5.62 The machine part shown in Fig. P5.62 is ¼ in. thick and is made of SAE 4340 heattreated steel (see Appendix D for properties). Determine the maximum safe load P if a factor of safety of 2 with respect to failure by yield is specified.
Fig. P5.62
Solution σ allow =
Fillet: D
d
=
P allow Hole: d
w
=
P allow
σ yield
FS
3.0 in. 2.0 in. =
=
At
K t
3.0 in.
=
K t
2
=
=
=
r d
66 ksi
=
0.4 in. 2.0 in.
=
0.2
(66 ksi)(2.0 in.)(0.25 in.) 1.72
0.1667
At
σ allow
132 ksi
1.5
σ allow
0.5 in.
=
=
K t ≅ 1.72
∴
=
19.19 kips
(a)
K t ≅ 2.48
∴
(66 (66 ksi)( ksi)(3.0 3.0 in. in. − 0.5 0.5 in.)( in.)(0. 0.25 25 in.) in.) 2.48
=
16.63 kips
(b)
Controlling Load: The hole controls in this case; therefore,
P allow
=
16.63 kips
Ans.
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5.63 The machine part shown in Fig. P5.63 is 10 mm thick and is made of cold-rolled 18-8 stainless steel (see Appendix D for properties). Determine the maximum safe load P load P if if a factor of safety of 1.5 with respect to failure by yield is specified.
Fig. P5.63
Solution σ allow =
σ yield
FS
Small Hole: d 16 mm
w
=
P allow
160 mm
=
=
P allow
At
K t
160 mm
=
=
σ allow
Large Hole: d 64 mm
w
=
=
At
σ allow
K t
1,138 MPa 1. 5
0 .1
=
758.67 MPa
K t ≅ 2.62
∴
(758 (758.6 .67 7 N/m N/mm2 )(16 )(160 0 mm mm − 16 mm) mm)(1 (10 0 mm) mm) 2.62
0 .4
=
=
=
416, 97 979 N = 417 kN kN
(a)
=
331,05 ,056 N = 331 kN
(b)
K t ≅ 2.20
∴
(758 (758.6 .67 7 N/m N/mm2 )(16 )(160 0 mm mm − 64 mm) mm)((10 mm) mm) 2.20
Controlling Load: The large hole controls in this case; therefore,
P allow
=
331 kN
Ans.
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5.64 A 1/8-in.-thick by 4-in.-wide steel bar is transmitting an axial tensile load of 500 pounds. After the load is applied, a 1/64-in.diameter hole is drilled through the bar, as shown in Fig. P5.64. (a) Determine the stress at point A A (on the edge of the hole) in the bar before and after the hole is drilled. (b) Does the axial stress at point B B on the edge of the bar increase or decrease as the hole is drilled? Explain.
Fig. P5.64
Solution (a) Stress at point A: Before hole is drilled: P 500 lb σ A =
A
=
(4 in.)(0.125 in.)
=
1, 000 psi
After hole is drilled: d 1/ 64-in. 64-in. = = 0.003906 w 4 in. σ A =
PK P K t At
=
Ans.
K t ≅ 3
∴
(500 lb)(3) (4 in. in. − 1/ 64-i 64-in.) n.)(0 (0.12 .125 5 in.) in.)
=
3,010 psi
Ans.
(b) Stress at point B: The axial stress at point B point B decreases. decreases. Since the average stress changes very little with the introduction of the small hole and since the stress at A at A is is larger than the average stress, the axial stress far from the hole must be less than the average stress.
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5.65 The machine part shown in Fig. P5.65 is 200 mm wide by 25 mm thick thick and is made o 2014-T4 aluminum (see Appendix D for properties). Determine the maximum safe load P if a factor of safety of 2.25 with respect to failure by yield is specified.
Fig. P5.65
Solution σ allow =
Hole: d
w
σ yield
FS
80 mm
=
200 mm
P allow
=
b
=
At
K t
100 mm
P allow
=
=
At
σ allow
K t
2.25
=
σ allow
Notches: r 25 mm
290 MPa
=
0. 4
=
128.89 MPa
K t ≅ 2.20
∴
(128. 128.89 89 N/m N/mm m2 )(20 )(200 0 mm mm − 80 mm mm)(25 )(25 mm) mm) 2.20
0.25
=
=
d r
=
50 mm 25 mm
=
2
(128.89 N/mm N/mm2 )(100 )(100 mm)(25 mm)(25 mm) mm) 1.95
=
175,759 ,759 N = 175.8 kN kN
(a)
K t ≅ 1.95
∴
=
165,244 ,244 N = 165.2 kN kN
(b)
Controlling Load: The notches control in this case; therefore,
P allow
=
165.2 kN
Ans.
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5.66 The machine part shown in Fig. P5.66 is ¼-in.thick ¼-in.thick and and is made made o structural steel (see Appendix D for properties). Determine the maximum safe load P if a factor of safety of 3 with respect to failure by yield is specified.
