf 12.345 − 1

F2 = F1 − F = 573.7 − 456.1 = 117.6 lbf Fi =

F1 + F2 573.7 + 117.6 − Fc = − 77.4 = 268.3 lbf 2 2

These are small reductions since f is close to f , but improvements nevertheless. dip = 17-4

3C 2 w 3(192/12) 2 (0.393) = 0.562 in = 2Fi 2(268.3)

From the last equation given in the Problem Statement, 1 1 − {2T /[d(a0 − a2 )b]} 2T 1− exp( f φ) = 1 d(a0 − a2 )b 2T exp( f φ) = exp( f φ) − 1 d(a0 − a2 )b 1 2T exp( f φ) b= a0 − a2 d exp( f φ) − 1 exp( f φ) =

But 2T /d = 33 000Hd /V Thus,

17-5

1 33 000Hd exp( f φ) b= a0 − a2 V exp( f φ) − 1

Q.E.D.

Refer to Ex. 17-1 on p. 878 for the values used below. (a) The maximum torque prior to slip is, T =

63 025Hnom K s n d 63 025(15)(1.25)(1.1) = = 742.8 lbf · in Ans. n 1750

The corresponding initial tension is, 742.8 11.17 + 1 T exp( f θ) + 1 Fi = = = 148.1 lbf Ans. D exp( f θ) − 1 6 11.17 − 1

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(b) See Prob. 17-4 statement. The final relation can be written 1 33 000Ha exp( f θ) bmin = Fa C p Cv − (12γ t/32.174)(V /60) 2 V [exp( f θ) − 1] 1 33 000(20.6)(11.17) = 100(0.7)(1) − {[12(0.042)(0.13)]/32.174}(2749/60) 2 2749(11.17 − 1) = 4.13 in Ans. This is the minimum belt width since the belt is at the point of slip. The design must round up to an available width. Eq. (17-1): −1

θd = π − 2 sin

D−d 2C

−1

= π − 2 sin

18 − 6 2(96)

= 3.016 511 rad −1 D − d −1 18 − 6 = π + 2 sin θ D = π + 2 sin 2C 2(96) = 3.266 674 Eq. (17-2): 1 L = [4(96) 2 − (18 − 6) 2 ]1/2 + [18(3.266 674) + 6(3.016 511)] 2 = 230.074 in Ans. (c)

F =

2T 2(742.8) = = 247.6 lbf d 6

( F1 ) a = bFa C p Cv = F1 = 4.13(100)(0.70)(1) = 289.1 lbf F2 = F1 − F = 289.1 − 247.6 = 41.5 lbf 0.271 Fc = 25.6 = 17.7 lbf 0.393 Fi =

F1 + F2 289.1 + 41.5 − Fc = − 17.7 = 147.6 lbf 2 2

Transmitted belt power H H= nfs =

F(V ) 247.6(2749) = = 20.6 hp 33 000 33 000 H Hnom K s

=

20.6 = 1.1 15(1.25)

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437

Chapter 17

If you only change the belt width, the parameters in the following table change as shown. Ex. 17-1

This Problem

6.00 0.393 25.6 420 172.4 270.6 0.33* 0.139

4.13 0.271 17.6 289 42 147.7 0.80** 0.176

b w Fc ( F1 ) a F2 Fi f dip

*Friction underdeveloped **Friction fully developed 17-6

The transmitted power is the same.

Fc Fi ( F1 ) a F2 Ha n fs f dip

b = 6 in

b = 12 in

n-Fold Change

25.65 270.35 420 172.4 20.62 1.1 0.139 0.328

51.3 664.9 840 592.4 20.62 1.1 0.125 0.114

2 2.46 2 3.44 1 1 0.90 0.34

If we relax Fi to develop full friction ( f = 0.80) and obtain longer life, then

Fc Fi F1 F2 f dip

b = 6 in

b = 12 in

n-Fold Change

25.6 148.1 297.6 50 0.80 0.255

51.3 148.1 323.2 75.6 0.80 0.503

2 1 1.09 1.51 1 2

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17-7

F1 ␣ d

y Rx

Ry

x

D

␣ F2 C

Find the resultant of F1 and F2 : α = sin−1

D−d 2C

D−d 2C 1 D−d 2 cos α = ˙ 1− 2 2C sin α =

1 R x = F1 cos α + F2 cos α = ( F1 + F2 ) 1 − 2 R y = F1 sin α − F2 sin α = ( F1 − F2 )

D−d 2C

D−d 2C

2 Ans.

Ans.

From Ex. 17-2, d = 16 in, D = 36 in, C = 16(12) = 192 in, F1 = 940 lbf, F2 = 276 lbf −1 36 − 16 = 2.9855◦ α = sin 2(192) 2 36 − 16 1 R x = (940 + 276) 1 − = 1214.4 lbf 2 2(192) 36 − 16 y = 34.6 lbf R = (940 − 276) 2(192) 16 d = (940 − 276) = 5312 lbf · in T = ( F1 − F2 ) 2 2 17-8

Begin with Eq. (17-10), F1 = Fc + Fi

2 exp( f θ) exp( f θ) − 1

Introduce Eq. (17-9): T exp( f θ) + 1 2T 2 exp( f θ) exp( f θ) = Fc + F1 = Fc + D exp( f θ) − 1 exp( f θ) + 1 D exp( f θ) − 1 exp( f θ) F1 = Fc + F exp( f θ) − 1

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Chapter 17

exp( f θ) Now add and subtract Fc exp( f θ) − 1 exp( f θ) exp( f θ) exp( f θ) + F − Fc F1 = Fc + Fc exp( f θ) − 1 exp( f θ) − 1 exp( f θ) − 1 exp( f θ) exp( f θ) + Fc − Fc F1 = ( Fc + F) exp( f θ) − 1 exp( f θ) − 1 Fc exp( f θ) − F1 = ( Fc + F) exp( f θ) − 1 exp( f θ) − 1 F1 =

( Fc + F) exp( f θ) − Fc exp( f θ) − 1

Q.E.D.

From Ex. 17-2: θd = 3.037 rad, F = 664 lbf, exp( f θ) = exp[0.80(3.037)] = 11.35 , and Fc = 73.4 lbf. F1 =

(73.4 + 664)(11.35 − 73.4) = 802 lbf (11.35 − 1)

F2 = F1 − F = 802 − 664 = 138 lbf 802 + 138 − 73.4 = 396.6 lbf 2 1 1 802 − 73.4 F1 − Fc = ln = 0.80 Ans. f = ln θd F2 − Fc 3.037 138 − 73.4

Fi =

17-9

This is a good class project. Form four groups, each with a belt to design. Once each group agrees internally, all four should report their designs including the forces and torques on the line shaft. If you give them the pulley locations, they could design the line shaft when they get to Chap. 18. For now you could have the groups exchange group reports to determine if they agree or have counter suggestions.

17-10

If you have the students implement a computer program, the design problem selections may differ, and the students will be able to explore them. For K s = 1.25, n d = 1.1, d = 14 in and D = 28 in, a polyamide A-5 belt, 8 inches wide, will do (bmin = 6.58 in)

17-11

An efficiency of less than unity lowers the output for a given input. Since the object of the drive is the output, the efficiency must be incorporated such that the belt’s capacity is increased. The design power would thus be expressed as Hd =

17-12

Hnom K s n d eff

Ans.

Some perspective on the size of Fc can be obtained from w V 2 12γ bt V 2 Fc = = g 60 g 60

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An approximate comparison of non-metal and metal belts is presented in the table below.

γ , lbf/in3 b, in t, in

Non-metal

Metal

0.04 5.00 0.20

0.280 1.000 0.005

The ratio w/wm is 12(0.04)(5)(0.2) w = ˙ 29 = wm 12(0.28)(1)(0.005) The second contribution to Fc is the belt peripheral velocity which tends to be low in metal belts used in instrument, printer, plotter and similar drives. The velocity ratio squared influences any Fc /( Fc ) m ratio. It is common for engineers to treat Fc as negligible compared to other tensions in the belting problem. However, when developing a computer code, one should include Fc . 17-13

Eq. (17-8): F = F1 − F2 = ( F1 − Fc )

exp( f θ) − 1 exp( f θ)

Assuming negligible centrifugal force and setting F1 = ab from step 3, F exp( f θ) bmin = a exp( f θ) − 1 Also,

Hd = Hnom K s n d =

(1)

(F)V 33 000

33 000Hnom K s n d V 1 33 000Hd exp( f θ) = a V exp( f θ) − 1

F = Substituting into (1), bmin

17-14

Ans.

The decision set for the friction metal flat-belt drive is: A priori decisions • Function:

Hnom = 1 hp , n = 1750 rev/min , N p = 106 belt passes.

V R = 2,

C= ˙ 15 in ,

K s = 1.2 ,

• Design factor: n d = 1.05 • Belt material and properties: Table 17-8:

301/302 stainless steel S y = 175 000 psi, E = 28 Mpsi, ν = 0.285

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Chapter 17

• Drive geometry: d = 2 in, D = 4 in • Belt thickness: t = 0.003 in Design variables: • Belt width b • Belt loop periphery Preliminaries Hd = Hnom K s n d = 1(1.2)(1.05) = 1.26 hp T =

63 025(1.26) = 45.38 lbf · in 1750

A 15 in center-to-center distance corresponds to a belt loop periphery of 39.5 in. The 40 in loop available corresponds to a 15.254 in center distance. 4−2 −1 = 3.010 rad θd = π − 2 sin 2(15.254) 4−2 −1 θ D = π + 2 sin = 3.273 rad 2(15.274) For full friction development exp( f θd ) = exp[0.35(3.010)] = 2.868 π(2)(1750) πdn = = 916.3 ft/s 12 12 S y = 175 000 psi V =

Eq. (17-15): S f = 14.17(106 )(106 ) −0.407 = 51 212 psi From selection step 3 a = Sf −

28(106 )(0.003) Et t = 51 212 − (0.003) (1 − ν 2 )d (1 − 0.2852 )(2)

= 16.50 lbf/in of belt width ( F1 ) a = ab = 16.50b For full friction development, from Prob. 17-13, bmin =

F exp( f θd ) a exp( f θd ) − 1

F =

2(45.38) 2T = = 45.38 lbf d 2

So bmin

45.38 = 16.50

2.868 2.868 − 1

= 4.23 in

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Decision #1: b = 4.5 in F1 = ( F1 ) a = ab = 16.5(4.5) = 74.25 lbf F2 = F1 − F = 74.25 − 45.38 = 28.87 lbf Fi =

74.25 + 28.87 F1 + F2 = = 51.56 lbf 2 2

Existing friction 1 f = ln θd

Ht = nfs =

F1 F2

74.25 1 ln = 0.314 = 3.010 28.87

45.38(916.3) (F)V = = 1.26 hp 33 000 33 000 Ht Hnom K s

=

1.26 = 1.05 1(1.2)

This is a non-trivial point. The methodology preserved the factor of safety corresponding to n d = 1.1 even as we rounded bmin up to b. Decision #2 was taken care of with the adjustment of the center-to-center distance to accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this. Remember to subsequently recalculate θd and θ D . 17-15

Decision set: A priori decisions ˙ 20 in, K s = 1.25, • Function: Hnom = 5 hp, N = 1125 rev/min , V R = 3, C = 6 N p = 10 belt passes • Design factor: n d = 1.1 • Belt material: BeCu, S y = 170 000 psi, E = 17(106 ) psi, ν = 0.220 • Belt geometry: d = 3 in, D = 9 in • Belt thickness: t = 0.003 in Design decisions • Belt loop periphery • Belt width b Preliminaries: Hd = Hnom K s n d = 5(1.25)(1.1) = 6.875 hp T =

63 025(6.875) = 385.2 lbf · in 1125

Decision #1: Choose a 60-in belt loop with a center-to-center distance of 20.3 in. 9−3 −1 θd = π − 2 sin = 2.845 rad 2(20.3) 9−3 −1 θ D = π + 2 sin = 3.438 rad 2(20.3)

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For full friction development: exp( f θd ) = exp[0.32(2.845)] = 2.485 V =

π(3)(1125) πdn = = 883.6 ft/min 12 12

S f = 56 670 psi From selection step 3 17(106 )(0.003) Et t = 56 670 − (0.003) = 116.4 lbf/in a = Sf − (1 − ν 2 )d (1 − 0.222 )(3) 2T 2(385.2) = = 256.8 lbf d 3 256.8 F exp( f θd ) 2.485 = = 3.69 in = a exp( f θd ) − 1 116.4 2.485 − 1

F = bmin

Decision #2: b = 4 in F1 = ( F1 ) a = ab = 116.4(4) = 465.6 lbf F2 = F1 − F = 465.6 − 256.8 = 208.8 lbf Fi =

465.6 + 208.8 F1 + F2 = = 337.3 lbf 2 2

Existing friction 1 f = ln θd

F1 F2

1 465.6 = ln = 0.282 2.845 208.8

256.8(883.6) (F)V = = 6.88 hp 33 000 33 000 6.88 H = = 1.1 = 5(1.25) 5(1.25)

H= nfs

Fi can be reduced only to the point at which f = f = 0.32. From Eq. (17-9) 385.2 2.485 + 1 T exp( f θd ) + 1 Fi = = = 301.3 lbf d exp( f θd ) − 1 3 2.485 − 1 Eq. (17-10):

F1 = Fi

2 exp( f θd ) 2(2.485) = 301.3 = 429.7 lbf exp( f θd ) + 1 2.485 + 1

F2 = F1 − F = 429.7 − 256.8 = 172.9 lbf and 17-16

f = f = 0.32

This solution is the result of a series of five design tasks involving different belt thicknesses. The results are to be compared as a matter of perspective. These design tasks are accomplished in the same manner as in Probs. 17-14 and 17-15 solutions.

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The details will not be presented here, but the table is provided as a means of learning. Five groups of students could each be assigned a belt thickness. You can form a table from their results or use the table below t, in

b CD a d D Fi F1 F2 nfs L f Fi F1 F2 f

0.002

0.003

0.005

0.008

0.010

4.000 20.300 109.700 3.000 9.000 310.600 439.000 182.200 1.100 60.000 0.309 301.200 429.600 172.800 0.320

3.500 20.300 131.900 3.000 9.000 333.300 461.700 209.000 1.100 60.000 0.285 301.200 429.600 172.800 0.320

4.000 20.300 110.900 3.000 9.000 315.200 443.600 186.800 1.100 60.000 0.304 301.200 429.600 172.800 0.320

1.500 18.700 194.900 5.000 15.000 215.300 292.300 138.200 1.100 70.000 0.288 195.700 272.700 118.700 0.320

1.500 20.200 221.800 6.000 18.000 268.500 332.700 204.300 1.100 80.000 0.192 166.600 230.800 102.400 0.320

The first three thicknesses result in the same adjusted Fi , F1 and F2 (why?). We have no figure of merit, but the costs of the belt and the pulleys is about the same for these three thicknesses. Since the same power is transmitted and the belts are widening, belt forces are lessening. 17-17

This is a design task. The decision variables would be belt length and belt section, which could be combined into one, such as B90. The number of belts is not an issue. We have no figure of merit, which is not practical in a text for this application. I suggest you gather sheave dimensions and costs and V-belt costs from a principal vendor and construct a figure of merit based on the costs. Here is one trial. Preliminaries: For a single V-belt drive with Hnom = 3 hp, n = 3100 rev/min, D = 12 in, and d = 6.2 in, choose a B90 belt, K s = 1.3 and n d = 1. L p = 90 + 1.8 = 91.8 in Eq. (17-16b):

  2   π π 91.8 − (12 + 6.2) − 2(12 − 6.2) 2 C = 0.25 91.8 − (12 + 6.2) +   2 2 = 31.47 in −1

θd = π − 2 sin

12 − 6.2 = 2.9570 rad 2(31.47)

exp( f θd ) = exp[0.5123(2.9570)] = 4.5489 V =

π(6.2)(3100) πdn = = 5031.8 ft/min 12 12

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Table 17-13:

180° 180° = (2.957 rad) = 169.42° Angle θ = θd π π The footnote regression equation gives K 1 without interpolation: K 1 = 0.143 543 + 0.007 468(169.42°) − 0.000 015 052(169.42°) 2 = 0.9767

The design power is Hd = Hnom K s n d = 3(1.3)(1) = 3.9 hp From Table 17-14 for B90, K 2 = 1. From Table 17-12 take a marginal entry of Htab = 4, although extrapolation would give a slightly lower Htab . Ha = K 1 K 2 Htab

Eq. (17-17):

= 0.9767(1)(4) = 3.91 hp The allowable Fa is given by Fa =

63 025(3.91) 63 025Ha = = 25.6 lbf n(d/2) 3100(6.2/2)

The allowable torque Ta is 25.6(6.2) Fa d = = 79.4 lbf · in 2 2 From Table 17-16, K c = 0.965. Thus, Eq. (17-21) gives, 5031.8 2 Fc = 0.965 = 24.4 lbf 1000 Ta =

At incipient slip, Eq. (17-9) provides: exp( f θ) + 1 79.4 4.5489 + 1 T = = 20.0 lbf Fi = d exp( f θ) − 1 6.2 4.5489 − 1 Eq. (17-10):

F1 = Fc + Fi Thus,

2 exp( f θ) 2(4.5489) = 24.4 + 20 = 57.2 lbf exp( f θ) + 1 4.5489 + 1

F2 = F1 − Fa = 57.2 − 25.6 = 31.6 lbf

Ha Nb (3.91)(1) = 1.003 Ans. = Hd 3.9 If we had extrapolated for Htab , the factor of safety would have been slightly less than one.

Eq. (17-26):

Life

nfs =

Use Table 17-16 to find equivalent tensions T1 and T2 . T1 = F1 + ( Fb ) 1 = F1 +

576 Kb = 57.2 + = 150.1 lbf d 6.2

T2 = F1 + ( Fb ) 2 = F1 +

576 Kb = 57.2 + = 105.2 lbf D 12

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From Eq. (17-27), the number of belt passes is: −1 1193 −10.929 1193 −10.929 NP = + = 6.76(109 ) 150.1 105.2 From Eq. (17-28) for N P > 109 , t=

NP L p 109 (91.8) > 720V 720(5031.8)

t > 25 340 h Ans. Suppose n f s was too small. Compare these results with a 2-belt solution. Htab = 4 hp/belt,

Ta = 39.6 lbf · in/belt,

Fa = 12.8 lbf/belt, nfs = Also,

Ha = 3.91 hp/belt

Nb Ha Nb Ha 2(3.91) = 2.0 = = Hd Hnom K s 3(1.3)

F1 = 40.8 lbf/belt,

F2 = 28.0 lbf/belt,

Fi = 9.99 lbf/belt,

Fc = 24.4 lbf/belt

( Fb ) 1 = 92.9 lbf/belt,

( Fb ) 2 = 48 lbf/belt

T1 = 133.7 lbf/belt,

T2 = 88.8 lbf/belt

N P = 2.39(10 ) passes, t > 605 600 h 10

Initial tension of the drive: ( Fi ) drive = Nb Fi = 2(9.99) = 20 lbf 17-18

Given: two B85 V-belts with d = 5.4 in, D = 16 in, n = 1200 rev/min, and K s = 1.25 L p = 85 + 1.8 = 86.8 in

Table 17-11: Eq. (17-17b):

  2   π π 2 C = 0.25 86.8 − (16 + 5.4) + 86.8 − (16 + 5.4) − 2(16 − 5.4)   2 2 = 26.05 in Ans.

Eq. (17-1): −1

θd = 180° − 2 sin

16 − 5.4 = 156.5° 2(26.05)

From table 17-13 footnote: K 1 = 0.143 543 + 0.007 468(156.5°) − 0.000 015 052(156.5°) 2 = 0.944 Table 17-14: Belt speed:

K2 = 1 π(5.4)(1200) V = = 1696 ft/min 12

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Use Table 17-12 to interpolate for Htab . 2.62 − 1.59 (1696 − 1000) = 2.31 hp/belt Htab = 1.59 + 2000 − 1000 Ha = K 1 K 2 Nb Htab = 1(0.944)(2)(2.31) = 4.36 hp Assuming n d = 1 Hd = K s Hnom n d = 1.25(1) Hnom For a factor of safety of one, Ha = Hd 4.36 = 1.25Hnom 4.36 = 3.49 hp Ans. Hnom = 1.25 17-19

Given: Hnom = 60 hp, n = 400 rev/min, K s = 1.4, d = D = 26 in on 12 ft centers. Design task: specify V-belt and number of strands (belts). Tentative decision: Use D360 belts. L p = 360 + 3.3 = 363.3 in Table 17-11: Eq. (17-16 b):

  2   π π C = 0.25 363.3 − (26 + 26) + 363.3 − (26 + 26) − 2(26 − 26) 2   2 2 = 140.8 in (nearly 144 in)

θd = π, V =

θ D = π,

exp[0.5123π] = 5.0,

π(26)(400) πdn = = 2722.7 ft/min 12 12

Table 17-13: For θ = 180°, K 1 = 1 Table 17-14: For D360, K 2 = 1.10 Table 17-12: Htab = 16.94 hp by interpolation Thus,

Ha = K 1 K 2 Htab = 1(1.1)(16.94) = 18.63 hp Hd = K s Hnom = 1.4(60) = 84 hp

Number of belts, Nb Nb =

K s Hnom Hd 84 = 4.51 = = K 1 K 2 Htab Ha 18.63

Round up to five belts. It is left to the reader to repeat the above for belts such as C360 and E360. 63 025(18.63) 63 025Ha Fa = = = 225.8 lbf/belt n(d/2) 400(26/2) Ta =

225.8(26) (Fa )d = = 2935 lbf · in/belt 2 2

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Eq. (17-21):

V Fc = 3.498 1000

2

2722.7 = 3.498 1000

2 = 25.9 lbf/belt

At fully developed friction, Eq. (17-9) gives 2935 5 + 1 T exp( f θ) + 1 = = 169.3 lbf/belt Fi = d exp( f θ) − 1 26 5−1 2 exp( f θ) 2(5) = 25.9 + 169.3 = 308.1 lbf/belt Eq. (17-10): F1 = Fc + Fi exp( f θ) + 1 5+1 F2 = F1 − Fa = 308.1 − 225.8 = 82.3 lbf/belt nfs =

Ha Nb (185.63) = 1.109 Ans. = Hd 84

Reminder: Initial tension is for the drive ( Fi ) drive = Nb Fi = 5(169.3) = 846.5 lbf A 360 belt is at the right-hand edge of the range of center-to-center pulley distances. D ≤ C ≤ 3( D + d) 26 ≤ C ≤ 3(26 + 26) 17-20

˙ 60 in, 14-in wide rim, Hnom = 50 hp, n = 875 rev/min, K s = 1.2, Preliminaries: D = n d = 1.1, m G = 875/170 = 5.147, d = ˙ 60/5.147 = 11.65 in (a) From Table 17-9, an 11-in sheave exceeds C-section minimum diameter and precludes D- and E-section V-belts. Decision: Use d = 11 in, C270 belts L p = 270 + 2.9 = 272.9 in Table 17-11:   2   π π 2 272.9 − (60 + 11) − 2(60 − 11) C = 0.25 272.9 − (60 + 11) +   2 2 = 76.78 in This fits in the range

D < C < 3( D + d) 60 < C < 3(60 + 11)

60 in < C < 213 in −1 60 − 11 θd = π − 2 sin = 2.492 rad 2(76.78) −1 60 − 11 θ D = π + 2 sin = 3.791 rad 2(76.78) exp[0.5123(2.492)] = 3.5846

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Chapter 17

For the flat on flywheel exp[0.13(3.791)] = 1.637 V =

πdn π(11)(875) = = 2519.8 ft/min 12 12

Table 17-13: Regression equation gives K 1 = 0.90 Table 17-14: Table 17-12:

K 2 = 1.15 Htab = 7.83 hp/belt

by interpolation

Eq. (17-17):

Ha = K 1 K 2 Htab = 0.905(1.15)(7.83) = 8.15 hp

Eq. (17-19):

Hd = Hnom K s n d = 50(1.2)(1.1) = 66 hp

Eq. (17-20):

Nb =

Hd 66 = 8.1 belts = Ha 8.15

Decision: Use 9 belts. On a per belt basis, 63 025(8.15) 63 025Ha = = 106.7 lbf/belt n(d/2) 875(11/2)

Fa =

106.7(11) Fa d = = 586.9 lbf per belt 2 2 2 V 2519.8 2 Fc = 1.716 = 1.716 = 10.9 lbf/belt 1000 1000

Ta =

At fully developed friction, Eq. (17-9) gives 586.9 3.5846 + 1 T exp( f θd ) + 1 Fi = = = 94.6 lbf/belt d exp( f θd ) − 1 11 3.5846 − 1 Eq. (17-10):

F1 = Fc + Fi

2 exp( f θd ) 2(3.5846) = 10.9 + 94.6 = 158.8 lbf/belt exp( f θd ) + 1 3.5846 + 1

F2 = F1 − Fa = 158.8 − 106.7 = 52.1 lbf/belt nfs =

Nb Ha 9(8.15) = 1.11 O.K. Ans. = Hd 66

Durability: ( Fb ) 1 = 145.45 lbf/belt, T1 = 304.4 lbf/belt, and Remember:

( Fb ) 2 = 76.7 lbf/belt T2 = 185.6 lbf/belt

t > 150 000 h ( Fi ) drive = 9(94.6) = 851.4 lbf

Table 17-9: C-section belts are 7/8" wide. Check sheave groove spacing to see if 14"-width is accommodating.

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(b) The fully developed friction torque on the flywheel using the flats of the V-belts is 1.637 − 1 exp( f θ) − 1 = 60(94.6) = 1371 lbf · in per belt Tflat = Fi exp( f θ) + 1 1.637 + 1 The flywheel torque should be Tfly = m G Ta = 5.147(586.9) = 3021 lbf · in per belt but it is not. There are applications, however, in which it will work. For example, make the flywheel controlling. Yes. Ans. 17-21 (a)

S is the spliced-in string segment length De is the equatorial diameter D is the spliced string diameter

S

De

␦

δ is the radial clearance S + π De = π D = π( De + 2δ) = π De + 2πδ

D⬘

From which δ=

S 2π

The radial clearance is thus independent of De . δ=

12(6) = 11.5 in 2π

Ans.

This is true whether the sphere is the earth, the moon or a marble. Thinking in terms of a radial or diametral increment removes the basic size from the problem. Viewpoint again! (b) and (c) ␦

␦ Dp

60"

dp

Pitch surface

Table 17-9: For an E210 belt, the thickness is 1 in. 1"

0.716"

4.5 210 + 4.5 210 − = π π π 4.5 2δ = π 4.5 = 0.716 in δ= 2π

d P − di =

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The pitch diameter of the flywheel is D P − 2δ = D D P = D + 2δ = 60 + 2(0.716) = 61.43 in We could make a table: Diametral Growth

A 1.3 π

2δ

Section B C D 1.8 π

2.9 π

3.3 π

E 4.5 π

The velocity ratio for the D-section belt of Prob. 17-20 is m G =

60 + 3.3/π D + 2δ = = 5.55 Ans. d 11

for the V-flat drive as compared to m a = 60/11 = 5.455 for the VV drive. The pitch diameter of the pulley is still d = 11 in, so the new angle of wrap, θd, is −1 D + 2δ − d Ans. θd = π − 2 sin 2C −1 D + 2δ − d Ans. θ D = π + 2 sin 2C Equations (17-16a) and (17-16b) are modified as follows ( D + δ − d) 2 π L p = 2C + ( D + 2δ + d) + 2 4C   π C p = 0.25 L − ( D + 2δ + d)  p 2 +

Lp −

π ( D + 2δ + d) 2

Ans.

2 − 2( D + 2δ − d) 2

  

Ans.

The changes are small, but if you are writing a computer code for a V-flat drive, remember that θd and θ D changes are exponential. 17-22

This design task involves specifying a drive to couple an electric motor running at 1720 rev/min to a blower running at 240 rev/min, transmitting two horsepower with a center distance of at least 22 inches. Instead of focusing on the steps, we will display two different designs side-by-side for study. Parameters are in a “per belt” basis with per drive quantities shown along side, where helpful.

