1,325 168 162MB
English Pages [1583] Year 2018
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1–1. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A.
0.3 m A 30� 0.1 m
80 N
45�
Solution Equations of Equilibrium: +
Q©Fx¿ = 0;
NA  80 cos 15° = 0 Ans.
NA = 77.3 N a+ ©Fy¿ = 0;
VA  80 sin 15° = 0 Ans.
VA = 20.7 N a+
©MA = 0;
MA + 80 cos 45°(0.3 cos 30°)  80 sin 45°(0.1 + 0.3 sin 30°) = 0 MA =  0.555 N # m
Ans.
or a+
©MA = 0;
MA + 80 sin 15°(0.3 + 0.1 sin 30°) 80 cos 15°(0.1 cos 30°) = 0 MA =  0.555 N # m
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished. Ans:
NA = 77.3 N, VA = 20.7 N, MA = 0.555 N # m
1
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1–2. Determine the resultant internal loadings on the cross section at point D.
C
1m
F
2m
1.25 kN/m
Solution
A
Support Reactions: Member BC is the two force member. 4 a+ ΣMA = 0; FBC (1.5)  1.875(0.75) = 0 5
D
E
0.5 m 0.5 m 0.5 m
B
1.5 m
FBC = 1.1719 kN + c ΣFy = 0; Ay +
4 (1.1719)  1.875 = 0 5
Ay = 0.9375 kN + ΣFx = 0; S
3 (1.1719)  Ax = 0 5 Ax = 0.7031 kN
Equations of Equilibrium: For point D + ΣFx = 0; ND  0.7031 = 0 S ND = 0.703 kN
Ans.
+ c ΣFy = 0; 0.9375  0.625  VD = 0 VD = 0.3125 kN
Ans.
a+ ΣMD = 0; MD + 0.625(0.25)  0.9375(0.5) = 0 MD = 0.3125 kN # m
Ans.
Ans: ND = 0.703 kN, VD = 0.3125 kN, MD = 0.3125 kN # m 2
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1–3.
C
1m
F
2m
1.25 kN/m
Solution
A
Support Reactions: Member BC is the twoforce member. 4 a+ ΣMA = 0; FBC (1.5)  1.875(0.75) = 0 5
D
E
0.5 m 0.5 m 0.5 m
B
1.5 m
FBC = 1.1719 kN + c ΣFy = 0; Ay +
4 (1.1719)  1.875 = 0 5
Ay = 0.9375 kN 3 + ΣFx = 0; (1.1719)  Ax = 0 S 5 Ax = 0.7031 kN Equations of Equilibrium: For point F + bΣFx′ = 0; NF  1.1719 = 0 Ans.
NF = 1.17 kN a + ΣFy′ = 0; VF = 0
Ans.
a+ ΣMF = 0; MF = 0
Ans.
Equations of Equilibrium: For point E + ΣFx = 0; NE  3 (1.1719) = 0 d 5 Ans.
NE = 0.703 kN + c ΣFy = 0; VE  0.625 +
4 (1.1719) = 0 5 Ans.
VE =  0.3125 kN a+ ΣME = 0;  ME  0.625(0.25) +
4 (1.1719)(0.5) = 0 5
ME = 0.3125 kN # m
Ans.
Negative sign indicates that VE acts in the opposite direction to that shown on FBD.
Ans: NF = 1.17 kN, VF = 0, MF = 0, NE = 0.703 kN, VE =  0.3125 kN, ME = 0.3125 kN # m 3
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*1–4. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Determine the resultant internal loadings acting on the cross section at C.
600 N/m A
B
D
C
1m
1m
1m
1.5 m
1.5 m 900 N
Solution Support Reactions: We will only need to compute By by writing the moment equation of equilibrium about A with reference to the freebody diagram of the entire shaft, Fig. a. a+ ΣMA = 0;
By(4.5)  600(2)(2)  900(6) = 0
By = 1733.33 N
Internal Loadings: Using the result of By, section CD of the shaft will be considered. Referring to the freebody diagram of this part, Fig. b, + ΣFx = 0; S
Ans.
NC = 0
+ c ΣFy = 0;
VC  600(1) + 1733.33  900 = 0
a+ ΣMC = 0;
1733.33(2.5)  600(1)(0.5)  900(4)  MC = 0
VC = 233 N MC = 433 N # m
Ans.
Ans.
The negative sign indicates that VC acts in the opposite sense to that shown on the freebody diagram.
Ans: NC = 0, VC =  233 N, MC = 433 N # m 4
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1–5. Determine the resultant internal loadings in the beam at cross sections through points D and E. Point E is just to the right of the 3kip 15kNload. load.
315kip kN 1.5 kip/ ft 25 kN/m
A
D 62 ft m
26m ft
B
E
4 ftm 1.5
4 ftm 1.5
C
Solution Support Reactions: For member AB a + ©MB = 0;
50(4/3) = 0= 0 9.00(4) –AA y(4) y(12)
+ : ©Fx = 0; + c ©Fy = 0;
kN Ayy ==16.67 3.00 kip
Bx = 0 By ++ 16.67 3.00 – 50 9.00 = 0= 0
Byy == 33.33 6.00 kip kN
Equations of Equilibrium: For point D + : ©Fx = 0; + c ©Fy = 0;
Ans.
ND = 0 3.00 – 12.5 2.25 –VV 16.67 DD==0 0
Ans.
V kN 0.750 kip VDD == 4.17 a + ©MD = 0;
2  3.00(6) MD ++ 12.25( 2.25(2) 0 3 ) – 16.67(2) = 0
kN #· ftm MD == 25.17 13.5 kip
Ans.
Equations of Equilibrium: For point E + : ©Fx = 0; + c ©Fy = 0;
Ans.
NE = 0 –33.33 6.00 – 15 3 – V VE = 00 E=
Ans.
 9.00 kN kip VEE == –48.33 a + ©ME = 0;
ME ++33.33(1.5) 6.00(4) ==00 M kN #· ft m ME = –50.00  24.0 kip
Ans.
Negative signs indicate that ME and VE act in the opposite direction to that shown on FBD. 1 2 (25)(4)
8 3 m
15 kN
= 50 kN
4 3 m
2m
1 2 (12.5)(2)
2m
4 3 m
= 12.5 kN
33.33 kN
15 kN
1.5 m
2 3 m
16.67 kN
Ans: ND = 0, VD = 4.17 kN,
MD = 25.0 kN # m, NE = 0, VE = 48.3 kN, ME = 50.0 kN # m 5
CH 01.indd 3
11/30/10 9:39:58 AM
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1–6. The shaft is supported by a smooth thrust bearing at B and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E.
B
A
1m
C
E
1m
1m
D
1m
1800 N 3600 N
Solution Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the freebody diagram of the entire shaft, Fig. a. a + ©MB = 0;
Cy(2) + 1800(1)  3600(3) = 0
Cy = 4500 N
Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the freebody diagram, Fig. b, + : ©Fx = 0;
NE = 0
+ c ©Fy = 0;
VE + 4500  3600 = 0
Ans. VE =  900 N
Ans.
a + ©ME = 0; 4500(1)  3600(2)  ME = 0 ME = 2700 N . m = 2.70 kN . m
Ans.
The negative signs indicates that VE and ME act in the opposite sense to that shown on the freebody diagram.
Ans: NE = 0, VE = 900 N, ME = 2.7 kN .m 6
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1–7. Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b, each of which passes through point A. The 2000N load is applied along the centroidal axis of the member.
a
b 30⬚
2000 N
2000 N
b
A a
(a) + : ©Fx = 0;
Na  2000 = 0 Na = 2000 N
Ans.
+ T©Fy = 0;
Va = 0
Ans.
R+ ©Fx = 0;
Nb  2000 cos 30° = 0
(b)
Nb = 1732 N +Q©F = 0; y
Ans.
Vb  2000 sin 30° = 0 Vb = 1000 N
Ans.
Ans: Na = 2000 N, Va = 0, Nb = 1732 N, Vb = 1000 N 7
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0.2 m 0.2 m
*1–8. The floor crane is used to lift a 600kg concrete pipe. Determine the resultant internal loadings acting on the cross section at G.
0.4 m
E B
0.6 m
G
F 0.3 m
H
C
D
0.5 m
A
75�
Solution Support Reactions: We will only need to compute FEF by writing the moment equation of equilibrium about D with reference to the freebody diagram of the hook, Fig. a. a + ©MD = 0;
FEF(0.3)  600(9.81)(0.5) = 0
FEF = 9810 N
Internal Loadings: Using the result for FEF, section FG of member EF will be considered. Referring to the freebody diagram, Fig. b, + : ©Fx = 0; + c ©Fy = 0; a + ©MG = 0;
9810  NG = 0
NG = 9810 N = 9.81 kN
Ans. Ans.
VG = 0
Ans.
MG = 0
Ans: NG = 9.81 kN, VG = 0, MG = 0 8
CH 01.indd 13
11/29/10 10:18:28 AM
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1–9. The floor crane is used to lift a 600kg concrete pipe. Determine the resultant internal loadings acting on the cross section at H.
0.2 m 0.2 m
0.4 m G
E B
0.6 m F 0.3 m
H
C
D
0.5 m
A
75�
Solution Support Reactions: Referring to the freebody diagram of the hook, Fig. a. a + ©MF = 0;
Dx(0.3)  600(9.81)(0.5) = 0
Dx = 9810 N
+ c ©Fy = 0;
Dy  600(9.81) = 0
Dy = 5886 N
Subsequently, referring to the freebody diagram of member BCD, Fig. b, a + ©MB = 0;
FAC sin 75°(0.4)  5886(1.8) = 0
+ : ©Fx = 0;
FAC = 27 421.36 N
Bx + 27 421.36 cos 75°  9810 = 0 Bx = 2712.83 N
+ c ©Fy = 0;
27 421.36 sin 75°  5886  By = 0
By = 20 601 N
Internal Loadings: Using the results of Bx and By, section BH of member BCD will be considered. Referring to the freebody diagram of this part shown in Fig. c, + : ©Fx = 0;
NH + 2712.83 = 0
NH = 2712.83 N =  2.71 kN
Ans.
+ c ©Fy = 0;
 VH  2060 = 0
VH = 20601 N =  20.6 kN
Ans.
a + ©MD = 0;
MH + 20601(0.2) = 0
MH =  4120.2 N # m = 4.12 kN # m Ans.
The negative signs indicates that NH, VH, and MH act in the opposite sense to that shown on the freebody diagram.
Ans: NH = 2.71 kN, VH = 20.6 kN, MH = 4.12 kN # m 9
CH 01.indd 14
11/29/10 10:18:29 AM
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1–10. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point C. Assume the reactions at the supports A and B are vertical.
4 kN/m
A
B D
C 1.5 m
3m
1.5 m
Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0; By(6) 
1 (4)(6)(2) = 0 2
By = 4.00 kN
Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through C, Fig. b, + ΣFx = 0; NC = 0 S + c ΣFy = 0; VC + 4.00 a+ ΣMC = 0;
4.00(4.5) 
Ans. 1 (3)(4.5) = 0 2
Ans.
VC = 2.75 kN
1 (3)(4.5)(1.5)  MC = 0 2
MC = 7.875 kN # m
Ans.
Ans: NC = 0, VC = 2.75 kN, MC = 7.875 kN # m 10
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1–11. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point D. Assume the reactions at the supports A and B are vertical.
4 kN/m
A
B D
C 1.5 m
3m
1.5 m
Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0; By(6) 
1 (4)(6)(2) = 0 2
By = 4.00 kN
Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through D, Fig. b, + ΣFx = 0; ND = 0 S
Ans.
+ c ΣFy = 0; VD + 4.00 a+ ΣMD = 0; 4.00(1.5) 
1 (1.00)(1.5) = 0 2
VD = 3.25 kN
Ans.
1 (1.00)(1.5)(0.5)  MD = 0 2
MD = 5.625 kN # m
Ans.
The negative sign indicates that VD acts in the sense opposite to that shown on the FBD.
Ans: ND = 0, VD =  3.25 kN, MD = 5.625 kN # m 11
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*1–12. a
The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section a–a that passes through point D.
225 mm 30 b B
A D b F E
Solution
a
150 mm F C
Internal Loadings: Referring to the freebody diagram of the section of the hacksaw shown in Fig. a, + ΣFx = 0; d
Na  a + 100 = 0
+ c ΣFy = 0;
Va  a = 0
a+ ΣMD = 0;  Ma  a  100(0.15) = 0
Na  a = 100 N
Ans. Ans.
Ma  a = 15 N # m
Ans.
The negative sign indicates that Na  a and Ma  a act in the opposite sense to that shown on the freebody diagram.
Ans: Na  a = 100 N, Va  a = 0, Ma  a =  15 N # m 12
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1–13. a
The blade of the hacksaw is subjected to a pretension force of F = 100 N. Determine the resultant internal loadings acting on section b–b that passes through point D.
225 mm 30 b B
A D b F E
Solution
a
150 mm F C
Internal Loadings: Referring to the freebody diagram of the section of the hacksaw shown in Fig. a, ΣFx′ = 0;
Nb  b + 100 cos 30° = 0 Nb  b =  86.6 N
Ans.
ΣFy′ = 0;
Vb  b = 50 N Vb  b  100 sin 30° = 0
Ans.
a+ ΣMD = 0;
Mb  b = 15 N # m  Mb  b  100(0.15) = 0
Ans.
The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that shown on the freebody diagram.
Ans: Nb  b =  86.6 N, Vb  b = 50 N, Mb  b = 15 N # m 13
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1–14. The boom DF of the jib crane and the column DE uniformweight wieghtofof the and hoistload andweigh load have aauniform 50750 lb/ft.N/m. If theIfhoist weigh N, determine the resultant internal in loadings in 300 lb, 1500 determine the resultant internal loadings the crane thecross cranesections on crossthrough sectionspoints through points A, B, and C. on A, B, and C.
D
2.4 8 ftm
2 ft 1.5 5 ftm
F
A
B
0.9 3 ftm
0.6 m C 1500 N 300 lb
2.1 7 ftm
Solution
E
Equations of Equilibrium: For point A + ; © Fx = 0; + c © Fy = 0;
0.750(0.9) = 0.675 kN
NA = 0
Ans.
150 300 == 00 VA –0.675 – 1.500
0.45 m 0.45 m
VA == 2.175 450 lbkN a + ©MA = 0;
Ans.
1.500 kN
1.500(0.9) –M MAA –0.675(0.45) 150(1.5) – 300(3) = 0= 0 0.750(3.3) = 2.475 kN
MA = 1.65 kN . m
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD. 1.65 m
Equations of Equilibrium: For point B
1.500 kN
+ ; © Fx = 0;
NB = 0
+ c © Fy = 0;
550 – 1.5 300==0 0 VBB – 2.475
Ans. 0.750(3.9) = 2.925 kN
VB = 3.975 kN a + © MB = 0;
1.65 m
Ans.
–M 1.500(3.3) MBB–2.475(1.65) 550(5.5) – 300(11) = =0 0
0.750(1.5) = 1.125 kN
MB = 9.03 kN . m
Ans.
1.95 m
1.95 m 1.500 kN
Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; © Fx = 0; + c © Fy = 0;
VC = 0
Ans.
–N NCC –1.125 250 – 2.925 650 – 1.500 300 ==00 kNlb =  1.20 kip  1200 NCC == –5.55
a + ©MC = 0;
Ans.
–M 1.500(3.9) MCC –2.925(1.95) 650(6.5) – 300(13) = =0 0 MC = 11.6 kN . m
Ans.
Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD.
Ans: NA 0, VA 2.175 kN, MA 1.65 kN · m, NB 0, VB 3.975 kN, MB 9.03 kN · m, VC 0, NC 5.55 kN, MC 11.6 kN · m 14
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1–15. The metal stud punch is subjected to a force of 120 N on the handle. Determine the magnitude of the reactive force at the pin A and in the short link BC. Also, determine the resultant internal loadings acting on the cross section at point D.
120 N
60
50 mm 100 mm E
B
30
50 mm D
100 mm 300 mm
A C
Solution
200 mm
Member: FBC cos 30°(50)  120(500) = 0
a+ΣMA = 0;
Ans.
FBC = 1385.6 N = 1.39 kN + c ΣFy = 0;
Ay  1385.6  120 cos 30° = 0 Ay = 1489.56 N
+ ΣFx = 0; d
Ax  120 sin 30° = 0; Ax = 60 N
FA = 21489.562 + 602
Ans.
= 1491 N = 1.49 kN
Segment: a+ ΣFx′ = 0;
ND  120 = 0 Ans.
ND = 120 N +Q ΣFy′ = 0; a+ ΣMD = 0;
Ans.
VD = 0 MD  120(0.3) = 0 MD = 36.0 N # m
Ans.
Ans: FBC = 1.39 kN, FA = 1.49 kN, ND = 120 N, VD = 0, MD = 36.0 N # m 15
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*1–16. Determine the resultant internal loadings acting on the cross section at point E of the handle arm, and on the cross section of the short link BC.
120 N
60
50 mm 100 mm E
B
30
50 mm D
100 mm 300 mm
A C
Solution
200 mm
Member: a+ ΣMA = 0; FBC cos 30°(50)  120(500) = 0 FBC = 1385.6 N = 1.3856 kN Segment: +bΣF = 0; x′
Ans.
NE = 0
a + ΣFy′ = 0; VE  120 = 0; VE = 120 N
Ans.
a+ ΣME = 0; ME  120(0.4) = 0; ME = 48.0 N # m
Ans.
Short link: + ΣFx = 0; V = 0 d
Ans.
+ c ΣFy = 0; 1.3856  N = 0; N = 1.39 kN
Ans.
a+ ΣMH = 0; M = 0
Ans.
Ans: NE = 0, VE = 120 N, ME = 48.0 N # m, Short link: V = 0, N = 1.39 kN, M = 0 16
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1–17. The forged steel clamp exerts a force of F = 900 N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A.
200 mm F � 900 N
a
30�
F � 900 N
A
a
Solution Internal Loadings: Referring to the freebody diagram of the section of the clamp shown in Fig. a, ©Fy¿ = 0;
900 cos 30°  Na  a = 0
Na  a = 779 N
Ans.
©Fx¿ = 0;
Va  a  900 sin 30° = 0
Va  a = 450 N
Ans.
a + ©MA = 0;
900(0.2)  Ma  a = 0
Ma  a =
180 N # m
Ans.
Ans: Naa = 779 N, Vaa = 450 N,
900(0.2)  Maa = 0, Maa = 180 N # m 17
CH 01.indd 12
11/29/10 10:18:26 AM
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1–18. Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is >ft22 acts N/m fixed to to the the ground groundand andaauniform uniformpressure pressureofof500 7 lb perpendicular to the face of the sign.
z ft 3m ft 2m
ft 3m 22
7 lb/ft 500 N/m
ft 6m B
Solution
A
©Fx = 0;
(VB)x – 7500 105 = 0; 0;
7500lbN = 7.5 kN (VB)x = 105
©Fy = 0;
(VB)y = 0
Ans.
©Fz = 0;
(NB)z = 0
Ans.
ft 4m
Ans.
©Mx = 0;
(MB)x = 0
Ans.
©My = 0;
# ft= 56.25 kN · m ·m (MB)y – 7500(7.5) 105(7.5) = 0; 0; (M(M = 788Nlb B)y B=)y56250
Ans.
©Mz = 0;
(TB)z 105(0.5) = 0; 0; (TB(T )z 3750 = 52.5 ft 3.75 kN · m – 7500(0.5) )zB= N ·lbm# =
Ans.
x
y
0.5 m
500(5)(3) = 7500 N
7.5 m
Ans: (VB)x = 7.5 kN, (VB)y = 0, (NB)z = 0, (MB)x = 0, (MB)y = 56.25 kN # m,
(TB)z = 3.75 kN # m 18
CH 01.indd 16
11/30/10 9:39:59 AM
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1–19. Determine the resultant internal loadings acting on the cross section at point C in the beam. The load D has a mass of 300 kg and is being hoisted by the motor M with constant velocity.
2m
2m
2m 0.1 m
0.1 m E
C
A
B 1m
1.5 m D
M
Solution + ΣFx = 0; NC + 2.943 = 0; NC =  2.94 kN d
Ans.
+ c ΣFy = 0; VC  2.943 = 0; VC = 2.94 kN
Ans.
a+ ΣMC = 0;  MC  2.943(0.6) + 2.943(0.1) = 0 MC =  1.47 kN # m
Ans.
Ans: NC = 2.94 kN, VC = 2.94 kN, MC = 1.47 kN # m 19
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*1–20. Determine the resultant internal loadings acting on the cross section at point E. The load D has a mass of 300 kg and is being hoisted by the motor M with constant velocity.
2m
2m
2m 0.1 m
0.1 m E
C
A
B 1m
1.5 m D
M
Solution + ΣFx = 0; NE + 2943 = 0 S +T
Ans.
NE =  2.94 kN ΣFy = 0;  2943  VE = 0
Ans.
VE =  2.94 kN a+ ΣME = 0; ME + 2943(1) = 0
ME =  2.94 kN # m
Ans.
Ans: NE =  2.94 kN, VE =  2.94 kN, ME =  2.94 kN # m 20
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1–21. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth.
20 N
40 mm
120 mm
15 mm C A
B
D 80 mm 20 N
30�
Solution + c ©Fy = 0; + : ©Fx = 0; +d ©MC = 0;
 VC + 60 = 0;
Ans.
VC = 60 N
Ans.
NC = 0  MC + 60(0.015) = 0;
Ans.
MC = 0.9 N.m
Ans: VC = 60 N, NC = 0, MC = 0.9 N # m 21
CH 01.indd 9
11/29/10 10:18:23 AM
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1–22. Determine the resultant internal loading on the cross section through point D of the pliers.
20 N
40 mm
120 mm
15 mm C A
B
D 80 mm 20 N
30�
Solution R+ ©Fy = 0;
VD  20 cos 30° = 0;
VD = 17.3 N
Ans.
+ b©Fx = 0;
ND  20 sin 30° = 0;
ND = 10 N
Ans.
+d ©MD = 0;
MD  20(0.08) = 0;
MD = 1.60 N.m
Ans.
Ans: VD = 17.3 N, ND = 10 N, MD = 1.60 N # m 22
CH 01.indd 9
11/29/10 10:18:23 AM
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1–23. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section at point C. The 400N forces act in the –z direction and the 200N and 80N forces act in the +y direction. The journal bearings at A and B exert only y and z components of force on the shaft.
z
400 mm 150 mm 150 mm 200 mm 200 mm 300 mm
80 N 80 N 200 N
Support Reactions:
A
160(0.4) + 400(0.7)  Ay(1.4) = 0
x
Ay = 245.71 N ΣFy = 0;
y
D
C
Solution ΣMz = 0;
B
200 N 400 N 400 N
 245.71  By + 400 + 160 = 0 By = 314.29 N
ΣMy = 0;
800(1.1)  Az(1.4) = 0
Az = 628.57 N
ΣFz = 0;
Bz + 628.57  800 = 0
Bz = 171.43 N
Equations of Equilibrium: For point C ΣFx = 0; ΣFy = 0;
Ans.
(NC)x = 0  245.71 + (VC)y = 0
Ans.
(VC)y =  246 N ΣFz = 0;
ΣMx = 0; ΣMy = 0;
ΣMz = 0;
628.57  800 + (VC)z = 0 (VC)z =  171 N
Ans.
(TC)x = 0
Ans.
(MC)y  628.57(0.5) + 800(0.2) = 0 (MC)y =  154 N # m
Ans.
(MC)z  245.71(0.5) = 0
(MC)z =  123 N # m
Ans.
Ans: (NC)x = 0, (VC)y = 246 N, (VC)z = 171 N, (TC)x = 0, (MC)y = 154 N # m, (MC)z = 123 N # m 23
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* 1.24.
FF N � 400 80 lb
The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a–a.
a
30 30�
5.75 0.23mm in. A 4 mm 0.16 in.
45 45� a
Solution Equations of Equilibrium: For section a–a +
Q©Fx¿ = 0;
VA –400 80 cos 15° = 0 Ans.
VA = 386.37 77.3 lb N a+ ©Fy¿ = 0;
NA –400 80 sin 15° = 0 Ans.
103.53 NA = 20.7 lb N a + ©MA = 0;
MAA –400 80 sin sin 15°(0.004) 15°(0.16) ++ 400 80 cos = 0= 0 –M cos15°(0.23) 15°(0.00575) 1.808lbN# in. ·m MA = 14.5
Ans.
400 N
0.00575 m
0.004 m
Ans: VA = 386.37 N, NA = 103.53 N, MA = 1.808 N . m 24
CH 01.indd 7
11/29/10 10:18:20 AM
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1–25. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section at point D. The 400N forces act in the z direction and the 200N and 80N forces act in the +y direction. The journal bearings at A and B exert only y and z components of force on the shaft.
z
400 mm 150 mm 150 mm 200 mm 200 mm 300 mm
80 N 80 N 200 N
Support Reactions:
A
160(0.4) + 400(0.7)  Ay (1.4) = 0
x
Ay = 245.71 N ΣFy = 0;
y
D
C
Solution ΣMz = 0;
B
200 N 400 N 400 N
 245.71  By + 400 + 160 = 0 By = 314.29 N
ΣMy = 0;
800(1.1)  Az(1.4) = 0
Az = 628.57 N
ΣFz = 0;
Bz + 628.57  800 = 0
Bz = 171.43 N
Equations of Equilibrium: For point D ΣFx = 0; ΣFy = 0;
Ans.
(ND)x = 0 (VD)y  314.29 + 160 = 0
Ans.
(VD)y = 154 N ΣFz = 0;
ΣMx = 0; ΣMy = 0;
ΣMz = 0;
171.43 + (VD)z = 0 (VD)z =  171 N
Ans.
(TD)x = 0
Ans.
171.43(0.55) + (MD)y = 0
(MD)y =  94.3 N # m
Ans.
314.29(0.55)  160(0.15) + (MD)z = 0 (MD)z =  149 N # m
Ans.
Ans: (ND)x = 0, (VD)y = 154 N, (VD)z =  171 N, (TD)x = 0, (MD)y =  94.3 N # m, (MD)z =  149 N # m 25
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1–26. The serving tray T used on an airplane is supported on each side by an arm. The tray is pin connected to the arm at A, and at B there is a smooth pin. (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown.
12 N 9N 15 mm B 60⬚
100 mm A
150 mm
T
500 mm
VC
C
MC NC
b+ ©Fx = 0;
NC + 9 cos 30° + 12 cos 30° = 0;
NC =  18.2 N
Ans.
a+ ©Fy = 0;
VC  9 sin 30°  12 sin 30° = 0;
VC = 10.5 N
Ans.
a + ©MC = 0;  MC  9(0.5 cos 60° + 0.115)  12(0.5 cos 60° + 0.265) = 0 MC =  9.46 N
#
m
Ans.
Ans: NC =  18.2 N VC = 10.5 N MC =  9.46 N 26
#
m
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1–27. z
The pipe has a mass of 12 kg>m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B.
300 N
400 N A C x
Solution
ΣFy = 0; ΣFz = 0;
2m
2m
Internal Loadings: Referring to the FBD of the right segment of the pipe assembly sectioned through B, Fig. a, ΣFx = 0;
1m
B
(VB)x + 300 = 0 4 (NB)y + 400 + 500 a b = 0 5
(VB)x = 300 N
Ans.
(NB)y = 800 N
Ans.
5
3 4
y 500 N
3 (VB)z  2 3 12(2)(9.81) 4  500 a b = 0 5
(VB)z = 770.88 N = 771 N Ans.
ΣMx = 0;
3 (MB)x  12(2)(9.81)(1)  12(2)(9.81)(2)  500 a b(2) 5 (MB)x = 2106.32 N # m = 2.11 kN # m
ΣMy = 0;
(TB)y + 300(2) = 0
ΣMz = 0;
(MB)z  300(2) = 0
 400(2) = 0 Ans.
(TB)y = 600 N # m (MB)z = 600 N # m
Ans. Ans.
The negative signs indicates that (VB)x, (NB)y, and (TB)y act in the sense opposite to those shown in the FBD.
Ans: (VB)x =  300 N, (NB)y =  800 N, (VB)z = 771 N, (MB)x = 2.11 kN # m, (TB)y =  600 N # m, (MB)z = 600 N # m 27
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*1–28. The brace and drill bit is used to drill a hole at O. If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A.
z
O x
�150 30 lb FFxx N
A
225 mm 9 in. 3 mm in. 225 75 9 in. mm 150 mm 6 in. 150 mm 6 in. 150 mm 6 in.
� 50 10 N lb Fz �250 50 lb FFyy N y
Solution Internal Loading: Referring to the freebody diagram of the section of the drill and brace shown in Fig. a, ©Fx = 0; ©Fy = 0; ©Fz = 0; ©Mx = 0; ©My = 0; ©Mz = 0;
30 =500 A VA B x  150
250=500 A NA B y  50
30 lb N A VA B x 5= 150
Ans.
= 50 10 N lb A VA B z 5
Ans.
50 lb N A NA B y 5= 250
50 5 = 00 A VA B z  10
Ans.
22.5 lb N #· ft m A MA B x5= 33.75
50(0.675)=500 A MA B x  10(2.25)
Ans.
= 33.75 22.5 lbN# ·ftm A TA B y 5
30(0.75) = 5 00 A TA B y  150(0.225)
Ans.
 37.5 lb N# ·ftm A MA B z 5= 256.25
30(1.25) = 5 00 A MA B z + 150(0.375)
Ans.
The negative sign indicates that (MA)Z acts in the opposite sense to that shown on the freebody diagram.
0.15 m 150 N 0.225 m
0.15 m 0.225 m
0.225 m
50 N
0.15 m
250 N
Ans: (VA)x = 150 N, (NA)y = 250 N, (VA)z = 50 N, (MA)x = 33.75 N # m, (TA)y = 33.75 N # m, (MA)z = −56.25 N # m
28
CH 01.indd 18
11/29/10 10:18:34 AM
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1–29. The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine the resultant internal loadings acting on the cross section at point B. Hint: The distance from the centroid C of segment AB to point O is CO = 0.9745r.
D
B
r C
45 22.5
O 90
A
Solution p rw = 0; 4
ΣFz = 0;
VB 
ΣFx = 0;
NB = 0
ΣMx = 0;
TB 
ΣMy = 0;
MB +
p rw(0.09968r) = 0; 4 p rw(0.3729 r) = 0; 4
Ans.
VB = 0.785 w r
Ans. TB = 0.0783 w r 2
Ans.
MB = 0.293 w r 2
Ans.
Ans: VB = 0.785wr, NB = 0, TB = 0.0783wr 2, MB =  0.293wr 2 29
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1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = N, dM>du =  T, and dT>du = M.
M dM V dV
T dT N dN
M V N
du
T
Solution ΣFx = 0; N cos
du du du du + V sin  (N + dN) cos + (V + dV) sin = 0 2 2 2 2
(1)
ΣFy = 0; N sin
du du du du  V cos + (N + dN) sin + (V + dV) cos = 0 2 2 2 2
(2)
ΣMx = 0; T cos
du du du du + M sin  (T + dT) cos + (M + dM) sin = 0 2 2 2 2
(3)
ΣMy = 0; T sin
du du du du  M cos + (T + dT) sin + (M + dM) cos = 0 2 2 2 2
Since
du du du du is can add, then sin = , cos = 1 2 2 2 2
Eq. (1) becomes Vdu  dN +
(4)
dVdu = 0 2
Neglecting the second order term, Vdu  dN = 0 dN = V QED du Eq. (2) becomes Ndu + dV +
dNdu = 0 2
Neglecting the second order term, Ndu + dV = 0 dV =  N QED du Eq. (3) becomes Mdu  dT +
dMdu = 0 2
Neglecting the second order term, Mdu  dT = 0 dT = M QED du Eq. (4) becomes Tdu + dM +
dTdu = 0 2
Neglecting the second order term, Tdu + dM = 0 dM =  T QED du
Ans: N/A 30
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The bar has a crosssectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at u from the horizontal. Plot the variation of these stresses as a function of u 10 … u … 90° 2 .
P
P u A
Solution Equations of Equilibrium: R + ΣFx = 0;
V  P cos u = 0
V = P cos u
Q + ΣFy = 0;
N  P sin u = 0
N = P sin u
Average Normal Stress and Shear Stress: Area at u plane, A′ = savg =
N P sin u P = = sin2 u A A′ A sin u
tavg =
V P cos u = A A′ sin u =
A . sin u Ans.
P P sin u cos u = sin 2u A 2A
Ans.
Ans: savg = 31
P P sin2 u, tavg = sin 2u A 2A
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*1–32. 4 kN
The builtup shaft consists of a pipe AB and solid rod BC. The pipe has an inner diameter of 20 mm and outer diameter of 28 mm. The rod has a diameter of 12 mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of these points.
B
A
6 kN
C
8 kN
6 kN E
D
Solution At D: sD =
P = A
4(103) p 2 4 (0.028
 0.02 2)
Ans.
= 13.3 MPa (C)
At E: sE =
P = A
8(103) p 4
(0.012 2)
Ans.
= 70.7 MPa (T)
Ans: sD = 13.3 MPa, sE = 70.7 MPa 32
CH 01.indd 22
11/29/10 10:18:36 AM
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1–33. The triangular blocks are glued along each side of the joint. A Cclamp placed between two of the blocks is used to draw the joint tight. If the glue can withstand a maximum average shear stress of 800 kPa, determine the maximum allowable clamping force F.
50 mm 45�
F glue 25 mm
F
Solution Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the freebody diagram of the triangular block, Fig. a. + : ©Fx = 0;
F cos 45°  V = 0
V =
2 F 2
Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10  3)m2. We obtain
tavg
V ; = A
2 F 2 800(10 ) = 1.25(10  3) 3
Ans.
F = 1414 N = 1.41 kN
Ans: F = 1.41 kN 33
CH 01.indd 45
11/29/10 10:19:24 AM
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1–34. The triangular blocks are glued along each side of the joint. A Cclamp placed between two of the blocks is used to draw the joint tight. If the clamping force is F = 900 N, determine the average shear stress developed in the glued shear plane.
50 mm 45�
F glue 25 mm
F
Solution Internal Loadings: The shear force developed on the glued shear plane can be obtained by writing the force equation of equilibrium along the x axis with reference to the freebody diagram of the triangular block, Fig. a. + : ©Fx = 0;
900 cos 45°  V = 0
V = 636.40 N
Average Normal and Shear Stress: The area of the glued shear plane is A = 0.05(0.025) = 1.25(10  3)m2. We obtain tavg =
V 636.40 = 509 kPa = A 1.25(10  3)
Ans.
Ans. V = 636.40 N, tavg = 509 kPa 34
CH 01.indd 45
11/29/10 10:19:24 AM
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1–35. Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average normal stress or the average shear stress at section b–b to exceed s = 15 MPa and t = 16 MPa, respectively. Member CB has a square cross section of 30 mm on each side.
B
b
w
b 3m
Solution
C
A
Support Reactions: FBD(a) 4m
4 a+ ΣMA = 0; FBC (3)  3w(1.5) = 0 FBC = 1.875w 5 Equations of Equilibrium: For section b–b, FBD(b) + ΣFx = 0; 4 (1.875w)  Vb  b = 0 Vb  b = 1.50w S 5 3 + c ΣFy = 0; (1.875w)  Nb  b = 0 Nb  b = 1.125w 5 Average Normal Stress and Shear Stress: The crosssectional area of section b–b, 5A ; where A = (0.03)(0.03) = 0.90 ( 103 ) m2. A′ = 3 5 Then A′ = (0.90) ( 103 ) = 1.50 ( 103 ) m2. 3 Assume failure due to normal stress. (sb  b)Allow =
Nb  b ′
A
;
15 ( 106 ) =
1.125w 1.50 ( 103 )
w = 20000 N>m = 20.0 kN>m Assume failure due to shear stress. (tb  b)Allow =
Vb  b A′
;
16(106) =
1.50w 1.50(103)
w = 16000 N>m = 16.0 kN>m (Controls !)
Ans.
Ans: w = 16.0 kN>m (Controls !) 35
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*1–36. The supporting wheel on a scaffold is held in place on the leg using a 4mmdiameter pin. If the wheel is subjected to a normal force of 3 kN, determine the average shear stress in the pin. Assume the pin only supports the vertical 3kN load.
Solution
+ c ΣFy = 0; 3 kN # 2V = 0; V = 1.5 kN
tavg
3 kN
1.5 ( 103 ) V = = p = 119 MPa 2 A 4 (0.004)
Ans.
Ans: tavg = 119 MPa 36
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1–37. If P = 5 kN, determine the average shear stress in the pins at A, B, and C. All pins are in double shear, and each has a diameter of 18 mm.
0.5 m
6P
3P
P 1.5 m
P 2m
1.5 m
0.5 m
B 5
C
3
A
4
Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, 3 5(0.5) + 30(2) + 15(4) + 5(5.5)  FBC a b(6) = 0 5
a+ ΣMA = 0;
FBC = 41.67 kN
a+ ΣMB = 0;
Ay(6)  5(0.5)  15(2)  30(4)  5(5.5) = 0 Ay = 30.0 kN
+ ΣFx = 0; S
4 41.67 a b  Ax = 0 5
Thus,
Ax = 33.33 kN
FA = 2 Ax2 + Ay2 = 233.332 + 30.02 = 44.85 kN
Average Shear Stress: Since all the pins are subjected to double shear, FBC 41.67 VB = VC = = kN = 20.83 kN (Fig. b) and VA = 22.42 kN (Fig. c) 2 2 For pins B and C 20.83 ( 103 ) VC = = 81.87 MPa = 81.9 MPa p A ( 0.0182 ) 4 ( 22.42 103 ) VA = tA = = 88.12 MPa = 88.1 MPa p A ( 0.0182 ) 4 tB = tC =
Ans.
Ans.
Ans: tB = tC = 81.9 MPa, tA = 88.1 MPa 37
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1–38. Determine the maximum magnitude P of the loads the beam can support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear, and each has a diameter of 18 mm.
0.5 m
6P
3P
P 1.5 m
P 2m
1.5 m
0.5 m
B 5
C
3
A
4
Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0;
3 P(0.5) + 6P(2) + 3P(4) + P(5.5)  FBC a b(6) = 0 5 FBC = 8.3333P
a+ ΣMB = 0;
Ay(6)  P(0.5)  3P(2)  6P(4)  P(5.5) = 0 Ay = 6.00P
+ ΣFx = 0; S
4 8.3333P a b  Ax = 0 5
Thus,
Ax = 6.6667P
FA = 2 Ax2 + Ay2 = 2(6.6667P)2 + (6.00P)2 = 8.9691P
Average Shear Stress: Since all the pins are subjected to double shear, FBC 8.3333P VB = VC = = = 4.1667P (Fig. b) and VA = 4.4845P (Fig. c). Since 2 2 pin A is subjected to a larger shear force, it is critical. Thus tallow =
VA ; A
80 ( 106 ) =
4.4845P p ( 0.0182 ) 4
P = 4.539 ( 103 ) N = 4.54 kN
Ans.
Ans: P = 4.54 kN 38
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1–39. Determine the average normal stress in each of the 20mmdiameter bars of the truss. Set P = 40 kN.
C
P
1.5 m
B
A
Solution Internal Loadings: The force developed in each member of the truss can be determined by using the method of joints. First, consider the equilibrium of joint C, Fig. a,
2m
4 + ΣFx = 0; 40  FBC a b = 0 FBC = 50 kN (C) S 5
3 + c ΣFy = 0; 50 a b  FAC = 0 5
FAC = 30 kN (T)
Subsequently, the equilibrium of joint B, Fig. b, is considered 4 + ΣFx = 0; 50 a b  FAB = 0 FAB = 40 kN (T) S 5
Average Normal Stress: The crosssectional area of each of the bars is A =
p (0.022) = 0.3142(10  3) m2. We obtain, 4 50(103) FBC (savg)BC = = = 159 MPa A 0.3142(10  3)
Ans.
(savg)AC =
30(103) FAC = = 95.5 MPa A 0.3142(10  3)
Ans.
(savg)AB =
40(103) FAB = = 127 MPa A 0.3142(10  3)
Ans.
Ans: (savg)BC = 159 MPa, (savg)AC = 95.5 MPa, (savg)AB = 127 MPa 39
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*1–40. If the average normal stress in each of the 20mmdiameter bars is not allowed to exceed 150 MPa, determine the maximum force P that can be applied to joint C.
C
P
1.5 m
B
A
Solution Internal Loadings: The force developed in each member of the truss can be determined by using the method of joints. First, consider the equilibrium of joint C, Fig. a. 4 + ΣFx = 0; P  FBC a b = 0 FBC = 1.25P(C) S 5 + c ΣFy = 0;
3 1.25P a b  FAC = 0 5
2m
FAC = 0.75P(T)
Subsequently, the equilibrium of joint B, Fig. b, is considered. + ΣFx = 0; S
4 1.25P a b  FAB = 0 5
FAB = P(T)
Average Normal Stress: Since the crosssectional area and the allowable normal stress of each bar are the same, member BC, which is subjected to the maximum normal force, is the critical member. The crosssectional area of each of the bars is p A = (0.022) = 0.3142(10  3) m2. We have, 4 (savg)allow =
FBC 1.25P ; 150(106) = A 0.3142(10  3) P = 37 699 N = 37.7 kN
Ans.
Ans: P = 37.7 kN 40
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Determine the maximum average shear stress in pin A of the truss. A horizontal force of P = 40 kN is applied to joint C. Each pin has a diameter of 25 mm and is subjected to double shear.
C
P
1.5 m
B
A
Solution Internal Loadings: The forces acting on pins A and B are equal to the support reactions at A and B. Referring to the freebody diagram of the entire truss, Fig. a, ΣMA = 0;
By(2)  40(1.5) = 0
By = 30 kN
+ ΣFx = 0; S
40  Ax = 0
Ax = 40 kN
2m
+ c ΣFy = 0; 30  Ay = 0 Ay = 30 kN Thus, FA = 2Ax2 + Ay 2 = 2402 + 302 = 50 kN
Since pin A is in double shear, Fig. b, the shear forces developed on the shear planes of pin A are VA =
FA 50 = = 25 kN 2 2
p Average Shear Stress: The area of the shear plane for pin A is AA = (0.0252) = 4 0.4909(10  3) m2. We have (tavg)A =
25(103) VA = = 50.9 MPa AA 0.4909(10  3)
Ans.
Ans: (tavg)A = 50.9 MPa 41
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1–42. The pedestal has a triangular cross section as shown. If it is subjected to a compressive force of 2250 N, specify the x and y coordinates for the location of point P(x, y), where the load must be applied on the cross section, so that the average normal stress is uniform. Compute the stress and sketch its distribution acting on the cross section at a location removed from the point of load application.
2250 N 300 mm
P(x,y) y
x
75 mm 150 mm
Solution x
2 (75) 1 (75)(300) + 75 + 1 (150) 1 (150)(300) 3 3 100 mm = 2 2 1 (75)(300) + 1 (150)(300) 2 2
Ans.
y=
1 (300 3
Ans.
σ=
P = A
mm) = 100 mm
2250
Ans.
1 (0.225)(0.3) 2 3
= 66.67(10 ) N/m 2 = 66.7 kPa 75 mm + 1 (150 mm ) 3
2 (75 3
mm )
1 (300 3
mm )
Ans: x = 100 mm, y = 100 mm, s = 66.7 kPa 42
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The plate has a width of 0.5 m. If the stress distribution at the support varies as shown, determine the force P applied to the plate and the distance d to where it is applied.
4m d
P
x
s (15x1/2) MPa
30 MPa
Solution The resultant force dF of the bearing pressure acting on the plate of area dA = b dx = 0.5 dx, Fig. a, 1
1
dF = sb dA = (15x2)(106)(0.5dx) = 7.5(106)x2 dx L
+ c ΣFy = 0;
dF  P = 0 L0
4m
1
7.5(106)x 2 dx  P = 0
P = 40(106) N = 40 MN
Ans.
Equilibrium requires L
a+ ΣMO = 0; L0
4m
xdF  Pd = 0 1
x[7.5(106)x2 dx]  40(106) d = 0 Ans.
d = 2.40 m
Ans: P = 40 MN, d = 2.40 m 43
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*1–44.
27 kN
The joint is subjected to the axial member force of 27 kN. Determine the average normal stress acting on sections AB and BC. Assume the member is smooth and is 40 mm thick. 60⬚ A
C
40 mm
20⬚ 112 mm
B
Solution + c ©Fy = 0;
 27 sin 60° + NBC cos 20° = 0 NBC = 24.88 kN
+ : ©Fx = 0;
NAB  27 cos 60°  24.88 sin 20° = 0 NAB = 22.01 kN
sAB =
NAB 22.01 (103) = = 13.76(106) N/m2 = 13.8 MPa AAB (0.04)(0.04)
Ans.
sBC =
NBC 24.88 (103) = = 5.554(106) N/m2 = 5.55 MPa (0.04)(0.112) ABC
Ans.
Ans: sAB = 13.8 MPa, sBC = 5.55 MPa 44
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1–45. The plastic block is subjected to an axial compressive force of 600 N. Assuming that the caps at the top and bottom distribute the load uniformly throughout the block, determine the average normal and average shear stress acting along section a–a.
600 N
a 150 mm 30 a 50 mm
Solution
50 mm 50 mm
Along a–a:
600 N
+bΣFx = 0; V  600 sin 30° = 0 V = 300 N +R ΣFy = 0;  N + 600 cos 30° = 0 N = 519.6 N sa  a = ta  a =
519.6 (0.05) 1 cos0.130° 2 300
(0.05) 1 cos0.130° 2
= 90.0 kPa
Ans.
= 52.0 kPa
Ans.
Ans: sa  a = 90.0 kPa, ta  a = 52.0 kPa 45
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1–46. The column is made of concrete having a density of 2.30 Mg>m3. At its top B it is subjected to an axial compressive force of 15 kN. Determine the average normal stress in the column as a function of the distance z measured from its base.
z
15 kN B
180 mm
Solution
4m
+ c ΣFy = 0 P  15  9.187 + 2.297z = 0
z
P = 24.187  2.297z s =
P 24.187  2.297z = = (238  22.6z) kPa A p(0.18)2
Ans.
x
y
Ans: s = (238  22.6z) kPa 46
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1–47. If P = 15 kN, determine the average shear stress in the pins at A, B, and C. All pins are in double shear, and each has a diameter of 18 mm.
0.5 m C 30 B
P
4P 1m
4P 1.5 m
1.5 m
2P 0.5 m
A
Solution For pins B and C: tB = tC = For pin A:
82.5 ( 103 ) V = p 18 2 = 324 MPa A 4 1 1000 2
Ans.
FA = 2(82.5)2 + (142.9)2 = 165 kN tA =
82.5 ( 103 ) V = p 18 2 = 324 MPa A 4 1 1000 2
Ans.
Ans: tB = 324 MPa, tA = 324 MPa 47
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* 1–48. The driver of the sports car applies his rear brakes and causes the tires to slip. If the normal force on each rear tire is 1800 N and the coefficient of kinetic friction between the tires and the pavement is µ k = 0.5, determine the average shear stress developed by the friction force on the tires. Assume the rubber of the tires is flexible and each tire is filled with an air pressure of 225 K P 1800 N
Solution From the tire pressure, p =
N ; A
225(103) =
1800 A
A = 0.008 m2
The friction is F = mkN = 0.5(1800) = 900 N Then the shear stress is tavg =
F 900 = = 112.5(103) N/m2 = 112.5 KPa 0.008 A
Ans.
Ans: tavg = 112.5 K Pa 48
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1–49. The beam is supported by two rods AB and CD that have crosssectional areas of 12 mm2 and 8 mm2, respectively. If d = 1 m, determine the average normal stress in each rod.
B
D 6 kN d
A
3m
C
Solution a+ ΣMA = 0; FCD(3)  6(1) = 0 FCD = 2 kN + c ΣFy = F 0; AB  6 + 2 = 0 FAB = 4 kN sAB =
4 ( 103 ) FAB = = 333 MPa AAB 12 ( 106 )
Ans.
sCD =
2 ( 103 ) FCD = = 250 MPa ACD 8 ( 106 )
Ans.
Ans: sAB = 333 MPa, sCD = 250 MPa 49
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1–50. The beam is supported by two rods AB and CD that have crosssectional areas of 12 mm2 and 8 mm2, respectively. Determine the position d of the 6kN load so that the average normal stress in each rod is the same.
B
D 6 kN d
A
3m
C
Solution a+ ΣMO = 0; FCD(3  d)  FAB(d) = 0 s =
(1)
FCD FAB = 12 8 (2)
FAB = 1.5 FCD From Eqs. (1) and (2), FCD(3  d)  1.5 FCD(d) = 0 FCD(3  d  1.5 d) = 0 3  2.5 d = 0
Ans.
d = 1.20 m
Ans: d = 1.20 m 50
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1–51. The uniform bar, having a crosssectional area of A and mass per unit length of m, is pinned at its center. If it is rotating in the horizontal plane at a constant angular rate of v, determine the average normal stress in the bar as a function of x.
L 2
L 2 V x
Solution Equation of Motion: + ΣFx = maN; d
1 1 N = m c (L  2x) d v2 c (L + 2x) d 2 4 =
mv2 2 ( L  4x2 ) 8
Average Normal Stress: s =
N mv2 2 ( L  4x2 ) = A 8A
Ans.
Ans: s = 51
mv2 2 ( L  4x2 ) 8A
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37.5 1.5 mm in.
*1–52. The two members used in the construction of an aircraft fuselage are joined together using a 30° fishmouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 2 kN.
4 kN 800 lb
30 30�
251 mm in.
4 kN 800 lb
251 mm in.
30 30�
Solution a+ ©Fy = 0;
N sin 30° = 0; = 0; N –2 400 sin 30°
kN lb N ==1200
+Q©Fx = 0;
2400 coscos 30°30° – V=V0; = 0;
kNlb V ==1.732 346.41
= A′
(0.0375)(0.025) = 1.875(10 −3 ) m 2 sin 30° 2 kN
N 1(10 3 ) = σ = = 533 kPa A′ 1.875(10 −3 ) m 2
= t
Ans.
2 kN
Ans.
V 1.732(10 3 ) = = 924 kPa A′ 1.875(10 −3 ) m 2
Ans:
σ= 533 kPa t 924 kPa = 52
CH 01.indd 26
11/29/10 10:18:39 AM
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1–53. P
The pier is made of material having a specific weight g. If it has a square cross section, determine its width w as a function of z so that the average normal stress in the pier remains constant. The pier supports a constant load P at its top where its width is w1.
w1 w1 z w
w L
Solution Assume constant stress s1, then at the top, P s1 = w1 2
(1)
For an increase in z the area must increase, dA =
g A dz g dW dA = or = dz s1 s1 s1 A
For the top section: A
g z dA = dz s1 L0 LA1 A In
g A = z s1 A1 A = A1e ( s ) z g
1
A = w2 A1 = w 21 w = w1 e ( 2 s ) z g
1
From Eq. (1), 2
w = w1 e
cw 1g d z 2P
Ans.
Ans: 2
w = w1 e 53
cw 1g d z 2P
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The 2Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the average normal stress in the cables. The diameters of AB and AC are 12 mm and 10 mm, respectively.
A
30
45 C
Solution
B
G
Internal Loadings: The normal force developed in cables AB and AC can be determined by considering the equilibrium of the hook for which the freebody diagram is shown in Fig. a. ΣFx′ = 0; 2000(9.81) cos 45°  FAB cos 15° = 0 FAB = 14 362.83 N (T) ΣFy′ = 0; 2000(9.81) sin 45°  14 362.83 sin 15°  FAC = 0 FAC = 10 156.06 N (T) Average Normal Stress: The crosssectional areas of cables AB and AC are p p AAB = (0.0122) = 0.1131(10  3) m2 and AAC = (0.012) = 78.540(10  6) m2 . 4 4 We have sAB =
FAB 14 362.83 = = 127 MPa AAB 0.1131(10  3)
Ans.
sAC =
FAC 10 156.06 = = 129 MPa AAC 78.540(10  6)
Ans.
Ans: sAB = 127 MPa, sAC = 129 MPa 54
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1–55. The 2Mg concrete pipe has a center of mass at point G. If it is suspended from cables AB and AC, determine the diameter of cable AB so that the average normal stress in this cable is the same as in the 10mmdiameter cable AC.
A
30
45 C
Solution
B
G
Internal Loadings: The normal force in cables AB and AC can be determined by considering the equilibrium of the hook for which the freebody diagram is shown in Fig. a. ΣFx′ = 0; 2000(9.81) cos 45°  FAB cos 15° = 0
FAB = 14 362.83 N (T)
ΣFy′ = 0; 2000(9.81) sin 45°  14 362.83 sin 15°  FAC = 0 FAC = 10 156.06 N (T) Average Normal Stress: The crosssectional areas of cables AB and AC are p p AAB = d AB2 and AAC = (0.012) = 78.540(10  6) m2. 4 4 Here, we require sAB = sAC FAC FAB = AAB AAC 14 362.83 10 156.06 = p 2 6 d 78.540(10 ) 4 AB Ans.
d AB = 0.01189 m = 11.9 mm
Ans: d AB = 11.9 mm 55
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*1–56. Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the 3 kN force is applied to the ring at B, determine the angle u so that the average normal stress in each rod is equivalent. What is this stress?
C
5 4
3
B
A u 3 kN
Solution Method of Joints: Referring to the FBD of joint B, Fig. a, 3 + c ΣFy = 0; FBC a b  3 cos u = 0 FBC = 5 cos u kN 5 + ΣFx = 0; S
4 (5 cos u)a b  3 sin u  FAB = 0 FAB = (4 cos u  3 sin u) kN 5
Average Normal Stress: sAB =
sBC =
[4 cos u  3 sin u] ( 103 ) 250 ( 106 ) FAB = = (4 cos u  3 sin u) p p AAB (0.004)2 4 (5 cos u) ( 103 ) 555.56 ( 106 ) FBC = = c d cos u p p ABC (0.006)2 4
It is required that sAB = sBC 250 ( 10 p
6
)
(4 cos u  3 sin u) = c
1.7778 cos u  3 sin u = 0 tan u =
555.56 ( 106 ) d cos u p
1.7778 3 Ans.
u = 30.65° = 30.7° Then s = sBC = c
555.56 ( 106 ) d cos 30.65° = 152.13 ( 106 ) Pa = 152 MPa p
Ans.
Ans: u = 30.7°, s = 152 MPa 56
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1–57. The bar has a crosssectional area of 400(10−6) m2. If it is subjected to a triangular axial distributed loading along its length which is 0 at x = 0 and 9 kN>m at x = 1.5 m, and to two concentrated loads as shown, determine the average normal stress in the bar as a function of x for 0 … x 6 0.6 m.
8 kN 4 kN x 0.6 m
0.9 m
Solution Internal Loading: Referring to the FBD of the right segment of the bar sectioned at x, Fig. a, + ΣFx = 0; S
8 + 4 + N =
1 (6x + 9)(1.5  x) = 0 2
5 18.75
Average Normal Stress: s =
N = A =
 3x2 6 kN
( 18.75  3x2 )( 103 ) 400 ( 106 )
5 46.9
 7.50x2 6 MPa
Ans.
57
Ans: s = 546.9  7.50x2 6 MPa
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The bar has a crosssectional area of 400(10−6) m2. If it is subjected to a uniform axial distributed loading along its length of 9 kN>m, and to two concentrated loads as shown, determine the average normal stress in the bar as a function of x for 0.6 m 6 x … 1.5 m.