Fig. P5.66
Solution σ allow =
σ yield
FS
Fillets: D 4.0 in. =
d
P allow Hole: d
w
=
P allow
2.0 in. =
=
4.0 in.
=
b
P allow
=
At
K t
=
At
σ allow
K t
=
=
2.0 in.
=
0 .5
1.40
K t ≅ 1.40
∴
=
4.29 kips
(a)
K t ≅ 2.45
∴
(12 ksi) ksi)(4. (4.0 0 in. in. − 0.75 in.)( in.)(0.25 0.25 in.) in.) 2.45
0 .3
=
1.0 in.
(12 ksi)(2.0 in.)(0.25 in.)
0.1875
σ allow
1.25 in.
12 ksi
d =
K t
=
r
2
At
Notches: r 0.375 in. =
3
σ allow
0.75 in.
=
36 ksi
=
d r
=
0.375 in. 0.375 in.
=
1
(12 ksi)(1.25 in.)(0.25 in.) 1.83
=
3.98 kips
(b)
K t ≅ 1.83
∴
=
2.05 kips
(c)
Controlling Load: The notches control in this case; therefore,
P allow
=
2.05 kips
Ans.
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5.67 The machine part shown in Fig. P5.67 is 20 mm thick, is made of AISI 1020 coldrolled steel (see Appendix D for properties), and is subjected to a tensile load of P of P = = 100 kN. Determine the minimum radius r that can be used between the two sections if a factor of safety of 2 with respect to failure by yield is specified.
Fig. P5.67
Solution σ allow =
K t = D d
=
σ yield
FS At
σ allow
P allow 80 mm 40 mm
=
=
=
427 MPa 2
=
213.5 MPa
(213.5 (213.5 N/mm N/mm2 )(40 )(40 mm)( mm)(20 20 mm) 100, 100, 000 N
=
1.708
2
Then, from Fig. 5.15 r r = = 0.25 d 40 mm
r
∴ min =
10 mm
Ans.
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5.68 The ½-in.-thick bar with semicircular (d/r = = 1) edge grooves, shown in Fig. P5.68, is made of structural steel (see Appendix D for properties) and will be subjected to an axial tensile load P load P of of 10 kips. Determine the minimum safe width B width B for for the bar if a factor of safety of 1.8 with respect to failure by yield must be maintained.
Fig. P5.68
Solution σ allow =
σ yield
FS
Edge Grooves: d =1 r A σ K t = allow t P
r b
=
=
=
36 ksi 1.8
=
20 ksi
(20 ksi)(0.5 in.)b 10 kips
=
b
∴
0.5 in.
∴
b
K t b b=
=
1
0.5 in. r /b 1.3 1.2
From Fig. 5.13 r
b=
K t
b 0.20 0.25 0.30 0.35
2.02 1.92 1.85 1.75
0.5 in.
K t
r /b
b
2.500 2.000 1.667 1.429
0.808 0.960 1.110 1.225
1.1 b 1.0 / t K 0.9
0.8 0.7 0.6 0.20
0.25
0.30
0.35
0.40
r/b
From plot at a value of K t / b = 1, r / b ≅ 0.265. b=
r 0.265
=
0.5 in. 0.265
=
1.887 in.
B = b + 1 in. = 1.887 in. + 1 in. = 2.89 in.
Ans.
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5.69 The stepped bar with a circular hole, shown in Fig. P5.69, is made of annealed 188 stainless steel (see Appendix D for properties). The bar is 25 mm thick and will be subjected subjec ted to an axial a xial tensile load P load P of of 125 kN. Determine the minimum safe width D width D for the bar if a factor of safety of 2 with respect to failure by yield must be maintained.
Fig. P5.69
Solution σ allow =
σ yield
FS
=
248 MPa 2
=
124 MPa
Hole: σ allow =
124 N/mm2
d D
=
d w
=
K t
=
P At Kt (125,000 N)
∴
( D − 30 mm)(25 mm)
30 mm
K t ( D − 30 mm)
D = w =
∴
w
=
1
−
0.02480 mm
30 mm d/w 0.06000 0.05000
From Fig. 5.14 d w 0.40 0.30 0.20 0.10
w = D = D 75 mm 100 mm 150 mm 300 mm
K t 2.20 2.30 2.42 2.62
K t D − 30 mm 0.048889 0.032857 0.020167 0.009704
) 0.04000 0 3
0.03000 D ( / t K
0.02000 0.01000 0.00000 0.00
0.10
0.20
0 .3 0
0.40
d/w
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From plot at a value of w=
d d/w
=
Fillet: r 16 mm
d
=
K t =
64 mm
30 mm 0 .2 5
=
At
σ allow
P
K t ( D − 30 mm) =
=
0.02480 mm 1 , d / w ≅ 0.25. −
120 mm
(a)
0.25
=
(124 N/mm N/mm2 )(64 mm)( mm)(25 25 mm) mm) 125,000 N
≅
1.6
From Fig. 5.15, D / d = 1.5 D = 1.5(64 mm) = 96 mm
(b)
Controlling Plate Width: The hole controls; therefore,
Dmin
=
120 mm
Ans.
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