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Parameter mG Ks nd K1 K2 d, in D, in θd , rad V, ft/min exp( f θd ) L p , in C, in Htab , uncorr. Nb Htab , uncorr. Ta , lbf · in Fa , lbf Ha , hp nfs F1 , lbf F2 , lbf ( Fb ) 1 , lbf ( Fb ) 2 , lbf Fc , lbf Fi , lbf T1 , lbf · in T2 , lbf · in N , passes t >h

Four A-90 Belts

Two A-120 Belts

7.33 1.1 1.1 0.877 1.05 3.0 22 2.333 1350.9 3.304 91.3 24.1 0.783 3.13 26.45(105.8) 17.6(70.4) 0.721(2.88) 1.192 26.28(105.2) 8.67(34.7) 73.3(293.2) 10(40) 1.024 16.45(65.8) 99.2 36.3 1.61(109 ) 93 869

7.142 1.1 1.1 0.869 1.15 4.2 30 2.287 1891 3.2266 101.3 31 1.662 3.326 60.87(121.7) 29.0(58) 1.667(3.33) 1.372 44(88) 15(30) 52.4(109.8) 7.33(14.7) 2.0 27.5(55) 96.4 57.4 2.3(109 ) 89 080

Conclusions: • Smaller sheaves lead to more belts. • Larger sheaves lead to larger D and larger V. • Larger sheaves lead to larger tabulated power. • The discrete numbers of belts obscures some of the variation. The factors of safety exceed the design factor by differing amounts. 17-23

In Ex. 17-5 the selected chain was 140-3, making the pitch of this 140 chain14/8 = 1.75 in. Table 17-19 confirms.

17-24 (a) Eq. (17-32): Eq. (17-33):

H1 = 0.004N11.08 n 01.9 p(3−0.07 p) H2 =

1000K r N11.5 p0.8 n 11.5

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Chapter 17

Equating and solving for n 1 gives 1/2.4 0.25(106 )K r N10.42 n1 = p(2.2−0.07 p) (b) For a No. 60 chain, p = 6/8 = 0.75 in, N1 = 17,

0.25(106 )(17)(17) 0.42 n1 = 0.75[2.2−0.07(0.75)]

Ans. K r = 17

1/2.4 = 1227 rev/min

Ans.

Table 17-20 confirms that this point occurs at 1200 ± 200 rev/min. (c) Life predictions using Eq. (17-40) are possible at speeds greater than 1227 rev/min. Ans. 17-25

Given: a double strand No. 60 roller chain with p = 0.75 in, N1 = 13 teeth at 300 rev/min, N2 = 52 teeth. Htab = 6.20 hp

(a) Table 17-20: Table 17-22:

K 1 = 0.75

Table 17-23:

K 2 = 1.7

Use Eq. (17-37):

Ks = 1 Ha = K 1 K 2 Htab = 0.75(1.7)(6.20) = 7.91 hp

Ans.

(b) Eqs. (17-35) and (17-36) with L/ p = 82 13 + 52 − 82 = −49.5 2   2 p 52 − 13  = 23.95 p C = 49.5 + 49.52 − 8 4 2π A=

[

C = 23.95(0.75) = 17.96 in,

round up to 18 in Ans.

(c) For 30 percent less power transmission, H = 0.7(7.91) = 5.54 hp T =

63 025(5.54) = 1164 lbf · in 300

Ans.

Eq. (17-29): D=

0.75 = 3.13 in sin(180◦ /13)

F=

1164 T = = 744 lbf r 3.13/2

Ans.

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17-26

Given: No. 40-4 chain, N1 = 21 teeth for n = 2000 rev/min, N2 = 84 teeth, h = 20 000 hours. ˙ 20 in. (a) Chain pitch is p = 4/8 = 0.500 in and C = Eq. (17-34):

2(20) 21 + 84 (84 − 21) 2 L = + + = 135 pitches (or links) p 0.5 2 4π 2 (20/0.5) L = 135(0.500) = 67.5 in Ans.

(b) Table 17-20:

Htab = 7.72 hp

(post-extreme power)

Eq. (17-40): Since K 1 is required, the N13.75 term is omitted. 7.722.5 (15 000) = 18 399 const = 135 18 399(135) 1/2.5 Htab = = 6.88 hp Ans. 20 000 (c) Table 17-22:

K1 =

Table 17-23:

21 17

1.5

= 1.37

K 2 = 3.3 = 1.37(3.3)(6.88) = 31.1 hp Ha = K 1 K 2 Htab

(d)

17-27

Ans.

V =

N1 pn 21(0.5)(2000) = = 1750 ft/min 12 12

F1 =

33 000(31.1) = 586 lbf Ans. 1750

This is our first design/selection task for chain drives. A possible decision set: A priori decisions • Function: Hnom , n 1 , space, life, K s • Design factor: n d • Sprockets: Tooth counts N1 and N2 , factors K 1 and K 2 Decision variables • Chain number • Strand count • Lubrication type • Chain length in pitches Function: Motor with Hnom = 25 hp at n = 700 rev/min; pump at n = 140 rev/min; m G = 700/140 = 5 Design Factor: n d = 1.1 Sprockets: Tooth count N2 = m G N1 = 5(17) = 85 teeth–odd and unavailable. Choose 84 teeth. Decision: N1 = 17, N2 = 84

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Chapter 17

Evaluate K 1 and K 2 Hd = Hnom K s n d Eq. (17-38): Ha = K 1 K 2 Htab Eq. (17-37): Equate Hd to Ha and solve for Htab : Htab =

K s n d Hnom K1 K2

Table 17-22:

K1 = 1

Table 17-23:

K 2 = 1, 1.7, 2.5, 3.3 for 1 through 4 strands Htab =

1.5(1.1)(25) 41.25 = (1)K 2 K2

Prepare a table to help with the design decisions:

Strands

K2

Htab

Chain No.

Htab

nfs

Lub. Type

1 2 3 4

1.0 1.7 2.5 3.3

41.3 24.3 16.5 12.5

100 80 80 60

59.4 31.0 31.0 13.3

1.58 1.40 2.07 1.17

B B B B

Design Decisions We need a figure of merit to help with the choice. If the best was 4 strands of No. 60 chain, then Decision #1 and #2: Choose four strand No. 60 roller chain with n f s = 1.17. nfs =

K 1 K 2 Htab 1(3.3)(13.3) = 1.17 = K s Hnom 1.5(25)

Decision #3: Choose Type B lubrication Analysis: Table 17-20:

Htab = 13.3 hp

Table 17-19:

p = 0.75 in

Try C = 30 in in Eq. (17-34): 2C N1 + N2 (N2 − N1 )2 L = + + p p 2 4π 2 C/ p = 2(30/0.75) +

(84 − 17) 2 17 + 84 + 2 4π 2 (30/0.75)

= 133.3 → 134 From Eq. (17-35) with p = 0.75 in,C = 30.26 in. Decision #4: Choose C = 30.26 in.

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Follow the decision set outlined in Prob. 17-27 solution. We will form two tables, the first for a 15 000 h life goal, and a second for a 50 000 h life goal. The comparison is useful. Function: Hnom = 50 hp at n = 1800 rev/min, n pump = 900 rev/min m G = 1800/900 = 2, K s = 1.2 life = 15 000 h, then repeat with life = 50 000 h Design factor: n d = 1.1 Sprockets: N1 = 19 teeth, N2 = 38 teeth Table 17-22 (post extreme):

K1 =

N1 17

1.5

=

19 17

1.5

= 1.18

Table 17-23: K 2 = 1, 1.7, 2.5, 3.3, 3.9, 4.6, 6.0 Decision variables for 15 000 h life goal: = Htab

nfs =

K s n d Hnom 1.2(1.1)(50) 55.9 = = K1 K2 1.18K 2 K2

(1)

K 1 K 2 Htab 1.18K 2 Htab = 0.0197K 2 Htab = K s Hnom 1.2(50)

Form a table for a 15 000 h life goal using these equations.

1 2 3 4 5 6 8

K2

Htab

1.0 1.7 2.5 3.3 3.9 4.6 6.0

55.90 32.90 22.40 16.90 14.30 12.20 9.32

Chain #

Htab

nfs

Lub

120 120 120 120 80 60 60

21.6 21.6 21.6 21.6 15.6 12.4 12.4

0.423 0.923 1.064 1.404 1.106 1.126 1.416

C C C C C C C

There are 4 possibilities where n f s ≥ 1.1 Decision variables for 50 000 h life goal From Eq. (17-40), the power-life tradeoff is: 2.5 2.5 ) 15 000 = (Htab ) 50 000 (Htab 15 000 2.5 1/2.5 (H ) Htab = = 0.618 Htab 50 000 tab

Substituting from (1),

Htab

55.9 = 0.618 K2

=

34.5 K2

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Chapter 17

The H notation is only necessary because we constructed the first table, which we normally would not do. nfs =

) K 1 K 2 Htab K 1 K 2 (0.618Htab = = 0.618[(0.0197)K 2 Htab ] = 0.0122K 2 Htab K s Hnom K s Hnom

Form a table for a 50 000 h life goal.

1 2 3 4 5 6 8

K2

Htab

1.0 1.7 2.5 3.3 3.9 4.6 6.0

34.50 20.30 13.80 10.50 8.85 7.60 5.80

Chain #

Htab

nfs

Lub

120 120 120 120 120 120 80

21.6 21.6 21.6 21.6 21.6 21.6 15.6

0.264 0.448 0.656 0.870 1.028 1.210 1.140

C C C C C C C

There are two possibilities in the second table with n f s ≥ 1.1. (The tables allow for the identification of a longer life one of the outcomes.) We need a figure of merit to help with the choice; costs of sprockets and chains are thus needed, but is more information than we have. Decision #1: #80 Chain (smaller installation) Ans. n f s = 0.0122K 2 Htab = 0.0122(8.0)(15.6) = 1.14

O.K .

Decision #2: 8-Strand, No. 80 Ans. Decision #3: Type C Lubrication Ans. Decision #4: p = 1.0 in, C is in midrange of 40 pitches 2C N1 + N2 ( N2 − N1 ) 2 L = + + p p 2 4π 2 C/ p = 2(40) +

19 + 38 (38 − 19) 2 + 2 4π 2 (40)

= 108.7

⇒

110 even integer Ans.

Eq. (17-36): N1 + N2 L 19 + 38 − = − 110 = −81.5 2 p 2   2 C 1 38 − 19  = 40.64 = 81.5 + 81.52 − 8 Eq. (17-35): p 4 2π A=

C = p(C/ p) = 1.0(40.64) = 40.64 in (for reference)

Ans.

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The objective of the problem is to explore factors of safety in wire rope. We will express strengths as tensions. (a) Monitor steel 2-in 6 × 19 rope, 480 ft long Table 17-2: Minimum diameter of a sheave is 30d = 30(2) = 60 in, preferably 45(2) = 90 in. The hoist abuses the wire when it is bent around a sheave. Table 17-24 gives the nominal tensile strength as 106 kpsi. The ultimate load is π(2) 2 = 333 kip Ans. Fu = (Su ) nom Anom = 106 4 The tensile loading of the wire is given by Eq. (17-46) a W Ft = + wl 1+ m g W = 4(2) = 8 kip,

m=1

Table (17-24): wl = 1.60d 2l = 1.60(22 )(480) = 3072 lbf or 3.072 kip Therefore,

2 = 11.76 kip Ans. Ft = (8 + 3.072) 1 + 32.2

Eq. (17-48): Fb =

Er dw Am D

and for the 72-in drum Fb =

12(106 )(2/13)(0.38)(22 )(10−3 ) = 39 kip 72

Ans.

For use in Eq. (17-44), from Fig. 17-21 ( p/Su ) = 0.0014 Su = 240 kpsi, Ff =

p. 908

0.0014(240)(2)(72) = 24.2 kip 2

(b) Factors of safety Static, no bending: n=

Fu 333 = 28.3 Ans. = Ft 11.76

Static, with bending: Eq. (17-49):

ns =

Fu − Fb 333 − 39 = 25.0 Ans. = Ft 11.76

Ans.

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Chapter 17

459

Fatigue without bending: nf =

Ff 24.2 = 2.06 Ans. = Ft 11.76

Fatigue, with bending: For a life of 0.1(106) cycles, from Fig. 17-21 ( p/Su ) = 4/1000 = 0.004 Ff = Eq. (17-50):

nf =

0.004(240)(2)(72) = 69.1 kip 2 69.1 − 39 = 2.56 Ans. 11.76

If we were to use the endurance strength at 106 cycles ( Ff = 24.2 kip) the factor of safety would be less than 1 indicating 106 cycle life impossible. Comments: • There are a number of factors of safety used in wire rope analysis. They are different, with different meanings. There is no substitute for knowing exactly which factor of safety is written or spoken. • Static performance of a rope in tension is impressive. • In this problem, at the drum, we have a finite life. • The remedy for fatigue is the use of smaller diameter ropes, with multiple ropes supporting the load. See Ex. 17-6 for the effectiveness of this approach. It will also be used in Prob. 17-30. • Remind students that wire ropes do not fail suddenly due to fatigue. The outer wires gradually show wear and breaks; such ropes should be retired. Periodic inspections prevent fatigue failures by parting of the rope. 17-30

Since this is a design task, a decision set is useful. A priori decisions • • • •

Function: load, height, acceleration, velocity, life goal Design Factor: n d Material: IPS, PS, MPS or other Rope: Lay, number of strands, number of wires per strand

Decision variables: • Nominal wire size: d • Number of load-supporting wires: m From experience with Prob. 17-29, a 1-in diameter rope is not likely to have much of a life, so approach the problem with the d and m decisions open. Function: 5000 lbf load, 90 foot lift, acceleration = 4 ft/s2 , velocity = 2 ft/s, life goal = 105 cycles Design Factor: n d = 2 Material: IPS Rope: Regular lay, 1-in plow-steel 6 × 19 hoisting

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Design variables Choose 30-in Dmin. Table 17-27: w = 1.60d 2 lbf/ft wl = 1.60d 2l = 1.60d 2 (90) = 144d 2 lbf, ea . Eq. (17-46):

Ft = =

W + wl m

a 5000 4 2 1+ = + 144d 1+ g m 32.2

5620 + 162d 2 lbf, each wire m

Eq. (17-47): ( p/Su )Su Dd 2 5 From Fig. 17-21 for 10 cycles, p/Su = 0.004; from p. 908, Su = 240 000 psi, based on metal area. Ff =

Ff =

0.004(240 000)(30d) = 14 400d lbf each wire 2

Eq. (17-48) and Table 17-27: Fb =

12(106 )(0.067d)(0.4d 2 ) E w dw Am = = 10 720d 3 lbf, each wire D 30

Eq. (17-45): Ff − Fb 14 400d − 10 720d 3 = Ft (5620/m) + 162d 2 We could use a computer program to build a table similar to that of Ex. 17-6. Alternatively, we could recognize that 162d 2 is small compared to 5620/m , and therefore eliminate the 162d 2 term. m 14 400d − 10 720d 3 nf = = (14 400d − 10 720d 3 ) ˙ 5620/m 5620 nf =

Maximize n f , ∂n f m =0= [14 400 − 3(10 720)d 2 ] ∂d 5620 From which

d* =

14 400 = 0.669 in 32 160

Back-substituting nf =

m [14 400(0.669) − 10 720(0.6693 )] = 1.14 m 5620

Thus n f = 1.14, 2.28, 3.42, 4.56 for m = 1, 2, 3, 4 respectively. If we choose d = 0.50 in, then m = 2. nf =

14 400(0.5) − 10 720(0.53 ) = 2.06 (5620/2) + 162(0.5) 2

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Chapter 17

This exceeds n d = 2 Decision #1: d = 1/2 in Decision #2: m = 2 ropes supporting load. Rope should be inspected weekly for any signs of fatigue (broken outer wires). Comment: Table 17-25 gives n for freight elevators in terms of velocity. 2 πd = 83 252d 2 lbf, each wire Fu = (Su ) nom Anom = 106 000 4 Fu 83 452(0.5) 2 n= = = 7.32 Ft (5620/2) + 162(0.5) 2 By comparison, interpolation for 120 ft/min gives 7.08-close. The category of construction hoists is not addressed in Table 17-25. We should investigate this before proceeding further. 17-31

2000 ft lift, 72 in drum, 6 × 19 MS rope. Cage and load 8000 lbf, acceleration = 2 ft/s2 . (a) Table 17-24: (Su ) nom = 106 kpsi; Su = 240 kpsi (p. 1093, metal area); Fig. 17-22: ( p/Su ) 106 = 0.0014 Ff = Table 17-24: Eq. (17-46):

0.0014(240)(72)d = 12.1d kip 2

wl = 1.6d 2 2000(10−3 ) = 3.2d 2 kip a Ft = (W + wl) 1 + g 2 2 = (8 + 3.2d ) 1 + 32.2 = 8.5 + 3.4d 2 kip

Note that bending is not included. n=

d, in 0.500 1.000 1.500 1.625 1.750 2.000

Ff 12.1d = Ft 8.5 + 3.4d 2 n 0.650 1.020 1.124 1.125 ← maximum n 1.120 1.095

Ans.

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(b) Try m = 4 strands

Ft =

8 + 3.2d 2 4

2 1+ 32.2

= 2.12 + 3.4d 2 kip Ff = 12.1d kip 12.1d n= 2.12 + 3.4d 2 d, in

n

0.5000 0.5625 0.6250 0.7500 0.8750 1.0000

2.037 2.130 2.193 2.250 ← maximum n Ans. 2.242 2.192

Comparing tables, multiple ropes supporting the load increases the factor of safety, and reduces the corresponding wire rope diameter, a useful perspective. 17-32 n=

ad b/m + cd 2

(b/m + cd 2 )a − ad(2cd) dn = =0 dd (b/m + cd 2 ) 2 From which

d* =

b mc

Ans.

√ a m a b/(mc) = n* = (b/m) + c[b/(mc)] 2 bc

Ans.

These results agree closely with Prob. 17-31 solution. The small differences are due to rounding in Prob. 17-31. 17-33

From Prob. 17-32 solution: n1 =

ad b/m + cd 2

Solve the above equation for m m=

b ad/n 1 − cd 2

[(ad/n 1 ) − ad 2 ](0) − b[(a/n 1 ) − 2cd] dm =0= ad [(ad/n 1 ) − cd 2 ]2

(1)

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Chapter 17

a Ans. 2cn 1 Substituting this result for d in Eq. (1) gives d* =

From which

m* =

4bcn 1 a2

Ans.

17-34 Am = 0.40d 2 = 0.40(22 ) = 1.6 in2 Er = 12 Mpsi, w = 1.6d 2 = 1.6(22 ) = 6.4 lbf/ft wl = 6.4(480) = 3072 lbf Treat the rest of the system as rigid, so that all of the stretch is due to the cage weighing 1000 lbf and the wire’s weight. From Prob. 5-6 δ1 = =

(wl)l Pl + AE 2AE 3072(480)(12) 1000(480)(12) + 1.6(12)(106 ) 2(1.6)(12)(106 )

= 0.3 + 0.461 = 0.761 in due to cage and wire. The stretch due to the wire, the cart and the cage is δ2 = 17-35 to 17-38

9000(480)(12) + 0.761 = 3.461 in Ans. 1.6(12)(106 )


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Chapter 18 Comment: This chapter, when taught immediately after Chapter 7, has the advantage of immediately applying the fatigue information acquired in Chapter 7. We have often done it ourselves. However, the disadvantage is that many of the items attached to the shaft have to be explained sufficiently so that the influence on the shaft is understood. It is the instructor’s call as to the best way to achieve course objectives. This chapter is a nice note upon which to finish a study of machine elements. A very popular first design task in the capstone design course is the design of a speed-reducer, in which shafts, and many other elements, interplay. 18-1

The first objective of the problem is to move from shaft attachments to influences on the shaft. The second objective is to “see” the diameter of a uniform shaft that will satisfy deflection and distortion constraints. (a) d P + (80/16)d P = 12 in 2 d P = 4.000 in Wt =

63 025(50) = 1313 lbf 1200(4/2)

W r = W t tan 25° = 1313 tan 25° = 612 lbf W =

Wt 1313 = = 1449 lbf cos 25° cos 25°

T = W t (d/2) = 1313(4/2) = 2626 lbf · in Reactions R A , R B , and load W are all in the same plane. w 2"

9"

R A = 1449(2/11) = 263 lbf R B = 1449(9/11) = 1186 lbf

11" RA

Mmax 2371 lbf • in

M

O

RB

0

9"

Mmax = R A (9) = 1449(2/11)(9) = 2371 lbf · in Ans.

11"

Components in xyz 1313

y 500.7 2" B x 612 1074.3

9"

111.3 z

A 238.7

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Chapter 18

(b) Using n d = 2 and Eq. (18-1) 32n d Fb(b2 − l 2 ) 1/4 d A = 3π Elθall 32(2)(1449)(2)(22 − 112 ) 1/4 = 3π(30)(106 )(11)(0.001) = 1.625 in A design factor of 2 means that the slope goal is 0.001/2 or 0.0005. Eq. (18-2): 32n d Fa(l 2 − a 2 ) 1/4 dB = 3π Elθall 32(2)(1449)(9)(112 − 92 ) 1/4 = 3π(30)(106 )(11)(0.001) = 1.810 in The diameter of a uniform shaft should equal or exceed 1.810 in. 18-2

Ans.

This will be solved using a deterministic approach with n d = 2. However, the reader may wish to explore the stochastic approach given in Sec. 7-17. Table A-20: Eq. (7-8): Eq. (7-18):

Sut = 68 kpsi and S y = 37.5 kpsi Se = 0.504(68) = 34.27 kpsi ka = 2.70(68) −0.265 = 0.883

Assume a shaft diameter of 1.8 in. Eq. (7-19):

kb =

1.8 0.30

−0.107

= 0.826

kc = kd = k f = 1 From Table 7-7 for R = 0.999, ke = 0.753. Eq. (7-17): From p. 444,

Se = 0.883(0.826)(1)(1)(1)(0.753)(34.27) = 18.8 kpsi K t = 2.14, K ts = 2.62

With r = 0.02 in, Figs. 7-20 and 7-21 give q = 0.60 and qs = 0.77, respectively. Eq. (7-31):

K f = 1 + 0.60(2.14 − 1) = 1.68 K f s = 1 + 0.77(2.62 − 1) = 2.25

(a) DE-elliptic from Eq. (18-21),   2 2 1/2 1/3  16(2) 2.25(2626) 1.68(2371) +3 = 1.725 in Ans. d= 4   π 18 800 37 500

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(b) DE-Gerber from Eq. (18-16),    2 1/2  1/3  2.25(2626)(18 800) 16(2)(1.68)(2371)  1+ 1+3 d=   π(18 800) 1.68(2371)(68 000) = 1.687 in Ans. From Prob. 18-1, deflection controls d = 1.81 in 18-3

It is useful to provide a cylindrical roller bearing as the heavily-loaded bearing and a ball bearing at the other end to control the axial float, so that the roller grooves are not subject to thrust hunting. Profile keyways capture their key. A small shoulder can locate the pinion, and a shaft collar (or a light press fit) can capture the pinion. The key transmits the torque in either case. The student should: • select rolling contact bearings so that the shoulder and fillet can be sized to the bearings; • build on the understanding gained from Probs. 18-1 and 18-2. Each design will differ in detail so no solution is presented here.

18-4

One could take pains to model this shaft exactly, using say finite element software. However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875 in. The reductions in diameter at the bearings will change the results insignificantly. Use E = 30(106 ) psi. To the left of the load: Fb (3x 2 + b2 − l 2 ) 6E I l 1449(2)(3x 2 + 22 − 112 ) = 6(30)(106 )(π/64)(1.8254 )(11)

θ AB =

= 2.4124(10−6 )(3x 2 − 117) At x = 0: At x = 9 in:

θ = −2.823(10−4 ) rad θ = 3.040(10−4 ) rad

At x = 11 in:

θ=

1449(9)(112 − 92 ) 6(30)(106 )(π/64)(1.8754 )(11)

= 4.342(10−4 ) rad Left bearing: Station 1 Allowable slope Actual slope 0.001 = 3.54 = 0.000 282 3

nfs =

Right bearing: Station 5 nfs =

0.001 = 2.30 0.000 434 2

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Chapter 18

Gear mesh slope: Section 18-2, p. 927, recommends a relative slope of 0.0005. While we don’t know the slope on the next shaft, we know that it will need to have a larger diameter and be stiffer. At the moment we can say 0.0005 = 1.64 nfs < 0.000 304 Since this is the controlling location on the shaft for distortion, n f s may be much less than 1.64. All is not lost because crowning of teeth can relieve the slope constraint. If this is not an option, then use Eq. (18-4) with a design factor of say 2. n(dy/dx) old 1/4 dnew = dold (slope) all 2(0.000 304) 1/4 = 2.341 in = 1.875 0.000 25 Technically, all diameters should be increased by a factor of 2.341/1.875, or about 1.25. However the bearing seat diameters cannot easily be increased and the overhang diameter need not increase because it is straight. The shape of the neutral surface is largely controlled by the diameter between bearings. The shaft is unsatisfactory in distortion as indicated by the slope at the gear seat. We leave the problem here. 18-5

Use the distortion-energy elliptic failure locus. The torque and moment loadings on the shaft are shown below. M, T (lbf • in)

2371

2626 Torque

Moment x

Candidate critical locations for strength: • Pinion seat keyway • Right bearing shoulder • Coupling keyway Preliminaries: ANSI/ASME shafting design standard uses notch sensitivities to estimate K f and K f s . Table A-20 for 1030 HR: Sut Se Eq. (7-8): ka Eq. (7-18): kc

= 68 kpsi, S y = 37.5 kpsi, = 0.504(68) = 34.27 kpsi = 2.70(68) −0.265 = 0.883 = kd = ke = 1

Pinion seat keyway See p. 444 for keyway stress concentration factors K t = 2.14 Profile keyway K ts = 2.62

H B = 137

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For an end-mill profile keyway cutter of 0.010 in radius, From Fig. 7-20:

q = 0.50

From Fig. 7-21:

qs = 0.65

Eq. (7-31):

K f s = 1 + qs (K ts − 1) = 1 + 0.65(2.62 − 1) = 2.05 K f = 1 + 0.50(2.14 − 1) = 1.57

Eq. (18-15): A = 2K f Ma = 2(1.57)(2371) = 7445 lbf · in √ √ B = 3K f s Tm = 3(2.05)(2626) = 9324 lbf · in 7445 σa A = 0.798 = = σm B 9324 1.875 −0.107 kb = = 0.822 0.30 r=

Eq. (7-19): Eq. (7-17): Eq. (18-22):

Se = 0.883(0.822)(0.753)(34.27) = 18.7 kpsi 1/2 16 1 2.05(2626) 2 1.57(2371) 2 = +3 4 n π(1.8753 ) 18 700 37 500 = 0.363,

Table 7-11:

from which n = 2.76 (0.798) 2 (37.5) 2 (18.7) 2 = 15.9 kpsi (0.798) 2 (37.5) 2 + 18.72

Sa = Sm =

15.9 Sa = = 19.9 kpsi r 0.798

15.9 = 0.736 < 0.7 37.5 − 15.9 Therefore, the threat lies in fatigue. rcrit =

Right-hand bearing shoulder The text does not give minimum and maximum shoulder diameters for 03-series bearings (roller). Use D = 1.75 in. 0.030 D 1.75 r = = 0.019, = = 1.11 d 1.574 d 1.574 From Fig. A-15-9, K t = 2.4 From Fig. A-15-8, K ts = 1.6

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Chapter 18

q = 0.65

From Fig. 7-20, From Fig. 7-21,

qs = 0.83 K f = 1 + 0.65(2.4 − 1) = 1.91 K f s = 1 + 0.83(1.6 − 1) = 1.50 0.453 M = 2371 = 537 lbf · in 2

Eq. (18-22):

1/2 1 16 1.50(262) 2 1.91(537) 2 = +3 4 n π(1.5743 ) 18 700 37 500 = 0.277,

from which n = 3.61

Overhanging coupling keyway There is no bending moment, thus Eq. (18-22) reduces to: √ √ 16 3K f s Tm 16 3(1.50)(2626) 1 = = n πd 3 S y π(1.53 )(37 500) = 0.275

from which n = 3.64

Summary of safety factors: Location Pinion seat RH BRG shoulder Coupling keyway Bearing slope Mesh slope

nfs

Solution Source

2.76 3.61 3.64 2.30 <1.64

Prob. 18-5 Prob. 18-5 Prob. 18-5 Prob. 18-4 Prob. 18-4

18-6

If you have assigned Probs. 18-1 through 18-5, offered suggestions, and discussed their work, you have provided a good experience for them. Since they are performing adequacy assessments of individual designs for this problem, each will be different. Thus no solution can be offered here.