8 kN 4 kN x 0.6 m
0.9 m
Solution Internal Loading: Referring to a FBD of the right segment of the bar sectioned at x, + ΣFx = 0; S
4 + 9 (1.5  x)  N = 0 N =
5 17.5
Average Normal Stress: s =
N = A =
 9x 6 kN
( 17.5  9x )( 103 ) 400 ( 106 )
5 43.75
 22.5x 6 MPa
Ans.
58
Ans: s = 5 43.75  22.5x 6 MPa
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The two steel members are joined together using a 30° scarf weld. Determine the average normal and average shear stress resisted in the plane of the weld.
15 kN
Internal Loadings: Referring to the FBD of the upper segment of the member sectioned through the scarf weld, Fig. a, ΣFx = 0;
N  15 sin 30° = 0
N = 7.50 kN
ΣFy = 0;
V  15 cos 30° = 0
V = 12.99 kN
A = 0.02 a
40 mm
15 kN
Average Normal and Shear Stress: The area of the scarf weld is
Thus,
20 mm
30
Solution
0.04 b = 1.6 ( 103 ) m2 sin 30°
s =
7.50 ( 103 ) N = = 4.6875 ( 106 ) Pa = 4.69 MPa An 1.6 ( 103 )
Ans.
t =
12.99 ( 103 ) V = = 8.119 ( 106 ) Pa = 8.12 MPa Av 1.6 ( 103 )
Ans.
Ans: s = 4.69 MPa, t = 8.12 MPa 59
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*1–60. The bar has a crosssectional area of 400(10−6) m2. If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads, determine the average normal stress in the bar as a function of x for 0 6 x … 0.5 m.
w 8 kN/m
6 kN 3 kN
x 0.5 m
0.75 m
Solution Equation of Equilibrium: + ΣFx = 0;  N + 3 + 6 + 8(1.25  x) = 0 S N = (19.0  8.00x) kN Average Normal Stress: s =
(19.0  8.00x)(103) N = A 400(106) Ans.
= (47.5  20.0x) MPa
Ans: s = (47.5  20.0x) MPa 60
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1–61. The bar has a crosssectional area of 400(10−6) m2. If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads, determine the average normal stress in the bar as a function of x for 0.5 m 6 x … 1.25 m.
w 8 kN/m
6 kN 3 kN
x 0.5 m
0.75 m
Solution Equation of Equilibrium: + ΣFx = 0;  N + 3 + 8(1.25  x) = 0 S N = (13.0  8.00x) kN Average Normal Stress: s =
(13.0  8.00x) ( 103 ) N = A 400 ( 106 ) Ans.
= (32.5  20.0x) MPa
Ans: s = (32.5  20.0x) MPa 61
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1–62. The prismatic bar has a crosssectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for 0 … x 6 a.
w0
x a
a
Solution Equation of Equilibrium: w + ΣFx = 0;  N + 1 a 0 x + w0 b (a  x) + 1 w0 a = 0 S 2 a 2 N =
w0 ( 2a2  x2 ) 2a
Average Normal Stress: s =
N = A
w0 2a
( 2a2  x2 ) A
=
w0 ( 2a2  x2 ) 2aA
Ans.
Ans: s = 62
w0 ( 2a2  x2 ) 2aA
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1–63. The prismatic bar has a crosssectional area A. If it is subjected to a distributed axial loading that increases linearly from w = 0 at x = 0 to w = w0 at x = a, and then decreases linearly to w = 0 at x = 2a, determine the average normal stress in the bar as a function of x for a 6 x … 2a.
w0
x a
a
Solution Equation of Equilibrium: w + ΣFx = 0;  N + 1 c 0 (2a  x) d (2a  x) = 0 S 2 a N =
w0 (2a  x)2 2a
Average Normal Stress: s =
N = A
w0 2a
(2a  x)2 A
=
w0 (2a  x)2 2aA
Ans.
Ans: s = 63
w0 (2a  x)2 2aA
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*1–64. Determine the greatest constant angular velocity v of the flywheel so that the average normal stress in its rim does not exceed s = 15 MPa. Assume the rim is a thin ring having a thickness of 3 mm, width of 20 mm, and a mass of 30 kg>m. Rotation occurs in the horizontal plane. Neglect the effect of the spokes in the analysis. Hint: Consider a freebody diagram of a semicircular segment of the ring. The center of mass for this segment is located at nr = 2r>π from the center.
v
0.8 mm
Solution + T ΣFn = m(aG)n;
2T = m( r )v2 2sA = ma
2r 2 bv p
2 ( 15 ( 106 ) ) (0.003)(0.020) = p(0.8)(30)a v = 6.85 rad>s
2(0.8) 2 bv p
Ans.
Ans: v = 6.85 rad>s 64
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1–65. Determine the largest load P that can be applied to the frame without causing either the average normal stress or the average shear stress at section a–a to exceed s = 150 MPa and t = 60 MPa, respectively. Member CB has a square cross section of 25 mm on each side.
B
2m
a a A
Solution
C 1.5 m
Analyze the equilibrium of joint C using the FBD shown in Fig. a,
P
4 + c ΣFy = 0; FBC a b  P = 0 FBC = 1.25P 5
Referring to the FBD of the cut segment of member BC Fig. b. + ΣFx = 0; S + c ΣFy = 0;
3 Na  a  1.25P a b = 0 5
4 1.25P a b  Va  a = 0 5
The crosssectional area of section 1.0417(10  3) m2. For Normal stress, sallow =
Na  a = 0.75P Va  a = P a–a
is
Aa  a = (0.025)a
0.025 b = 3>5
Na  a 0.75P ; 150(106) = Aa  a 1.0417(10  3)
P = 208.33(103) N = 208.33 kN For Shear Stress tallow =
Va  a P ; 60(106) = Aa  a 1.0417(10  3)
P = 62.5(103) N = 62.5 kN (Controls!)
Ans.
Ans: P = 62.5 kN 65
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1–66. The bars of the truss each have a crosssectional area of 780 mm2. Determine the average normal stress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.
B
C
3 ftm 0.9
A
4 ftm 1.2 P
E
D
4 ft 1.2 m
0.75 P
Solution Joint A: FAB 13.33 66.67(10 ) = = 85.5 MPa A AB 780(10−6)
(T)
FAE 53.33(103) = 68.4 MPa = = A AE 780(10−6)
(C)
sAB = sAE
FAB = 66.67 kN
3
FAE = 53.33 kN
Ans. 40 kN
Ans.
Joint E:
FEB =30 kN FED = 53.33 kN
53.33 kN
sED =
FED 53.33(103) = 68.4 MPa = A ED 780(10−6)
(C)
Ans.
sEB =
FEB 6.0 30(103) = = 38.5 MPa A EB 780(10−6)
(T)
Ans.
30 kN
FBC =146.67 kN 66.67 kN
Joint B: sBC =
FBC 29.33 146.67(103) = 188 MPa = A BC 780(10−6)
(T)
Ans.
sBD =
3 FBD 23.33 116.67(10 ) = = 150 MPa −6 A BD 780(10 )
(C)
Ans.
30 kN
FBD =116.67 kN
Ans: Joint A: sAB sAE Joint E: sED sEB Joint B: sBC
= = = = =
85.5 MPa (T), 68.4 MPa (C) 68.4 MPa (C), 38.5 MPa (T) 188 MPa (T),
66
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1–67. The bars of the truss each have a crosssectional the maximum maximum average average normal normal stress in area of 780 1.25mm in2.2.IfIfthe determine the the maximum any bar bar isisnot nottotoexceed exceed140 20MPa, ksi, determine magnitude P of the loads that can be applied to the truss.
B
0.9 3 ftm
A
1.2 4 ftm
Solution P
Joint A: + c ©Fy = 0;
P +
E
D
1.2 m 4 ft
0.75 P
3 F = 0 5 AB
FAB = 1.6667P + : ©Fx = 0;
C
4 FAE + (1.6667P)a b = 0 5 FAE = 1.3333P
(T)
FEB
(C) FED
1.3333 P
Joint E: + c ©Fy = 0;
0.75 P
FEB  0.75P = 0 FEB = 0.75P
(T) FBC
+ : ©Fx = 0;
1.3333P  FED = 0 FED = 1.3333P
1.6667 P
0.75 P
FBD
(C)
Joint B: + c ©Fy = 0;
3 3 FBD  0.75P  (1.6667P)a b = 0 5 5 FBD = 2.9167P
+ : ©Fx = 0;
(C)
4 4 FBC – (2.9167P)a b  (1.6667P)a b = 0 5 5
FBC = 3.6667P
(T)
The highest stressed member is BC:
3.6667 P = σ BC σ= 140(106 ) max ; 780(10 −6 ) P = 29.78(10 3 ) N = 29.8 kN
Ans.
Ans: P = 29.8 kN 67
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*1–68. 2
The radius of the pedestal is defined by r = (0.5e−0.08 y ) m, where y is in meters. If the material has a density of 2.5 Mg>m3, determine the average normal stress at the support. 2
3m
r 0.5e0.08y
y
Solution
r
2
2
A = p(0.5) = 0.7854 m dV = p ( r
2
) dy = p(0.5) ( e
V = 10 p(0.5) ( e 3
0.5 m
2
2
2
 0.08y
)
 0.08y2 2
) 2 dy = 0.7854 103 ( e  0.08y ) 2dy 2
W = rg V = (2500)(9.81)(0.7854) 10 ( e 3
W = 19.262 ( 103 ) 10 ( e 3
s =
2
 0.08y
 0.08y2 2
) dy
) 2dy = 38.849 kN
W 38.849 = = 49.5 kPa A 0.7854
Ans.
Ans: s = 49.5 kPa 68
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1–69. The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt Determine thethe maximum load P has a diameter of 0.3 7.5 in. mm. Determine maximum load that can P that canbebeapplied appliedtotothe themember member ifif the the allowable shear for the the bolts bolts isistallow ksi and tallow==8412MPa stress for and the allowable 20 MPa. ksi. average normal stress is sallow ==140
60� P
P
Solution a + ©Fy = 0;
N  P sin 60° = 0 N = 0.8660 P
b+ ©Fx = 0;
V  P cos 60° = 0 V = 0.5P
Assume failure due to shear:
V 0.5P ; 84(106 ) tallow = = 2Ab 2 π4 (0.00752 ) P = 14.84(10 3 ) N = 14.84 kN Assume failure due to normal force:
N 0.8660 P ; 140(106 ) sallow = = 2Ab 2 π4 (0.00752 ) P = 14.28(10 3 ) N = 14.3 kN (contπols)
Ans.
Ans: P = 14.3 kN (controls) 69
CH 01.indd 54
11/29/10 10:19:35 AM
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1–70. Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are 10 mm thick, determine to the nearest multiples of 5 mm the smallest dimension h of the horizontal segment so that it does not fail in shear. The average shear stress for the segment is tallow = 2.1 MPa.
B 13
800 4 kNlb
5
12
h A
Solution V 1.538(10 3 ) ; 2.1(106 ) tallow = = A (0.01)h
4 kN = 3.692 kN
h = 0.07326 m = 73.26 mm
V = 1.538 kN
Use h = 75 mm
Ans.
Ans: h = 0.07326 m, use h = 75 mm 70
CH 01.indd 50
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1–71. The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is tallow = 35 MPa.
a A
d a 20 mm
500 mm 200 N
Solution a+ ΣMA = 0;
Fa  a (20)  200(500) = 0 Fa  a = 5000 N
tallow =
Fa  a ; Aa  a
35(106) =
5000 d(0.025) Ans.
d = 0.00571 m = 5.71 mm
Ans: d = 5.71 mm 71
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*1–72. The lapbelt assembly is to be subjected to a force of 800 N. Determine (a) the required thickness t of the belt if the allowable tensile stress for the material is (st)allow = 10 MPa, (b) the required lap length dl if the glue can sustain an allowable shear stress of (tallow)g = 0.75 MPa, and (c) the required diameter dr of the pin if the allowable shear stress for the pin is (tallow)p = 30 MPa.
800 N 45 mm
t dl
dr 800 N
Solution Allowable Normal Stress: Design of belt thickness. (st)allow =
P ; A
10 A 106 B =
800 (0.045)t Ans.
t = 0.001778 m = 1.78 mm Allowable Shear Stress: Design of lap length. VA 400 ; 0.750 A 106 B = (tallow)g = A (0.045) dt
Ans.
Allowable Shear Stress: Design of pin size. VB 400 ; 30 A 106 B = p 2 (tallow)P = A 4 dr
Ans.
dt = 0.01185 m = 11.9 mm
dr = 0.004120 m = 4.12 mm
Ans: t = 1.78 mm, dt = 11.9 mm, dr = 4.12 mm 72
CH 01.indd 51
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1–73. The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter and the smallest diameter d of the rods. All parts are made of steel for which the failure normal stress is sfail = 500 MPa and the failure shear stress is t fail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 in tension and (F.S.)s = 1.75 in shear.
30 kN d
t
40 mm
10 mm
Solution Allowable Normal Stress: Design of rod size sallow =
sfail P = ; F.S A
500 ( 10
6
)
2.5
d
30 ( 10
3
=
)
30 kN
p 2 4d
Ans.
d = 0.01382 m = 13.8 mm Allowable Shear Stress: Design of cotter size. tallow =
tfail V = ; F.S A
375 ( 106 ) 1.75
=
15.0 ( 103 ) (0.01)t Ans.
t = 0.0070 m = 7.00 mm
Ans: d = 13.8 mm, t = 7.00 mm 73
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1–74. The truss is used to support the loading shown Determine the required crosssectional area of member BC if the allowable normal stress is sallow = 165 MPa.
4000 N
2000 N
2m
2m
F
B 30⬚ A
Solution
E
60⬚
2m
2m
C D
45⬚
For the entire truss, a + © MA = 0;
2000(2)  4000(2.8284) + 2(2.8284)(Dy) = 0 Dy = 2707.11 N
For method of section, a + © MF = 0;
2707.11(2.8284)  FBC (1.7932) = 0 F
FBC = 4270.06 N
FEF
Using the given allowable stress, sallow =
FBC ; A
165(106) =
2.0705sin 60° = 1.7932 m
4270.06 A
2 A = 25.88(10−6) m2 = 25.9 mm
Ans.
FCF
2 = 2.0705 m cos 15°
FBC 2.8284 m
Ans: A = 25.9 mm2 74
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1–75. If the allowable tensile stress for wires AB and AC is sallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 6 kN.
C B
5
45
4 3
A
Solution Normal Forces: Analyzing the equilibrium of joint A, Fig. a, + ΣFx = 0; S + c ΣFy = 0;
3 FAC a b  FAB sin 45° = 0 5
(1)
4 FAC a b + FAB cos 45°  6 = 0 5
P
(2)
Solving Eqs. (1) and (2)
FAC = 4.2857 kN FAB = 3.6365 kN Average Normal Stress: For wire AB, sallow =
FAB ; AAB
200 ( 106 ) =
3.6365 ( 103 ) p 2 d 4 AB
d AB = 0.004812 m = 4.81 mm
Ans.
For wire AC, sallow =
FAC ; AAC
200 ( 106 ) =
4.2857 ( 103 ) p 2 d 4 AC
d AC = 0.005223 m = 5.22 mm
Ans.
Ans: d AB = 4.81 mm, d AC = 5.22 mm 75
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*1–76. If the allowable tensile stress for wires AB and AC is sallow = 180 MPa, and wire AB has a diameter of 5 mm and AC has a diameter of 6 mm, determine the greatest force P that can be applied to the chain.
C B
5
45
4 3
A
Solution Normal Forces: Analyzing the equilibrium of joint A, Fig. a, + ΣFx = 0; S + c ΣFy = 0;
3 FAC a b  FAB sin 45° = 0 5
(1)
4 FAC a b + FAB cos 45°  P = 0 5
P
(2)
Solving Eqs. (1) and (2)
FAC = 0.7143P FAB = 0.6061P Average Normal Stress: Assuming failure of wire AB, sallow =
FAB ; AAB
180 ( 106 ) =
0.6061P p ( 0.0052 ) 4
P = 5.831 ( 103 ) N = 5.83 kN Assume the failure of wire AC, sallow =
FAC ; AAC
180 ( 106 ) =
0.7143P p ( 0.0062 ) 4
P = 7.125 ( 103 ) N = 7.13 kN Choose the smaller of the two values of P, Ans.
P = 5.83 kN
Ans: P = 5.83 kN 76
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1–77. The spring mechanism is used as a shock absorber for a load applied to the drawbar AB. Determine the force in each spring when the 50kN force is applied. Each spring is originally unstretched and the drawbar slides along the smooth guide posts CG and EF. The ends of all springs are attached to their respective members. Also, what is the required diameter of the shank of bolts CG and EF if the allowable stress for the bolts is sallow = 150 MPa?
k 80 kN/m C H
A
B
k¿ 60 kN/m
Solution
k¿ 60 kN/m
G
F 200 mm
Equations of Equilibrium: a+ ΣMH = 0;
E
200 mm D
 FBF (200) + FAG(200) = 0 FBF = FAG = F
+ c ΣFy = 0;
(1)
2F + FH  50 = 0
50 kN
Required, ∆ H = ∆ B ;
FH F = 80 60 (2)
F = 0.75 FH Solving Eqs. (1) and (2) yields, FH = 20.0 kN
Ans.
FBF = FAG = F = 15.0 kN
Ans.
Allowable Normal Stress: Design of bolt shank size. s allow =
P ; A
150 ( 106 ) =
15.0 ( 103 ) p 2 4d
d = 0.01128 m = 11.3 mm Ans.
d EF = d CG = 11.3 mm
Ans: FH = 20.0 kN, FBF = FAG = 15.0 kN, d EF = d CG = 11.3 mm 77
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1–78. The softride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load P = 1500 N, determine the required minimum diameter of pins B and C. Use a factor of safety of 2 against failure. The pins are made of material having a failure shear stress of tfail = 150 MPa, and each pin is subjected to double shear.
P A
300 mm
30 mm
100 mm
B 60�
C
D
Solution Internal Loadings: The forces acting on pins B and C can be determined by considering the equilibrium of the freebody diagram of the softride suspension system shown in Fig. a. a + ©MC = 0;
1500(0.4)  FBD sin 60°(0.1)  FBD cos 60°(0.03) = 0 FBD = 5905.36 N
+ : ©Fx = 0;
Cx  5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60°  1500  Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus, FB = FBD = 5905.36 N
FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N
Since both pins are in double shear, FB 5905.36 VB = = = 2952.68 N 2 2
VC =
FC 4666.98 = = 2333.49 N 2 2
Allowable Shear Stress: tfail 150 tallow = = = 75 MPa F.S. 2 Using this result, VB tallow = ; AB
75(106) =
2952.68 p 2 d 4 B Ans.
dB = 0.007080 m = 7.08 mm tallow =
VC ; AC
75(106) =
2333.49 p 2 d 4 C Ans.
dC = 0.006294 m = 6.29 mm
Ans: dB = 7.08 mm, dC = 6.29 mm 78
CH 01.indd 61
11/29/10 10:19:42 AM
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1–79. The softride suspension system of the mountain bike is pinned at C and supported by the shock absorber BD. If it is designed to support a load of P = 1500 N, determine the factor of safety of pins B and C against failure if they are made of a material having a shear failure stress of tfail = 150 MPa. Pin B has a diameter of 7.5 mm, and pin C has a diameter of 6.5 mm. Both pins are subjected to double shear.
P A
300 mm
30 mm
100 mm
B 60�
C
D
Solution Internal Loadings: The forces acting on pins B and C can be determined by considerning the equilibrium of the freebody diagram of the softride suspension system shown in Fig. a. + ©MC = 0;
1500(0.4)  FBD sin 60°(0.1)  FBD cos 60°(0.03) = 0 FBD = 5905.36 N
+ : ©Fx = 0;
Cx  5905.36 cos 60° = 0
+ c ©Fy = 0;
5905.36 sin 60°  1500  Cy = 0 Cy = 3614.20 N
Cx = 2952.68 N
Thus, FB = FBD = 5905.36 N
FC = 2 Cx 2 + Cy 2 = 2 2952.682 + 3614.202 = 4666.98 N
Since both pins are in double shear, VB =
FB 5905.36 = = 2952.68N 2 2
VC =
FC 4666.98 = = 2333.49 N 2 2
Allowable Shear Stress: The areas of the shear plane for pins B and C are p p A B = (0.00752) = 44.179(10  6)m2 and A C = (0.00652) = 33.183(10  6)m2. 4 4 We obtain
A tavg B B = A tavg B C =
VB 2952.68 = 66.84 MPa = AB 44.179(10  6) VC 2333.49 = 70.32 MPa = AC 33.183(10  6)
Using these results, tfail = (F.S.)B = t A avg B B tfail (F.S.)C = = A tavg B C
150 = 2.24 66.84
Ans.
150 = 2.13 70.32
Ans.
Ans: (F.S.)B = 2.24, (F.S.)C = 2.13 79
CH 01.indd 62
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*1–80. Determine the required diameter of the pins at A and B if the allowable shear stress for the material is tallow = 100 MPa. Both pins are subjected to double shear.
2 kN/m B A
3m
Solution Support Reactions: Member BC is a two force member. a+ ΣMA = 0;
FBC sin 45°(3)  6(1.5) = 0
C
FBC = 4.243 kN + c ΣFy = 0;
Ay + 4.243 sin 45°  6 = 0 Ay = 3.00 kN
+ ΣFx = 0; d
Ax  4.243 cos 45° = 0 Ax = 3.00 kN
Allowable Shear Stress: Pin A and pin B are subjected to double shear. FA = 23.002 + 3.002 = 4.243 kN and FB = FBC = 4.243 kN. Therefore, VA = VB = t allow =
4.243 = 2.1215 kN 2 V ; A
100 ( 106 ) =
2.1215 ( 103 ) p 2 4d
d = 0.005197 m = 5.20 mm Ans.
d A = d B = d = 5.20 mm
Ans: d A = d B = 5.20 mm 80
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1–81. The steel pipe is supported on the circular base plate and concrete pedestal. If the thickness of the pipe is t = 5 mm and the base plate has a radius of 150 mm, determine the factors of safety against failure of the steel and concrete. The applied force is 500 kN, and the normal failure stresses for steel and concrete are (sfail)st = 350 MPa and (sfail)con = 25 MPa, respectively.
t
500 kN 100 mm r
Solution Average Normal and Bearing Stress: The crosssectional area of the steel pipe and the bearing area of the concrete pedestal are Ast = p(0.12  0.0952) = 0.975(10  3)p m2 and (Acon)b = p(0.152) = 0.0225p m2. We have (savg)st =
500(103) P = = 163.24 MPa Ast 0.975(10  3)p
(savg)con =
500(103) P = = 7.074 MPa (Acon)b 0.0225p
Thus, the factor of safety against failure of the steel pipe and concrete pedestal are (F.S.)st =
(sfail)st (savg)st
(F.S.)con =
=
(sfail)con (savg)con
350 = 2.14 163.24 =
Ans.
25 = 3.53 7.074
Ans.
Ans: (F.S.)st = 2.14, (F.S.)con = 3.53 81
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1–82. The steel swivel bushing in the elevator control of an airplane is held in place using a nut and washer as shown in Fig. (a). Failure of the washer A can cause the push rod to separate as shown in Fig. (b). If the average shear stress is tavg = 145 MPa, determine the force F that must be applied to the bushing that will cause this to happen. The washer is 1.5 mm thick.
20 mm F
F A (a)
(b)
Solution From the given average shear stress, tavg =
V ; A
145(106) =
F 2p(0.01)(0.0015)
F = 13.67(103) = 13.7 kN
Ans.
Ans: F = 13.7 kN 82
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1–83. Determine the required minimum thickness t of member AB and edge distance b of the frame if P = 40 kN and the factor of safety against failure is 2. The wood has a normal failure stress of s fail = 42 MPa, a nd shear failure stress of tfail = 10.5 MPa.
P 75 mm B
75 7 mm t
A
30⬚ b
30⬚
C
Solution Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A. Fig. a. + : ©Fx = 0;
FAB cos 30°  FAC cos 30° = 0
FAC = FAB
+ c ©Fy = 0;
2FAB sin 30°  40 = 0
FAB = 40 kN
Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. b. + : ©Fx = 0;
(FB)x  40 cos 30° = 0
(FB)x = 34.64 kN
Referring to the freebody diagram shown in Fig. c, the shear force developed on the shear plane a–a is + : ©Fx = 0;
Va  a  34.64 = 0
Va  a = 34.64 kN
Allowable Normal Stress: sfail 42 = = 21 M Pa sallow = F.S. 2 tallow =
tfail 10.5 = = 5.25 M Pa F.S. 2
Using these results, sallow =
tallow =
FAB ; AAB Va  a ; Aa  a
40(103) (0.075)t t = 0.02540 m = 25.4 mm 21(106) =
5.25(106) =
Ans.
34.64(103)
(0.075)b b = 0.08798 m = 88.0 mm
Ans.
Ans: t = 25.4 mm, b = 88.0 mm 83
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*1–84. Determine the maximum allowable load P that can be safely supported by the frame if t = 30 mm and b = 90 mm. The wood has a normal failure stress of sfail = 42 MPa, and shear failure stress of tfail = 10.5 MPa. Use a factor of safety against failure of 2.
P 75 mm B
75 mm t
A
30⬚ b
30⬚
C
Solution Internal Loadings: The normal force developed in member AB can be determined by considering the equilibrium of joint A. Fig. a. + : ©Fx = 0; FAB cos 30°  FAC cos 30° = 0 FAC = FAB 2FAB sin 30°  P = 0
+ c ©Fy = 0;
FAB = P
Subsequently, the horizontal component of the force acting on joint B can be determined by analyzing the equilibrium of member AB, Fig. b. + (FB)x = 0.8660P : ©Fx = 0; (FB)x  P cos 30° = 0 Referring to the freebody diagram shown in Fig. c, the shear force developed on the shear plane a–a is + Va  a = 0.8660P : ©Fx = 0; Va  a  0.8660P = 0 Allowable Normal and Shear Stress: sallow =
sfail 42 = = 21 MPa F.S. 2
tallow =
tfail 10.5 = = 5.25 MPa F.S. 2
Using these results, sallow =
FAB ; AAB
21(106) =
P 0.075(0.03)
P = 47.25(103) N = 47.25 kN tallow =
Va  a ; Aa  a
5.25(106) =
0.8660P 0.075(0.09)
P = 40.92(103) N = 40.9 kN (controls)
Ans.
Ans: P = 40.9 kN 84
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The hanger is supported using the rectangular pin. Determine the magnitude of the allowable suspended load P if the allowable bearing stress is (sb)allow = 220 MPa, the allowable tensile stress is (st)allow = 150 MPa, and the allowable shear stress is tallow = 130 MPa. Take t = 6 mm, a = 5 mm and b = 25 mm.
20 mm 75 mm 10 mm a
37.5 mm
Allowable Normal Stress: For the hanger P ; A
150 1 106 2 =
b
37.5 mm
t
Solution
(st)allow =
a
P
P (0.075)(0.006)
P = 67.5 kN
Allowable Shear Stress: The pin is subjected to double shear. Therefore, V = tallow =
V ; A
130 1 106 2 =
P 2
P>2 (0.01)(0.025)
P = 65.0 kN
Allowable Bearing Stress: For the bearing area (sb)allow =
P ; A
220 1 106 2 =
P>2 (0.005)(0.025) Ans.
P = 55.0 kN (Controls!)
Ans: P = 55.0 kN 85
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1–86. The hanger is supported using the rectangular pin. Determine the required thickness t of the hanger, and dimensions a and b if the suspended load is P = 60 kN. The allowable tensile stress is (st)allow = 150 MPa, the allowable bearing stress is (sb)allow = 290 MPa, and the allowable shear stress is tallow = 125 MPa.
20 mm 75 mm 10 mm a
37.5 mm
Allowable Normal Stress: For the hanger P ; A
150 1 106 2 =
b
37.5 mm
t
Solution
(st)allow =
a
P
60(103) (0.075)t Ans.
t = 0.005333 m = 5.33 mm
Allowable Shear Stress: For the pin tallow =
V ; A
125 1 106 2 =
30(103) (0.01)b Ans.
b = 0.0240 m = 24.0 mm
Allowable Bearing Stress: For the bearing area (sb)allow =
P ; A
290 1 106 2 =
30(103) (0.0240) a Ans.
a = 0.00431 m = 4.31 mm
Ans: t = 5.33 mm, b = 24.0 mm, a = 4.31 mm 86
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187. The assembly is used to support the distributed loading ofww==500 10 lb>ft kN/m. Determine factor of safety of safety with loading of . Determine thethe factor with respect to yielding for the steel rod BC and the pins at A and B if the yield stress for the steel in tension is s 250 MPa shear 125aMPa. The of rod hasin., a y =in y = has and shear ty and . Thetrod diameter 0.40 = 18inksi diameter of 13 mm, and the pins each have a diameter of and the pins each have a diameter of 0.30 in. 10 mm.
C
1.2 4 ftm A
Solution For rod BC: s =
P = A
B 3
10(10 ) π
4
(0.0132 )
= 75.34 MPa
0.9 3 ftm w
sy
250 36 ==3.32 F. S. = = 2.71 s 13.26 75.34
0.3 1 ftm
Ans. 0.9 m
For pin B: tB =
VB 5(10 ) = π = 63.66 MPa A (0.012 ) 4 tY
F. S. =
tB
= =
FBC = 10 kN
Az = 6 kN
3
VB = 5 kN Ay = 4 kN
125 18 ==1.96 1.53 63.66 11.79
Ans.
VB = 5 kN 10(1.2) = 12 kN 10 kN
0.6 m
Pin B
For pin A:
FA =
4 2 + 6 2 = 7.211 kN
VA = 3.606 kN VA = 3.606 kN
VA = tA =
7.211 kN = 3.606 kN 2
7.211 kN Pin A
VA 3.606(10 3 ) = = 45.91 MPa π (0.012 ) A 4
F. S. =
tY tA
=
125 = 2.72 45.91
Ans.
Ans: (F.S.)rod = 3.32, (F.S.)pinB = 1.96 (F.S.)pinA = 2.72 87
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* 1–88. If the allowable allowableshear shearstress stressfor foreach each 0.30of of thethe 10mm12.5 and ksi, in.diameter steel at B, A,and B, and is tallow diameter steel pinspins at A, C is Ctallow = 90=MPa, and the allowable normal for13mmdiameter the 0.40in.diameter the allowable normal stressstress for the rod is rod , determine largest intensity of = 22 ksi sallowis =sallow 150 MPa, determine the the largest intensity w ofwthe the uniform distributed be suspended uniform distributed loadload that that can can be suspended fromfrom the the beam. beam.
C
4 ftm 1.2 A
B 3 ftm 0.9 w 1 ftm 0.3
Solution Assume failure of pins B and C: 3 1.667w 0.5w(10 ) tallow == 90 12.5 = =p p 2 2 ) ) 4 (10 4 (0.3
0.5w
0.3605w 0.3605w
0.5w
1.0w
w == 14.14 0.530kN/m kip>ft
0.721w
Ans.
(controls) (controls)
Assume Assumefailure failureof ofpins pinsA: A:
FBC = 1.0w
Ax = 0.6w
FFAA = 2(0(2w) = 2.404 w .6 w)2 + ((1.333w) 0.4 w)2 =2 0.721w
Ay = 0.4w 1.2w
1.202w 3) 0.3605w(10 12.5 ttallow = = pp 22 allow = 90 (10 )) 44(0.3 ww = 19.61 0.735 kN/m kip>ft Assumefailure failureof ofrod rodBC: BC: Assume ssallow 22 == allow = 150
3.333w 3) 1.0w(10 p p 22 4 (13 )) 4 (0.4
ww = 19.91 0.829 kN/m kip>ft
Ans: w = 14.14 kN/m (controls) 88
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The compound wooden beam is connected together by a bolt at B. Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is 1st 2 allow = 150 MPa and the allowable bearing stress for the wood is 1sb 2 allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt.
2m
3 kN
2m
2 kN 1.5 kN 1.5 m 1.5 m 1.5 m
1.5 m C
A
D B
Solution From FBD (a): a + ΣMD = 0;
FB(4.5) + 1.5(3) + 2(1.5)  FC(6) = 0 (1)
4.5 FB  6 FC =  7.5 From FBD (b): a + ΣMA = 0;
FB(5.5)  FC(4)  3(2) = 0 (2)
5.5 FB  4 FC = 6 Solving Eqs. (1) and (2) yields FB = 4.40 kN;
FC = 4.55 kN
For bolt: sallow = 150(106) =
4.40(103) p 2 4 (d B)
d B = 0.00611 m Ans.
= 6.11 mm For washer: sallow = 28 (104) =
4.40(103) p 2 4 (d w
 0.006112) Ans.
d w = 0.0154 m = 15.4 mm
Ans: dB = 6.11 mm, dw = 15.4 mm 89
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The two aluminum rods support the vertical force of P = 20 kN. Determine their required diameters if the allowable tensile stress for the aluminum is sallow = 150 MPa.
B
C
A
45
P
Solution + c ΣFy = 0; FAB sin 45°  20 = 0; FAB = 28.284 kN + ΣFx = 0; 28.284 cos 45°  FAC = 0; FAC = 20.0 kN S For rod AB: sallow =
FAB ; AAB
150(106) =
28.284(103) p 2 4 dAB
Ans.
dAB = 0.0155 m = 15.5 mm For rod AC: sallow =
FAC ; AAC
150(106) =
20.0(103) p 4
2 dAC
Ans.
dAC = 0.0130 m = 13.0 mm
Ans: dAB = 15.5 mm, dAC = 13.0 mm 90
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The two aluminum rods AB and AC have diameters of 10 mm and 8 mm, respectively. Determine the largest vertical force P that can be supported. The allowable tensile stress for the aluminum is sallow = 150 MPa.
B
C
A
45
P
Solution + c ΣFy = 0; FAB sin 45°  P = 0; P = FAB sin 45°
(1)
+ ΣFx = 0; FAB cos 45°  FAC = 0 S
(2)
Assume failure of rod AB: sallow =
FAB FAB ; 150(106) = p 2 AAB (0.01) 4
FAB = 11.78 kN From Eq. (1), P = 8.33 kN Assume failure of rod AC: sallow =
FAC FAC ; 150(106) = p 2 AAC 4 (0.008)
FAC = 7.540 kN Solving Eqs. (1) and (2) yields: FAB = 10.66 kN; P = 7.54 kN Choose the smallest value Ans.
P = 7.54 kN
Ans: P = 7.54 kN 91
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*1–92. The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the largest diameter d2 of the opening, and the largest diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (sb)allow = 350 MPa and allowable shear stress is tallow = 125 MPa.
140 kN d1 B
20 mm A
10 mm
C d3 d2
Solution Allowable Shear Stress: Assume shear failure for disk C. tallow =
140 ( 103 ) V ; 125 ( 106 ) = A pd 2(0.01) Ans.
d 2 = 0.03565 m = 35.7 mm Allowable Bearing Stress: Assume bearing failure for disk C. (sb)allow =
P ; 350 ( 106 ) = A
140 ( 103 ) p 4
1 0.035652  d 23 2
Ans.
d 3 = 0.02760 m = 27.6 mm
Allowable Bearing Stress: Assume bearing failure for disk B. (sb)allow =
140 ( 103 ) P ; 350 ( 106 ) = p 2 A 4 d1 d 1 = 0.02257 m = 22.6 mm
Since d 3 = 27.6 mm 7 d 1 = 22.6 mm, disk B might fail due to shear. t =
140 ( 103 ) V = = 98.7 MPa 6 tallow = 125 MPa (O.K!) A p(0.02257)(0.02)
Therefore,
Ans.
d 1 = 22.6 mm
Ans: d 2 = 35.7 mm, d 3 = 27.6 mm, d 1 = 22.6 mm 92
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1–93. The aluminium bracket A is used to support the centrally applied load of If 40itkN. it has a thickness constant centrally applied load of 8 kip. has If a constant thickness 12 mm, determine theheight smallest height in of 0.5 in.,ofdetermine the smallest h in orderh to order to prevent failure. failure shearstress stress is prevent a sheara shear failure. The The failure shear tfail == 160 MPa. Use factorofofsafety safetyfor forshear shearofofF.S. F.S. == 2.5. Use a afactor 23 ksi.
A
h
Solution Equation of Equilibrium: + c ©Fy = 0;
V V –408 ==00
840kip kN
V –=408.00 kNkip
Allowable Shear Stress: Design of the support size tallow =
tfail V = ; F.S A
40 kN
160(106 ) 40(10 3 ) = 2.5 h(0.012)
V = 40 kN
Ans.
h = 0.05208 m = 52.1 mm
Ans: h = 52.1 mm 93
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The rods AB and CD are made of steel. Determine their smallest diameter so that they can support the dead loads shown. The beam is assumed to be pin connected at A and C. Use the LRFD method, where the resistance factor for steel in tension is f = 0.9, and the dead load factor is gD = 1.4. The failure stress is sfail = 345 MPa.
B
D 6 kN
5 kN
4 kN
A
Solution
C 2m
2m
3m
3m
Support Reactions: a+ ΣMA = 0;
FCD(10)  5(7)  6(4)  4(2) = 0 FCD = 6.70 kN
a+ ΣMC = 0;
4(8) + 6(6) + 5(3)  FAB(10) = 0 FAB = 8.30 kN
Factored Loads: FCD = 1.4(6.70) = 9.38 kN FAB = 1.4(8.30) = 11.62 kN For rod AB 0.9[345(106)] p a
d AB 2 b = 11.62(103) 2
Ans.
d AB = 0.00690 m = 6.90 mm
For rod CD 0.9[345(106)] p a
d CD 2 b = 9.38(103) 2
Ans.
d CD = 0.00620 m = 6.20 mm
Ans: d AB = 6.90 mm, d CD = 6.20 mm 94
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1–95. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the size of square bearing plates A¿ and B¿ required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. Take P = 100 kN.
40 kN/m
A 1.5 m
P
A¿
B¿ 3m
B 1.5 m
Solution Referring to the FBD of the bean, Fig. a a + ©MA = 0;
NB(3) + 40(1.5)(0.75)  100(4.5) = 0
NB = 135 kN
a + ©MB = 0;
40(1.5)(3.75)  100(1.5)  NA(3) = 0
NA = 25.0 kN
For plate A¿ , NA (sb)allow = ; A A¿
1.5(106) =
25.0(103) a2A¿ Ans.
aA¿ = 0.1291 m = 130 mm For plate B¿ , sallow =
NB ; A B¿
1.5(106) =
135(103) a2B¿ Ans.
aB¿ = 0.300 m = 300 mm
Ans: aA′ = 130 mm, aB′ = 300 mm 95
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*1–96. If the allowable bearing stress for the material under the supports at A and B is 1sb2allow = 1.5 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A¿ and B¿ have square cross sections of 150 mm * 150 mm and 250 mm * 250 mm, respectively.
40 kN/m
A 1.5 m
P
A¿
B¿ 3m
B 1.5 m
Solution Referring to the FBD of the beam, Fig. a, a + ©MA = 0;
NB(3) + 40(1.5)(0.75)  P(4.5) = 0
NB = 1.5P  15
a + ©MB = 0;
40(1.5)(3.75)  P(1.5)  NA(3) = 0
NA = 75  0.5P
For plate A¿ , NA (sb)allow = ; A A¿
1.5(106) =
(75  0.5P)(103) 0.15(0.15)
P = 82.5 kN For plate B¿ , NB ; (sb)allow = A B¿
1.5(106) =
(1.5P  15)(103) 0.25(0.25) Ans.
P = 72.5 kN (Controls!)
Ans: P = 72.5 kN (Controls!) 96
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R1–1. The circular punch B exerts a force of 2 kN on the top of the plate A. Determine the average shear stress in the plate due to this loading.
2 kN B 4 mm
A
2 mm
Solution Average Shear Stress: The shear area A = p(0.004)(0.002) = 8.00(10  6)p m2 tavg =
2(103) V = 79.6 MPa = A 8.00(10  6)p
Ans.
Ans: tavg = 79.6 MPa 97
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R1–2. C
Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC is sallow = 200 MPa and the allowable shear stress for the pins is ta llow = 70 MPa.
1.5mm in. 40
Solution
B
Referring to the FBD of member AB, Fig. a, a + ©MA = 0;
2(8)(4)  F sinsin 60°60° (8)(2.4) = 0= 0 FBC F=BC9.238 kipkN 30(2.4)(1.2) –BC FBC = 41.57
+ : ©Fx = 0;
9.238 41.57 cos 60° –AAx x= = 0 0
+ c ©Fy = 0;
9.238 2(8) + +AAyy = 0 0 41.57 sin 60° – 30(2.4)
60� 60
8 ftm 2.4
A
2 kip/ft 30 kN/m
A x = 4.619 Ax kip = 20.785 kN AA kip 36.00 kN y y==8.00
Thus, the force acting on pin A is 2 2 2 2 FA == 2Ax2A+2x A+y2 A 4.619 +.008.00 = 9.238 F = 41.57 kN kip = 2y =202 .785 + 36
Pin A is subjected to single shear, Fig. c, while pin B is subjected to double shear, Fig. b. 9.238 FBCFBC 41.57 kN kip V 41.57 kN = = = 20.785 VA ==FF 9.238 kip VBV= = 4.619 AA= = B = 2 2 2 2 For member BC FBC 41.57(10 3 ) ;= sallow = 200(106 ) = t 0.005196 m = 5.196 mm A BC (0.04)t 1 6 mm UseUse t = t =in. 4 For pin A, VA ; tallow = AA
For pin B, VB ; tallow = AB
= 70(106 )
= 70(106 )
Ans.
41.57(10 3 ) = dA 0.02750 = m 27.50 mm π d2 4 A 1 UseUse dA d=A = 1 28inmm 8 20.785(10 3 ) = dB 0.01944 = m 19.44 mm π d2 4 B 13 UseUse dB d=B = 20 inmm 16
Ans.
Ans. FBC = 41.57 kN
1.2 m
1.2 m
FA = 41.57 kN
30(2.4) kN
Ans: t = 6 mm, dA = 28 mm, dB = 20 mm 98
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R1–3. The long bolt passes through the 30mmthick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a–a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b–b.
8 mm
a 7 mm
18 mm
b
8 kN
b a 30 mm
Solution P = A
8 (103)
= 208 MPa
Ans.
(tavg)a =
8 (103) V = = 4.72 MPa A p (0.018)(0.030)
Ans.
(tavg)b =
8 (103) V = = 45.5 MPa A p (0.007)(0.008)
Ans.
ss =
p 4
(0.007)2
Ans: ss = 208 MPa, (tavg)a = 4.72 MPa, (tavg)b = 45.5 MPa 99
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*R1–4.
The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 2.0 kN/m and the column FC has a weight of 3.0 kN/m, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation. kN Given: wb := 2.0 L := 3.6m d := 1.8m m kN wc := 3.0 m a := 3.6m
H := 4.8m e := 1.2m b := 3.6m
c := 1.2m
Solution Beam AB: L c :=
+ ΣΜA=0;
+ ΣF y=0;
2
2
L +c
vb :=
c
hb :=
L
Lc Lc By⋅ ( L) − wAB⋅ ( L ) ( 0.5 ⋅ L ) = 0 L By := wb⋅ ( L) ⋅ ⎛⎜ 0.5 ⎞ ⎝ L⎠ m By = 7936.64 lb 2 s Ay := −By + wb⋅ ( L ) −Ay − By + wb⋅ ( L) = 0 By By = FBC⋅ vb FBC := vb −FBC⋅ ( h) + Ax = 0 Ax := FBC⋅ hb
( )
+
ΣF x=0;
( )
Ay = 3.6 kN FBC = 11.38 kN Ax = 10.8 kN
Segment AD:
+
ΣF x=0;
+
ΣF y=0;
+ ΣΜD=0;
ND + Ax = 0 −Ay + wb⋅ ( d) + VD = 0
ND := −Ax
ND = −10.8 kN
Ans.
VD := Ay − wb⋅ ( d)
VD = 0 kN
Ans.
MD + ⎡wb⋅ ( d)⎤ ⋅ ( 0.5 ⋅ d) − Ay⋅ ( d) = 0 ⎣ ⎦ MD := −⎡wb⋅ ( d)⎤ ⋅ ( 0.5 ⋅ d) + Ay⋅ ( d) ⎣ ⎦
100
MD = 3.24 kN⋅ m
Ans.
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* R1–4. Continued
Member CG:
Hb :=
2
H +b
2
Column FC:
+ ΣΜC=0;
H
vc :=
Fx⋅ ( H) − Ax⋅ c = 0
hc :=
Hb
b Hb
c Fx := Ax⋅ ⎛⎜ ⎞ H
⎝ ⎠
Fx = 2.7 kN
( )
( )
+
ΣF x=0;
FBC⋅ hb − Ax + Fx − FCG⋅ hc = 0
+
ΣF y=0;
FCG = 4.5 kN hc −Fy + By + wc⋅ ( H) + FBC⋅ vb + FCG⋅ vc = 0 FCG :=
( )
FBC⋅ hb − Ax + Fx
( ) ( ) Fy := By + wc⋅ ( H) + FBC⋅ ( vb) + FCG⋅ ( vc) Fy = 25.2 kN
Segment FE:
+ +
ΣF x=0;
VE − Fx = 0
VE := Fx
VE = 2.7 kN
Ans.
ΣF y=0;
NE + wc⋅ ( e) − Fy = 0
NE := −wc⋅ ( e) + Fy
NE = 21.6 kN
Ans.
−ME + Fy⋅ ( e) = 0
ME := Fy⋅ ( e)
+ ΣΜE=0;
ME = 30.24 kN⋅m
Ans.
Ans: ND = −10.8 kN, VD = 0 kN, MD = 3.24 kN ⋅ m NE = 21.6 kN, VE = 2.7 kN, ME = 30.24 kN ⋅ m 101
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R1–5. Determine the average punching shear stress the circular shaft creates in the metal plate through section AC and BD. Also, what is the average bearing stress developed on the surface of the plate under the shaft?
40 kN
50 mm A
B 10 mm
C 60 mm
Solution Average Shear and Bearing Stress: The area of the shear plane and the bearing area on the punch are AV = p(0.05)(0.01) = 0.5(10  3)p m2 and Ab = p ( 0.122  0.062 ) = 2.7(10  3)p m2. We obtain 4 tavg = sb =
D
40(103) P = = 25.5 MPa AV 0.5(10  3)p
120 mm
Ans.
40(103) P = 4.72 MPa = Ab 2.7(10  3)p
Ans.
Ans: tavg = 25.5 MPa, sb = 4.72 MPa 102
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R1–6. The 150 mm by 150 mm block of aluminum supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section a–a. Show the results on a differential volume element located on the plane.
6 kN
a
30
Solution a
Equation of Equilibrium: + QΣFx = 0;
Va  a  6 cos 60° = 0
Va  a = 3.00 kN
a + ΣFy = 0;
Na  a  6 sin 60° = 0
Na  a = 5.196 kN
150 mm
Average Normal Stress and Shear Stress: The cross sectional Area at section a–a is A = a
0.15 b(0.15) = 0.02598 m2. sin 60°
sa  a =
5.196(103) Na  a = = 200 kPa A 0.02598
Ans.
ta  a =
3.00(103) Va  a = = 115 kPa A 0.02598
Ans.
Ans: sa  a = 200 kPa, ta  a = 115 kPa 103
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R1–7. The yokeandrod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members.
5 kN
40 mm
30 mm A 25 mm
Solution
5 kN
For the 40  mm  dia. rod: s40 =
5 (103) P = p = 3.98 MPa 2 A 4 (0.04)
Ans.
For the 30  mm  dia. rod: s30 =
5 (103) V = p = 7.07 MPa 2 A 4 (0.03)
Ans.
Average shear stress for pin A: tavg =
2.5 (103) P = p = 5.09 MPa 2 A 4 (0.025)
Ans.
Ans: s40 = 3.98 MPa, s30 = 7.07 MPa, tavg = 5.09 MPa 104
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*R1–8. The cable has a specific weight g (weight/volume) and crosssectional area A. Assuming the sag s is small, so that the cable’s length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C.
A
B s C L/2
L/2
Solution Equation of Equilibrium: a+ ΣMA = 0;
Ts 
gAL L a b = 0 2 4 T =
gAL2 8s
Average Normal Stress: gAL2
gL2 T 8s s = = = A A 8s
Ans.
Ans: s = 105
gL2 8s
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2–1. An airfilled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until the ball’s diameter becomes 175 mm, determine the average normal strain in the rubber.
Solution d0 = 150 mm d = 175 mm e =
pd  pd 0 175  150 = = 0.167 mm>mm 150 pd 0
Ans.
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.
Ans: e = 0.167 mm>mm
106
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2–2. A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an outer diameter of 125 mm, determine the average normal strain in the strip.
Solution L0 = 375 mm L = p(125 mm) e =
L  L0 125p  375 = = 0.0472 mm/mm 375 L0
Ans.
Ans. e = 0.0472 mm/mm 107
CH 02.indd 73
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2–3. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain in wires CE and BD.
D
E
4m
P A
Solution ∆LCE ∆LBD = 3 7 ∆LBD =
3 (10) 7
B
3m
C
2m
2m
= 4.286 mm
PCE =
∆LCE 10 = = 0.00250 mm>mm L 4000
Ans.
PBD =
∆LBD 4.286 = = 0.00107 mm>mm L 4000
Ans.
Ans: PCE = 0.00250 mm>mm, PBD = 0.00107 mm>mm 108
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*2–4. The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 2°. Determine the average normal strain in each wire. The wires are unstretched when the lever is in the horizontal position.
G 200 mm
F 200 mm 300 mm
300 mm
E
B
A
C
200 mm
D
H
Solution
2° bp rad = 0.03491 rad. 180 Since u is small, the displacements of points A, C, and D can be approximated by Geometry: The lever arm rotates through an angle of u = a dA = 200(0.03491) = 6.9813 mm dC = 300(0.03491) = 10.4720 mm dD = 500(0.03491) = 17.4533 mm Average Normal Strain: The unstretched length of wires AH, CG, and DF are LAH = 200 mm, LCG = 300 mm, and LDF = 300 mm. We obtain (Pavg)AH = (Pavg)CG = (Pavg)DF =
dA 6.9813 = = 0.0349 mm>mm LAH 200
Ans.
dC 10.4720 = = 0.0349 mm > mm LCG 300
Ans.
dD 17.4533 = = 0.0582 mm >mm LDF 300
Ans.
109
Ans: (Pavg)AH = 0.0349 mm>mm (Pavg)CG = 0.0349 mm > mm (Pavg)DF = 0.0582 mm >mm
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P
The pinconnected rigid rods AB and BC are inclined at u = 30° when they are unloaded. When the force P is applied u becomes 30.2°. Determine the average normal strain in wire AC.
B
u
Solution
A
Geometry: Referring to Fig. a, the unstretched and stretched lengths of wire AD are
u
600 mm
C
LAC = 2(600 sin 30°) = 600 mm LAC ′ = 2(600 sin 30.2°) = 603.6239 mm Average Normal Strain: (Pavg)AC =
LAC ′  LAC 603.6239  600 = = 6.04(10  3) mm>mm LAC 600
Ans.
Ans: (Pavg)AC = 6.04(10  3) mm>mm 110
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2–6. The wire AB is unstretched when u = 45°. If a load is applied to the bar AC, which causes u to become 47°, determine the normal strain in the wire.
B
u
L
Solution 2
2
L = L +
2 L′AB
C
 2LL′AB cos 43°
A L
L′AB = 2L cos 43° PAB = =
L′AB  LAB LAB 2L cos 43°  22L
= 0.0343
22L
Ans.
Ans: PAB = 0.0343 111
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If a horizontal load applied to the bar AC causes point A to be displaced to the right by an amount ∆L, determine the normal strain in the wire AB. Originally, u = 45°.
B
u
L
Solution
C
L′AB = 4( 22L ) + ∆L  2 ( 22L ) (∆L) cos 135° 2
2
A L
= 22L2 + ∆L2 + 2L∆L
PAB = =
=
L′AB  LAB LAB
22L2 + ∆L2 + 2L∆L  22L 22L
C
1 +
∆L2 ∆L +  1 L 2L2
Neglecting the higher  order terms, 1
PAB
∆L 2 = a1 + b  1 L = 1 + =
1 ∆L + .....  1 2 L
(binomial theorem)
0.5∆L L
Ans.
Also, PAB =
∆L sin 45° 22L
=
0.5 ∆L L
Ans.
Ans: PAB = 112
0.5∆L L
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* y
The rectangular plate is subjected to the deformation shown by the dashed line. Determine the average shear strain gxy in the plate.
150 mm B
3 mm
200 mm
Solution x
A
Geometry: u′ = tan1 u = a
3 mm
3 = 0.0200 rad 150
p + 0.0200b rad 2
Shear Strain:
gxy =
p p p  u =  a + 0.0200b 2 2 2
Ans.
=  0.0200 rad
Ans: gxy =  0.0200 rad 113
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y
The square deforms into the position shown by the dashed lines. Determine the shear strain at each of its corners, A, B, C, and D, relative to the x, y axes. Side D′B′ remains horizontal.
D¿
B¿
3 mm B
D
53 mm
50 mm 91.5
Solution
C A
Geometry:
x
50 mm
B′C′ = 2(8 + 3)2 + (53 sin 88.5°)2 = 54.1117 mm 2
C¿ 8 mm
2
C′D′ = 253 + 58  2(53)(58) cos 91.5° = 79.5860 mm
B′D′ = 50 + 53 sin 1.5°  3 = 48.3874 mm cos u =
(B′D′)2 + (B′C′)2  (C′D′)2 2(B′D′)(B′C′) 2
=
48.3874 + 54.11172  79.58602 =  0.20328 2(48.3874)(54.1117)
u = 101.73° b = 180°  u = 78.27° Shear Strain: (gA)xy = (gB)xy =
p 91.5° b =  0.0262 rad  pa 2 180°
Ans.
p p 101.73° b =  0.205 rad  u =  pa 2 2 180°
(gC)xy = b (gD)xy = p a
Ans.
p 78.27° p b = pa =  0.205 rad 2 180° 2
Ans.
88.5° p b =  0.0262 rad 180° 2
Ans.
Ans: (gA)xy (gB)xy (gC)xy (gD)xy 114
= = = =
0.0262 rad 0.205 rad 0.205 rad 0.0262 rad
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Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB. If a force is applied to the end B of the member and causes it to rotate by u = 0.5°, determine the normal strain in the cable. Originally the cable is unstretched.
u P
B
800 mm
Solution Geometry: Referring to the geometry shown in Fig. a, the unstretched and stretched lengths of cable AB are LAB = 26002 + 8002 = 1000 mm
A
C
600 mm
LAB′ = 26002 + 8002  2(600)(800) cos 90.5° = 1004.18 mm
Average Normal Strain: PAB =
LAB′  LAB 1004.18  1000 = = 0.00418 mm>mm LAB 1000
Ans.
Ans: PAB = 0.00418 mm>mm 115
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Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB. If a force is applied to the end B of the member and causes a normal strain in the cable of 0.004 mm>mm, determine the displacement of point B. Originally the cable is unstretched.
u P
B
800 mm
Solution Geometry: Referring to the geometry shown in Fig. a, the unstretched and stretched lengths of cable AB are LAB = 26002 + 8002 = 1000 mm
A
C
600 mm
LAB′ = 26002 + 8002  2(600)(800) cos (90° + u) LAB′ = 21(106)  0.960(106) cos (90° + u)
Average Normal Strain: PAB =
Thus,
21(106)  0.960(106) cos (90° + u)  1000 LAB′  LAB ; 0.004 = LAB 1000 u = 0.4784° a
p b = 0.008350 rad 180°
∆ B = uLBC = 0.008350(800) = 6.68 mm
Ans.
Ans: ∆ B = 6.68 mm 116
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*2–12. y
Determine the shear strain gxy at corners A and B if the plastic distorts as shown by the dashed lines.
