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4–5. The assembly consists of a steel rod CB and an aluminum rod BA, each having a diameter diameter of 12 mm.If the rod is subjected to the axial loadings at A and at the coupling B, determine the displacement of the coupling B and the end A. The unstretched length of each segment is shown in the figure. Neglect the size of the connections connections at B and C , and assume that they are rigid. Est = 200 GPa, Eal = 70 GPa.
PL = AE
dB
=
dA
= ©
12(103)(3) p
2 9 4 (0.012) (200)(10 )
PL = AE
2 9 4 (0.012) (200)(10 )
+
B
A
18 kN 6 kN 3m
= 0.00159 m = 1.59 mm
12(103)(3) p
C
2m
Ans.
18(103)(2) p
2 9 4 (0.012) (70)(10 )
= 0.00614 m = 6.14 mm
Ans.
Ans: dB
187
= 1.59 mm, dA = 6.14 mm
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4–10. The assembly consists of two 10-mm diameter red brass C83400 copper rods AB and CD, a 15-mm 15-mm diameter diameter 304 stainless steel rod EF , an and d a rig rigid id bar bar G. If the horizontal displacement of end F of rod EF is 0.45 mm, determine the magnitude of P .
300 mm
A
450 mm
B
P E
4P F
C
D
P G
Internal Loading: The normal forces developed in rods EF , AB, an and d CD are shown on the free-body diagram s in Figs. a and b. Displacement: The cross-sectional areas of rods EF and AB are AEF =
56.25(10 - 6 )p m2 and
AAB =
dF
= ©
0.45 =
p
4
p
4
(0.0152 ) =
(0.012 ) = 25(10 - 6 )p m2.
P L P L PL = EF EF + AB AB AE AEF Est AAB Ebr 4P(450) -6
9
56.25(10 )p(193)(10 )
+
P(300) -6
25(10 )p(101)(109)
P = 4967 N = 4.97 kN
Ans.
Ans: P = 49.7 kN 1 92
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4–13. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm 2 and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied.
C
300 N/ N/ m 1.5 m D A
B
2m
2m
a+ © MA = 0; 1200(2) - TCB(0.6)(2) = 0
TCB = 2000 N
>
dB C
=
(2000)(2.5) PL = = 0.0051835 AE 14(10 - 6)(68.9)(109)
(2.5051835)2 = (1.5)2 + (2)2 - 2(1.5)(2) cos u u u
= 90.248°
= 90.248° - 90° = 0.2478° = 0.004324 rad
dD
=
u
r = 0.004324(4000) = 17.3 mm
Ans.
Ans: dD
195
= 17.3 mm
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4–17. The hanger consists of three 2014-T6 aluminum AC C and BD, an alloy rods rods,, rigid beams A and d a sp sprin ring. g. If th the e vertical displacement of end F is 5 mm, mm, det determ ermine ine the the magnitude of the load P. Ro Rods ds AB and CD each have a diameterr of 10 mm, and rod EF has a diameter of 15 mm. diamete The spring has a stiffness of k = 100 MN m and is unstretched when P = 0.
C
A
>
450 mm
E
450 mm
Internal Loading: The normal forces developed in rods EF , AB , an and d CD and the spring are shown in their respective free-body diagram s shown in Figs. a, b, an and d c.
B
D
F
Displacements: The cross-sectional areas of the rods a re
AEF =
p
4
(0.0152) = 56.25(10 - 6)p m2 and p
AAB = ACD =
P
(0.012) = 25(10 - 6)p m2.
4
>
=
>
=
dF E
dB A
P(450) FEF LEF = = 34.836(10 - 6)P T AEF Eal 56.25(10 - 6)p(73.1)(109)
>
(P 2)(450) FAB LAB = = 39.190(10 - 6)P T AAB Eal 25(10 - 6)p(73.1)(109)
The positive signs indicate that ends F and B move away from E and A, respectively. Applying the spring formula with
c
k = 100(103)
N 1m d a 1000 b a b = 100(10 ) N>mm. 1 kN 1000 mm
kN m
>
dE B
3
=
Fsp k
=
-P
= - 10(10 - 6)P = 10(10 - 6)P T
100(103)
The negative sign indicates that E moves towards B. Thu Thus, s, the vertical displacement displacement of F is (+ T )
>
dF A
=
>
dB A
+
>
dF E
+
>
dE B
5 = 34.836(10 - 6)P + 39.190(10 - 6)P + 10(10 - 6)P
P = 59 505.71 N = 59.5 kN
Ans.
Ans:
P = 59.5 kN 199
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4–19. Collar A can slide freely along the smooth vertical guide. If the vertical displacement of the collar is 0.035 in. and the supporti supporting ng 0.75 in. diamete diameterr rod AB is made of 304 stainless steel, determine the magnitude of P.