18-7

Preliminaries:

20 16 = 80 rev/min n c = 1200 80 60 50 T = 63 025 = 39 391 lbf · in 80 d + (20/60)d = 16 in, d = 24 in 2

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Wt =

39 391 T = = 3283 lbf d/2 12

W r = 3283 tan 25° = 1531 lbf W =

Wt 3283 = = 3622 lbf cos 25° cos 25°

Loads W, R1 and R2 are all in the same plane. 3622 8 = 2634 lbf R1 = 3622 3" 8" 11 R2 R1 3 R2 = 3622 = 988 lbf 11 417.5 Components 3283

895.4

8"

1113.5 3"

1531

2387.6

The steepest slope will be at the left bearing. Eq. (18-1): 32n d Fb(b2 − l 2 ) 1/4 d = 3π Elθall 32(2)(3622)(8)(82 − 112 ) 1/4 = 2.414 in = 3π(30)(106 )(11)(0.001) The design should proceed in the same manner as Probs. 18-1 to 18-6. 18-8

From the shaft forces in Probs. 18-1 and 18-7 solutions, we can construct the force portions of the free-body diagram for shaft b of Fig. P18-1. y 395 612 1531 2276

657

2" x

6" 1313

3" 3283

z 875

50 T = 63 025 240

= 13 130 lbf · in

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471

Chapter 18

The resulting forces at each location are not in the same plane; therefore, we must work in terms of components. Horizontal plane

Vertical plane y 1531 z

C

3"

H G

875

3283

657

2" D

C

z

x

3"

H G

2276

1313

D 2"

612

395

The slope at the left support is the greatest. 32(2) [1531(8)(82 − 112 ) − 1313(2)(22 − 112 )]2 d = 6 3π(30)(10 )(11)(0.001)

1/4 2 1/2

+ [3283(8)(8 − 11 ) − 612(2)(2 − 11 )] 2

2

2

2

= 2.320 in with d = 2.320, determine the slopes at C, G, H, and D. 1 [1531(8)(82 − 112 ) − 1313(2)(22 − 112 )]2 θC = 6(30)(106 )(π/64)(2.3204 )(11) 1/2 + [3283(8)(82 − 112 ) − 612(2)(22 − 112 )]2 = 0.000 500 rad (checks) 1 [1531(8)(3(32 ) + 82 − 112 ) θG = 6 4 6(30)(10 )(π/64)(2.320 )(11) − 1313(2)(3(32 ) + 22 − 112 )]2 + [3283(8)(3(32 ) + 82 − 112 ) 1/2 − 612(2)(3(32 ) + 22 − 112 )]2

= 0.000 245 rad θH =

1 6(30)(106 )(π/64)(2.3204 )(11)

= 0.000 298 rad

[1531(3)(112 − 3(22 ) − 32 ) − 1313(9)(112 − 3(22 ) − 92 )]2 + [3283(3)(112 − 3(22 ) − 32 ) 1/2 − 612(9)(112 − 3(22 ) − 92 )]2

1 [1531(3)(112 − 32 ) − 1313(9)(112 − 92 )]2 6 4 6(30)(10 )(π/64)(2.320 )(11) 1/2 + [3283(3)(112 − 32 ) − 612(9)(112 − 92 )]2 = 0.000 314 rad

θD =

The shaft diameter should be increased for the same reason given in Problem 18-4 (gear mesh slope). 2(0.000 298) 1/4 = 2.883 in dnew = 2.32 0.000 25

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Strength constraints MG = 3 8752 + 22762 = 7315 lbf · in M H = 2 6572 + 3952 = 1533 lbf · in Point G is more critical. Assume d = 2.88 in. Sut = 68 kpsi, S y = 37.5 kpsi kb = 0.91(2.88)−0.157 = 0.771 kc = kd = k f = 1

Similar to Prob. 8-2, Eq. (7-19):

and

ka = 0.883

For R = 0.995, Table A-10 provides z = 2.576. Eq. (7-28): Eq. (7-17): Eq. (7-31):

ke = 1 − 0.08(2.576) = 0.794 Se = 0.883(0.771)(1)(1)(0.794)(1)(0.504)(68) = 18.5 kpsi K f = 1.68, K f s = 2.25

Using DE-elliptic theory, Eq. (18-21)   2 2 1/2 1/3  16(2) 2.25(13 130) 1.68(7315) +3 d= 4   π 18 500 37 500 = 2.687 in O.K. Students will approach the design differently from this point on. 18-9

The bearing ensemble reliability is related to the six individual reliabilities by R = R1 R2 R3 R4 R5 R6 For an ensemble reliability of R,the individual reliability goals are Ri = R 1/6 The radial loads at bearings A through F were found to be A

B

C

D

E

263

1186

2438

767

2634

F 988 lbf

It may be useful to make the bearings at A, D, and F one size and those at B, C, and E another size, to minimize the number of different parts. In such a case R 1 = R B RC R E , R 2 = R A R D R F R = R1 R2 = ( R B RC R E )( R A R D R F ) where the reliabilities R A through R F are the reliabilities of Sec. 11-10, Eqs. (11-18), (11-19), (11-20), and (11-21). A corollary to the bearing reliability description exists and is given as R = Ra R b R c In this case you can begin with

Ri = R 1/3

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473

Chapter 18

18-10

This problem is not the same as Prob. 11-9, although the figure is the same. We have a design task of identifying bending moment and torsion diagrams which are preliminary to an industrial roller shaft design. y F yB F zB

z

F zC F yC

y

FC = 30(8) = 240 lbf FCz = 0.4(240) = 96 lbf T = FCz (2) = 96(2) = 192 lbf · in FBz =

192 T = = 128 lbf 1.5 1.5

y

FB = FBz tan 20° = 128 tan 20° = 46.6 lbf (a) xy-plane y FOy O

FAy 5.75"

11.5" C

46.6

A 2.75" B

x

240

y

M O = 240(5.75) − FA (11.5) − 46.6(14.25) = 0 y

FA =

240(5.75) − 46.6(14.25) = 62.3 lbf 11.5 y

M A = FO (11.5) − 46.6(2.75) − 240(5.75) = 0 y

FO =

240(5.75) + 46.6(2.75) = 131.1 lbf 11.5

Bending moment diagram Mxy (lbf • in) C

O

A B 128

754

x

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xz-plane 128

96 5.75" O

C

A

x

B

11.5"

F zO

2.75"

F zA

z

MO = 0 = 96(5.75) − FAz (11.5) + 128(14.25)

96(5.75) + 128(14.25) = 206.6 lbf 11.5 MA = 0 FAz =

= FOz (11.5) + 128(2.75) − 96(5.75) FOz =

96(5.75) − 128(2.75) = 17.4 lbf 11.5

Bending moment diagram: Mxz (lbf • in)

100 A B

O

C

x

352

1002 + (−754) 2 = 761 lbf · in M A = (−128) 2 + (−352) 2 = 375 lbf · in

MC =

This approach over-estimates the bending moment at C, torque at C but not at A. (b) xy-plane y 131.1

62.3

46.6

30 lbf/in

x

Mx y = −131.1x + 15x − 1.752 − 15x − 9.752 − 62.3x − 11.51 Mxy (lbf • in) O

1.75

C

A B

x

128

229 514

Mmax occurs at 6.12 in Mmax = −516 lbf · in MC = 131.1(5.75) − 15(5.75 − 1.75) 2 = 514

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475

Chapter 18

Reduced from 754 lbf · in. The maximum occurs at x = 6.12 in rather than C, but it is close enough. xz-plane x

128

12 lbf/in

x 17.4

206.6

Mx z = 17.4x − 6x − 1.752 + 6x − 9.752 + 206.6x − 11.51 Mxz (lbf • in) O

30.5

4.1

1.75

C

A B

x

352 Mnet (lbf • in)

516

Let Mnet = Mx2y + Mx2z Plot Mnet (x) 1.75 ≤ x ≤ 11.5 in Mmax = 516 lbf · in at x = 6.25 in

374 231 x

O

Torque: In both cases the torque rises from 0 to 192 lbf · in linearly across the roller and is steady until the coupling keyway is encountered; then it falls linearly to 0 across the key. Ans. 18-11

To size the shoulder for a rolling contact bearing at location A, the fillet has to be less than 1.0 mm (0.039 in). Choose r = 0.030 in. Given: n d = 2, K f = K f s = 2. From Prob. 18-10, . M = 375 lbf · in, Tm = 192 lbf · in r 0.030 in

A

Table A-20 for 1035 HR: Sut Se Eq. (7-8): kc ka Eq. (7-18):

= 72 kpsi, S y = 39.5 kpsi, = 0.504(72) = 36.3 kpsi = kd = ke = k f = 1 = 2.70(72)−0.265 = 0.869

H B = 143

Solve for the bearing seat diameter using Eq. (18-21) for the DE-elliptic criterion, d = 1 in Trial #1: 1 −0.107 = 0.879 kb = Eq. (7-19): 0.30 Eq. (7-17):

Se = 0.869(0.879)(36.3) = 27.7 kpsi   2 2 1/2 1/3  16(2) 2(192) 2(375) +3 d= 4   π 27 700 39 500 = 0.833 in

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d = 0.833 in 0.833 −0.107 = 0.896 kb = 0.30

Trial #2:

Se = 0.869(0.896)(36.3) = 28.3 kpsi   2 2 1/2 1/3  16(2) 2(192) 2(375) +3 d= 4  π  28 300 39 500 = 0.830 in d = 0.830 in 0.830 −0.107 kb = = 0.897 0.30

Trial #3:

Se = 0.869(0.897)(36.3) = 28.3 kpsi No further change in Se, therefore, d = 0.830 in Ans. The example is to show the nature of the strength-iterative process, with some simplification to reduce the effort. Clearly the stress concentration factors K t , K ts , K f , K f s and the shoulder diameter would normally be involved. 18-12

From Prob. 18-10, integrate Mx y and Mx z xy plane, with dy/dx = y E I y = − EIy = −

131.1 2 62.3 (x ) + 5x − 1.753 − 5x − 9.753 − x − 11.52 + C1 2 2

5 5 62.3 131.1 3 (x ) + x − 1.754 − x − 9.754 − x − 11.53 + C1 x + C2 6 4 4 6 y = 0 at x = 0 y = 0 at x = 11.5

From (1)

(1)

x = 0: x = 11.5:

⇒ ⇒

C2 = 0 C1 = 1908.4 lbf · in3

E I y = 1908.4 E I y = −2153.1

xz plane (treating z ↑ +) 17.4 2 206.6 (x ) − 2x − 1.753 + 2x − 9.753 + x − 11.52 + C3 E I z = (2) 2 2 17.4 3 1 1 206.6 EIz = (x ) − x − 1.754 + x − 9.754 + x − 11.53 + C3 x + C4 6 2 2 6 z = 0 at x = 0 z = 0 at x = 11.5 From (2)

x = 0: x = 11.5:

⇒ ⇒

C4 = 0 C3 = 8.975 lbf · in3

E I z = 8.975 E I z = −683.5

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477

Chapter 18

At O: A:

1908.42 + 8.9752 = 1908.4 lbf · in3 E I θ = (−2153.1)2 + (−683.5)2 = 2259 lbf · in3 EIθ =

With I = (π/64)d 4 ,

64 d= (2259) π Eθ

1/4

64(2259) = π(30)(106 )(0.001)

(dictates size) 1/4

= 1.113 in Ans. For θ = 0.0005 rad

d=

18-13

0.001 0.0005

1/4

(1.113) = 1.323 in Ans.

From Prob. 18-12, Eqs. (1) and (2), At B, x = 14.25" : E I y = −2328 lbf · in3 E I z = −1167 lbf · in3 E I θ = (−2328)2 + (−1167)2 = 2604 lbf · in3 1/4 64(2604) d= = 1.371 in Ans. π(30)(106 )(0.0005) With n d = 2

1/4 2 (1.371) = 1.630 in Ans. d= 1

Note: If 0.0005" is divided in the mesh to 0.000 25"/gear then for n d = 1, d = 1.630 in and for n d = 2, d = (2/1) 1/4 (1.630) = 1.938 in. 18-14

Similar to earlier design task; each design will differ.

18-15

Based on the results of Probs. 18-12 and 18-13, the shaft is marginal in deflections (slopes) at the bearings and gear mesh. In the previous edition of this book, numerical integration of general shape beams was used. In practice, finite elements is predominately used. If students have access to finite element software, have them model the shaft. If not, solve a simpler version of shaft. The 1" diameter sections will not affect the results much, so model the 1" diameter as 1.25" . Also, ignore the step in AB. y

C O z

A

x B

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The deflection equations developed in Prob. 18-12 still apply to section OCA. E I θ = 1908.4 lbf · in3 E I θ = 2259 lbf · in3 (still dictates)

O: A:

θ=

2259 = 0.000 628 rad 30(106 )(π/64)(1.254 )

n=

0.001 = 1.59 0.000 628

At gear mesh, B xy plane y 46.6 lbf

C A

x

O B

From Prob. 18-12, with I = I1 in section OCA, y A = −2153.1/E I1 Since y B/A is a cantilever, from Table A-9-1, with I = I2 in section AB y B/A = ∴ yB

F x(x − 2l) 46.6 = (2.75)[2.75 − 2(2.75)] = −176.2/E I2 2E I2 2E I2

= y A + y B/A =−

176.2 2153.1 − 30(106 )(π/64)(1.254 ) 30(106 )(π/64)(0.8754 )

= −0.000 803 rad

(magnitude greater than 0.0005 rad)

xz plane 128 lbf A

O

x

C B z

z A = − z B = − θB =

128(2.752 ) 484 683.5 =− , z B/A = − 2E I2 E I2 E I1 683.5 30(106 )(π/64)(1.254 )

−

484 30(106 )(π/64)(0.8754 )

= −0.000 751 rad

(−0.000 803) 2 + (0.000 751) 2 = 0.001 10 rad

Crowned teeth must be used. Finite element results: θ O = 5.47(10−4 ) rad θ A = 7.09(10−4 ) rad θ B = 1.10(10−3 ) rad

Error in simplified model 3.0% 11.4% 0.0%

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Chapter 18

479

The simplified model yielded reasonable results. Sut = 72 kpsi, S y = 39.5 kpsi

Strength

At the shoulder at A, x = 10.75 in. From Prob. 18-10, Mx y = −209.3 lbf · in, Mx z = −293.0 lbf · in, T = 192 lbf · in M = (−209.3) 2 + (−293) 2 = 360.0 lbf · in Se = 0.504(72) = 36.29 kpsi ka = 2.70(72) −0.265 = 0.869 −0.107 1 = 0.879 kb = 0.3 kc = kd = ke = k f = 1 Se = 0.869(0.879)(36.29) = 27.7 kpsi From Fig. A-15-8 with D/d = 1.25 and r/d = 0.03, K ts = 1.8. From Fig. A-15-9 with D/d = 1.25 and r/d = 0.03, K t = 2.3 From Fig. 7-20 with r = 0.03 in, q = 0.65. From Fig. 7-21 with r = 0.03 in, qs = 0.83 K f = 1 + 0.65(2.3 − 1) = 1.85 Eq. (7-31): K f s = 1 + 0.83(1.8 − 1) = 1.66 Using DE-elliptic, 2K f Ma 2(1.85)(360) = 2.413 r=√ =√ 3K f s Tm 3(1.66)(192) Sa =

2S y Se2 2(39.5)(27.72 ) = 26.0 kpsi = Se2 + S y2 (27.72 ) + (39.52 )

Sm = S y − Sa = 39.5 − 26.0 = 13.5 kpsi rcrit =

Sa 26 = 1.926 = Sm 13.5 r > rcrit

Therefore, the threat is fatigue. Eq. (18-22),

1/2 16 1.66(192) 2 1 1.85(360) 2 = +3 4 n π(13 ) 27 700 39 500 n = 3.92

Perform a similar analysis at the profile keyway under the gear. The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem.

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18-16

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Treat car, truck frame, and wheels as a free body: Tipping moment = 17 100(72) = 1.231(106 ) lbf · in Force of resisting couple, Fc = 1.231(106 )/59.5 = 20 690 lbf

42 750

Force supporting weight, Fw = 42 750/2 = 21 375 lbf

17 100

Fr = 17 100 lbf R1 = Fw + Fc = 21 375 + 20 690 = 42 065 lbf R2 = Fw − Fc = 21 375 − 20 690 = 685 lbf

72" Fr Fw

Fw

Fc

Fc

For the car and truck as a free body: Tipping moment = 17 100(72 − 16.5) = 949 050 lbf · in Gc =

949 050 = 11 863 lbf 80

42 750 17 100

The force at the journal is G w = 42 750/2 = 21 375 lbf G r = 17 100 lbf R1 = 21 375 + 11 863 = 33 238 lbf R2 = 21 375 − 11 863 = 9512 lbf Wheels and axle as a free body

332

33 238

72 16.5 Gr Gw

Gw

Gc

Gc

9512 17 100

17 100 42 065

685

Axle as a free body: Couple due to flange force = (17 100)(33/2) = 282 150 lbf · in

33 238 9512 282 150 lbf • in 17 100 17 100 42 065

685

M (lbf • in)

97 498 340 690 622 840

Midspan moment: M = 33 238(40) + 282 150 − 42 065(59.5/2) = 360 240 lbf · in Since the curvature and wind loads can be from opposite directions, the axle must resist 622 840 lbf · in at either wheel seat and resist 360 240 lbf · in in the center. The bearing load could be 33 238 lbf at the other bearing. The tapered axle is a consequence of this. Brake forces are neglected because they are small and induce a moment on the perpendicular plane.

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Chapter 18

18-17

481

Some information is brought forward from the solution of Prob. 18-16. At the wheel seat σ =

32M 32(622 840) = 18 496 psi = πd 3 π(73 )

At mid-axle σ =

32(360 240) = 23 630 psi π(5.375) 3

The stress at the wheel seat consists of the bending stress plus the shrink fit compression combining for a higher von Mises stress. 18-18

This problem has to be done by successive trials, since Se is a function of shaft size in Eq. (18-21). The material is SAE 2340 for which Sut = 1226 MPa, S y = 1130 MPa, and H B ≥ 368. ka = 4.51(1226) −0.265 = 0.685

Eq. (7-18):

Trial #1: Choose dr = 22 mm 22 −0.107 = 0.893 kb = 7.62 Se = 0.685(0.893)(0.504)(1226) = 378 MPa dr = d − 2r = 0.75D − 2D/20 = 0.65D D=

22 dr = = 33.8 mm 0.65 0.65

r=

D 33.8 = = 1.69 mm 20 20

Fig. A-15-14: d = dr + 2r = 22 + 2(1.69) = 25.4 mm d 25.4 = 1.15 = dr 22 r 1.69 = 0.077 = dr 22 K t = 1.9 Fig. A-15-15:

K ts = 1.5

Fig. 7-20:

r = 1.69 mm, q = 0.90

Fig. 7-21:

r = 1.69 mm, qs = 0.97

Eq. (7-31):

K f = 1 + 0.90(1.9 − 1) = 1.81 K f s = 1 + 0.97(1.5 − 1) = 1.49

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Eq. (18-21) with d as dr ,  1/2 1/3   16(2.5) 3 2 3 2 1.49(45)(10 ) 1.81(70)(10 ) +3 dr = 4   π 378 1130 = 20.5 mm Trial #2: Choose dr = 20.5 mm 20.5 −0.107 = 0.900 kb = 7.62 Se = 0.685(0.900)(0.504)(1226) = 381 MPa D=

20.5 dr = = 31.5 mm 0.65 0.65

r=

31.5 D = = 1.58 mm 20 20

Figs. A-15-14 and A-15-15: d = dr + 2r = 20.5 + 2(1.58) = 23.7 mm 23.7 d = 1.16 = dr 20.5 r 1.58 = 0.077 = dr 20.5 We are at the limit of readability of the figures so K t = 1.9, ∴ K f = 1.81

K ts = 1.5 q = 0.9, K f s = 1.49

qs = 0.97

Using Eq. (18-21) produces no changes. Therefore we are done. Decisions: dr = 20.5 D= Use D = 32 mm, 18-19

20.5 = 31.5 mm, 0.65

d = 24 mm,

d = 0.75(31.5) = 23.6 mm

r = 1.6 mm Ans.

Refer to Prob. 18-18. Trial #1, n d = 2.5, dr = 22 mm, Se = 378 MPa, K f = 1.81, K f s = 1.49 Eq. (18-30)

 1/2 1/3  32(2.5)  3 2 3 2 70(10 ) 45(10 ) 1.81 + 1.49 dr =  π  (378) (1130) = 20.5 mm

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483

Chapter 18

Referring to Trial #2 of Prob. 18-18, dr = 20.5 mm and Se = 381 MPa. Substitution into Eq. (18-30) yields dr = 20.5 mm again. Solution is the same as Prob. 18-18; therefore use D = 32 mm, F cos 20°(d/2) = T,

18-20

d = 24 mm,

r = 1.6 mm

Ans.

F = 2T /(d cos 20°) = 2(3000)/(6 cos 20°) = 1064 lbf MC = 1064(4) = 4257 lbf · in

(a) Static analysis using fatigue stress concentration factors and Eq. (6-45): 1/3 16n 2 2 1/2 4(K f M) + 3(K f s T ) d= π Sy 1/3 16(2.5) 2 2 1/2 4(1.8)(4257) + 3(1.3)(3000) = π(60 000) = 1.526 in

Ans.

ka = 2.70(80) −0.265 = 0.845

(b) Assume d = 2.00 in

kb =

2 0.3

−0.107

= 0.816

Se = 0.845(0.816)(0.504)(80) = 27.8 kpsi (1) DE-Gerber, Eq. (18-16):    2 1/2 1/3  16(2.5)(1.8)(4257) 1 + 1 + 3 1.3(3000)(27 800)  d=   π(27 800) 1.8(4257)(80 000) = 1.929 in Revising kb results in d = 1.927 in

Ans.

(2) DE-elliptic using d = 2 in for Se with Eq. (18-21),   2 2 1/2 1/3  16(2.5) 1.3(3000) 1.8(4257) +3 = 1.927 in d= 4   π 27 800 60 000 Revising kb results in d = 1.926 in

Ans.

(3) Soderberg, Eq. (18-30):   2 2 1/2 1/3  32(2.5) 4257 3000 1.8 + 1.3 = 1.932 in d=   π 27 800 60 000 Revising kb results in d = 1.930 in

Ans.

(4) DE-Goodman, Eq. (18-34): 1/3 √ 1.3(3000) 1.8(4257) 16(2.5) 2 + 3 = 2.008 in Ans. d= π 27 800 80 000

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18-21

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Page 484


(a) DE-Gerber, Eqs. (18-13) and (18-15): 1/2 = 2920 lbf · in A = 4[2.2(600)]2 + 3[1.8(400)]2 1 / 2 = 2391 lbf · in B = 4[2.2(500)]2 + 3[1.8(300)]2    2 1/2 1/3  8(2)(2920) 1 + 1 + 2(2391)(30 000)  d=   π(30 000) 2920(100 000) = 1.016 in Ans. (b) DE-elliptic, Equation prior to Eq. (18-19): 1/3 B2 16n A2 + 2 d= π Se2 Sy   2 2 1/3 16(2) 2920 2391  = + = 1.012 in π 30 000 80 000

Ans.

(c) MSS-Soderberg, Eq. (18-28):   2 2 1/2 1/3  32(2) 600 400 500 300 d= + + + 1.82 2.22  π  80 000 30 000 80 000 30 000 = 1.101 in

Ans.

(d) DE-Goodman: Eq. (18-32) can be shown to be 1/3 B 16n A d= + π Se Sut 1/3 2391 16(2) 2920 + = = 1.073 in π 30 000 100 000

18-22

Criterion

d(in)

DE-Gerber DE-elliptic MSS-Soderberg DE-Goodman

1.016 1.012 1.101 1.073

Compared to DE-Gerber 0.4% lower 8.4% higher 5.6% higher

less conservative more conservative more conservative

We must not let the basis of the stress concentration factor, as presented, impose a viewpoint on the designer. Table A-16 shows K ts as a decreasing monotonic as a function of a/D. All is not what it seems. Let us change the basis for data presentation to the full section rather than the net section. τ = K ts τ0 = K ts τ0 32T 32T K ts = = K ts π AD 3 π D3

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485

Chapter 18

Therefore K ts A

K ts = Form a table: (a/D)

A

K ts

K ts

0.050 0.075 0.100 0.125 0.150 0.175 0.200

0.95 0.93 0.92 0.89 0.87 0.85 0.83

1.77 1.71 1.68 1.64 1.62 1.60 1.58

1.86 1.84 1.83 ← minimum 1.84 1.86 1.88 1.90

K ts has the following attributes: • • • •

It exhibits a minimum; It changes little over a wide range; . Its minimum is a stationary point minimum at a/D = 0.100; Our knowledge of the minima location is 0.075 ≤ (a/D) ≤ 0.125

We can form a design rule: in torsion, the pin diameter should be about 1/10 of the shaft diameter, for greatest shaft capacity. However, it is not catastrophic if one forgets the rule. 18-23

Preliminaries: 63 025(4.5) = 2532 lbf · in 112 2532 T Wt = = = 633 lbf r 8/2 633 W = = 674 lbf cos 20° T =

674 lbf 7" R1

3" 10"

R2

3 R1 = 674 = 202 lbf 10 7 = 472 lbf R2 = 674 10

Mmax = 202(7) = 1414 lbf · in

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Is the task strength-controlled or distortion-controlled? With regards to distortion use n = 2 in Eq. (18-1): 32(2)(674)(3)(32 − 102 ) 1/4 = 1.43 in d L = 3π(30)(106 )(10)(0.001) Eq. (18-2):

32(2)(674)(7)(102 − 72 ) 1/4 = 1.53 in dR = 3π(30)(106 )(10)(0.001)

For the gearset, use θ AB developed in Prob. 18-4 solution. To the left of the load, θ AB =

Fb (3x 2 + b2 − l 2 ) 6E I l

Incorporating I = πd 4 /64 and n d , 1/4 32n d Fb 2 2 2 (3x + b − l ) d = 3π Elθall At the gearset, x = 7 with θall = 0.000 25 (apportioning gearmesh slope equally). 32(2)(674)(3)[3(72 ) + 32 − 102 ] 1/4 ≤ 1.789 in d≤ 3π(30)(106 )(10)(0.000 25) Since 1.789 > 1.75 in, angular deflection of the matching gear should be less than 0.000 25 rad. Crowned gears should thus be used, or n d scutinized for reduction. . . . . Concerning strength: For K f = 2, K f s = 1.5, Se = Sut /4, S y = Sut /2 in Eq. (18-1): 2 2 1/2 16n 1.5(2532) 2(1414) d3 = +3 4 π Sut /4 Sut /2 Solving for Sut and using n = 2 and d = 1.75 in Sut =

16(2) {4[4(2)(1414)]2 + 3[2(1.5)(2532)]2 }1/2 = 49 740 psi 3 π(1.75 )

This gives an approximate idea of the shaft material strength necessary and helps identify an initial material. With this perspective students can begin. 18-24

This task is a change of pace. Let s be the scale factor of the model, and subscript m denote ‘model.’ lm = sl σ = Mm =

Mc , I

σm = σ

σm Im σ s4 I = s3 M = cm sc

Ans.

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Chapter 18

The load that causes bending is related to reaction and distance. M m = R m am =

Fm bm am lm

Solving for Fm gives Mm l m s 3 M(sl) = s2 F Fm = = am bm (sa)(sb)

Ans.