12 mm 4 mm 3 mm
B
8 mm
C 300 mm
Solution
D
Geometry: Referring to the geometry shown in Fig. a, the smallangle analysis gives 7 a = c = = 0.022876 rad 306 5 b = = 0.012255 rad 408 2 u = = 0.0049383 rad 405
400 mm
A
2 mm
x
5 mm
Shear Strain: By definition, (gA)xy = u + c = 0.02781 rad = 27.8(103) rad
Ans.
(gB)xy = a + b = 0.03513 rad = 35.1(103) rad
Ans.
Ans: (gA)xy = 27.8(103) rad (gB)xy = 35.1(103) rad 117
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2–13. y
Determine the shear strain gxy at corners D and C if the plastic distorts as shown by the dashed lines.
12 mm 4 mm 3 mm
B
8 mm
C 300 mm
Solution
D
Geometry: Referring to the geometry shown in Fig. a, the smallangle analysis gives 4 a = c = = 0.013201 rad 303 2 u = = 0.0049383 rad 405 5 b = = 0.012255 rad 408
400 mm
A
2 mm
x
5 mm
Shear Strain: By definition (gxy)C = a + b = 0.02546 rad = 25.5(103) rad
Ans.
(gxy)D = u + c = 0.01814 rad = 18.1(103) rad
Ans.
Ans: (gxy)C = 25.5(103) rad (gxy)D = 18.1(103) rad 118
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2–14. y
The material distorts into the dashed position shown. Determine the average normal strains Px, Py and the shear strain gxy at A, and the average normal strain along line BE.
15 mm C
30 mm D
50 mm B 200 mm E
Solution
A
150 mm
50 mm F
x
Geometry: Referring to the geometry shown in Fig. a, tan u =
15 p ; u = (3.4336°) a rad b = 0.05993 rad 250 180°
LAC ′ = 2152 + 1502 = 262725 mm
EE′ 50 BB′ 200 = ; BB′ = 12 mm = ; EE′ = 6 mm 15 250 30 250
x′ = 150 + EE′  BB′ = 150 + 6  12 = 144 mm LBE = 21502 + 1502 = 15022 mm LB ′E ′ = 21442 + 1502 = 243236 mm
Average Normal and Shear Strain: Since no deformation occurs along x axis,
Ans.
(Px)A = 0 (Py)A =
LAC′  LAC 262725  250 = = 1.80(103) mm>mm LAC 250
Ans.
By definition, Ans.
(gxy)A = u = 0.0599 rad PBE =
LB′E′  LBE 243236  15022 = =  0.0198 mm>mm LBE 15022
Ans.
Ans: (Px)A = 0 (Py)A = 1.80(103) mm>mm (gxy)A = 0.0599 rad PBE = 0.0198 mm>mm 119
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y
The material distorts into the dashed position shown. Determine the average normal strains along the diagonals AD and CF.
15 mm C
30 mm D
50 mm B 200 mm E
Solution
A
150 mm
50 mm F
x
Geometry: Referring to the geometry shown in Fig. a, LAD = LCF = 21502 + 2502 = 285000 mm
LAD′ = 2(150 + 30)2 + 2502 = 294900 mm LC′F = 2(150  15)2 + 2502 = 280725 mm
Average Normal Strain: PAD = PCF =
LAD′  LAD 294900  285000 = = 0.0566 mm>mm LAD 285000
LC′F  LCF 280725  285000 = = 0.0255 mm>mm LCF 285000
Ans. Ans.
Ans: PAD = 0.0566 mm>mm PCF = 0.0255 mm>mm 120
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*2–16. The nylon cord has an original length L and is tied to a bolt at A and a roller at B. If a force P is applied to the roller, determine the normal strain in the cord when the roller is at C, and at D. If the cord is originally unstrained when it is at C, determine the normal strain P′D when the roller moves to D. Show that if the displacements ∆ C and ∆ D are small, then P′D = PD − PC .
D C P
B
C
L
Solution LC = 2L2 + ∆ 2C
=
A
2L2 + ∆ 2C  L L
PC =
L 41 +
For small ∆ C, PC = 1 +
2 C 2
1 ∆L 2
 L
L
=
C
1 + a
∆ 2C L2
b  1
1 ∆ 2C 1 ∆ 2C a 2b  1 = 2 L 2 L2
Ans.
In the same manner, PD =
1 ∆ 2D 2 L2
PD ′ =
2L2 + ∆ 2D  2L2 + ∆ 2C
Ans.
2L2 + ∆ 2C
=
41 +
For small ∆ C and ∆ D, PD ′ =
PD ′ =
11
+
2 1 ∆C 2 L2
11
∆ 2C  ∆ 2D 2
2L 
2
∆ 2C

11 ∆ 2C 2
+
2
2 1 ∆D 2 L2
2
=
1 2L2 1 2L2
∆ 2D L2
 41 +
41 +
∆ 2C L2
∆ 2C L2
( ∆ 2C  ∆ 2D ) ( 2L2 + ∆ 2C )
+
1 2 L
=
1 ( ∆ 2C  ∆ 2D ) = PC  PD 2L2
QED
Also this problem can be solved as follows: AC = L sec uC ; AD = L sec uD PC =
L sec uC  L = sec uC  1 L
PD =
L sec uD  L = sec uD  1 L
Expanding sec u sec u = 1 +
u2 5 u4 + ...... 2! 4!
121
D
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*2–16. Continued
For small u neglect the higher order terms sec u = 1 +
u2 2
Hence, PC = 1 +
u C2 u C2  1 = 2 2
PD = 1 +
u D2 u D2  1 = 2 2
PD ′ =
L sec uD  L sec uC sec uD =  1 = sec uD cos uC  1 L sec uC sec uC
Since cos u = 1 
u2 u4 + ...... 2! 4!
sec uD cos uC = a1 + = 1 
u C2 u D2 ......ba1 ......b 2 2
u 2C u 2C u 2D u 2D + 2 2 4
Neglecting the higher order terms sec uD cos uC = 1 + PD ′ = c 1 +
u 2C u 2D 2 2
u C2 u 22 u 12 u D2 d  1 = 2 2 2 2
QED
= PD  PC
Ans: 1 2 1 = 2
PC = PD 122
∆ 2C L2 ∆ 2D L2
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2–17. P
A thin wire, lying along the x axis, is strained such that each point on the wire is displaced ∆x = kx2 along the x axis. If k is constant, what is the normal strain at any point P along the wire?
x
x
Solution P =
d(∆x) dx
Ans.
= 2 k x
Ans: P = 2kx 123
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2–18. y
The block is deformed into the position shown by the dashed lines. Determine the average normal strain along line AB.
15 mm
30 mm 70 mm 30 mm 55 mm B¿ B
100 mm
110 mm
Solution Geometry:
x
A
AB = 21002 + (70  30)2 = 107.7033 mm
30 mm
70 mm
AB′ = 3(70  30  15)2 + ( 1102  152 ) = 111.8034 mm
Average Normal Strain: PAB = =
AB′  AB AB 111.8034  107.7033 107.7033
= 0.0381 mm>mm = 38.1 ( 103 ) mm
Ans.
Ans: PAB = 38.1 ( 103 ) mm 124
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2–19. Nylon strips are fused to glass plates. When moderately heated the nylon will become soft while the glass stays approximately rigid. Determine the average shear strain in the nylon due to the load P when the assembly deforms as indicated.
y 2 mm P
3 mm 5 mm 3 mm 5 mm 3 mm
x
Solution 1 mm
From the geometry shown in Fig. a 1 θ = tan 1 a b = 11.31° = 0.1974 rad 5 The shear strain is
π
π g =− + θ = −0.1974 πad = − 0.197 πad 2 2
Ans.
5 mm
θ
(a)
Ans: g = −0.197 rad 125
CH 02.indd 75
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*2–20.
u � 2�
The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt u = 2°. Determine the approximate normal strain in the wire when the frame is in this position. Assume the columns are rigid and rotate about their lower supports.
u � 2�
B
3m
1m
A
4m
Solution Geometry: The vertical displacement is negligible xA = (1) ¢
2° ≤ p = 0.03491 m 180°
xB = (4) ¢
2° ≤ p = 0.13963 m 180°
x = 4 + xB  xA = 4.10472 m Ans.
A¿B¿ = 232 + 4.104722 = 5.08416 m AB = 232 + 42 = 5.00 m Average Normal Strain: eAB = =
A¿B¿  AB AB 5.08416  5 = 16.8 A 10  3 B m>m 5
Ans: A¿B¿ = 5.08416 m 126
CH 02.indd 86
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2–21. y
The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal AC, and the average shear strain at corner A relative to the x, y axes.
6 mm 400 mm 2 mm
2 mm
6 mm C
D
300 mm
Solution
2 mm
Geometry: The unstretched length of diagonal AC is
A
LAC = 23002 + 4002 = 500 mm
400 mm
x
B 3 mm
Referring to Fig. a, the stretched length of diagonal AC is LAC′ = 2(400 + 6)2 + (300 + 6)2 = 508.4014 mm
Referring to Fig. a and using small angle analysis, f =
2 = 0.006623 rad 300 + 2
a =
2 = 0.004963 rad 400 + 3
Average Normal Strain: Applying Eq. 2, (Pavg)AC =
LAC′  LAC 508.4014  500 = = 0.0168 mm>mm LAC 500
Ans.
Shear Strain: Referring to Fig. a, Ans.
(gA)xy = f + a = 0.006623 + 0.004963 = 0.0116 rad
Ans: (Pavg)AC = 0.0168 mm>mm, (gA)xy = 0.0116 rad 127
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2–22.
y
The corners B and D of the square plate are given the displacements indicated. Determine the shear strains at A and B.
A
16 mm D
B
x
3 mm 3 mm 16 mm
16 mm
Solution
C
16 mm
Applying trigonometry to Fig. a f = tan  1 a a = tan  1 a
13 p rad b = 39.09° a b = 0.6823 rad 16 180°
16 p rad b = 50.91° a b = 0.8885 rad 13 180°
By the definition of shear strain,
A gxy B A = A gxy B B =
p p  2f =  2(0.6823) = 0.206 rad 2 2
Ans.
p p  2a =  2(0.8885) =  0.206 rad 2 2
Ans.
Ans:
Agxy BA = 0.206 rad A gxy B B = 0.206 rad 128
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2–23. y
Determine the shear strain gxy at corners A and B if the plate distorts as shown by the dashed lines.
5 mm
2 mm 2 mm
B
C
4 mm
300 mm
D
Solution
400 mm
A
2 mm
x
3 mm
Geometry: For small angles, a = c =
2 = 0.00662252 rad 302
b = u =
2 = 0.00496278 rad 403
Shear Strain: (gB)xy = a + b = 0.0116 rad = 11.6 1 10  3 2 rad
Ans.
(gA)xy = u + c
= 0.0116 rad = 11.6 1 10  3 2 rad
Ans.
Ans: (gB)xy = 11.6(10  3) rad, (gA)xy = 11.6(10  3) rad 129
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*2–24. y
Determine the shear strain gxy at corners D and C if the plate distorts as shown by the dashed lines.
5 mm
2 mm 2 mm
B
C
4 mm
300 mm
D
Solution
400 mm
A
2 mm
x
3 mm
Geometry: For small angles, a = c =
2 = 0.00496278 rad 403
b = u =
2 = 0.00662252 rad 302
Shear Strain: (gC)xy = a + b = 0.0116 rad = 11.6 1 10  3 2 rad
Ans.
(gD)xy = u + c
= 0.0116 rad = 11.6 1 10  3 2 rad
Ans.
Ans: (gC)xy = 11.6(10  3) rad, (gD)xy = 11.6(10  3) rad 130
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2–25. y
Determine the average normal strain that occurs along the diagonals AC and DB.
5 mm
2 mm 2 mm
B
C
4 mm
300 mm
D
Solution
400 mm
A
2 mm
x
3 mm
Geometry: AC = DB = 24002 + 3002 = 500 mm
DB′ = 24052 + 3042 = 506.4 mm
A′C′ = 24012 + 3002 = 500.8 mm
Average Normal Strain: PAC =
A′C′  AC 500.8  500 = AC 500
= 0.00160 mm>mm = 1.60 1 10  3 2 mm>mm
Ans.
= 0.0128 mm>mm = 12.8 1 10  3 2 mm>mm
Ans.
PDB =
DB′  DB 506.4  500 = DB 500
Ans: PAC = 1.60 1 10  3 2 mm>mm 131
PDB = 12.8 1 10  3 2 mm>mm
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2–26. If the unstreched length of the bowstring is 887.5 mm, determine the average normal strain in the string when it is streched to the position shown.
18 mm in. 450
6 in. 150 mm 18 mm in. 450
Geometry: Referring to Fig. a, the stretched length of the string is 4502 + 150 948.68in. mm 62 2= =37.947 L = 2L¿ = 2 218 Average Normal Strain: eavg =
L  L0 37.947 –887.5 35.5 948.68 == 0.0689 0.0689mm/mm in.>in. = L0 35.5 887.5
Ans.
450 mm
450 mm
150 mm
Ans: eavg = 0.0689 mm/mm 132
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2–27. y
The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the shear strain, gxy, at A.
45
45
x¿
A¿ 5 mm
800 mm 2
L = 2800 + 5  2(800)(5) cos 135° = 803.54 mm sin 135° sin u = ; 803.54 800
gxy =
A
45
Solution 2
800 mm
x
u = 44.75° = 0.7810 rad
p p  2u =  2(0.7810) 2 2 Ans.
= 0.00880 rad
Ans: gxy = 0.00880 rad 133
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*2–28. y
The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px along the x axis.
45
45
x¿
A¿ 5 mm
800 mm 2
L = 2800 + 5  2(800)(5) cos 135° = 803.54 mm Px =
A
45
Solution 2
800 mm
803.54  800 = 0.00443 mm>mm 800
Ans.
x
Ans: Px = 0.00443 mm>mm 134
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2–29. y
The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Determine the average normal strain Px′ along the x′ axis.
45
800 mm
45
x¿
A
45
Solution
A¿ 5 mm
800 mm
L = 800 cos 45° = 565.69 mm Px′ =
5 = 0.00884 mm>mm 565.69
Ans.
x
Ans: Px′ = 0.00884 mm>mm 135
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2–30.
y y� x2
The rubber band AB has an unstretched length of 1 m. If it is fixed at B and attached to the surface at point A¿, determine the average normal strain in the band.The surface is defined by the function y = (x2) m, where x is in meters.
A¿ ft 1m B
Solution
ft 1m
x
A
Geometry: L =
L0
1m ft
A
1 + a
dy 2 b dx dx
However y = x2 then L = =
L0
dy = 2x dx
1m ft
21 + 4 x2 dx
1 1m ft C 2x21 + 4 x2 + ln A 2x + 21 + 4x2 B D �0 4
= 1.47894 ft m
Average Normal Strain: L  L0 1.47894  1 = = 0.479 m/m ft>ft eavg = L0 1
Ans.
Ans: eavg = 0.479 m/m 136
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2–31. y
The rectangular plate is deformed into the shape shown by the dashed lines. Determine the average normal strain along diagonal BD, and the average shear strain at corner B relative to the x, y axes.
6 mm 400 mm 2 mm
2 mm
6 mm C
D
300 mm
Solution
2 mm
Geometry: The unstretched length of diagonal BD is
A
LBD = 23002 + 4002 = 500 mm
400 mm
x
B 3 mm
Referring to Fig. a, the stretched length of diagonal BD is LB′D′ = 2(300 + 2  2)2 + (400 + 3  2)2 = 500.8004 mm
Referring to Fig. a and using small angle analysis, f =
2 = 0.004963 rad 403
a =
3 = 0.009868 rad 300 + 6  2
Average Normal Strain: Applying Eq. 2, (Pavg)BD =
LB′D′  LBD 500.8004  500 = = 1.60(10  3) mm>mm Ans. LBD 500
Shear Strain: Referring to Fig. a, (gB)xy = f + a = 0.004963 + 0.009868 = 0.0148 rad
Ans.
Ans: (Pavg)BD = 1.60(10  3) mm>mm, (gB)xy = 0.0148 rad 137
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*2–32. The nonuniform loading causes a normal strain in the shaft p that can be expressed as Px = k sin a xb , where k is a L constant. Determine the displacement of the center C and the average normal strain in the entire rod.
C A
B L — 2
L — 2
Solution Px = k sin a (∆x)C =
L0
p xb L L>2
Px dx =
L0
L>2
k sin a
p xb dx L
L>2 p L L p = k a b cos a xb ` =  k a bacos  cos 0b p p L 2 0
= (∆x)B =
kL p L0
Ans.
L
k sin a
p xbdx L
L L p L 2kL =  k a b cos a xb ` =  k a b(cos p  cos 0) = p p p L 0
Pavg =
(∆x)B L
=
2k p
Ans.
Ans:
kL p 2k = p
(∆x)C = Pavg 138
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2–33. y
The fiber AB has a length L and orientation u. If its ends A and B undergo very small displacements u A and vB respectively, determine the normal strain in the fiber when it is in position A′ B′.
B¿ vB B L A
uA A¿
u x
Solution Geometry: LA′B′ = 2(L cos u  u A)2 + (L sin u + vB)2
= 2L2 + u 2A + v2B + 2L(vB sin u  u A cos u)
Average Normal Strain: PAB =
=
LA′B′  L L
C
1 +
u 2A + v2B 2
L
+
2(vB sin u  u A cos u) L
 1
Neglecting higher terms u 2A and v2B PAB = J1 +
2(vB sin u  u A cos u) L
1 2
R  1
Using the binomial theorem: PAB = 1 + =
2u A cos u 1 2vB sin u a b + c 1 2 L L
vB sin u u Acos u L L
Ans.
Ans: PAB = 139
vB sin u u Acos u L L
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2–34. If the normal strain is defined in reference to the final length ∆s′, that is, P= = lim a ∆s′ S 0
∆s′  ∆s b ∆s′
instead of in reference to the original length, Eq. 2–2, show that the difference in these strains is represented as a secondorder term, namely, P  P= = P P′.
Solution P =
∆s′  ∆s ∆s
P  P= =
∆s′  ∆s ∆s′  ∆s ∆s ∆s′
=
∆s′2  ∆s∆s′  ∆s′∆s + ∆s2 ∆s∆s′
=
∆s′2 + ∆s2  2∆s′∆s ∆s∆s′
=
(∆s′  ∆s)2 ∆s∆s′
= ¢
∆s′  ∆s ∆s′  ∆s ≤¢ ≤ ∆s ∆s′
= P P=
(Q.E.D)
Ans: N/A 140
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3–1. A tension test was performed on a steel specimen having an original diameter of 12.5 mm and gauge length of 50 mm. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 25 mm = 140 MPa and 25 mm= 0.05 mm >mm. Redraw the elastic region, using the same stress scale but a strain scale of 25 mm = 0.001 mm >mm.
= A
Load (kN) Elongation (mm) 0 7.0 21.0 36.0 50.0 53.0 53.0 54.0 75.0 90.0 97.0 87.8 83.3
1 = π (0.01252 ) 122.72(10 −6 ) m 2 4
L = 50 mm. s(M Pa)
e(mm> mm)
0
0
57.07
0.00025
171.21
0.00075
293.51
0.00125
407.66
0.00175
432.12
0.0025
432.12
0.0040
440.27
0.010
611.49
0.020
733.79
0.050
790.86
0.140
715.85
0.200
679.16 Eapprox =
0 0.0125 0.0375 0.0625 0.0875 0.125 0.2 0.5 1.0 2.5 7.0 10.0 11.5
0.230 6
336(10 ) 9 ) N/m 2 224 GPa = 224(10= 0.0015
Ans.
n sio ver is y r ina M). It ry lim S a I e ( n r i l p nua elim t a me sen ns Ma his pr e so s. e r r p t o a i e t r a ut tion e ons s' Sol olu ll be hat her i s t t t u r y wi sol SM cto kel raft se tru ven li the I he d anual The e Ins t e m g s th and parin rs in thi of nd re ro ible r a p s e s ed po of and rrect ge co sta sions e b is om e will . s d e e Th blish u p e r
Ans: (sult)approx = 770 MPa, (sR)approx = 652 MPa, (sY)approx = 385 MPa, Eapprox = 224 GPa 141
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3–2. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience.
Solution
00 232.4 33.2 45.5 318.5 49.4 345.8 51.5 360.5 53.4 373.8
0 0 0.0006 0.0006 0.0010 0.0010 0.0014 0.0014 0.0018 0.0018 0.0022
0.0022
420
232.4(106 ) − 0 9 = ) N/m 2 387 GPa = 387.33(10 0.0006 − 0
Ans. 350
Modulus of Resilience: The modulus of resilience is equal to the area under the linear portion of the stress–strain diagram (shown shaded). = (U i )r
eP (mm/mm) (in./in.)
σ (MPa)
Modulus of Elasticity: From the stress–strain diagram E =
S(MPa) (ksi)
1 N m 3 N⋅m 232.4(106 ) = ) 69.7 kJ/m 3 = 0.0006 69.72(10 2 m m 2 m3
Ans.
280 232.4 210
140
70
e (mm/mm)
Ans: E = 387 GPa , ur = 69.7 kJ/m 3 142
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3–3. Data taken from a stress–strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is sr = 53.4 373.8ksi. MPa
(MPa) S (ksi) 00 232.4 33.2 45.5 318.5 49.4 345.8 51.5 360.5 53.4 373.8
Solution Modulus of Toughness: The modulus of toughness is equal to the area under the stress–strain diagram (shown shaded).
= (U i )t
1 N m 232.4(106 ) 2 (0.0004 + 0.0010) 2 m m
s (MPa) 420
350 318.5 280
N m 1 55.3(106 ) 2 0.0012 2 m m +
0 0 0.0006 0.0006 0.0010 0.0010 0.0014 0.0014 0.0018 0.0018 0.0022 0.0022
373.8
m N + 318.5(106 ) 2 0.0012 m m +
e P(mm/mm) (in./in.)
232.4 210
N m 1 86.1(106 ) 2 0.0004 2 m m
140
3 N⋅m = 595.28(10 = ) 595 kJ/m 3 m3
Ans.
70
e (mm/mm)
Ans: (ui)t 595 kJ/m3 143
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*3–4. The stress–strain diagram for a metal alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.
Solution From the stress–strain diagram, Fig. a, = E
283(106 ) 9 = 283(10= ) N/m 2 283 GPa 0.001
Ans.
= σ Y 283 = MPa σ σ /t 547 GPa
Thus, 6 π 2 = PY σ= = ) 32.01(10 3= ) N 32.0 kN Y A 283(10 ) 4 (0.012
Ans.
6 π 2 3 = = ) 61.86(10= ) N 61.9 kN Pσ /t σ= σ /t A 547(10 ) 4 (0.012
Ans.
Ans: E 283 GPa, PY 32.0 kN, Pult 61.9 kN 144
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3–5. The stress–strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. If the specimen is loaded until it is stressed to 500 MPa, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded.
Solution From the stress–strain diagram Fig. a, the modulus of elasticity for the steel alloy is = E
283(106 ) 9 = 283(10= ) N/m 2 283 GPa 0.001
when the specimen is unloaded, its normal strain recovered along line AB, Fig. a, which has a gradient of E. Thus
Elasticity Recovery =
500(106 ) = = 0.001767 mm/mm E 283(109 )
σ
Amount of Elastic Recovery = (0.001767 mm/mm)(50 mm) = 0.08834 mm = 0.0883 mm
Ans.
PP = 0.05 Thus, the permanent set0.0 is eP = 0.08 – 0.001767 = 0.07823 mm/mm = 0.047(2) = PPL Then, the ¢L increase in gauge length= is0
Ans.
Ans: Elastic Recovery 0.0883 mm 욼L 3.91 mm 145
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36. The stress–strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material.
Solution The Modulus of resilience is equal to the area under the stress–strain diagram up to the proportional limit. sPL = 283 MPa
PPL = 0.001 m/m
Thus, (Ui)r =
1 N m 1 3 N⋅m sPLPPL = 283(106 ) = 0.001 141.5(10 ) 2 2 m 2 m m3 = 141.5 kJ/m 3
Ans.
The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number is 32. Thus, N m 6 N⋅m (U i ) = 32 100(106 ) = ) 128 MJ/m 3 = 0.04 128(10 ult aprox m m 2 m3
Ans.
Ans: (ui)r = 141.5 kJ>m3, [(Ui)ult]approx = 128 MJ>m3 146
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3–7. A specimen isisoriginally long, hashas a diameter of originally300 1 ftmm long, a diameter 0.5 12 in.,mm, and isand subjected to a force 500 lb. theWhen force of is subjected to aofforce ofWhen 2.5 kN. is increased from 500 lb to 1800 lb, the the force is increased from 2.5 kN to 9specimen kN,, the elongates specimen 0.009 in. 0.225 Determine the modulus of elasticity for the elongates mm. Determine the modulus of elasticity for material if it if remains linear elastic. the material it remains linear elastic.
Solution Normal Stress and Strain: Applying s =
= σ1 = σ2 = ∆ε
dL P and e = . A L
2.5(10 3 ) 6 = 22.10(10 = ) N/m 2 22.10 MPa π (0.012 2 ) 4 9(10 3 ) 6 = 79.58(10 = ) N/m 2 79.58 MPa 2 (0.012 ) 4
π
0.225 = 0.000750 mm/mm 300
Modulus of Elasticity: E =
∆σ (79.58 − 22.10)(106 ) Pa 76.6 GPa = = 76.63(109 ) = 0.000750 ∆ε
Ans.
Ans: E 76.6 GPa 147
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*3–8. The strut is supported by a pin at C and an A36 AB. If the the wire wire has has aa diameter diameter of of 0.2 5 mm, steel guy wire AB. in., determine how much it stretches when the distributed load acts on the strut.
A
60� 60 200 lb/ft 3.4 kN/m
Solution C
Here, we are only interested in determining the force in wire AB. a + ©MC = 0;
11 FAB cos 60°(2.7) 60°(9) – (3.4)(2.7)(0.9) 600kN lb (200)(9)(3) = 0= 0 F FAB AB cos AB ==3.06 22
B 9 ftm 2.7
The normal stress the wire is FAB 3.06(10 3 ) 6 155.84(10 = ) N/m 2 155.84 MPa = π sAB = = 2 AAB (0.005 ) 4 250ksi MPa, Hooke’s Lawcan canbe beapplied applied to determine , Hooke’s Law determine the thestrain strain Since sAB 6 sY = 36 in wire. sAB = EPAB;
155.84(106) = 200(109)PAB 0.7792(10–3  3) mm/mm AB = PeAB = 0.6586(10 ) in>in
9(12)3) 2.7(10 The unstretched length of the wire is LAB = = 3117.69 mm. Thus, the wire sin sin 60° 60° stretches 3 0.7792(10–3 )(3117.69) )(124.71) dAB = PAB LAB = 0.6586(10
= 2.429 mm = 2.43 mm
Ans.
1 (3.4)(2.7) kN 2
0.9 m
1.8 m
Ans: dAB = 2.43 mm 148
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3–9. The s P diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience.
s (MPa) 0.385
0.077 1
0.385
0.077 = 0.0385 MPa 2
(U i )r =
Ans.
1 0.077(106 ) N/m 2 (2 m/m) 2 3
3
= 77.0(10 ) N ⋅ m/= m 77.0 kJ/m
(U i )t = +
P (mm/mm)
s (MPa)
Solution E =
2 2.25
0.077 3
1
Ans.
2 2.25
P (mm/mm)
1 0.077(106 ) N/m 2 (2 m/m) 2
{
}
1 0.077(106 ) + 0.385(106 ) N/m 2 [(2.25 − 2) m/m ] 2
3 = 134.75 N ⋅ m/m= 135 kJ/m 3
Ans.
Ans: E = 0.0385 MPa, (ui)r = 77.0 kJ>m3, (ui)t = 135 kJ>m3 149
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3–10. A structural member in a nuclear reactor is made of a zirconium alloy. If an axial load of 20 kN is to be supported by the member, determine its required crosssectional area. Use a factor of safety of 3 relative to yielding. What is the load on the member if it is 1 m long and its elongation is 0.5 mm? Ezr = 100 GPa, sY = 400 MPa. The material has elastic behavior.
Solution Allowable Normal Stress: F.S. = 3 =
sy sallow 57.5 400 sallow
sallow = 19.17 133.33ksi MPa sallow = 19.17 5 = 133.33
P A 3 4 20(10 ) AA
2 2 A == 150 0.2087 = 0.209 in2 mmin
Ans.
Stress–Strain Relationship: Applying Hooke’s law with e =
0.02 d 0.5 == 0.0005 mm/mm = 0.000555 in.>in. 3 L 3 (12) 1(10 ) 33 )(0.0005) = 50 MPa s == Ee Ee==100(10 14 A 10 = 7.778 ksi B (0.000555)
Normal Force: Applying equation s =
P . A
P ==sA sA= =50(150) 7.778 (0.2087) kip P = 7500 N== 1.62 7.5 kN
Ans.
Ans: A 150 mm 2, P 7.5 kN 150
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3–11.
s (ksi) (MPa)
A tension test was performed on an aluminum 2014T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the proportional limit,(b) the modulus of elasticity, and (c) the yield strength based on a 0.2% strain offset method.
70 490 60 420 50 350 40 280 30 210 20 140 10 70 00
Solution
0.02 0.02 0.002 0.002
0.04 0.04 0.004 0.004
0.06 0.06 0.006 0.006
0.08 0.08 0.008 0.008
0.10 0.10 0.010 0.010
(in./in.) PP (mm/mm)
Proportional Limit and Yield Strength: From the stress–strain diagram, Fig. a, 44 ksi spl ==308 MPa
Ans.
60 ksi sYY ==420 MPa
Ans.
Modulus of Elasticity: From the stress–strain diagram, the corresponding strain for is is Thus, sPL == 308 44 ksi eplepl= =0.004 MPa 0.004in.>in. mm/mm. Thus, E5 =
44 – 00 308 33 5=77.0(10 ) )MPa 11.0(10 ksi = 77.0 GPa 0.004 –00 0.004
Ans.
Ans: spl = 308 MPa sY = 420 MPa E = 77.0 GPa 1 51
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*3–12.
(ksi) ss(MPa)
A tension test was performed on an aluminum 2014T6 alloy specimen. The resulting stress–strain diagram is shown in the figure. Estimate (a) the modulus of resilience; and (b) modulus of toughness.
70 490 60 420 50 350 40 280 30 210 20 140 10 70 0
Solution
0.02 0.002
0.04 0.004
0.06 0.006
0.08 0.008
0.10 0.010
(in./in.) PP(mm/mm)
Modulus of Resilence The modulus of resilence is equal to the area under the stress–strain diagram up to the propotional limit. From the stress–strain diagram, spl ==308 44 ksi MPa
e pl == 0.004 0.004mm/mm in.>in.
Thus, 1 2
1 2
A Ui B r = splepl = [308(106)](0.004) = 0.616 MJ/m3
Ans.
Modulus of Toughness: The modulus of toughness is equal to the area under the entire stress–strain diagram. This area can be approximated by counting the number of squares. The total number of squares is 65. Thus,
C A Ui Bt Dapprox = 65[70(106)](0.01) = 45.5 MJ/m3
Ans.
s (MPa)
490 420 350 280 210 140 70 e (mm/mm)
Ans:
AUi Br = 0.616 MJ/m3
152
CH 03.indd 106
C A Ui Bt Dapprox = 45.5 MJ/m3
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3.13. 8000 40 kNlb
A bar having a length of 125 mm and crosssectional area of 437.5 mm2 is subjected to an axial force of 40 kN. If the bar streches 0.05 mm, determine the modulus of elasticity of the material. The material has linearelastic behavior.
8000 40 kNlb
5 in. 125 mm
Solution Normal Stress and Strain:
s5
40(103) P 5 5 91.43(106) N/m2 5 91.43 MPa 437.5(10−6) A
e5
dL 0.05 5 5 0.000400 mm/mm L 125
Modulus of Elasticity: E5
91.43(106) s 5 5 228.57(109) N/m2 5 229 GPa e 0.000400
Ans.
Ans: E 5 229 GPa 1 53
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3–14. The rigid pipe is supported by a pin at A and an A36 steel guy wire BD. If the wire has a diameter of 6.5 mm, determine how much it streches when a load of P ==3600 kN acts onon thethe pipe. pipe. lb acts
B
1.2 4 ftm
P A
D C 0.9 3 ftm
0.9 3 ftm
Solution Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a
FBD
a + ©MA = 0;
= ( 45 ) (0.9) − 3(1.8)
0
= 7.50 kN FBD
The normal stress developed in the wire is
σ= BD
FBD = ABD
7.50(10 3 ) 2 = 226.02(106 ) N/m = 226.02 MPa 2 (0.0065 ) 4
π
Since sBD 6 sy = 36 , Hooke’s Law cancan bebe applied totodetermine 250ksi MPa, Hooke’s Law applied determinethe thestrain straininin the wire. sBD = EPBD;
226.02(106 ) = 200(109 ) PAB PBD = 1.1301(10 −3 ) mm/mm
1.5=m.60 Thus, the the in. Thus, The unstretched length of the wire is LBD = 23 0.922 ++ 14.222==5ft wire stretches dBD = PBD LBD = 1.1301(10 −3 )[1.5(10 3 )]
= mm 1.70 mm = 1.6951
Ans.
3 kN
0.9 m
0.9 m
Ans: dBD = 1.70 mm 1 54
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3–15. The rigid pipe is supported by a pin at A and an A36 guy wire BD. BD. If If the the wire wire has hasaadiameter diameterofof0.25 6 mm, in., determine the load load PP ifif the the end end CCisisdisplaced displaced1.875 0.075mm in. downward.
B
4 ftm 1.2
P A
D C 3 ftm 0.9
Solution
3 ftm 0.9
Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a FBD A 45 B (0.9) (3) –P(6) P(1.8)= =0 0
a + ©MA = 0;
FBD = 2.50 P
The unstretched length for wire BD is LBD = 23 ft m. = 60 in. From From the the 0.922 ++ 14.222==51.5 geometry shown in Fig. b, the stretched length of wire BD is LBD¿ =
15002 + 1.6752  2(1500)(1.675) cos 143.13° = 1501.3403 mm
Thus, the normal strain is PBD =
LBD¿  LBD 1501.3403 – 1500 5 0.8936(10−3) mm/mm = 1500 LBD
Then, the normal stress can be obtain by applying Hooke’s Law. 9 6 = )[0.8936(10 −3 )] 178.71(10 = ) N−m 2 178.71 MPa sBD = EPBD 200(10=
MPa, result is valid. 36 ksi Since sBD 6 sy = 250 , thethe result is valid. sBD =
FBD ; ABD
178.71(106 ) =
2.50 P π
(0.00652 ) 4
P = 2.3721(10 3 ) N = 2.37 kN
Ans.
LBD = 1500 mm
0.9 m
0.9 m
1.675 mm
Ans:
P = 2.37 kN
155
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*3–16. Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used for connections. If a nut on the bolt is tightened so that the six 3mm high heads of the indicator are strained 0.1 mm>mm, and leave a contact area on each head of 1.5 mm2, determine the tension in the bolt shank. The material has the stress–strain diagram shown.
s (MPa) 3 mm
600 450
0.0015
0.3
P (mm/ mm)
Solution Stress–Strain Relationship: From the stress–strain diagram with P = 0.1 mm>mm 7 0.0015 mm>mm s  450 600  450 = 0.1  0.0015 0.3  0.0015 s = 499.497 MPa Axial Force: For each head P = sA = 499.4971 ( 106 ) (1.5) ( 106 ) = 749.24 N Thus, the tension in the bolt is Ans.
T = 6 P = 6(749.24) = 4495 N = 4.50 kN
Ans: T = 4.50 kN 156
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3–17.
B
The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm.
2m P
A
Solution
C 0.75 m 0.75 m
D
0.5 m
From the stress–strain diagram, E =
32.2(10)6 = 3.22(109) Pa 0.01
s (MPa) 100 95
Thus,
70 60
40(10 ) FAB = p = 31.83 MPa 2 AAB 4 (0.04)
sAB =
eAB
31.83(106) sAB = 0.009885 mm>mm = = E 3.22(109)
sCD
40(103) FCD = = p = 7.958 MPa 2 ACD 4 (0.08)
50 tension
40 32.2 20 0
6
eCD =
compression
80 3
7.958(10 ) sCD = 0.002471 mm>mm = E 3.22(109)
0
0.01 0.02 0.03 0.04
P (mm/mm)
dAB = eABLAB = 0.009885(2000) = 19.771 mm dCD = eCDLCD = 0.002471(500) = 1.236 mm Angle of tilt a: tan a =
18.535 ; 1500
Ans.
a = 0.708°
Ans. a = 0.708° 157
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3–18.
B
The stress–strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm.
2m P
A
C
Solution
0.75 m 0.75 m
D
0.5 m
Rupture of strut AB: sR =
FAB ; AAB
50(106) =
s (MPa)
P>2
; p 2 4 (0.012)
100 95
Ans.
P = 11.3 kN (controls)
70 60
Rupture of post CD: sR =
FCD ; ACD
compression
80
95(106) =
50
P>2
tension
40 32.2
p 2 4 (0.04)
20
P = 239 kN
0
0
0.01 0.02 0.03 0.04
P (mm/mm)
Ans. P = 11.3 kN (controls) 1 58
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3–19. The stress–strain diagram for a bone is shown, and can be P = 0.451106 2 s + described by the equation 12 3 0.36110 2 s , where s is in kPa. Determine the yield strength assuming a 0.3% offset.
P
s
P 0.45(106)s + 0.36(1012)s3 P
P
Solution P = 0.45(106)s + 0.36(1012)s3, dP = E =
1 0.45(106)
+ 1.08(1012) s2 2 ds
ds 1 ` = = 2.22(106) kPa = 2.22 GPa dP 0.45(10  6) s=0
The equation for the recovery line is s = 2.22(106)(P  0.003). This line intersects the stress–strain curve at sYS = 2027 kPa = 2.03 MPa
Ans.
Ans: sYS = 2.03 MPa 159
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*3–20. The stress–strain diagram for a bone is shown and can be described by the equation P = 0.451106 2 s + 0.3611012 2 s3, where s is in kPa. Determine the modulus of toughness and the amount of elongation of a 200mmlong region just before it fractures if failure occurs at P = 0.12 mm>mm.
P
s
P 0.45(106)s + 0.36(1012)s3 P
P
Solution When P = 0.12 120(103) = 0.45 s + 0.36(106)s3 Solving for the real root: s = 6873.52 kPa ut = ut =
LA L0
dA =
L0
6873.52
(0.12  P)ds
6873.52
(0.12  0.45(106)s  0.36(1012)s3)ds 6873.52
= 0.12 s  0.225(106)s2  0.09(1012)s4 0 = 613 kJ>m3
Ans. Ans.
d = PL = 0.12(200) = 24 mm
Ans: u t = 613 kJ>m3, d = 24 mm 160
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3–21. The two bars are made of polystyrene, which has the stress–strain diagram shown. If the crosssectional area of bar AB is 975 mm2 and BC is 2600 mm2, determine the largest force P that can be supported before any member ruptures. Assume that buckling does not occur.
P 1.2 m
C
B
1m
A (MPa)
Solution
175
5 + c gFy = 0; FAB  P = 0; 61 61 6 + FBC = P = ; ©F x = 0; 0 5 61 Assuming failure of bar BC:
61 P 5
(C)
FBC = 1.20 P
(T)
FAB =
140
(1)
105
(2)
FBC ; ABC
35(106 ) =
tension
35 0
From the stress–strain diagram (sR)t = 35 M Pa s =
compression
70
0
0.20
0.40
0.60
0.80
(mm/mm)
FBC 3 = FBC 91.0(10 = ) N 91.0 kN ; 2600(10 −6 )
From Eq. (2), P = 75.83 kN Assuming failure of bar AB: From stress–strain diagram (sR)c = 175 M Pa s =
FAB ; AAB
175(106 ) =
FAB 3 ; = FAB 170.625(10 = ) N 170.625 kN 975(10 −6 )
From Eq. (1), P ⫽ 109.23 kN Choose the smallest value P = 75.8 kN
Ans.
Ans: P = 75.8 kN 161
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3–22. The two bars are made of polystyrene, which has the stress–strain diagram shown. Determine the crosssectional area of each bar so that the bars rupture simultaneously when the load P = 13.5 kN. Assume that buckling does not occur.
P 1.2 m
C
B
1m
A (MPa)
Solution
175
+ c ©Fy = 0;
5 FAB= − 13.5= 0 61
FAB= 21.09 kN
+ : ©Fx = 0;
6 − FBC + 21.06 = 0 61
FBC 16.2 kN =
140 105
compression
70 tension
35
For member BC:
0
(smax)t =
FBC ; ABC
35(106 ) =
0
0.20
0.40
0.60
0.80
(mm/mm)
16.2(10 3 ) ; ABC
−3 = ) m 2 462.9 = mm 2 463 mm 2 ABC 0.4629(10 =
Ans.
For member BA: (smax)c =
FBA ; ABA
175(106 ) =
21.09(10 3 ) ; AAB
−3 = ) m 2 120.5 = mm 2 121 mm 2 AAB 0.1205(10 =
Ans.
Ans: ABC = 463 mm2, ABA = 121 mm2 162
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3–23. The stress–strain diagram for many metal alloys can be described analytically using the RambergOsgood three parameter equation P = s>E + ksn, where E, k, and n are determined from measurements taken from the diagram. Using the stress–strain diagram shown in the 3 E == 30110 210 GPa figure, take E and determine determine the the other two 2 ksiand parameters k and n and thereby obtain an analytical expression for the curve.
ss(MPa) (ksi) 560 80 420 60 280 40 140 20 0.1 0.2 0.3 0.4 0.5
–6) P (10–6
Solution Choose, s == 280 MPa, e e= =0.1 0.1 40 ksi, s == 420 MPa, e e= =0.3 0.3 60 ksi, 0.1== 0.1
40 280 nn k(280) ++ k(40) 33 30(10 )) 210(10
0.3== 0.3
60 420 nn k(420) ++ k(60) 33 30(10 )) 210(10
0.098667 == k(280) 0.098667 k(40)nn 0.29800 == k(420) 0.29800 k(60)nn 0.3310962 = (0.6667)n ln (0.3310962) = n ln (0.6667) n = 2.73
Ans.
k = 21.0(109)
Ans.
Ans: n = 2.73 k = 21.0(109)
1 63
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*3–24. The s–P diagram for a collagen fiber bundle from which a human tendon is composed is shown. If a segment of the Achilles Achilles tendon tendon at atAAhas hasaalength lengthofof165 6.5mm in. and an approximate crosssectional area of 145 mm 0.229 in2, determine its elongation if the foot supports a load of 125 625 lb, N, which causes a tension in the tendon of 1718.75 N. 343.75 lb.
ss(MPa) (ksi) 31.50 4.50
A
26.25 3.75 21.00 3.00 15.75 2.25 11.85 10.50 1.50
625 125 N lb
5.25 0.75 0.035 0.05
0.10
P (mm/mm) (in./in.)
Solution σ=
P 1718.75 = = 11.85 MPa A 145(10 −6 )
From the graph e == 0.035 0.035mm/mm in.>in. Ans.
0.035(165) = 5.775 d = eL = 0.035(6.5) 0.228 mm in.
Ans: d = 5.775 mm
164
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3–25. The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, np = 0.4.
300 N
300 N 200 mm
Solution s =
P = A
Plong =
300 p 2 4 (0.015)
= 1.678 MPa
1.678 ( 106 ) s = = 0.0006288 E 2.70 ( 109 ) Ans.
d = Plong L = 0.0006288(200) = 0.126 mm Plat =  nPlong =  0.4(0.0006288) = 0.0002515 ∆d = Platd = 0.0002515(15) = 0.00377 mm
Ans.
Ans: d = 0.126 mm, ∆d =  0.00377 mm 165
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3–26. The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve are 50 mm long. Determine the axial pressure p that must be applied to the top of the plug to cause it to contact the sides of the sleeve. Also, how far must the plug be compressed downward in order to do this? The plug is made from a material for which E = 5 MPa, n = 0.45.
p
Solution Plat =
d′  d 32  30 = = 0.06667 mm>mm d 30
v = 
Plat Plat 0.06667 ; Plong = = =  0.1481 mm>mm Plong v 0.45
p = s = E Plong = 5 ( 106 ) (0.1481) = 741 kPa
Ans.
d = Plong L = 0.1481(50) = 7.41 mm
Ans.
Ans: p = 741 kPa, d = 7.41 mm 166
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3–27. The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. When the applied load on the specimen is 50 kN, the diameter is 12.67494 mm. Determine Poisson’s ratio for the material.
s (MPa) 490
Solution
0.007
P (mm/ mm)
Average Normal Stress: 50 ( 10 ) N = = 394.71 ( 106 ) Pa = 394.71 MPa p A 2 ( 0.0127 ) 4 3
s =
Average Normal Strain: Referring to the stress–strain diagram, the modulus of 490 ( 106 ) elasticity is E = = 70.0 ( 109 ) Pa = 70.0 GPa. 0.007 Plong = Plat =
394.71 ( 106 ) s = = 0.0056386 mm>mm E 70.0 ( 109 )
d  d0 12.67494  12.7 = =  0.0019732 d0 12.7
Poisson’s Ratio: The lateral and longitudinal strain can be related using Poisson’s ratio, that is n = 
(  0.019732) Plat = = 0.350 Plong 0.0056386
Ans.
Ans: n = 0.350 167
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*3–28. The elastic portion of the stress–strain diagram for an aluminum alloy is shown in the figure. The specimen from which it was obtained has an original diameter of 12.7 mm and a gage length of 50.8 mm. If a load of P = 60 kN is applied to the specimen, determine its new diameter and length. Take n = 0.35.
s (MPa) 490
Solution
0.007
P (mm/ mm)
Average Normal Stress: 60 ( 10 ) N = = 473.65 ( 106 ) Pa = 473.65 MPa p A 2 ( 0.0127 ) 4 3
s =
Average Normal Strain: Referring to the stress–strain diagram, the modulus of 490 ( 106 ) elasticity is E = = 70.0 ( 109 ) Pa = 70.0 GPa. 0.007 Plong =
473.65 ( 106 ) s = = 0.0067664 mm>mm E 70.0 ( 109 )
Thus,
dL = Plong L0 = 0.0067664(50.8) = 0.34373 mm
Then
Ans.
L = L0 + dL = 50.8 + 0.34373 = 51.1437 mm
Poisson’s Ratio: The lateral strain can be related to the longitudinal strain using Poisson’s ratio.
Plat =  nPlong =  0.35(0.0067664) =  0.0023682 mm>mm
Thus,
dd = Plat d =  0.0023682(12.7) = 0.030077 mm
Then d = d 0 + dd = 12.7 + (  0.030077) = 12.66992 mm Ans.
= 12.67 mm
Ans: L = 51.1437 mm, d = 12.67 mm 168
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3–29. The brake pads for a bicycle tire are made of rubber. If a frictional force of 50 N is applied to each side of the tires, determine the average shear strain in the rubber. Each pad has crosssectional dimensions of 20 mm and 50 mm. Gr = 0.20 MPa. 50 mm
Solution
10 mm
10 mm
Average Shear Stress: The shear force is V = 50 N. t =
V 50 = = 50.0 kPa A 0.02(0.05)
ShearStress – Strain Relationship: Applying Hooke’s law for shear t = Gg 50.0 ( 10
3
) = 0.2 ( 106 ) g Ans.
g = 0.250 rad
Ans: g = 0.250 rad 169
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3–30.
P 2
The lap joint is connected together using a 30 mm diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the shear strain developed in the shear plane of the bolt when P = 340 kN.
P 2
P
t (MPa) 525 350
Solution Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the freebody diagram shown in Fig. a. + : ©Fx = 0;
340  2V = 0
0.005
0.05
g (rad)
V = 170 kN
Shear Stress and Strain: V 170(10 3 ) 6 240.50(10 = ) N/m 2 240.50 MPa t = = = π (0.032 ) A 4 Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 240.50 350 = ; g 0.005
g = 3.4357(103) rad = 3.44(103) rad
Ans.
Ans: g = 3.44(103) rad 170
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3–31
P 2
The lap joint is connected together using a 30 mm diameter bolt. If the bolt is made from a material having a shear stress–strain diagram that is approximated as shown, determine the permanent shear strain in the shear plane of the bolt when the applied force P = 680 kN is removed.
P 2
P
t (MPa) 525
Solution
350
Internal Loadings: The shear force developed in the shear planes of the bolt can be determined by considering the equilibrium of the freebody diagram shown in Fig. a. + : ©Fx = 0;
680  2V = 0
V = 340 kN 0.005
Shear Stress and Strain:
0.05
g (rad)
340(10 3 ) V 6 481.00(10 = ) N/m 2 481.00 MPa = π t= = 2 A (0.03 ) 4
Using this result, the corresponding shear strain can be obtained from the shear stress–strain diagram, Fig. b. 481.00 350 = 525  350 ; g  0.005 0.05  0.005
g = 0.03869 rad
When force P is removed, the shear strain recovers linearly along line BC, Fig. b, with a slope that is the same as line OA. This slope represents the shear modulus. = G
350(10 3 ) 9 = 700.0(10 = ) Pa 70.0 GPa 0.005
Thus, the elastic recovery of shear strain is t = Ggr;
481.00(106) = 70.0(109)gr
gr = 6.8715(103) rad
And the permanent shear strain is gP = g  gr = 0.03869  6.874(103) = 0.031815 rad = 0.0318 rad
Ans.
Ans: gP = 0.0318 rad 171
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*3–32. A shear spring is made by bonding the rubber annulus to a rigid fixed ring and a plug. When an axial load P is placed on the plug, show that the slope at point y in the rubber is dy>dr =  tan g =  tan1P>12phGr22. For small angles we can write dy>dr =  P>12phGr2. Integrate this expression and evaluate the constant of integration using the condition that y = 0 at r = ro. From the result compute the deflection y = d of the plug.
P
h
ro
y
d
ri r y
Shear Stress–Strain Relationship: Applying Hooke’s law with tA
g =
P = . 2p r h
tA P = G 2p h G r
dy P =  tan g =  tan a b dr 2p h G r
(Q.E.D)
If g is small, then tan g = g. Therefore, dy P = dr 2p h G r
At r = ro,
y = 
dr P 2p h G L r
y = 
P ln r + C 2p h G
0 = 
P ln ro + C 2p h G
y = 0
C =
Then, y =
ro P ln r 2p h G
At r = ri,
y = d d =
P ln ro 2p h G
ro P ln ri 2p h G
Ans.
Ans. d =
172
ro P ln ri 2p h G
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3–33. The support consists of three rigid plates, which are connected together using two symmetrically placed rubber pads. If a vertical force of 5 N is applied to plate A, determine the approximate vertical displacement of this plate due to shear strains in the rubber. Each pad has crosssectional dimensions of 30 mm and 20 mm. Gr = 0.20 MPa.
C
B
40 mm
40 mm
A
Solution tavg
V 2.5 = = = 4166.7 Pa A (0.03)(0.02)
g =
4166.7 t = 0.02083 rad = G 0.2(106)
5N
Ans.
d = 40(0.02083) = 0.833 mm
Ans. d = 0.833 mm 173
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3–34. A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A when the vertical load P is applied. Assume that the displacement is small so that d = a tan g ≈ ag.
P
d A
h
Solution P Average Shear Stress: The rubber block is subjected to a shear force of V = . 2
a
a
P
t =
V P 2 = = A bh 2bh
Shear Strain: Applying Hooke’s law for shear P
g =
t P 2bh = = G G 2bhG
Thus, d = ag =
Pa 2bhG
Ans.
Ans: d = 174
Pa 2bhG
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R3–1.
s (ksi) (MPa)
The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. When the applied load is 45 kN, the new diameter of the specimen is 12.4780 mm. Calculate the shear modulus Gal for the aluminum.
350
P (mm/mm) (in./in.)
0.00480
Solution From the stress–strain diagram,
σ 350(106 ) 2 72.92 GPa = = 72.92(109 ) N/m= ε 0.00480
Eal=
When specimen is loaded with a 45kN load,
σ=
= ε lonε
ε lat = = V
= Gal
45(10 3 )
P = A
2 = 366.69(106 ) N/m= 366.69 MPa 2 (0.0125 ) 4
π
σ
= Eal
366.69(106 ) = 5.0289(10 −3 ) mm/mm 72.92(109 )
d′ − d 12.4780 − 12.5 = = −1.76(10 −3 ) mm/mm d 12.5
ε ε lonε
lat =
−1.76(10 −3 ) = 0.3500 5.0289(10 −3 )
72.92(109 ) Eat = = 27.00(109 )= N/m 2 27.0 GPa 2(1 + v) 2(1 + 0.3500)
Ans.
Ans:
Gal = 27.0 GPa 175
CH 03.indd 116
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R3–2.
s (ksi) (MPa)
The elastic portion of the tension stress–strain diagram for an aluminum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. If the applied load is 40 kN, determine the new diameter of the specimen. The shear modulus is Gal = 27 GPa.
350
0.00480
P (mm/mm) (in./in.)
Solution σ=
40(10 3 )
P = A
2 = 325.95(106 ) N/m= 325.95 MPa 2 (0.0125 ) 4
π
From the stress–strain diagram, E =
350(106 ) 9 ) N/m 2 72.92 GPa = 72.92(10= 0.00480
ε lonε= = G
σ
= E
325.95(106 ) = 4.4702(10 −3 ) mm/mm 72.92(109 )
E 72.92(109 ) = ; 27(109 ) = ; v 0.3503 2(1 + v) 2(1 + v)
−vε lonε = −0.3503[4.4702(10 −3 )] = −1.5659(10 −3 ) mm/mm ε lat = ∆d =ε lat d =[−1.5659(10 −3 )](12.5) =−0.01957 mm d′= d + ∆d= 12.5 − 0.01957= 12.4804 mm
Ans.
Ans: d′ = 12.4804 mm 176
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R3–3. The rigid beam rests in the horizontal position on two 2014T6 aluminum cylinders having the unloaded lengths shown. If each cylinder has a diameter of 30 mm, determine the placement x of the applied 80kN load so that the beam remains horizontal. What is the new diameter of cylinder A after the load is applied? nal = 0.35.
80 kN x
220 mm
A
B
210 mm
3m
Solution a+ΣMA = 0; FB(3)  80(x) = 0; FB =
80x 3
a+ΣMB = 0;  FA(3) + 80(3  x) = 0; FA =
(1) 80(3  x)
(2)
3
Since the beam is held horizontally, dA = dB P
s =
s P A ; P = = A E E P
PL A d = PL = a b L = E AE dA = dB;
80(3  x) 3
(220)
AE
=
80x 3
(210) AE
80(3  x)(220) = 80x(210) Ans.
x = 1.53 m From Eq. (2), FA = 39.07 kN sA =
39.07(103) FA = = 55.27 MPa p A (0.032) 4
Plong =
55.27(106) sA = =  0.000756 E 73.1(109)
Plat =  nPlong =  0.35(  0.000756) = 0.0002646 d′A = d A + d Plat = 30 + 30(0.0002646) = 30.008 mm
Ans.
Ans: x = 1.53 m, d A ′ = 30.008 mm 177
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*R3–4.
C P
When the two forces are placed on the beam, the diameter of the A36 steel rod BC decreases from 40 mm to 39.99 mm. Determine the magnitude of each force P.