P
A
2 ft
B
1.5 ft
Internal Loading: The normal force developed in rod AB can be determined by considering the equilibrium of collar A with reference to its free-body diagram, Fig. a.
+ c © Fy = 0;
- FAB
a 45 b - P = 0
FAB = - 1.25 P
Displacements: The cross-sectional area of rod AB is
AAB =
p
4
initial length of rod AB is (0.752) = 0.4418 in2, and the initial
LAB = 2 22 + 1.52 = 2.5 ft. The axial deformation of rod AB is dAB
=
- 1.25P(2.5)(12) FABLAB = = - 0.003032P AABEst 0.4418(28.0)(103)
The negative sign indicates that end A moves towards B. From the the geometry shown 1.5 = 36.87°. Thu in Fig. b, we obtain u = tan - 1 Thus, s, 2
a b
dAB
= (dA)V cos u
0.003032P = 0.035 cos 36.87°
P = 9.24 kip
Ans.
Ans: P = 9.24 kip 201
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The steel pipe is filled with concrete and subjected to a compressive compressive force of 80 kN. Determ Determine ine the average normal stress in the concrete and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm. Est = 200 GPa, Ec = 24 GPa. 4–33.
80 kN
500 mm
+ c © Fy = 0;
Pst + Pcon - 80 = 0 dst
=
dcon
9
=
p
Pst L p
2
4 (0.08
2
- 0.07 ) (200) (10 )
(1)
Pcon L 2 4 (0.07 )
(24) (109)
Pst = 2.5510 Pcon
(2)
Solving Eqs. Eqs. (1) and (2) yields Pst = 57.47 kN sst
scon
=
=
Pst Ast
=
Pcon Acon
Pcon = 22.53 kN
57.47 (103) p
=
4 (0.08
2
- 0.072)
22.53 (103) p
4 (0.07
2
)
= 48.8 MPa
Ans.
= 5.85 MPa
Ans.
Ans: sst
215
= 48.8 MPa, scon = 5.85 MPa
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The load of 2800 lb is to be supported by the two essentially vertical A-36 steel wires. If originally wire AB is 60 in. long and and wire A AC C is 40 in. in. long, determ determine ine the cross cross-sectional area of AB if the load is to be shared equally between both wires. Wire A AC C has a cross-sectional area of 0.02 in2. 4–39.
B
C
60 in. 40 in. A
TAC = TAB =
2800 = 1400 lb 2 dAC = dAB
1400(40) (0.02)(29)(106)
=
1400(60) AAB(29)(106)
AAB = 0.03 in2
Ans.
Ans:
AAB = 0.03 in2 221
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The bolt has a diameter of 20 mm and passes through a tube that has an inner diameter of 50 mm and an outer diameter of 60 mm. If the bolt and tube are made of A-36 steel, determine the normal stress in the tube and bolt when a force of 40 kN is applied to the bolt.Assume the end caps are rigid. 4–45.
160 mm
40 kN
40 kN 150 mm
Referring to the FBD of left portion of the cut assembly, assembly, Fig. a
+ © F = 0; x
40(103) - Fb - Ft = 0
:
(1)
Here, it is required that the bolt and the tube have have the same deformation. Thus dt = db
Ft(150) p
2 4 (0.06
- 0.052) C 200(109) D
=
Fb(160) p
2 4 (0.02 )
C 200(109) D
Ft = 2.9333 Fb
(2)
Solving Eqs (1) and (2) yields Fb = 10.17 (103) N
Ft = 29.83 (103) N
Thus, sb =
st =
Fb Ab Ft At
=
=
10.17(103) p
2 4 (0.02 )
= 32.4 MPa
29.83 (103) p
2 4 (0.06
- 0.052)
Ans.
= 34.5 MPa
Ans.
Ans:
sb = 32.4 MPa, st = 34.5 MPa 227
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The support consists of a solid red brass C83400 copper post surrounded by a 304 stainless steel tube. Before the load is applied the gap between these two parts is 1 mm. Given the dimensions dimensions shown, determ determine ine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials. 4–47.
P
A
1 mm
0.25 m
60 mm 80 mm
Require,
10 mm
dst = dbr + 0.001
Fst(0.25) 2
2
9
p[(0.05) - (0.04) ]193(10 )
=
Fbr(0.25) p(0.03)2(101)(109)
+ 0.001
0.45813 Fst = 0.87544 Fbr + 106
(1)
+ c © Fy = 0;
(2)
Fst + Fbr - P = 0
Assume brass brass yields, yields, then (Fbr)max = sg Abr = 70(106)(p)(0.03)2 = 197 920.3 N (Pg)br = sg> E =
70.0(106) 9
101(10 )
= 0.6931(10 - 3) mm> mm
dbr = (eg)brL = 0.6931(10 - 3)(0.25) = 0.1733 mm < 1 mm
Thus only the brass is loaded. P = Fbr = 198 kN
Ans.