For deflection use Table A-9-6 for section AB, ym = =

" Fm bm xm ! 2 2 − lm2 xm + bm 6E m Im lm (s 2 F)(sb)(sx) 2 2 (s x + s 2 b2 − s 2l 2 ) 6E(s 4 I )(sl)

= sy

Ans.

(as expected)

For section BC, the same is expected. For slope, consider section AB y AB = θ AB = θm =

Fb (3x 2 + b2 − l 2 ) 6E I l

s 2 F(sb) (3s 2 x 2 + s 2 b2 − s 2l 2 ) = θ 6E(s 4 I )(sl)

The same will apply to section BC Summary: Slope:

ym = y

Deflection:

ym = sy =

y 2

Moment:

Mm = s 3 M =

M 8

Force:

Fm = s 2 F =

F 4

These relations are applicable for identical materials and stress levels. 18-25

If you have a finite element program available, it is highly recommended. Beam deflection programs can be implemented but this is time consuming and the programs have narrow applications. Here we will demonstrate how the problem can be simplified and solved using singularity functions. Deflection: First we will ignore the steps near the bearings where the bending moments are low. Thus let the 30 mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10 mm. The full bending stresses will not develop at the outer fibers so full stiffness will not develop either. Thus, ignore this step and let the diameter be 45 mm.

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Statics: Left support: R1 = 7(315 − 140)/315 = 3.889 kN Right support: R2 = 7(140)/315 = 3.111 kN Determine the bending moment at each step. x(mm) M(N · m)

0 0

40 155.56

100 388.89

140 544.44

210 326.67

275 124.44

315 0

I35 = (π/64)(0.0354 ) = 7.366(10−8 ) m4 , I40 = 1.257(10−7 ) m4 , I45 = 2.013(10−7 ) m4 Plot M/I as a function of x. x(m)

M/I (109 N/m3 )

0 0.04 0.04 0.1 0.1 0.14 0.14 0.21 0.21 0.275 0.275 0.315

0 2.112 1.2375 3.094 1.932 2.705 2.705 1.623 2.6 0.99 1.6894 0

Step

Slope

Slope

52.8 −0.8745

30.942

−21.86

−1.162

19.325

−11.617

0

−15.457

0.977

−24.769

0.6994

−42.235

−34.78 −9.312 −17.47

4 MI (109 Nm3)

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3 2 1 0

0

0.05

0.1

0.15

0.2 x (mm)

0.25

0.3

0.35

The steps and the change of slopes are evaluated in the table. From these, the function M/I can be generated: M/I = 52.8x − 0.8745x − 0.040 − 21.86x − 0.041 − 1.162x − 0.10 − 11.617x − 0.11 − 34.78x − 0.141 + 0.977x − 0.210 − 9.312x − 0.211 + 0.6994x − 0.2750 − 17.47x − 0.2751 109 Integrate twice: dy = 26.4x 2 − 0.8745x − 0.041 − 10.93x − 0.042 − 1.162x − 0.11 E dx − 5.81x − 0.12 − 17.39x − 0.142 + 0.977x − 0.211 − 4.655x − 0.212 + 0.6994x − 0.2751 − 8.735x − 0.2752 + C1 109 (1) E y = 8.8x 3 − 0.4373x − 0.042 − 3.643x − 0.043 − 0.581x − 0.12 − 1.937x − 0.13 − 5.797x − 0.143 + 0.4885x − 0.212 − 1.552x − 0.213 + 0.3497x − 0.2752 − 2.912x − 0.2753 + C1 x + C2 109

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Chapter 18

Boundary conditions: y = 0 at x = 0 yields C2 = 0; y = 0 at x = 0.315 m yields C1 = −0.295 25 N/m2 . Equation (1) with C1 = −0.295 25 provides the slopes at the bearings and gear. The following table gives the results in the second column. The third column gives the results from a similar finite element model. The fourth column gives the result of a full model which models the 35 and 55 mm diameter steps. x (mm)

θ (rad) −0.001 4260 −0.000 1466 0.001 3120

0 140 315

F.E. Model

Full F.E. Model

−0.001 4270 −0.000 1467 0.001 3280

−0.001 4160 −0.000 1646 0.001 3150

The main discrepancy between the results is at the gear location (x = 140 mm) . The larger value in the full model is caused by the stiffer 55 mm diameter step. As was stated earlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere between the full model and the simplified model which in any event is a small value. As expected, modeling the 30 mm dia. as 35 mm does not affect the results much. It can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load has to be reduced or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the maximum load should be Fmax = (0.001/0.001 46)7 = 4.79 kN . With a design factor this would be reduced further. To increase the stiffness of the shaft, increase the diameters by (0.001 46/0.001) 1/4 = 1.097. Form a table: Old d, mm New ideal d, mm Rounded up d, mm

20.00 21.95 22.00

30.00 32.92 34.00

35.00 38.41 40.00

40.00 43.89 44.00

45.00 49.38 50.00

55.00 60.35 62.00

Repeating the full finite element model results in x = 0: x = 140 mm: x = 315 mm:

θ = −9.30 × 10−4 rad θ = −1.09 × 10−4 rad θ = 8.65 × 10−4 rad

Well within our goal. Have the students try a goal of 0.0005 rad at the bearings. Strength: Due to stress concentrations and reduced shaft diameters, there are a number of locations to look at. A table of nominal stresses is given below. Note that torsion is only to the right of the 7 kN load. Using σ = 32M/(πd 3 ) and τ = 16T /(πd 3 ), x (mm) σ (MPa) τ (MPa) σ (MPa)

0 0 0 0

15 22.0 0 22.0

40 37.0 0 37.0

100 61.9 0 61.9

110 47.8 0 47.8

140 60.9 6 61.8

Table A-20 for AISI 1020 CD steel: Sut = 470 MPa,

210 52.0 8.5 53.1

275 39.6 12.7 45.3

S y = 390 MPa

300 17.6 20.2 39.2

330 0 68.1 118.0

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At x = 210 mm: ka = 4.51(470) −0.265 = 0.883, kb = (40/7.62) −0.107 = 0.837 Se = 0.883(0.837)(0.504)(470) = 175 MPa D/d = 45/40 = 1.125, r/d = 2/40 = 0.05. From Figs. A-15-8 and A-15-9, K t = 1.9 and K ts = 1.32. From Figs. 7-20 and 7-21, q = 0.75 and qs = 0.92, K f = 1 + 0.75(1.9 − 1) = 1.68, and K f s = 1 + 0.92(1.32 − 1) = 1.29. From Eq. (18-22),

1/2 16 1 1.29(107) 2 1.68(544.44) 2 = +3 4 n π(0.04)3 175(106 ) 390(106 ) n = 1.20

Depending on the application, this may be too low. At x = 330 mm: The von Mises stress is the highest but it comes from the steady torque only. D/d = 30/20 = 1.5, r/d = 2/20 = 0.1 ⇒ K ts = 1.42, qs = 0.92 ⇒ !√ " 1.39(107) 16 1 = 3 n π(0.02) 3 390(106 )

K f s = 1.39

n = 2.38 Check the other locations. If worse-case is at x = 210 mm, the changes discussed for the slope criterion will improve the strength issue. 18-26

In Eq. (18-37) set I = to obtain

πd 4 , 64

A=

πd 2 4

2 gE d π ω= l 4 γ

or 4l 2 ω d= 2 π

#

γ gE

(a) From Eq. (1) and Table A-5, 2 # 1 386(30)(106 ) π = 868 rad/s Ans. ω= 24 4 0.282 (b) From Eq. (2), 4(24) 2 (2)(868) 0.282 d= = 2 in Ans. π2 386(30)(106 )

(1)

(2)

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Chapter 18

(c) From Eq. (2), lω =

π d 4 l 2

gE γ

Since d/l is the same regardless of the scale. lω = constant = 24(868) = 20 832 ω=

20 832 = 1736 rad/s 12

Ans.

Thus the first critical speed doubles. 18-27

From Prob. 18-26, ω = 868 rad/s A = 0.7854 in2 , I = 0.04909 in4 , γ = 0.282 lbf/in3 , E = 30(106 ) psi, w = Aγ l = 0.7854(0.282)(24) = 5.316 lbf One element: δ11 =

Eq. (18-37)

12(12)(242 − 122 − 122 ) = 1.956(10−4 ) in/lbf 6 6(30)(10 )(0.049 09)(24)

y1 = w1 δ11 = 5.316(1.956)(10−4 ) = 1.0398(10−3 ) in

y12 = 1.0812(10−6 ) wy = 5.316(1.0398)(10−3 ) = 5.528(10−3 )

wy 2 = 5.316(1.0812)(10−6 ) = 5.748(10−6 ) $ wy 5.528(10−3 ) ω1 = g $ 2 = 386 = 609 rad/s (30% low) wy 5.748(10−6 )

Two elements: 2.658 lbf 6"

2.658 lbf 6"

6"

6"

δ11 = δ22 =

18(6)(242 − 182 − 62 ) = 1.100(10−4 ) in/lbf 6 6(30)(10 )(0.049 09)(24)

δ12 = δ21 =

6(6)(242 − 62 − 62 ) = 8.556(10−5 ) in/lbf 6 6(30)(10 )(0.049 09)(24)

y1 = w1 δ11 + w2 δ12 = 2.658(1.100)(10−4 ) + 2.658(8.556)(10−5 ) = 5.198(10−4 ) in = y2 , y12 = y22 = 2.702(10−7 ) in2 wy = 2(2.658)(5.198)(10−4 ) = 2.763(10−3 )

wy 2 = 2(2.658)(2.702)(10−7 ) = 1.436(10−6 ) 2.763(10−3 ) ω1 = 386 = 862 rad/s (0.7% low) 1.436(10−6 )

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Three elements: 1.772 lbf

1.772 lbf

4"

8"

δ11 = δ33 = δ22 =

4"

20(4)(242 − 202 − 42 ) = 6.036(10−5 ) in/lbf 6 6(30)(10 )(0.049 09)(24)

12(12)(242 − 122 − 122 ) = 1.956(10−4 ) in/lbf 6 6(30)(10 )(0.049 09)(24)

δ12 = δ32 = δ13 =

1.772 lbf 8"

12(4)(242 − 122 − 42 ) = 9.416(10−5 ) in/lbf 6 6(30)(10 )(0.049 09)(24)

4(4)(242 − 42 − 42 ) = 4.104(10−5 ) in/lbf 6(30)(106 )(0.049 09)(24)

y1 = 1.772[6.036(10−5 ) + 9.416(10−5 ) + 4.104(10−5 )] = 3.465(10−4 ) in y2 = 1.772[9.416(10−5 ) + 1.956(10−4 ) + 9.416(10−5 )] = 6.803(10−4 ) in y3 = 1.772[4.104(10−5 ) + 9.416(10−5 ) + 6.036(10−5 )] = 3.465(10−4 ) in wy 2 = 1.246(10−6 ) wy = 2.433(10−3 ), 2.433(10−3 ) = 868 rad/s ω1 = 386 (same as in Prob. 18-26) 1.246(10−6 ) The point was to show that convergence is rapid using a static deflection beam equation. The method works because: • If a deflection curve is chosen which meets the boundary conditions of moment-free and deflection-free ends, and in this problem, of symmetry, the strain energy is not very sensitive to the equation used. • Since the static bending equation is available, and meets the moment-free and deflectionfree ends, it works. 18-28

(a) For two bodies, Eq. (18-39) is (m 1 δ11 − 1/ω2 ) m 1 δ21

m 2 δ12 =0 (m 2 δ22 − 1/ω2 )

Expanding the determinant yields, 2 1 1 + m 1 m 2 (δ11 δ22 − δ12 δ21 ) = 0 − (m 1 δ11 + m 2 δ22 ) 2 ω ω12 Eq. (1) has two roots 1/ω12 and 1/ω22 . Thus

1 1 − ω2 ω12

1 1 − ω2 ω22

=0

(1)

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Chapter 18

or,

1 ω2

2

1 1 + + 2 2 ω1 ω2

2 1 1 1 =0 + 2 ω ω1 ω22

(2)

Equate the third terms of Eqs. (1) and (2), which must be identical. 1 1 = m 1 m 2 (δ11 δ22 − δ12 δ21 ) ω12 ω22 and it follows that

⇒

1 = ω12 m 1 m 2 (δ11 δ22 − δ12 δ21 ) 2 ω2

ω2 =

1 ω1

g2 w1 w2 (δ11 δ22 − δ12 δ21 )

Ans.

(b) In Ex. 18-5, Part (b) the first critical speed of the two-disk shaft (w1 = 35 lbf, w2 = 55 lbf) is ω1 = 124.7 rad/s. From part (a), using influence coefficients 1 3862 = 466 rad/s Ans. ω2 = 124.7 35(55)[2.061(3.534) − 2.2342 ](10−8 ) 18-29

In Eq. (18-35) the term I /A appears. For a hollow unform diameter shaft, ! ! # " "! " 4 4 /64 π d − d 1 2 I 1 do2 + di2 do2 − di2 o i " ! = = = d + di2 A 16 4 o do2 − di2 π do2 − di2 /4 This means that when a solid shaft is hollowed out, the critical speed increases beyond that of the solid shaft. By how much? 2 1 2 + d2 d o di i 4 = 1+ 1 2 do 4 do The possible values of di are 0 ≤ di ≤ do , so the range of critical speeds is √ √ ωs 1 + 0 to about ωs 1 + 1 √ or from ωs to 2ωs . Ans.

18-30

All steps will be modeled using singularity functions with a spreadsheet (see next page). Programming both loads will enable the user to first set the left load to 1, the right load to 0 and calculate δ11 and δ21 . Then setting left load to 0 and the right to 1 to get δ12 and δ22 . The spreadsheet shown on the next page shows the δ11 and δ21 calculation. Table for M/I vs x is easy to make. The equation for M/I is: M/I = D13x + C15x − 10 + E15x − 11 + E17x − 21 + C19x − 90 + E19x − 91 + E21x − 141 + C23x − 150 + E23x − 151

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Integrating twice gives the equation for E y. Boundary conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C2 = 0). Substitution back into the deflection equation at x = 2, 14 inches provides the δ ’s. The results are: δ11 − 2.917(10−7 ), δ12 = δ21 = 1.627(10−7 ), δ22 = 2.231(10−7 ). This can be verified by finite element analysis. y1 = 20(2.917)(10−7 ) + 35(1.627)(10−7 ) = 1.153(10−5 ) y2 = 20(1.627)(10−7 ) + 35(2.231)(10−7 ) = 1.106(10−5 )

y12 = 1.329(10−10 ),

y22 = 1.224(10−10 ) wy 2 = 6.942(10−9 )

wy = 6.177(10−4 ),

Neglecting the shaft, Eq. (18-36) gives ω1 =

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

6.177(10−4 ) = 5860 rad/s or 55 970 rev/min Ans. 386 6.942(10−9 )

B

C

F1 = 1

D

E

F2 = 0

x 0 1 2 9 14 15 16

0 0.875 1.75 0.875 0.25 0.125 0

x 0 1 1 2 2 9 9 14 14 15 15 16

0 1.114 082 0.477 36 0.954 719 0.954 719 0.477 36 0.305 837 0.087 382 0.087 382 0.043 691 0.159 155 0

step

M/I

−0.636 722 477 0 −0.171 522 4

G

H

R1 = 0.875 (left reaction) I1 = I4 = 0.7854 I2 = 1.833 I3 = 2.861

M

F

slope 1.114 082

slope

0.477 36

−0.636 72

−0.068 19

−0.545 55

−0.043 69

0.024 503

0

−0.043 69

0

0.115 463 554

−0.159 15

−0.115 46

I

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495

Chapter 18

A 26 27 28 29 30

B

C

D

E

F

G

H

I

C1 = −4.906 001 093 δ11 = 2.91701E-07 δ21 = 1.6266E-07 1.2 1 M (lbf • in)

shi20396_ch18.qxd

0.8 0.6 0.4 0.2 0

0

2

4

6

8 x (in)

10

12

14

16

Repeat for F1 = 0 and F2 = 1. Modeling the shaft separately using 2 elements gives approximately 11.32 lbf 4.5"

11 lbf 4.5"

R1

3.5"

3.5"

9"

R2

The spreadsheet can be easily modified to give δ11 = 9.605(10−7 ),

δ12 = δ21 = 5.718(10−7 ),

δ22 = 5.472(10−7 )

y1 = 1.716(10−5 ),

y2 = 1.249(10−5 ), y12 = 2.946(10−10 ), y22 = 1.561(10−10 ), wy = 3.316(10−4 ), wy 2 = 5.052(10−9 ) 3.316(10−4 ) = 5034 rad/s Ans. ω1 = 386 5.052(10−9 ) A finite element model of the exact shaft gives ω1 = 5340 rad/s. The simple model is 5.7% low. Combination Using Dunkerley’s equation, Eq. (18-45): 1 1 1 = + 2 58602 50342 ω1

⇒

3819 rad/s Ans.

18-31 and 18-32 With these design tasks each student will travel different paths and almost all details will differ. The important points are • The student gets a blank piece of paper, a statement of function, and some constraints–explicit and implied. At this point in the course, this is a good experience. • It is a good preparation for the capstone design course.

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• The adequacy of their design must be demonstrated and possibly include a designer’s notebook. • Many of the fundaments of the course, based on this text and this course, are useful. The student will find them useful and notice that he/she is doing it. • Don’t let the students create a time sink for themselves. Tell them how far you want them to go. 18-33

I used this task as a final exam when all of the students in the course had consistent test scores going into the final examination; it was my expectation that they would not change things much by taking the examination. This problem is a learning experience. Following the task statement, the following guidance was added. • Take the first half hour, resisting the temptation of putting pencil to paper, and decide what the problem really is. • Take another twenty minutes to list several possible remedies. • Pick one, and show your instructor how you would implement it. The students’ initial reaction is that he/she does not know much from the problem statement. Then, slowly the realization sets in that they do know some important things that the designer did not. They knew how it failed, where it failed, and that the design wasn’t good enough; it was close, though. Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and the problem may not be solved. To many students’s credit, they chose to keep the shaft geometry, and selected a new material to realize about twice the Brinell hardness.

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A

Appendix

Useful Tables Appendix Outline A–1

Standard SI Prefixes

A–2

Conversion Factors

A–3

Optional SI Units for Bending, Torsion, Axial, and Direct Shear Stresses

A–4

Optional SI Units for Bending and Torsional Deflections

A–5

Physical Constants of Materials

A–6

Properties of Structural-Steel Angles

A–7

Properties of Structural-Steel Channels

A–8

Properties of Round Tubing

A–9

Shear, Moment, and Deflection of Beams

961 962 963

963

963 964 966

968 969

A–10

Cumulative Distribution Function of Normal (Gaussian) Distribution

A–11

A Selection of International Tolerance Grades—Metric Series

A–12

Fundamental Deviations for Shafts—Metric Series

A–13

A Selection of International Tolerance Grades—Inch Series

A–14

Fundamental Deviations for Shafts—Inch Series

A–15

Charts of Theoretical Stress-Concentration Factors K t

A–16

Approximate Stress-Concentration Factors K t and K ts for Bending a Round Bar or Tube with a Transverse Round Hole 987

A–17

Preferred Sizes and Renard (R-series) Numbers

A–18

Geometric Properties

A–19

American Standard Pipe

A–20

Deterministic ASTM Minimum Tensile and Yield Strengths for HR and CD Steels 994

A–21

Mean Mechanical Properties of Some Heat-Treated Steels

A–22

Results of Tensile Tests of Some Metals

A–23

Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels

A–24

Mechanical Properties of Three Non-Steel Metals

A–25

Stochastic Yield and Ultimate Strengths for Selected Materials

A–26

Stochastic Parameters from Finite Life Fatigue Tests in Selected Metals

977

978

979 980

981 982

989

990 993

995

997 998

1000 1002 1003 959

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Mechanical Engineering Design

A–27

Finite Life Fatigue Strengths of Selected Plain Carbon Steels

A–28

Decimal Equivalents of Wire and Sheet-Metal Gauges

A–29

Dimensions of Square and Hexagonal Bolts

A–30

Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws

A–31

Dimensions of Hexagonal Nuts

A–32

Basic Dimensions of American Standard Plain Washers

A–33

Dimensions of Metric Plain Washers

A–34

Gamma Function

1012

1004

1005

1007

1009

1011

1010

1008

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Useful Tables

Table A–1

Name ∗†

Standard SI Prefixes

Symbol

Factor

exa

E

1 000 000 000 000 000 000 = 1018

peta

P

1 000 000 000 000 000 = 1015

tera

T

1 000 000 000 000 = 1012

giga

G

1 000 000 000 = 109

mega

M

1 000 000 = 106

kilo

k

1 000 = 103

hecto‡

h

100 = 102

deka

da

10 = 101

deci‡

d

0.1 = 10−1

centi

c

0.01 = 10−2

milli

m

0.001 = 10−3

micro

µ

0.000 001 = 10−6

nano

n

0.000 000 001 = 10−9

pico

p

0.000 000 000 001 = 10−12

femto

f

0.000 000 000 000 001 = 10−15

atto

a

0.000 000 000 000 000 001 = 10−18

‡

‡

∗ If possible use multiple and submultiple prefixes in steps of 1000. † Spaces are used in SI instead of commas to group numbers to avoid confusion with the practice in some European countries

of using commas for decimal points. ‡ Not recommended but sometimes encountered.

961

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Table A–2 Conversion Factors A to Convert Input X to Output Y Using the Formula Y = AX ∗ Multiply Input X

By Factor A

To Get Output Y

Multiply Input X

By Factor A

To Get Output Y

British thermal unit, Btu

1055

joule, J

mile/hour, mi/h

1.61

kilometer/hour, km/h

mile/hour, mi/h

0.447

meter/second, m/s

moment of inertia, lbm ·ft2

0.0421

kilogram-meter2, kg · m2

Btu/second, Btu/s

1.05

kilowatt, kW

calorie

4.19

joule, J

centimeter of mercury (0◦ C)

1.333

kilopascal, kPa

moment of inertia, lbm · in2

centipoise, cP

0.001

pascal-second, Pa · s

degree (angle)

0.0174

radian, rad

moment of section (second moment of area), in4

foot, ft

0.305

meter, m

ounce-force, oz

0.278

newton, N

foot2, ft2

0.0929

meter2, m2

ounce-mass

0.0311

kilogram, kg

4.45

newton, N

1.36

newton-meter, N·m

foot/minute, ft/min

0.0051

meter/second, m/s

pound, lbf †

foot-pound, ft · lbf

1.35

joule, J

pound-foot, lbf · ft

foot-pound/ second, ft · lbf/s

1.35

watt, W

foot/second, ft/s

0.305

meter/second, m/s

gallon (U.S.), gal

3.785

liter, L

horsepower, hp

0.746

kilowatt, kW

pound/inch, lbf/in

inch, in

0.0254

meter, m

pound/inch , psi (lbf/in2)

inch, in

25.4

inch2, in2

645

millimeter, mm millimeter2, mm2

inch of mercury (32◦ F)

3.386

kilopascal, kPa

kilopound, kip

4.45

kilonewton, kN

kilopound/inch2, kpsi (ksi)

6.89

megapascal, MPa (N/mm2)

2

mass, lbf · s /in mile, mi

175 1.610

kilogram, kg kilometer, km

pound/foot2, lbf/ft2

293 41.6

47.9

kilogram-millimeter2, kg · mm2 centimeter4, cm4

pascal, Pa

pound-inch, lbf · in

0.113

joule, J

pound-inch, lbf · in

0.113

newton-meter, N·m

175

2

6.89

newton/meter, N/m kilopascal, kPa

pound-mass, lbm

0.454

kilogram, kg

pound-mass/ second, lbm/s

0.454

kilogram/second, kg/s

quart (U.S. liquid), qt 946

milliliter, mL

section modulus, in3

16.4

centimeter3, cm3

slug

14.6

kilogram, kg

ton (short 2000 lbm) 907 yard, yd 0.914

∗ Approximate. † The U.S. Customary system unit of the pound-force is often abbreviated as lbf to distinguish it from the pound-mass, which is abbreviated as lbm.

kilogram, kg meter, m

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963

Useful Tables

Table A–3

Axial and Direct Shear

Bending and Torsion

Optional SI Units for Bending Stress σ = Mc/l, Torsion Stress τ = Tr/J, Axial Stress σ = F/A, and Direct Shear Stress τ = F/A

M,T

I,J

c, r

σ, τ

F

A

σ, τ

N · m∗

m4

m

Pa

N∗

m2

Pa

N·m

cm4

cm

MPa (N/mm2)

N†

mm2

MPa (N/mm2)

2

4

†

N·m

mm

mm

GPa

kN

m

kPa

kN · m

cm4

cm

GPa

kN†

mm2

GPa

4

N · mm

mm

†

2

mm

MPa (N/mm )

∗ Basic relation. † Often preferred.

Table A–4

Bending Deflection

Optional SI Units for Bending Deflection y = f (Fl 3/El) or y = f (wl 4/El) and Torsional Deflection θ = Tl/GJ

F, wl

l

I

N∗

m

m4

kN

mm

†

mm 4

4

Torsional Deflection l

J

G

θ

N · m∗

m

m4

Pa

rad

N·m

mm

mm4

GPa

4

E

y

T

Pa

m

GPa

mm

†

kN

m

m

GPa

µm

N · mm

mm

mm

N

mm

mm4

kPa

m

N·m

cm

cm4

rad 2

MPa (N/mm )

rad

MPa (N/mm2)

rad

∗ Basic relation. † Often preferred.