1m
A
P 1m
1m
1m
B 0.75 m
Solution Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a. 4 FBC a b (3)  P(2)  P(1) = 0 5
a + ©MA = 0;
FBC = 1.25P
Normal Stress and Strain: The lateral strain of rod BC is Plat =
d  d0 39.99  40 = =  0.25(10  3) mm>mm d0 40
Plat =  nPa;
 0.25(103) =  (0.32)Pa Pa = 0.78125(103) mm>mm
Assuming that Hooke’s Law applies, sBC = EPa;
sBC = 200(109)(0.78125)(103) = 156.25 MPa
Since s 6 sY, the assumption is correct. sBC =
FBC ; ABC
156.25(106) =
1.25P p A 0.042 B 4
P = 157.08(103)N = 157 kN
Ans.
Ans: P = 157 kN 178
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R3–5.
C P
If P = 150 kN, determine the elastic elongation of rod BC and the decrease in its diameter. Rod BC is made of A36 steel and has a diameter of 40 mm.
1m
P 1m
1m
1m
B
A
0.75 m
Solution Equations of Equilibrium: The force developed in rod BC can be determined by writing the moment equation of equilibrium about A with reference to the freebody diagram of the beam shown in Fig. a. a + ©MA = 0;
4 FBC a b (3)  150(2)  150(1) = 0 5
FBC = 187.5 kN
Normal Stress and Strain: The lateral strain of rod BC is sBC =
187.5(103) FBC = = 149.21 MPa p ABC A 0.042 B 4
Since s 6 sY, Hooke’s Law can be applied. Thus, sBC = EPBC;
149.21(106) = 200(109)PBC PBC = 0.7460(103) mm>mm
The unstretched length of rod BC is LBC = 27502 + 10002 = 1250 mm. Thus the elongation of this rod is given by dBC = PBCLBC = 0.7460(103)(1250) = 0.933 mm
Ans.
We obtain, Plat =  nPa ;
Plat =  (0.32)(0.7460)(103) =  0.2387(103) mm>mm
Thus, dd = Plat dBC =  0.2387(103)(40) =  9.55(103) mm
Ans.
Ans. dBC = 0.933 mm dd = 9.55(103) mm 179
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R3–6. The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 4 kN, determine the normal strain in the bolts. Each bolt has a diameter of 5 mm. If σY = 280 MPa and Est = 200 GPa, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released?
C L
H
Solution Normal Stress:
σ=
P = A
4(10 3 )
2 = 203.72(106 ) N/m= 203.72 MPa < sg = 280 MPa 2 (0.005 ) 4
π
Normal Strain: Since s 6 sg, Hooke’s law is still valid.
= ε
σ
= E
203.72(106 ) = 1.0186(10 −3 ) mm/mm = 1.02(10 −3 ) mm/mm 200(109 )
If the nut is unscrewed, the load is zero. Also, the nut is not stressed beyond sg and so no permanent strain will be developed. Therefore, the strain e = 0
Ans.
Ans.
Ans: e = 0.0010186 mm>mm, eunscr = 0 180
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R3–7. The stress–strain diagram for polyethylene, which is used to sheath coaxial cables, is determined from testing a specimen that has a gauge length of 250 mm. If a load P on the specimen develops a strain of P = 0.024 mm/mm, determine the approximate length of the specimen, measured between the gauge points, when the load is removed. Assume the specimen recovers elastically.
Solution From the graph s = 14 Mpa, P = 0.004 mm/mm, E =
s 14(106) = 3.50(109) Pa = 3.50 GPa = P 0.004
P = 0.024 mm/mm, s = 25.67 Mpa L' = 250 mm + 0.024(250) = 256 mm Elastic strain recovery: Prec =
s 25.67(106) = = 7.3333(10−3) mm/mm E 3.50(109)
δ = PrecL = [7.3333(10−3)](250) mm/mm = 1.8333 mm L = L' − δ = 256 mm − 1.8333 mm = 254.167 mm
Ans.
Ans: 254.167 mm 181
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*R3–8. The pipe with two rigid caps attached to its ends is subjected to an axial force P. If the pipe is made from a material having a modulus of elasticity E and Poisson’s ratio n, determine the change in volume of the material.
ri ro L P
Section a – a
a
a
Solution Normal Stress: The rod is subjected to uniaxial loading. Thus, slong =
P and slat = 0. A
P
dV = AdL + 2prLdr = APlong L + 2prLPlatr Using Poisson’s ratio and noting that AL = pr 2L = V, dV = PlongV  2nPlongV = Plong (1  2n)V =
slong E
(1  2n)V
Since slong = P>A, dV = =
P (1  2n)AL AE PL (1  2n) E
Ans.
Ans: dV = 182
PL (1  2n) E
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R3–9. The 8mmdiameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa, Emg = 45 GPa.
50 mm A
30 mm
Solution Normal Stress: 8(103)
sb =
P = Ab
p 2 4 (0.008 )
ss =
P = As
p 2 4 (0.02
= 159.15 MPa
8(103)  0.0122)
= 39.79 MPa
Normal Strain: Applying Hooke’s Law Pb =
159.15(106) sb = = 0.00227 mm>mm Eal 70(109)
Ans.
Ps =
39.79(106) ss = = 0.000884 mm>mm Emg 45(109)
Ans.
Ans: Pb = 0.00227 mm>mm, Ps = 0.000884 mm>mm 183
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R3–10. An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2 mm horizontally when it is subjected to a horizontal force P = 2 kN, determine the shear modulus of the polymer. The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysis.
400 mm P 2 kN
200 mm
Solution Normal and Shear Stress: t =
2(103) V = = 50 kPa A 0.4(0.1)
Referring to the geometry of the undeformed and deformed shape of the block shown in Fig. a, g =
2 = 0.01 rad 200
Applying Hooke’s Law, t = Gg;
50(103) = G(0.01) Ans.
G = 5 MPa
Ans: G = 5 MPa 184
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4–1. The A992 steel rod is subjected to the loading shown. If the crosssectional area of the rod is 60 mm2, determine the displacement of B and A, Neglect the size of the couplings at B, C, and D.
D 0.75 m C 60⬚
60⬚
1.50 m 3.30 kN
3.30 kN
Solution
B 5
3
3
3
4
3
10.4(10 )(1.50) 16.116(10 )(0.75) PL + = dB = © 6 9 AE 60(10 )(200)(10 ) 60(10  6)(200)(109)
2 kN
= 0.00231 m = 2.31 mm
Ans.
5 4
A
0.50 m 2 kN
8 kN
3
dA = dB +
8(10 )(0.5) 60(10  6)(200)(109)
= 0.00264 m = 2.64 mm
Ans.
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.
Ans:
dB = 2.31 mm, dA = 2.64 mm 185
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4–2. The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D if the diameters of each segment are dAB = 20 mm, dBC = 25 mm, and dCD = 12 mm. Take Ecu = 126 GPa.
2m
3.75 m
22.5 kN
36 kN
2.5 m 9 kN
A 22.5 kN B
C 9 kN
27 kN D
Solution PL −36(10 3 )(2000) 9(10 3 )(3750) 27(10 3 )(2500) = + + AE π (0.02 2 ) 126(109 ) π (0.0252 ) 126(109 ) π (0.012 2 ) 126(109 ) 4 4 4
δ A −D = ∑
= 3.4635 = mm 3.46 mm
Ans.
The positive sign indicates that end A moves away from end D.
Ans: dA>D = 3.46 mm away from end D. 186
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4–3. The composite shaft, consisting of aluminum, copper, and steel sections, is subjected to the loading shown. Determine the displacement of end A with respect to end D and the normal stress in each section. The crosssectional area and modulus of elasticity for each section are shown in the figure. Neglect the size of the collars at B and C.
Aluminum Eal = 70 GPa AAB = 58 mm2
Copper Ecu = 126 GPa ABC = 77 mm2
A
B 450 mm
16 kN
ACD = 39 mm2 8 kN
16 kN
9 kN
Steel Est = 200 GPa
C
300 mm
8 kN
7 kN D
400 mm
Solution σ= AB
9(10 3 ) PAB 2 = = 155.17(106 ) N−m = 155 MPa AAB 58(10 −6 )
(T)
Ans.
σ= BC
23(10 3 ) PBC 2 = = 298.70(106 ) N−m = 299 MPa ABC 77(10 −6 )
(C)
Ans.
σ = CD
7(10 3 ) PBC 2 = = 179.49(106 ) N−m = 179 MPa ABC 39(10 −6 )
(C)
Ans.
−23(10 3 )(300) 9(10 3 )(450) −7(10 3 )(400) PL = + + AE 58(10 −6 ) 70(109 ) 77(10 −6 ) 126(109 ) 39(10 −6 ) 200(109 )
δ ND = ∑
= −0.07263 mm = −0.0726 mm
Ans.
The negative sign indicates end A moves towards end D.
Ans: sAB = 155 MPa 1T2, sBC = 299 MPa 1C2, sCD = 179 MPa 1C2, dA>D = 0.0726 mm towards end D. 187
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*4–4. Determine the displacement of B with respect to C of the composite shaft in Prob. 4–3.
Aluminum Eal = 70 GPa AAB = 58 mm2
Copper Ecu = 126 GPa ABC = 77 mm2
A
B 450 mm
dB>C =
−23(10 3 )(300) PL = −0.7112 mm = −0.711 mm = AE 77(10 −6 ) 126(109 )
16 kN
ACD = 39 mm2 8 kN
16 kN
9 kN
Steel Est = 200 GPa
C
300 mm
8 kN
7 kN D
400 mm
Ans.
The negative sign indicates end B moves towards end C.
Ans.
dB>C = −0.711 mm 188
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4–5. The 2014T6 aluminium rod has a diameter of 30 mm and supports the load shown. Determine the displacement of end A with respect to end E. Neglect the size of the couplings.
A
B
8 kN 4 kN 4m
C
D
6 kN 2m
E 2 kN
2m
2m
Solution dA>E = Σ =
PL 1 = [8(4) + 4(2)  2(2) + 0(2)] ( 103 ) AE AE 36 ( 103 )
p (0.03)2(73.1) ( 109 ) 4
= 0.697 ( 10  3 ) = 0.697 mm
Ans.
Ans: dA>E = 0.697 mm 189
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4–6. The A992 steel drill shaft of an oil well extends 3600 m into the ground. Assuming that the pipe used to drill the well is suspended freely from the derrick at A, determine the maximum average normal stress in each pipe segment and the elongation of its end D with respect to the fixed end at A. The shaft consists of three different sizes of pipe, AB, BC, and CD, each having the length, weight per unit length, and crosssectional area indicated.
A 2
AAB = 1600 mm wAB = 50 N/m
1500 m B
2
ABC = 1125 mm wBC = 40 N/m ACD = 800 mm2 wCD = 30 N/m
1500 m C D
600 m
Solution
σA =
P 50(1500) + 78000 2 = = 95.625(106 ) N−m= 95.6 MPa A 1600(10 −6 )
Ans.
σ= B
P 40(1500) + 18000 2 = = 69.33(106 ) N−m= 69.3 MPa A 1125(10 −6 )
Ans.
σ C=
30(600) P 2 = = 22.5(106 ) N−m= 22.5 MPa A 800(10 −6 )
Ans.
dD = ©
600 m 1500 m 1500 m P(x) dx (40 x + 18000)dx 30 x dx (50 x + 78000)dx + + = 1125(10 −6 ) 200(109 ) 800(10 −6 ) 200(109 ) 1600(10 −6 ) 200(109 ) L0 L0 L A(x) E L0
= 0.8952 = m 0.895 m
Ans.
Ans. sA = 95.6 MPa, sB = 69.3 MPa, sC = 22.5 MPa, dD = 0.895 m 190
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4–7. P
The truss is made of three A36 steel members, each having a crosssectional area of 400 mm2. Determine the horizontal displacement of the roller at C when P = 8 kN.
B
5 kN
0.8 m
A
C
Solution By observation the horizontal displacement of roller C is equal to the displacement of point C obtained from member AC.
0.8 m
0.6 m
FCA = 5.571 kN dC =
5.571 ( 103 ) (1.40) FCAL = = 0.0975 mm S AE (400) ( 10  6 ) (200) ( 106 )
Ans.
Ans: dC = 0.0975 mm S 191
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*4–8. The truss is made of three A36 steel members, each having a crosssectional area of 400 mm2. Determine the magnitude P required to displace the roller to the right 0.2 mm.
P B
5 kN
0.8 m
A
C
Solution 0.8 m
a+ MA = 0;  P(0.8)  5(0.8) + Cy(1.4) = 0
0.6 m
Cy = 0.5714 P + 2.857 4 + c ΣFy = C 0; y  FBC a b = 0 5 FBC = 1.25 Cy
+ ΣFx = 0; S
 FAC + 1.25 Cy (0.6) = 0 FAC = 0.75 Cy = 0.4286 P + 2.14286
Require, dCA = 0.0002 =
(0.4286 P + 2.14286) ( 103 ) (1.4) (400) ( 106 ) (200) ( 109 ) Ans.
P = 21.7 kN
Ans: P = 21.7 kN 192
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4–9. The assembly consists of two 10mm diameter red brass C83400 copper rods AB and CD, a 15mm diameter 304 stainless steel rod EF, and a rigid bar G. If P = 5 kN, determine the horizontal displacement of end F of rod EF.
300 mm A
450 mm B
P E
4P F
C
DG
P
Solution Internal Loading: The normal forces developed in rods EF, AB, and CD are shown on the freebody diagrams in Figs. a and b. p Displacement: The crosssectional areas of rods EF and AB are AEF = (0.0152) = 4 p 56.25(10  6)p m2 and AAB = (0.012) = 25(10  6)p m2. 4 dF = Σ =
PEF LEF PAB LAB PL = + AE AEF Est AAB Ebr 20(103)(450) 6
9
56.25(10 )p(193)(10 )
+
5(103)(300) 25 ( 10  6)p(101)(109) Ans.
= 0.453 mm The positive sign indicates that end F moves away from the fixed end.
Ans: dF = 0.453 mm 193
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4–10. The assembly consists of two 10mm diameter red brass C83400 copper rods AB and CD, a 15mm diameter 304 stainless steel rod EF, and a rigid 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
DG
P
Solution Internal Loading: The normal forces developed in rods EF, AB, and CD are shown on the freebody diagrams in Figs. a and b. Displacement: The crosssectional areas of rods EF and AB are AEF = 56.25(10  6 )p m2 and AAB =
p (0.012 ) = 25(10  6 )p m2. 4
dF = Σ
PEF LEF PAB LAB PL = + AE AEF Est AAB Ebr
0.45 =
4P(450) 6
9
56.25(10 )p(193)(10 )
+
p (0.0152) = 4
P(300) 6
25(10 )p(101)(109) Ans.
P = 4967 N = 4.97 kN
Ans: P = 4.97 kN 194
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4–11. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of of the the 2.5kN 500lb load if the members were originally horizontal when the load was 16 mmin2.2. applied. Each wire has a crosssectional area of 0.025
E
F
0.9 3 ftm
Solution
0.54 1.8 ftm
C 0.6 2 ftm
1 ftm 0.3
I
Internal Forces in the wires:
A
0.9 3 ftm
FBD (b) FBG(1.2) – 2.5(0.9) = 0
+ c ©Fy = 0;
FAH + 1.875 – 2.5 = 0
1.5 5 ftm
H
D
a + ©MA = 0;
G
B 1 ft 0.3 m 2.5 500kN lb
FFBC 375.0 kN lb BC = 1.875 = 125.0 FF = 0.625 kN lb AH AH
FBD (a) a + ©MD = 0;
0.625(0.3)= =00 FF = 0.2083 FCF (3) – 125.0(1) = 41.67kN lb CF CF(0.9) CF
+ c ©Fy = 0;
0.2083– 125.0 0.625 = 0 0 FDE = 0.4167 kN lb FF FDE = 83.33 DE++ 41.67 DE
0.3 m
0.6 m
Displacement:
δD =
FDE LDE 0.4167(10 3 )(0.9)(10 3 ) = = 0.121438 mm ADE E [16(10 −6 )][193(109 )]
δC =
FCF LCF 0.2083(10 3 )(0.9)(10 3 ) = = 0.060719 mm ACF E [16(10 −6 )][193(109 )]
0.3 m
0.9 m
2.5 kN
0.060719 mm
′ δH
0.060719 ′ = = ; δH 0.040479 mm 0.6 0.9
0.3 m
0.6 m
δ H = 0.040479 + 0.060719 = 0.101198 mm δA H =
FAH LAH 0.625(10 3 )(0.54)(10 3 ) = = 0.109294 mm AAH E [16(10 −6 )][193(109 )]
0.101198 + 0.109294 = 0.210492 mm δA = δ H + δ A −H =
δB =
FBG LBG 1.875(10 3 )(1.5)(10 3 ) = = 0.910784 mm ABG E [16(10 −6 )][193(109 )]
δ I′ 0.700291 = = ; δ I′ 0.525219 mm 0.9 1.2 δ I = 0.210492 + 0.525219 = 0.7357 mm = 0.736 mm
0.9 m
0.3 m
Ans.
0.210492 mm
0.700291 mm
Ans: dl = 0.736 mm 195
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*4–12. The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the 2.5kN load load isis applied. applied. The The members were originally 500lb horizontal, and each wire has a crosssectional area of 16 mmin2.2. 0.025
E
F
G
3 ftm 0.9
Solution
1.8 ft 0.54 m
5 ftm 1.5
H
D
C 1 ft 0.3 m
2 ftm 0.6 I
Internal Forces in the wires:
A
B 3 ftm 0.9
FBD (b) a + ©MA = 0;
FBG 2.5(0.9)==00 F (4) – 500(3) BG(1.2)
+ c ©Fy = 0;
+ 1.875 = 0 FF 375.0 –2.5 500 = 0 AH + AH
1 ft 0.3 m 500kN lb 2.5
FFBG 375.0kN lb BG = 1.875 FF = 0.625 kN lb = 125.0 AH AH
FBD (a) a + ©MD = 0;
0.625(0.3)= =00 FF = 0.2083 FCF (3) – 125.0(1) = 41.67kN lb CF CF(0.9) CF
+ c ©Fy = 0;
0.2083– 125.0 0.625 = 0 0 FDE = 0.4167 kN lb FF FDE = 83.33 DE++ 41.67 DE
0.3 m
0.6 m
Displacement:
δD =
FDE LDE 0.4167(10 3 )(0.9)(10 3 ) = = 0.121438 mm ADE E [16(10 −6 )][193(109 )]
δC =
FCF LCF 0.2083(10 3 )(0.9)(10 3 ) = = 0.060719 mm ACF E [16(10 −6 )][193(109 )]
0.3 m
0.9 m
2.5 kN
0.060719 mm
′ δH
0.060719 ′ = = ; δH 0.040479 mm 0.6 0.9
0.3 m
0.6 m
′ + δC = 0.040479 + 0.060719 = 0.101198 mm δH = δH tan α =
δA H =
0.060719 ; α 0.003865 = = ° 0.00387° 0.9(10 3 )
Ans.
FAH LAH 0.625(10 3 )(0.54)(10 3 ) = = 0.109294 mm AAH E [16(10 −6 )][193(109 )]
δA = δH + δA H = 0.101198 + 0.109294 = 0.210492 mm δB =
FBG LBG 1.875(10 3 )(1.5)(10 3 ) = = 0.910784 mm ABG E [16(10 −6 )][193(109 )]
β = tan
0.700291 1.2(10 3 )
0.9 m
β 0.03344 ;= = ° 0.0334°
0.3 m
0.210492 mm
0.700291 mm
Ans. α = 0.00387° β = 0.0334° 196
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4–13. A springsupported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have a total weight of 4 kN, determine the displacement of the pipe when it is attached to the support.
F B
D
k G 0.75 m
Solution
0.75 m k H
E
A
C
Internal Force in the Rods: 0.25 m 0.25 m
FBD (a) a + ©MA = 0;
FCD (0.5)  4(0.25) = 0
+ c ©Fy = 0;
FAB + 2.00  4 = 0
FCD = 2.00 kN
FAB = 2.00 kN
FBD (b) + c ©Fy = 0;
FEF  2.00  2.00 = 0
FEF = 4.00 kN
Displacement: dD = dE =
FEFLEF = AEFE
dA>B = dC>D =
4.00(103)(750) p 4
(0.012)2(193)(109)
PCDLCD = ACDE
= 0.1374 mm
2(103)(750) p 4
(0.005)2(193)(109)
= 0.3958 mm
dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm Displacement of the spring dsp =
Fsp k
=
2.00 = 0.0333333 m = 33.3333 mm 60
dlat = dC + dsp Ans.
= 0.5332 + 33.3333 = 33.9 mm
Ans. dD = 0.1374 mm, dA>B = 0.3958 mm, dC = 0.5332 mm, dtot = 33.9 mm 1 97
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4–14. A springsupported pipe hanger consists of two springs, which are originally unstretched and have a stiffness of k = 60 kN>m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when it is filled with fluid, determine the weight of the fluid.
F B
D
k G 0.75 m
Solution
0.75 m k H
E
A
C
Internal Force in the Rods: 0.25 m 0.25 m
FBD (a) a + ©MA = 0; + c ©Fy = 0;
FCD(0.5)  W(0.25) = 0
FCD =
W  W = 0 2
W 2
FAB +
FAB =
W 2
FBD (b) + c ©Fy = 0;
FEF 
W W = 0 2 2
FEF = W
Displacement: dD = dE =
FEFLEF = AEFE
W(750) p 2 9 4 (0.012) (193)(10 )
= 34.35988(10  6) W dA>B = dC>D =
FCDLCD = ACDE
W 2 (750) p 2 (0.005) (193)(109) 4
= 98.95644(10  6) W dC = dD + dC>D = 34.35988(10  6) W + 98.95644(10  6) W = 0.133316(10  3) W Displacement of the spring dsp =
Fsp k
=
W 2
60(103)
(1000) = 0.008333 W
dlat = dC + dsp 82 = 0.133316(10  3) W + 0.008333W Ans.
W = 9685 N = 9.69 kN
Ans. W = 9.69 kN 1 98
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4–15. The steel bar has the original dimensions shown in the figure. If it is subjected to an axial load of 50 kN, determine the change in its length and its new crosssectional dimensions at section a–a. Est = 200 GPa, nst = 0.29.
50 kN
a
A
60 mm 20 mm 20 mm
B 200 mm 350 mm
a
C
20 mm 50 mm
D
50 kN
200 mm
Solution dA>D = Σ
2(50) ( 103 ) (200) 50 ( 103 ) (350) PL = + AE (0.02)(0.05)(200) ( 109 ) (0.06)(0.05)(200) ( 109 ) Ans.
= 0.129 mm dB>C =
50 ( 10 ) (350) PL = = 0.02917 mm AE (0.06)(0.05)(200) ( 109 )
PBC =
dB>C 0.02917 = 0.00008333 LBC 350
3
Plat =  v Plong =  (0.29)(0.00008333) =  0.00002417 h′ = 50  50(0.00002417) = 49.9988 mm
Ans.
w′ = 60  60(0.00002417) = 59.9986 mm
Ans.
Ans: dA>D = 0.129 mm, h′ = 49.9988 mm, w′ = 59.9986 mm 199
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*4–16. The ship is pushed through the water using an A36 steel propeller shaft that is 8 m long, measured from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings. A
B
C
D
5 kN
Solution
8m
Internal Force: As shown on FBD. Displacement: dA =
PL = AE
 5.00 (103)(8) p 4
(0.42  0.32) 200(109)
=  3.638(10  6) m =  3.64 A 10  3 B mm
Ans.
Negative sign indicates that end A moves towards end D.
Ans. dA = 3.64(10 3) mm 200
CH 04.indd 122
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4–17. The bar has a length L and crosssectional area A. Determine its elongation due to the force P and its own weight.The material has a specific weight g (weight>volume) and a modulus of elasticity E. L
Solution d = =
L P(x) dx 1 = (gAx + P) dx AE L0 L A(x) E
gAL2 gL2 1 PL a + PL b = + AE 2 2E AE
P
Ans.
Ans. d =
L gL 2 1 PL (gAx + P) dx = + AE L0 2E AE
2 01
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4–18. The assembly consists of three titanium (Ti6A14V) rods and a rigid bar AC. The crosssectional area of each rod figure.If If aaforce forceof of30 6 kip is given in the figure. kN is applied to the ring F, determine the horizontal displacement of point F.
D
4 ftm 1.2
C in2 2
ACD 1 mm ACD � 600
20.6 ft m E
10.3ftm F
6 kip 30 kN 2
�2 1200 in mm EFEF 1 ftmAA 0.3
1.5mm in2 2 AA AB�900 AB
Solution
B
6 ftm 1.8
2
A
Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x
FCD (3) – 6(1) = 0= 0 FCD = =2.00 kip F 30(0.3) FCD 10 kN CD(0.9) 6 30  2.00 – 10 – FF == 0 0 AB AB
FABFAB == 4.00 20 kip kN
0.6 m
FEF = 30 kN
Displacement:
0.3 m
3
= δ F −E
0.6
0.1667 mm
FAB LAB 20(10 3 )(1.8)(10 3 ) = = 0.3333 mm AAB E [900(10 −6 )][120(109 )]
= δA
δ E′
0.3 m
3
FCDLCD 10(10 )(1.2)(10 ) = = 0.1667 mm ACDE [600(10 −6 )][120(109 )]
= δC
0.6 m
=
FEF LEF 30(10 3 )(0.3)(10 3 ) = = 0.0625 mm AEF E [1200(10 −6 )][120(109 )] 0.1667 , 0.9
δ E′ = 0.1111 mm
δ E = δC + δ E′ = 0.1667 + 0.1111 = 0.2778 mm δ= δ E + δ F −E F = 0.2778 + 0.0625 = 0.3403 mm = 0.340 mm
Ans.
Ans. = δ F 0.3403 mm 202
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4–19.
D
1.2 4 ftm
C 2 2 mm
ACD � 600 ACD 1 in
The assembly consists of three titanium (Ti6A14V) rods and a rigid bar AC. The crosssectional area of each rod is given in the figure. If a forceof 30 kN is applied to the ring F, determine the angle of tilt of bar AC.
20.6 ft m E
F 10.3ftm
30 kN 6 kip 2
AA 1.5mm in2 2 AB AB�900 B
1.8 6 ftm
A
Solution Internal Force in the Rods: a + ©MA = 0; + ©F = 0; : x
F (3) – 6(1) = 0= 0 FCD = =2.00 kip FCD 30(0.3) FCD 10 kN CD(0.9) 6 30  –2.00 FAB = 0 10 –FAB = 0
0.6 m
FAB kip FAB= =4.00 20 kN
δC =
10(10 3 )(1.2)(10 3 ) FCDLCD = = 0.1667 mm ACDE [600(10 −6 )][120(109 )]
δA =
20(10 3 )(1.8)(10 3 ) FAB LAB = = 0.3333 mm AAB E [900(10 −6 )][120(109 )]
0.6 m
FEF = 30 kN 0.3 m
Displacement:
0.3 m
0.16666667 mm
0.3333 − 0.1667 −1 δ A − δ C tan −1 θ tan = = LAC 0.9(10 3 ) = 0.01061 = ° 0.0106°
Ans.
Ans.
θ 0.0106° = 203
2
�2 1200 in mm 0.3 EFEF 1 ftmAA
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* 4–20. The pipe is stuck in the ground so that when it is pulled upward the frictional force along its length varies linearly from zero at B to fmax (force/length) at C. Determine the initial force P required to pull the pipe out and the pipe’s elongation just before it starts to slip. The pipe has a length L, crosssectional area A, and the material from which it is made has a modulus of elasticity E.
P B
L
Solution From FBD (a) + c ΣFy = P 0;
1 (F L) = 0 2 max P =
fmax
Fmax L 2
C
Ans.
From FBD (b) +T ΣFy = 0;
P(x) +
Fmax L 1 Fmax x a bx = 0 2 L 2 P(x) =
d = =
L P(x)
dx
L0 A(x)E
=
L0
L
Fmax L 2AE
Fmax L Fmax x2 2 2L L
dx 
Fmax x
2
L0 2AEL
dx
Fmax L2 3AE
Ans.
Ans:
204
P =
Fmax L , 2
d =
Fmax L2 3AE
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4–21. The post is made of Douglas fir and has a diameter of 100 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance distributed around the post that is triangular along its sides; that is, it varies from w = 0 at y = 0 to w = 12 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.
20 kN A y w
2m
12 kN/m B
Solution
F
Equation of Equilibrium: Referring to the FBD of the entire post, Fig. a, + c ΣFy = 0;
F +
1 (12)(2)  20 = 0 2
Ans.
F = 8.00 kN
Normal Force: Referring to the FBD of the upper segment of the post sectioned at arbitrary distance y, Fig. b + c ΣFy = 0;
1 (6y)(y)  20  P(y) = 0 2
Py = ( 3y2  20 ) kN
Displacement: For Douglas Fir, E = 13.1 GPa L
dA>B = =
N(y)dy
L0 A(y)E
=
1 AE L0
2 meters
( 3y2  20 ) dy
2 meters 1 ( y3  20y ) ` AE 0
= = 
32 kN # m AE
32 ( 103 ) p ( 0.12 ) [13.1 ( 109 ) ] 4
=  0.3110 ( 103 ) m =  0.311 mm
Ans.
The sign indicates that end A moves toward end B.
Ans: F = 8.00 kN, dA>B =  0.311 mm 205
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4–22. The post is made of Douglas fir and has a diameter of 100 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 4 kN>m at y = 0 to w = 12 kN>m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post.
20 kN A y w
2m
12 kN/m B
Solution
F
Equation of Equilibrium: Referring to the FBD of the entire post, Fig. a, + c ΣFy = 0;
F +
1 (4 + 12)(2)  20 = 0 2
Ans.
F = 4.00 kN
Normal Force: Referring to the FBD of the upper segment of the post sectioned at arbitrary distance y, Fig. b, + c ΣFy = 0;
(4 + 2y)y  20  P(y) = 0
P(y) = ( 2y2 + 4y  20 ) kN
Displacement: For Douglas Fir, E = 13.1 GPa L
dA>B =
N(y)dy
L0 A(y)E
= =
1 AE L0
2m
( 2y2 + 4y  20 ) dy
2m 1 2 3 a y + 2y2  20yb ` AE 3 0
= 
80 kN # m 3 AE
80(103)
= 3c
p ( 0.12 ) d [13.1 ( 109 ) ] 4
=  0.2592 ( 103 ) m =  0.259 mm
Ans.
Ans: F = 4.00 kN, dA>B = 0.259 mm 206
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4–23. The rigid bar is supported by the pinconnected rod CB that has a crosssectional area of 14 mm2 and is made from 6061T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied.
C 300 N/m 1.5 m A
2m
B
D 2m
Solution a+ ΣMA = 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 = 90.248° u = 90.248°  90° = 0.2478° = 0.004324 rad Ans.
dD = u r = 0.004324(4000) = 17.3 mm
Ans: dD = 17.3 mm 207
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*4–24. The weight of the kentledge exerts an axial force of P = 1500 kN on the 300mm diameter highstrength concrete bore pile. If the distribution of the resisting skin friction developed from the interaction between the soil and the surface of the pile is approximated as shown, determine the resisting bearing force F for equilibrium. Take p0 = 180 kN>m. Also, find the corresponding elastic shortening of the pile. Neglect the weight of the pile.
P p0
12 m
Solution
F
Internal Loading: By considering the equilibrium of the pile with reference to its entire freebody diagram shown in Fig. a. We have + c ©Fy = 0;
F +
1 (180)(12)  1500 = 0 2
F = 420 kN
Ans.
Also, p(y) =
180 y = 15y kN>m 12
The normal force developed in the pile as a function of y can be determined by considering the equilibrium of the sectional of the pile with reference to its freebody diagram shown in Fig. b. + c ©Fy = 0;
1 (15y)y + 420  P(y) = 0 2
P(y) = (7.5y2 + 420) kN
p Displacement: The crosssectional area of the pile is A = (0.32) = 0.0225p m2. 4 We have 12 m P(y)dy (7.5y2 + 420)(103)dy = 0.0225p(29.0)(109) L0 A(y)E L0 L
d =
12 m
=
L0
c 3.6587(10  6)y2 + 0.2049(10  3) d dy
= c 1.2196(10  6)y3 + 0.2049(10  3)y d
12 m 0
= 4.566(10  3) m = 4.57 mm
Ans.
Ans: F = 420 kN, d = 4.57 mm 208
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4–25. Determine the elongation of the aluminum strap when it is subjected to an axial force of 30 kN. Eal = 70 GPa.
15 mm 30 kN 250 mm
50 mm
6 mm 15 mm 800 mm
30 kN 250 mm
Solution d = (2) =
d2 Ph PL ln + Et(d 2  d 1) d 1 AE 2(30) ( 103 ) (250)
(70) ( 10 ) (0.006)(0.05  0.015) 9
= 2.37 mm
aln
30 ( 103 ) (800) 50 b + 15 (0.006)(0.05)(70) ( 109 )
Ans.
Ans: d = 2.37 mm 209
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4–26. The ball is truncated at its ends and is used to support the bearing load P. If the modulus of elasticity for the material is E, determine the decrease in the ball’s height when the load is applied.
P
r 2
r r 2
Solution Displacement: Geometry: A(y) = px2 = p ( r 2  y2 ) Displacement: When x = L
d = = =
r , 2
y = {
23 r 2
P(y) dy
L0 A(y) E 23
r 2 dy P 23 pE L 2 r r 2  y2 23
r + y 2r P 1 c ln d` r p E 2r r  y  23 2
=
P [ ln 13.9282  ln 0.07180 ] 2p r E
=
2.63 P prE
Ans.
Ans: d = 210
2.63 P prE
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4–27. The linkage is made of two pinconnected A36 2 members,each each having a crosssectional steel members, having a crosssectional area ofarea 1.5 inof . 1000 mm2. If force a vertical P is = 250 kN istoapplied to If a vertical of Pforce = 50ofkip applied point A, point A, determine itsdisplacement vertical displacement at A. determine its vertical at A.
P A 0.6 2 ftm B
Solution
C 0.45 1.5 ftm
0.45 1.5 ftm
Analysing the equilibrium of Joint A by referring to its FBD, Fig. a, + ©F = 0 ; : x
3 3 FAC a b  FAB a b = 0 5 5
4 250== 00 2F a b  50 5
+ c ©Fy = 0
The
initial
FAC = FAB = F F  31.25kN kip F ==–156.25
of members AB and AC is 12 in 0.45 ++0.26 == 0.75 m.ft)a The axialb deformation members AB and AC is deformation of members (2.50 = 30 in. Theofaxial L == 21.5 1 ft AB and AC is 22
δ=
length
22
FL (−156.25)(10 3 )(750) = = −0.58594 mm AE [1000(10 −6 )][200(109 )]
The negative sign indicates that end A moves toward B and C. From the geometry 1.5 0.45 shown in Fig. b, u = tan  1 a Thus, bb = 36.87°. 36.87°.Thus, 2 0.6
(δ= A )ν
0.58594 δ = = 0.7324 = mm 0.732 mm ↓ coθθ coθ 36.87°
Ans.
d = 0.58594 mm
P = 250 kN
0.6 m
0.45 m
Ans. (dA)v = 0.732 mm T 211
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*4–28. The linkage is made of two pinconnected A36 2 steel members, members,each each having a crosssectional having a crosssectional area ofarea 1.5 inof . 1000 mm2. Determine the magnitude forcetoPdisplace needed Determine the magnitude of the forceofP the needed to displace point A 0.625 mm downward. point A 0.025 in. downward.
P A
Solution
2 ftm 0.6 B
Analysing the equilibrium of joint A by referring to its FBD, Fig. a + ©F = 0; : x + c ©Fy = 0;
The
initial
3 3 FAC a b  FAB a b = 0 5 5 4 2F a b  P = 0 5
C 1.5 ftm 0.45
FAC = FAB = F
1.5 ftm 0.45
F =  0.625 P
length
of members AB and AC are 12 in deformation of members (2.50 = 30 in. Theofaxial L == 21.5 0.452 2++0.2622 ==0.75 m.ft)a The axialb deformation members AB and AC is 1 ft AB and AC is FL −0.625P(750) δ= = = −2.34375(10 −6 )P AE [1000(10 −6 )][200(109 )] The negative sign indicates that end A moves toward B and C. From the geometry 1.5 0.45 shown in Fig. b, we obtain u = tan  1 a Thus bb = 36.87°. 36.87°.Thus, 2 0.6
(δ= A )ν
0.025 = 0.625 5
δ
coθθ 3 ))P P 0.4310(10 –6 2.34375(10 cos 36.87°
3 = P 213.33(10 = ) N 213 kN
Ans.
(dA)v = 0.625 mm d = 2.34375(10–6)P
0.6 m
0.45 m
Ans. = P 213 kN
212
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4–29. The bar has a crosssectional area of 1800 mm2, and E = 250 GPa. Determine the displacement of its end A when it is subjected to the distributed loading.
x
A
Solution P(x) =
L0
4 ftm 1.5
x
w dx = 500
P(x) dx = AE L0
L0
x
0
1
(
)
x3 dx = 375x 4/3 N
)1.5 m
L
dA =
� 500x1/3 N/m lb/in. w
x 4/3 ) dx ( 375=
[1800(10 −6 )][250(109 )]
1.5 m
3 −6 0.8333(10 −6= ) x 7/3 0.9199(10 = ) m 0.920 mm 7 0
–7 0.0128 in. dA ==9.199(10 )m
Ans.
= 0.00092 mm
Ans: dA = 2.990 mm 213
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4–30. Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.
P d2 t
h
d1
Solution
P
d 1h + (d 2  d 1)x d2  d1 w = d1 + x = h h d =
P(x) dx
L A(x)E
=
P E L0
h
dx [d 1h + (d 2  d 1)x]t h
h
= = = =
Ph dx E t L0 d 1h + (d 2  d 1)x h h d1 h d2  d1 Ph dx Ph = a b c ln a1 + xb d 0 E t d 1 h L0 1 + d 2  d 1 x E t d1 h d2  d1 d1 h 0 d1 h
d2  d1 d1 + d2  d1 Ph Ph c ln a1 + bd = c ln a bd E t(d 2  d 1) d1 E t(d 2  d 1) d1 d2 Ph c ln d E t (d 2  d 1) d1
Ans.
Ans: d =
214
d2 Ph c ln d E t (d 2  d 1) d1
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4–31. The column is constructed from highstrength concrete and six A36 steel reinforcing rods. If it is subjected kN, determine determine the average normal to an axial force force of of 150 30 kip, stress in the concrete and in each rod. Each rod has a 20 mm. diameter of 0.75 in.
1004mm in. 150 kN 30 kip
Solution Equations of Equilibrium: + c ©Fy = 0;
[1]
150== 00 6Pst + Pcon  30
1.2 3 ftm
Compatibility: dst = dcon Pst (1.2) Pcon (1.2) = π (0.02 2 ) [200(109 )] π (0.2 2 ) − 6 π (0.02 2 ) [29.0(109 )] 4 4 4
( )
150 kN
[2]
Pstst ==0.073368 0.064065PP con con Solving Eqs. [1] and [2] yields: Pstst ==7.641 1.388kN kip
kN Pcon ==104.152 21.670 kip
Average Normal Stress:
σ= st
σ= con
Pcon = Acon
Pst 7.641(10 3 ) 2 = = 24.32(106 ) N−m = 24.3 MPa π (0.02 2 ) Ast 4 104.15(10 3 ) π
(0.2 2 ) − 6 4
= 3.527(106 ) N− = m 2 3.53 MPa
( π4 ) (0.022 )
Ans.
Ans.
Ans. sst = 24.3 MPa, scon = 3.53 MPa 215
CH 04.indd 145
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*4–32. The column is constructed from highstrength concrete and six A36 steel reinforcing rods. If it is subjected to an axial force force of of 150 30 kip, kN,determine determine the required diameter of each rod so that onefourth of the load is carried by the concrete and threefourths by the steel.
1004mm in. 150 kN 30 kip
Solution kNisisrequired requiredto todistribute distributein insuch such aa manner manner that that 3/4 3/4 Equilibrium: The force of 150 30 kip of the force is carried by steel and 1/4 of the force is carried by concrete. Hence Pst =
3 (30) (150)== 22.5 112.5kip kN 4
Pcon =
1.2 3 ftm
1 (30) == 7.50 (150) 37.5 kip kN 4
Compatibility: dst = dcon PstL Pcon L = AstEst Acon Econ
Ast =
Pst Acon Econ Pcon Est
( )
(112.5) π4 (0.2 2 ) − 6 π4 (0.02 2 ) 29(109 ) π 6 d2 = 4 (37.5) 200(109 ) d = 0.05221 m = 52.2 mm
Ans.
Ans. d = 52.2 mm 216
CH 04.indd 145
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4–33. The A36 steel pipe has a 6061T6 aluminum core. It is subjected to a tensile force of 200 kN. Determine the average normal stress in the aluminum and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm.
400 mm 200 kN
200 kN
Solution Equations of Equilibrium: + d ΣFx = 0; Pal + Pst  200 = 0
(1)
Compatibility: dal = dst Pal(400) p 4
( 0.072 ) (68.9) ( 109 )
=
Pst(400) p 4
( 0.082  0.072 ) (200) ( 109 ) (2)
Pal = 1.125367 Pst Solving Eqs. (1) and (2) yields:
Pst = 94.10 kN Pal = 105.90 kN
Average Normal Stress: sal =
105.90 ( 103 ) Pal = p = 27.5 MPa 2 Aal 4 ( 0.07 )
sst =
Pst = Ast
94.10 ( 103 ) p 4
( 0.082  0.072 )
Ans. Ans.
= 79.9 MPa
Ans: sal = 27.5 MPa, sst = 79.9 MPa 217
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4–34. The 304 stainless steel post A has a diameter of mmand andisissurrounded surroundedby byaared red brass brass C83400 C83400 tube B. d == 50 2 in. kN is applied Both rest on the rigid rigid surface. surface. If If aa force forceof of25 5 kip to the rigid cap, determine the average normal stress developed in the post and the tube.
5 kip 25 kN
B in. 2008mm
A
B A
3 in. 75 mm
Solution d
Equations of Equilibrium: + c ©Fy = 0;
[1]
Pst + Pbr  25 5 == 0
0.5mm in. 12 25 kN
Compatibility: dst = dbr
Pst (200) Pbπ(200) 5 2 9 2 π (0.05 ) [193(10 )] 4 (0.15 − 0.126 2 ) [101(109 )] 4 π
[2]
Pstst 5 P = 0.72120 0.69738 P Pbr br Solving Eqs. [1] and [2] yields: Pbr ==14.525 2.9457kN kip
= 2.0543 PP = 10.475 kNkip st st
Average Normal Stress:
14.525(10 3 ) 2 = 2.792(106 ) N−m = 2.79 MPa π (0.152 − 0.126 2 ) 4
σ= bπ
Pbπ = Abπ
σ= θt
Pθt 10.475(10 3 ) 2 = = 5.335(106 ) N−m = 5.34 MPa π (0.052 ) Aθt 4
Ans.
Ans.
Ans. sbr = 2.79 MPa, sst = 5.34 MPa 218
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4–35. The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a kN isis applied applied to to the the rigid rigid cap, cap, determine the force of 25 5 kip required diameter d of the steel post so that the load is shared equally between the post and tube.
25 kN 5 kip
B
B 200 8mm in.
A
A
75 mm 3 in.
Solution The force force of of525 is shared equally brass steel. Hence Equilibrium: The kipkN is shared equally by by thethe brass andand steel. Hence
d
12 0.5mm in.
Pst = Pbr = P = 12.5 2.50 kN kip Compatibility: dst = dbr PL PL = AstEst AbrEbr Ast = a
AbrEbr Est
π (0.152 − 0.126 2 ) 101(109 ) p 2 4 bd = 4 193(109 )
Ans.
d 5 0.05888 m = 58.9 mm
Ans. d 5 58.9 mm 219
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*4–36. The A36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm. If it fits snugly between the fixed walls before it is loaded, determine the reaction at the walls when it is subjected to the load shown.
B
A
300 mm
C
8 kN 8 kN 700 mm
Solution + ΣFx = 0; FA + FC  16 = 0 S
(1)
By superposition: + ) 0 =  ∆ + d (S C C 0 =
 16 (300) AE
+
FC (1000) AE Ans.
FC = 4.80 kN
From Eq. (1), Ans.
FA = 11.2 kN
Ans: FC = 4.80 kN, FA = 11.2 kN 220
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The 10mmdiameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the yield stress for the steel is (sY)st = 640 MPa, and for the bronze (sY)br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est = 200 GPa, Ebr = 100 GPa.
P
10 mm
20 mm
Solution + c ΣFy = 0; Pst + Pbr  P = 0
(1)
Assume failure of bolt: p Pst = (sY)st(A) = 640 ( 106 ) a b ( 0.012 ) 4
P
= 50265.5 N
dst = dbr PstL p 4
( 0.012 ) (200) ( 109 )
=
PbrL p 4
( 0.022  0.012 ) (100) ( 109 )
Pst = 0.6667 Pbr 50 265.5 = 0.6667 Pbr Pbr = 75398.2 N From Eq. (1) P = 50265.5 + 75398.2 = 125663.7 N = 126 kN
Ans.
(controls)
Assume failure of sleeve: p Pbr = (sY)br(A) = 520 ( 106 ) a b ( 0.022  0.012 ) = 122 522.11 N 4 Pst = 0.6667 Pbr = 0.6667(122 522.11) = 81 681.4 N From Eq. (1), P = 122 522.11 + 81 681.4 = 204 203.52 N = 204 kN
Ans: P = 126 kN 221
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The 10mmdiameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the bolt is subjected to a compressive force of P = 20 kN, determine the average normal stress in the steel and the bronze. Est = 200 GPa, Ebr = 100 GPa.
P
10 mm
20 mm
Solution + c ΣFy = 0; Pst + Pbr  20 = 0
(1)
dst = dbr PstL p 4
( 0.012 ) (200) ( 109 )
=
PbrL p 4
P
( 0.022  0.012 ) (100) ( 109 ) (2)
Pst = 0.6667 Pbr Solving Eqs. (1) and (2) yields Pst = 8 kN Pbr = 12 kN sst =
8 ( 103 ) Pst = p = 102 MPa 2 Ast 4 ( 0.01 )
sbr =
Pbr = Abr
12 ( 103 ) p 4
( 0.022  0.012 )
Ans.
Ans.
= 50.9 MPa
Ans: sst = 102 MPa, sbr = 50.9 MPa 222
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4–39. If column AB is made from high strength precast concrete and reinforced with four 20 mm diameter A36 steel rods, determine the average normal stress developed in the concrete and in each rod. Set P = 350 kN.
P
P
A
a
225 mm
a 225 mm
3m
Section aa
Solution Equation of Equilibrium: Referring to the freebody diagram of the cut part of the concrete column shown in Fig. a, + c ©Fy = 0;
Pcon + 4Pst  2(350) = 0
B
(1)
Compatibility Equation: Since the steel bars and the concrete are firmly bonded, their deformation must be the same. Thus, dcon = dst Pcon (3) Pst (3) = 0.225(0.225) − 4 π (0.02 2 ) [29.0(109 )] π (0.2 2 ) [200(109 )] 4 4
( )
Pcon = 22.7859Pst
(2)
Solving Eqs. (1) and (2), Pst = 26.13 kN
Pcon = 595.47 kN
Normal Stress: Applying Eq. (16), scon =
Pcon 595.47(10 3 ) 6 = 12.06(10 = ) N m 2 12.1 Mpa = Acon 0.225(0.225) − 4 π (0.02 2 )
(4)
Pst 26.13(10 3 ) 6 83.18(10 = ) N m 2 83.2 Mpa = π sst = = 2 Ast (0.02 ) 4
( )
Ans.
Ans.
Ans: scon = 12.1 MPa , sst = 83.2 MPa 223
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*4–40. If column AB is made from high strength precast concrete and reinforced with four 20 mm diameter A36 steel rods, determine the maximum allowable floor loadings P. The allowable normal stress for the high strength concrete and the steel are (sallow)con = 18 MPa and (sallow)st = 170 MPa, respectively.
P
P
A
a
225 mm
a 225 mm
3m
Solution Equation of Equilibrium: Referring to the freebody diagram of the cut part of the concrete column shown in Fig. a, + c ©Fy = 0;
Pcon + 4Pst  2P = 0
Section aa
B
(1)
Compatibility Equation: Since the steel bars and the concrete are firmly bonded, their deformation must be the same. Thus, dcon = dst Pcon (3)
0.225(0.225) − 4
( 4 ) (0.02 π
2
) [29.0(10 )] 9
=
Pst (3) (0.2 ) [200(109 )] 4
π
2
Pcon = 22.7859Pst
(2)
Solving Eqs. (1) and (2), Pst = 0.07467P
Pcon = 1.7013P
Allowable Normal Stress: (scon)allow =
Pcon ; Acon
18(106) =
1.7013P p 0.225(0.225)  4 a b (0.022) 4
P = 522.31(103) N = 522 kN (controls) (sst)allow =
Pst ; Ast
170(106) =
Ans.
0.07467P p (0.022) 4
P = 715.28(103) N = 715.28 kN
Ans: P = 522 kN 224
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4–41. Determine the support reactions at the rigid supports A and C. The material has a modulus of elasticity of E.
3 d 4
d P B
A 2a
C a
Solution Equation of Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a, + ΣFx = 0; S
(1)
P  FA  FC = 0
Compatibility Equation: Using the method of superposition, Fig. b, + ) d = d  d (S P FC
0 =
FC =
P(2a) p a d 2 bE 4 9 P 17
 ≥
FCa 2 p 3 a db E 4 4
+
FC(2a) p a d 2 bE 4
¥ Ans.
Substituting this result into Eq. (1), FA =
8 P 17
Ans.
Ans:
9 P, 17 8 FA = P 17 FC =
225
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If the supports at A and C are flexible and have a stiffness k, determine the support reactions at A and C. The material has a modulus of elasticity of E.
3 d 4
d P B
A 2a
C a
Solution Equation of Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a, + ΣFx = 0; P  FA  FC = 0 S
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + ) d = d  d (S C P FC FC(2a) P(2a) FC FCa FC P + = ≥ ¥ + ¥  ≥ + 2 k k k p 2 p 2 p 3 a d bE a d bE a db E 4 4 4 4 9(8ka + pd 2E)
136ka + 18pd 2E
FA = a
64ka + 9pd 2E bP 136ka + 18pd 2E
Substituting this result into Eq. (1),
Ans.
d P
FC = c
Ans.
Ans: FC = c FA = a 226
9(8ka + pd 2E) 136ka + 18pd 2E
d P,
64ka + 9pd 2E bP 136ka + 18pd 2E
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4–43. The tapered member is fixed connected at its ends A and B B and and isissubjected subjectedto toaaload loadPP= =357kN kipatatx x= 750 = 30mm in. Determine the reactions at the supports. The material is 50 mm thick is made froom 2014T6 aluminum. 2 in. thick andand is made from 2014T6 aluminum.
A
B P
1506 mm in.
3 in. 75 mm
x
Solution
1500 mm 60 in.
Equilibrium. Refering to the FBD of the member, Fig. a + ©F = 0; : x
FA + FB  735== 00
(1)
Compatibility. From the geometry shown in Fig. b 1.5 − x y= 2 0.0375 + [0.05(3 − x)] m (0.0375) = 1.5
Thus, the crosssectional area of the member as a function of x is A= (0.05)[ 0.05(3 − x)] = [ 0.0025(3 − x)] m 2
It is required that dA>B = 0 0.75 m
– 
0
dx FAA dx ++ 0.0025(3 – x)(E)
0.75 m

0
FA dx 3–x
1.5 m
+
0.75 m
FA ln(3 − x) 0
0.75 m
1.5 m
0.75 m
FB dx 3–x
F FBBdx == 00 0.0025(3 – x)(E)
=0 1.5 m
0 − FB ln(3 − x) 0.75 m =
2.25 1.5 0 FA ln − FB ln = 3 2.25 FA = 1.4094 FB
(2)
Solving Eqs. (1) and (2) FA = 20.47 = 20.5 kN
Ans.
FB = 14.53 = 14.5 kN
Ans.
35 kN FA
FB (a)
y 0.0375 m 0.0375 m
x
1.5 − x (b)
Ans. y = 75  0.025x, FA = 20.47 kN, FB = 14.53 kN 227
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*4–44. The tapered member is fixed connected at its ends A and B and is subjected to a load P. Determine the location x of the load and its greatest magnitude so that the average normal stress in the bar does not exceed σallow = 28 MPa. The number is 50 mm thick.
A
B P
1506 mm in.
3 in. 75 mm
x
Solution
1500 mm 60 in.
Equilibrium. Refering to the FBD of the member, Fig. a + ©F = 0; : x
(1)
FA + FB  P = 0
Compatibility. From the geometry shown in Fig. b 1.5 − x y= 2 0.0375 + [0.05(3 − x)] m (0.0375) = 1.5
Thus, the crosssectional area of the member as a function of x is A= (0.05)[ 0.05(3 − x)] = [ 0.0025(3 − x)] m 2
It is required that dA>B = 0 x
– 
0 x

dx FAA dx ++ 0.0025(3 – x)(E) FA dx
0
3–x
1.5 m
+
1.5 m x
FB dx 3–x
x
x
F FBBdx == 00 0.0025(3 – x)(E)
=0 1.5 m
FA ln(3 − x) 0 − FB ln(3 − x) x
0 =
3 − x 1.5 0 FA ln − FB ln 3 − x = 3
(2)
The maximum normal stress in segment AC occurs at position x and in segment BC occurs at B. Then = σ allow
σ allow =
FA FA ; = 28(106 ) = FA [70(10 3 )(3 − x)] N A 0.0025(3 − x) FB FB ; 28(106 ) = = FB 105(10 3 ) N (0.05)(0.075) A
P
FB (a)
Substitute these results into Eq. (2) 1.5 3− x 3 70(10 3 )(3 − x)ln 0 − 105(10 )ln = 3 3−x 3−x ln 3 3− x 3
3− x
3− x
1.5 = ln 3−x
1.5 = 3−x
C
FA
y
0.0375 m
0.0375 m
1.5 1.5 − x
x (b)
1.5
Solving numerically, x = 0.721698 m = 722 m
Ans.
Then, FA =[70(10 3 )(3 − 0.7216980) =159.48(10 3 ) N
Substitute the results of FA and FB into Eq. (1) 159.48(10 3 ) + 105(10 3 ) − P = 0 3 P 264.48(10 = = ) 264 N
Ans.
Ans. x = 722 m, P = 264 N
228
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4–45. The rigid bar supports the uniform distributed load kN/m.Determine Determinethe theforce force in in each each cable cable if each cable of 90 6 kip>ft. 3 mm 200 GPa. has a crosssectional crosssectional area areaof of36 andEE= = 31110 0.05 in22, ,and 2 ksi.
C
ft 26 m
6 kip/ft 90 kN/m
Solution
A
Equilibrium: a + ©MA = 0;
TCB a
2 25
(1) –270(1.5) b (3) 54(4.5) + TCD a
Geometry and compatibility: u = tan  1
D
B ft 13 m
2 25
(3)==00 b9
ft 13 m
ft 13 m
(1) 270 kN
62 = 45° 62
1.5 m
1.5 m
2 (8.4853)22– 2(1)(2.8284) 2(3)(8.4853) cos cos u9 LB¿C = (1) (( 2 22++ (2.8284)
Also, 2.8284 m
(3)2 ++ (2.8284) cos u9 (8.4853)22– 2(3)(2.8284) 2(9)(8.4853) cos L2D¿C = (9)
(2)
Thus, eliminating cos u¿ . L2B¿C (0.176778)
2m 5m
5m 1m
1m
+ 1.590978 =
L2B¿C (0.176778) L2B¿C
=
=
–L2D'C(0.058926)
2 0.058926LD'C
2 0.3333LD'C
+ 1.001735
+ 0.589243
+ 3.333
But, LB¿C = 5 + δBC,
LD¿C = 5 + δDC,
Neglect squares or d¿ B since small strain occurs. 2 2 + d+BCd)BC =) 5 =+ 45 2 5+ d2BC L2B¿C = ( 545 2 2 245 dDC L2D¿C = ( 245 =) 5=+ 45 2 5+ d2DC 5 + d+DCd)DC
5 + 2 5δ BC = 0.3333(5 + 2 5δ DC ) + 3.333 2 5δ BC = 0.3333(2 5δ DC )
dDC = 3dBC Thus, TCD 245 TCBTCB 5 5 245 5 3 = 3 AE AE AE AE TCD = 3 TCB From Eq. (1). TCD = 135.84 kN = 136 kN
Ans. Ans.