Ans:
P = 198 kN 229
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The rigid bar supports the uniform distributed load of 6 kip ft. Determine the force in each cable if each cable has a cross-sectional area of 0.05 in2, and E = 31 103 ksi. 4–51.
>
C
1 2
6 ft
6 kip/ kip/ ft ft A D
B
3 ft
a + © MA
u = 2 LB¿C¿ =
=
0;
tan - 1 (3)2
TCB
6 6
+
=
a 2 25 b (3)
-
54(4.5)
+ TCD
a 2 2 5 b 9
=
0
3 ft
3 ft
(1)
45°
(8.4853)2
2(3)(8.4853) cos u ¿
-
Also, 2 LD¿C¿ =
(9)2
+
(8.4853)2
-
2(9)(8.4853) cos u ¿
(2)
Thus, Thu s, elimina eliminating ting cos u ¿ . 2 - LB¿C¿(0.019642) + 2 LB¿C¿(0.019642) = 2 LB¿C¿ =
2 = - LD¿C¿(0.0065473) +
1.5910
0.0065473 L2D¿C¿
0.333 L2D¿C¿
+
1.001735
0.589256
30
+
But,
2 45 45
LB¿C =
+ dBC BC¿¿ ,
2 45 45
LD¿C =
+ dDC DC¿¿
Neglect squares or d ¿ B since small strain occurs. 2 LD¿C =
( 2 45 45
+ dBC)
2 LD¿C =
( 2 45 45
+ dDC)
45
+
2 2 45 45 dBC
2 2 45 45 dBC dDC =
=
2
=
45
+
2 2 45 45 dBC
2
=
45
+
2 2 45 45 dDC
=
0.333(45
+
2 2 45 45 dDC)
+
30
0.333(2 2 45 45 dDC)
3dBC
Thus, TCD
2 45 45
AE TCD =
=
3
TCB
2 45 45
AE
3 TCB
From Fro m Eq. (1). TCD =
27.1682 kip
TCB =
9.06 kip
=
27.2 kip
Ans. Ans. Ans: TCD =
233
27.2 kip, TCB
=
9.06 kip
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4–57. The rigid bar is originally horizontal and is supported by two A-36 steel cables each having a crosssectional area of 0.04 in2 . Determine the rotation of the bar when the 800-lb load is applied.
C
12 ft
800 lb B
A
Referring to the FBD of the rigid bar Fig. a, a + © MA
=
0;
FBC
a 1213 b (5)
a 35 b (16)
+ FCD
The unstretched lengths of wires BC and CD are LCD =
2 12 122
dBC =
+
162
FBC LBC AE
800(10)
-
=
0
2 12 122
LBC =
5 ft
D
5 ft
6 ft
(1)
+
52
=
13 ft and
20 ft. The stretch of wires BC and CD are
=
FBC (13)
=
dCD =
AE
FCD LCD AE
=
FCD(20) AE
Referring to the geometry shown in Fig. b, the vertical vertical displacement displacement of a point on 12 3 d the rigid bar is dv = . Fo Forr points B and D, cos uB = and cos uD = . Thus, cos u 13 5 the vertical displacements of points B and D are
Ad B
=
Ad B
=
B v
D v
dBC
cos uB dCD
cos uD
=
=
>
=
169 FBC 12AE
>
=
100 FCD 3 AE
FBC (13) AE
>
12 13
FCD (20) AE
>
3 5
The similar triangles shown in Fig. c gives
Ad B
B v
5 1 5
a 16912 b FBC
AE
Ad B
D v
=
=
FBC =
16 1 16
a 1003 b FCD
AE
125 FCD 169
(2)
Solving Eqs (1) and (2), yields FCD =
614.73 lb
FBC =
454.69 lb
Thus,
Ad B
D v
=
100(614.73) 3(0.04) C 29.0 (106) D
=
0.01766 ft
=
0.0633°
Then u =
ft 180° a 0.01766 ba b 16 ft p
Ans.
2 40
Ans: u = 0.0633°
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4–69. The assembly has the diameters and material makeup indicated. If it fits securely between its fixed supports when the temperature is T1 = 70°F, determine the average normal stress in each material when the temperature reaches T2 = 110°F.
2014-T6 Aluminum C 86100 Bronze 12 in.
A
© Fx =
>
dA
0;
D
=
D
8 in. B
4 ft
304 Stainless steel
C
4 in.