Table A–5

Physical Constants of Materials Modulus of Elasticity E

Modulus of Rigidity G

Unit Weight w

Material

Mpsi

GPa

Mpsi

GPa

Poisson’s Ratio v

Aluminum (all alloys)

10.4

71.7

3.9

26.9

0.333

0.098

169

26.6

Beryllium copper

18.0

124.0

7.0

48.3

0.285

0.297

513

80.6

Brass

15.4

106.0

5.82

40.1

0.324

0.309

534

83.8

lbf/in3

lbf/ft 3

kN/m3

Carbon steel

30.0

207.0

11.5

79.3

0.292

0.282

487

76.5

Cast iron (gray)

14.5

100.0

6.0

41.4

0.211

0.260

450

70.6

Copper

17.2

119.0

6.49

44.7

0.326

0.322

556

87.3

1.6

11.0

0.6

4.1

0.33

0.016

28

4.3

Douglas fir Glass Inconel

6.7

46.2

2.7

18.6

0.245

0.094

162

25.4

31.0

214.0

11.0

75.8

0.290

0.307

530

83.3

Lead

5.3

36.5

1.9

13.1

0.425

0.411

710

111.5

Magnesium

6.5

44.8

2.4

16.5

0.350

0.065

112

17.6

Molybdenum

48.0

331.0

17.0

117.0

0.307

0.368

636

100.0

Monel metal

26.0

179.0

9.5

65.5

0.320

0.319

551

86.6

Nickel silver

18.5

127.0

7.0

48.3

0.322

0.316

546

85.8

Nickel steel

30.0

207.0

11.5

79.3

0.291

0.280

484

76.0

Phosphor bronze

16.1

111.0

6.0

41.4

0.349

0.295

510

80.1

Stainless steel (18-8)

27.6

190.0

10.6

73.1

0.305

0.280

484

76.0

Titanium alloys

16.5

114.0

6.2

42.4

0.340

0.160

276

43.4

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Table A–6 Properties of StructuralSteel Angles∗†

w = weight per foot, lbf/ft m = mass per meter, kg/m A = area, in2 (cm2) I = second moment of area, in4 (cm4) k = radius of gyration, in (cm) y = centroidal distance, in (cm) Z = section modulus, in3, (cm3) Size, in

w

A

1 × 1 × 18 × 14 1 12 × 1 12 × 18 × 14 2 × 2 × 18 × 14 × 38 1 1 2 2 × 2 2 × 14 × 38 3 × 3 × 14 × 38 × 12 3 12 × 3 12 × 14 × 38 × 12 4 × 4 × 14 × 38 × 12 × 58 6 × 6 × 38 × 12 × 58 × 34

0.80

3 1

1 y 3

l1−1

k1−1

Z1−1

y

k3−3

0.234

0.021

0.298

0.029

0.290

0.191

1.49

0.437

0.036

0.287

0.054

0.336

0.193

1.23

0.36

0.074

0.45

0.068

0.41

0.29

2.34

0.69

0.135

0.44

0.130

0.46

0.29

1.65

0.484

0.190

0.626

0.131

0.546

0.398

3.19

0.938

0.348

0.609

0.247

0.592

0.391

4.7

1.36

0.479

0.594

0.351

0.636

0.389

4.1

1.19

0.703

0.769

0.394

0.717

0.491

5.9

1.73

0.984

0.753

0.566

0.762

0.487

4.9

1.44

1.24

0.930

0.577

0.842

0.592

7.2

2.11

1.76

0.913

0.833

0.888

0.587

9.4

2.75

2.22

0.898

1.07

0.932

0.584

5.8

1.69

2.01

1.09

0.794

0.968

0.694

8.5

2.48

2.87

1.07

1.15

1.01

0.687

11.1

3.25

3.64

1.06

1.49

1.06

0.683

6.6

1.94

3.04

1.25

1.05

1.09

0.795

9.8

2.86

4.36

1.23

1.52

1.14

0.788

12.8

3.75

5.56

1.22

1.97

1.18

0.782

15.7

4.61

6.66

1.20

2.40

1.23

0.779

14.9

4.36

15.4

1.88

3.53

1.64

1.19

19.6

5.75

19.9

1.86

4.61

1.68

1.18

24.2

7.11

24.2

1.84

5.66

1.73

1.18

28.7

8.44

28.2

1.83

6.66

1.78

1.17

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Useful Tables

Table A–6 Properties of StructuralSteel Angles∗† (Continued)

Size, mm

m

A

25 × 25 × 3

1.11

1.42

×4

1.45

1.85

l1−1

965

k1−1

Z1−1

y

k3−3

0.80

0.75

0.45

0.72

0.48

1.01

0.74

0.58

0.76

0.48

×5

1.77

2.26

1.20

0.73

0.71

0.80

0.48

40 × 40 × 4

2.42

3.08

4.47

1.21

1.55

1.12

0.78

×5

2.97

3.79

5.43

1.20

1.91

1.16

0.77

×6

3.52

4.48

6.31

1.19

2.26

1.20

0.77

50 × 50 × 5

3.77

4.80

11.0

1.51

3.05

1.40

0.97

×6

4.47

5.59

12.8

1.50

3.61

1.45

0.97

×8

5.82

7.41

16.3

1.48

4.68

1.52

0.96

60 × 60 × 5

4.57

5.82

19.4

1.82

4.45

1.64

1.17

×6

5.42

6.91

22.8

1.82

5.29

1.69

1.17

×8

7.09

9.03

29.2

1.80

6.89

1.77

1.16

× 10 80 × 80 × 6 ×8 × 10 100 ×100 × 8

8.69 7.34 9.63 11.9

11.1

34.9

1.78

8.41

1.85

1.16

55.8

2.44

9.57

2.17

1.57

12.3

72.2

2.43

12.6

2.26

1.56

15.1

87.5

2.41

15.4

2.34

1.55

9.35

12.2

15.5

145

3.06

19.9

2.74

1.96

× 12

17.8

22.7

207

3.02

29.1

2.90

1.94

× 15

21.9

27.9

249

2.98

35.6

3.02

1.93

150 × 150 × 10

23.0

29.3

624

4.62

56.9

4.03

2.97

× 12

27.3

34.8

737

4.60

67.7

4.12

2.95

× 15

33.8

43.0

898

4.57

83.5

4.25

2.93

× 18

40.1

51.0

1050

4.54

98.7

4.37

2.92

∗ Metric sizes also available in sizes of 45, 70, 90, 120, and 200 mm. † These sizes are also available in aluminum alloy.

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Page 966


Table A–7 Properties of Structural-Steel Channels∗ 2

a, b = size, in (mm) w = weight per foot, lbf/ft m = mass per meter, kg/m t = web thickness, in (mm) A = area, in2 (cm2) I = second moment of area, in4 (cm4) k = radius of gyration, in (cm) x = centroidal distance, in (cm) Z = section modulus, in3 (cm3)

t 1 a

1

x 2 b

a, in

b, in

t

A

w

3

1.410

0.170

1.21

4.1

3

1.498

0.258

1.47

5.0

3

1.596

0.356

1.76

4

1.580

0.180

1.57

4

1.720

0.321

5

1.750

0.190

5

1.885

6

1.920

6 6

l1−1

k1−1

Z1−1

l2−2

k2−2

Z2−2

x

1.66

1.17

1.10

0.197

0.404

0.202

0.436

1.85

1.12

1.24

0.247

0.410

0.233

0.438

6.0

2.07

1.08

1.38

0.305

0.416

0.268

0.455

5.4

3.85

1.56

1.93

0.319

0.449

0.283

0.457

2.13

7.25

4.59

1.47

2.29

0.433

0.450

0.343

0.459

1.97

6.7

7.49

1.95

3.00

0.479

0.493

0.378

0.484

0.325

2.64

9.0

8.90

1.83

3.56

0.632

0.489

0.450

0.478

0.200

2.40

8.2

13.1

2.34

4.38

0.693

0.537

0.492

0.511

2.034

0.314

3.09

10.5

15.2

2.22

5.06

0.866

0.529

0.564

0.499

2.157

0.437

3.83

13.0

17.4

2.13

5.80

1.05

0.525

0.642

0.514

7

2.090

0.210

2.87

9.8

21.3

2.72

6.08

0.968

0.581

0.625

0.540

7

2.194

0.314

3.60

12.25

24.2

2.60

6.93

1.17

0.571

0.703

0.525

7

2.299

0.419

4.33

14.75

27.2

2.51

7.78

1.38

0.564

0.779

0.532

8

2.260

0.220

3.36

11.5

32.3

3.10

8.10

1.30

0.625

0.781

0.571

8

2.343

0.303

4.04

13.75

36.2

2.99

9.03

1.53

0.615

0.854

0.553

8

2.527

0.487

5.51

18.75

44.0

2.82

11.0

1.98

0.599

1.01

0.565

9

2.430

0.230

3.91

13.4

47.7

3.49

10.6

1.75

0.669

0.962

0.601

9

2.485

0.285

4.41

15.0

51.0

3.40

11.3

1.93

0.661

1.01

0.586

9

2.648

0.448

5.88

20.0

60.9

3.22

13.5

2.42

0.647

1.17

0.583

10

2.600

0.240

4.49

15.3

67.4

3.87

13.5

2.28

0.713

1.16

0.634

10

2.739

0.379

5.88

20.0

78.9

3.66

15.8

2.81

0.693

1.32

0.606

10

2.886

0.526

7.35

25.0

91.2

3.52

18.2

3.36

0.676

1.48

0.617

10

3.033

0.673

8.82

30.0

103

3.43

20.7

3.95

0.669

1.66

0.649

12

3.047

0.387

7.35

25.0

144

4.43

24.1

4.47

0.780

1.89

0.674

12

3.170

0.510

8.82

30.0

162

4.29

27.0

5.14

0.763

2.06

0.674

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Page 967

Useful Tables

967

Table A–7 Properties of Structural-Steel Channels (Continued) a × b, mm

∗

m

t

A

76 × 38

6.70

5.1

8.53

102 × 51

10.42

6.1

13.28

I1−1 74.14 207.7

k1−1

Z1−1

I2−2

k2−2

Z2−2

x

2.95

19.46

10.66

1.12

4.07

1.19

3.95

40.89

29.10

1.48

8.16

1.51

127 × 64

14.90

6.4

18.98

482.5

5.04

152 × 76

17.88

6.4

22.77

851.5

6.12

75.99 111.8

67.23 113.8

1.88

15.25

1.94

2.24

21.05

2.21

152 × 89

23.84

7.1

30.36

1166

6.20

153.0

215.1

2.66

35.70

2.86

178 × 76

20.84

6.6

26.54

1337

7.10

150.4

134.0

2.25

24.72

2.20

178 × 89

26.81

7.6

34.15

1753

7.16

197.2

241.0

2.66

39.29

2.76

203 × 76

23.82

7.1

30.34

1950

8.02

192.0

151.3

2.23

27.59

2.13

203 × 89

29.78

8.1

37.94

2491

8.10

245.2

264.4

2.64

42.34

2.65

229 × 76

26.06

7.6

33.20

2610

8.87

228.3

158.7

2.19

28.22

2.00

229 × 89

32.76

8.6

41.73

3387

9.01

296.4

285.0

2.61

44.82

2.53

254 × 76

28.29

8.1

36.03

3367

9.67

265.1

162.6

2.12

28.21

1.86

254 × 89

35.74

9.1

45.42

4448

350.2

302.4

2.58

46.70

2.42

305 × 89

41.69

10.2

53.11

7061

11.5

463.3

325.4

2.48

48.49

2.18

305 × 102

46.18

10.2

58.83

8214

11.8

539.0

499.5

2.91

66.59

2.66

These sizes are also available in aluminum alloy.

9.88

shi20361_app_A.qxd

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Page 968


Table A–8 Properties of Round Tubing

wa = unit weight of aluminum tubing, lbf/ft ws = unit weight of steel tubing, lbf/ft m = unit mass, kg/m A = area, in2 (cm2) I = second moment of area, in4 (cm4) J = second polar moment of area, in4 (cm4) k = radius of gyration, in (cm) Z = section modulus, in3 (cm3) d, t = size (OD) and thickness, in (mm) Size, in 1× 1× 1 12 × 1 12

×

2× 2× 2 12 × 2 12

×

3× 3× 4× 4×

1 8 1 4 1 8 1 4 1 8 1 4 1 8 1 4 1 4 3 8 3 16 3 8

Size, mm

wa

ws

A

l

k

Z

J

0.416

1.128

0.344

0.034

0.313

0.067

0.067

0.713

2.003

0.589

0.046

0.280

0.092

0.092

0.653

1.769

0.540

0.129

0.488

0.172

0.257

1.188

3.338

0.982

0.199

0.451

0.266

0.399

0.891

2.670

0.736

0.325

0.664

0.325

0.650

1.663

4.673

1.374

0.537

0.625

0.537

1.074

1.129

3.050

0.933

0.660

0.841

0.528

1.319

2.138

6.008

1.767

1.132

0.800

0.906

2.276

2.614

7.343

2.160

2.059

0.976

1.373

4.117

3.093

2.718

0.938

1.812

5.436

2.246

4.090

1.350

2.045

8.180

4.271

7.090

1.289

3.544

14.180

3.742 2.717 5.167 m

10.51 7.654 14.52 A

l

k

Z

J

12 × 2

0.490

0.628

0.082

0.361

0.136

0.163

16 × 2

0.687

0.879

0.220

0.500

0.275

0.440

16 × 3

0.956

1.225

0.273

0.472

0.341

0.545

20 × 4

1.569

2.010

0.684

0.583

0.684

1.367

25 × 4

2.060

2.638

1.508

0.756

1.206

3.015

25 × 5

2.452

3.140

1.669

0.729

1.336

3.338

30 × 4

2.550

3.266

2.827

0.930

1.885

5.652

30 × 5

3.065

3.925

3.192

0.901

2.128

6.381

42 × 4

3.727

4.773

8.717

1.351

4.151

17.430

42 × 5

4.536

5.809

10.130

1.320

4.825

20.255

50 × 4

4.512

5.778

15.409

1.632

6.164

30.810

50 × 5

5.517

7.065

18.118

1.601

7.247

36.226

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Useful Tables

Table A–9 Shear, Moment, and Deflection of Beams (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

969

1 Cantilever—end load y

R1 = V = F l

M = F(x − l) F x

M1

M1 = Fl

R1

y=

F x2 (x − 3l) 6E I

ymax = −

Fl 3 3E I

V

+ x M x –

2 Cantilever—intermediate load y

R1 = V = F l a

M1 = Fa

M A B = F(x − a)

b

MB C = 0

F A

B

C x

yA B =

F x2 (x − 3a) 6E I

yB C =

Fa 2 (a − 3x) 6E I

ymax =

Fa 2 (a − 3l) 6E I

M1 R1 V

+ x M –

x

(continued)

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Page 970


Table A–9 Shear, Moment, and Deflection of Beams (Continued) (Note: Force and moment reactions are positive in the directions shown; equations for shear force V and bending moment M follow the sign conventions given in Sec. 4–2.)

3 Cantilever—uniform load y

R1 = wl

l

M1 =

w

V = w(l − x)

x M1

y=

R1 V

wl 2 2 M =−

w (l − x)2 2

wx 2 (4lx − x 2 − 6l 2 ) 24E I

ymax = −

wl 4 8E I

+ x M x

–

4 Cantilever—moment load y

R1 = 0

l M1

MB

y=

A x

B R1 V x M

x

MB x 2 2E I

M1 = M B ymax =

M = MB MB l2 2E I

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Useful Tables


971

5 Simple supports—center load y

R1 = R2 =

l l/2

F

A

B

V A B = R1

C x

R1

F 2

V A B = R1

VB C = −R2

Fx F M B C = (l − x) 2 2 Fx = (4x 2 − 3l 2 ) 48E I

MA B =

R2

yA B

V

ymax = −

+

Fl 3 48E I

x –

M

+ x

6 Simple supports—intermediate load y

R1 =

l a

b F

A

B

V A B = R1

C x

R1

R2 =

Fa l

VB C = −R2

Fbx Fa MB C = (l − x) l l Fbx 2 = (x + b2 − l 2 ) 6E I l Fa(l − x) 2 = (x + a 2 − 2lx) 6E I l

MA B =

R2

yA B

V

yB C

+ –

Fb l

x

M

+ x

(continued)

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Page 972



7 Simple supports—uniform load y

R1 = R2 =

l w

V =

wl − wx 2

wx (l − x) 2 wx y= (2lx 2 − x 3 − l 3 ) 24E I

M=

x R1

wl 2

R2

V

ymax = −

5wl 4 384E I

+ x

–

M

+ x

8 Simple supports—moment load y

R1 = R2 =

l b

a

A

C x

B

yA B

R1

yB C

V

+ x M

+ –

x

MB l MB = (x − l) l

V =

MB x MB C l MB x 2 = (x + 3a 2 − 6al + 2l 2 ) 6E I l MB 3 = [x − 3lx 2 + x(2l 2 + 3a 2 ) − 3a 2 l] 6E I l

MA B =

R2

MB

MB l

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Useful Tables


973

9 Simple supports—twin loads y

R1 = R2 = F

l a

F

A

F

VB C = 0

VC D = −F

a

B

VA B = F

C

MA B = F x

D x R2

R1

M B C = Fa

MC D = F(l − x)

Fx 2 (x + 3a 2 − 3la) 6E I Fa = (3x 2 + a 2 − 3lx) 6E I Fa = (4a 2 − 3l 2 ) 24E I

yA B =

V

yB C +

ymax x –

M

+ x

10 Simple supports—overhanging load y a

l

F

R1 B

A

C x

R2

VA B MA B

V

yA B +

yB C

x

–

Fa F R2 = (l + a) l l Fa =− VB C = F l Fax =− M B C = F(x − l − a) l Fax 2 = (l − x 2 ) 6E I l F(x − l) = [(x − l)2 − a(3x − l)] 6E I

R1 =

yc = −

M

Fa 2 (l + a) 3E I

x –

(continued)

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Page 974



11 One fixed and one simple support—center load y l

R1 =

F

l/2 A

V A B = R1

C

B

11F 16

R2 =

5F 16

M1 =

3Fl 16

VB C = −R2

x

MA B =

R2

M1 R1

F (11x − 3l) 16

MB C =

5F (l − x) 16

F x2 (11x − 9l) 96E I F(l − x) = (5x 2 + 2l 2 − 10lx) 96E I

yA B =

V

yB C + x –

M

+ x

–

12 One fixed and one simple support—intermediate load y l

F

a A

C

B

x R2

M1 R1 V

+ –

Fb 2 (3l − b2 ) 2l 3 Fb M1 = 2 (l 2 − b2 ) 2l R1 =

b

x

V A B = R1

+ –

x

Fa 2 (3l − a) 2l 3

VB C = −R2

MA B =

Fb 2 [b l − l 3 + x(3l 2 − b2 )] 2l 3

MB C =

Fa 2 2 (3l − 3lx − al + ax) 2l 3

yA B =

Fbx 2 [3l(b2 − l 2 ) + x(3l 2 − b2 )] 12E I l 3

yB C = y A B −

M

R2 =

F(x − a)3 6E I

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Useful Tables


975

13 One fixed and one simple support—uniform load y

5wl 3wl R2 = 8 8 5wl − wx V = 8 w M = − (4x 2 − 5lx + l 2 ) 8

R1 =

l

x M1

0.4215l R1

R2 ymax

y=

V 5l / 8

M1 =

wl 2 8

wx 2 (l − x)(2x − 3l) 48E I

ymax = − +

wl 4 185E I

x

–

M l /4 + x

–

14 Fixed supports—center load y

R1 = R2 =

l l/2

F

A

B

C x

M1

M2 R1

R2

V A B = −VB C = MA B = yA B =

V

x

M1 = M2 =

Fl 8

F 2

F (4x − l) 8

MB C =

F (3l − 4x) 8

F x2 (4x − 3l) 48E I

ymax = −

+

F 2

Fl 3 192E I

–

M

+ –

–

x

(continued)

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Page 976



15 Fixed supports—intermediate load y l a

Fb2 (3a + b) l3

R2 =

M1 =

Fab2 l2

Fa 2 b l2

b

F

A

R1 =

B

C x

M1

V A B = R1

M2 R1

R2

MA B =

V

M2 =

Fa 2 (3b + a) l3

VB C = −R2

Fb2 [x(3a + b) − al] l3

M B C = M A B − F(x − a) + x

–

M

yA B =

Fb2 x 2 [x(3a + b) − 3al] 6E I l 3

yB C =

Fa 2 (l − x)2 [(l − x)(3b + a) − 3bl] 6E I l 3

+ –

–

x

16 Fixed supports—uniform load y

R1 = R2 =

l

x M1

M2 R1

R2

V

x

–

M 0.2113l + –

–

x

M1 = M2 =

w (l − 2x) 2 w M= (6lx − 6x 2 − l 2 ) 12 V =

y=− +

wl 2

ymax = −

wx 2 (l − x)2 24E I wl 4 384E I

wl 2 12

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Page 977

Useful Tables

977

Table A–10 Cumulative Distribution Function of Normal (Gaussian) Distribution (z α ) =

−∞

=

zα

2 u 1 du exp − √ 2 2π

α 1−α

f(z) ⌽(z␣)

zα ≤ 0 zα > 0

␣

0 z␣

Zα

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681

0.4641

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.4286

0.4247

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3897

0.3859

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.3557

0.3520

0.3483

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3238

0.3192

0.3156

0.3121

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.2877

0.2843

0.2810

0.2776

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

0.2546

0.2514

0.2483

0.2451

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

0.9

0.1841

0.1814

0.1788

0.1762

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

1.0

0.1587

0.1562

0.1539

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1379

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1170

1.2

0.1151

0.1131

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

0.0985

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

0.0838

0.0823

1.4

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0.0708

0.0694

0.0681

1.5

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0594

0.0582

0.0571

0.0559

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

0.0475

0.0465

0.0455

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

0.0392

0.0384

0.0375

0.0367

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

0.0322

0.0314

0.0307

0.0301

0.0294

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

0.0256

0.0250

0.0244

0.0239

0.0233

2.0

0.0228

0.0222

0.0217

0.0212

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

2.1

0.0179

0.0174

0.0170

0.0166

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

2.2

0.0139

0.0136

0.0132

0.0129

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

2.3

0.0107

0.0104

0.0102

0.00990

0.00964

0.00939

0.00914

0.00889

0.00866

0.00842

2.4

0.00820

0.00798

0.00776

0.00755

0.00734

0.00714

0.00695

0.00676

0.00657

0.00639

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

2.7

0.00347

0.00336

0.00326

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00264

2.8

0.00256

0.00248

0.00240

0.00233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

2.9

0.00187

0.00181

0.00175

0.00169

0.00164

0.00159

0.00154

0.00149

0.00144

0.00139

(continued)

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Page 978


Table A–10 Cumulative Distribution Function of Normal (Gaussian) Distribution (Continued) Zα 3

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.00135

0.03968

0.03687

0.03483

0.03337

0.03233

0.03159

0.03108

0.04723

0.04481

4

4

0.0 317

4

0.0 207

4

0.0 133

5

0.0 854

5

0.0 541

5

0.0 340

5

0.0 211

5

0.0 130

6

0.0 793

0.06479

5

0.06287

0.06170

0.07996

0.07579

0.07333

0.07190

0.07107

0.08599

0.08332

0.08182

6

9

0.0 987

9

0.0 530

9

0.0 282

9

0.0 149

10

0.0 777

10

0.0 402

10

0.0 206

10

0.0 104

11

0.011260

zα

−1.282

−1.643

−1.960

−2.326

−2.576

−3.090

−3.291

−3.891

F(zα)

0.10

0.05

0.025

0.010

0.005

0.001

0.005

0.00005

0.000005

R(zα)

0.90

0.95

0.975

0.999

0.995

0.999

0.9995

0.9999

0.999995

Table A–11

0.0 523 −4.417

Tolerance Grades

Basic Sizes

IT6

IT7

IT8

IT9

IT10

IT11

0–3

0.006

0.010

0.014

0.025

0.040

0.060

A Selection of International Tolerance Grades—Metric Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters) Source: Preferred Metric Limits and Fits, ANSI B4.2-1978. See also BSI 4500.

80–120

0.022

0.035

0.054

0.087

0.140

0.220

120–180

0.025

0.040

0.063

0.100

0.160

0.250

180–250

0.029

0.046

0.072

0.115

0.185

0.290

250–315

0.032

0.052

0.081

0.130

0.210

0.320

315–400

0.036

0.057

0.089

0.140

0.230

0.360

3–6

0.008

0.012

0.018

0.030

0.048

0.075

6–10

0.009

0.015

0.022

0.036

0.058

0.090

10–18

0.011

0.018

0.027

0.043

0.070

0.110

18–30

0.013

0.021

0.033

0.052

0.084

0.130

30–50

0.016

0.025

0.039

0.062

0.100

0.160

50–80

0.019

0.030

0.046

0.074

0.120

0.190

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979

Useful Tables

Table A–12 Fundamental Deviations for Shafts—Metric Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Millimeters) Source: Preferred Metric Limits and Fits , ANSI B4.2-1978. See also BSI 4500.

Upper-Deviation Letter

Lower-Deviation Letter

Basic Sizes

c

d

f

g

h

0–3

−0.060

−0.020

−0.006

−0.002

0

k 0

n

p

s

u

+0.004

+0.006

+0.014

+0.018

3–6

−0.070

−0.030

−0.010

−0.004

0

+0.001

+0.008

+0.012

+0.019

+0.023

6–10

−0.080

−0.040

−0.013

−0.005

0

+0.001

+0.010

+0.015

+0.023

+0.028

10–14

−0.095

−0.050

−0.016

−0.006

0

+0.001

+0.012

+0.018

+0.028

+0.033

14–18

−0.095

−0.050

−0.016

−0.006

0

+0.001

+0.012

+0.018

+0.028

+0.033

18–24

−0.110

−0.065

−0.020

−0.007

0

+0.002

+0.015

+0.022

+0.035

+0.041

24–30

−0.110

−0.065

−0.020

−0.007

0

+0.002

+0.015

+0.022

+0.035

+0.048

30–40

−0.120

−0.080

−0.025

−0.009

0

+0.002

+0.017

+0.026

+0.043

+0.060

40–50

−0.130

−0.080

−0.025

−0.009

0

+0.002

+0.017

+0.026

+0.043

+0.070

50–65

−0.140

−0.100

−0.030

−0.010

0

+0.002

+0.020

+0.032

+0.053

+0.087

65–80

−0.150

−0.100

−0.030

−0.010

0

+0.002

+0.020

+0.032

+0.059

+0.102

80–100

−0.170

−0.120

−0.036

−0.012

0

+0.003

+0.023

+0.037

+0.071

+0.124

100–120

−0.180

−0.120

−0.036

−0.012

0

+0.003

+0.023

+0.037

+0.079

+0.144

120–140

−0.200

−0.145

−0.043

−0.014

0

+0.003

+0.027

+0.043

+0.092

+0.170

140–160

−0.210

−0.145

−0.043

−0.014

0

+0.003

+0.027

+0.043

+0.100

+0.190

160–180

−0.230

−0.145

−0.043

−0.014

0

+0.003

+0.027

+0.043

+0.108

+0.210

180–200

−0.240

−0.170

−0.050

−0.015

0

+0.004

+0.031

+0.050

+0.122

+0.236

200–225

−0.260

−0.170

−0.050

−0.015

0

+0.004

+0.031

+0.050

+0.130

+0.258

225–250

−0.280

−0.170

−0.050

−0.015

0

+0.004

+0.031

+0.050

+0.140

+0.284

250–280

−0.300

−0.190

−0.056

−0.017

0

+0.004

+0.034

+0.056

+0.158

+0.315

280–315

−0.330

−0.190

−0.056

−0.017

0

+0.004

+0.034

+0.056

+0.170

+0.350

315–355

−0.360

−0.210

−0.062

−0.018

0

+0.004

+0.037

+0.062

+0.190

+0.390

355–400

−0.400

−0.210

−0.062

−0.018

0

+0.004

+0.037

+0.062

+0.208

+0.435

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Page 980


Table A–13 A Selection of International Tolerance Grades—Inch Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Inches, Converted from Table A–11)

Basic Sizes

Tolerance Grades IT6

IT7

IT8

IT9

IT10

IT11

0–0.12

0.0002

0.0004

0.0006

0.0010

0.0016

0.0024

0.12–0.24

0.0003

0.0005

0.0007

0.0012

0.0019

0.0030

0.24–0.40

0.0004

0.0006

0.0009

0.0014

0.0023

0.0035

0.40–0.72

0.0004

0.0007

0.0011

0.0017

0.0028

0.0043

0.72–1.20

0.0005

0.0008

0.0013

0.0020

0.0033

0.0051

1.20–2.00

0.0006

0.0010

0.0015

0.0024

0.0039

0.0063

2.00–3.20

0.0007

0.0012

0.0018

0.0029

0.0047

0.0075

3.20–4.80

0.0009

0.0014

0.0021

0.0034

0.0055

0.0087

4.80–7.20

0.0010

0.0016

0.0025

0.0039

0.0063

0.0098

7.20–10.00

0.0011

0.0018

0.0028

0.0045

0.0073

0.0114

10.00–12.60

0.0013

0.0020

0.0032

0.0051

0.0083

0.0126

12.60–16.00

0.0014

0.0022

0.0035

0.0055

0.0091

0.0142

−0.0075 −0.0075 −0.0083 −0.0083

−0.0118 −0.0130 −0.0142 −0.0157

10.00–11.20

11.20–12.60

12.60–14.20

14.20–16.00

−0.0067 −0.0067

−0.0102 −0.0110

8.00–9.00

−0.0057 −0.0067

−0.0091 −0.0094

6.40–7.20

7.20–8.00

9.00–10.00

−0.0057 −0.0057

−0.0079 −0.0083

4.80–5.60

5.60–6.40

−0.0047 −0.0047

−0.0067 −0.0071

3.20–4.00

4.00–4.80

−0.0039 −0.0039

−0.0055 −0.0059

2.00–2.60

2.60–3.20

−0.0031 −0.0031

−0.0047 −0.0051

1.20–1.60

1.60–2.00

−0.0026 −0.0026

−0.0043 −0.0043

0.72–0.96

0.96–1.20

−0.0016 −0.0020

−0.0031 −0.0037

0.24–0.40

0.40–0.72

−0.0008 −0.0012

−0.0024 −0.0028

0–0.12

0.12–0.24

d

−0.0024

−0.0024

−0.0022

−0.0022

−0.0020

−0.0020

−0.0020

−0.0017

−0.0017

−0.0017

−0.0014

−0.0014

−0.0012

−0.0012

−0.0010

−0.0010

−0.0008

−0.0008

−0.0006

−0.0005

−0.0004

−0.0002

f

g

−0.0007

−0.0007

−0.0007

−0.0007

−0.0006

−0.0006

−0.0006

−0.0006

−0.0006

−0.0006

−0.0005

−0.0005

−0.0004

−0.0004

−0.0004

−0.0004

−0.0003

−0.0003

−0.0002

−0.0002

−0.0002

−0.0001

Upper-Deviation Letter

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

h 0

+0.0002

+0.0002

+0.0002

+0.0002

+0.0002

+0.0002

+0.0002

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

+0.0001

0

0

0

k

+0.0015

+0.0015

+0.0013

+0.0013

+0.0012

+0.0012

+0.0012

+0.0011

+0.0011

+0.0011

+0.0009

+0.0009

+0.0008

+0.0008

+0.0007

+0.0007

+0.0006

+0.0006

+0.0005

+0.0004

+0.0003

+0.0024

+0.0024

+0.0022

+0.0022

+0.0020

+0.0020

+0.0020

+0.0017

+0.0017

+0.0017

+0.0015

+0.0015

+0.0013

+0.0013

+0.0010

+0.0010

+0.0009

+0.0009

+0.0007

+0.0006

+0.0005

+0.0002

p

s

+0.0082

+0.0075

+0.0067

+0.0062

+0.0055

+0.0051

+0.0048

+0.0043

+0.0039

+0.0036

+0.0031

+0.0028

+0.0023

+0.0021

+0.0017

+0.0017

+0.0014

+0.0014

+0.0011

+0.0009

+0.0007

+0.0006

Lower-Deviation Letter +0.0002

n

+0.0171

+0.0154

+0.0130

+0.0124

+0.0112

+0.0102

+0.0093

+0.0083

+0.0075

+0.0067

+0.0057

+0.0049

+0.0040

+0.0034

+0.0028

+0.0024

+0.0019

+0.0016

+0.0013

+0.0011

+0.0009

+0.0007

u

3:43 PM

c

6/3/03

Basic Sizes

Fundamental Deviations for Shafts—Inch Series (Size Ranges Are for Over the Lower Limit and Including the Upper Limit. All Values Are in Inches, Converted from Table A–12)

Table A–14

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981

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Table A–15 Charts of Theoretical Stress-Concentration Factors K*t Figure A–15–1

3.0 d

Bar in tension or simple compression with a transverse hole. σ0 = F/A, where A = (w − d )t and t is the thickness.