TCB = 45.28 kN = 45.3 kN
Ans. TCD = 136 kN, TCB = 45.3 kN
229
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4–46. The rigid bar is originally horizontal and is supported by two cables each having a crosssectional area and EE ==31110 200 32GPa. of 36 0.05mm in2,2, and ksi. Determine the slight rotation of the bar when the uniform load is applied.
C
ft 26 m
Solution See solution of Prob. 451.
6 kip/ft 90 kN/m
TCD = 135.84 27.1682kN kip
A D
B
(135.84)(10 3 ) 5 TCDLCD δ DC = = = 0.04219 m AE [36(10 −6 )][200(109 )]
ft 13 m
ft 13 m
ft 13 m
Using Eq. (2) of Prob. 451, ( 5 + 0.04219)2 =+ (3)2 (2.8284)2 − 2(3)(2.8284)coθθ ′ = θ ′ 45.9023°
Thus, Ans.
¢u = 45.9023° – 45° = 0.9023° = 0.902°
Ans. ¢u = 0.902° 2 30
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4–47. The specimen represents a filamentreinforced matrix system made from plastic (matrix) and glass (fiber). If there are n fibers, each having a crosssectional area of Af and modulus of Ef, embedded in a matrix having a crosssectional area of Am and modulus of Em, determine the stress in the matrix and in each fiber when the force P is applied on the specimen.
P
P
Solution + c ΣFy = 0;
(1)
 P + Pm + Pf = 0
dm = df PfL PmL = ; AmEm nAfEf
Pm =
AmEm P nAfEf f
(2)
Solving Eqs. (1) and (2) yields Pm =
AmEm P; nAfEf + AmEm
Pf =
nAfEf nAfEf + AmEm
P
Normal stress:
sm =
sf =
Pm = Am Pf nAf
=
a a
AmEm  Pb nAfEf + AmEm Am
nAfEf nAfEf + AmEm nAf
Pb
=
=
Em P nAfEf + AmEm Ef
nAfEf + AmEm
Ans.
Ans.
P
Ans: sm = sf =
231
Em P, nAfEf + AmEm Ef nAfEf + AmEm
P
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*4–48. The rigid beam is supported by the three suspender bars. Bars AB and EF are made of aluminum and bar CD is made of steel. If each bar has a crosssectional area of 450 mm2, determine the maximum value of P if the allowable stress is (sallow)st = 200 MPa for the steel and (sallow)al = 150 MPa for the aluminum. Est = 200 GPa, Eal = 70 GPa.
B al
D
F
st A
2m
al C
E
0.75 m 0.75 m 0.75 m 0.75 m P
Solution
2P
Equation of Equilibrium: Referring to the FBD of the rigid beam Fig. a, a+ ΣMA = 0;
FCD(1.5) + FEF(3)  P(0.75)  2P(2.25) = 0 (1)
1.5 FCD + 3 FEF = 5.25 P a+ ΣME = 0;
2 P(0.75) + P(2.25)  FCD(1.5)  FAB(3) = 0 (2)
1.5 FCD + 3 FAB = 3.75 P Compatibility: Referring to the displacement diagram of the rigid beam, Fig. b, dCD  dAB dEF  dAB = 1.5 3 2dCD = dEF + dAB 2a
FCDL A 3 200 ( 10
9
)4
b =
FCD =
FEFL A 3 70 ( 10
9
)4
+
FABL A 3 70 ( 109 )4
10 (F + FAB) 7 EF
(3)
Solving Eqs. (1), (2) and (3), FEF = 0.8676 P
FAB = 0.3676 P
FCD = 1.7647 P
Assuming that bar EF fails. Then (sallow)al =
FEF ; A
150 ( 106 ) =
0.8676 P 450 ( 106 )
P = 77.80 ( 103 ) N = 77.80 kN Assuming that bar CD fails. Then (sallow)st =
FCD ; A
200 ( 106 ) =
1.7647 P 450 ( 106 )
P = 51.0 ( 103 ) N = 51.0 kN(control!) Ans.
Ans: P = 51.0 kN 232
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If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force P = 200 kN is applied. The assembly is made of solid A36 steel cylinders.
600 mm
600 mm
0.15 mm
P A
50 mm
D B
25 mm
C
Solution Equation of Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a, + ΣFx = 0; 200 ( 103 )  FD  FA = 0 S
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + ) d = dP  dFD (S 0.15 =
200(103)(600) p (0.052)(200)(109) 4
FD (600) FD (600)  C + S p p (0.052)(200)(109) (0.0252)(200)(109) 4 4 Ans.
FD = 20 365.05 N = 20.4 kN Substituting this result into Eq. (1),
Ans.
FA = 179 634.95 N = 180 kN
Ans: FD = 20.4 kN, FA = 180 kN 233
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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 shown, determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials.
P A
1 mm
0.25 m
Solution
60 mm 80 mm
Require,
10 mm
dst = dbr + 0.001 Fst(0.25) p[(0.05)2  (0.04)2]193(109)
=
Fbr(0.25) p(0.03)2(101)(109)
+ 0.001
0.45813 Fst = 0.87544 Fbr + 106
(1)
+ c ΣFy = 0;
(2)
Fst + Fbr  P = 0
Assume brass yields, then (Fbr)max = sg Abr = 70(106)(p)(0.03)2 = 197 920.3 N (Pg)br = sg >E =
70.0(106) 101(109)
= 0.6931(10  3) mm>mm
dbr = (Pg)brL = 0.6931(10  3)(0.25) = 0.1733 mm 6 1 mm Thus only the brass is loaded. Ans.
P = Fbr = 198 kN
Ans: P = 198 kN 234
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The assembly consists of two red brass C83400 copper rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rod EF of diameter 40 mm, and a rigid cap G. If the supports at A, C, and F are rigid, determine the average normal stress developed in the rods.
300 mm
450 mm 40 kN
A
30 mm
B
E
F 40 mm
C
30 mm
D
40 kN G
Solution Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the freebody diagram of the assembly shown in Fig. a, + ΣFx = 0; S
2F + FEF  2 3 40 ( 103 )4 = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b, + ) 0 =  dP + dEF (S 0 = 
40(103)(300) p (0.032)(101)(109) 4
+ ≥
1 FEF >2 2 (300) FEF (450) ¥ + p p (0.042)(193)(109) (0.032)(101)(109) 4 4
FEF = 42 483.23 N
Substituting this result into Eq. (1), F = 18 758.38 N Normal Stress: We have, sAB = sCD =
sEF =
F 18 758.38 = = 26.5 MPa p ACD (0.032) 4
Ans.
FEF 42 483.23 = = 33.8 MPa p AEF (0.042) 4
Ans.
Ans: sAB = sCD = 26.5 MPa, sEF = 33.8 MPa 235
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* The bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter of 40 mm and an outer diameter of 50 mm. The bolt and sleeve are made of A36 steel and are secured to the rigid brackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm, determine the tension in the bolt when a force of 50 kN is applied to the brackets.
200 mm 25 kN
25 kN B
A
25 kN
25 kN 220 mm
Solution Equation of Equilibrium: + ΣFx = 0; S
Pb + Ps  25  25 = 0 (1)
Pb + Ps  50 = 0 Compatibility: db = ds Pb(220) p 4
( 0.02 ) 200 ( 10 ) 2
9
=
Ps(200) p 4
( 0.05  0.042 ) (200) ( 109 ) 2
(2)
Pb = 0.40404 Ps Solving Eqs. (1) and (2) yields: Ps = 35.61 kN
Ans.
Pb = 14.4 kN
Ans: Pb = 14.4 kN 236
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4–53. The 2014T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. If the assembly fits snugly between the rigid supports so that there is no gap at C, determine the support reactions when the axial force of 400 kN is applied. The assembly is attached at D.
D A 400 mm
400 kN B
A992 steel
800 mm a
Solution
a
50 mm 25 mm
Equation of Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a,
C
FD + (FC)al + (FC)st  400(103) = 0
+ c ΣFy = 0;
2014–T6 aluminum alloy
Section a–a
(1)
Compatibility Equation: Using the method of superposition, Fig. b, (+ T) 0 = +
0 = dp  dFC 400(103)(400) 2
9
p(0.025 )(73.1)(10 )
 £
(FC)al(800) 2
9
p(0.025 )(73.1)(10 )
+
[(FC)al + (FC)st](400) p(0.0252)(73.1)(109)
400(103) = 3(FC)al + (FC)st
§ (2)
Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their deformation must be the same. Thus, (dBC)st = (dBC)al (FC)st(800) 2
2
9
p(0.05  0.025 )(200)(10 )
=
(FC)al(800) p(0.0252)(73.1)(109)
(FC)st = 8.2079(FC)al
(3)
Solving Eqs. (1) and (2), (FC)al = 35.689 kN
(FC)st = 292.93 kN
Substituting these results into Eq. (1), Ans.
FD = 71.4 kN Also, FC = (FC)st + (FC)al = 35.689 + 292.93
Ans.
= 329 kN
Ans: FD = 71.4 kN, FC = 329 kN 237
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4–54. The 2014T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. When no load is applied to the assembly, the gap between end C and the rigid support is 0.5 mm. Determine the support reactions when the axial force of 400 kN is applied.
D A 400 mm
400 kN B
A992 steel
800 mm a
Solution
a
50 mm 25 mm
Equation of Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a,
C
FD + (FC)al + (FC)st  400(103) = 0
+ c ΣFy = 0;
2014–T6 aluminum alloy
Section a–a
(1)
Compatibility Equation: Using the method of superposition, Fig. b, (+ T)
dC = dP  dFC
0.5 = +
400(103)(400) 2
9
p(0.025 )(73.1)(10 )
 ≥
(FC)al (800) 2
9
p(0.025 )(73.1)(10 )
+
[(FC)al + (FC)st](400) p(0.0252)(73.1)(109)
220.585(103) = 3(FC)al + (FC)st
¥ (2)
Also, since the aluminum rod and steel tube of segment BC are firmly bonded, their deformation must be the same. Thus, (dBC)st = (dBC)al (FC)st (800) 2
2
9
p(0.05  0.025 )(200)(10 )
=
(FC)al (800) p(0.0252)(73.1)(109) (3)
(FC)st = 8.2079(FC)al Solving Eqs. (2) and (3), (FC)al = 19.681 kN
(FC)st = 161.54 kN
Substituting these results into Eq. (1), Ans.
FD = 218.777 kN = 219 kN Also, FC = (FC)al + (FC)st = 19.681 + 161.54
Ans.
= 181 kN
Ans: FD = 219 kN, FC = 181 kN 238
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4–55. The three suspender bars are made of A992 steel and have equal crosssectional areas of 450 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to the loading shown.
A 2m
C
B 80 kN
50 kN E
D 1m
1m
1m
F 1m
Solution Referring to the FBD of the rigid beam, Fig. a, + c ΣFy = 0; a+ ΣMD = 0;
FAD + FBE + FCF  50(103)  80(103) = 0
(1)
FBE(2) + FCF(4)  50(103)(1)  80(103)(3) = 0
(2)
Referring to the geometry shown in Fig. b, dBE = dAD + a dBE =
dCF  dAD b(2) 4
1 1 d + dCF 2 2 AD
FCF L FBE L 1 FADL = a + b AE 2 AE AE
(3)
FAD + FCF = 2 FBE
Solving Eqs. (1), (2), and (3) yields FBE = 43.33(103) N
FAD = 35.83(103) N
FCF = 50.83(103) N
Thus, sBE =
43.33(103) FBE = = 96.3 MPa A 0.45(10  3)
Ans.
sAD =
35.83(103) FAD = = 79.6 MPa A 0.45(10  3)
Ans. Ans.
sCF = 113 MPa
Ans: sBE = 96.3 MPa, sAD = 79.6 MPa, sCF = 113 MPa 239
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*4–56. The three A36 steel wires each have a diameter of 2 mm and unloaded lengths of LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the force in each wire after the 150kg mass is suspended from the ring at A.
B
C
D
5
5
4
4 3
3
A
Solution Equations of Equilibrium: + ΣFx = 0; S + c ΣFy = 0;
3 3 F  FAB = 0 5 AD 5
FAD = FAB = F
4 2a F b + FAC  150(9.81) = 0 5
(1)
1.6F + FAC  1471.5 = 0
Compatibility: dAD = dAC cos u Since the displacement is very small, cos u = dAD = F(2) AE
=
4 5
4 d 5 AC 4 FAC (1.6) c d 5 AE
(2)
F = 0.640 FAC Solving Eqs. (1) and (2) yields: FAC = 727 N
Ans.
FAB = FAD = F = 465 N
Ans.
Ans: FAC = 727 N, FAB = 465 N 240
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4–57. The A36 steel wires AB and AD each have a diameter of 2 mm and the unloaded lengths of each wire are LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the required diameter of wire AC so that each wire is subjected to the same force when the 150kg mass is suspended from the ring at A.
B
C
D
5
5
4
4 3
3
A
Solution Equations of Equilibrium: Each wire is required to carry the same amount of load. Hence FAB = FAC = FAD = F Compatibility: dAD = dAC cos u Since the displacement is very small, cos u = dAD = F(2) p 4
( 0.0022 ) E
=
4 5
4 d 5 AC F(1.6) p 2 4 dACE
Ans.
d AC = 0.001789 m = 1.79 mm
Ans: d AC = 1.79 mm 241
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4–58. The post is made from 606lT6 aluminum and has a diameter of 50 mm. It is fixed supported at A and B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally uncompressed, determine the reactions at A and B when the force P = 40 kN is applied to the collar.
A 0.25 m
P C
0.25 m
Solution
k 200 MN/m B
Equations of Equilibrium: FA + FB + Fsp  40 ( 103 ) = 0
+ c ΣFy = 0;
(1)
Compatibility: 0 = dp  dB 0 =
40 ( 103 ) (0.25) p 4
( 0.052 ) 68.9 ( 109 ) ( FB + Fsp ) (0.25) + 2 9 4 ( 0.05 ) 68.9 ( 10 )
£p
FB + p 2 ( ) 0.05 68.9 ( 109 ) 4 0.25
Fsp + 200 ( 106 )
§ (2)
FB + Fsp = 23119.45 Also, dsp = dBC Fsp 200 ( 10
6
)
=
FB + p 2 9 4 ( 0.05 ) 68.9 ( 10 ) 0.25
Fsp + 200 ( 106 ) (3)
FB = 2.7057 Fsp Solving Eq. (2) and (3) yields Fsp = 6238.9 N
Ans.
FB = 16880.6 N = 16.9 kN Substitute the results into Eq. (1)
Ans.
FA = 16880.6 N = 16.9 kN
Ans: FB = 16.9 kN, FA = 16.9 kN 242
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4–59. The post is made from 606lT6 aluminum and has a diameter of 50 mm. It is fixed supported at A and B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally uncompressed, determine the compression in the spring when the load of P = 50 kN is applied to the collar.
A 0.25 m
P C
0.25 m
Solution
k 200 MN/m B
Compatibility: 0 = dp  dB 0 =
50 ( 103 ) (0.25) p 4
( 0.052 ) 68.9 ( 109 ) ( FB + Fsp ) (0.25) + 2 9 4 ( 0.05 ) 68.9 ( 10 )
£p
FB + p 2 ( 0.05 ) 68.9 ( 109 ) 4 0.25
Fsp + 200 ( 106 )
FB + Fsp = 28899.31
§
(1)
Also, dsp = dBC Fsp 200 ( 10
6
)
=
FB + p 2 9 4 ( 0.05 ) 68.9 ( 10 ) 0.25
Fsp + 200 ( 106 ) (2)
FB = 2.7057 Fsp Solving Eqs. (1) and (2) yield Fsp = 7798.6 N FB = 21100.7 N Thus, dsp =
Fsp k
=
7798.6 200 ( 106 )
= 0.0390 ( 10  3 ) m = 0.0390 mm
Ans.
Ans: dsp = 0.0390 mm 243
2018 Pearson Education, Ltd. All rights This is protected all copyright laws as they currently laws exist.asNo portion ©© 2010 Pearson Education, Inc., Upper Saddlereserved. River, NJ. Allmaterial rights reserved. Thisunder material is protected under all copyright they currently this material be reproduced, in any form any or means, without in writingin from the publisher. exist. Noofportion of thismay material may be reproduced, in or anybyform by any means,permission without permission writing from the publisher. *4–60. The press consists of two rigid heads that are held 1 together by 12mmdiameter rods.A A 6061bythe thetwo twoA36 A36steel steel 2 in.diameter rods. T6solidaluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. If it is then tightened onehalf turn, determine the average normal stress in the rods and in the cylinder. The singlethreaded screw screw on on the thebolt bolthas hasaalead leadofof0.25 0.01mm in. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.
300 12 mm in.
50 mm 2 in.
250 10 mm in.
Solution Equilibrium: + ©F = 0; : x
2Fst  Fal = 0
Compatibility: dst = 0.125(10−3)  dal Fst (0.3) Fal (0.25) = 0.125(10 −3 ) − π (0.012 2 ) [200(109 )] π (0.052 ) [68.9(109 )] 4 4 = 13.2629 Fst 125(10 3 ) − 1.8480 Fal
Solving, Fst = 7.3708(103) N Fal = 14.7416(103) N Normal stress: srod =
7.3708(10 3 ) Fst 6 = 65.17(10 = ) N/m 2 65.2 MPa = π 2 (0.012 ) Ast 4
Ans.
scyl =
14.7416(10 3 ) Fal 6 = 7.508(10 = ) N/m 2 7.51 MPa = π 2 Aal (0.05 ) 4
Ans.
( )
( )
Ans. srod = 65.2 MPa, scyl = 7.51 MPa
0.125 (10−3) m
2 44
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300 12 mm in.
4–61. The press consists of two rigid heads that are held together by the two A36 steel 12mmdiameter rods. A 6061T6solidaluminum cylinder is placed in the press and the screw is adjusted so that it just presses up against the cylinder. Determine the angle through which the screw can be turned before the rods or the specimen begin to yield. The singlethreaded screw on the bolt has a lead of 0.25 mm. Note: The lead represents the distance the screw advances along its axis for one complete turn of the screw.
50 mm 2 in.
Solution
250 10 mm in.
Equilibrium: + ©F = 0; : x
2Fst  Fal = 0
Compatibility: dst = d  dal Fst (0.3) Fal (0.25) = d− 2 9 π (0.052 ) [68.9(109 )] (0.012 ) [200(10 )] 4 4 π
= Fst (109 )d − 1.8480 Fal 13.2629
(1)
Assume steel yields first, 6 = σ Y 250(10 = )
Fst
; Fst 28.27(10 3 ) N =
( π4 ) (0.0122 )
Then Fal = 56.55(10 3 ) N = σ al
56.55(10 3 ) 6 = 28.80(10 = ) N/m 2 28.80 MPa π (0.052 ) 4
( )
4.50 6 37 ksi MPa, steel yields first asfirst assumed. From From Eq. (1), 28.80ksi MPa < 255 steel yields as assumed. Eq. (1), d = 0.4795(10 −3 ) m = 0.4795 mm Thus, u 0.47950 = 360° 0.01 0.25 Ans.
u = 690.48° = 690°
Ans. u = 690° 245
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4–62. The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a crosssectional area of 600 mm2, and the spring has a stiffness of k = 2 MN >m and an unstretched length of 1.02 m, determine the force in each post after the load is applied to the bar.
50 kN/m
A
C
B 1m
k
Solution
1m
1m
Equations of Equilibrium: a+ ΣMC = 0; + c ΣFy = 0;
FB (1)  FA(1) = 0
FA = FB = F
2F + Fsp  100 ( 10
3
) =0
(1)
Compatibility:
(+T)
dA + 0.02 = dsp F(1) 600 ( 10  6 ) 9.65 ( 109 )
+ 0.02 =
Fsp 2.0 ( 106 )
0.1727F + 20 ( 103 ) = 0.5 Fsp
(2)
Solving Eqs. (1) and (2) yields: Ans.
FA = FB = F = 25581.7 N = 25.6 kN Fsp = 48 836.5 N
Ans: FA = FB = 25.6 kN 246
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4–63. The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a crosssectional area of 600 mm2, and the spring has a stiffness of k = 2 MN >m and an unstretched length of 1.02 m, determine the vertical displacement of A and B after the load is applied to the bar.
50 kN/m
A
C
B 1m
k
Solution
1m
1m
Equations of Equilibrium: a+ ΣMC = 0;
FB (1)  FA(1) = 0 2F + Fsp  100 ( 10
3
+ c ΣFy = 0;
FA = FB = F
) =0
(1)
Compatibility:
(+T)
dA + 0.02 = dsp F(1)
600 ( 10  6 ) 9.65 ( 109 )
+ 0.02 =
Fsp 2.0 ( 106 )
0.1727F + 20 ( 103 ) = 0.5 Fsp
(2)
Solving Eqs. (1) and (2) yields: F = 25 581.7 N Fsp = 48 836.5 N Displacement: dA = dB = =
FL AE
25 581.7(1000) 600 ( 10  6 ) (9.65) ( 109 )
Ans.
= 4.42 mm
Ans: dA = dB = 4.42 mm 247
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*4–64. P
The assembly consists of two posts AB and CD each made from material 1 having a modulus of elasticity of E1 and a crosssectional area A1, and a central post made from material 2 having a modulus of elasticity E2 and crosssectional area A2. If a load P is applied to the rigid cap, determine the force in each material.
d
d
A
E 1
B
C 1
2 F
L D
Solution Equilibrium: + c ΣFy = 0;
(1)
2F1 + F2  P = 0
Compatibility: d = d1 = d2 F1 L F2L = A1 E1 A2 E2 Solving Eq. (1) and (2) yields: F1 = a F2 = a
F1 = a
A1 E1 bF A2 E2 2
(2)
A1 E1 bP 2 A1 E1 + A2 E2
Ans.
A2 E2 bP 2A1 E1 + A2 E2
Ans.
Ans:
A1 E1 bP, 2 A1 E1 + A2 E2 A2 E2 F2 = a bP 2A1 E1 + A2 E2 F1 = a
248
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4–65. P
The assembly consists of two posts AB and CD each made from material 1 having a modulus of elasticity of E1 and a crosssectional area A1, and a central post EF made from material 2 having a modulus of elasticity E2 and a crosssectional area A2. If posts AB and CD are to be replaced by those having a material 2, determine the required crosssectional area of these new posts so that both assemblies deform the same amount when loaded.
d
d
A
E 1
B
C 1
2 F
L D
Solution + c ΣFy = 0;
(1)
2F1 + F2  P = 0
Compatibility: din = d1 = d2 F1 L F2L = A1 E1 A2 E2
F1 = a
Solving Eq. (1) and (2) yields: F1 = a din
A1 E1 bP 2 A1 E1 + A2 E2
F2 L = = A2 E2
F2 = a
1 2A EA +E A E 2 PL 2
1
2
1
2
2
A2 E2
=
A1 E1 bF A2 E2 2
(2)
A2 E2 bP 2A1 E1 + A2 E2 PL 2A1 E1 + A2 E2
Compatibility: When material 1 has been replaced by material 2 for two side posts, then dfinal = d1 = d2 F1 L F2 L = A′1 E2 A2 E2
F1 = a
A′1 bF A2 2
(3)
Solving for F2 from Eq. (1) and (3) F2 = a dfinal = Requires,
A2 bP 2A′1 + A2 F2 L = A2 E2
1 2A′A+ 2
1
A2
2 PL
A2 E2
=
PL E2 ( 2A′1 + A2 )
din = dfinal PL PL = 2A1 E1 + A2 E2 E2 ( 2A′1 + A2 ) A′1 = a
E1 bA E2 1
Ans.
Ans: A′1 = a 249
E1 bA E2 1
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4–66. P
The assembly consists of two posts AB and CD each made from material 1 having a modulus of elasticity of E1 and a crosssectional area A1, and a central post EF made from material 2 having a modulus of elasticity E2 and a crosssectional area A2. If post EF is to be replaced by one having a material 1, determine the required crosssectional area of this new post so that both assemblies deform the same amount when loaded.
d
d
A
E 1
B
C 1
2 F
L D
Solution + c ΣFy = 0;
(1)
2F1 + F2  P = 0
Compatibility: din = d1 = d2 F1 L F2L = A1 E1 A2 E2
F1 = a
Solving Eq. (1) and (2) yields: F1 = a din
A1 E1 bP 2 A1 E1 + A2 E2
F2 L = = A2 E2
F2 = a
1 2A EA +E A E 2 P 2
1
2
1
2
2
A2 E2
=
A1 E1 bF A2 E2 2
(2)
A2 E2 bP 2A1 E1 + A2 E2
PL 2A1 E1 + A2 E2
Compatibility: When material 2 has been replaced by material 1 for central posts, then dfinal = d1 = d2 F2 L F1 L = A1 E1 A′2E1
F2 = a
A′2 bF A2 1
(3)
Solving for F1 from Eq. (1) and (3) F1 = a dfinal Requires,
A1 bP 2A1 + A2′
F1 L = = A1E1
1 2A A+
1
1
A′2
2 PL
A1 E1
=
PL E1 ( 2A1 + A′2 )
din = dfinal PL PL = 2A1E1 + A2 E2 E1 ( 2A1 + A′2 ) A′2 = a
E2 bA E1 2
Ans.
Ans: A′2 = a 250
E2 bA E1 2
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4–67. The wheel is subjected to a force of 18 kN from the axle. Determine the force in each of the three spokes. Assume the rim is rigid and the spokes are made of the same material, and each has the same crosssectional area.
B
120
C
0.4 m
A
120
18 kN
D
Solution Equations of Equilibrium: + ΣFx = 0; S
FAC cos 30°  FAD cos 30° = 0 FAC = FAD = F
+ c Σ Fy = 0; FAB + 2F sin 30°  18 = 0 (1)
FAB + F = 18 Compatibility: dAC = dAB cos 60° F(0.4) AE
=
FAB (0.4)
AE F = 0.5FAB
cos 60° (2)
Solving Eq. (1) and (2) yields: FAB = 12.0 kN (T)
Ans.
FAC = FAD = F = 6.00 kN (C)
Ans.
Ans: FAB = 12.0 kN (T) FAC = FAD = 6.00 kN (C) 251
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*4–68. The C83400redbrass rod AB and 2014T6aluminum rod BC are joined at the collar B and fixed connected at their ends. If there is no load in the members when T1 = 10°C, determine the average normal stress in each member when T2 = 45°C. Also, how far will the collar be displaced? The crosssectional area of each member is 1130 mm2.
B
A
1m
C
0.6 m
Solution Equilibrium: ©Fx = 0;
Fbr = Fal = F
Compatibility:
dA>C = 0 
FalLBC Fbr LAB + abr¢T LAB + aal ¢T LBC = 0 AAB Ebr ABCEal F(1)

[(1130)(10−6)][101(109)]

[(1130)(10−6)][73.1(109)]
+ 18(10 6)(45  10)(1)
F(0.6)
+ 23(10  6)(45  10)(0.6) = 0
F = 69.45(103) N
sbr = sal =
69.45(103) = 61.46(106) N/m2 = 61.5 MPa 1130(10−6)
61.46 MPa < (sg)al and (sg)br
dB = 
[69.45(103)][(1)(103)] [1130(10−6)][101(109)]
Ans. OK
+ 18(10  6)(45  10)[1(103)]
dB = 0.02147 mm = 0.0215 mm :
Ans.
Ans. sbr = 61.5 MPa dB = 0.0215 mm 252
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4–69. Three bars each made of different materials are connected together and placed between two walls when the temperature is T1 = 12°C. Determine the force exerted on the (rigid) supports when the temperature becomes T2 = 18°C. The material properties and crosssectional area of each bar are given in the figure.
Steel Copper Brass Est � 200 GPa Ebr � 100 GPa Ecu � 120 GPa ast � 12(10�6)/�C abr � 21(10�6)/°C acu � 17(10�6)/�C Ast � 200 mm2
300 mm
Solution + ) (;
Abr � 450 mm2
200 mm
Acu � 515 mm2
100 mm
0 = ¢T  d
0 = 12(10  6)(6)(0.3) + 21 (10  6)(6)(0.2) + 17 (10  6)(6)(0.1) 
F(0.3) 6
9
200(10 )(200)(10 )

F(0.2) 6
9
450(10 )(100)(10 )

F(0.1) 515(10  6)(120)(109) Ans.
F = 4203 N = 4.20 kN
Ans. 0 = ∆T  d, F = 4.20 kN 253
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4–70. The steel bolt has a diameter of 7 mm and fits through an aluminum sleeve as shown. The sleeve has an inner diameter of 8 mm and an outer diameter of 10 mm. The nut at A is adjusted so that it just presses up against the sleeve. If the assembly is originally at a temperature of T1 = 20°C and then is heated to a temperature of T2 = 100°C, determine the average normal stress in the bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast = 14(106)>°C, aal = 23(106)>°C.
A
Compatibility: (ds)T  (db)T = (ds)F + (db)F 23(10  6)(100  20)L  14(10  6)(100  20)L =
p 2 4 (0.01
FL +  0.0082)70(109)
FL p 2 9 4 (0.007 )200(10 )
F = 1133.54 N Average Normal Stress: 1133.54 = 40.1 MPa  0.0082)
ss =
F = As
sb =
F 1133.54 = 29.5 MPa = p 2 Ab 4 (0.007 )
Ans.
p 2 4 (0.01
Ans.
Ans. ss = 40.1 MPa, sb = 29.5 MPa 254
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4–71. The AM1004T61 magnesium alloy tube AB is capped with a rigid plate E. The gap between E and end C of the 6061T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the normal stress developed in the tube and the rod if the temperature rises to 80° C. Neglect the thickness of the rigid cap.
25 mm a
Section aa
E B
A a
Solution
300 mm
20 mm
C
25 mm
D
0.2 mm 450 mm
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10  6)(80  30)(300) = 0.39 mm and
A dT)CD = aal ¢TLCD = 24(10  6)(80  30)(450) = 0.54 mm. Referring deformation diagram of the tube and the rod shown in Fig. a, d =
C A dT B AB  A dF B AB D + C A dT B CD  A dF B CD D
0.2 = C 0.39 
F(300)
p A 0.025  0.02 B (44.7)(10 ) 2
2
9
S + C 0.54 
F = 32 017.60 N
to
the
F(450) p 4
A 0.0252 B (68.9)(109)
S
Normal Stress: sAB = sCD =
F 32 017.60 = = 45.3 MPa AAB p A 0.0252  0.022 B
Ans.
F 32 017.60 = = 65.2 MPa p 2 ACD 4 A 0.025 B
Ans.
F = 107 442.47 N
Ans. sAB = 45.3 MPa, sCD = 65.2 MPa, F = 107,442.47 N 255
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*4–72. The AM1004T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is at 30° C. Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap.
25 mm a
Section aa
E B
A a
Solution
300 mm
20 mm
C
25 mm
D
0.2 mm 450 mm
Then sCD =
F 107 442.47 = = 218.88MPa 6 (sY)al p 2 ACD 4 A 0.025 B
(O.K.!)
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a free expansion of A dT B AB = amg ¢TLAB = 26(10  6)(T  30)(300) = 7.8(10  6) (T  30) and
A dT B CD = aal ¢TLCD = 24(10  6)(T  30)(450) = 0.0108(T  30).
Referring to the deformation diagram of the tube and the rod shown in Fig. a, d =
C A dT B AB  A dF B AB D + C A dT B CD  A dF B CD D
0.2 = C 7.8(10  3)(T  30) 
+ C 0.0108(T  30) T = 172° C
107 442.47(300) p A 0.0252  0.022 B (44.7)(109)
107 442.47(450) p 4
A 0.0252 B (68.9)(109)
S
S
Ans.
Ans. T = 172° C 256
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4–73. The pipe is made of A992 steel and is connected to the collars at A and B. When the temperature is 15°C, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to vary along the pipe defined by T = (35 + 30x)C, where x is in meter, determine the average normal stress in the pipe. The inner diameter is 50 mm, the wall thickness is 4 mm.
A
B 2.4 m
Solution Compatibility: The crosssectional area is A 5 0 = dT  dF
Where
p 2 2 –6 2 (8 – )= 216(10 ) p m . 4
dT =
L0
L
a ¢T dx and ¢T x x C.
2.4m 8ft
10)  6 B 0 == 12(10 6.60 A–6 0
F (2.4) + 15 (20(40 + 30x) dxx)– −6 L0 [216(10 )π ][200(109 )]
2 2 15(8) 30(2.4) F (2.4) 6 6.60–6 00==12(10 )c 20(2.4) + + d – A 10 B B 40(8) 2 2 [216(10 −6 )π ][200(109 )]
F = 91.20(10 3 ) N
Average Normal Stress:
91.20(10 3 ) 6 134.40(10 = ) N−m 2 134 MPa = s5 = 216(10 −6 )π
Ans.
Ans: s = 134 MPa 257
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4–74. 4–74. The bronze C86100 pipe has an inner radius of 0.5 and a awall in. IfIf the 12.5in. mm and wallthickness thicknessofof0.2 5 mm. the gas gas flowing through it changes the temperature of the pipe uniformly from T at to A TtoB = TB15°C B, determine the TAA ==60°C 200°F = 60°F at A at B,atdetermine the axial axial it exerts onwalls. the walls. fitted between forceforce it exerts on the The The pipepipe was was fitted between the the walls when = 60°F. walls when T =T 15°C.
A
B
8 ftm 2.4
Solution Temperature Gradient:
45°C
2.48– x x T(x) ==1560+ +a a b 140= 60 = 200  17.5x T(x) b (45) – 18.75x 2.4 8 Compatibility: 0 = dT  dF
Where
15°C
2.4 m
dT = 1 a¢Tdx
2.4 m
2.4m 2ft
10)  6 B 0 == 17(10 9.60 A–6 0
F (2.4) [(60 – 18.75x) – 15]dx – [(200  17.5x) 60] π (0.0352 − 0.0252 ) [103(109 )] L0 4
2.4m 2ft
F (2.4) 9.60–6 00==17(10 )  6 B (45 − 18.75 x) dx − A 10 L 0 π (0.0352 − 0.0252 ) [103(109 )] 4
= F 18.57(10 3 ) N = 18.6 kN
Ans.
Ans: F = 18.6 kN 258
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4–75. 40ftlong A36 4–75. The 12mlong A36steel steelrails rails on on a train track are laid with a small gap between them to allow for thermal expansion. Determine the required gap d so that the rails just touch one another when the temperature is increased from Using thisgap, gap,what whatwould wouldbe be the T11 ==–30°C  20°FtotoTT 90°F. T 30°C. Using this 2 2= = axial force in the rails if the temperature were to rise to T The crosssectional area of of each railrail is 3200 mm The crosssectional area each is 5.10 in2. T33 ==40°C? 110°F?
d
d
40m ft 12
Solution Thermal Expansion: Note that since adjacent rails expand, each rail will be d required to expand on each end, or d for the entine rail. 2
12 m
8.64 mm
3 6 12(10–6)[30 – (–30)](12)(10 ) )[90  (  20)](40)(12) d = a¢TL = 6.60(10
= 8.64 mm
Ans.
Compatibility: + B A:
8.64 mm
= dTT – ddFF 80.34848 .64 mm =
= 8.64 12(10 −6 )[40 − (−30)]12(10 3 ) −
F (12)(10 3 ) [3200(10 −6 )][200(109 )]
3 F 76.8(10 = ) N 76.8 kN =
Ans.
Ans: d = 8.64 mm, F = 76.8 kN 259
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*4–76. The device is used to measure a change in temperature. Bars AB and CD are made of A36 steel and 2014T6 aluminum alloy respectively. When the temperature is at at 75°F, 40°C,ACE ACEis isin in horizontal position. Determine thethe horizontal position. Determine the the vertial displacement pointeratat EE when when the vertical displacement of ofthethepointer temperature rises rises to to 150°F. 80°C.
6 mm 0.25 in.
A
723 mm in.
E
C
36 1.5mm in.
Solution
B
D
Thermal Expansion: –3 6 3 23(10–6)(80 – 40)(36) = 33.12(10 ) mm )(150  75)(1.5) = 1.44(10 ) in. A dT B CD = aal ¢TLCD = 12.8(10
–3 6 12(10–6)(80 – 40)(36) = 17.28(10 ) mm  3) in. )(150  75)(1.5) = 0.7425(10 A dT B AB = ast ¢TLAB = 6.60(10
From the geometry of the deflected bar AE shown Fig. b, dE = A dT B AB + C
A dT B CD  A dT B AB 6 0.25
–3  3) 1.44(10
(78) S (3.25)
6 mm
72 mm
3 –3 0.7425(10 )
) –3  3 32.12(10 ) – 17.28(10 = 17.28(10 ) +) c+ B d (78) R (3.25) = 0.7425(10 6 0.25 = 0.2102 0.00981mm in.
Ans.
Ans. dE = 0.2102 mm 260
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4–77. The bar has a crosssectional area A, length L, modulus of elasticity E, and coefficient of thermal expansion a. The temperature of the bar changes uniformly along its length from TA at A to TB at B so that at any point x along the bar T = TA + x(TB  TA)>L. Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar and the bar has a temperature of TA.
x A
B TB
TA
Solution + S
0 = ∆ T  dF
(1)
However, d∆ T = a∆ T dx = a aTA + L
∆ T = a
TB  TA x  TA bdx L
TB  TA TB  TA 2 L x dx = a c x d` L 2L 0 L0
= ac
TB  TA aL Ld = (TB  TA) 2 2
From Eq. (1). 0 =
aL FL (TB  TA) 2 AE
F =
a AE (TB  TA) 2
Ans.
Ans: F = 261
aAE (TB  TA) 2
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4–78. When the temperature is at 30°C, the A36 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, respectively. If the temperature drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume each tank provides a rigid support at A and B.
150 mm 10 mm Section a – a 6m x
A
Solution
a a
B
Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a, the temperature gradient can be expressed as a function of x as T(x) = 80 +
50 50 (6  x) = a130 xb°C 6 6
Thus, the change in temperature as a function of x is ∆T = T(x)  30° = a130 
50 50 xb  30 = a100 xb°C 6 6
Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of dT = a
L
∆Tdx = 12(10  6)
L0
6m
a100 
50 xbdx = 0.0054 m = 5.40 mm 6
Using the method of superposition, Fig. b, +) (S
0 = dT  dF 0 = 5.40 
F(6000) 2
p(0.16  0.152)(200)(109)
F = 1 753 008 N Normal Stress: s =
F 1 753 008 = = 180 MPa A p(0.162  0.152)
Ans.
Ans: s = 180 MPa 262
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4–79. When the temperature is at 30°C, the A36 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, respectively. If the temperature drop along the pipe is linear, determine the average normal stress developed in the pipe. Assume the walls of each tank act as a spring, each having a stiffness of k = 900 MN>m.
150 mm 10 mm Section a – a 6m x
A
Solution
a a
B
Temperature Gradient: Since the temperature varies linearly along the pipe, Fig. a, the temperature gradient can be expressed as a function of x as T(x) = 80 +
50 50 (6  x) = a130 xb°C 6 6
Thus, the change in temperature as a function of x is ∆T = T(x)  30° = a130 
50 50 xb  30 = a100 xb°C 6 6
Compatibility Equation: If the pipe is unconstrained, it will have a free expansion of dT = a
L
∆Tdx = 12(10  6)
L0
6m
a100 
50 xbdx = 0.0054 m = 5.40 mm 6
Using the method of superposition, Fig. b, +) (S
d = dT  dF F 6
900(10 )
(1000) = 5.40  C
F(6000) 2
p(0.16  0.152)(200)(109)
+
F 900(106)
(1000) S
F = 1 018 361 N
Normal Stress: s =
F 1 018 361 = = 105 MPa A p(0.162  0.152)
Ans.
Ans: s = 105 MPa 263
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*4–80. When the temperature is at 30°C, the A36 steel pipe fits snugly between the two fuel tanks. When fuel flows through the pipe, it causes the temperature to vary along the pipe as T = (53 x2  20x + 120)°C, where x is in meters. Determine the normal stress developed in the pipe. Assume each tank provides a rigid support at A and B.
150 mm 10 mm Section a – a 6m x
A
Solution
a a
B
Compatibility Equation: The change in temperature as a function of x is 5 5 ∆T = T  30° = a x2  20x + 120b  30 = a x2  20x + 90b°C. If the pipe 3 3 is unconstrained, it will have a free expansion of dT = a
L
∆Tdx = 12(10  6)
L0
6m
5 a x2  20x + 90bdx = 0.0036 m = 3.60 mm 3
Using the method of superposition, Fig. b, +) (S
0 = dT  dF 0 = 3.60 
F(6000) 2
p(0.16  0.152)(200)(109)
F = 1 168 672.47 N Normal Stress: s =
F 1 168 672.47 = = 120 MPa A p ( 0.162  0.152 )
Ans.
Ans: s = 120 MPa 264
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4–81. The 50mmdiameter cylinder is made from Am 1004T61 magnesium and is placed in the clamp when the temperature is T1 = 20° C. If the 304stainlesssteel carriage bolts of the clamp each have a diameter of 10 mm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the force in the cylinder when the temperature rises to T2 = 130°C.
100 mm
150 mm
Solution + c ΣFy = 0; Fst = Fmg = F dmg = dst amg Lmg ∆T 26(10  6)(0.1)(110) 
FmgLmg EmgAmg
= astLst ∆T +
FstLst EstAst
F(0.1) F(0.150) = 17(10  6)(0.150)(110) + p p 44.7(109) (0.05)2 193(109)(2) (0.01)2 4 4 Ans.
F = 904 N
Ans: F = 904 N 265
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4–82. The 50mmdiameter cylinder is made from Am 1004T61 magnesium and is placed in the clamp when the temperature is T1 = 15°C. If the two 304stainlesssteel carriage bolts of the clamp each have a diameter of 10 mm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the temperature at which the average normal stress in either the magnesium or the steel first becomes 12 MPa.
100 mm
150 mm
Solution + c ΣFy = 0; Fst = Fmg = F dmg = dst amg Lmg ∆T 26(10  6)(0.1)(∆T) 
FmgLmg EmgAmg
= astLst ∆T +
FstLst EstAst
F(0.1) F(0.150) = 17(10  6)(0.150)(∆T) + p p 44.7(109) (0.05)2 193(109)(2) (0.01)2 4 4
The steel has the smallest crosssectional area. p F = sA = 12(106)(2) a b (0.01)2 = 1885.0 N 4
Thus,
∆T = 229° Ans.
T2 = 229° + 15° = 244° C
Ans: T2 = 244° C 266
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4–83. T he rigid block has a weight of 400 kN and is to be supported by posts A and B, which are made of A36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before they are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased 20°F. Each in2. mm2. by 10°C. Each post post has has aa crosssectional crosssectional area area of of 85000
A
C
B
31 mft
31 mft
Solution Equations of Equilibrium: a + ©MC = 0;
400 kN
FB(3) (3) = 0 (1)  FA(1)
+ c ©Fy = 0;
FA = FB = F
2F + FC  400(103) = 0
[1] 1m
Compatibility: (+ T )
1m
(dC)F  (dC)T = dF FC L FL − 18(10 −6 )(10)L = 5000(10 −6 ) [101(109 )] 5000(10 −6 ) [200(109 )]
1.9802 FC − F = 180(10 3 )
[2]
Solving Eqs. [1] and [2] yields: F = 123.393(103) N
FC = 153.214(103) N
average Normal Sress: sA = sB = sC =
123.393(103) F = = 24.68(106) N/m2 = 24.7 MPa 5000(10−6) A
153.214(103) FC = 30.64(106) N/m2 = 30.6 MPa = 5000(10−6) A
Ans. Ans.
Ans. sA = sB = 24.7 MPa, sC = 30.6 MPa 267
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*4–84. The cylinder CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a crosssectional area of 125 mm2. CD is made of aluminum and has a crosssectional area of 375 mm2. Est = 200 GPa, Eal = 70 GPa, and aal = 23(106)>°C.
0.7 mm B
F C
– 240 mm 300 mm
D A
+ E
Solution dst = (dg)al  dal  0.0007 Fst(0.3) 6
9
(125)(10 )(200)(10 )
= 23(10  6)(150)(0.24) 
F(0.24) (375)(10  6)(70)(109)
 0.0007
12Fst = 128 000  9.1428F
(1)
+ c ΣFy = 0;
(2)
F  2Fst = 0
Solving Eqs. (1) and (2) yields, FAB = FEF = Fst = 4.23 kN Ans.
FCD = F = 8.45 kN
Ans: FCD = 8.45 kN 268
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4–85. The cylinder CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a crosssectional area of 125 mm2. CD is made of aluminum and has a crosssectional area of 375 mm2. Est = 200 GPa, Eal = 70 GPa, ast = 12(106)>°C, and aal = 23(106)>°C.
0.7 mm B
F C
– 240 mm 300 mm
D A
+ E
Solution dst + (dT)st = (dT)al  dal  0.0007 Fst(0.3) (125)(10  6)(200)(109)
+ 12(10  6)(50  30)(0.3)
= 23(10  6)(180  30)(0.24) 
Fal(0.24) 375(10  6)(70)(109)
 0.0007
12.0Fst + 9.14286Fal = 56000
(1)
+ c ΣFy = 0;
(2)
Fal  2Fst = 0
Solving Eqs. (1) and (2) yields: Ans.
FAB = FEF = Fst = 1.85 kN FCD = Fal = 3.70 kN
Ans: FAB = FEF = 1.85 kN 269
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4–86. t
The metal strap has a thickness t and width w and is subjected to a temperature gradient T1 to T2 (T1 6 T2). This causes the modulus of elasticity for the material to vary linearly from E1 at the top to a smaller amount E2 at the bottom. As a result, for any vertical position y, measured from the top surface, E = [(E2  E1)>w]y + E1. Determine the position d where the axial force P must be applied so that the bar stretches uniformly over its cross section.
T1 P
w d
P
T2
Solution P = constant = P0 P0 =
s = E
s = P0 a
s E2  E1 aa by + E1 b w
E2  E1 y + E1 b w
+ ΣFx = 0: S
P =
L0
w
s t dy =
P = P0t a
P0 t a a
L0
LA
s dA = 0
m
P0 a
E2  E1 y + E1 b t dy w
E2  E1 E2 + E1 + E1w b = P0 t a bw 2 2
a+ ΣM0 = 0: P0 t a
P 
P(d) 
LA
y sdA = 0
w E2 + E1 E2  E1 2 bwd = P0 aa by + E1yb t dy 2 w L0
E1 2 E2 + E1 E2  E1 2 bwd = P0 t a w + w b 2 3 2
E2 + E1 1 bd = (2E2 + E1)w 2 6
d = a
2 E2 + E1 bw 3(E2 + E1)
Ans.
Ans: d = a 270
2E2 + E1 bw 3(E2 + E1)
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4–87. 5 mm
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN.
40 mm
20 mm P
P r 10 mm 20 mm
Solution For the fillet: w 40 = = 2 h 20 From Fig. 4–23,
r 10 = = 0.5 h 20
K = 1.4 smax = Ksavg = 1.4 a
8 (103) 0.02 (0.005)
= 112 MPa
b
For the hole: 2r 20 = = 0.5 w 40 From Fig. 4–24,
K = 2.1 smax = Ksavg = 2.1 a
8 (103) (0.04  0.02)(0.005)
= 168 MPa
b
Ans.
Ans: smax = 168 MPa 271
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*4–88. If the allowable normal stress for the bar is s allow = 120 MPa, determine the maximum axial force P that can be applied to the bar.
5 mm 40 mm
20 mm P
P r 10 mm 20 mm
Solution Assume failure of the fillet. w 40 = = 2; h 20 From Fig. 4–23.
r 10 = = 0.5 h 20
K = 1.4 sallow = sm ax = Ksavg 120 (106) = 1.4 a
P b 0.02 (0.005)
P = 8.57 kN
Assume failure of the hole. 2r 20 = = 0.5 w 40 From Fig. 4–24.
K = 2.1 sallow = smax = Ksavg 120 (104) = 2.1 a
P b (0.04  0.02) (0.005)
Ans.
P = 5.71 kN (controls)
Ans: P = 5.71 kN 272
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4–89. The steel bar has the dimensions shown. Determine the maximum axial force P that can be applied so as not to exceed an allowable tensile stress of sallow = 150 MPa.
20 mm 60 mm
30 30 mm mm P
P r 15 mm 24 mm
Solution Assume failure occurs at the fillet: w 60 = = 2 h 30 From the text,
and
r 15 = = 0.5 h 30
K = 1.4 smax = sallow = Ksavg 150 (106) = 1.4 c
P d 0.03 (0.02)
P = 64.3 kN
Assume failure occurs at the hole: 2r 24 = = 0.4 w 60 From the text,
K = 2.2 smax = sallow = Ksavg 150 (106) = 2.2 c
P d (0.06  0.024) (0.02)
P = 49.1 kN (controls!)
Ans.
Ans: P = 49.1 kN 273
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4–90. r = 30 mm
The A36 steel plate has a thickness of 12 mm. If sallow = 150 MPa, determine the maximum axial load P that it can support. Calculate its elongation, neglecting the effect of the fillets.
r = 30 mm
120 mm
C
B 60 mm P A
800 mm
60 mm P D
200 mm
200 mm
Solution Maximum Normal Stress at fillet: r 30 = = 0.5 and h 60
w 120 = = 2 h 60
From the text, K = 1.4 smax = sallow = Ksavg 150(106) = 1.4J
P R 0.06(0.012) Ans.
P = 77142.86 N = 77.1 kN Displacement: d = Σ =
PL AE 77142.86(400) 9
(0.06)(0.012)(200)(10 )
+
77142.86(800) (0.12)(0.012)(200)(109) Ans.
= 0.429 mm
Ans: P = 77.1 kN, d = 0.429 mm 274
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4–91. Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and has an allowable stress of sallow = 147 21 ksi. MPa.
40.125 mm in. 25 mm 1.25 in.
37.5 1.875mm in.
P
P
Solution 15 mm 0.75 in.
Assume failure of the fillet. r 0.25 5 = 0.2 = = 0.2 h 1.25 25
r mmin. � 50.25
37.5 w 1.875 ==1.51.5 = 25 h 1.25
From Fig. 424, K = 1.73 sallow = smax = Ksavg
P 147(106 ) = 1.73 (0.025)(0.004) P = 8.497(10 3 ) N Assume failure of the hole. r 0.375 7.5 ==0.20 = 0.20 w 1.875 37.5 From Fig. 425, K = 2.45 sallow = smax = Ksavg
P 147(106 ) = 2.45 (0.0375 − 0.015)(0.004) P = 5.400(10 3 ) N
(controls)
Ans. Ans.
= 5.40 kN
Ans. P = 5.40 kN 275
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*4–92.
0.125 4 mm in. 1.25 in. 25 mm
1.875 in. 37.5 mm
Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN. P
P
Solution 0.75 in. 15 mm
At fillet: r 0.25 5 = 0.2 = = 0.2 h 1.25 25
r� 0.25 5 mmin.
w 1.875 37.5 ==1.5 = 1.5 h 1.25 25
From Fig. 424, K = 1.73
8(10 3 ) P 6 = = σ max K= N−m 2 138 MPa 138.4(10 ) = 1.73 (0.025)(0.004) A At hole: r 0.375 7.5 ==0.20 = 0.20 w 1.875 37.5 From Fig. 425, K = 2.45
8(10 3 ) 6 = = σ max 2.45 = ) N−m 2 218 MPa (Controls) 217.78(10 (0.0375 − 0.015)(0.004) Ans.
Ans. = σ max 2.45 MPa 276
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4–93. The member is to be made from a steel plate that is 60.25 mmin.thick. hole isis drilled drilled through through its center, thick.If Ifa a25mm 1in. hole determine the approximate width w of the plate so that it can support an axial force of 3350 The 16.75lb. kN. Theallowable allowablestress stressis s ksi.MPa. isallow sallow= =22150
60.25 mmin. w
3350kN lb 16.75
3350 lb 16.75 kN
1 in. 25 mm
Solution sallow = smax = Ksavg
16.75(10 3 ) 150(106 ) = K (w − 0.025)(0.006) = w 0.025 + 0.01861K
By trial and error, from Fig. 425, choose
r = 0.2; 0.17; w
K ==2.50 2.45 K
0.025 + 0.01861(2.50) = 0.07153 m = 71.5 mm w=
Since
0.5 r 12.5 0.17 = == 0.2 w 2.49 71.53
Ans.
OK
Ans. K = 2.50, w = 71.5 mm 277
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4–94.
12.5 mm 0.5 in.
A
The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stressconcentration factor for this geometry?
15 0.6mm in. 20 0.8mm in.
5 mm 0.2 in.
P
15 0.6mm in. B
Solution
42 MPa 6 ksi
36 ksi 252 MPa
Number of squares = 28 3 = = = 73.5(10 ) N 73.5 kN P 28[42(106 )](0.005)(0.0125)
σ avg = = K
Ans.
73.5(10 3 ) P 2 = 196(106 ) N−m= = 196 MPa A 2(0.015)(0.0125)
σ max 252 = = 1.286 = 1.29 σ avg 196
Ans.
Ans. P = 73.5 kN, K = 1.29 278
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4–95. The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress concentration factor?
10 mm A 20 mm 80 mm
P
B 5 MPa 30 MPa
Solution Number of squares = 19 P = 19(5)(106)(0.02)(0.01) = 19 kN savg = K =
Ans.
19(103) P = = 23.75 MPa A 0.08(0.01)
smax 30 MPa = = 1.26 savg 23.75 MPa
Ans.
Ans: P = 19 kN, K = 1.26 279
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*4–96.
B
The three bars are pinned together and subjected to the load P. If each bar has a crosssectional area A, length L, and is made from an elastic perfectly plastic material, for which the yield stress is sY, determine the largest load (ultimate load) that can be supported by the bars, i.e., the load P that causes all the bars to yield. Also, what is the horizontal displacement of point A when the load reaches its ultimate value? The modulus of elasticity is E.
L C
u L D
A
P
u L
Solution When all bars yield, the force in each bar is, FY = sYA + ©F = 0; : x
P  2sYA cos u  sYA = 0 Ans.
P = sYA(2 cos u + 1) Bar AC will yield first followed by bars AB and AD. dAB = dAD = dA =
FY(L) sYAL sYL = = AE AE E
dAB sYL = cos u E cos u
Ans.
Ans. dA =
dAB sYL = cos u E cos u
280
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4–98. The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m and crosssectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with (sY)st = 120 MPa and (sY)al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and (b) 720 N. Eal = 70 GPa, Est = 200 GPa.
Aluminum
Steel
Solution + c ΣFy = 0;
(1)
Fal + Fst  W = 0
Assume both wires behave elastically. dal = dst;
FalL FstL = A(70) A(200) (2)
Fal = 0.35 Fst (a) When W = 600 N, solving Eqs. (1) and (2) yields: Fst = 444.44 N = 444 N
Ans.