6 ft
3 ft
FA = FB = F
0;
-
F(4)(12) 2
p(6)
-
-
6
(10.6)(10 )
F(6)(12) 2
p(4)
6
(15)(10 )
F(3)(12) 2 6 p(2) (28)(10 )
F =
+
12.8(10 - 6)(110
-
70)(4)(12)
+
9.60(10 - 6)(110
-
70)(6)(12)
+
9.60(10 - 6)(110
-
70)(3)(12)
=
0
277.69 kip
sal =
277.69 2 p(6)
=
2.46 ksi
Ans.
sbr =
277.69 2 p(4)
=
5.52 ksi
Ans.
sst =
277.69 2 p(2)
=
22.1 ksi
Ans.
Ans: sal =
2 52
2.46 ksi, sbr
=
5.52 ksi, sst
=
22.1 ksi
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4–78. When the temperatur temperature e is at 30°C, the A-36 A-36 steel pipe fits snugly between the two fuel tanks.When fuel flows through the pipe, the temperatures at ends A and B rise to 130°C and 80°C, 80°C, respectively. If the temperature drop along the pipe is linear, linear, determi determine ne the average average normal stress stress developed in the pipe. pipe. Assume each tank provides a rigid support at A and B.
150 mm
10 mm
Section a - a 6m x a
A
a
B
Temperature Gradient: Since the temperature varies linearly along the pipe, pipe, Fig. a, a, the temperature gradient can be expressed as a function of x as T(x) =
80
+
50 (6 6
- x) =
a 130
-
b
50 x °C 6
Thus,, the change in temperature as a function of x is Thus ¢ T = T(x) -
30°
=
a 130
50 x 6
-
b
30
-
=
a 100
-
b
50 x °C 6
Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of
dT = a
L
6m
-6
12(10 )
¢ Tdx =
L a 100
-
0
50 x 6
b
dx =
0.0054 m
=
5.40 mm
Using the method of superposition, superposition, Fig. b, (+) :
0
= dT - dF
0
=
F =
5.40
-
F(6000) 2
p(0.16
-
0.152)(200)(109)
1 753 008 N
Normal Stress: s =
F A
=
1 753 008 2 2 p(0.16 - 0.15 )
=
180 MPa
Ans.
Ans: s =
261
180 MPa
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4–81. The 50-mm-diameter cylinder is made from Am 1004-T61 magnesium and is placed in the clamp when the temperature is T1 = 20° C. If the 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mm, and they hold the cylinder snug with negligible force agai nst the rigid jaws, determine the force in the cylinder when the temperature rises to T2 = 130° C.
+ c © Fy = 0;
150 mm
Fst = Fmg = F dmg
amg Lmg ¢ T
26(10 - 6)(0.1)(110) -
100 mm
-
p
4
dst
FmgLmg EmgAmg
F(0.1)
44.7(109)
=
=
astLst ¢ T
+
FstLst EstAst
= 17(10 - 6)(0.150)(110) +
(0.05)2
F(0.150)
193(109)(2)
F = 904 N
p
4
(0.01)2
Ans.
Ans: F = 904 N 2 64
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4–83. The wires AB and AC are made of steel, and wire AD is made of copper. Before the 150-lb force is applied, AB and AC are each each 60 in. long and and AD is 40 40 in. in. lo long ng.. If the the temperature tempera ture is incre increased ased by 80°F , determi determine ne the force in each each wire wire needed needed to suppo support rt the the load. load. Take Est = 29(103) ksi, Ecu = 17(103) ksi, ast = 8(10 - 6)> °F, acu = 9.60(10 - 6)> °F. Each wire has a cross-sectional area of 0.0123 in2.
C
D
B
40 in. 60 in.
45
45
60 in.
A
150 lb
Equations of Equilibrium:
+
:
© Fx = 0;
FAC cos 45° - FAB cos 45° = 0 FAC = FAB = F
+ c © Fy = 0;
2F sin 45° + FAD - 150 = 0
(1)
Compatibility:
(dAC)T = 8.0(10 - 6)(80)(60) = 0.03840 in. (dAC)T2 =
(dAC)T 0.03840 = = 0.05431 in. cos 45° cos 45°
(dAD)T = 9.60(10 - 6)(80)(40) = 0.03072 in. d0
= (dAC)T2 - (dAD)T = 0.05431 - 0.03072 = 0.02359 in.
(dAD)F = (dAC)Fr + FAD(40) 6
0.0123(17.0)(10 )
=
d0
F(60)
0.0123(29.0)(106) cos 45°
+ 0.02359
0.1913FAD - 0.2379F = 23.5858
(2)
Solving Eq. (1) and (2) yields: FAC = FAB = F = 10.0 lb
Ans.
FAD = 136 lb
Ans.
Ans: FAC = FAB = 10.0 lb, FAD = 136 lb 2 66