2.8

w

2.6 Kt 2.4

2.2

2.0

Figure A–15–2

0

0.1

0.2

0.3

0.4 d/w

0.5

0.6

3.0

Rectangular bar with a transverse hole in bending. σ0 = Mc/I, where 3 I = (w − d )h /12.

0.7

0.8

d d/h = 0

w

2.6 0.25

M

M

0.5

2.2

h

1.0

Kt

2.0

1.8

⬁ 1.4

1.0

Figure A–15–3

0

0.1

0.2

0.3

0.4 d/w

0.5

3.0

0.6

0.7

0.8

r w /d = 3

Notched rectangular bar in tension or simple compression. σ0 = F/A, where A = dt and t is the thickness.

w

2.6

d

1.5 2.2 1.2 Kt

1.1 1.8 1.05 1.4

1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

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Useful Tables

983

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued) Figure A–15–4

3.0 w/d = ⬁

1.10

Notched rectangular bar in bending. σ0 = Mc/I, where c = d/2, I = td 3 /12, and t is the thickness.

2.6

r M

1.5

1.05

w

M

d

2.2 1.02

Kt 1.8

1.4

1.0

Figure A–15–5

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

0.25

0.30

3.0

Rectangular filleted bar in tension or simple compression. σ0 = F/A, where A = dt and t is the thickness.

r

D/d = 1.50 2.6

d

D 1.10 2.2 Kt

1.05 1.8 1.02 1.4

1.0

Figure A–15–6

0

0.05

0.10

0.15 r/d

0.20

3.0 r

Rectangular filleted bar in bending. σ0 = Mc/I, where 3 c = d/2, I = td /12, t is the thickness.

2.6

M

1.05

d

D

M

3 2.2

1.1 1.3

Kt 1.8

D/d = 1.02

1.4

1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

(continued)

*Factors from R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951, p. 169; no. 3, March 1951, p. 161, no. 5, May 1951, p. 159; no. 6, June 1951, p. 173; no. 7, July 1951, p. 155. Reprinted with permission from Machine Design, a Penton Media Inc. publication.

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Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued) Figure A–15–7

2.6

Round shaft with shoulder fillet in tension. σ0 = F/A, where A = πd 2 /4.

r 2.2

d

D

Kt 1.8

D/d 1.0

= 1.

50

1.10

5

1.4

1.02 1.0

0

Figure A–15–8

3.0

Round shaft with shoulder fillet in torsion. τ0 = Tc/J, where 4 c = d/2 and J = πd /32.

2.6

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r d

D

T

T

2.2 Kts 1.8

D/d =

1.4

1.0

Figure A–15–9

1.20 1.33

2

1.09

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

3.0 r

Round shaft with shoulder fillet in bending. σ0 = Mc/I, where c = d/2 and I = πd 4 /64.

2.6 M

d

D

M

2.2 Kt 1.8

D/d

=3

1.5 1.4

1.10

1.02

1.05 1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

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Useful Tables

985

Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued)

Figure A–15–10

4.0 d

Round shaft in torsion with transverse hole.

3.6

Kts, B

2.8

Figure A–15–11

B A

J ␲D3 dD2 c = 16 – 6 (approx)

Kts, A

Kts 3.2

2.4

D

T

0

0.05

0.10

0.15 d/D

0.20

0.25

0.30

3.0 d

Round shaft in bending with a transverse hole. σ0 = 2 M/[(πD3 /32) − (dD /6)], approximately.

D

2.6 M

M

2.2 Kt 1.8

1.4

1.0

Figure A–15–12 Plate loaded in tension by a pin through a hole. σ0 = F/A, where A = (w − d)t . When clearance exists, increase Kt 35 to 50 percent. (M. M. Frocht and H. N. Hill, “Stress Concentration Factors around a Central Circular Hole in a Plate Loaded through a Pin in Hole,” J. Appl. Mechanics, vol. 7, no. 1, March 1940, p. A-5.)

0

0.05

0.10

0.15 d/D

0.20

0.25

0.30

11

h

9

d

h/w = 0.35 w

7

t

Kt 5 h/w = 0.50 3

1

h/w ⱖ 1.0

0

0.1

0.2

0.3

0.4 d/w

0.5

0.6

0.7

0.8

(continued)


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Table A–15 Charts of Theoretical Stress-Concentration Factors K*t (Continued)

Figure A–15–13

3.0 r

1.15

Grooved round bar in tension. σ0 = F/A, where A = πd 2 /4.

2.6 D

1.05

d

2.2 Kt 1.02

D/d = 1.50

1.8

1.4

1.0

Figure A–15–14

3.0

Grooved round bar in bending. σ0 = Mc/l, where 4 c = d/2 and I = πd /64.

2.6

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r

M

D

M

d

1.05 2.2 Kt D/d = 1.50

1.02 1.8

1.4

1.0

0

Figure A–15–15

2.6

Grooved round bar in torsion. τ0 = Tc/J, where c = d/2 4 and J = πd /32.

2.2

0.05

0.10

0.15 r/d

0.20

0.25

0.30

r T

T D

1.8

d

1.05

Kts

D/d = 1.30 1.4 1.02 1.0

0

0.05

0.10

0.15 r/d

0.20

0.25

0.30


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Useful Tables

a

Table A–16 Approximate StressConcentration Factor Kt for Bending of a Round Bar or Tube with a Transverse Round Hole

987

D

d

M

M

The nominal bending stress is σ0 = M/Z net where Z net is a reduced value of the section modulus and is defined by

Source: R. E. Peterson, Stress Concentration Factors, Wiley, New York, 1974, pp. 146, 235.

Z net =

πA (D 4 − d 4 ) 32D

Values of A are listed in the table. Use d = 0 for a solid bar

d/D 0.9

0.6

0

a/D

A

Kt

A

Kt

A

Kt

0.050

0.92

2.63

0.91

2.55

0.88

2.42

0.075

0.89

2.55

0.88

2.43

0.86

2.35

0.10

0.86

2.49

0.85

2.36

0.83

2.27

0.125

0.82

2.41

0.82

2.32

0.80

2.20

0.15

0.79

2.39

0.79

2.29

0.76

2.15

0.175

0.76

2.38

0.75

2.26

0.72

2.10

0.20

0.73

2.39

0.72

2.23

0.68

2.07

0.225

0.69

2.40

0.68

2.21

0.65

2.04

0.25

0.67

2.42

0.64

2.18

0.61

2.00

0.275

0.66

2.48

0.61

2.16

0.58

1.97

0.30

0.64

2.52

0.58

2.14

0.54

1.94 (continued)

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Page 988


Table A–16 (Continued) Approximate Stress-Concentration Factors Kts for a Round Bar or Tube Having a Transverse Round Hole and Loaded in Torsion Source: R. E. Peterson, Stress Concentration Factors, Wiley, New York, 1974, pp. 148, 244.

D

a

d

T

T

The maximum stress occurs on the inside of the hole, slightly below the shaft surface. The nominal shear stress is τ0 = T D/2Jnet , where Jnet is a reduced value of the second polar moment of area and is defined by Jnet =

π A(D 4 − d 4 ) 32

Values of A are listed in the table. Use d = 0 for a solid bar. d/D 0.9

0.8 A

0.6 Kts

A

0.4 Kts

A

0

a/D

A

Kts

Kts

A

Kts

0.05

0.96

1.78

0.95

1.77

0.075

0.95

1.82

0.93

1.71

0.10

0.94

1.76

0.93

1.74

0.92

1.72

0.92

1.70

0.92

1.68

0.125

0.91

1.76

0.91

1.74

0.90

1.70

0.90

1.67

0.89

1.64

0.15

0.90

1.77

0.89

1.75

0.87

1.69

0.87

1.65

0.87

1.62

0.175

0.89

1.81

0.88

1.76

0.87

1.69

0.86

1.64

0.85

1.60

0.20

0.88

1.96

0.86

1.79

0.85

1.70

0.84

1.63

0.83

1.58

0.25

0.87

2.00

0.82

1.86

0.81

1.72

0.80

1.63

0.79

1.54

0.30

0.80

2.18

0.78

1.97

0.77

1.76

0.75

1.63

0.74

1.51

0.35

0.77

2.41

0.75

2.09

0.72

1.81

0.69

1.63

0.68

1.47

0.40

0.72

2.67

0.71

2.25

0.68

1.89

0.64

1.63

0.63

1.44

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Page 989

Useful Tables

Table A–17 Preferred Sizes and Renard (R-Series) Numbers (When a choice can be made, use one of these sizes; however, not all parts or items are available in all the sizes shown in the table.)

989

Fraction of Inches 1 , 1 , 1 , 3 , 1 , 5 , 3 , 1 , 5 , 3 , 7 , 1 , 9 , 5 , 11 , 3 , 7 , 1, 1 14 , 1 12 , 1 34 , 2, 2 14 , 64 32 16 32 8 32 16 4 16 8 16 2 16 8 16 4 8 2 12 , 2 34 , 3, 3 14 , 3 12 , 3 34 , 4, 4 14 , 4 12 , 4 34 , 5, 5 14 , 5 12 , 5 34 , 6, 6 12 , 7, 7 12 , 8, 8 12 , 9, 9 12 , 10, 10 12 , 11, 11 12 , 12, 12 12 , 13, 13 12 , 14, 14 12 , 15, 15 12 , 16, 16 12 , 17, 17 12 , 18,

18 12 , 19, 19 12 , 20 Decimal Inches 0.010, 0.012, 0.016, 0.020, 0.025, 0.032, 0.040, 0.05, 0.06, 0.08, 0.10, 0.12, 0.16, 0.20, 0.24, 0.30, 0.40, 0.50, 0.60, 0.80, 1.00, 1.20, 1.40, 1.60, 1.80, 2.0, 2.4, 2.6, 2.8, 3.0, 3.2, 3.4, 3.6, 3.8, 4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 7.0, 7.5, 8.5, 9.0, 9.5, 10.0, 10.5, 11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0, 16.5, 17.0, 17.5, 18.0, 18.5, 19.0, 19.5, 20 Millimeters 0.05, 0.06, 0.08, 0.10, 0.12, 0.16, 0.20, 0.25, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, 0.90, 1.0, 1.1, 1.2, 1.4, 1.5, 1.6, 1.8, 2.0, 2.2, 2.5, 2.8, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 8.0, 9.0, 10, 11, 12, 14, 16, 18, 20, 22, 25, 28, 30, 32, 35, 40, 45, 50, 60, 80, 100, 120, 140, 160, 180, 200, 250, 300 Renard Numbers* 1st choice, R5: 1, 1.6, 2.5, 4, 6.3, 10 2d choice, R10: 1.25, 2, 3.15, 5, 8 3d choice, R20: 1.12, 1.4, 1.8, 2.24, 2.8, 3.55, 4.5, 5.6, 7.1, 9 4th choice, R40: 1.06, 1.18, 1.32, 1.5, 1.7, 1.9, 2.12, 2.36, 2.65, 3, 3.35, 3.75, 4.25, 4.75, 5.3, 6, 6.7, 7.5, 8.5, 9.5 *May be multiplied or divided by powers of 10.

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Page 990


Table A–18 Geometric Properties

Part 1 Properties of Sections A = area G = location of centroid Ix = x 2 d A = second moment of area about x axis Ix y = x y d A = mixed moment of area about x and y axes JG = r 2 d A = (x 2 + y 2 ) d A = Ix + I y = second polar moment of area about axis through G k x2

= Ix /A = squared radius of gyration about x axis

Rectangle

y b 2

h

h 2

G

x

b

A = bh

bh 3 Ix = 12

b3 h Iy = 12

Circle

Ix y = 0

y

D x

G

A=

π D2 4

Ix = I y =

π D4 64

Ix y = 0

Hollow circle

y

d

D

G

A=

π 2 (D − d 2 ) 4

Ix = I y =

π (D 4 − d 4 ) 64

x

Ix y = 0

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Useful Tables

991

Table A–18 Geometric Properties (Continued)

y

Right triangles

y

b 3

b h 3 G

h h 3

G

x b 3

b

A=

bh 2

Ix =

bh 3 36

Iy =

x

h

b3 h 36

−b2 h 2 72

Ix y =

y

y

Right triangles

b 3

b h 3 h

h 3

h x

G

b 3

b

A=

bh 2

Ix =

bh 3 36

Iy =

x

G

b3 h 36

Ix y =

b2 h 2 72 y

y

Quarter-circles 4r 3␲

r 4r 3␲ G

4r 3␲

r

A=

πr 2 4

Ix = I y = r 4

π 4 − 16 9π

Ix y = r 4

1 4 − 8 9π

y

y

Quarter-circles

x

G

x

4r 3␲

4r 3␲

r 4r 3␲ G

x 4r 3␲

4r 3␲

πr 2 A= 4

Ix = I y = r

4

π 4 − 16 9π

G

x

r

Ix y = r

4

4 1 − 9π 8

(continued)

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Table A–18

Part 2 Properties of Solids (ρ = Density, Weight per Unit Volume)

Geometric Properties (Continued)

Rods

y

d z

l x

m=

πd 2 lρ 4g

I y = Iz =

ml 2 12 y

Round disks t

d

x

z

m=

πd 2 tρ 4g

Ix =

md 2 8

md 2 16

I y = Iz =

y

Rectangular prisms

b

c

z

m=

abcρ g

Ix =

m 2 (a + b2 ) 12

a

Iy =

x

m 2 m 2 (a + c2 ) Iz = (b + c2 ) 12 12

y

Cylinders

d z

m=

πd 2 lρ 4g

Ix =

md 2 8

l

x

m (3d 2 + 4l 2 ) 48

I y = Iz = y

Hollow cylinders

di do z

m=

π do2 − di2 lρ 4g

Ix =

l

m 2 do + di2 8

x

I y = Iz =

m 2 3do + 3di2 + 4l 2 48

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Useful Tables

Table A–19 American Standard Pipe

993

Wall Thickness, in Nominal Size, in 1 8 1 4 3 8 1 2 3 4

Outside Diameter, in

Threads per inch

Standard No. 40

Extra Strong No. 80

Double Extra Strong

0.405

27

0.070

0.098

0.540

18

0.090

0.122

0.675

18

0.093

0.129

0.840

14

0.111

0.151

0.307

1.050

14

0.115

0.157

0.318

1.315

11 12

0.136

0.183

0.369

1 14 1 12

1.660

0.143

0.195

0.393

0.148

0.204

0.411

1

2

2.375

11 12 11 12 11 12

0.158

0.223

0.447

2 12

2.875

8

0.208

0.282

0.565 0.615

1.900

3

3.500

8

0.221

0.306

3 12

4.000

8

0.231

0.325

4

4.500

8

0.242

0.344

0.690

5

5.563

8

0.263

0.383

0.768

6

6.625

8

0.286

0.441

0.884

8

8.625

8

0.329

0.510

0.895

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Table A–20 Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels [The strengths listed are estimated ASTM minimum values in the size range 18 to 32 mm ( 34 to 1 14 in). These strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform to ASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numbering system is not a specification. See Table 1–1 for certain ASTM steels.] Source: 1986 SAE Handbook, p. 2.15. 1

2

3

UNS No.

SAE and/or AISI No.

Processing

G10060

1006

HR

G10100

1010

G10150

1015

G10180

1018

G10200

1020

G10300

1030

G10350

1035

G10400

1040

G10450

1045

G10500

1050

G10600

1060

4 5 Tensile Yield Strength, Strength, MPa (kpsi) MPa (kpsi) 300 (43)

170 (24)

6

7

8

Elongation in 2 in, %

Reduction in Area, %

Brinell Hardness

30

55

86

CD

330 (48)

280 (41)

20

45

95

HR

320 (47)

180 (26)

28

50

95

CD

370 (53)

300 (44)

20

40

105

HR

340 (50)

190 (27.5)

28

50

101

CD

390 (56)

320 (47)

18

40

111

HR

400 (58)

220 (32)

25

50

116

CD

440 (64)

370 (54)

15

40

126

HR

380 (55)

210 (30)

25

50

111

CD

470 (68)

390 (57)

15

40

131

HR

470 (68)

260 (37.5)

20

42

137

CD

520 (76)

440 (64)

12

35

149

HR

500 (72)

270 (39.5)

18

40

143

CD

550 (80)

460 (67)

12

35

163

HR

520 (76)

290 (42)

18

40

149

CD

590 (85)

490 (71)

12

35

170

HR

570 (82)

310 (45)

16

40

163

CD

630 (91)

530 (77)

12

35

179

HR

620 (90)

340 (49.5)

15

35

179

CD

690 (100)

580 (84)

10

30

197

HR

680 (98)

370 (54)

12

30

201

G10800

1080

HR

770 (112)

420 (61.5)

10

25

229

G10950

1095

HR

830 (120)

460 (66)

10

25

248

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995

Useful Tables

Table A–21 Mean Mechanical Properties of Some Heat-Treated Steels [These are typical properties for materials normalized and annealed. The properties for quenched and tempered (Q&T) steels are from a single heat. Because of the many variables, the properties listed are global averages. In all cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gauge length 2 in. unless noted, all specimens were oil-quenched.] Source: ASM Metals Reference Book, 2d ed., American Society for Metals, Metals Park, Ohio, 1983.

1

2

3

AISI No.

Treatment

Temperature °C (°F)

1030

1040

1050

1060

1095

1141

4 5 Tensile Yield Strength Strength, MPa (kpsi) MPa (kpsi)

6

7

8

Elongation, %


Brinell Hardness

Q&T*

205 (400)

848 (123)

648 (94)

17

47

495

Q&T*

315 (600)

800 (116)

621 (90)

19

53

401

Q&T*

425 (800)

731 (106)

579 (84)

23

60

302

Q&T*

540 (1000)

669 (97)

517 (75)

28

65

255

Q&T*

650 (1200)

586 (85)

441 (64)

32

70

207

Normalized

925 (1700)

521 (75)

345 (50)

32

61

149

Annealed

870 (1600)

430 (62)

317 (46)

35

64

137

Q&T

205 (400)

779 (113)

593 (86)

19

48

262

Q&T

425 (800)

758 (110)

552 (80)

21

54

241

Q&T

650 (1200)

634 (92)

434 (63)

29

65

192

Normalized

900 (1650)

590 (86)

374 (54)

28

55

170

Annealed

790 (1450)

519 (75)

353 (51)

30

57

149

Q&T*

205 (400)

1120 (163)

807 (117)

9

27

514

Q&T*

425 (800)

1090 (158)

793 (115)

13

36

444

Q&T*

650 (1200)

717 (104)

538 (78)

28

65

235

Normalized

900 (1650)

748 (108)

427 (62)

20

39

217

Annealed

790 (1450)

636 (92)

365 (53)

24

40

187

Q&T

425 (800)

1080 (156)

765 (111)

14

41

311

Q&T

540 (1000)

965 (140)

669 (97)

17

45

277

Q&T

650 (1200)

800 (116)

524 (76)

23

54

229

Normalized

900 (1650)

776 (112)

421 (61)

18

37

229

Annealed

790 (1450)

626 (91)

372 (54)

22

38

179

Q&T

315 (600)

1260 (183)

813 (118)

10

30

375

Q&T

425 (800)

1210 (176)

772 (112)

12

32

363

Q&T

540 (1000)

1090 (158)

676 (98)

15

37

321

Q&T

650 (1200)

896 (130)

552 (80)

21

47

269

Normalized

900 (1650)

1010 (147)

500 (72)

9

13

293

Annealed

790 (1450)

658 (95)

380 (55)

13

21

192

Q&T

315 (600)

1460 (212)

1280 (186)

9

32

415

Q&T

540 (1000)

896 (130)

765 (111)

18

57

262 (continued)

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Table A–21 (Continued) Mean Mechanical Properties of Some Heat-Treated Steels [These are typical properties for materials normalized and annealed. The properties for quenched and tempered (Q&T) steels are from a single heat. Because of the many variables, the properties listed are global averages. In all cases, data were obtained from specimens of diameter 0.505 in, machined from 1-in rounds, and of gauge length 2 in. Unless noted, all specimens were oil-quenched.] Source: ASM Metals Reference Book, 2d ed., American Society for Metals, Metals Park, Ohio, 1983.

1

2

3

AISI No.

Treatment

Temperature °C (°F)

4130

4140

4340

*Water-quenched

4 5 Tensile Yield Strength Strength, MPa (kpsi) MPa (kpsi)

6

7

8

Elongation, %


Brinell Hardness

Q&T*

205 (400)

1630 (236)

1460 (212)

10

41

467

Q&T*

315 (600)

1500 (217)

1380 (200)

11

43

435

Q&T*

425 (800)

1280 (186)

1190 (173)

13

49

380

Q&T*

540 (1000)

1030 (150)

910 (132)

17

57

315

Q&T*

650 (1200)

814 (118)

703 (102)

22

64

245

Normalized

870 (1600)

670 (97)

436 (63)

25

59

197

Annealed

865 (1585)

560 (81)

361 (52)

28

56

156

Q&T

205 (400)

1770 (257)

1640 (238)

8

38

510

Q&T

315 (600)

1550 (225)

1430 (208)

9

43

445

Q&T

425 (800)

1250 (181)

1140 (165)

13

49

370

Q&T

540 (1000)

951 (138)

834 (121)

18

58

285

Q&T

650 (1200)

758 (110)

655 (95)

22

63

230

Normalized

870 (1600)

1020 (148)

655 (95)

18

47

302

Annealed

815 (1500)

655 (95)

417 (61)

26

57

197

Q&T

315 (600)

1720 (250)

1590 (230)

10

40

486

Q&T

425 (800)

1470 (213)

1360 (198)

10

44

430

Q&T

540 (1000)

1170 (170)

1080 (156)

13

51

360

Q&T

650 (1200)

965 (140)

855 (124)

19

60

280

Aluminum alloy Aluminum alloy

2024

7075

T6

T4

T6

Annealed

Annealed

Q&T 600°F

Q&T 600°F

HR

Annealed

Annealed

Condition

542 (78.6)

296 (43.0)

169 (24.5)

276 (40.0)

241 (35.0)

1720 (250)

1520 (220)

193 (28.0)

358 (52.0)

220 (32.0)

593 (86.0)

446 (64.8)

324 (47.0)

568 (82.4)

601 (87.3)

1930 (210)

1580 (230)

424 (61.5)

646 (93.7)

341 (49.5)

0.24 0.041 0.048

†

758 (110) 1880 (273)†

†

1760 (255) 1410 (205) 1270 (185) 620 (90) 689 (100) 882 (128)

2340 (340) 1520 (221)† 1600 (233)† 325 (47.2)† 533 (77.3)† 706 (102)†

2380 (345)

729 (106)

992 (144)

898 (130)†

0.13

0.15

0.28

0.45

0.51

0.14

0.25

620 (90.0)

628 (91.1)†

Strain Strength, Exponent m

Coefficient σ0, MPa (kpsi)

Fracture, σf, MPa (kpsi)

*Values from one or two heats and believed to be attainable using proper purchase specifications. The fracture strain may vary as much as 100 percent. † Derived value.

Aluminum alloy

2011

Steel

4142

Stainless steel

Steel

1045

304

Steel

1212

Stainless steel

Steel

1144

303

Steel

Material

1018

Number

Ultimate Su, MPa (kpsi)

0.18

0.18

0.10

1.67

1.16

0.43

0.81

0.85

0.49

1.05

Fracture Strain εf

3:43 PM

Yield Sy, MPa (kpsi)

Strength (Tensile)

Source: J. Datsko, “Solid Materials,” chap. 7 in Joseph E. Shigley and Charles R. Mischke (eds.-in-chief), Standard

Handbook of Machine Design, 2nd ed., McGraw-Hill, New York, 1996, pp. 7.47–7.50.

Results of Tensile Tests of Some Metals*

6/3/03

Table A–22


997

998

LT L L L L L L L L L L L L

1005-1009

1005-1009

1005-1009

1015

1020

1040

1045

1045

1045

1045

1045

1045

1144

L

RQC-100 (c) L

LT

RQC-100 (c)

LT

L

H-11

1005-1009

L

Gainex (c)

10B62

L

L

AM-350 (c) LT

L

A538C (b)

Gainex (c)

L

A538B (b)

CDSR

Q&T

Q&T

Q&T

Q&T

Q&T

Q&T

As forged

HR plate

Normalized

HR sheet

CD sheet

CD sheet

HR sheet

Q&T

HR plate

HR plate

Ausformed

HR sheet

HR sheet

CD

HR, A

STA

STA

STA

265

595

500

450

390

410

225

225

108

80

90

125

125

90

430

290

290

660

496

480

460

405

74

77

90

64

60

50

60

68

52

930 135

2240 325

1825 265

1585 230

1345 195

1450 210

725 105

620

440

415

345

415

470

360

1640 238

930 135

940 136

2585 375

510

530

1905 276

1315 191

2000 290

1860 270

1515 220

33

41

51

55

59

51

65

60

62

68

80

64

66

73

38

67

43

33

64

58

20

52

55

56

67

0.51

0.52

0.71

0.81

0.89

0.72

1.04

0.93

0.96

1.14

1.6

1.02

1.09

1.3

0.89

1.02

0.56

0.40

1.02

0.86

0.23

0.74

0.81

0.82

1.10

195

205

205

205

205

200

200

200

205

205

200

200

205

205

195

205

205

205

200

200

180

195

180

185

185

28.5

30

30

30

30

29

29

29

29.5

30

29

29

30

30

28

30

30

30

29.2

29.2

26

28

26

27

27

93

78

75

84

1000 145

2725 395

2275 330

1795 260

1585 230

1860 270

1225 178

1540 223

895 130

825 120

640

540

515

580

1780 258

1240 180

1240 180

3170 460

805 117

805 117

2690 390

2800 406

2240 325

2135 310

1655 240

0.86 0.08 0.66 0.66 0.32 0.15 0.30 0.11 0.10 0.95 0.41 0.61 1.00 0.60 0.45 0.35

−0.071 −0.077 −0.07 −0.07 −0.067 −0.09 −0.059 −0.073 −0.109 −0.11 −0.12 −0.14 −0.095 −0.073 −0.074 −0.07

0.32

0.86

−0.07

−0.08

0.10

−0.102

0.25

0.33

−0.14

0.07

0.60

−0.07

−0.081

0.80

−0.071

−0.08

0.30

−0.065

−0.58

−0.60

−0.68

−0.69

−0.68

−0.70

−0.66

−0.57

−0.51

−0.64

−0.39

−0.41

−0.51

−0.43

−0.56

−0.69

−0.69

−0.74

−0.68

−0.65

−0.42

−0.84

−0.75

−0.71

−0.62

3:43 PM

AM-350 (c)

L

A538A (b)

6/3/03

Grade (a)

Tensile Strength Sut

Source: ASM Metals Reference Book, 2nd ed., American Society for Metals, Metals Park,

Fatigue True Strength Strain Fatigue Fatigue Fatigue Modulus of Coefficient Strength HardReduction at Ductility Ductility σ f Orienta- Description ness in Area Fracture Elasticity E Exponent Coefficient Exponent 4 tion (e) (f) HB MPa ksi % εf GPa 10 psi MPa ksi b εF c

Ohio, 1983, p. 217.