Fal = 155.55 N = 156 N
Ans.
sal =
Fal 155.55 = = 38.88 MPa 6 (sg)al = 70 MPa Ast 4(10  6)
OK
sst =
Fst 444.44 = = 111.11 MPa 6 (sg)st = 120 MPa Ast 4(10  6)
OK
The elastic analysis is valid for both wires. (b) When W = 720 N, solving Eqs. (1) and (2) yields: Fst = 533.33 N:
Fst = 186.67 N
sal =
Fal 186.67 = = 46.67 MPa 6 (sg)al = 70 MPa Aal 4(10  6)
sst =
Fst 533.33 = = 133.33 MPa 7 (sg)st = 120 MPa Ast 4(10  6)
OK
Therefore, the steel wire yields. Hence, Fst = (sg)stAst = 120 ( 106 ) (4) ( 10  6 ) = 480 N
Ans.
From Eq. (1), Fal = 240 N
Ans.
sal =
240 4(10  6)
OK
= 60 MPa 6 (sg)al
Ans: (a) Fst = 444 N, Fal = 156 N, (b) Fst = 480 N, Fal = 240 N 282
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4–99. The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m and crosssectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with (sY)st = 120 MPa and (sY)al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and (b) 720 N. Eal = 70 GPa, Est = 200 GPa.
Fal
Aluminum
Fst
Steel
Solution Equations of Equilibrium: + c ΣFy = 0;
(1)
Fal + Fst  w = 0
Elastic Analysis: Assume both wires behave elastically. dal = dst Fal L A(70) ( 10
9
)
=
Fst L A(200) ( 109 ) (2)
Fal = 0.350 Fst a) When W = 600 N, solving Eq. (1) and (2) yields: Fst = 444.44 N = 444 N
Ans.
Fal = 155.55 N = 156 N
Ans.
Average Normal Stress: sal =
Fal 155.55 = = 38.88 MPa < (sg)al = 70.0 MPa Asl 4.00 ( 106 )
(OK!)
sst =
Fst 444.44 = = 111.11 MPa < (sg)st = 120 MPa Ast 4.00 ( 106 )
(OK!)
The average normal stress for both wires do not exceed their respective yield stress. Therefore, the elastic analysis is vaild for both wires b) When W = 720 N, solving Eq. (1) and (2) yields: Fst = 533.33 N
Fal = 186.67 N
Average Normal Stress: sal =
Fal 186.67 = = 46.67 MPa < (sg)al = 70.0 MPa Aal 4.00 ( 106 )
sst =
Fst 533.33 = = 133.33 MPa > (sg)st = 120 MPa Ast 4.00 ( 106 )
(OK!)
Therefore, the steel wire yields. Hence, Fst = (sg)stAst = 120 ( 106 )( 4.00 )( 106 ) = 480 N From Eq. (1),
Ans. Ans.
Fal = 240 N sal =
240 4.00 ( 106 )
(OK!)
= 60.00 MPa < (sg)al
Ans: Fst = Fal = Fst = Fal = 283
444 N, 156 N, 480 N, 240 N
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*4–100.
1.2 4 ftm
The distributed loading is applied to the rigid beam, (MPa) which is supported by the three bars. Each bar has a ss(ksi) crosssectional area of 780 mm2 and is made from a 60 material having a stress–strain diagram that can be 420 approximated by the two line segments shown. If a load of w = 400 kN/m is applied to the beam, determine the 252 36 stress in each bar and the vertical displacement of the beam.
Solution
0.0012 0.0012
a + ©MB = 0;
A A
B B
420 252
0.2 0.2
C
A
s (MPa)
1.2 4 ftm 1.5 m
w B
∋∋(in./in.) (mm/mm)
1.5 5 ftm C
w
FC(1.2) – FA(1.2) = 0; FA = FC = F
+ c ©Fy = 0;
0.0012
0.2
∋ (mm/mm)
(1)
2F + FB  960 200 = 0
Since the loading and geometry are symmetrical, the bar will remain horizontal. Therefore, the displacement of the bars is the same and hence, the force in each bar is the same. From Eq. (1). 320 kN F = FB = 66.67 kip Thus,
σ= A σ= B σ= C
320(10 3 ) 2 = 410.26(106 ) N−m= 410 MPa 780(10 −6 )
Ans.
From the stressstrain diagram:
410.26 − 252 420 − 252 = = ; ε 0.18847 mm−mm ε − 0.0012 0.2 − 0.0012 3 δ ε= L 0.18847(1.5)(10= = ) 282.71 mm = 283 mm
1.2 m
Ans.
1.2 m
2.4(400) = 960 kN
s (MPa) 420 252
e mm/mm)
Ans. δ = 283 mm
284
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4–101.
4 ft 1.2 m
The distributed loading is applied to the rigid beam, which is supported by the three bars. Each bar has a crosssectional area of 468 mm2 and is made from a material having a stress–strain diagram that can be approximated by the two line segments shown. Determine the intensity of the distributed loading w needed to cause the beam to be displaced downard 37.5 mm.
36 252
+ c ©Fy = 0;
C
A A
s (MPa) 420
0.0012
FC(1.2) 0; (4) – FAA(1.2) (4) == 0;
B
60 420
Solution a + ©MB = 0;
A
s s(ksi) (MPa)
4 ft 1.2 m 1.5 m
252
0.2
w B B
∋ (in./in.) (mm/mm)
5 ft 1.5 m C C
w w
FA = FC = F
2F + FB  2.4w 8 w ==00
0.0012
(1)
0.2
∋ (mm/mm)
Since the system and the loading are symmetrical, the bar will remain horizontal. Hence the displacement of the bars is the same and the force supported by each bar is the same. From Eq. (1), (2)
FB = F = 0.8 2.6667 w w From the stressstrain diagram: e =
37.5 1.5 = 50.025 0.025in.>in. mm/mm 51.5(10 (12) 3)
ss–252 420 60– 252 36 36 5= ; ; s 5 272.11 MPa 0.025 –0.0012 0.0012 0.2 0.2– 0.0012 0.0012 Hence F = sA = [272.11(106)][468(10−6)] = 127.35(103) N From Eq. (2), w = 159.19(103) N/m = 159 kN/m
Ans.
1.2 m
1.2 m
2.42v
s (MPa) 420 252
e mm/mm)
Ans. w = 159 kN/m 285
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4–102. The rigid lever arm is supported by two A36 steel wires having the same diameter of 4 mm. Determine the smallest force P that will cause (a) only one of the wires to yield; (b) both wires to yield. Consider A36 steel as an elastic perfectly plastic material.
P
450 mm 150 mm 150 mm 30 A
Solution
E
300 mm
Equation of Equilibrium: Referring to the freebody diagram of the lever arm shown in Fig. a, a+ ΣME = 0;
C
B
D
FAB (300) + FCD (150)  P(450) = 0 (1)
2FAB + FCD = 3P
(a) Elastic Analysis: The compatibility equation can be written by referring to the geometry of Fig. b. dAB = a
300 bd 150 CD
dAB = 2dCD
FCD L FAB L = 2a b AE AE FCD =
1 F 2 AB
(2)
Assuming that wire AB is about to yield first, FAB = (sY)st AAB = 250 1 106 2 c
From Eq. (2),
FCD =
p 1 0.0042 2 d = 3141.59 N 4
1 (3141.59) = 1570.80 N 2
Substituting the result of FAB and FCD into Eq. (1), Ans.
P = 2618.00 N = 2.62 kN (b) Plastic Analysis: Since both wires AB and CD are required to yield, FAB = FCD = (sY)st A = 250 1 106 2 c
Substituting this result into Eq. (1),
p 1 0.0042 2 d = 3141.59 N 4 Ans.
P = 3141.59 N = 3.14 kN
Ans: (a) P = 2.62 kN, (b) P = 3.14 kN 286
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4–103. The 1500kN weight is slowly set on the top of a post made of 2014T6 aluminum with an A36 steel core. If both materials can be considered elastic perfectly plastic, determine the stress in each material. Aluminum 251mm in.
Solution
mm 250in.
Steel
Equations of Equilibrium: + c ©Fy = 0;
[1]
Pst + Pal  1500 300 == 0
Elastic Analysis: Assume both materials still behave elastically under the load. dst = dal Pal L Pst L = π (0.052 ) [200(109 )] π (0.12 − 0.052 ) [73.1(109 )] 4 4
Pst = 0.9120 Pal Solving Eqs. [1] and [2] yields: Pal = 784.52 kN
Pst = 715.48 kN
Average Normal Stress:
784.52(10 3 ) = 133.18 MPa < (σ γ= )al 414 MPa π (0.12 − 0.052 ) 4
σ= al
Pal = Aal
σ= st
Pst 715.48(10 3 ) = = 364.39 MPa > (σ γ= )st 250 MPa π (0.052 ) Ast 4
(OK!)
Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the steel is sst = (sγ)st = 250 36.0MPa ksi
Ans.
6 π 3 ) 4 (0.052 ) 490.87(10 = ) N 490.87 kN Pst = (sg)stAst = 250(10 =
From Eq. [1] Pal = 186.90 1009.13kip kN Pal 1009.13(10 3 ) 6 = = 171.31(10 ) N−m 2 = 171.31 MPa < (σ γ )al 414 MPa sal = = π 2 2 Aal (0.1 − 0.05 ) 4 Ans.
Then sal = 171.31 MPa = 171 MPa
Ans. sst = 250 MPa, sal = 171 MPa 287
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*4–104. The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is sY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of the distributed load w that can be placed on the beam that will cause only one of the wires to start to yield and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the calculation, assume that the steel is elastic perfectly plastic.
E
D
800 mm A
B
C G
Solution
400 mm
250 mm
w 150 mm
Equations of Equilibrium: a + ©MA = 0;
FBE(0.4) + FCD(0.65)  0.8w (0.4) = 0 [1]
0.4 FBE + 0.65 FCD = 0.32w (a) By observation, wire CD will yield first. p Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN. 4 From the geometry dCD dBE = ; 0.4 0.65
dCD = 1.625dBE FCDL FBEL = 1.625 AE AE [2]
FCD = 1.625 FBE Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields: FBE = 4.099 kN
Ans.
w = 18.7 kN>m (b) When both wires yield FBE = FCD = (sg)A p = 530 A 106 B a b A 0.0042 B = 6.660 kN 4
Substituting the results into Eq. [1] yields:
Ans.
w = 21.9 kN>m
Ans. w = 21.9 kN>m 288
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4–105. The rigid beam is supported by three 25mm diameter A36 steel rods. If the beam supports the force of P = 230 kN, determine the force developed in each rod. Consider the steel to be an elastic perfectly plastic material.
D
F
E
600 mm P A
400 mm
Solution Equation of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, FAD + FBE + FCF  230(103) = 0
+ c ΣFy = 0;
(1)
FBE(400) + FCF(1200)  230(103)(800) = 0
a+ ΣMA = 0;
FBE + 3FCF = 460 ( 103 )
(2)
Elastic Analysis: Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =
dCF  dAD b(400) 1200
2 1 d + dCF 3 AD 3
FBEL 2 FADL 1 FCF L = a b + a b AE 3 AE 3 AE FBE =
2 1 F + FCF 3 AD 3
(3)
Solving Eqs. (1), (2), and (3) FCF = 131 428.57 N
FBE = 65 714.29 N FAD = 32 857.14 N
Normal Stress: sCF =
sBE =
sAD =
FCF 131428.57 = = 267.74 MPa 7 (sY)st p ACF (0.0252) 4
(N.G.)
FBE 65714.29 = = 133.87 MPa 6 (sY)st p ABE (0.0252) 4 FAD 32857.14 = = 66.94 MPa 6 (sY)st p AAD (0.0252) 4
(O.K.)
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using FCF = (sY)st ACF = 250 (106) c
p (0.0252) d = 122 718.46 N = 123 kN 4
289
Ans.
B
400 mm
C
400 mm
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4–105.
Continued
Substituting this result into Eq. (2), Ans.
FBE = 91844.61 N = 91.8 kN Substituting the result for FCF and FBE into Eq. (1),
Ans.
FAD = 15436.93 N = 15.4 kN sBE =
sAD =
FBE 91844.61 = = 187.10 MPa 6 (sY)st p ABE ( 0.0252 ) 4 FAD 15436.93 = = 31.45 MPa 6 (sY)st p AAD ( 0.0252 ) 4
(O.K.)
(O.K.)
Ans: FCF = 123 kN, FBE = 91.8 kN, FAD = 15.4 kN 290
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4–106. The rigid beam is supported by three 25mm diameter A36 steel rods. If the force of P = 230 kN is applied on the beam and removed, determine the residual stresses in each rod. Consider the steel to be an elastic perfectly plastic material.
D 600 mm
P A
400 mm
Solution Equation of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, FAD + FBE + FCF  230 ( 103 ) = 0
+ c ΣFy = 0;
(1)
FBE(400) + FCF(1200)  230 ( 103 ) (800) = 0
a+ ΣMA = 0;
FBE + 3FCF = 460(103)
(2)
Elastic Analysis: Referring to the deflection diagram of the beam shown in Fig. b, the compatibility equation can be written as dBE = dAD + a dBE =
dCF  dAD b(400) 1200
2 1 d + dCF 3 AD 3
(3)
FBE L 2 FAD L 1 FCF L = a b + a b AE 3 AE 3 AE FBE =
2 1 F + FCF 3 AD 3
(4)
Solving Eqs. (1), (2), and (4)
FCF = 131428.57 N
FBE = 65714.29 N FAD = 32857.14 N
Normal Stress: sCF =
sBE =
sAD =
FCF 131428.57 = = 267.74 MPa (T) 7 (sY)st p ACF ( 0.0252 ) 4 FBE 65714.29 = = 133.87 MPa (T) 6 (sY)st p ABE ( 0.0252 ) 4 FAD 32857.14 = = 66.94 MPa (T) 6 (sY)st p AAD ( 0.0252 ) 4
(N.G.)
(O.K.)
(O.K.)
Since rod CF yields, the elastic analysis is not valid. The solution must be reworked using sCF = (sY)st = 250 MPa (T) FCF = sCF ACF = 250 ( 106 ) c
F
E
p ( 0.0252 ) d = 122718.46 N 4
291
B
400 mm
C
400 mm
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4–106.
Continued
Substituting this result into Eq. (2), FBE = 91844.61 N Substituting the result for FCF and FBE into Eq. (1), FAD = 15436.93 N sBE =
sAD =
FBE 91844.61 = = 187.10 MPa (T) 6 (sY)st p ABE ( 0.0252 ) 4 FAD 15436.93 = = 31.45 MPa (T) 6 (sY)st p AAD ( 0.0252 ) 4
(O.K.)
(O.K.)
Residual Stresses: The process of removing P can be represented by applying the force P′, which has a magnitude equal to that of P but is opposite in sense. Since the process occurs in a linear manner, the corresponding normal stress must have the same magnitude but opposite sense to that obtained from the elastic analysis. Thus, = = = sCF = 267.74 MPa (C) sBE = 133.87 MPa (C) sAD = 66.94 MPa (C)
Considering the tensile stress as positive and the compressive stress as negative, = (sCF)r = sCF + sCF = 250 + (  267.74) =  17.7 MPa = 17.7 MPa (C)
Ans.
= (sBE)r = sBE + sBE = 187.10 + (  133.87) = 53.2 MPa (T)
Ans.
(sAD)r = sAD +
= sAD
= 31.45 + (  66.94) =  35.5 MPa = 35.5 MPa (C)
Ans.
Ans: (sCF)r = 17.7 MPa (C), (sBE)r = 53.2 MPa (T), (sAD)r = 35.5 MPa (C) 292
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4–107. CC
The wire BC has a diameter of 3.4 mm and the material has the stress–strain characteristics shown in the figure. Determine the vertical displacement of the handle at D if the pull at the grip is slowly increased and reaches a magnitude of (a) P = 2250 N, (b) P = 3000 N.
40 in.mm 1000 AA
DD
BB 50 in.mm 1250
30 in. 750 mm
Solution
PP
s (ksi) (MPa)
Equations of Equilibrium: a + ©MA = 0;
[1]
F (50)  –P(80) = 0= 0 FBC P(2000) BC(1250)
2250lb, N, (a) From Eq. [1] when P = 450
720 lbN FBC = 3600
80 560 70 490
Average Normal Stress and Strain: sBC =
FBC = ABC
3600 720 pp 2 44 (0.0034 )
5 396.51(106) N/m2 5 396.51 MPa
0.007 0.007
0.12 0.12
(in./in.) P (mm/mm)
From the Stress–Strain diagram 396.51 490 70 = 5 ; ; 0.007 eBC 0.007
0.005867mm/mm in.>in. eeBC BC5=0.005664 1250 mm
Displacement:
750 mm
dBC = eBCLBC = 0.005664(1000) = 5.6644 mm dD dD dBCdBC = 5 ; ; 80 2000 501250
dD 5 (1.6)(5.664) = 9.0631 mm = 9.06 mm
600 lb, (b) From Eq. [1] when P == 3000 N,
560 490
Ans.
(εBC, 396.51)
F 960 lb FBC == 4800 N
Average Normal Stress and Strain: sBC =
FBC = ABC
4800 960 p 2 44 (0.0034 )
(0.007, 490)
(0.12, 560) x (εBC, 528.68)
e(mm/mm)
5 528.68(106) N/m2 5 528.68 MPa
From Stress–Strain diagram 56080– 490 528.68 − 490  70 5 eBC − 0.007 0.12 0.12– 0.007  0.007
eBC = 0.06944 mm/mm
Displacement: dBC = eBCLBC = 0.06944(1000) = 69.44 mm ddDD dBCdBC 8 = 5 ; ; dD dD 5 (1.6)(69.44) = 111.11 mm = 111 mm 80 501250 2000
1250 mm
Ans.
750 mm
Ans. (a) dD = 9.06 mm, (b) dD = 111 mm
293
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*4–108.
P
The bar having a diameter of 50 mm is fixed connected at its ends and supports the axial load P. If the material is elastic perfectly plastic as shown by the stress–strain diagram, determine the smallest load P needed to cause segment CB to yield. If this load is released, determine the permanent displacement of point C.
A
B
C
2 ftm 0.6
3 ftm 0.9
(ksi) ss(MPa)
20 140
Solution
(in./in.) P (mm/mm)
0.001
When P is increased, region AC will become plastic first, then CB will become plastic. Thus,
F= A 140(106 ) π4 (0.052= ) 274.89(10 3 ) = N 274.89 kN A F= B σ= + ©F = 0; : x
(1)
FA + FB  P = 0 kN P ==2(274.89) 2(62.832)= =549.78 125.66 kip P = 550 kN
Ans.
The deflection of point C is, dC = eL = (0.001)(3)(12) in. ; (0.9)(103=) =0.036 0.9 mm d Consider the reverse of P on the bar.
FB′ (0.6) F ′ (0.9) 5 = B AE AE FA ¿ = 1.5 FB ¿ So that from Eq. (1) FB ¿ = 0.4P FA ¿ = 0.6P 3 FB′L 0.4 549.78(10 ) 0.9(10 ) = = 0.720 mm S AE π (0.052 ) 140(106 )−0.001 4 ¢d = 0.036  0.0288 = 0.00720 mm din. ; 0.9 – 0.720 5 0.180 3
= δC′
Ans.
Ans.
P = 550 kN, ¢d = 0.180 mm. ; 294
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4–109. The rigid beam is supported by the three posts A, B, and C of equal length. Posts A and C have a diameter of 75 mm and are made of a material for which E = 70 GPa and sY = 20 MPa. Post B has a diameter of 20 mm and is made of a material for which E′ = 100 GPa and sY ′ = 590 MPa. Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield.
P
A
P
B
2m
2m
C
2m
2m
Solution ΣMB = 0;
FA = FC = Fal
+ c ΣFy = 0;
(1)
Fat + 2Fat  2P = 0
(a) Post A and C will yield, Fal = (st)alA p = 20(104)a b(0.075)2 a = 88.36 kN
(Eal)r =
(sr)al Eal
=
20(104) 70(104)
= 0.0002857
Compatibility condition: dbr = dal = 0.0002857(L) Fbr (L) = 0.0002857 L p (0.02)2(100) ( 104 ) 4 Fbr = 8.976 kN sbr =
8.976(103) = 28.6 MPa 6 sg p (0.023) 4
OK.
From Eq. (1), 8.976 + 2(88.36)  2P = 0 Ans.
P = 92.8 kN (b) All the posts yield: Fbr = (sg)brA p = (590)(104)a b(0.022) 4 = 185.35 kN
Fal = 88.36 kN From Eq. (1); 185.35 + 2(88.36)  2P = 0 Ans.
P = 181 kN
Ans: P = 92.8 kN, P = 181 kN 295
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4–110. The rigid beam is supported by the three posts A, B, and C. Posts A and C have a diameter of 60 mm and are made of a material for which E = 70 GPa and sY = 20 MPa. Post B is made of a material for which E′ = 100 GPa and sY ′ = 590 MPa. If P = 130 kN, determine the diameter of post B so that all three posts are about to yield.
P
A
P
B
2m
2m
C
2m
2m
Solution + c ΣFy = 0;
(1)
2(Fg)al + Fbr  260 = 0 (Fal)g = (sg)al A
From Eq. (1),
p = 20(106)a b(0.06)2 = 56.55 kN 4 2(56.55) + Fbr  260 = 0 Fbr = 146.9 kN (sg)br = 590 ( 106 ) =
146.9 ( 103 ) p (d )3 4 B Ans.
d B = 0.01779 m = 17.8 mm
Ans: d B = 17.8 mm 296
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R4–1. The assembly consists of two A992 steel bolts AB and EF and an 6061T6 aluminum rod CD. When the temperature is at 30° C, the gap between the rod and rigid member AE is 0.1 mm. Determine the normal stress developed in the bolts and the rod if the temperature rises to 130° C. Assume BF is also rigid.
A
C
0.1 mm E 25 mm
25 mm 300 mm
400 mm 50 mm B
D
F
Solution Equation of Equilibrium: Referring to the freebody diagram of the rigid cap shown in Fig. a, + c ΣFy = 0;
(1)
Fr  2Fb = 0
Compatibility Equation: If the bolts and the rod are unconstrained, they will have a free expansion of (dT)b = ast ∆TLb = 12(10  6)(130  30)(400) = 0.48 mm and (dg)r = aal ∆TLr = 24(10  6)(130  30)(300) = 0.72 mm. Referring to the initial and final position of the assembly shown in Fig. b, (dT)r  dFr  0.1 = (dT)b + dFb 0.72 
Fr (300) Fb(400)  0.1 = 0.48 + p p (0.052)(68.9)(109) (0.0252)(200)(109) 4 4 (2)
Fb + 0.5443Fr = 34361.17 Solving Eqs. (1) and (2). Fb + 16 452.29 N
Fr = 32 904.58 N
Normal Stress: sb =
sr =
Fb 16 452.29 = = 33.5 MPa p Ab 2 (0.025 ) 4
Ans.
Fr 32 904.58 = = 16.8 MPa p Ar (0.052) 4
Ans.
Ans: sb = 33.5 MPa, sr = 16.8 MPa 297
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R4–2. The assembly shown consists of two A992 steel bolts AB and EF and an 6061T6 aluminum rod CD. When the temperature is at 30° C, the gap between the rod and rigid member AE is 0.1 mm. Determine the highest temperature to which the assembly can be raised without causing yielding either in the rod or the bolts. Assume BF is also rigid.
A
C
0.1 mm E 25 mm
25 mm 300 mm
400 mm 50 mm B
D
F
Solution Equation of Equilibrium: Referring to the freebody diagram of the rigid cap shown in Fig. a, + c ΣFy = 0;
(1)
Fp  2Fb = 0
Normal Stress: Assuming that the steel bolts yield first, then p Fb = (sg)stAb = 250(106) c (0.0252) d = 122 718.46 N 4
Substituting this result into Eq. (1), Fp = 245 436.93 N Then, sp =
Fp Ap
=
245 436.93 = 125 MPa 6 (sg)al p (0.052) 4
(O.K!)
Compatibility Equation: If the assembly is unconstrained, the bolts and the post will have free expansion of (dT)b = ast ∆TLb = 12(10  6)(T  30)(400) = 4.8(10  3)(T  30) and (dT)p = aal ∆TLp = 24(10  6)T  30)(300) = 7.2(10  3)(T  30). Referring to the initial and final position of the assembly shown in Fig. b, (dT)p  dFp  0.1 = (dT)b + dFb 7.2(10  3)(T  30) 
245 436.93(300) p (0.052)(68.9)(109) 4
 0.1 = 4.8(10  3)(T  30) +
122 718.46(400) p (0.0252)(200)(109) 4 Ans.
T = 506.78° C = 507°C
Ans: T = 507°C 298
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R4–3. The rods each have the same 25mm diameter and 600mm length. If they are made of A992 steel, determine the forces developed in each rod when the temperature increases by 50° C.
C
600 mm 60 B
Solution
A
60
600 mm
Equation of Equilibrium: Referring to the freebody diagram of joint A shown in Fig. a, + c ΣFx = 0;
FAD sin 60°  FAC sin 60° = 0
+> ΣFx = 0;
FAB  2F cos 60° = 0
D
FAC = FAD = F
(1)
FAB = F
Compatibility Equation: If AB and AC are unconstrained, they will have a free expansion of 1 dT 2 AB = 1 dT 2 AC = ast ∆TL = 12(10  6)(50)(600) = 0.36 mm. Referring to the initial and final position of joint A, dFAB 
1 dT 2 AB
= adT ′ b
AC
 dFAC ′
dAC = 2dAC. cos 60° = 2(dT)AC and dFAC ′ = 2dFAC. Thus, this equation becomes
Due to symmetry, joint A will displace horizontally, and dAC ′ = Thus, adT ′ b dFAB 
AC
1 dT 2 AB
FAB (600)
= 2 1 dT 2 AC  2dAC
p 1 0.0252 2 (200)(109) 4
 0.36 = 2(0.36)  2C
FAB + 2F = 176 714.59
F(600) p 1 0.0252 2 (200)(109) 4
S
(2)
Solving Eqs. (1) and (2), FAB = FAC = FAD = 58 904.86 N = 58.9 kN (C)
Ans.
Ans: FAB = FAC = FAD = 58.9 kN (C) 299
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*R4–4. B
A
Two A36 steel pipes, each having a crosssectional area of 200 mm2, are screwed together using a union at B as shown. Originally the assembly is adjusted so that no load is on the pipe. If the union is then tightened so that its screw, having a lead of 0.55 mm, undergoes two full turns, determine the average normal stress developed in the pipe. Assume that the union at B and couplings at A and C are rigid. Neglect the size of the union. Note: The lead would cause the pipe, when unloaded, to shorten shorten 0.55 mm when the union is rotated one revolution.
3 ftm 0.9
C 2 ftm 0.6
2(0.55) = 1.10 mm
The loads acting on both segments AB and BC are the same since no external load acts on the system. 1.10 = dB>A + dB>C 1.10 5
P(0.9)(1000) [200(10−6)][200(109)]
1
P(0.6)(1000) [200(10−6)][200(109)]
P 5 29.33(103) 5 29.33 kN sAB = sBC =
29.33(103) P 5 5 146.67(106) N/m2 5 147 MPa A 200(10−6)
Ans.
Ans. sAB = sBC = 147 MPa 300
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R45. The force P is applied to the bar, which is composed of an elastic perfectly plastic material. Construct a graph to show how the force in each section AB and BC (ordinate) varies as P (abscissa) is increased. The bar has crosssectional areas of 625 mm2 in region AB and 2500 mm2 in region BC, and σY = 210 MPa. Given: L AB := 150mm AAB := 625mm2 2
L BC := 50mm
ABC := 2500mm σ Y := 210MPa
Solution: Equations of equilibrium:
+ ΣF x=0;
P − FA − FC = 0
[1]
Elastic behavior:
+
0 = ∆ C − δC 0=
( P) ⋅ L AB E ⋅ AAB
−
⎡ ( FC) ⋅ LBC ( FC) ⋅ LAB⎤ + ⎢ ⎥ E⋅ AAB ⎣ E⋅ ABC ⎦
( )
0 = 6P − FC ( 0.5 + 6) Substituting [2] into [1]:
FC =
12 P 13
[2]
FA =
1 P 13
[3]
By comparison, segment BC will yield first. Hence,
( )
FC := σ Y ⋅ ABC From [2]: From [3]:
13 ⋅F 12 C 1 FA := ⋅P 13 P :=
FC = 525 kN P = 568.75 kN FA = 43.75 kN
When segment AB yields,
( ) FC := ( σ Y) ⋅ ABC
FA := σ Y ⋅ AAB
From [1]:
P := FA + FC
FA = 131.25 kN FC = 525 kN P = 656.25 kN
301
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2014T6aluminum aluminum a diameter of rodrod has has a diameter of 0.5 in. R4–6. The 2014T6 12 mm and isattached lightly attached to the rigid supports and and is lightly to the rigid supports at A andatBA when B1 when T1 = If the temperature T2 = –20°, T2 = 10°F, If 25°C. the temperature becomesbecomes and T = 70°F. andaxial an axial force = 80 appliedtotothe therigid rigid collar collar as lb Nis isapplied an force of Pof=P 16 shown, determine the reactions at A and B.
A
B
P/2 P/2 5 in. 125 mm
8 in. 200 mm
40 N
Solution
40 N
Compatibility:
40 N
40 N
+ 0 = ¢  ¢ + d : B T B = 0
FB (325) 80(125) − 23(10 −6 )[25° − (−20°)](325) + π (0.012 2 ) [73.1(109 )] π (0.012 2 ) [73.1(109 )] 4 4 3
= ) 8.53 kN = FB 8.526(10
+ ©F = 0; : x
Ans.
40 N
FA = 8.526 kN
40 N
2(0.008) 2.1251– FA F 2(0.040) + 8.526 =A0 = 0
FA = 8.606 kN = 8.61 kN
Ans.
Ans. FB = 8.53 kN, FA = 8.61 kN 302
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R4–7. The 2014T6 aluminum rod has a diameter of 0.5mm in. and isis lightly lightlyattached attachedtoto rigid supports A 12 thethe rigid supports at Aatand and B when Determine the Pforce thatbemust B when T1 = T 40°C. Determine the force that P must ap1 = 70°F. be applied to the so that, when = 0°F, atthe plied to the collar so collar that, when T = 0°C, the Treaction B reaction is zero. at B is zero.
A
B
P/2 P/2 125 mm 5 in.
200 mm 8 in.
Solution + : = 0
0 = ¢ B  ¢ T + dB P(125) − 23(10 −6 )(40)(325) + 0 π (0.012 2 ) [73.1(109 )] 4
3 = = P 19.776(10 ) N 19.8 kN
Ans.
Ans. P = 19.8 kN 303
CH 04.indd 209
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*R4–8. The rigid link is supported by a pin at A and two A36 steel steelwires, wires,each each having an unstretched length of having an unstretched length of 12 in. 300 mm and crosssectional area in of2. 7.8 mm2. Determine and crosssectional area of 0.0125 Determine the force the force developed in the wires the link supports the developed in the wires when thewhen link supports the vertical vertical loadlb.of 1.75 kN. load of 350
300 12 mm in. C 125 5mm in. B 100 4mm in.
A
Solution 150 mm 6 in.
Equations of Equilibrium: a + ©MA = 0;
[1]
 FCC(225) (9) – F (4) + +350(6) = 0= 0 –F FBB(100) 1.75(150)
1.75 350 kN lb
Compatibility: ddCC ddBB = 5 4 9 100 225
125 mm
(L) F F FB(L) FC(L) = 5 4AE 9AE 100AE 225AE
125 mm 100 mm
2.25FB − FC = 0‚
[2]
100 mm 150 mm
Solving Eqs. [1] and [2] yields:
1.75 kN
FB = 0.4330 kN = 433 N
Ans.
FC = 0.9742 kN = 974 N
Ans.
Ans. FB = 433 N‚ FC = 974 N‚ 304
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R4–9. The joint is made from three A992 steel plates that are bonded together at their seams. Determine the displacement of end A with respect to end B when the joint is subjected to the axial loads. Each plate has a thickness of 5 mm.
100 mm 23 kN
46 kN
B 23 kN
A 600 mm 200 mm
800 mm
Solution dA>B = Σ
46(103)(600) 46(103)(200) 23(103)(800) PL = + + AE (0.005)(0.1)(200)(109) 3(0.005)(0.1)(200)(109) (0.005)(0.1)(200)(109) Ans.
= 0.491 mm
Ans: dA>B = 0.491 mm 305
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5–1. The solid shaft of radius r is subjected to a torque T. Determine the radius r′ of the inner core of the shaft that resists onehalf of the applied torque (T>2). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shearstress distribution.
r¿ r
T
Solution Tc Tr 2T =p 4 = 3 J r pr 2
a) tmax =
t =
a T2 br′
Since t = r′ =
b)
L0 L0 L0
r 2
r 2
r 2
=
p 4 2 (r′)
r 1
24
T p(r′)3
r′ t ; r max
T r′ 2T = a b 3 r pr 3 p(r′)
Ans.
= 0.841r
dT = 2p
L0
r′
tr2 dr r′
dT = 2p
r tmax r2 dr L0 r r′
dT = 2p
r 2T a 3 b r2 dr L0 r pr r′
4T T = 4 r3 dr 2 r L0 r′ =
r′ 1
24
Ans.
= 0.841r
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.
306
Ans: r′ = 0.841r
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5–2. The solid shaft of radius r is subjected to a torque T. Determine the radius r′ of the inner core of the shaft that resists onequarter of the applied torque (T>4). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shearstress distribution.
r¿ r
T
Solution T(r) Tc 2T = p 4 = 3 J (r ) pr 2
a) tmax =
Since t = t′ = r′ = b) t =
r′ 2Tr′ t = r max pr 4
(T4 )r′ T′c′ 2Tr′ ; 4 = p 4 J′ pr 2 (r′) r 1
44
r r 2T 2T t = a 3b = r; c max r pr pr 4
dT = rt dA = r c L0
T 4
Ans.
= 0.707 r
dT =
r L
4T 4
r′ T 4T r4  ; = 4 4 r 4 0
r′
dA = 2pr dr
2T 4T r d (2pr dr) = 4 r3dr pr 4 r
r3dr (r′)4 1 = 4 r4 Ans.
r′ = 0.707 r
Ans: r′ = 0.707 r 307
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5–3. A shaft is made of an aluminum alloy having an allowable shear stress of tallow = 100 MPa. If the diameter of the shaft is 100 mm, determine the maximum torque T that can be transmitted. What would be the maximum torque T′ if a 75mmdiameter hole were bored through the shaft? Sketch the shearstress distribution along a radial line in each case.
T T¿
Solution Allowable Shear Stress: Torsion formula can be applied. For the solid shaft, tmax = tallow =
T(0.05) Tc ; 100 ( 106 ) = p 4 J 2 (0.05 )
T = 19.63 ( 103 ) N # m = 19.6 kN # m
Ans.
For the hollow shaft, tmax = tallow =
Tc ; 100 ( 106 ) = J
T′(0.05) p 2
(0.054  0.03754)
T′ = 13.42 ( 103 ) N # m = 13.4 kN # m
Ans.
The shear stress at the inner surface where r = 0.0375 m is tr = 0.0375 m =
T′r J
=
13.42 ( 103 ) (0.0375) p 4 2 (0.05
 0.03754)
= 75.0 ( 106 ) Pa = 75.0 MPa
The shear stress distribution along the radius of the crosssection of the solid and hollow shafts are shown in Figs. a and b respectively.
Ans: T = 19.6 kN # m, T′ = 13.4 kN # m 308
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The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm and a wall thickness of 5 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shearstress distribution over the cross section.
600 N 75 mm
75 mm
5 mm 25 mm
600 N
Solution
T = 600(0.15) = 90 N # m tmax =
ti =
Tc = J
Tr = J
90(0.0175) p 2
[(0.0175)4  (0.0125)4] 90(0.0125)
p 2
[(0.0175)4  (0.0125)4]
Ans.
= 14.5 MPa
= 10.3 MPa
Ans: tmax = 14.5 MPa 309
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The solid shaft is fixed to the support at C and subjected to the torsional loadings. Determine the shear stress at points A and B on the surface, and sketch the shear stress on volume elements located at these points.
35 mm
C A
B 20 mm
35 mm
300 Nm 800 Nm
Solution tB =
800(0.02) TB r = p = 6.79 MPa 4 J 2 (0.035 )
Ans.
tA =
500(0.035) TA c = 7.42 MPa = p 4 J 2 (0.035 )
Ans.
Ans: tB = 6.79 MPa, tA = 7.42 MPa 310
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The solid 30mmdiameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress in the shaft.
300 Nm
500 Nm
A
200 Nm
C
400 Nm
300 mm
Solution
D B
400 mm
Internal Torque: As shown on torque diagram. Maximum Shear Stress: From the torque diagram Tmax = 400 N # m. Then, applying torsion formula, abs = t max
=
500 mm
Tmax c J 400(0.015) p 2
(0.0154)
Ans.
= 75.5 MPa
Ans: abs = 75.5 MPa t max 311
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The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall and three torques are applied to it, determine the absolute maximum shear stress developed in the pipe. 30 Nm 20 Nm 80 Nm
Solution tmax =
Tmax c = J
90(0.02) p 4 2 (0.02
 0.01854) Ans.
= 26.7 MPa
Ans: tmax = 26.7 MPa 312
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*5–8. The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm. If it is tightly secured to the wall at A and three torques are applied to it as shown, determine the absolute maximum shear stress developed in the pipe.
A
30 Nm 20 Nm
Solution 80 Nm
tmax =
Tmax c = J
90(0.02) p 2
(0.024  0.01854)
= 26.7 MPa
Ans..
Ans: tmax = 26.7 MPa 313
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The solid aluminum shaft has a diameter of 50 mm and an allowable shear stress of tallow = 60 MPa. Determine the largest torque T1 that can be applied to the shaft if it is also subjected to the other torsional loadings. It is required that T1 act in the direction shown. Also, determine the maximum shear stress within regions CD and DE.
E D C
300 Nm
B
600 Nm
A
T1 900 Nm
Solution Internal Torque: Assuming that failure occurs at region BC of the shaft, where the torque will be greatest. Referring to the FBD of the right segment of the shaft sectioned through region BC, Fig. a g M x = 0; T1  900  TBC = 0 TBC = T1  900
Maximum Shear Stress: Applying the torsion formula, tmax = tallow =
(T1  900)(0.025) TBC c ; 60 ( 106 ) = p 4 J 2 ( 0.025 )
(T1)max = T1 = 2372.62 N # m = 2.37 kN # m
Ans.
using this result the torque diagram shown in Fig. b can be plotted. This indicates that region BC indeed is subjected to maximum internal torque, thus, the critical region. From the torque diagram, the internal torques in regions CD and DE are TCD = 872.62 N # m and TDE = 572.62 N # m respectively. (tmax)CD =
872.62 (0.025) TCD C = p = 35.55 ( 106 ) Pa = 35.6 MPa 4 J 2 ( 0.025 )
Ans.
(tmax)DE =
572.62 (0.025) TDC C = p = 23.33 ( 106 ) Pa = 23.3 MPa 4 J ( ) 0.025 2
Ans.
Ans: (T1)max = 2.37 kN # m, (tmax)CD = 35.6 MPa, (tmax)DE = 23.3 MPa 314
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The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in the shaft and sketch the shearstress distribution along a radial line of the shaft where the shear stress is maximum. Set T1 = 2000 N # m.
E D C
300 Nm
B
600 Nm
A
T1 900 Nm
Solution Internal Torque: The torque diagram plotted in Fig. a indicates that region BC of the shaft is subjected to the greatest internal torque; thus, the critical region where the absolute maximum shear stress occurs . Here TBC = 1100 N # m. Applying the torsion formula, tmax abs =
1100 (0.025) TBC c = p = 44.82 ( 106 ) Pa = 44.8 MPa 4 J 2 ( 0.025 )
Ans.
The shear stress distribution along the radius of the crosssection of the shaft is shown in Fig. b.
Ans: tabs = 44.8 MPa max
315
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5–11. The 60mmdiameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the shaft’s surface and specify their locations, measured from the free end.
400 Nm
C
4 kNm/m B
A
0.5 m 800 Nm
d
0.2 m 0.2 m
0.5 m
Solution Internal Torque: The torque diagram plotted in Fig. a indicates that region 1.0 m 6 x 6 1.2 m of the shaft is subjected to the greatest internal torque where Tmax = 1200 N # m, whereas the minimum internal torque, Tmin = 0, occurs at x = 0.7 m. Shear Stress: Applying the torsion formula, abs = tmax
1200 (0.03) Tmax C = p = 28.29 ( 106 ) Pa = 28.3 MPa 4 J ( ) 0.03 2
Ans.
occurs within the region 1.0 m < x < 1.2 m abs = tmin
Ans.
Tmin C occurs at x = 0.700 m J
Ans.
Ans: abs = 28.3 MPa, tmax for 1.0 m 6 x 6 1.2 m, abs = 0, tmin at x = 0.700 m 316
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*5–12. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft if the allowable shear stress for the material is tallow = 60 MPa.
C
400 Nm 4 kNm/m B
A
0.5 m 800 Nm
d
0.2 m 0.2 m
0.5 m
Solution Internal Torque: The torque diagram plotted in Fig. a indicates that region 1.0 m 6 x 6 1.2 m of the shaft is subjected to the greatest internal torque Tmax = 1200 N # m; thus, that is the critical region where the absolute maximum shear stress occurs. Allowable Shear Stress: Applying the torsion formula, abs = tallow = tmax
60 ( 106 ) =
Tmax C J
1 d2 2 p d 4 2 122
1200
Ans.
d = 0.04670 m = 46.7 mm
Ans: d = 46.7 mm 317
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5–13. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller smallerpipe pipehas has outer diameter of in. 18.75 and anan outer diameter of 0.75 andmm an inner an inner diameter ofwhereas 17 mm, the whereas larger has diameter of 0.68 in., largerthe pipe has pipe an outer an outer of diameter mmdiameter and an inner diameter 21.5 diameter 1 in. andofan25inner of 0.86 in. If theof pipe is mm. If secured the pipetoisthe tightly to the wall at C, determine tightly wall secured at C, determine the maximum shear the maximum shear stress developed each section of the stress developed in each section of the in pipe when the couple pipe when the couple applied to the handles of the shown is applied to theshown handlesisof the wrench. wrench.
C
B
A 75 N in. 15 lb1506 mm 2008mm in.
Solution tAB
15Nlb 75
26.25(0.009375) Tc 6 = 62.54(10 = ) N−m 2 62.5 MPa = = π 4 4 J (0.009375 − 0.0085 ) 2
Ans.
26.25(0.0125) Tc 6 = 18.89(10 = ) N−m 2 18.9 MPa = π 4 4 J (0.0125 − 0.01075 ) 2
Ans.
tBC =
26.25 N · m TAB = 26.25 N · m 26.25 N · m TBC = 26.25 N · m
Ans: tAB 62.5 MPa, tBC 18.9 MPa 318
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5–14. A steel tube having an outer diameter of 60 mm is used to transmit 6.75 kW when turning at 27 revNmin. Determine the inner diameter d of the tube to the nearest mm if the allowable shear stress is tallow = 70 MPa.
d 60 mm
Solution
ω = 27
πev 2π πad 1 min = 0.9π πad−s min 1 πev 60 s
P = Tω 6.75(10 3 ) = T(0.9π ) = T 2.387(10 3 ) N ⋅ m tmax = tallow =
70(106 ) =
Tc J
2.387(10 3 )(0.03) π
2
( 0.034 − ci4 )
ci = 0.01996 m = = 0.03992 = d 2= ci 2(0.01996) m 39.92 mm Use d = 40 mm
Ans.
Ans: Use d = 40 mm 319
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5–15. The 60mmdiameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the shear stress at points A and B, and sketch the shear stress on volume elements located at these points.
2 kNm/m
C 0.4 m
600 Nm
B
A
0.4 m 0.3 m 0.3 m
400 Nm
d
Solution Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula tA =
400(0.03) TA c = p = 9.43 MPa 4 J 2 ( 0.03 )
Ans.
tB =
600(0.03) TB c = p = 14.1 MPa 4 J 2 ( 0.03 )
Ans.
Ans: tA = 9.43 MPa, tB = 14.1 MPa 320
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*5–16. The 60mmdiameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses on the shaft’s surface, and specify their locations, measured from the fixed end C.
2 kNm/m
C 0.4 m
600 Nm
B
A
0.4 m 0.3 m 0.3 m
400 Nm
d
Solution
Internal Torque: From the torque diagram, the maximum torque Tmax = 1400 N # m occurs at the fixed support and the minimum torque Tmin = 0 occurs at x = 0.700 m. Shear Stress: Applying the torsion formula tabs
min
=
tabs = max
Tmin c = 0 occurs at x = 0.700 m J
Ans.
1400(0.03) Tmax c = p = 33.0 MPa 4 J 2 ( 0.03 )
Ans. Ans.
occurs at x = 0
According to SaintVenant’s principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Therefore, tabs obtained is not valid. max
Ans: tabs max
= 0 occurs at x = 0.700 m,
min
= 33.0 MPa occurs at x = 0
tabs 321
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5–17. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft if the allowable shear stress for the material is tallow = 1.6 MPa.
2 kNm/m
C 0.4 m
600 Nm
B
A
0.4 m 0.3 m 0.3 m
400 Nm
d
Solution
Internal Torque: From the torque diagram, the maximum torque Tmax = 1400 N # m occurs at the fixed support. Allowable Shear Stress: Applying the torsion formula tabs = tallow = Tmax c max J 175 ( 106 ) =
1400 1 d2 2
1 2
p d 4 2 2
Ans.
d = 0.03441 m = 34.4 mm
According to SaintVenant’s principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Therefore, the above analysis is not valid.
Ans: d = 34.4 mm 322
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5–18. The motor delivers a torque of 50 N # m to the shaft AB. This torque is transmitted to shaft CD using the gears at E and F. Determine the equilibrium torque Tⴕ on shaft CD and the maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts.
A
50 mm 30 mm B
Solution
35 mm T¿
Equilibrium: a + ©ME = 0;
50  F(0.05) = 0
a + ©MF = 0;
T¿  1000(0.125) = 0
C
E
125 mm D
F
F = 1000 N
T¿ = 125 N # m
Ans.
Internal Torque: As shown on FBD. Maximum Shear Stress: Applying torsion Formula. (tAB)max =
50.0(0.015) TAB c = 9.43 MPa = p 4 J 2 (0.015 )
Ans.
(tCD)max =
125(0.0175) TCDc = 14.8 MPa = p 4 J 2 (0.0175 )
Ans.
Ans. (tAB)max = 9.43 MPa (tCD)max = 14.8 MPa 223
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5–19. If the applied torque on shaft CD is T¿ = 75 N # m, determine the absolute maximum shear stress in each shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating.
A
50 mm 30 mm B 35 mm T¿
Solution
C
E
125 mm D
F
Equilibrium: a + ©MF = 0;
75  F(0.125) = 0;
a + ©ME = 0;
600(0.05)  TA = 0
F = 600 N
TA = 30.0 N # m Internal Torque: As shown on FBD. Maximum Shear Stress: Applying the torsion formula (tEA)max =
30.0(0.015) TEA c = 5.66 MPa = p 4 J 2 (0.015 )
Ans.
(tCD)max =
75.0(0.0175) TCDc = 8.91 MPa = p 4 J 2 (0.0175 )
Ans.
Ans. (tEA)max = 5.66 MPa (tCD)max = 8.91 MPa 224
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* The shaft has an outer diameter of 100 mm and an inner diameter of 80 mm. If it is subjected to the three torques, determine the absolute maximum shear stress in the shaft. The smooth bearings A and B do not resist torque.
E
A
D
10 kNm
Maximum Shear Stress: Referring to the torque diagram shown in Fig. a, region ED is subjected to the largest internal torque, which is Tmax = 10 kN # m. Thus, the absolute maximum shear stress occurs in this region. Applying the torsion formula, Tmax C = J
C
15 kNm
Solution
tabs = max
B
10 ( 103 ) (0.05) p 2
( 0.054  0.044 )
= 86.26 ( 106 ) Pa = 86.3 MPa
5 kNm
Ans.
Ans: abs = 86.3 MPa tmax 325
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The shaft has an outer diameter of 100 mm and an inner diameter of 80 mm. If it is subjected to the three torques, plot the shear stress distribution along a radial line for the cross section within region CD of the shaft. The smooth bearings at A and B do not resist torque.
E
A
D
10 kNm
C
15 kNm
Solution Shear Stress: Referring to the torque diagram shown in Fig. a, Region CD of the shaft is subjected to an internal torque of TCD = 5 kN # m. The torsion formula will be applied. The maximum shear stress is (tmax)CD =
B
TCD C = J
5 ( 103 ) (0.05) p 2
( 0.054  0.044 )
5 kNm
= 43.13 ( 106 ) Pa = 43.1 MPa
The shear stress at the inner surface of the hollow shaft is
(tp = 0.04 m)CD =
TCDr = J
5 ( 103 ) (0.04) p 2
( 0.054  0.044 )
= 34.51 ( 106 ) Pa = 34.5 MPa
Also, tmax C
=
tp r
r
; (tp = 0.04 m)CD = a C btmax = a
0.04 b(43 # 13) 0.05
= 34.51 MPa = 34.5 MPa
The shear stress distribution along the radius of the crosssection of the shaft is shown in Fig. b.
Ans: N/A 326
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5–22. If the gears are subjected to the torques shown, determine the maximum shear stress in the segments AB and BC of the A36 steel shaft. The shaft has a diameter of 40 mm.
300 Nm 100 Nm A 200 Nm B
Solution The internal torque developed in segments AB and BC are shown in their respective FBDs, Figs. a and b. The polar moment of inertia of the shaft is J =
C
p ( 0.024 ) = 80 ( 109 ) p m4. Thus, 2
(tAB)max =
300(0.02) TAB c = = 23.87 ( 106 ) Pa = 23.9 MPa J 80 ( 109 ) p
Ans.
(tBC)max =
200(0.02) TBC c = = 15.92 ( 106 ) Pa = 15.9 MPa J 80 ( 109 ) p
Ans.
Ans: (tAB)max = 23.9 MPa, (tBC)max = 15.9 MPa 327
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5–23. If the gears are subjected to the torques shown, determine the required diameter of the A36 steel shaft to the nearest mm if tallow = 60 MPa.
300 Nm 100 Nm A 200 Nm B
Solution The internal torque developed in segments AB and BC are shown in their respective FBDs, Fig. a and b.
C
Here, segment AB is critical since its internal torque is the greatest. The polar p d 4 pd 4 moment of inertia of the shaft is J = a b = . Thus, 2 2 32 tallow =
TC ; J
60 ( 106 ) =
300 ( d>2 ) pd4 >32
Ans.
d = 0.02942 m = 30 mm
Ans: d = 30 mm 328
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*5–24. The rod has a diameter of 25 mm and a weight of 10 lb/ft. Determine thethe maximum torsional stress in the 150 N/m. Determine maximum torsional stress in rod the at a at section located at Aatdue to the weight. rod a section located A due to rod’s the rod’s weight.
1.35 m 4.5 ft B
A
0.45 1.5 ftm
0.45 m 1.5 ft
1.2 4 ftm
Solution Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;
TAA –150(1.2)(0.6) 10(4)(2) = 0= 0
12in # ft· m 108 TATA==80 lb N a
0.6 m 150(1.2) N
The polar moment of inertia of the cross section at A is
= J
π = (0.01254 ) 38.35(10 −9 ) m 4 . Thus 2 tmax =
TA c 108(0.0125) 6 = 35.20(10 = ) N−m 2 35.2 MPa = J 38.35(10 −9 )
Ans.
Ans: tmax 35.2 MPa 329
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5–25. The rod has a diameter of 25 mm and a weight of 15 lb/ft. Determine thethe maximum torsional stress in the 225 N/m. Determine maximum torsional stress in rod the at a at section located at Bat due to the weight. rod a section located B due to rod’s the rod’s weight.
4.5 ft 1.35 m B
A
0.45 1.5 ftm
1.5 ft 0.45 m
4 ftm 1.2
Solution Here, we are only interested in the internal torque. Thus, other components of the internal loading are not indicated in the FBD of the cut segment of the rod, Fig. a. ©Mx = 0;
TBB –225(1.2)(0.6) 162lbN# ft · ma 15(4)(2) = 0= 0 TBT= 120 B=
12 in
The polar moment of inertia of the crosssection at B is
= J
π = (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2 tmax =
162(0.0125) TB c 6 52.80(10 = ) N−m 2 52.8 MPa = = J 38.35(10 −9 )
Ans.
TB = 162 N · m
0.6 m
Tmax = 52.8 MPa
12.5 mm
225(1.2) N
Ans: tmax 52.8 MPa 330
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The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F, which delivers 12 kW of power to the shaft while it is turning at 50 rev>s. If gears A, B, and C remove 3 kW, 4 kW, and 5 kW respectively, determine the maximum shear stress in the shaft within regions CF and BC. The shaft is free to turn in its support bearings D and E.
3 kW 4 kW D
A
5 kW 25 mm
B C
E
12 kW
F
Solution v = 50
rev 2p rad c d = 100 p rad>s s rev
TF =
12(103) P = = 38.20 N # m v 100 p
TA =
3(103) P = = 9.549 N # m v 100 p
TB =
4(103) P = = 12.73 N # m v 100 p
(tmax)CF =
38.20(0.0125) TCF c = p = 12.5 MPa 4 J 2 (0.0125 )
Ans.
(tmax)BC =
22.282(0.0125) TBC c = p = 7.26 MPa 4 J 2 (0.0125 )
Ans.
Ans: (tmax)CF = 12.5 MPa, (tmax)BC = 7.26 MPa 331
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5–27. The solid steel shaft DF has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at F, which delivers 12 kW of power to the shaft while it is turning at 50 rev>s. If gears A, B, and C remove 3 kW, 4 kW, and 5 kW respectively, determine the absolute maximum shear stress in the shaft.
3 kW 4 kW D
A
5 kW 25 mm
B C
E
12 kW
F
Solution v = 50
rev 2p rad c d = 100 p rad>s s rev
TF =
12(103) P = = 38.20 N # m v 100p
TA =
3(103) P = = 9.549 N # m v 100p
TB =
4(103) P = = 12.73 N # m v 100p
From the torque diagram, Tmax = 38.2 N # m tabs = max
38.2(0.0125) Tc = 12.5 MPa = p 4 J 2 (0.0125 )
Ans.
Ans: tabs max
332
= 12.5 MPa
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**5–28. The drive shaft AB of an automobile is made of a having an an allowable allowableshear shearstress stressofoftallow tallow==856 MPa. If steel having If the ksi. the outer diameter ofshaft the shaft thedelivers engine outer diameter of the is 2.5isin.62.5 andmm the and engine delivers kW to the shaft is turning 1140rev>min, rev/min, 200 hp 165 to the shaft whenwhen it is itturning atat1140 determine the minimum minimum required required thickness thicknessof of the shaft’s wall. determine
B
A
Solution πev 2π πad 1 min ω = 1140 = 38π πad−s min 1 πev 60 s P = Tω 165(10 3 ) = T(38π ) = T 1.382(10 3 ) N ⋅ m tallow =
Tc J
56(106 ) =
[1.382(10 3 )](0.03125) π
2
(0.031254 − ri4 )
ri = 0.02608 m t = r0 − ri = 0.03125 − 0.02608 = 0.005170 m = 5.17 mm
Ans.
Ans. t = 5.17 mm 333
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5–29. The drive shaft AB of an automobile is to be thinwalled tube. tube.The Theengine enginedelivers delivers125 150kW hp designed as aa thinwalled when the shaft is turning at 1500 rev>min. rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is in.mm. The The material has an allowable shear stress is 2.5 62.5 material has an allowable shear of tallow 7 ksi. stress of = tallow = 50 MPa.
B
A
Solution
ω = 1500
πev 2π πad 1 min = 50π πad−s min 1 πev 60 s
P = Tω 125(10 3 ) = T(50π ) = T 795.77 N ⋅ m tallow =
Tc J
50(106 ) =
795.77(0.03125) π (0.031254 2
− ri4 )
ri = 0.02825 m t = r0 − ri = 0.03125 − 0.02825 = 0.002998 m = 3.00 mm
Ans.
Ans. t 3.00 mm 334
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5–30. A ship has a propeller drive shaft that is turning at 1500 rev>min rev/min while developing 1800 1500 hp. kW.IfIfit itis is8 ft 2.4long m long and and a diameter ofdetermine 100 mm, the determine theshear maximum has ahas diameter of 4 in., maximum stress shear in the shaft caused by torsion. in the stress shaft caused by torsion.
Solution Internal Torque:
ω = 1500
πev 2π πad 1 min = 50.0π πad−s min 1 πev 60 s
550 ft # lb>s P ==1500 1800kW hp a= 1500(103) N b ·=m/s 990 000 ft # lb>s 1 hp
T=
P 1500(10 3 ) = = 9549.30 N ⋅ m ω 50.0π
Maximum Shear Stress: Applying torsion formula tmax =
9549.30(0.05) Tc 3 = 48.63(10 = ) N−m 2 48.6 MPa = π (0.054 ) J
Ans.
2
Ans. tmax = 48.6 MPa 335
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5–31. The motor A develops a power of 300 W and turns its connected pulley at 90 rev>min. Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is tallow = 85 MPa.
60 mm 90 rev/min
A B
150 mm
Solution Internal Torque: For shafts A and B vA = 90
rev 2p rad 1 min a b = 3.00p rad>s min rev 60 s
P = 300 W = 300 N # m>s P 300 = = 31.83 N # m vA 3.00p
TA =
vB = vA a
rA 0.06 b = 3.00pa b = 1.20p rad>s rB 0.15
P = 300 W = 300 N # m>s TB =
P 300 = = 79.58 N # m vB 1.20p
Allowable Shear Stress: For shaft A tmax = tallow = 85 A 106 B =
TA c J 31.83 A d2A B
A B
p dA 4 2 2
dA = 0.01240 m = 12.4 mm
Ans.
For shaft B tmax = tallow = 85 A 106 B =
TB c J 79.58 A d2B B
A B
p dB 4 2 2
dB = 0.01683 m = 16.8 mm
Ans.
Ans. dA = 12.4 mm, dB = 16.8 mm 336
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*5–32. When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsional resistance TA. Also, soil along the sides of the pipe creates a distributed frictional torque along its length, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB that must be supplied by the drive unit to overcome the resisting torques, and calculate the maximum shear stress in the pipe. The pipe has an outer radius ro and an inner radius ri.
TB B
L
Solution TA +
1 t L  TB = 0 2 A
TB =
2TA + t AL 2
tA A
TA
Ans.
Maximum shear stress: The maximum torque is within the region above the distributed torque. tmax =
Tc J
tmax =
c
(2TA + t AL) 2 p 4 2 (r o

d (ro)
r 4i )
=
(2TA + t AL)ro
Ans.
p(r 4o  r 4i )
Ans:
2TA + t AL , 2 (2TA + t AL)ro = p(r 4o  r 4i )
TB = tmax
337
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5–33. The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev>s. If gears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress in the shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E.
3 kW
2 kW 25 mm
1 kW A D
B
E
C
Solution TC =
3(103) P = 9.549 N # m = v 50(2p)
TA =
1 T = 3.183 N # m 3 C
(tAB)max =
3.183 (0.0125) TC = p = 1.04 MPa 4 J 2 (0.0125 )
Ans.
(tBC)max =
9.549 (0.0125) TC = 3.11 MPa = p 4 J 2 (0.0125 )
Ans.
Ans: (tAB)max = 1.04 MPa, (tBC)max = 3.11 MPa 338
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5–34. The shaft is subjected to a distributed torque along its length of t = (10x2) N # m>m, where x is in meters. If the maximum stress in the shaft is to remain constant at 80 MPa, determine the required variation of the radius c of the shaft for 0 … x … 3 m.
3m
x
c
t (10x2) Nm/m
Solution T = t =
L
t dx =
L0
x
10 x2dx =
10 3 x 3
3 (10 Tc 3 )x c ; 80(106) = p 4 J 2 c
c 3 = 26.526(109) x3 Ans.
c = (2.98 x) mm
Ans: c = (2.98 x) mm 339
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5–35. The motor delivers 12 kW to the pulley A while turning turning at a constant rate of 1800 rpm. Determine to the 1 nearest of shaft of BC if BC the nearest multiples 5 mm thediameter smallest diameter shaft 8 in. theofsmallest allowable shearshear stressstress for steel is t=allow The does belt = 12The ksi. belt if the allowable is tallow 84 MPa. does noton slip onpulley. the pulley. not slip the
B
C 3 in. 75 mm
A
1.5 37.5in.mm
Solution The angular velocity of shaft BC can be determined using the pulley ratio that is vBC = a
rA 0.0375 rev 2p rad 1 min b vA = a b a 1800 ba ba b = 30p rad>s 0.075 rC min 1 rev 60 s
The power is 550 ft # n>s P ==12 (15 hp) b = 8250 ft # lb>s kW =a12(103) N · m/s 1 hp Thus, T =
P 12(10 8250 3) = = 127.32 N . m v 30p 30p
The polar moment of inertia of the shaft is J = tallow =
Tc ; J
84(106 ) =
pd4 p d 4 a b = . Thus, 2 2 32
127.32(d−2)
π d 4 −32
= d 0.01976 = m 19.76 mm d = 0.7639 in = Use d 20 mm
Ans.
Ans. d 20 mm 240
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*5–36. The gear motor can develop 1.6 kW when it turns at 450 rev/min. ofof 251mm, rev>min.IfIfthe theshaft shafthas hasa diameter a diameter in., determine the maximum shear stress developed in the shaft.
Solution The angular velocity of the shaft is v = ¢ 450
rev 2p rad 1 min ≤ ¢ ≤ ¢ ≤ = 15p rad>s min 1 rev 60 s
and the power is 550 ft # lb>s P ==1.6 2 hp 1100 ft # lb>s kW¢ = 1.6(103) N≤ · =m/s 1 hp Then
T=
P 1.6(10 3 ) = = 33.95 N ⋅ m ω 15π
The polar moment of inertia of the= shaft is J tmax =
π = (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2
33.95(0.0125) Tc 6 = 11.07(10 ) N−m 2 11.1 MPa == J 38.35(10 −9 )
Ans.
Ans. tmax = 11.1 MPa 341
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5–37. The gear motor can develop 2.4 kW when it turns 150150 rev>min. If Ifthe at rev/min. theallowable allowableshear shearstress stressfor for the the shaft shaft is 12MPa, ksi, determine tallow ==84 determine the the smallest smallest diameter diameter of the shaft 1 to the nearest multiples used. of 5be mm that can be used. 8 in. that can
Solution The angular velocity of the shaft is v = a 150
rev 2p rad 1 min ba ba b = 5p rad>s min 1 rev 60 s
and the power is 550 ft # lb>s P == 2.4 (3 kW hp) a= 2.4(103) N b· m/s = 1650 ft # lb>s 1 hp Then T =
P 1650 3) 12 in 2.4(10 # ft)a 152.79 ·m = = (105.04 lb N v 5p 5p
The polar moment of inertia of the shaft is J = tallow =
Tc ; J
84(106 ) =
pd4 p d 4 a b = . Thus, 2 2 32
152.79(d−2)
π d 4 −32 Ans.
= d 0.02100 = m 21.00 mm d = 0.8118 in. = use d 25 mm
Ans. d = 21.00 mm 342
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5–38. The 25mmdiameter shaft on the motor is made of a material having an allowable shear stress of tallow = 75 MPa. If the motor is operating at its maximum power of 5 kW, determine the minimum allowable rotation of the shaft.
Solution Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.01254 B = 38.3495(109) m4. 2 tallow =
Tc ; J
75(106) =
T(0.0125) 38.3495(109)
T = 230.10 N # m Internal Loading: T =
P ; v
230.10 =
5(103) v
v = 21.7 rad>s
Ans.
Ans. v = 21.7 rad>s 343
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5–39. The drive shaft of the motor is made of a material having an allowable shear stress of tallow = 75 MPa . If the outer diameter of the tubular shaft is 20 mm and the wall thickness is 2.5 mm, determine the maximum allowable power that can be supplied to the motor when the shaft is operating at an angular velocity of 1500 rev>min.
Solution Internal Loading: The angular velocity of the shaft is v = a 1500
rev 2p rad 1 min ba ba b = 50p rad>s min 1 rev 60 s
We have T =
P P = v 50p
Allowable Shear Stress: The polar moment of inertia of the shaft is p J = A 0.014  0.00754 B = 10.7379(109) m4. 2
tallow =
Tc ; J
75(106) =
a
P b (0.01) 50p
10.7379(109)
P = 12 650.25 W = 12.7 kW
Ans.
Ans. P = 12.7 kW 344
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*5–40. The pump operates using the motor that has a power of 85 W. If the impeller at B is turning at 150 rev>min, determine the maximum shear stress in the 20mmdiameter transmission shaft at A.
150 rev/min A
B
Solution Internal Torque: v = 150
rev 2p rad 1 min ¢ ≤ = 5.00p rad>s min rev 60 s
P = 85 W = 85 N # m>s T =
P 85 = = 5.411 N # m v 5.00p
Maximum Shear Stress: Applying torsion formula tmax = =
Tc J 5.411 (0.01) p 4 2 (0.01 )
Ans.
= 3.44 MPa
Ans: tmax = 3.44 MPa 345
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5–41. Two wrenches are used to tighten the pipe. If P = 300 N is applied to each wrench, determine the maximum torsional shear stress developed within regions AB and BC. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm. Sketch the shear stress distribution for both cases.
P 250 mm
C B A
Solution
250 mm
Internal Loadings: The internal torque developed in segments AB and BC of the pipe can be determined by writing the moment equation of equilibrium about the x axis by referring to their respective free  body diagrams shown in Figs. a and b. ©Mx = 0; TAB  300(0.25) = 0
P
TAB = 75 N # m
And ©Mx = 0; TBC  300(0.25)  300(0.25) = 0
TBC = 150 N # m
Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254  0.014 B = 22.642(10  9)m4. 2
A tmax B AB =
75(0.0125) TAB c = 41.4 MPa = J 22.642(10  9)
A tAB B r = 0.01 m = A tmax B BC =
Ans.
TAB r 75(0.01) = 33.1 MPa = J 22.642(10  9)
150(0.0125) TBC c = 82.8 MPa = J 22.642(10  9)
A tBC B r = 0.01 m =
Ans.
TBC r 150(0.01) = 66.2 MPa = J 22.642(10  9)
The shear stress distribution along the radial line of segments AB and BC of the pipe is shown in Figs. c and d, respectively.
Ans: (tmax)AB = 41.4 MPa, (tmax)BC = 82.8 MPa 346
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5–42. Two wrenches are used to tighten the pipe. If the pipe is made from a material having an allowable shear stress of tallow = 85 MPa, determine the allowable maximum force P that can be applied to each wrench. The pipe has an outer diameter of 25 mm and inner diameter of 20 mm.
P 250 mm
C B A
Solution
250 mm
Internal Loading: By observation, segment BC of the pipe is critical since it is subjected to a greater internal torque than segment AB. Writing the moment equation of equilibrium about the x axis by referring to the freebody diagram shown in Fig. a, we have ©Mx = 0; TBC  P(0.25)  P(0.25) = 0
P
TBC = 0.5P
Allowable Shear Stress: The polar moment of inertia of the pipe is p J = A 0.01254  0.014 B = 22.642(10  9)m4 2 tallow =
TBC c ; J
85(106) =
0.5P(0.0125) 22.642(10  9)
P = 307.93N = 308 N
Ans.
Ans. P = 308 N 347
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5–43. The solid shaft has a linear taper from rA at one end to rB at the other. Derive an equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis.
T rB T B A
x
L
rA
Solution r = rB + =
rBL + (rA  rB)(L  x) rA  rB (L  x) = L L
rA(L  x) + rBx L
tmax =
Tc Tr 2T = p 4 = J pr3 2 r
= pc
2T rA(L  x) + rBx L
d
3
=
2TL3 p[rA(L  x) + rBx]3
Ans.
Ans: tmax =
348
2TL3 p[rA(L  x) + rBx]3
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*5–44. A motor delivers 375 kW to the shaft, which is tubular and has an outer diameter of 50 mm. If it is rotating at 200 radNs, determine its largest inner diameter to the nearest mm if the allowable shear stress for the material is tallow = 175 MPa.
A
B
150 mm
Solution P 375(10 3 ) N ⋅ m−s = T=
P 375(10 3 ) = = 1.875(10 3 ) N ⋅ m ω 200
tmax = tallow =
175(106 ) =
Tc J
1.875(10 3 )(0.025) π
2
( 0.0254 − ci4 )
ci = 0.02166 m = 0.04332 = m 43.32 mm = d 2= ci 2(0.02166) Use di = 43 mm
Ans.
Ans: Use di = 43 mm 349
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The A36 steel tubular shaft is 2 m long and has an outer diameter of 50 mm. When it is rotating at 40 rad>s, it transmits 25 kW of power from the motor M to the pump P. Determine the smallest thickness of the tube if the allowable shear stress is tallow = 80 MPa.
P
M
Solution The internal torque in the shaft is 25(103) P = = 625 N # m v 40 p The polar moment of inertia of the shaft is J = (0.0254  ci 4). Thus, 2 T =
tallow =
Tc ; 80(106) = J
625(0.025) p 4 2 (0.025
 ci 4)
ci = 0.02272 m So that t = 0.025  0.02272 Ans.
= 0.002284 m = 2.284 mm = 2.28 mm
Ans: t = 2.28 mm 350
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The A36 solid steel shaft is 2 m long and has a diameter of 60 mm. It is required to transmit 60 kW of power from the moor M to the pump P. Determine the smallest angular velocity the shaft if the allowable shear stress is tallow = 80 MPa.
P
M
Solution The polar moment of inertia of the shaft is J = tallow =
Tc ; J
80(106) =
p (0.034) = 0.405(106)p m4. Thus, 2
T(0.03) 0.405(106)p
T = 3392.92 N # m P = Tv;
60(103) = 3392.92 v Ans.
v = 17.68 rad>s = 17.7 rad>s
Ans: v = 17.7 rad>s 351
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5–47. The propellers of a ship are connected to an A36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaft rotates at 20 rad>s, determine the maximum torsional stress in the shaft and its angle of twist.
Solution T =
4.5(106) P = = 225(103) N # m v 20
tmax =
f =
225(103)(0.170) Tc = = 44.3 MPa p J [(0.170)4  (0.130)4] 2
Ans.
225 1 103 2 (60) TL = = 0.2085 rad = 11.9° p JG [(0.170)4  (0.130)4)75(109) 2
Ans.
Ans: tmax = 44.3 MPa, f = 11.9° 352
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*5–48. The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain in the shaft is gmax = Tc>JG. What is the shear strain on an element located at point A, c>2 from the center of the shaft? Sketch the shear strain distortion of this element.
T
c/ 2
c
A
T
L
Solution From the geometry: gL = r f; g = Since f = g =
rf L
TL , then JG
Tr JG
(1)
However the maximum shear strain occurs when r = c gmax =
Tc JG
QED
Shear strain when r = g =
T(c>2) JG
=
c is from Eq. (1), 2
Tc 2 JG
Ans.
Ans: g = 353
Tc 2 JG
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The A36 steel shaft has a diameter of 50 mm and is subjected to the distributed and concentrated loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x.
200 Nm/m
A
250 Nm x 0.5 m
C 0.5 m
B
Solution Internal Torque: As shown on FBD. Maximum Shear Stress: The maximum torque occurs at x = 0.5 m where Tmax = 150 + 200(0.5) = 250 N # m. tmax = ABS
250(0.025) Tmax c = 10.2 MPa = p 4 J 2 (0.025 )
Ans.
Angle of Twist: For 0 … x 6 0.5 m L0
L
=
L0
x
=
150x + 100x2 JG
f(x) =
= =
T(x) dx JG
(150 + 200x) dx JG
150x + 100x2
p 2
( 0.0254 ) 75.0 ( 109 )
3 3.26x
+ 2.17x2 4 (103) rad
At x = 0.5 m, f = f C = 0.00217 rad
For 0.5 m 6 x 6 1 m Since T(x) = 0, then f(x) = f C = 0.00217 rad
Ans: tmax = 10.2 MPa ABS
354
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The 60mmdiameter shaft is made of 6061T6 aluminum having an allowable shear stress of tallow= 80 MPa. Determine the maximum allowable torque T. Also, find the corresponding angle of twist of disk A relative to disk C.
1.20 m A B 2T 3
Solution
T
Internal Loading: The internal torques developed in segments AB and BC of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: Segment AB is critical since it is subjected to a greater p internal torque. The polar moment of inertia of the shaft is J = (0.034) = 2 0.405 ( 10  6 ) p m4. We have tallow =
TAB c ; J
80(106) =
1.20 m
C
1 3T
(23 T)(0.03) 0.405(10  6)p
T = 5089.38 N # m = 5.09 kN # m
Ans.
Angle of Twist: The internal torques developed in segments AB and BC of the shaft 2 1 are TAB = (5089.38) = 3392.92 N # m and TBC =  (5089.38) = 1696.46 N # m. 3 3 We have f A>C = g f A>C =
TBCLBC TiLi TABLAB = + JiGi JGal JGal
3392.92(1.20) 6
9
0.405(10 )p(26)(10 )
+
 1696.46(1.20) 0.405(10  6)p(26)(109) Ans.
= 0.06154 rad = 3.53°
Ans: T = 5.09 kN # m, f A>C = 3.53° 355
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5–51. The 60mmdiameter shaft is made of 6061T6 aluminum. If the allowable shear stress is tallow = 80 MPa, and the angle of twist of disk A relative to disk C is limited so that it does not exceed 0.06 rad, determine the maximum allowable torque T.
1.20 m A B 2T 3
Solution
T
Internal Loading: The internal torques developed in segments AB and BC of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: Segment AB is critical since it is subjected to a greater p internal torque. The polar moment of inertia of the shaft is J = (0.034) = 2 6 4 0.405(10 )p m . We have tallow =
TAB c ; J
80(103) =
1.20 m
C
1 3T
(23T)(0.03) 0.405(10  6)p
T = 5089.38 N # m = 5.089 kN # m Angle of Twist: It is required that f A>C = 0.06 rad. We have f A>C = Σ 0.06 =
TiLi TBC LBC TAB LAB = + JiGi JGal JGal (23T)(1.2)
0.405(10  6)p(26)(109)
+
( 13T)(1.2) 0.405(10  6)p(26)(109)
T = 4962.14 N # m = 4.96 kN # m (controls)
Ans.
Ans: T = 4.96 kN # m 356
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The splined ends and gears attached to the A992 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The shaft has a diameter of 40 mm.
1200 Nm
400 Nm B
600 Nm D C
600 mm A
Solution Internal Torque: The torque diagram shown in Fig. a can be plotted. From this diagram, TAC = 200 N # m, TCD = 800 N # m and TDB = 400 N # m.
400 mm
Angle of Twist: f B>A = g
500 mm
200 Nm
TL JG
1 (T L + TCD LCD + TDB LDB) JG AC AC 1 = [200(0.4) + 800(0.5) + (  400)(0.6)] JG 240 N # m2 = JG =
For A992 steel, G = 75 GPa. Then f B>A =
240 p 2
( 0.02 ) 3 75 ( 109 ) 4 4
= (0.01273 rad) a
180° b = 0.730° G p rad
Ans.
Ans: f B>A = 0.730° G 357
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5–53. The hydrofoil boat has an A36 steel propeller shaft shaftisthat is long. 100 ftItlong. It is connected to an diesel inlineengine diesel that 30 m is connected to an inline engine that delivers a maximum power 2500 hp and that delivers a maximum power of 2000 kWofand causes the causes the shaft to rpm. rotate at outer 1700 diameter rpm. If ofthe shaft to rotate at 1700 If the theouter shaft diameter thethe shaft is 8thickness in. and the wall thickness is 38 the in., is 200 mmofand wall is 10 mm, determine determine shear the maximum shear in stress developed in the maximum stress developed the shaft. Also, what is shaft. Also, what is the of “wind orshaft angleatof twist in the the “wind up,” or angle twistup,” in the full power? shaft at full power?
30 mft 100
Solution Internal Torque:
ω = 1700
πev 2π πad 1 min = 56.667π πad−s min 1 πev 60 s
= = P 2000 kW 2000(10 3 ) N ⋅ m−s T=
P 2000(10 3 ) = = 11.23(10 3 ) N ⋅ m ω 56.667π
Maximum Shear Stress: Applying torsion Formula. tmax =
Tc J
11.23(10 3 )(0.1) 6 = 20.80(10 ) N−m 2 20.8 MPa == π (0.14 − 0.09 4 ) 2
Ans.
Angle of Twist: f =
11.23(10 3 )(30) TL = π JG [ 2 (0.14 − 0.09 4 )][75(109 )] = 0.08319 πad = 4.77°
Ans.
358
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5–54. The turbine develops 300 kW of power, which is transmitted to the gears such that both B and C receive an equal amount. If the rotation of the 100mmdiameter A992 steel shaft is v = 600 rev>min., determine the absolute maximum shear stress in the shaft and the rotation of end D of the shaft relative to A. The journal bearing at D allows the shaft to turn freely about its axis.
A B
v
C
1.5 m
D
Solution
2m 1m
External Applied Torque: Gears B and C withdraw equal amount of power, 300 PB = PC = P = = 150 kw. Here, the angular velocity of the shaft is 2 v = a600
rev 2p rad 1 min ba ba b = 20p rad>s. Then, the torque exerted on gears min 1 rev 60 s
B and C can be determined from TB = TC
150 ( 103 ) p 7500 = = N#m p v 20p
Using this result, the torque diagram shown in Fig. a can be plotted. Maximum Shear Stress: From the torque diagram, we notice that the maximum 15000 torque Tmax = N # m occurs at region BA. Thus it is the critical region where p the absolute maximim shear stress occurs. Applying the torsion formula, t abs = max
Tmax C = J
15000 p p 2
(0.05)
( 0.054 )
= 24.32 ( 106 ) Pa = 24.3 MPa
Angle of twist: From the torque diagram, TDC = 0, TCB = 15000 TBA = N#m p f D>A = g = =
Ans. 7500 N # m and p
TL 1 = (T L + TCB LCB + TBA LBA) JG JG DC DC
1 7500 15000 c0 + a b(2) + a b(1.5)d p p JG 37500 N # m2 p JG
For A992 steel, G = 75 GPa. Thus f D>A =
37500 p p c ( 0.054 ) d [75 ( 109 ) ] 2
= (0.016211 rad) a
= 0.9288°
180° b p rad
Ans.
= 0.929° Ans: t abs = 24.3 MPa max
f D>A = 0.929° 359
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5–55. The shaft is made of A992 steel. It has a diameter of 25 mm and is supported by bearings at A and D, which allow free rotation. Determine the angle of twist of B with respect to D.
A
B
90 Nm C
0.6 m 90 Nm
0.75 m
D 0.9 m
Solution The internal torques developed in segments BC and CD are shown in Figs. a and b. The polar moment of inertia of the= shaft is J
π
= (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2
TiLi TBC LBC TCD LCD = + FB>D = a JiGi J Gst J Gst
=
90(0.75) [38.35(10 −9 )]75(106 )
+0
= 0.02347 πad = 1.34°
Ans.
TCD = 0
(b)
TBC = 90 N ⭈ m
90 N ⭈ m
(a)
Ans: fB>D = 1.34° 360
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5–56. The shaft is made of A36 steel. It has a diameter of 25 mm and supportedbybybearings bearingsatatAAand andD, D, which allow 1 in. and is is supported free rotation. Determine the angle of twist of gear C with respect to B.
A
B
90 60 Nm lbft C
0.6 2 ftm 90 lbft Nm 60
0.75 m 2.5 ft
D 0.9 3 ftm
Solution The internal torque developed in segment BC is shown in Fig. a The polar moment of inertia of the= shaft is J fC>B =
π
= (0.01254 ) 38.35(10 −9 ) m 4 . Thus, 2
TBC LBC 90(0.75) = J Gst [38.35(10 −9 )]75(106 )
= 0.02347 πad = 1.34°
Ans.
TBC = 90 N ⭈ m
90 N ⭈ m
(a)
Ans: fC>B = 1.34° 361
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5–57. The rotating flywheelandshaft, when brought to a sudden stop at D, begins to oscillate clockwisecounterclockwise such that a point A on the outer edge of the flywheel is displaced through a 6mm arc. Determine the maximum shear stress developed in the tubular A36 steel shaft due to this oscillation. The shaft has an inner diameter of 24 mm and an outer diameter of 32 mm. The bearings at B and C allow the shaft to rotate freely, whereas the support at D holds the shaft fixed.
D 1.5 m
C
B A
Solution
75 mm
s = ru 6 = 75 f f = 0.08 rad f = 0.08 =
TL JG T(1.5) J(75) ( 109 )
T = 4 ( 109 ) J tmax = =
TC J 4(109)(J)(0.016) J Ans.
= 64.0 MPa
Ans: tmax = 64.0 MPa 362
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5–58. The A992 steel shaft has a diameter of 50 mm and is subjected to the distributed loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x.
500 Nm
A
1000 Nm/m x 0.5 m
C 0.5 m
B
Solution Internal Torque: Referring to the FBD of the right segment of the shaft shown in Fig. a (0.5 m < x … 1 m) T(x) = {1000  1000x} N # m
ΣMx = 0; T(x)  1000 (1 x) = 0 And Fig. b (0 … x 6 0.5 m)
ΣMx = 0; T(x) + 500 (0.5x)  1000 (0.5) = 0 TAc = {500x + 250} N # m
Maximum Shear Stress: The maximum torque at x = 0.5 m, Tmax = 1000  1000(0.5) = 500 N # m. Applying the torsion formula, abs = tmax
500 (0.025) TmaxC = p = 20.37 ( 106 ) Pa = 20.4 MPa 4 J 2 (0.025 )
where
Ans.
Angle of Twist: For A992 steel G = 75 GPa. For region 0 … x 6 0.5 m (between B and C). f(x) = =
L
T(x)dx JG
=
3 ( 0.025 )4 3 75 ( 10 )4 Lo 1
4
p 2
5 0.005432 ( x2
9
x (500 x + 250) dx
+ x )6 rad
Ans.
At x = 0.5 m, f = f C = 0.004074 rad. For region 0.5 m 6 x … 1 m f (x) = f C +
L
T(x) dx
= 0.004074 +
JG x
3 ( 0.0254 ) 3 75 ( 109 ) 4 L0.5 m 1
p 2
(1000  1000x) dx
= {  0.01086x2 + 0.02173 x  0.004074} rad The maximum f occurs at where
Ans.
df = 0. dx
df =  0.02173x + 0.02173 = 0 dx x = 1m Thus, f max = f B =  0.01086 ( 12 ) + 0.02173(1)  0.004074 = 0.006791 rad The plot of f vs x is as shown in Fig. c.
Ans: abs = 20.4 MPa tmax For 0 … x 6 0.5 m, f(x) = 5 0.005432 ( x2 + x )6 rad For 0.5 m 6 x … 1 m, f(x) = { 0.01086x2 + 0.02173 x  0.004074} rad 363
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The shaft is made of A992 steel with the allowable shear stress of tallow = 75 MPa. If gear B supplies 15 kW of power, while gears A, C and D withdraw 6 kW, 4 kW and 5 kW, respectively, determine the required minimum diameter d of the shaft to the nearest millimeter. Also, find the corresponding angle of twist of gear A relative to gear D. The shaft is rotating at 600 rpm.
A B C
600 mm
D
600 mm 600 mm
Solution Internal Loading: The angular velocity of the shaft is v = a600
rev 1 min 2p rad ba ba b = 20p rad>s min 60 s 1 rev
Thus, the torque exerted on gears A, C, and D are TA =
6(103) PA = = 95.49 N # m v 20p
TC =
4(103) PC = = 63.66 N # m v 20p
TD =
5(103) PD = = 79.58 N # m v 20p
The internal torque developed in segments AB, CD, and BC of the shaft are shown in Figs. a, b, and c, respectively. Allowable Shear Stress: Segment BC of the shaft is critical since it is subjected to a greater internal torque.
tallow
TBC c = ; 75(106) = J
d 143.24 a b 2 p d 4 a b 2 2
d = 0.02135 m = 21.35 mm Ans. p Angle of Twist: The polar moment of inertia of the shaft is J = (0.0114) = 2 7.3205(10  9)p m4. We have Use
d = 22 mm
f A>D = Σ f A>D =
TiLi TBCLBC TCDLCD TABLAB = + + JiGi JGst JGst JGst
0.6 ( 95.49 + 143.24 + 79.58) 7.3205(10 )p(75)(109) 9
Ans.
= 0.04429 rad = 2.54°
Ans: Use d = 22 mm, f A>D = 2.54° F 364
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Gear B supplies 15 kW of power, while gears A, C and D withdraw 6 kW, 4 kW and 5 kW, respectively. If the shaft is made of steel with the allowable shear stress of tallow = 75 MPa, and the relative angle of twist between any two gears cannot exceed 0.05 rad, determine the required minimum diameter d of the shaft to the nearest millimeter. The shaft is rotating at 600 rpm.
A B C
600 mm
D
600 mm 600 mm
Solution Internal Loading: The angular velocity of the shaft is v = a600
rev 1 min 2p rad ba ba b = 20p rad>s min 60 s 1 rev
Thus, the torque exerted on gears A, C, and D are TA =
6(103) PA = = 95.49 N # m v 20p
TC =
4(103) PC = = 63.66 N # m v 20p
TD =
5(103) PD = = 79.58 N # m v 20p
The internal torque developed in segments AB, CD, and BC of the shaft are shown in Figs. a, b, and c, respectively. Allowable Shear Stress: Segment BC of the shaft is critical since it is subjected to a greater internal torque.
tallow
TBC c = ; J
75(106) =
d 143.24 a b 2 p d 4 a b 2 2
d = 0.02135 m = 21.35 mm Angle of Twist: By observation, the relative angle of twist of gear D with respect to gear B is the greatest. Thus, the requirement is f D>B = 0.05 rad. f D>B = Σ
TiLi TBC LBC TCD LCD = + = 0.05 JiGi JGst JGst
0.6 p d 4 9 2 ( 2 ) (75)(10 )
(143.24 + 79.58) = 0.05
d = 0.02455 m = 24.55 mm = 25 mm (controls!) Use
Ans.
d = 25 mm
Ans: Use d = 25 mm 365
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The turbine develops 150 kW of power, which is transmitted to the gears such that C receives 70% and D receives 30%. If the rotation of the 100mmdiameter A36 steel shaft is v = 800 rev>min., determine the absolute maximum shear stress in the shaft and the angle of twist of end E of the shaft relative to B. The journal bearing at E allows the shaft to turn freely about its axis.
B
v
C
D
3m
E
4m
Solution
2m
rev 1 min 2p rad P = Tv; 150(10 ) W = T a800 ba ba b min 60 sec 1 rev 3
T = 1790.493 N # m
TC = 1790.493(0.7) = 1253.345 N # m TD = 1790.493(0.3) = 537.148 N # m Maximum torque is in region BC. tmax =
1790.493(0.05) TC = = 9.12 MPa p 4 J 2 (0.05)
f E>B = Σa =
Ans.
TL 1 b = [1790.493(3) + 537.148(4) + 0] JG JG 7520.171
p 4 9 2 (0.05) (75)(10 )
Ans.
= 0.0102 rad = 0.585°
Ans: tmax = 9.12 MPa, f E>B = 0.585° 366
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5–62. The turbine develops 150 kW of power, which is transmitted to the gears such that both C and D receive an equal amount. If the rotation of the 100mmdiameter A36 steel shaft is v = 500 rev>min., determine the absolute maximum shear stress in the shaft and the rotation of end B of the shaft relative to E. The journal bearing at E allows the shaft to turn freely about its axis.
B
v
C
D
3m
E
4m
Solution
2m
rev 1 min 2p rad P = Tv; 150(10 ) W = T a500 ba ba b min 60 sec 1 rev 3
T = 2864.789 N # m
TC = TD =
T = 1432.394 N # m 2
Maximum torque is in region BC. tmax =
2864.789(0.05) TC = = 14.6 MPa p 4 J 2 (0.05)
f B>E = Σa =
Ans.
TL 1 b = [2864.789(3) + 1432.394(4) + 0] JG JG 14323.945
p 4 9 2 (0.05) (75)(10 )
Ans.
= 0.0195 rad = 1.11°
Ans: tmax = 14.6 MPa, f B>E = 1.11° 367
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5–63. The 50mmdiameter A992 steel shaft is subjected to the torques shown. Determine the angle of twist of the end A. 400 Nm B
200 Nm C
300 mm 600 mm
600 mm
Solution
800 Nm D A
Internal Torque: The torque diagram shown in Fig. a can be plotted. From this diagram, TAD = 800 N # m, TDC =  1000 N # m and TCB = 600 N # m. Angle of Twist: f B>A = g
TL JG
=
1 (T L + TDC LDC + TCB LCB) JG AD AD
=
1 [(  800)(0.6) + ( 1000)(0.6) + (  600)(0.3)] JG
= 
1260 N # m2 JG
For A992 steel, G = 75 GPa. Then fA =
1260 p 2
( 0.0254 ) 3 75 ( 109 ) 4
= ( 0.02738 rad) a
180° b =  1.569° = 1.57° F p rad
Ans.
Ans: f A = 1.57° F 368
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*5–64. The 60mmdiameter solid shaft is made of 2014T6 aluminum and is subjected to the distributed and concentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft.
1.5 kNm 2 kNm/m
B 0.4 m 0.6 m
A
Solution Internal Torque: Referring to the FBD of the right segments of the shaft shown in Fig. a and b gMx = 0; TAC + 2000x = 0 TAC = (  2000x) N # m
And
gMx = 0; TCB + 2000 (0.6)  1500 = 0 TCB = 300 N # m
Angle of Twist: For 2014  T6 Aluminum, G = 27 GPa. f A = g
x
1 TL c T dx + TCB LCB d = JG JG L0 AC =
1
3
p 4 2 (0.03)
4 3 27 ( 10 ) 4
= (  0.006986 rad)a
9
c
L0
180° b p rad
0.6 m
(  2000 x)dx + 300 (0.4) d
Ans .
=  0.400° = 0.400° F
Ans: f A = 0.400° F 369
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5–65. The two shafts are made of A36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when the torques are applied to the assembly as shown.
D
10 in. 250 mm
C
80 lbft 120 Nm A
30 in. 750 mm
40 60 lbft Nm
8 in. 200 mm 10 in. 250 mm
Solution
4 in. 100 mm 12 in. 300 mm
6 in.mm 150 B
Internal Torque: As shown on FBD. Angle of Twist: TL fE = a JG
60 F = 0.100 = 600 N
1 {−90(0.75) + 30(0.25)} = π 4 [ 2 (0.0125 )][75(109 )]
0.100 m 60 N · m
TAF = –60 N · m
 0.01778 = rad = 0.01778 == –0.020861 0.020861 rad rad
150 150 φF = = = 0.031291 πad φE (0.020861) 100 100 fA>F =
0.150 m
60 N · m
TCA = –600(0.150) = –90 N · m F = 600 N
TGF LGF JG
0.150 m TBA = 30 N · m
−60(0.25) = π [ 2 (0.01254 )][75(109 )]
120 N · m
F = 600 N
rad rad = –0.0052152  0.004445 rad = 0.0052152 0.004445 rad
fF fE
fA = fF + fA>F
fA>F
0.0052152 = 0.031291 0.02667 ++ 0.004445 = 0.036506 rad = 2.09°
Ans.
Ans. fA = 2.09° 370
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5–66. The A36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (kx2) N # m>m, where x is in meters. If a torque of T = 50 N # m is applied to the bolt head, determine the constant k and the amount of twist in the 50mm length of the shank. Assume the shank has a constant radius of 4 mm. B
t
Solution
T = 50 Nm 50 mm
dT = t dx
T =
L0
x
A
0.05
0.05 m
x3 kx dx = k ` 3
= 41.667(10  6)k
2
0
50  41.6667(10  6) k = 0 k = 1.20(106) N>m2
In the general position, T =
f =
L
T(x)dx JG
=
1 JG L0
1.875 = JG
L0
x
1.20(106)x2dx = 0.4(106)x3
0.05 m
0.4(106)x4 1 = c 50x d JG 4 =
Ans.
[50  0.4(106)x3]dx 0.05 m
`
0
1.875 p 4 9 2 (0.004 )(75)(10 )
Ans.
= 0.06217 rad = 3.56°
Ans: k = 1.20(106) N>m2, f = 3.56° 371
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5–67. The A36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (kx2>3) N # m>m, where x is in meters. If a torque of T = 50 N # m is applied to the bolt head, determine the constant k and the amount of twist in the 50mm length of the shank. Assume the shank has a constant radius of 4 mm. B
t
Solution
T = 50 Nm 50 mm
dT = t dx T =
L0
0.05
x
A
2 3 5 0.05 kx3 dx = k x3 ` = (4.0716)(10  3) k 5 0
50  4.0716 (10  3) k = 0
2
k = 12.3(103) N>m(3)
Ans.
In the general position, T =
L0
x
5
2
12.28(103)x3 dx = 7.368(103) x3
Angle of twist: f = = =
L
T(x) dx JG
=
1 JG L0
0.05 m
8
[50  7.3681(103)x3]dx
3 8 0.05 m 1 c 50x  7.3681(103)a bx3 d JG 8 0 1.5625
p 4 9 2 (0.004 )(75)(10 )
Ans.
= 0.0518 rad = 2.97°
Ans: k = 12.3(103) N>m2>3, f = 2.97° 372
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*5–68. The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the angle of twist at end A. The shear modulus is G.
B t0
x t = t0 1 + — L
( ( ) 2)
x L
Solution TB TB =
L
L
f =
L
2 t0
t dx = 0
t dx = t 0
= t0 c x +
A
L
3
a1 +
x2 b dx L2
L x L 4 d ` = t 0 aL + b = t 0 L 2 3 3 3L 0
T (x) dx JG L
1 4 x3 c t 0 L  t 0 ax + b d dx JG L0 3 3L2
= =
L t0 4 7 t 0 L2 x2 x4 c Lx  a + bd ` = 2 JG 3 2 12 JG 12L 0
However J =
f =
p 4 c, 2
7 t 0 L2
Ans.
6 p c4 G
Ans: f = 373
7 t 0 L2 6 p c4 G
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5–69. The tubular drive shaft for the propeller of a hovercraft is 6 m long. If the motor delivers 4 MW of power to the shaft when the propellers rotate at 25 rad>s, determine the required inner diameter of the shaft if the outer diameter is 250 mm. What is the angle of twist of the shaft when it is operating? Take tallow = 90 MPa and G = 75 GPa.
6m
Solution Internal Torque: P = 4 ( 106 ) W = 4 ( 106 ) N # m>s T =
4 ( 106 ) P = = 160(103) N # m v 25
Maximum Shear Stress: Applying torsion formula. tmax = tallow =
Tc J 160 ( 103 ) ( 0.125 )
90 ( 106 ) = p 2
c 0.1254  a
dt 4 b d 2
Ans.
d t = 0.2013 m = 201 mm Angle of Twist: f =
TL = JG
160 ( 103 ) ( 6 ) p 2
( 0.1254  0.100654 ) 75 ( 109 ) Ans.
= 0.0576 rad = 3.30°
Ans: d t = 201 mm, f = 3.30° 374
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5–70. The A36 steel assembly consists of a tube having an outer outerradius radiusofof1 in. 25 and mma and wall thickness mm. wall athickness of 0.125of in.3Using Using rigidat plate it is connected the 1indiameter solid 25mma rigidaplate B, itatisB, connected to theto solid diameter AB. Determine the rotation of the end tube’s shaft AB.shaft Determine the rotation of the tube’s C end if a C if a torque appliedtotothe thetube tubeat atthis this end. The torque of 200oflb25# N in.· m is is applied end A of the shaft is fixed supported.
B C 25 Nm 200 lbin.
100 mm 4 in.
A 150 mm 6 in.
Solution fB =
TABL 25(0.25) = = 0.0021730 πad JG [ π2 (0.01254 )][75(109 )]
fC>B =
25 N · m TAB = 25 N · m
TCBL −25(0.1) = JG [ π2 (0.0254 − 0.022 4 )][75(109 )]
TCB = −25 N · m 25 N · m
0.0001357 πad G = −0.0001357 πad = (+ G) fC = fB + fC>B = 0.0021730 + 0.0001357 rad== 0.113° 0.132° = 0.0023087 0.001964 rad
Ans.
Ans. fC = 0.113° 375
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5–71. The A36 hollow steel shaft is 2 m long and has an outer diameter of 40 mm. When it is rotating at 80 rad>s, it transmits 32 kW of power from the engine E to the generator G. Determine the smallest thickness of the shaft if the allowable shear stress is tallow = 140 MPa and the shaft is restricted not to twist more than 0.05 rad.
E
G
Solution P = Tv 32(103) = T(80) T = 400 N # m Shear stress failure t =
Tc J
tallow = 140 (106) =
400 (0.02) p 4 2 (0.02
 ri 4 )
ri = 0.01875 m Angle of twist limitation: f = 0.05 =
TL JG 400(2) p 4 2 (0.02
 r i 4)(75)(109)
ri = 0.01247 m
(controls)
t = ro  ri = 0.02  0.01247 = 0.00753 m Ans.
= 7.53 mm
Ans: t = 7.53 mm 376
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*5–72. The A36 solid steel shaft is 3 m long and has a diameter of 50 mm. It is required to transmit 35 kW of power from the engine E to the generator G. Determine the smallest angular velocity of the shaft if it is restricted not to twist more than 1°.
E
G
Solution f = 1°(p) 180°
=
TL JG T(3) p 4 9 2 (0.025 )(75)(10 )
T = 267.73 N # m P = Tv
35(103) = 267.73(v) Ans.
v = 131 rad>s
Ans: v = 131 rad>s 377
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5–73. The motor produces a torque of T = 20 N # m on gear A. If gear C is suddenly locked so it does not turn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to D of the L2steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also, calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings at G at H.
A 100 mm 30 mm B
G
60 mm D
0.2 m
Solution
E C
0.8 m
H 40 mm
0.4 m
F
0.2 m
F (0.03) = 20 F = 666.67 N
T′ = (666.67) (0.1) = 66.67 N # m Since shaft is held fixed at C, the torque is only in region EF of the shaft. f F>E =
66.67(0.6) TL = = 0.999 ( 10 )  3 rad p JG 3 ( 0.025 ) 4  ( 0.015 ) 4 4 75 ( 109 ) 2
Ans.
Since the torque in region ED is zero, f F>D = 0.999 (10)  3 rad tmax =
Ans.
66.67(0.025) Tc = p J ( ( 0.025 ) 4  ( 0.015 ) 4 ) 2 Ans.
= 3.12 MPa
Ans: f F>E = 0.999(10) 3 rad, f F>D = 0.999(10) 3 rad, tmax = 3.12 MPa 378
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5–74. The shaft has a radius c and is subjected to a torque per unit length of t0, which is distributed uniformly over the shaft’s entire length L. If it is fixed at its far end A, determine the angle of twist f of end B. The shear modulus is G.
A
f L
c B
t0
Solution f = 3 = =
T (x) dx JG
 t0 L x dx JG 30
 t 0 x2 L  t 0 L2 c d ` = JG 2 JG 2 0  t 0 L2 2JG
However, J = f =
=
 t 0 L2 4
pc G
=
p 4 c 2 t 0 L2
Ans.
p c4 G
Ans: f = 379
t 0L2 pc 4G
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5–75. The 60mmdiameter solid shaft is made of A36 steel and is subjected to the distributed and concentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft due to these loadings.
2 kNm/m 600 N · m 400 Nm
B 0.8 m 0.6 m
A
Solution Internal Torque: As shown on FBD. Angle of Twist: fA = Σ =
TL JG  400 (0.6)
p 2
( 0.03 ) 75.0 ( 10 ) 4
9
+ 3
0.8 m
0
(200  2000x)dx p 2
( 0.034 ) 75.0 ( 109 ) Ans.
=  0.007545 rad = 0.432° F
Ans: f A = 0.432° F 380
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*5–76. The contour of the surface of the shaft is defined by the equation y = e ax, where a is a constant. If the shaft is subjected to a torque T at its ends, determine the angle of twist of end A with respect to end B. The shear modulus is G.
B
x
y = e ax T
y A L
T
Solution f =
T dx J(x)G L
where, J(x) =
p ax 4 (e ) 2
L
= = =
L 2T dx 2T 1 = ab ` pG 4 a e 4ax 0 pG L0 e 4ax
2T 1 1 T e 4aL  1 a+ b = a b pG 4 a e 4aL 4a 2apG e 4aL T (1  e  4aL) 2apG
Ans.
Ans: f = 381
T (1  e  4aL) 2apG
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5–77. The steel shaft is made from two segments: AC has a diameter of 0.5 12 mm, andCB CBhas hasa adiameter diameterofof1 25 in, and in. mm. If it If is it is fixed at ends its ends and andsubjected subjectedtotoaa torque torque of fixed at its A A and B Band 300 N · m, determine the maximum shear stress in the shaft. Gst == 75 GPa. 32 ksi. 10.8110
A
0.5mm in. 12 C D 300 N m
5 in. 125 mm
in. 251 mm
8 in. 200 mm
B 12 mm in. 300
Solution Equilibrium: TA + TB  300 = 0
(1)
Compatibility condition: fD>A = fD>B
TA (0.125) π [ 2 (0.006 4 )]G
300 N · m
+
TA (0.2) π [ 2 (0.01254 )]G
TB (0.3) = π [ 2 (0.01254 )]G
TA = 0.11743TB
(2)
Solving Eqs. (1) and (2) yields
TA = 31.53 N ⋅ m
TB = 268.47 N ⋅ m
(τ AC )max =
TC 31.53(0.06) 2 = = 92.92(106 ) N−m= 92.9 MPa (Max) π (0.006 4 ) J 2
(τ DB )max =
TC 268.47(0.0125) 6 87.51(10 = ) N−m 2 87.5 MPa == π (0.01254 ) J 2
Ans.
Ans: (τ AC )max 92.9 MPa 382
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5–78. The steel shaft has a diameter of 40 mm and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB of the shaft. Gst = 75 GPa.
3 kN
A 3 kN C 400 mm
50 mm 50 mm
B
600 mm
Solution Equilibrium: (1)
TA + TB  3000(0.1) = 0 Compatibility condition: f C>A = f C>B TA (400) JG
=
TB(600) JG (2)
TA = 1.5 TB Solving Eqs (1) and (2) yields: TB = 120 N # m
TA = 180 N # m (tAC)max =
180(0.02) Tc = p = 14.3 MPa 4 J 2 (0.02 )
Ans.
(tCB)max =
120(0.02) Tc = p = 9.55 MPa 4 J 2 (0.02 )
Ans.
Ans: (tAC)max = 14.3 MPa, (tCB)max = 9.55 MPa 383
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5–79. The A992 steel shaft has a diameter of 60 mm and is fixed at its ends A and B. If it is subjected to the torques shown, determine the absolute maximum shear stress in the shaft.
200 Nm B 500 Nm
D
1m
1.5 m C A
1m
Solution Referring to the FBD of the shaft shown in Fig. a, ΣMx = 0; TA + TB  500  200 = 0
(1)
Using the method of superposition, Fig. b f A = (f A)TA  (f A)T 0 =
TA (3.5) JG
 c
500 (1.5) JG
+
700 (1) JG
TA = 414.29 N # m
d
Substitute this result into Eq (1), TB = 285.71 N # m Referring to the torque diagram shown in Fig. c, segment AC is subjected to maximum internal torque. Thus, the absolute maximum shear stress occurs here. tabs = max
414.29 (0.03) TAC c = = 9.77 MPa p J 4 (0.03) 2
Ans.
Ans: tabs = 9.77 MPa max 384
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*5–80. The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the torque, determine the maximum shear stress in regions AC and CB.
B
2 kNm C
600 mm A 800 mm
Solution Equilibrium: Referring to the FBD of the shaft, Fig. a, gMx = 0; TA + TB  2000 = 0
(1)
Compatibility: It is required that f CA = f CB TA LCA TB LCB = JG JG TA (0.8) JG
=
TB (0.6) JG (2)
TA = 0.75 TB Solving Eqs (1) and (2)
TB = 1142.86 N # m TA = 857.14 N # m
Maximum Shear Stress: Applying the torsion formula, (tmax)AC =
857.14 (0.02) TA c = p = 68.21 ( 106 ) Pa = 68.2 MPa 4 J 2 ( 0.02 )
Ans.
(tmax)B C =
1142.86 (0.02) TB c = = 90.946 ( 106 ) Pa = 90.9 MPa p 4 J ( ) 0.02 2
Ans.
Ans: (tmax)AC = 68.2 MPa, (tmax)BC = 90.9 MPa 385
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5–81. The Am1004T61 magnesium tube is bonded to the A36 steel rod. If the allowable shear stresses for the magnesium and steel are (tallow)mg = 45 MPa and (tallow)st = 75 MPa, respectively, determine the maximum allowable torque that can be applied at A. Also, find the corresponding angle of twist of end A.
900 mm B
Solution
A
Equilibrium: Referring to the freebody diagram of the cut part of the assembly shown in Fig. a, ΣMx = 0; Tmg + Tst  T = 0
80 mm 40 mm
T
(1)
Compatibility Equation: Since the steel rod is bonded firmly to the magnesium tube, the angle of twist of the rod and the tube must be the same. Thus, (f st)A = (f mg)A TstL p (0.024)(75)(109) 2
=
TmgL p (0.044  0.024)(18)(109) 2 (2)
Tst = 0.2778Tmg Solving Eqs. (1) and (2), Tmg = 0.7826T
Tst = 0.2174T
Allowable Shear Stress: (tallow)mg =
Tmg c J
; 45(106) =
0.7826T(0.04) p (0.044  0.024) 2 T = 5419.25 N # m
(tallow) st =
Tstc ; J
75(106) =
0.2174T(0.02) p (0.024) 2 T = 4335.40 N # m = 4.34 kN # m (control!)
Ans.
Angle of Twist: Using the result of T, Tst = 942.48 N # m. We have fA =
942.48(0.9) TstL = = 0.045 rad = 2.58° p JstGst (0.024)(75)(109) 2
Ans.
Ans: T = 4.34 kN # m, f A = 2.58° 386
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5–82. The Am1004T61 magnesium tube is bonded to the A36 steel rod. If a torque of T = 5 kN # m is applied to end A, determine the maximum shear stress in each material. Sketch the shear stress distribution.
900 mm B
Solution
A
Equilibrium: Referring to the freebody diagram of the cut part of the assembly shown in Fig. a, ΣMx = 0; Tmg + Tst  5(103) = 0
80 mm 40 mm
T
(1)
Compatibility Equation: Since the steel rod is bonded firmly to the magnesium tube, the angle of twist of the rod and the tube must be the same. Thus, (f st)A = (f mg)A TstL p (0.024)(75)(109) 2
=
TmgL p (0.044  0.024)(18)(109) 2 (2)
Tst = 0.2778Tmg Solving Eqs. (1) and (2), Tmg = 3913.04 N # m
Tst = 1086.96 N # m
Maximum Shear Stress: (tst)max =
(tmg)max =
1086.96(0.02) Tstcst = = 86.5 MPa p Jst (0.024) 2 Tmgcmg Jmg
(tmg) r = 0.02 m =
=
Tmg r Jmg
Ans.
3913.04(0.04) = 41.5 MPa p (0.044  0.024) 2 =
3913.04(0.02) = 20.8 MPa p (0.044  0.024) 2
Ans.
Ans.
Ans: (tst)max = 86.5 MPa, (tmg)max = 41.5 MPa, (tmg) r = 0.02 m = 20.8 MPa 387
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5–83. A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N # m. If the steel portion has a diameter of 30 mm, determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa.
C
1.60 m
B 680 Nm A 0.75 m
Solution Compatibility Condition: f B> C = f B>A T(1.60) p 4 9 2 (c )(39)(10 )
=
T(0.75) p 4 9 2 (0.015 )(75)(10 )
c = 0.02134 m Ans.
d = 2c = 0.04269 m = 42.7 mm
Ans: d = 42.7 mm 388
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*5–84. Determine the absolute maximum shear stress in the shaft of Prob. 5–88.
C
1.60 m
B 680 Nm A 0.75 m
Solution Equilibrium, 2T = 680
T = 340 N # m
tabs occurs in the steel. See solution to Prob. 5–88. max
tabs = max
340(0.015) Tc = p 4 J 2 (0.015) Ans.
= 64.1 MPa
Ans: tabs = 64.1 MPa max 389
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5–85. The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T = 75 N · m is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel. Take Gst = 75 GPa, Gbr = 38 GPa.
0.9 3 ftm 0.6 2 ftm A A 12 0.5mm in.
Solution 25 mm C 1 in. C
Equilibrium:
B B T 50 75 lbft Nm T
(1)
Tbr + Tst  50 75 = 0 Both the steel tube and brass core undergo the same angle of twist fC>B fC>B =
Tbr (0.6) Tst (0.6) TL = = π JG [ 2 (0.01254 )][38(109 )] [ π2 (0.0254 − 0.01254 )][75(109 )]
75 N · m
(2)
Tbr ==0.033778 0.032464TTstst Solving Eqs. (1) and (2) yields:
# ft; 48.428Nlb· m; Tstst ==72.549 fC = ©
1.572 TTbrbr==2.451 Nlb · m# ft
2.451(0.6) 75(0.9) TL + = π 4 9 π JG [ 2 (0.0125 )][38(10 )] [ 2 (0.0254 )][75(109 )] = 0.002476 0.002019 rad ==0.142° 0.116°
(tst)max AB =
Ans.
TABc 75(0.025) 6 = 3.056(10 = ) N−m 2 3.056 MPa = π J (0.0254 ) 2
(tst)max BC =
Tst c 72.549(0.025) = π J [ 2 (0.0254 − 0.01254 )]
6 = ) N−m 2 3.15 MPa (Max) = 3.153(10
Ans.
6
(gst)max =
(tst)max 3.153(10 ) = 42.0(10 −6 ) πad = 9 G 75(10 )
Ans.
(tbr)max =
2.451(0.0125) Tbr c 6 = 0.7988(10 = ) N−m 2 0.799 MPa = π 4 J (0.0125 ) 2
Ans.
(gbr)max =
(tbr)max 0.7988(106 ) = 21.0(10 −6 ) πad = G 38(109 )
Ans.
390
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5–86. The shafts are made of A36 steel and have the diameterof of100 4 in. If aIftorque ofof 1525kip same diameter mm. a torque kN# ·ftmisis applied to gear B, determine the absolute maximum shear stress developed in the shaft.
0.75 2.5mft 0.75 2.5mft
AA
Solution
BB
Equilibrium: Referring to the free  body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, we have (1)
F(0.15)– 25 15 ==00 ©Mx = 0; TA + F(0.5)
1506mm in.
2515kN mft kip DD 30012 mm in.