Mean Monotonic and Cyclic Stress-Strain Properties of Selected Steels

Table A–23


L

950X (d)

DAT

Plate channel

HR plate

Plate channel

HR bar

HR plate

Q&T

Q&T

A

SH, Q&T

Q&T

Q&T

Q&T

HR, A

Q&T

Q&T

Q&T and deformed

Q&T and deformed

Q&T

Q&T and deformed

Q&T

310

225

156

150

150

159

410

280

260

518

430

350

409

243

560

475

450

475

450

400

380

335

310

77

64

82

82

695 101

530

440

565

565

565 227

1000 145

925 134

2015 292

1670 242

1240 180

1470 213

825 120

2240 325

1930 280

1930 280

2035 295

1760 255

1550 225

1415 205

1250 181

1060 154

1075 156

1425 207

895 130

890 129

950 138

1035 150

68

72

65

69

64

32

33

14

11

42

57

38

43

27

35

37

20

42

47

48

28

29

60

55

67

60

49

25

1.15

1.24

1.06

1.19

1.03

0.38

0.41

0.16

0.12

0.87

0.84

0.48

0.57

0.31

0.43

0.46

0.22

0.54

0.63

0.66

0.34

0.35

0.69

0.79

1.12

0.93

0.68

0.29

195

205

205

205

205

200

195

205

205

195

195

200

195

205

205

200

200

205

200

205

200

200

200

200

220

205

205

200

28.2

29.5

30

30

29.6

29

28

30

30

28

28

29

28

30

30

29

29

30

29

30

28.9

29

29.2

29

32

29.9

29.9

28.8

91 1055 153

1005 146

625

970 141

1170 170

1855 269

1220 177

1040 151

2585 375

1930 280

1655 240

2000 290

1200 174

2655 385

2170 315

2105 305

2070 300

2000 290

1895 275

1825 265

1250 181

1450 210

1825 265

1695 246

1275 185

1275 185

1275 185

1585 230

0.95 0.85 0.35 0.85 0.21

−0.12 −0.11 −0.075 −0.10 −0.08

0.41 0.38

−0.073 −0.057

0.18 0.16

−0.09 −0.071

0.73 0.40

−0.076 −0.071

0.45 0.48

−0.095 −0.091

0.09 0.07

−0.081

0.60

−0.09

−0.089

0.40 0.20

−0.08

0.50

−0.09

−0.082

0.06 0.45

−0.08 −0.08

1.2 0.22

−0.08 −0.10

0.92 0.89

−0.083 −0.081

0.68 0.93

−0.076 −0.071

0.27

−0.09

−0.53

−0.61

−0.54

−0.59

−0.61

−0.65

−0.60

−0.47

−0.56

−0.57

−0.62

−0.60

−0.54

−0.76

−0.61

−0.76

−0.77

−0.73

−0.75

−0.75

−0.62

−0.51

−0.59

−0.69

−0.63

−0.65

−0.65

−0.53

Notes: (a) AISI/SAE grade, unless otherwise indicated. (b) ASTM designation. (c) Proprietary designation. (d) SAE HSLA grade. (e) Orientation of axis of specimen, relative to rolling direction; L is longitudinal (parallel to rolling direction); LT is long transverse (perpendicular to rolling direction). (f) STA, solution treated and aged; HR, hot rolled; CD, cold drawn; Q&T, quenched and tempered; CDSR, cold drawn strain relieved; DAT, drawn at temperature; A, annealed. From ASM Metals Reference Book, 2nd edition, 1983; ASM International, Materials Park, OH 44073-0002; table 217. Reprinted by permission of ASM International ®, www.asminternational.org.

L L

L

950X (d)

LT

950C (d)

950C (d)

950X (d)

L L

L

9262

L

52100

9262

9262

L L

4340

5160

L L

4340

4340

L L

4142

L

4142

4142

L L

4142

L

4142

Q&T, DAT DAT

365

258

260

290

305

3:43 PM

4142

L L

4142

4142

L L

4140

Q&T

Q&T

Q&T forging

Q&T forging

DAT

6/3/03

4142

L L

4130

4130

L L

1541F

1541F

L

1144


999

1000

22 26 31 36.5 42.5 52.5 62.5

20

25

30

35

40

50

60

187.5

164

140

124

109

97

83

Compressive Strength Suc, kpsi

88.5

73

57

48.5

40

32

26

20.4–23.5

18.8–22.8

16–20

14.5–17.2

13–16.4

11.5–14.8

9.6–14

Tension†

7.8–8.5

7.2–8.0

6.4–7.8

5.8–6.9

5.2–6.6

4.6–6.0

3.9–5.6

Torsion

Modulus of Elasticity, Mpsi

24.5

21.5

18.5

16

14

11.5

10

Endurance Limit* Se, kpsi

*Polished or machined specimens. † The modulus of elasticity of cast iron in compression corresponds closely to the upper value in the range given for tension and is a more constant value than that for tension.

Tensile Strength Sut, kpsi

ASTM Number

302

262

235

212

201

174

156

Brinell Hardness HB

1.50

1.35

1.25

1.15

1.10

1.05

1.00

Fatigue StressConcentration Factor Kf

3:43 PM

Shear Modulus of Rupture Ssu, kpsi

6/3/03

Mechanical Properties of Three Non-Steel Metals (a) Typical Properties of Gray Cast Iron [The American Society for Testing and Materials (ASTM) numbering system for gray cast iron is such that the numbers correspond to the minimum tensile strength in kpsi. Thus an ASTM No. 20 cast iron has a minimum tensile strength of 20 kpsi. Note particularly that the tabulations are typical of several heats.]

Table A–24


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Table A–24 Mechanical Properties of Three Non-Steel Metals (Continued) (b) Mechanical Properties of Some Aluminum Alloys [These are typical properties for sizes of about 12 in; similar properties can be obtained by using proper purchase specifications. The values given for fatigue strength correspond to 50(107) cycles of completely reversed stress. Alluminum alloys do not have an endurance limit. Yield strengths were obtained by the 0.2 percent offset method.] Aluminum Association Number

Strength Temper

Yield, Sy, MPa (kpsi)

Tensile, Su, MPa (kpsi)

Fatigue, Sf, MPa (kpsi)


Brinell Hardness HB

O

70 (10)

179 (26)

90 (13)

22

45

Wrought: 2017 2024

O

76 (11)

186 (27)

90 (13)

22

47

T3

345 (50)

482 (70)

138 (20)

16

120

H12

117 (17)

131 (19)

55 (8)

20

35

H16

165 (24)

179 (26)

65 (9.5)

14

47

H34

186 (27)

234 (34)

103 (15)

12

63

H38

234 (34)

276 (40)

110 (16)

6

77

H32

186 (27)

234 (34)

117 (17)

18

62

H36

234 (34)

269 (39)

124 (18)

10

74

319.0*

T6

165 (24)

248 (36)

69 (10)

2.0

80

†

T5

172 (25)

234 (34)

83 (12)

1.0

100

T6

207 (30)

289 (42)

103 (15)

1.5

105

T6

172 (25)

241 (35)

62 (9)

3.0

80

T7

248 (36)

262 (38)

62 (9)

0.5

85

3003 3004 5052 Cast: 333.0

335.0*

*Sand casting. † Permanent-mold casting.

(c) Mechanical Properties of Some Titanium Alloys Yield, Sy (0.2% offset) MPa (kpsi)

Strength Tensile, Sut MPa (kpsi)


Hardness (Brinell or Rockwell)

Titanium Alloy

Condition

Ti-35A†

Annealed

210 (30)

275 (40)

30

135 HB

Ti-50A†

Annealed

310 (45)

380 (55)

25

215 HB

Ti-0.2 Pd

Annealed

280 (40)

340 (50)

28

200 HB

Ti-5 Al-2.5 Sn

Annealed

760 (110)

790 (115)

16

36 HRC

Ti-8 Al-1 Mo-1 V

Annealed

900 (130)

965 (140)

15

39 HRC

Ti-6 Al-6 V-2 Sn

Annealed

970 (140)

1030 (150)

14

38 HRC

900 (130)

830 (120)

14

36 HRC

1207 (175)

1276 (185)

8

40 HRC

Ti-6Al-4V

Annealed

Ti-13 V-11 Cr-3 Al

Sol. + aging

†

Commercially pure alpha titanium 1001

1002

87.6

CD HR CD CD CD CD HT bolts

1018

1035

1045

1117

1137

12L14

1038

53.4

T6 .025”

75.5

67.5

T6

7075

64.9

T4

2024

28.1

175.4 0

2024

Ti-6AL-4V

198.8 149.1

AM350SS

A

17-7PSS

84.8 105.3

A

85.0

195.9

403SS

310SS

105.0

A A

CD

301SS

304SS

191.2

CD

201SS

64.8 122.2

Nodular Nodular

100-70-04

93.9

604515

Malleable Pearlitic

Malleable

35018

32510

44.5 53.3

Malleable

ASTM40

133.4

79.6

106.5

2.10

1.50

1.64

1.73

7.91

8.29

9.51

3.09

4.23

4.14

5.68

5.82

7.76

7.65

3.77

3.83

2.68

1.59

4.34

3.38

6.92

6.15

5.25

7.13

3.92

68.8

55.9

60.2

24.2

141.8

101.8

163.3

95.7

71.6

66.6

92.3

151.9

180.7

47.6

53.7

80.1

44.7

48.7

27.7

122.3

70.3

96.2

73.0

90.2

72.6

30.8

x0

76.2

68.1

65.5

28.7

178.5

152.4

202.3

106.4

86.3

86.6

106.6

193.6

197.9

125.6

66.1

95.3

54.3

53.8

46.2

134.6

80.4

107.7

84.4

120.5

87.5

90.1

θ

3.53

9.26

3.16

2.43

4.85

6.68

4.21

3.44

3.45

5.11

2.38

8.00

2.06

11.84

3.23

4.04

3.61

3.18

4.38

3.64

1.36

1.72

2.01

4.38

3.86

12

b

63.7

53.4

40.8

163.7

63.0

189.4

78.5

37.9

46.8

166.8

79.3

49.0

60.2

34.9

38.5

78.1

98.1

81.4

95.5

49.6

78.4

µSy

1.98

1.17

1.83

9.03

5.05

11.49

3.91

3.76

4.70

9.37

4.51

4.20

2.78

1.47

1.42

8.27

4.24

4.71

6.59

3.81

5.90

σSy

58.9

51.2

38.4

101.5

38.0

144.0

64.8

30.2

26.3

139.7

64.1

33.8

50.2

30.1

34.7

64.3

92.2

72.4

82.1

39.5

56

x0

64.3

53.6

41.0

167.4

65.0

193.8

79.9

38.9

48.7

170.0

81.0

50.5

61.2

35.5

39.0

78.8

98.7

82.6

97.2

50.8

80.6

2.63

1.91

1.32

8.18

5.73

4.48

3.93

2.17

4.99

3.17

3.77

4.06

4.02

3.67

2.93

1.72

1.41

2.00

2.14

2.88

4.29

b

0.0278

0.0222

0.0253

0.0616

0.0451

0.0556

0.0478

0.0293

0.0499

0.0487

0.0541

0.0304

0.0396

0.0626

0.0582

0.0408

0.0502

0.0298

0.0975

0.0253

0.0869

0.0577

0.0632

0.0606

0.0455

0.0655

CSut

0.0311

0.0219

0.0449

0.0552

0.0802

0.0607

0.0498

0.0992

0.1004

0.0562

0.0569

0.0857

0.0462

0.0421

0.0369

0.1059

0.0432

0.0579

0.0690

0.0768

0.0753

CSy

3:43 PM

83.1

117.7

5.74

σSut

6/3/03

86.2

µSut

Material

(March 1992), pp. 29–34.

θ

Source: Data compiled from “Some Property Data and

Corresponding Weibull Parameters for Stochastic Mechanical Design,” Trans. ASME Journal of Mechanical Design, vol. 114

Stochastic Yield and Ultimate Strengths for Selected Materials

Table A–25


510 (74) 588 (85.4) 3.4 455 (66)

579 (84) 699 (101.5) 4.3 510 (74) 604 (87.7) 5.2

θ b x0 θ b

712 (108)

684 (99.3)

38.1 (5.53)

116 (16.9)

21.4 (3.11)

5.0

528 (76.7)

657 (95.4)

36.6 (5.31)

95 (13.8)

17.4 (2.53)

5.5

463 (67.2)

393 (57)

4.1

496 (72.0)

420 (61)

2.85

425 (61.7)

391 (56.7)

106

493 (71.6)

35.1 (5.10)

77 (11.2)

14.0 (2.03)

107

9

Statistical parameters from a large number of fatigue tests are listed. Weibull distribution is denoted W and the parameters are x0, “guaranteed” fatigue strength; θ, characteristic fatigue strength; and b, shape factor. Normal distribution is denoted N and the parameters are µ, mean fatigue strength; and σ, standard deviation of the fatigue strength. The life is in stress-cycles-to-failure. TS = tensile strength, YS = yield strength. All testing by rotating-beam specimen.

992 (144)

µ

39.6 (5.75)

1040 (151)

365 (53)

143 (20.7)

HT-46

489 (71)

µ

W

N σ

599 (87)

2.75

2.60

b W x0

Ti-6A1-4V

T-4

2024

744 (108)

661 (96)

503 (73.0)

594 (86.2)

462 (67)

544 (79)

θ

105

7 8 Stress Cycles to Failure

x0

104

6

Aluminum

OQ&T, 1300°F

3140

799 (116)

W

Distribution

5

26.3 (3.82)

OQ&T 1200°F

2340

565 (82)

YS MPa (kpsi)

TS MPa (kpsi) 723 (105)

4

3

N σ

WQ&T, 1210°F

Condition

Number

1046

2

1

Source: E. B. Haugen, Probabilistic Mechanical Design, Wiley, New York, 1980,

3:43 PM

Appendix 10–B.

Stochastic Parameters for Finite Life Fatigue Tests in Selected Metals

6/3/03

Table A–26


1003

1004 Source: Compiled from Table 4 in H. J. Grover, S. A. Gordon,

860

OQT

1200

OQT

As Rec.

369

224

227

162

67 Rb

277

1200

193

WQT

196

N

1200

164

*BHN = Brinell hardness number; RA = fractional reduction in area.

10120

1095

1060

.56 MN

N, AC

1050 WQT

HR, N

1045

209 195

WQT Forged

1040

132

Normal

1035

135

Air-cooled

1030

BHN*

Furnace cooled

Condition

180

117

115

84

134

111

98

97

92

107

92

103

72

80

58

130

59

65

33

65

84

47

70

47

63

53

87

35

45

30

0.15

0.12

0.40

0.37

0.20

0.57

0.42

0.58

0.40

0.49

0.23

0.65

0.54

0.62

0.63

RA*

77

50

65

94

61

50

80

10

4

102

60

68

43

60

81

55

60

48

70

80

51

4(10 )

4

95

56

64

40

55

73

51

57

46

56

72

44

47

37

105

91

51

57

34

50

62

47

52

40

47

40

65

40

42

34

4(105)

91

50

56

31

48

57

43

50

38

47

47

60

37

38

30

106

91

50

56

30

48

55

41

50

34

47

33

57

34

38

28

4(106)

Stress Cycles to Failure

91

50

56

30

48

55

41

50

34

47

33

57

33

38

25

107

56

30

55

41

50

57

33

108

3:43 PM

1020

Material

Yield Strength kpsi

6/3/03

Tensile Strength kpsi

and L. R. Jackson, Fatigue of Metals and Structures, Bureau of Naval Weapons Document NAVWEPS 00-25-534, 1960.

Finite Life Fatigue Strengths of Selected Plain Carbon Steels

Table A–27


0.324 9 0.289 3 0.257 6 0.229 4 0.204 3 0.181 9 0.162 0 0.144 3 0.128 5 0.114 4 0.101 9 0.090 74 0.080 81 0.071 96 0.064 08 0.057 07 0.050 82 0.045 26

0 1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17

0.065 0.058

0.120 0.109 0.095 0.083 0.072

0.203 0.180 0.165 0.148 0.134

0.340 0.300 0.284 0.259 0.238 0.220

0.454 0.425 0.380

Principal Use:

0.062 5 0.056 25

0.125 0.109 357 0.093 75 0.078 125 0.070 312 5

0.203 125 0.187 5 0.171 875 0.156 25 0.140 625

0.312 5 0.281 25 0.265 625 0.25 0.234 375 0.218 75

0.500 0.468 75 0.437 5 0.406 25 0.375 0.343 75

Nonferrous Sheet, Wire, and Rod

0.059 8 0.053 8

0.119 6 0.104 6 0.089 7 0.074 7 0.067 3

0.194 3 0.179 3 0.164 4 0.149 5 0.134 5

0.239 1 0.224 2 0.209 2

Ferrous Sheet

0.062 5 0.054 0

0.120 5 0.105 5 0.091 5 0.080 0 0.072 0

0.192 0 0.177 0 0.162 0 0.148 3 0.135 0

0.306 5 0.283 0 0.262 5 0.243 7 0.225 3 0.207 0

0.490 0.461 5 0.430 5 0.393 8 0.362 5 0.331 0

Ferrous Wire Except Music Wire

Steel Wire or Washburn & Moen

0.037 0.039

0.026 0.029 0.031 0.033 0.035

0.016 0.018 0.020 0.022 0.024

0.009 0.010 0.011 0.012 0.013 0.014

0.004 0.005 0.006 0.007 0.008

Music Wire

Music Wire

0.175 0.172

0.188 0.185 0.182 0.180 0.178

0.201 0.199 0.197 0.194 0.191

0.227 0.219 0.212 0.207 0.204

Steel Drill Rod

Stubs Steel Wire

(continued)

0.177 0 0.173 0

0.191 0 0.189 0 0.185 0 0.182 0 0.180 0

0.204 0 0.201 0 0.199 0 0.196 0 0.193 5

0.228 0 0.221 0 0.213 0 0.209 0 0.205 5

Twist Drills and Drill Steel

Twist Drill

3:43 PM

0.580 0 0.516 5 0.460 0 0.409 6 0.364 8

Ferrous Sheet and Plate, 480 lbf/ft3

Tubing, Ferrous Strip, Flat Wire, and Spring Steel

Manufacturers Standard

6/3/03

7/0 6/0 5/0 4/0 3/0 2/0

United States Standard †

Birmingham or Stubs Iron Wire

American or Brown & Sharpe

Name of Gauge:

Decimal Equivalents of Wire and Sheet-Metal Gauges* (All Sizes Are Given in Inches)

Table A–28


1005

1006

Nonferrous Sheet, Wire, and Rod 0.040 30 0.035 89 0.031 96 0.028 46 0.025 35 0.022 57 0.020 10 0.017 90 0.015 94 0.014 20 0.012 64 0.011 26 0.010 03 0.008 928 0.007 950 0.007 080 0.006 305 0.005 615 0.005 000 0.004 453 0.003 965 0.003 531 0.003 145

Principal Use:

18 19 20

21 22 23 24 25

26 27 28 29 30

31 32 33 34 35

36 37 38 39 40

0.007 031 25 0.006 640 625 0.006 25

0.010 937 5 0.010 156 25 0.009 375 0.008 593 75 0.007 812 5

0.018 75 0.017 187 5 0.015 625 0.014 062 5 0.012 5

0.034 375 0.031 25 0.028 125 0.025 0.021 875

*Specify sheet, wire, and plate by stating the gauge number, the gauge name, and the decimal equivalent in parentheses. † Reflects present average and weights of sheet steel.

0.004

0.010 0.009 0.008 0.007 0.005

0.018 0.016 0.014 0.013 0.012

0.032 0.028 0.025 0.022 0.020

0.006 7 0.006 4 0.006 0

0.010 5 0.009 7 0.009 0 0.008 2 0.007 5

0.017 9 0.016 4 0.014 9 0.013 5 0.012 0

0.032 9 0.029 9 0.026 9 0.023 9 0.020 9

0.047 8 0.041 8 0.035 9

Ferrous Sheet

0.009 0 0.008 5 0.008 0 0.007 5 0.007 0

0.013 2 0.012 8 0.011 8 0.010 4 0.009 5

0.018 1 0.017 3 0.016 2 0.015 0 0.014 0

0.031 7 0.028 6 0.025 8 0.023 0 0.020 4

0.047 5 0.041 0 0.034 8

Ferrous Wire Except Music Wire

Steel Wire or Washburn & Moen

0.085 0.090 0.095

0.063 0.067 0.071 0.075 0.080

0.047 0.049 0.051 0.055 0.059

0.041 0.043 0.045

Music Wire

Music Wire

0.106 0.103 0.101 0.099 0.097

0.120 0.115 0.112 0.110 0.108

0.146 0.143 0.139 0.134 0.127

0.157 0.155 0.153 0.151 0.148

0.168 0.164 0.161

Steel Drill Rod

Stubs Steel Wire

0.106 5 0.104 0 0.101 5 0.099 5 0.098 0

0.120 0 0.116 0 0.113 0 0.111 0 0.110 0

0.147 0 0.144 0 0.140 5 0.136 0 0.128 5

0.159 0 0.157 0 0.154 0 0.152 0 0.149 5

0.169 5 0.166 0 0.161 0

Twist Drills and Drill Steel

Twist Drill

3:43 PM

0.05 0.043 75 0.037 5

Ferrous Sheet and Plate, 480 lbf/ft3

Tubing, Ferrous Strip, Flat Wire, and Spring Steel

Manufacturers Standard

6/3/03

0.049 0.042 0.035

United States Standard †

Birmingham or Stubs Iron Wire

American or Brown & Sharpe

Name of Gauge:

Decimal Equivalents of Wire and Sheet-Metal Gauges* (All Sizes Are Given in Inches) (Continued)

Table A–28


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6/3/03

3:43 PM

Page 1007

Useful Tables

1007

Table A–29 Dimensions of Square and Hexagonal Bolts H

W R

Head Type Square

Regular Hexagonal

Heavy Hexagonal

Structural Hexagonal

W

H

Rmin

W

H

Rmin

0.01

7 8

5 16

0.009

25 64

0.021

Nominal Size, in

W

H

W

H

Rmin

1 4

3 8

11 64

7 16

11 64

0.01

5 16

1 2

13 64

1 2

7 32

0.01

3 8

9 16

1 4

9 16

1 4

0.01

7 16

5 8

19 64

5 8

19 64

0.01

1 2

3 4

21 64

3 4

11 32

0.01

7 8

11 32

5 8

15 16

27 64

15 16

27 64

0.02

1 1 16

27 64

0.02

1 1 16

1 18

1 2

1 18

1 2

0.02

1 14

1 2

0.02

1 14

15 32

0.021

1

1 12

21 32

1 12

43 64

0.03

1 58

43 64

0.03

1 58

39 64

0.062

1 18

1 11 16

3 4

0.03

1 13 16

3 4

0.03

1 13 16

11 16

0.062

1 14

1 78

27 32

1 78

27 32

0.03

2

27 32

0.03

2

25 32

0.062

1 38

1 2 16

29 32

1 2 16

29 32

0.03

3 2 16

29 32

0.03

3 2 16

27 32

0.062

1 12

2 14

1

2 14

1

0.03

2 38

1

0.03

2 38

15 16

0.062

8

3.58

8

3.58

0.2

M6

10

4.38

0.3

M8

13

5.68

0.4

M10

16

6.85

0.4

M12

18

7.95

0.6

21

7.95

0.6

M14

21

9.25

0.6

24

9.25

0.6

M16

24

10.75

0.6

27

10.75

0.6

27

10.75

0.6

M20

30

13.40

0.8

34

13.40

0.8

34

13.40

0.8

M24

36

15.90

0.8

41

15.90

0.8

41

15.90

1.0

M30

46

19.75

1.0

50

19.75

1.0

50

19.75

1.2

M36

55

23.55

1.0

60

23.55

1.0

60

23.55

1.5

3 4

3 4

1 11 16

Nominal Size, mm M5

shi20361_app_A.qxd

1008

6/3/03

3:43 PM

Page 1008


Table A–30 Dimensions of Hexagonal Cap Screws and Heavy Hexagonal Screws (W = Width across Flats; H = Height of Head; See Figure in Table A–29)

Nominal Size, in

Minimum Fillet Radius

Type of Screw Cap W

Heavy W

Height H

1 4 5 16

0.015

7 16

5 32

0.015

1 2

3 8

0.015

9 16

7 16

0.015

5 8

13 64 15 64 9 32

1 2

0.015

5 8

0.020

3 4 15 16

3 4 7 8

0.020

1 18

1 14

0.040

5 1 16

7 1 16

1

0.060

1 12

1 18

1 14

0.060

1 78

2

35 64 39 64 25 32

1 38

0.060

1 2 16

3 2 16

27 32

1 12

0.060

2 14

2 38

15 16

7 8

5 16

1 1 16

25 64 15 32

Nominal Size, mm M5

0.2

8

3.65

M6

0.3

10

4.15

M8

0.4

13

5.50

M10

0.4

16

6.63

M12

0.6

18

21

7.76

M14

0.6

21

24

9.09

M16

0.6

24

27

10.32

M20

0.8

30

34

12.88

M24

0.8

36

41

15.44

M30

1.0

46

50

19.48

M36

1.0

55

60

23.38

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3:43 PM

Page 1009

Useful Tables

Table A–31 Dimensions of Hexagonal Nuts

Height H Nominal Size, in

Width W

Regular Hexagonal

Thick or Slotted

JAM

1 4 5 16

7 16

7 32

9 32

5 32

1 2 9 16

21 64 13 32

3 16

3 8 7 16

11 16

17 64 21 64 3 8

1 2

1 4 5 16

9 16

3 4 7 8

29 64 9 16

5 16

5 8

15 16

39 64 23 32

3 4 7 8

1 18

13 16

7 16 31 64 35 64 41 64 3 4 55 64 31 32

5 1 32

7 32

3 8

1

1 12

1 18

1 11 16

1 14

1 78

1 1 16

1 14

27 64 31 64 35 64 39 64 23 32

1 38

1 2 16

1 11 64

1 38

25 32

1 12

2 14

9 1 32

1 12

27 32

5 1 16

29 32

1

Nominal Size, mm M5

8

4.7

5.1

2.7

M6

10

5.2

5.7

3.2

M8

13

6.8

7.5

4.0

M10

16

8.4

9.3

5.0

M12

18

10.8

12.0

6.0

M14

21

12.8

14.1

7.0

M16

24

14.8

16.4

8.0

M20

30

18.0

20.3

10.0

M24

36

21.5

23.9

12.0

M30

46

25.6

28.6

15.0

M36

55

31.0

34.7

18.0

1009

shi20361_app_A.qxd

6/3/03

Table A–32 Basic Dimensions of American Standard Plain Washers (All Dimensions in Inches)

3:43 PM

Page 1010

Fastener Size

ID

OD

Thickness

#6

0.138

0.156

0.375

0.049

#8

0.164

0.188

0.438

0.049

#10

0.190

0.219

0.500

0.049

3 16

0.188

0.250

0.562

0.049

#12

0.216

0.250

0.562

0.065

1 N 4 1 W 4 5 N 16 5 W 16 3 N 8 3 W 8 7 N 16 7 W 16 1 N 2 1 W 2 9 N 16 9 W 16 5 N 8 5 W 8 3 N 4 3 W 4 7 N 8 7 W 8

0.250

0.281

0.625

0.065

1N

0.250

0.312

0.734

0.065

0.312

0.344

0.688

0.065

0.312

0.375

0.875

0.083

0.375

0.406

0.812

0.065

0.375

0.438

1.000

0.083

0.438

0.469

0.922

0.065

0.438

0.500

1.250

0.083

0.500

0.531

1.062

0.095

0.500

0.562

1.375

0.109

0.562

0.594

1.156

0.095

0.562

0.625

1.469

0.109

0.625

0.656

1.312

0.095

0.625

0.688

1.750

0.134

0.750

0.812

1.469

0.134

0.750

0.812

2.000

0.148

0.875

0.938

1.750

0.134

0.875

0.938

2.250

0.165

1.000

1.062

2.000

0.134

1W

1.000

1.062

2.500

0.165

1 18 N

1.125

1.250

2.250

0.134

1 18 W

1.125

1.250

2.750

0.165

1 14 N

1.250

1.375

2.500

0.165

1 14 1 38 1 38 1 12 1 12 1 58 1 34 1 78

W

1.250

1.375

3.000

0.165

N

1.375

1.500

2.750

0.165

W

1.375

1.500

3.250

0.180

N

1.500

1.625

3.000

0.165

W

1.500

1.625

3.500

0.180

1.625

1.750

3.750

0.180

1.750

1.875

4.000

0.180

1.875

2.000

4.250

0.180

2

2.000

2.125

4.500

0.180

2 14

2.250

2.375

4.750

0.220

2 12 2 34

2.500

2.625

5.000

0.238

2.750

2.875

5.250

0.259

3

3.000

3.125

5.500

0.284

N = narrow; W = wide; use W when not specified. 1010

Diameter

Washer Size

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Useful Tables

1011

Table A–33 Dimensions of Metric Plain Washers (All Dimensions in Millimeters) Washer Size*

Minimum ID

Maximum OD

Maximum Thickness

Washer Size*

Minimum ID

Maximum OD

Maximum Thickness

1.6 N

1.95

4.00

0.70

10 N

10.85

20.00

2.30

1.6 R

1.95

5.00

0.70

10 R

10.85

28.00

2.80

1.6 W

1.95

6.00

0.90

10 W

10.85

39.00

3.50

2N

2.50

5.00

0.90

12 N

13.30

25.40

2.80

2R

2.50

6.00

0.90

12 R

13.30

34.00

3.50

2W

2.50

8.00

0.90

12 W

13.30

44.00

3.50

2.5 N

3.00

6.00

0.90

14 N

15.25

28.00

2.80

2.5 R

3.00

8.00

0.90

14 R

15.25

39.00

3.50

2.5 W

3.00

10.00

1.20

14 W

15.25

50.00

4.00

3N

3.50

7.00

0.90

16 N

17.25

32.00

3.50

3R

3.50

10.00

1.20

16 R

17.25

44.00

4.00

3W

3.50

12.00

1.40

16 W

17.25

56.00

4.60

3.5 N

4.00

9.00

1.20

20 N

21.80

39.00

4.00

3.5 R

4.00

10.00

1.40

20 R

21.80

50.00

4.60

3.5 W

4.00

15.00

1.75

20 W

21.80

66.00

5.10

4N

4.70

10.00

1.20

24 N

25.60

44.00

4.60

4R

4.70

12.00

1.40

24 R

25.60

56.00

5.10

4W

4.70

16.00

2.30

24 W

25.60

72.00

5.60

5N

5.50

11.00

1.40

30 N

32.40

56.00

5.10

5R

5.50

15.00

1.75

30 R

32.40

72.00

5.60

5W

5.50

20.00

2.30

30 W

32.40

90.00

6.40

6N

6.65

13.00

1.75

36 N

38.30

66.00

5.60

6R

6.65

18.80

1.75

36 R

38.30

90.00

6.40

6W

6.65

25.40

2.30

36 W

38.30

110.00

8.50

8N

8.90

18.80

2.30

8R

8.90

25.40

2.30

8W

8.90

32.00

2.80

N = narrow; R = regular; W = wide. *Same as screw or bolt size.

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Table A–34

Values of (n) =

e−x x n−1 dx; (n + 1) = n(n)

0

Gamma Function* Source: Reprinted with permission from William H. Beyer (ed.), Handbook of Tables for Probability and Statistics, 2nd ed., 1966. Copyright CRC Press, Boca Raton, Florida.

∞

n

(n)

n

(n)

n

(n)

n

(n)

1.00

1.000 00

1.25

.906 40

1.50

.886 23

1.75

.919 06

1.01

.994 33

1.26

.904 40

1.51

.886 59

1.76

.921 37

1.02

.988 84

1.27

.902 50

1.52

.887 04

1.77

.923 76

1.03

.983 55

1.28

.900 72

1.53

.887 57

1.78

.926 23

1.04

.978 44

1.29

.899 04

1.54

.888 18

1.79

.928 77

1.05

.973 50

1.30

.897 47

1.55

.888 87

1.80

.931 38

1.06

.968 74

1.31

.896 00

1.56

.889 64

1.81

.934 08

1.07

.964 15

1.32

.894 64

1.57

.890 49

1.82

.936 85

1.08

.959 73

1.33

.893 38

1.58

.891 42

1.83

.939 69

1.09

.955 46

1.34

.892 22

1.59

.892 43

1.84

.942 61

1.10

.951 35

1.35

.891 15

1.60

.893 52

1.85

.945 61

1.11

.947 39

1.36

.890 18

1.61

.894 68

1.86

.948 69

1.12

.943 59

1.37

.889 31

1.62

.895 92

1.87

.951 84

1.13

.939 93

1.38

.888 54

1.63

.897 24

1.88

.955 07

1.14

.936 42

1.39

.887 85

1.64

.898 64

1.89

.958 38

1.15

.933 04

1.40

.887 26

1.65

.900 12

1.90

.961 77

1.16

.929 80

1.41

.886 76

1.66

.901 67

1.91

.965 23

1.17

.936 70

1.42

.886 36

1.67

.903 30

1.92

.968 78

1.18

.923 73

1.43

.886 04

1.68

.905 00

1.93

.972 40

1.19

.920 88

1.44

.885 80

1.69

.906 78

1.94

.976 10

1.20

.918 17

1.45

.885 65

1.70

.908 64

1.95

.979 88

1.21

.915 58

1.46

.885 60

1.71

.910 57

1.96

.983 74

1.22

.913 11

1.47

.885 63

1.72

.912 58

1.97

.987 68

1.23

.910 75

1.48

.885 75

1.73

.914 66

1.98

.991 71

1.24

.908 52

1.49

.885 95

1.74

.916 83

1.99

.995 81

2.00 1.000 00 *For large positive values of x, (x) approximates the asymptotic series 2x 139 571 1 1 x x e−x − − + ··· 1+ + 2 3 4 x 12x 288x 51 840x 2 488 320x

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Appendix

Answers to Selected Problems B–1

Chapter 1

1–8 10 ≤ F ≤ 10.5 lbf, 13.5 ≤ N ≤ 14.2 lbf, K = 0.967, 19.3 ≤ T ≤ 20.3 lbf · in 1–12 (a) e1 = 0.006 067 977, e2 = 0.009 489 743, e = 0.015 557 720, (b) e1 = −0.003 932 023, e2 = −0.000 510 257, e = −0.004 442 280 1–15 (a) σ = 13.1 MPa, (b) σ = 70 MPa, (c) y = 15.5 mm, (d) θ = 5.18◦

B–2

Chapter 2

2–1 x¯ = 122.9 kilocycles, sx = 30.3 kilocycles 2–2 x¯ = 198.55 kpsi, sx = 9.55 kpsi 2–3 x¯ = 78.4 kpsi, sx = 6.57 kpsi 2–11 (a) F¯i = 5.979 lbf, s Fi = 0.396 lbf; (b) k¯ = 9.766 lbf/in, sk = 0.390 lbf/in 2–19 L 10 = 84.1 kcycles 2–23 R = 0.987 2–25 (a) w = 0.020 ± 0.015 in, (b) w = 0.020 ± 0.005 in 2–32 D¯ 0 = 4.012 in, t D0 = 0.036 in; D0 = 4.012 ± 0.036 in 2–39 x¯ = 98.26 kpsi, sx = 4.30 kpsi 2–46 µn = 122.9 kcycles, σˆ n = 34.8 kcycles

B–3

Chapter 3

3–9 E = 30 Mpsi, Sy = 45.5 kpsi, Sut = 85.5 kpsi, area reduction = 45.8 percent . 3–11 (Sy )0.001 = 35 kpsi, E σ =0 = 25 Mpsi, E σ =20 kpsi = 14 Mpsi . 3–14 G = 77.3 GPa, Sys = 200 MPa 3–18 S¯ut = 125.2 kpsi, σˆ Sut = 1.9 kpsi . . 3–20 (a) u R = 34.5 in · lbf/in3 , (b) u T = 3 3 66.7(10 ) in · lbf/in

B–4

B

Chapter 4

4–4 (a) V (x) = −1.43 − 40x − 40 + 30x − 80 + 71.43x − 140 − 60x − 180 lbf M(x) = −1.43x − 40x − 41 + 30x − 81 + 71.43x − 141 − 60x − 181 lbf · in 4–6 (a) Mmax = 253 lbf · in, (b) (a/l)∗ = 0.207, M ∗ = 214 lbf · in 4–8 (a) σ1 = 14, σ2 = 4, σ3 = 0, 2θ = 53.1◦ cw; (b) σ1 = 18.6, σ2 = 6.4, σ3 = 0, 2θ = 55◦ ccw; (c) σ1 = 26.2, σ2 = 7.78, σ3 = 0, 2θ = 139.7◦ cw; (d) σ1 = 23.4, σ2 = 4.57, σ3 = 0, 2θ = 122◦ cw 4–13 σ = 10.2 kpsi, δ = 0.0245 in, 1 = 0.000 340, ν = 0.292, 2 = −0.000 099 1, d = −0.000 049 6 in 4–18 σ1 = 30 MPa, σ2 = 10 MPa, σ3 = −20 MPa, τmax = 25 MPa 4–22 (a) Mmax = 21 600 kip · in, (b) xmax = 523 in from left or right supports 4–23 (a) σ A = 42 kpsi, σ B = 18.5 kpsi, σC = 2.7 kpsi, σ D = −52.7 kpsi 4–27 Mmax = 219 lbf · in, σ = 17.8 kpsi, τmax = 3.4 kpsi, both models 4–33 The same 1 -in-thick strips: Tmax = 29.95 lbf · in, 4–37 Two 16 θ = 0.192 rad, kt = 156 lbf · in/rad. One 18 -in-thick strip: Tmax = 59.90 lbf · in, θ = 0.0960 rad, kt = 624 lbf · in/rad 4–43 dC = 45 mm 4–47 σmax = 11.79 kpsi, τmax = 7.05 kpsi 4–53 pi = 639 psi 4–57 (σr )max = 3656 psi 4–65 δmax = 0.038 mm, δmin = 0.0175 mm, pmax = 147.5 MPa, pmin = 67.9 MPa 4–69 For δmax , p = 33.75 kpsi, (σt )o = 56.25 kpsi, (σt )i = −33.75 kpsi, δo = 0.001 11 in, δi = −0.000 398 in 4–72 σi = 26.3 kpsi, σo = −15.8 kpsi 1013

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4–77 σi = 71.3 kpsi, σo = −34.2 kpsi 4–81 pmax = 399F 1/3 MPa, σmax = 399F 1/3 MPa, τmax = 120F 1/3 MPa

B–5

Chapter 5

5–1 (a) k = (1/k1 + 1/k2 + 1/k3 ), (b) k = k1 + k2 + k3 , (c) k = [1/k1 + 1/(k2 + k3 )]−1 5–10 λ = 8a 2 /(3l) 5–12 σmax = −20.4 kpsi, y = −0.908 in 5–15 yleft = −0.0506 in, yright = −0.0506 in, ymidspan = 0.0190 in 5–18 ymax = −0.0130 in 5–20 z A = 0.0368 in, z B = 0.00430 in 5–26 Use d = 1 38 in 5–30 y B = 0.0459 in 5–37 y A = −0.101 in, yx=20 in = −0.104 in 5–45 y A = −0.133 in 5–48 yx=10 in = −0.0167 in 5–51 (a) σb = 76.5 kpsi, σc = −15.2 kpsi, (b) σb = 78.4 kpsi, σc = −13.3 kpsi 5–56 R O = 3.89 kip, RC = 1.11 kip, both in same direction 5–59 σ B E = 140 MPa, σ D F = 71.2 MPa, y B = −0.670 mm, yC = −2.27 mm, y D = −0.341 mm 5–64 δ A = (π + 4)P R 3 /(4E I ), δ B = π P R 3 /(4E I ) 5–67 δ = 0.476 mm 5–73 (a) t = 0.5 in, (b) No 5–81 ymax = 2k1 a/(k1 + k2 )

B–6

Chapter 6

6–2 (a) MSS: n = 4.17, DE: n = 4.17, (b) MSS: n = 4.17, DE: n = 4.81, (c) MSS: n = 2.08, DE: n = 2.41, (c) MSS: n = 4.17, DE: n = 4.81 6–3 (a) MSS: n = 2.17, DE: n = 2.50, (b) MSS: n = 1.45, DE: n = 1.56, (c) MSS: n = 1.52, DE: n = 1.65, (c) MSS: n = 1.27, DE: n = 1.50 6–9 (a) DE: σ = 12.29 kpsi, n = 3.42 6–10 (a) DCM: σ1 = 90 kpsi, σ2 = 0, σ3 = −50 kpsi, r = −0.56, n = 1.77 6–12 (a) M2M: n = 3.89 6–13 (a) σ A = σ B = 20 kpsi, r = 1, n = 1.5

6–18 (σt )max 13.21 kpsi, σl = 6.42 kpsi, σr = −500 psi, σ = 11.9 kpsi, n = 3.87 6–21 Using BCM, select d = 1 38 in 6–25 d = 18 mm 6–32 (a) δ = 0.0005 in, p = 3516 psi, (σt )i = −5860 psi, (σr )i = −3516 psi, (σt )o = −9142 psi, (σr )o = −3516 psi 6–35 n o = 2.81, n i = 2.41 6–40 p = 29.2 MPa

B–7

Chapter 7

7–1 Se = 94.4 kpsi 7–3 Se = 33.4 kpsi, σ F = 112.4 kpsi, b = −0.0836, f = 0.899, a = 106.1 kpsi, S f = 48.2 kpsi, N = 409 530 cycles 7–5 (S f )ax = 162N −0.0851 kpsi, 103 ≤ N ≤ 106 cycles 7–6 Se = 243 MPa 7–10 Se = 221.8 MPa, ka = 0.899, kb = 1, kc = 0.85, Se = 169.5 MPa, K t = 2.5, K f = 2.09, Fa = 21.6 kN, Fy = 98.7 kN 7–12 Yield: n y = 1.18. Fatigue: (a) n f = 1.06, (b) n f = 1.31, (c) n f = 1.32 7–17 n y = 5.06, (a) n f = 2.44, (b) n f = 2.55 7–23 At the fillet n f = 1.70 7–24 (a) T = 3.42 N · m, (b) T = 4.21 N · m, (c) n y = 1.91 7–27 (a) Pall = 16.1 kN, n y = 5.69, (b) Pall = 51.4 kN, n y = 3.87 7–29 (a) 24 900 cycles, (b) 27 900 cycles 7–34 Rotation presumed. Se = 55.7 LN(1, 0.138) kpsi, ka = 0.768 LN(1, 0.058), kb = 0.879, Se = 37.6 LN(1, 0.150) kpsi, K f = 1.598 LN(1, 0.15), = 22.8 LN(1, 0.15) kpsi, z = −2.373, R = 0.991

B–8

Chapter 8

8–1 (a) Thread depth 2.5 mm, thread width 2.5 mm, dm = 22.5 mm, dr = 20 mm, l = p = 5 mm 8–4 TR = 16.23 N · m, TL = 6.62 N · m, e = 0.294 8–8 T = 16.5 lbf · in, dm = 0.5417 in, l = 0.1667 in, sec α = 1.033, T = 0.0696F , Tc = 0.0328F, Ttotal = 0.1024F , F = 161 lbf

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Answers to Selected Problems

8–11 L T = 1.25 in, L G = 1.109 in, H = 0.4375 in, L G + H = 1.5465 in, use 1.75 in, ld = 0.500 in, lt = 0.609 in 8–13 L T = 1.25 in, L > h + 1.5d = 1.625 in, use 1.75 in, ld = 0.500 in, lt = 0.625 in 8–15 (a) Ad = 0.442 in2 , Atube = 0.552 in2 , kb = 1.02(106 ) lbf/in, km = 1.27(106 ) lbf/in, C = 0.445, (b) Fi = 11 810 lbf 8–18 Frusta to Wileman ratio is 1.11/1.08 8–22 n = 4.73 8–23 n = 5.84 8–27 kb = 4.63 Mlbf/in, km = 7.99 Mlbf/in using frustums 8–34 (a) L = 2.5 in, (b) kb = 6.78 Mlbf/in, km = 14.41 Mlbf/in, C = 0.320 8–37 Load: n = 3.19. Separation: n = 4.71. Fatigue: n = 3.27 8–43 Bolt shear: n = 3.26. Bolt bearing: n = 5.99. Member bearing: n = 3.71. Member tension: n = 5.36 8–48 F = 2.22 kN 8–50 Bearing on bolt, n = 9.58; shear of bolt, n = 5.79; bearing on members, n = 5.63; bending of members, n = 2.95

B–9

Chapter 9

9–1 F = 17.7 kip 9–3 F = 11.3 kip 9–5 (a) τ = 1.13F kpsi, τx = τ y = 5.93F kpsi, τmax = 9.22F kpsi, F = 2.17 kip; (b) τall = 11 kpsi, Fall = 1.19 kip 9–8 τ = 0 (why?), F = 49.2 kN 9–9 A two-way tie for first, vertical parallel beads, and square beads 9–10 First: horizontal parallel beads. Second: square beads 9–11 Decisions: Pattern; all-around square Electrode: E60XX Type: two parallel fillets, two transverse fillets Length of beads: 12 in Leg: 14 in 9–20 τmax = 18 kpsi 9–22 n = 3.57

B–10

1015

Chapter 10

10–3 (a) L 0 = 5.17 in, (b) FSsy = 45.2 lbf, (c) k = 11.55 lbf/in, (d) (L 0 )cr = 5.89 in, guide spring 10–5 (a) L 0 = 47.7 mm, (b) p = 5.61 mm, (c) Fs = 81.1 N, (d) k = 2643 N/m, (e) (L 0 )cr = 105.2 mm, needs guidance 10–9 Not solid safe, L 0 ≤ 0.577 in 10–15 Not solid safe, L 0 ≤ 66.6 mm 10–19 (a) p = 10 mm, L s = 44.2 mm, Na = 12 turns, (b) k = 1080 N/m, (c) Fs = 81.9 N, (d) τs = 271 MPa 10–29 (a) L 0 = 16.12 in, (b) τi = 14.95 kpsi, (c) k = 4.855 lbf/in, (d) F = 85.8 lbf, (e) y = 14.4 in 10–33 (a) k = 24.7 lbf · in/turn each, (b) 297 kpsi 10–34 k = 2E I /[R 2 (19π R + 18l)]

B–11

Chapter 11

11–1 x D = 540, FD = 2.278 kN, C10 = 18.59 kN, 02–30 mm deep-groove ball bearing, R = 0.919 11–4 R = R1 R2 = 0.927(0.942) = 0.873, goal not met 11–8 x D = 180, C10 = 57.0 kN 11–11 C10 = 8.88 kN 11–13 R O = 195 N, R E = 196 N, deep-groove 02–25 mm at O and C 11–18 l2 = 0.267(106 ) rev

B–12

Chapter 12

12–1 cmin = 0.000 75 in, r = 0.500 in, r/c = 667, N j = 18.3 r/s, S = 0.261, h 0 /c = 0.595, r f /c = 5.8, Q/(rcNl) = 3.98, Q s /Q = 0.5, h 0 = 0.000 446 in, H = 0.0134 Btu/s, Q = 0.0274 in3/s, Q s = 0.0137 in3/s 12–3 SAE 10: h 0 = 0.000 275 in, pmax = 847 psi, cmin = 0.0025 in 12–7 h 0 = 0.0165 mm, f = 0.007 65, Q = 1263 mm3/s 12–9 h 0 = 0.010 mm, H = 34.3 W, Q = 1072 mm3/s, Q s = 793 mm3/s 12–11 Tav = 65◦ C, h 0 = 0.0272 mm, H = 45.2 W, Q s = 1712 mm3/s 12–20 15.2 mPa · s

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B–13

Chapter 13

Page 1016

13–1 35 teeth, 3.25 in 13–2 400 rev/min, p = 3π mm, C = 112.5 mm 13–4 a = 0.3333 in, b = 0.4167 in, c = 0.0834 in, p = 1.047 in, t = 0.523 in, d1 = 7 in, d1b = 6.578 in, d2 = 9.333 in, d2b = 8.77 in, pb = 0.984 in 13–5 d P = 2.333 in, dG = 5.333 in, γ = 23.63◦ , = 66.37◦ , A0 = 2.911 in, F = 0.873 in 13–8 (a) 13, (b) 15, (c) 18, (d) 16 13–10 10:20 and higher 13–13 (a) pn = 3π mm, pt = 10.40 mm, px = 22.30 mm, (b) m t = 3.310 mm, φt = 21.88◦ , (c) d p = 59.58 mm, dG = 105.92 mm 13–15 e = 4/51, n d = 47.06 rev/min cw 13–22 n A = 68.57 rev/min cw 13–27 n b /n a = 11/36 same sense 13–33 F A = 71.5 i + 53.4 j + 350.5 k lbf, F B = −149.5 i − 590.4 k lbf 13–40 FC = 1565 i + 672 j lbf; F D = 1610 i − 425 j + 154 k lbf

B–14

Chapter 14

14–1 σ = 7.63 kpsi 14–4 σ = 82.6 MPa 14–7 F = 2.5 in 14–10 m = 2 mm, F = 25 mm 14–14 σc = −617 MPa 14–17 W t = 16 890 N, H = 97.2 kW (pinion bending); W t = 3433 N, H = 19.8 kW (pinion and gear wear) 14–18 W t = 1356 lbf, H = 34.1 hp (pinion bending); W t = 1720 lbf, H = 43.3 hp (gear bending), W t = 265 lbf; H = 6.67 hp (pinion and gear wear) 14–22 W t = 775 lbf, H = 19.5 hp (pinion bending); W t = 300 lbf, H = 7.55 hp (pinion wear) AGMA method accounts for more conditions 14–24 Rating power = min(157.5, 192.9, 53.0, 59.0) = 53 hp 14–28 Rating power = min(270, 335, 240, 267) = 240 hp 14–34 H = 69.7 hp

B–15

Chapter 15

= 690 lbf, H1 = 16.4 hp, WGt = 620 lbf, 15–1 H2 = 14.8 hp 15–2 W Pt = 464 lbf, H3 = 11.0 hp, WGt = 531 lbf, H4 = 12.6 hp 15–8 Pinion core 300 Bhn, case, 373 Bhn; gear core 339 Bhn, case, 345 Bhn 15–9 All four W t = 690 lbf 15–11 Pinion core 180 Bhn, case, 266 Bhn; gear core, 180 Bhn, case, 266 Bhn W Pt

B–16

Chapter 16

16–1 (a) Right shoe: pa = 111.4 psi cw rotation, (b) Right shoe: T = 2530 lbf · in; left shoe: 1310 lbf · in; total T = 3840 lbf · in, (c) RH shoe: R x = −229 lbf, R y = 940 lbf, R = 967 lbf; LH shoe: R x = 130 lbf, R y = 171 lbf, R = 215 lbf 16–3 LH shoe: T = 161.4 N · m, pa = 610 kPa; RH shoe: T = 59.0 N · m, pa = 222.8 kPa, Ttotal = 220.4 N · m 16–5 pa = 203 kN, T = 38.76 N · m 16–8 a = 1.209r , a = 1.170r 16–10 P = 1560 lbf, T = 29 980 lbf · in 16–14 (a) T = 8200 lbf · in, P = 504 lbf, H = 26 hp; (b) R = 901 lbf; (c) p|θ=0 = 70 psi, p|θ=270◦ = 27.3 psi 16–17 (a) F = 1885 lbf, T = 7125 lbf · in; (c) torque capacity exhibits a stationary point maximum √ d ∗ = 3.75 in, T ∗ = 7173 16–18 (a) d ∗ = D/ 3; (b) √ ∗ lbf · in; (c) (d/D) = 1/ 3 = 0.577 16–19 (a) Uniform wear: pa = 82.2 kPa, F = 949 N; (b) Uniform pressure: pa = 79.1 kPa, F = 948 N 16–23 Cs = 0.08, t = 5.30 in 16–26 (b) Ie = I M + I P + n 2 I P + I L /n 2 ; (c) Ie = 10 + 1 + 102 (1) + 100/102 = 112 16–27 (c) n ∗ = 2.430, m ∗ = 4.115, which are independent of I L

B–17

Chapter 17

17–1 (a) Fc = 0.913 lbf, Fi = 101.1 lbf, F1a = 147 lbf, F2 = 57 lbf; (b) Ha = 2.5 hp, n f s = 1.0; (c) 0.151 in

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Answers to Selected Problems

17–3 A-3 polyamide belt, b = 6 in, Fc = 77.4 lbf, T = 10 946 lbf · in, F1 = 573.7 lbf, F2 = 117.6 lbf, Fi = 268.3 lbf, dip = 0.562 in 17–5 (a) T = 742.8 lbf · in, Fi = 148.1 lbf; (b) b = 4.13 in; (c) F1 = 293.4 lbf, Fc = 17.7 lbf, Fi = 147.6 lbf, F2 = 41.5 lbf, H = 20.6 hp, n f s = 1.1 17–7 R x = (F1 + F2 ){1 − 0.5[(D − d)/(2C)]2 }, R y = (F1 − F2 )(D − d)/(2C). From Ex. 17–2, R y = 1214.4 lbf, R x = 34.6 lbf 17–14 With d = 2 in, D = 4 in, life of 106 passes, b = 4.5 in, n f s = 1.05 17–17 Select one B90 belt 17–20 Select nine C270 belts, life > 109 passes, life > 150 000 h 17–24 (b) n 1 = 1227 rev/min. Table 17–20 confirms this point occurs in the range 1200 ± 200 rev/min, (c) Eq. (17–40) applicable at speeds exceeding 1227 rev/min for No. 60 chain 17–25 (a) Ha = 7.91 hp; (b) C = 18 in; (c) T = 1164 lbf · in, F = 744 lbf 17–27 Four-strand No. 60 chain, N1 = 17 teeth, N2 = 84 teeth, rounded L/ p = 134, n f s = 1.17, life 15 000 h (pre-extreme)

B–18

1017

Chapter 18

18–1 (a) Maximum bending moment 2371 lbf · in, (b) d A = 1.625 in, d B = 1.810 in, average diameter ≥ 1.810 in 18–2 (a) d = 1.725 in, (b) d = 1.687 in 18–13 d = 1.371 in; d = 1.630 in for n d = 2 18–18 d = 24 mm, D = 32 mm, r = 1.6 mm 18–20 (a) Static: d = 1.526 in; (b) DE-Gerber: d = 1.929 in; ASME-elliptic: d = 1.927 in; MSS-Soderberg: d = 1.932 in; DE-Goodman: d = 2.008 in 18–24 Slope: ym = y ; deflection: ym = sy = y/2; moment: Mm = s 3 M = M/8; Force: Fm = s 2 F = F/4; same material, same stress levels 18–26 (a) ω = 868 rad/s; (b) d = 2 in; (c) ω = 1736 rad/s (doubles) 18–28 (b) ω = 466 rad/s = 4450 rev/min

[Solution Manual] Shigley's Mechanical Engineering Design 7th com

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