CC EE
0.93mft
and (2)
©Mx = 0; F(1) = 00 F(0.3)– T TEE =
Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: fCrC = fDrD
¢
TABLAB TDE LDE TBCLBC + ≤ rC = ¢ ≤ rD JGst JGst JGst
(0.9)(0.3)  TAA(0.75) (2.5) + F(0.15)(0.75)](0.15) F(0.5)(2.5) D (0.5) = = –T TEE(3)(1) C[–T 0.5F==2.4T 2.4T TAA –0.15F EE
(3)
Solving Eqs. (1), (2), and (3), we have F ==24.51 F 4.412kN kip
TE = 7.353 m TEkN= · 4.412 kip # ft
# ft TAA ==21.32 12.79kN kip· m
Maximum Shear Stress: By inspection, segment AB of shaft ABC is subjected to the greater torque.
A tmax B abs =
0.150 m
3
TAB c [21.32(10 )](0.05) = π (0.054 ) Jst 2 6 = ) N−m 2 109 MPa = 108.60(10
Ans.
TBC = F(0.150)
0.3 m
0.15 m
25 kN · m
Ans.
Atmax Babs = 109 MPa
391
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5–87. The shafts are made of A36 steel and have the same diameter mm. a torque kN# ·ftmisisapplied applied to diameterof of100 4 in. If aIftorque ofof 1525kip gear B, determine the angle of twist of gear B.
2.5 ft 0.75 m
Solution
2.5 ft 0.75 m
Equilibrium: Referring to the free  body diagrams of shafts ABC and DE shown in Figs. a and b, respectively, F(0.15) – 25 ©Mx = 0; TA + F(0.5) 15 ==00
(1)
and
A B 15 kN kipm ft 25
6 in. 150 mm D
12mm in. 300
F(0.3)– T TEE = ©Mx = 0; F(1) = 00
C E
(2) 3 ftm 0.9
Internal Loadings: The internal torques developed in segments AB and BC of shaft ABC and shaft DE are shown in Figs. c, d, and e, respectively. Compatibility Equation: It is required that fCrC = fDrD
¢
TAB LAB TDE LDE TBC LBC + ≤ rC = ¢ ≤ rD JGst JGst JGst
(0.9)(0.3)  TAA(0.75) (2.5) + F(0.15)(0.75)](0.15) F(0.5)(2.5) D (0.5) = = –T TEE(3)(1) C[–T TAA–0.15F 0.5F==2.4T 2.4T T EE
(3)
Solving Eqs. (1), (2), and (3), TE =T7.353 kN · mkip # ft E = 4.412
F ==24.51 4.412kN kip F
# ft · m TA =TA12.79 kipkN = 21.32
 12.79 kN kip·# m ft Angle of Twist: Here, TAB =  TA = –21.32 fB =
TAB LAB [−21.32(10 3 )](0.75) = π JGst [ 2 (0.054 )][75(109 )]
= –0.02172  0.01666rad rad==1.24° 0.955° F
Ans.
0.150 m 0.15 m
0.3 m TBC = F(0.150)
25 kN · m
392
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*5–88. The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB.
2 kN A
2 kN
C
400 mm
50 mm
50 mm B
600 mm
Solution Equilibrium: (1)
TA + TB  2(0.1) = 0 Compatibility: f C>A = f C>B TA(0.4) JG
=
TB(0.6) JG (2)
TA = 1.50TB Solving Eqs. (1) and (2) yields: TB = 0.080 kN # m
TA = 0.120 kN # m
Maximum Shear Stress:
( tAC ) max =
0.12 ( 103 ) (0.02) TA c = = 9.55 MPa p 4 J 2 ( 0.02 )
Ans.
( tCB ) max =
0.08 ( 103 ) (0.02) TB c = = 6.37 MPa p 4 J 2 ( 0.02 )
Ans.
Ans: (tAC)max = 9.55 MPa, (tCB)max = 6.37 MPa 393
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5–89. The two shafts are made of A36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is applied to the gear at E, determine the reactions at A and B.
B F
D
50 mm
0.75 m
100 mm
500 Nm E
C 1.5 m A
Solution Equilibrium: TA + F(0.1)  500 = 0
(1)
TB  F(0.05) = 0
(2)
TA + 2TB  500 = 0
(3)
From Eqs. (1) and (2)
Compatibility: 0.1f E = 0.05f F f E = 0.5f F TA(1.5) JG
= 0.5c
TB(0.75) JG
TA = 0.250TB
d
(4)
Solving Eqs. (3) and (4) yields: TB = 222 N # m
Ans.
TA = 55.6 N # m
Ans.
Ans: TB = 22.2 N # m, TA = 55.6 N # m 394
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5–90. The two shafts are made of A36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N # m is applied to the gear at E, determine the rotation of this gear.
B F
D
50 mm
0.75 m
100 mm
500 Nm E
C 1.5 m A
Solution Equilibrium: TA + F(0.1)  500 = 0
(1)
TB  F(0.05) = 0
(2)
From Eqs. (1) and (2) (3)
TA + 2TB  500 = 0 Compatibility: 0.1f E = 0.05f F f E = 0.5f F TA(1.5) JG
= 0.5c
TB(0.75) JG
TA = 0.250TB
d
(4)
Solving Eqs. (3) and (4) yields: TB = 222.22 N # m
TA = 55.56 N # m
Angle of Twist: fE =
TAL = JG
55.56(1.5) p 2
(0.01254)(75.0)(109) Ans.
= 0.02897 rad = 1.66°
Ans: f E = 1.66° 395
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5–91. The two 1mlong shafts are made of 2014T6 aluminum. Each has a diameter of 30 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 900 N · m is applied to the top gear as shown, determine the maximum shear stress in each shaft.
A B C
900 600 Nm lbft E 80 4mm in. 40 2mm in.
13 m ft
D
F
Solution TA + F (0.08) − 900 = 0
(1)
TB − F (0.04) = 0
(2)
900 N · m
80 mm
40 mm
From Eqs. (1) and (2) TA + 2TB  600 900 = 0 80(f 40(f 4(f ) )== 2(f F); EE F);
(3)
fE = 0.5fF
TAL TBL = 0.5 a b; JG JG
(4)
TA = 0.5TB
Solving Eqs. (3) and (4) yields:
# ft; 240Nlb· m; TBB ==360 (tBD)max =
120 TTAA= =180 N lb · m# ft
TB c 360(0.015) 6 = 67.91(10 = ) N−m 2 67.9 MPa = π J (0.0154 )
Ans.
TA c 180(0.015) 6 = 33.95(10 = ) N−m 2 34.0 MPa = π (0.0154 ) J
Ans.
2
(tAC)max =
2
Ans. (tBD)max = 67.9 MPa, (tAC)max = 34.0 MPa 396
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*5–92. If the shaft is subjected to a uniform distributed torque of t = 20 kN # m>m, determine the maximum shear stress developed in the shaft. The shaft is made of 2014T6 aluminum alloy and is fixed at A and C.
400 mm
20 kNm/m 600 mm a A 80 mm 60 mm
Solution Equilibrium: Referring to the freebody diagram of the shaft shown in Fig. a, we have ΣMx = 0; TA + TC  20(103)(0.4) = 0
B a C
Section a–a
(1)
Compatibility Equation: The resultant torque of the distributed torque within the region x of the shaft is TR = 20(103)x N # m. Thus, the internal torque developed in the shaft as a function of x when end C is free is T(x) = 20(103)x N # m, Fig. b. Using the method of superposition, Fig. c, fC = 0 = 0 =
1 fC 2 t
L0
0.4 m
L0
0.4 m

1 fC 2 T
T(x)dx JG

C
TCL JG
20(103)xdx JG
0 = 20(103) ¢

TC(1) JG
x2 0.4 m ≤`  TC 2 0
TC = 1600 N # m
Substituting this result into Eq. (1), TA = 6400 N # m Maximum Shear Stress: By inspection, the maximum internal torque occurs at support A. Thus,
1 tmax 2 abs
=
6400(0.04) TA c = = 93.1 MPa J p a0.044  0.034 b 2
Ans.
Ans: t abs = 93.1 MPa max
397
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5–93. The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its midpoint, determine the reactions at the supports. A
T
2c
B c
L/2 L/ 2
Solution Equilibrium: (1)
TA + TB  T = 0 Section Properties: r(x) = c + J(x) = Angle of Twist: fT =
c c x = (L + x) L L
4 p c pc 4 c (L + x) d = (L + x)4 2 L 2L4
Tdx = L L J(x)G L2
L
Tdx pc 4 2L4
(L + x)4 G L
=
2TL4 dx pc 4 G LL2 (L + x)4 L
= = fB =
2TL4 1 c d ` 4 3pc G (L + x)3 L2
37TL 324 pc 4 G
Tdx = L J(x)G L0 =
TBdx pc 4 2L4
(L + x)4G
2TBL4
L
dx pc G L0 (L + x)4 4
= =
L
2TBL4
L 1 d ` 3 3pc G (L + x) 0 4
7TB L
c
12pc 4G
Compatibility: 0 = fT  fB 0 =
7TBL 37TL 324pc 4G 12pc 4G
TB =
37 T 189
Ans.
Substituting the result into Eq. (1) yields: TA =
152 T 189
Ans.
Ans: TB =
398
37 152 T, TA = T 189 189
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5–94. The shaft of radius c is subjected to a distributed torque t, measured as torque>length of shaft. Determine the reactions at the fixed supports A and B.
B t0
(
t t0 1
( Lx ) 2 )
x L
A
Solution
2t0
T(x) =
L0
By superposition:
x
t 0 a1 +
x2 x3 b dx = t ax + b 0 L2 3L2
(1)
0 = f  fB
0 =
L0
L
x3 b dx 2 7t 0L TB(L) 3L2 =  TB(L) JG JG 12
t 0 ax +
TB =
7t 0 L 12
Ans.
From Eq. (1), T = t 0 aL + TA +
4t 0 L L3 b = 3 3L2
7t 0 L 4t 0 L = 0 12 3 TA =
3t 0 L 4
Ans.
Ans: TB = 399
7t 0L 3t 0L , TA = 12 4
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5–95. If end B of the shaft, which has an equilateral triangle cross section, is subjected to a torque of T = 1200 N # m, determine the maximum shear stress developed in the shaft. Also, find the angle of twist of end B. The shaft is made from 6061T1 aluminum.
0.6 m A
75 mm
B T
Solution Maximum Shear Stress: 20T 20(1200) 6 = 56.89(10 ) N−m 2 56.9 MPa tmax = = = 0.0753 a3
Ans.
Angle of Twist: f =
=
46TL a4G
46(1200)(0.6) (0.0754 )[26(109 )]
= 0.04026 πad = 2.31°
Ans.
Ans. tmax = 56.9 MPa f = 2.31° 400
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*5–96. If the shaft has an equilateral triangle cross section and is made from 606 T aluminum alloy that has an allowable shear stress of tallow = 84 MPa, determine the maximum allowable torque T that can be applied to end B. Also, find the corresponding angle of twist of end B.
0.6 m A
75 mm
B
Solution
T
Allowable Shear Stress: tallow =
20T ; a3
84(106 ) =
20T 0.0753
= T 1.772(10 3 ) N= ⋅ m 1.77 kN ⋅ m
Ans.
Angle of Twist: f =
46TL a4G
46[1.772(10 3 )](0.6) =
(0.0754 )[26(109 )]
= 3.41° = 0.05945 πad
Ans.
Ans:
T = 1.77 kN # m, f = 3.41⬚
401
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5–97. The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to the torsional loading, determine the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to end A.
A 20 Nm
50 Nm
30 Nm 2m C 1.5 m
50 mm 20 mm B
Solution Maximum Shear Stress: (tBC)max =
2TBC 2
pab
=
2(30.0) p(0.05)(0.022) Ans.
= 0.955 MPa (tAC)max =
2TAC 2
pab
=
2(50.0) p(0.05)(0.022) Ans.
= 1.59 MPa Angle of Twist:
f B>A = a =
(a2 + b2)T L p a3b3 G (0.052 + 0.022)
p(0.053)(0.023)(37.0)(109)
[(  30.0)(1.5) + ( 50.0)(2)] Ans.
= 0.003618 rad = 0.207°
Ans: (tBC)max = 0.955 MPa, (tAC)max = 1.59 MPa, f B>A = 0.207° 402
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5–98. Solve Prob. 5–98 for the maximum shear stress within regions AC and BC, and the angle of twist f of end B relative to C.
A 20 Nm
50 Nm
30 Nm 2m C 1.5 m
50 mm 20 mm B
Solution Maximum Shear Stress: (tBC)max =
2TBC 2
pab
=
2(30.0) p(0.05)(0.022) Ans.
= 0.955 MPa (tAC)max =
2TAC 2
pab
=
2(50.0) p(0.05)(0.022) Ans.
= 1.59 MPa Angle of Twist: f B>C = =
(a2 + b2) TBC L p a3 b3 G (0.052 + 0.022)(  30.0)(1.5) p(0.053)(0.023)(37.0)(109) Ans.
=  0.001123 rad = 0.0643°
Ans: (tBC)max = 0.955 MPa, (tAC)max = 1.59 MPa, f B>C = 0.0643° 403
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5–99. If a = 25 mm and b = 15 mm, determine the maximum shear stress in the circular and elliptical shafts when the applied torque is T = 80 N # m. By what percentage is the shaft of circular cross section more efficient at withstanding the torque than the shaft of elliptical cross section?
b
a
a
Solution For the circular shaft: (tmax)c =
80(0.025) Tc = p = 3.26 MPa 4 J 2 (0.025 )
Ans.
For the elliptical shaft: (tmax)e =
2(80) 2T = = 9.05 MPa 2 pab p(0.025)(0.0152)
, more efficient = =
(tmax)e  (tmax)c (tmax)c
Ans.
(100,)
9.05  3.26 (100,) = 178, 3.26
Ans.
Ans: (tmax)c = 3.26 MPa, (tmax)e = 9.05 MPa, , more efficient = 178, 404
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*5–100. It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in the process of manufacturing, with one dimension smaller than the other by a factor k as shown. Determine the factor by which the maximum shear stress is increased.
kd
d
d
Solution For the circular shaft: (tmax)c =
T 1 d2 2 Tc 16T = p d 4 = J p d3 2 122
For the elliptical shaft: (tmax)e =
2T 2T 16T = = 2 d kd 2 pab p k2 d3 p 1 2 21 2 2
Factor of increase in maximum shear stress = =
(tmax)e (tmax)c
=
16T p k 2d 3 16T p d3
1 k2
Ans.
Ans: Factor of increase = 405
1 k2
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5–101. The brass wire has a triangular cross section, 2 mm on a side. If the yield stress for brass is tY = 205 MPa, determine the maximum torque T to which it can be subjected so that the wire will not yield. If this torque is applied to the 4mlong segment, determine the greatest angle of twist of one end of the wire relative to the other end that will not cause permanent damage to the wire. Gbr = 37 GPa.
T
4m T
Solution Allowable Shear Stress: tmax = tg = 205(106) =
20T a3 20T 0.0023
T = 0.0820 N # m
Ans.
Angle of Twist: f =
46(0.0820)(4) 46TL = a4G (0.0024)(37)(109) Ans.
= 25.5 rad
Ans: T = 0.0820 N # m, f = 25.5 rad 406
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5–102. If the solid shaft is made from red brass C83400 copper having an allowable shear stress of tallow = 28 MPa, determine the maximum allowable torque T that can be applied at B.
0.6 m
A
100 mm
50 50 mm 1.2 m mm
B T
C
Solution Equilibrium: Referring to the freebody diagram of the square bar shown in Fig. a, we have ©Mx = 0;
TA + TC  T = 0
(1)
Compatibility Equation: Here, it is required that fB>A = fB>C
7.10TA (0.6) 4
aG
=
7.10TC (1.2) a 4G
TA = 2TC
(2)
Solving Eqs. (1) and (2), TC =
1 T 3
TA =
2 T 3
Allowable Shear Stress: Segment AB is critical since it is subjected to the greater internal torque.
tallow =
4.81TA a3
;
28(106 ) =
( )
4.81 23 T 0.13
T 8.732(10 3 ) N= = ⋅ m 8.73 kN ⋅ m
Ans.
Ans: T = 8.73 kN ⋅ m 407
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5–103. If the solid shaft is made from red brass C83400 copper and it is subjected to a torque T = 8 kN · m at B, determine the maximum shear stress developed in segments AB and BC.
0.6 m
A
100 mm
Solution
50 50 mm 1.2 m mm
B T
C
Equilibrium: Referring to the freebody diagram of the square bar shown in Fig. a, we have ©Mx = 0;
(1)
TA + TC  8 = 0
Compatibility Equation: Here, it is required that fB>A = fB>C
7.10TA (0.6) a 4G
=
7.10TC (1.2) a 4G
TA = 2TC
(2)
Solving Eqs. (1) and (2),
TC = 2.667 kN ⋅ m
TA = 5.333 kN ⋅ m
Maximum Shear Stress: (tmax)AB = (tmax)BC =
4.81TA 3
a
4.81TC a
3
4.81[5.333(10 3 )] 6 25.65(10 = ) N−m 2 25.7 MPa Ans. = = 0.13 4.81[2.667(10 3 )] 6 12.83(10 ) N−m 2 12.8 MPa Ans. = = = 0.13
Ans. (τ max ) AB 25.7 MPa, (τ max )BC 12.8 MPa = = 408
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*5–104. If the shaft is subjected to the torque of 3 kN # m, determine the maximum shear stress developed in the shaft. Also, find the angle of twist of end B. The shaft is made from A36 steel. Set a = 50 mm.
600 mm
a
A
Solution
a
a
B 3 kNm
Maximum Shear Stress: 3
tmax =
2(3)(10 ) 2T = = 61.1 MPa 2 pab p(0.05)(0.0252)
Ans.
Angle of Twist: f = =
(a2 + b2)TL pa3b3G (0.052 + 0.0252)(3)(103)(0.6) p(0.053)(0.0253)(75)(109) Ans.
= 0.01222 rad = 0.700°
Ans: tmax = 61.1 MPa, f B = 0.700° 409
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If the shaft is made from A36 steel having an allowable shear stress of tallow = 75 MPa, determine the minimum dimension a for the cross section to the nearest millimeter. Also, find the corresponding angle of twist at end B.
600 mm
a
A
Solution
a
2T ; pab2
B 3 kNm
Allowable Shear Stress: tallow =
a
75(106) =
2(3)(103) p(a)(a2)2
a = 0.04670 m Use
Ans.
a = 47 mm
Angle of Twist: f =
=
(a2 + b2)TL pa3b3G c 0.0472 + a
0.047 2 b d (3)(10 3)(0.6) 2
p(0.047 3)a
0.047 3 b (75)(10 9) 2
Ans.
= 0.01566 rad = 0.897°
Ans: Use a = 47 mm, f B = 0.897° 410
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The plastic tube is subjected to a torque of 150 N # m. Determine the mean dimension a of its sides if the allowable shear stress is tallow = 60 MPa. Each side has a thickness of t = 3 mm.
a
a
150 Nm
Solution a2 1 a a Am = 4 c a ba b d = 2 2 2 2 tavg =
T ; 2 tAm
60(106) =
150 2(0.003) 12 a2 Ans.
a = 0.0289 m = 28.9 mm
Ans: a = 28.9 mm 411
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The plastic tube is subjected to a torque of 150 N # m. Determine the average shear stress in the tube if the mean dimension a = 200 mm. Each side has a thickness of t = 3 mm.
a
a
150 Nm
Solution 1 0.2 0.2 Am = 4 c a ba b d = 0.02 m2 2 2 2 tavg =
T 150 = 2 t Am 2(0.003)(0.02)
Ans.
= 1.25 MPa
Ans: tavg = 1.25 MPa 412
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*5–108. For a given maximum shear stress, determine the factor by which the torque carrying capacity is increased if the halfcircular section is reversed from the dashedline position to the section shown. The tube is 0.1 2.5 in. mmthick. thick.
45 mm 1.80 in. 15 in. mm 0.6
30 mm 1.20 in. 12.5 0.5 mm in.
Solution A = m (0.0275)(0.04375) −
π (0.013752 ) = 0.9061(10 −3 ) m 2 2
′ (0.0275)(0.04375) + A = m
π (0.013752 ) = 1.5001(10 −3 ) m 2 2
tmax =
T 2t Am
T = 2 t Am tmax Factor = =
2t Am ¿ tmax 2t Am tmax Am ¿ 1.5001(10 −3 ) = 1.655 = 1.66 = Am 0.9061(10 −3 )
Ans.
Ans: Factor 1.66 413
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5–109. A torque of 200 N · m is applied to the tube. If the wall thickness is 2.5 mm, determine the average shear stress in the tube. 50 mm 50 mm
48.75 mm
Solution = Am
π (0.048752 ) = 1.8665(10 −3 ) m 2 4
tavg =
200 T 6 = 21.43(10 = ) N−m 2 21.4 MPa = 2t Am 2(0.0025)[1.8665(10 −3 )]
Ans.
Ans: tavg = 21.44 MPa 414
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5–110. The 6061T6 aluminum bar has a square cross section of 25 mm by 25 mm. If it is 2 m long, determine the maximum shear stress in the bar and the rotation of one end relative to the other end.
C 1.5 m 20 Nm B
0.5 m A 80 Nm
60 N·m
Solution
25 mm
25 mm
Maximum Shear Stress: tmax =
4.81Tmax a
3
=
4.81(80.0) (0.0253)
Ans.
= 24.6 MPa
Angle of Twist: 7.10(  20.0)(1.5) 7.10( 80.0)(0.5) 7.10TL f A>C = a 4 = + 4 9 aG (0.025 )(26.0)(10 ) (0.0254)(26.0)(109) Ans.
=  0.04894 rad = 2.80°
Ans: tmax = 24.6 MPa, f A>C = 2.80° 415
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5–111. The aluminum strut is fixed between the two walls at A and B. If it has a 50 mm by 50 mm square cross section, and it is subjected to the torque of 120 N · m at C, determine the reactions at the fixed supports. Also, what is the angle of twist at C? Gal 5 27 GPa.
A C 0.6 2 ftm m 120 Nft 80 lb
B
0.9 3 ftm
Solution By superposition: 0 = f  fB 0 =
7.10(T 7.10(80)(2) 7.10(120)(0.6) 7.10(T B)(5) B)(1.5)  4 4 4 G a4aG a Ga G
120 N · m
32 N lb ·# m ft TB = 48
Ans.
32 –120 80 == 00 TA + 48 TA = 48 72 lb N #· ft m fC =
Ans.
7.10(48)(0.9) = 0.0018176 = πad 0.104° (0.054 )[27(109 )]
120 N · m 48 N · m
Ans.
Ans: T B 48 N · m, T A 72 N · m, fC 0.104° 416
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*5–112. Determine the constant thickness of the rectangular tube if average stress is not to exceed 84 MPa when a torque of T = 2 kN · m is applied to the tube. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown.
T 100 mm
Solution
50 mm
= Am (0.05)(0.1) = 0.005 m 2 tavg =
T 2t Am
84(106 ) =
2(10 3 ) 2t(0.005)
= t 0.002381 = m 2.38 mm
Ans.
Ans: t = 2.38 mm 417
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5–113. Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to exceed 84 MPa. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown and the tube has a thickness of 3 mm.
T 100 mm
Solution 50 mm
= Am (0.05)(0.1) = 0.005 m 2 tavg =
T 2t Am ;
84(106 ) =
2(10 3 ) 2(0.003)(0.005)
= t 2.52(10 3 ) N= ⋅ m 2.52 kN ⋅ m
Ans.
Ans: T = 2.52 kN ⋅ m 418
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5–114. Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. By what percentage is the torsional strength reduced when the eccentricity e is onefourth of the difference in the radii?
ab 2 a
b e 2
e 2
Solution Average Shear Stress: For the aligned tube tavg =
T T = 2 t Am 2(a  b)(p) 1 a +2 b 2 2
T = tavg (2)(a  b)(p) a
For the eccentric tube
tavg =
a + b 2 b 2
T′ 2 t Am
t = a = a 
e e  a + bb = a  e  b 2 2 1 3 (a  b)  b = (a  b) 4 4
3 a + b 2 T′ = tavg (2) c (a  b) d (p) a b 4 2 Factor =
tavg (2) 3 34 (a  b) 4 (p) 1 a +2 b 2 2 T′ 3 = = T 4 tavg (2)(a b)(p) 1 a +2 b 2 2
Percent reduction in strength = a1 
3 b * 100, = 25, 4
Ans.
Ans: Percent reduction in strength = 25, 419
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5–115. The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of t = 5 mm. If the allowable shear stress is tallow = 56 MPa, and the tube is to resist a torque of T = 340 N · m, determine the necessary dimension b. The mean area for the ellipse is Am = pb(0.5b).
b 0.5b
340 N.m
Solution tavg = tallow =
56(106 ) =
T 2tAm
340 2(0.005)[π b(0.5b)]
= b 0.01966 = m 19.7 mm
Ans.
Ans: b = 19.7 mm 420
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The 304 stainless steel tube has a thickness of 10 mm. If the allowable shear stress is tallow = 80 MPa, determine the maximum torque T that it can transmit. Also, what is the angle of twist of one end of the tube with respect to the other if the tube is 4 m long? The mean dimensions are shown.
30 mm
70 mm
T
Solution Section Properties: Am = 0.07(0.03) = 0.00210 m2 L
ds = 2(0.07) + 2(0.03) = 0.200 m
Allowable Average Shear Stress: tavg = tallow = 80 ( 106 ) =
T 2 t Am T 2(0.01)(0.00210)
T = 3360 N # m = 3.36 kN # m
Ans.
Angle of Twist: f =
3360(4)(0.200) TL ds = 4 A2mG L t 4 ( 0.002102 ) (75.0) ( 109 ) (0.01) Ans.
= 0.2032 rad = 11.6°
Ans: T = 3.36 kN # m, f = 11.6° 421
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The 304 stainless steel tube has a thickness of 10 mm. If the applied torque is T = 50 N # m, determine the average shear stress in the tube. The mean dimensions are shown. 30 mm
70 mm
T
Solution Section Properties: Am = 0.07(0.03) = 0.00210 m2 Average Shear Stress: tavg = =
T 2 t Am 50 2(0.01)(0.00210) Ans.
= 1.19 MPa
Ans: tavg = 1.19 MPa 422
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5–118. The plastic hexagonal tube is subjected to a torque of 150 N # m. Determine the mean dimension a of its sides if the allowable shear stress is tallow = 60 MPa. Each side has a thickness of t = 3 mm.
a
t = 3 mm
T = 150 Nm
Solution 1 Am = 4c (a cos 30°)(a sin 30°) d + (a)(2a)cos 30° = 2.5981 a2 2 tavg = tallow = 60 ( 106 ) =
T 2 t Am
150 (2)(0.003) ( 2.5981 a2 ) Ans.
a = 0.01266 m = 12.7 mm
Ans: a = 12.7 mm 423
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5–119. The symmetric tube is made from a highstrength steel, having the mean dimensions shown and a thickness of 5 mm. If it is subjected to a torque of T = 40 N # m, determine the average shear stress developed at points A and B. Indicate the shear stress on volume elements located at these points.
20 mm
30 mm
60 mm A B
40 Nm
Solution Am = 4(0.04)(0.06) + (0.04)2 = 0.0112 m2 tavg =
T 2 t Am
(tavg)A = (tavg)B =
40 = 357 kPa 2(0.005)(0.0112)
Ans.
Ans: (tavg)A = (tavg)B = 357 kPa 424
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The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diameter segment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglect the stress concentration at the fillet.
T
60 mm 55 mm T
Solution Maximum Elastic Torque: For the larger diameter segment tY =
TY c J
TY =
tY J c
=
tY ( p2 )( 0.034 ) 0.03
= 13.5 ( 106 ) p tY Elastic  Plastic Torque: For the smaller diameter segment 2p 3 3 6 6 TP = 2p 3 tY c = 3 tY ( 0.0275 ) = 13.86 ( 10 ) ptY 7 13.5 ( 10 ) ptY . Applying Eq. 5–26 of the text, we have T = 13.5 ( 106 ) ptY =
p tY ( 4 c 3  r3Y ) 6 p tY 3 4 ( 0.02753 )  r3Y 4 6
Ans.
rY = 0.01298 m = 13.0 mm
Ans: rY = 13.0 mm 425
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5–121. The steel step shaft has an allowable shear stress of tallow = 8 MPa. If the transition between the cross sections has a radius r = 4 mm, determine the maximum torque T that can be applied.
50 mm
20 mm
T 2
T
20 mm
T 2
Solution Allowable Shear Stress: D 50 = = 2.5 d 20
and
r 4 = = 0.20 d 20
From the text, K = 1.25 tmax = tallow = K
Tc J t 2 (0.01) § 4 2 (0.01 )
8(106) = 1.25£ p T = 20.1 N # m
Ans.
Ans: T = 20.1 N # m 426
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5–122. The shaft is fixed to the wall at A and is subjected to the torques shown. Determine the maximum shear stress in the shaft. A fillet weld having a radius of 2.75 mm is used to connect the shafts at B.
900 Nm
A 100 Nm 50 mm
300 Nm
B 25 mm
Solution
C
Internal Torque: The internal torque in regions CD, BD and AE are indicated in their respective FBD, Fig. a, Maximum Shear Stress: For region CD, (tmax)CD =
300(0.0125) TCDC = = 97.78 ( 106 ) Pa = 97.8 MPa (Max.) p J ( 0.01254 ) 2
Ans.
For Region AE, (tmax)AE =
1100(0.025) TAE C = = 44.82 ( 106 ) Pa = 44.8 MPa p J ( 0.0254 ) 2
D 50 r 2.75 For the fillet, enter = = 2 and = = 0.11 into Fig. 532, we get d 25 d 25 K = 1.375. (tmax)f = K a
200(0.0125) TBDC b = 1.375£ § = 89.64 ( 106 ) Pa = 89.6 MPa p J 4 ( 0.0125 ) 2
Ans: ( tmax ) = 97.8 MPa 427
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5–123. The steel shaft is made from two segments: AB and BC, which are connected using a fillet weld having a radius of 2.8 mm. Determine the maximum shear stress developed in the shaft.
C 50 mm
D
100 Nm
20 mm B A
60 Nm
Solution (tmax)CD =
40 Nm
100(0.025) TCDc = p 4 J 2 (0.025 )
= 4.07 MPa For the fillet: D 50 = = 2.5; d 20
r 2.8 = = 0.14 d 20
From Fig. 532, K = 1.325 (tmax)f = K
60(0.01) TABc d = 1.325 c p 4 J 2 (0.01 ) = 50.6 MPa (max)
Ans.
Ans. (tmax)f = 50.6 MPa (max) 428
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The builtup shaft is to be designed to rotate at 450 rpm while transmitting 230 kW of power. Is this possible? The allowable shear stress is tallow = 150 MPa.
100 mm
60 mm
Solution Internal Torque: Here, v = a450 Then,
T =
rev 2p rad 1 min ba ba b = 15p rad>s min 1 rev 60 s
230 ( 103 ) P = = 4880.75 N # m v 15 p
Maximum Shear Stress: At the fillet, tmax = tallow = K a 150 ( 106 ) = K£
TC b J 4880.75 (0.03) § p ( 0.034 ) 2
K = 1.30 Enter this value and r = 0.14 ; 60
D 100 r = = 1.67 into Fig. 532, we get = 0.14. Thus, d 60 d r = 8.4 mm
Dd 100 60 = = 20 mm > r = 8.4 mm, it is possible to construct the 2 2 transition. Ans.
Since
Ans: It is possible. 429
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The builtup shaft is designed to rotate at 450 rpm. If the radius of the fillet weld connecting the shafts is r = 13.2 mm, and the allowable shear stress for the material is tallow = 150 MPa, determine the maximum power the shaft can transmit. 100 mm
60 mm
Solution Internal Torque: Here, v = a450 Then,
rev 2p rad 1 min ba ba b = 15p rad>s min 1 rev 60 s
T =
P P = v 15 p
D 100 r Maximum Shear Stress: Enter = = 1.67 and = 0.22 into Fig 532, we get d 60 d K = 1.2. Then tmax = tallow = K a
TC b J
150 ( 106 ) = 1.2£
P 15 p (0.03) § p 4 2 0.03
(
)
P = 249.82 ( 103 ) W = 250 kW
Ans.
Ans: P = 250 kW 430
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A solid shaft has a diameter of 40 mm and length of 1 m. It is made from an elasticplastic material having a yield stress of tY = 100 MPa. Determine the maximum elastic torque TY and the corresponding angle of twist. What is the angle of twist if the torque is increased to T = 1.2TY? G = 80 GPa.
Solution Maximum elastic torque TY, tY =
TY c J
TY =
100 ( 106 )( p2 )( 0.024 ) tY J = = 1256.64 N # m = 1.26 kN # m c 0.02
Ans.
Angle of twist: gY = f =
100 ( 106 ) tY = = 0.00125 rad G 80 ( 109 ) gY 0.00125 L = (1) = 0.0625 rad = 3.58° rY 0.02
Ans.
Also, f =
TY L = JG
1256.64(1) p 2
( 0.024 ) (80) ( 109 )
= 0.0625 rad = 3.58°
From Eq. 5–26 of the text, T =
p(100) ( 106 ) ptY 14 c 3  r3Y 2 ; 1.2(1256.64) = 3 4 ( 0.023 )  r3Y 4 6 6 rY = 0.01474 m
f′ =
gY 0.00125 L = (1) = 0.0848 rad = 4.86° rY 0.01474
Ans.
Ans: TY = 1.26 kN # m, f = 3.58°, f′ = 4.86° 431
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5–127. Determine the torque needed to twist a short 2mmdiameter steel wire through several revolutions if it is made from steel assumed to be elastic perfectly plastic and having a yield stress of tY = 50 MPa. Assume that the material becomes fully plastic.
Solution Fully plastic torque is applied. TP =
2p 2p t c3 = (50)(106)(0.0013) = 0.105 N # m 3 Y 3
Ans.
Ans: TP = 0.105 N # m 432
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5–128. A solid shaft is subjected to the torque T, which causes the material to yield. If the material is elastic plastic, show that the torque can be expressed in terms of the angle of twist f of the shaft as T = 43 TY11  f3Y>4f32, where TY and fY are the torque and angle of twist when the material begins to yield.
Solution gY gL = L r rY
f = rY =
gYL f
(1)
When rY = c, f = fY From Eq. (1), c =
gYL fY
(2)
Dividing Eq. (1) by Eq. (2) yields: rY fY = c f
(3)
Use Eq. 526 from the text. T =
r3Y p tY 2p tYc3 (4 c3  r3Y) = a1 )b 6 3 4 c3
Use Eq. 524, TY = T =
p t c3 from the text and Eq. (3) 2 Y
f3Y 4 TY a 1 b 3 4 f3
QED
433
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5–129. The solid shaft is made of an elastic perfectly plastic material. Determine the torque T needed to form an elastic core in the shaft having a radius of rY = 20 mm. If the shaft is 3 m long, through what angle does one end of the shaft twist with respect to the other end? When the torque is removed, determine the residual stress distribution in the shaft and the permanent angle of twist.
80 mm T
T
t (MPa) 160
Solution 0.004
ElasticPlastic Torque: T = =
p tY 1 4c 3  r3Y 2 6
p(160)(106) 6
3 4 1 0.043 2
 0.023 4
= 20776.40 N # m = 20.8 kN # m
g (rad)
Ans.
Angle of Twist: f =
gY 0.004 L = a b(3) = 0.600 rad = 34.4° rY 0.02
Ans.
When the reverse T = 20776.4 N # m is applied, G = f′ =
160(106) 0.004 TL = JG
= 40 GPa 20776.4(3)
p 4 9 2 (0.04 )(40)(10 )
= 0.3875 rad
The permanent angle of twist is, f r = f  f′ Ans.
= 0.600  0.3875 = 0.2125 rad = 12.2° Residual Shear Stress: (t′)r = c =
20776.4(0.04) Tc = = 206.67 MPa p 4 J 2 (0.04 )
(t′)r = 0.02 m =
20776.4(0.02) Tc = = 103.33 MPa p 4 J 2 (0.04 )
(tr)r = c =  160 + 206.67 = 46.7 MPa Ans.
(tr)r = 0.02 m =  160 + 103.33 =  56.7 MPa
Ans: T = 20.8 kN # m, f = 34.4°, (tr)max = 56.7 MPa, f r = 12.2° 434
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5–130. The shaft is subjected to a maximum shear strain of 0.0048 rad. Determine the torque applied to the shaft if the material has strain hardening as shown by the shear stress–strain diagram. 50 mm 2 in.
T
Solution
t (MPa) (ksi)
From the shear  strain diagram,
84 12
rY 0.05 = ; 0.0006 0.0048
426
rY = 0.00625 m
From the shear stress–strain diagram, 0.0006
0.0048
g (rad)
42(106 ) 9 τ 1 = = ρ 6.72(10 ) ρ 0.00625
τ 2 − 42(106 ) 84(106 ) − 42(106 ) = ρ − 0.00625 0.05 − 0.00625 τ 2 0.96(109 )ρ + 36(106 ) = c
T = 2π ∫ τ ρ 2d ρ 0
= 2π ∫
0.00625 m
0
50 mm
[6.72(109 ) ρ ]ρ 2d ρ + 2π ∫
0.05 m
0.00625 m
= 2π [1.68(109 ) ρ 4 ]
0.00625 m 0
[0.96(109 ) ρ + 36(106 )]ρ 2d ρ
+ 2π [0.24(109 ) ρ 4 + 12(106 ) ρ 3 ]
0.05 m
84 MPa
42 MPa
0.00625 m
3
= 18.84(10 ) N= ⋅ m 18.8 kN ⋅ m
Ans.
0.00625 m
0.05 m
Ans: T 18.8 kN · m 435
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5–131. A solid shaft having a diameter of 50 mm is made of of elasticplastic material a yield of elasticplastic material having ahaving yield stress of tY =stress 112 MPa 3 tand 16 ksimodulus 2 ksi. and shear of G = 12110 of G =modulus 84 GPa. Determine the torque Y =shear Determine torquean required develop elastic corea required to the developo elastic to core in the an shaft having in the shaft having a diameter in. Also, what is the diameter of 25 mm. Also, what is of the1plasstic torque? plastic torque?
Solution Use Eq. 526 from the text: T =
p tY π [112(106 )] [4(0.0253 ) − 0.01253 ] (4 c3  rY 3) = 6 6
= 3.551(10 3 ) N= ⋅ m 3.55 kN ⋅ m
Ans.
Use Eq. 527 from the text: TP =
2p 2π [112(106 )](0.0253 ) t c3 = 3 Y 3
= 3.665(10 3 ) N= ⋅ m 3.67 kN ⋅ m
Ans.
Ans. T = 3.55 kN ⋅ m, TP = 3.67 kN ⋅ m 336
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*5–132. The hollow shaft has the cross section shown and is made of an elastic perfectly plastic material having a yield shear stress of tY . Determine the ratio of the plastic torque Tp to the maximum elastic torque TY .
c c 2
Solution Maximum Elastic Torque: In this case, the torsion formula is still applicable. tY =
TY c J
TY =
J t c Y
= =
c 4 p 4 Jc  a b R tY 2 2 c
15 3 pc tY 32
Plastic Torque: Using the general equation, with t = tY , c
TP = 2ptY
r2dr Lc>2 c
= 2ptY ¢ =
r3 ≤2 3 c>2
7 pc 3tY 12
The ratio is 7 pc 3tY TP 12 = = 1.24 TY 15 3 pc tY 32
Ans.
Ans: TP = 1.24 TY 437
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150 mm
5–133. The hollow shaft has inner and outer diameters of 60 mm and 80 mm, respectively. If it is made of an elasticperfectly plastic material, which has the tg diagram shown, determine the reactions at the fixed supports A and C.
450 mm B
C 15 kNm
A
t (MPa)
Solution
120
Equation of Equilibrium. Refering to the free  body diagram of the shaft shown in Fig. a, ©Mx = 0; TA + TC  15 A 103 B = 0
(1)
Elastic Analysis. It is required that fB>A = fB>C. Thus, the compatibility equation is fB>A = fB>C TCLBC TALAB = JG JG TA (0.45) = TC(0.15) TC = 3TA
(2)
Solving Eqs. (1) and (2), TA = 3750 N # m
TC = 11 250N # m
The maximum elastic torque and plastic torque in the shaft can be determined from p A 0.044  0.034 B J 2 T (120) A 106 B = 8246.68 N # m TY = tY = D c 0.04
co
TP = 2ptY
Lci
r2dr
= 2p(120) A 106 B ¢
0.04 m
r3 = 9299.11 N # m ≤` 3 0.03 m
Since TC 7 TY, the results obtained using the elastic analysis are not valid. Plastic Analysis. Assuming that segment BC is fully plastic, TC = TP = 9299.11 N # m = 9.3 kN # m
Ans.
Substituting this result into Eq. (1), TA = 5700 N # m = 5.70 kN # m
Ans.
438
0.0016
g (rad)
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5–133. Continued Since TA 6 TY, segment AB of the shaft is still linearly elastic. Here, 120 A 106 B = 75GPa. G = 0.0016 fB>C = fB>A =
fB>C =
gi L ; ci BC
5700.89(0.45) TALAB = 0.01244 rad = p JG A 0.044  0.034 B (75) A 109 B 2 0.01244 =
gi (0.15) 0.03
gi = 0.002489 rad Since gi 7 gY, segment BC of the shaft is indeed fully plastic.
Ans. TA = 5.70 kN # m, TC = 9.3 kN # m 439
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5–134. The 2mlong tube is made of an elastic perfectly plastic material as shown. Determine the applied torque T, which subjects the material at the tube’s outer edge to a shear strain of gmax = 0.006 rad. What would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube.
T
60 mm
50 mm
Solution
t (MPa)
Plastic Torque: The tube will be fully plastic if gi Ú gY = 0.003 rad. From Fig. a, gi 0.006 = ; 0.05 0.06
gi = 0.005 rad 7 gY 0.003
The tube indeed is fully plastic. Then Tp = 2p = =
Lci
c0
tYr2 dp
2 pt ( C 30  C 3i ) 3 Y 2 p 3 180 ( 106 ) 4 ( 0.063  0.053 ) 3
= 34.306 ( 103 ) N # m = 34.3 kN # m
Ans.
Angle of Twist: fP = Here, G =
gmax C0
L = a
0.006 b(2) = 0.2 rad 0.06
180 ( 106 ) tY = = 60 ( 109 ) Pa. When a reverse torque of gY 0.003
T ′P = 34.306 ( 103 ) N # m is applied f′P =
180
34.306 ( 103 ) (2) TP ′L = = 0.1085 rad p JG ( 0.064  0.054 ) 3 60 ( 109 ) 4 2
Thus, the permanent angle of twist is
f r = f p  f′p = 0.2  0.1085 = (0.09151 rad) a
180° b = 5.24° Ans. p rad
440
g (rad)
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5–134. Continued
Residual Shear Stress:
( t′P ) o =
( t′P ) o =
34.306 ( 103 ) (0.06) T′PC = 195.29 ( 106 ) Pa = 195.29 MPa = p J ( 0.064  0.054 ) 2 34.306 ( 103 ) (0.05) T′Pr = = 162.74 ( 106 ) Pa = 162.74 MPa p J 4 4 ( 0.06  0.05 ) 2
Thus (tr)o =  tY + ( t′p ) o =  180 + 195.29 = 15.3 MPa
Ans.
(tr)i =  tY + ( t′P ) i =  180 + 162.74 = 17.3 MPa
Ans.
The sketch of the residual shear stress distribution is shown in Fig. b.
Ans: TP = 34.3 kN # m, f r = 5.24°, (tr)o = 15.3 MPa, (tr)i =  17.3 MPa 441
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5–135. The shaft is made of an elastic perfectly plastic material as shown. Plot the shearstress distribution acting along a radial line if it is subjected to a torque of T = 20 kN # m. What is the residual stress distribution in the shaft when the torque is removed?
T
40 mm
t (MPa) 170
Solution 0.00227
Elastic  Plastic Torque: The maximum elastic torque is
g (rad)
p p t C3 = 3 170 ( 106 ) 4 ( 0.043 ) = 17.09 ( 103 ) N # m 2 Y 2
TY =
And the plastic torque is Tp =
2 2 p t C 3 = p 3 170 ( 106 ) 4 ( 0.043 ) = 22.79 ( 103 ) N # m 3 Y 3
Since TY 6 T 6 TP, the shaft exhibits elasticplastic behavior. T =
20 ( 103 ) =
ptY ( 4C 3  r3Y ) 6 p 3 170 ( 106 ) 4 6
3 4 ( 0.043 )  r3Y 4
rY = 0.03152 m = 31.5 mm Residual shear stress: When the reverse torque T′ = 20 kN # m is applied, t′C =
t ′rY =
20 ( 103 ) (0.04) T′C = = 198.94 ( 106 ) Pa = 198.94 MPa p J 4 ( 0.04 ) 2 20 ( 103 ) (0.03152) T′rY = = 156.76 ( 106 ) Pa = 156.76 MPa p J 4 ( 0.04 ) 2
Thus, (tr)c = tY + t′C =  170 + 198.94 = 28.9 MPa
Ans.
(tr)rY =  tY + t′rY =  170 + 156.76 =  13.2 MPa
Ans.
The sketch of the residual shear stress distribution is shown in Fig. b.
Ans: (tr)c = 28.9 MPa, (tr)rY =  13.2 MPa 442
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*5–136. The tube has a length of 2 m and is made of an elastic perfectly plastic material as shown. Determine the torque needed to just cause the material to become fully plastic. What is the permanent angle of twist of the tube when this torque is removed?
80 mm
60 mm t (MPa) 130
Solution Plastic Torque:
0.005
Tp = 2p
= =
Lci
g (rad)
co
tY r2 dp
2 pt ( C 30  C 3i ) 3 Y 2 p 3 130 ( 106 ) 4 ( 0.083  0.063 ) 3
= 80.592 ( 103 ) N # m = 80.6 KN # m
Ans.
Angle of Twist: Here, rY = Ci = 0.06 m. fP =
gY 0.005 L = a b(2) = 0.16667 rad rY 0.06
130 ( 106 ) tY = = 26 ( 109 ) Pa. When a reverse torque of T P′ = 80.592 gY 0.005 N # m is applied,
Here, G =
( 103 )
f′P =
80.592 ( 103 ) (2) T′P L = = 0.14095 rad p JG ( 0.084  0.064 ) 3 26 ( 109 ) 4 2
Thus, the permanent angle of twist is
f r = f p  f′p = 0.16667  0.14095 = (0.02571 rad) a
180° b = 1.47° p rad
Ans.
Ans: TP = 80.6 kN # m, f r = 1.47° 443
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5–137. The tube has a length of 2 m and is made of an elastic perfectly plastic material as shown. Determine the torque needed to just cause the material to become fully plastic. What is the permanent angle of twist of the tube when this torque is removed?
100 mm
60 mm t (MPa) 350
Solution Plastic Torque:
0.007
TP = 2p = =
Lci
g (rad)
co
tY r2 dr
2p tY 3 1 c o  c 3i 2 3 2p(350)(106) 3
( 0.053  0.033 )
= 71837.75 N # m = 71.8 kN # m
Ans.
Angle of twist: fr =
gY L rY
= a
Where rY = ci = 0.03 m
0.007 b(2) = 0.4667 rad 0.03
When a reverse TP = 71837.75 N # m is applied. G = f P′ =
350(106) 0.007
= 50 GPa
71837.75(2) TPL = = 0.3363 rad p JG ( 0.054  0.034 ) 50 ( 109 ) 2
Permanent angle of twist: f r = f P  f P′ = 0.4667 0.3363 Ans.
= 0.1304 rad = 7.47°
Ans: TP = 71.8 kN # m, f r = 7.47° 444
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5–138. A torque is applied to the shaft having a radius of 80 mm. If the material obeys a shear stress–strain relation of t = 500 g ¼ MPa, determine the torque that must be applied to the shaft so that the maximum shear strain becomes 0.008 rad.
80 mm T t (MPa)
Solution
0.008
tr Function: First, relate g to r. g = Then
g (rad)
r r gmax = a b(0.008) = 0.1r c 0.08
t = 500 ( 106 ) (0.1r)1>4 = 281.17 ( 106 ) r1>4 Torque: T = 2p
= 2p
Lo
Lo
c
2
tr dp
0.08m
6
9
281.17 ( 10 ) r 4 dp
= 1.7666 ( 109 ) a
4 13>4 0.08 m r b` 13 0
= 148.02 ( 103 ) N # m = 148 kN # m
Ans.
Ans: T =  148 kN # m 445
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5–139. A tubular shaft has an inner diameter of 60 mm, an outer diameter of 80 mm, and a length of 1 m. It is made of an elastic perfectly plastic material having a yield stress of tY = 150 MPa. Determine the maximum torque it can transmit. What is the angle of twist of one end with respect to the other end if the inner surface of the tube is about to yield? G = 75 GPa.
1m
30 mm 40 mm
Solution Plastic Torque: The plastic torque is the maximum torque a circular shaft can transmit. Co
Tp = 2p
LCi
= 2p tY a = =
2
tY p dp p3 3
b`
Co Ci
2 p tY ( C 30  C 3i ) 3 2 p 3 150 ( 106 ) 4 ( 0.043  0.033 ) 3
= 11.624 ( 103 ) N # m = 11.6 kN # m
Ans.
Angle of Twist: The yield strain at the inner surface of the tube can be determined from tY = G gY ;
150 ( 106 ) = 75 ( 109 ) gY gY = 0.002 rad
Then, the angle of twist is f =
gY rY
L = a
0.002 b(1) 0.03
= (0.06667 rad) a
180° b = 3.82° p rad
Ans.
Ans: TP = 11.6 kN # m, f = 3.82° 446
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*5–140. The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diameter segment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglect the stress concentration at the fillet.
80 mm
T
75 mm T
Solution Maximum Elastic Torque: For the larger diameter segment, the torsion formula can be applied. TYc tY = J tY c
tYJ T = = C
p ( 0.044 ) d 2 = 32.0 ( 106 ) p tY 0.04
Elastic  Plastic Torque: For the smaller diameter segment, the plastic torque can be determined from Tp = 2p
L0
c
tY r2 dp
=
2p 3 tc 3 Y
=
2p t ( 0.03753 ) 3 Y
= 35.16 ( 106 ) p tY Since T 6 TP, the smaller segment is not fully plastic T =
p tY ( 4c 3  rY3 ) 6
32.0 ( 106 ) p tY =
p tY 3 4 ( 0.03753 )  r3Y 4 6
rY = 0.02665 m
Ans.
= 26.7 mm
Ans: rY = 26.7 mm 447
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5–141. The shear stress–strain diagram for a solid 50mmdiameter shaft can be approximated as shown in the figure. Determine the torque required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 3 m long, what is the corresponding angle of twist?
(MPa) 125
50
0.0025
0.010
(rad)
Solution g =
r g c max
gmax = 0.01 When g = 0.0025 r = =
cg gmax 0.025(0.0025) 0.010
= 0.00625
50(106) t  0 = r  0 0.00625 t = 8000(106)(r) t  50(106) r  0.00625
=
125(106)  50(106) 0.025  0.00625
t = 4000(106)(r) + 25(106) T = 2p = 2p
L0
c
L0
0.00625
tr2 dr 8000(106)r3 dr 0.025
+ 2p
[4000(106)r + 25(106)]r2 dr L0.00625
T = 3269 N # m = 3.27 kN # m f =
Ans.
gmax 0.01 L = (3) c 0.025 Ans.
= 1.20 rad = 68.8°
Ans: T = 3.27 kN # m, f = 68.8° 448
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5–142. The shear stress–strain diagram for a solid 50mmdiameter shaft can be approximated as shown in the figure. Determine the torque T required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 1.5 m long, what is the corresponding angle of twist?
T
1.5 m T t (MPa) 125
Solution
50
Strain Diagram: rg 0.0025
=
0.025 ; 0.01
0.0025
rg = 0.00625 m
0.010
g (rad)
Stress Diagram: t1 =
50(106) r = 8(109) r 0.00625
t2  50(106) 125(106)  50(106) = r  0.00625 0.025  0.00625 t2 = 4 A 109 B r + 25 A 106 B The Ultimate Torque: c
T = 2p
L0
t r2dr 0.00625 m
= 2p
L0
8 A 109 B r3 dr 0.025 m
+ 2p
L0.00625 m
9 6 C 4 A 10 B r + 25 A 10 B D r2dr
m = 2p C 2 A 109 B r4 D 0.00625 0
+ 2p B 1 A 109 B r4 +
25(106)r3 0.025 m R 2 3 0.00625 m
= 3269.30 N # m = 3.27 kN # m
Ans.
Angle of Twist: f =
gmax 0.01 L = a b (1.5) = 0.60 rad = 34.4° c 0.025
Ans.
Ans. T = 3.27 kN # m, f = 34.4° 449
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5–143. The shaft consists of two sections that are rigidly connected. If the material is elastic plastic as shown, determine the largest torque T that can be applied to the shaft. Also, draw the shearstress distribution over a radial line for each section. Neglect the effect of stress concentration.
T 1 in. 25 mm
0.75 in. 20 mm T
Solution t (ksi) (MPa)
0.75 in. diameter segment will be fully plastic. From Eq. 527 of the text: 20mm
12 84
2π τY c3 T = Tp = 3 2π = [84(106 )](0.013 ) 3
0.005
= 175.93 N ⋅ = m 176 N ⋅ m
Ans.
84 MPa
g (rad)
57.3 MPa
For 125mm – in. diameter diametersegment: segment: tmax =
175.93(0.0125) Tc = 57.34(106 ) N−m 2 = π 4 J (0.0125 ) 2
= 57.3 6.75 MPa ksi 6< ttYY
Ans. T = Tp = 176 N ⋅ m 450
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*5–144. A steel alloy core is bonded firmly to the copper alloy tube to form the shaft shown. If the materials have the t g diagrams shown, determine the torque resisted by the core and the tube.
450 mm A
100 mm 60 mm B
15 kNm
t (MPa) 180
Solution Equation of Equilibrium. Referring to the freebody diagram of the cut part of the assembly shown in Fig. a, ΣMx = 0; Tc + Tt  15 1 103 2 = 0
(1) 180 1 10
6
Elastic Analysis. The shear modulus of steel and copper are Gst = 36 1 106 2 and G ∞ = = 18 GPa. Compatibility requires that 0.002 fC = ft
0.0024
2
= 75 GPa
Tc
1 0.034 2 (75) 1 109 2
=
Tc = 0.6204Tt
1 0.054
 0.034 2 (18) 1 109 2
(2)
Tc = 5743.05 N # m
The maximum elastic torque and plastic torque of the core and the tube are 1 3 1 pc (tY)st = p 1 0.033 2 (180) 1 106 2 = 7634.07 N # m 2 2 2 3 2 (TP)c = pc (tY)st = p 1 0.033 2 (180) 1 106 2 = 10 178.76 N # m 3 3 (TY)c =
and
p 1 0.054  0.034 2 J 2 (TY)t = tY = D T c (36) 1 106 2 d = 6152.49 N # m c 0.05 (TP)t = 2p(tY) ∞
Lci
co
r2dr = 2p(36) 1 106 2 ¢
t (MPa)
Copper Alloy
Solving Eqs. (1) and (2), Tt = 9256.95 N # m
Steel Alloy
0.002
Tt p 2
r3 0.05 m ≤` = 7389.03 N # m 3 0.03 m
Since Tt 7 (TY)t, the results obtained using the elastic analysis are not valid.
451
g (rad)
36
TcL TtL = JcGst JtG ∞ p 2
0.0024
g (rad)
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5–144. Continued
Plastic Analysis. Assuming that the tube is fully plastic, Tt = (TP)t = 7389.03 N # m = 7.39 kN # m
Ans.
Substituting this result into Eq. (1), Tc = 7610.97 N # m = 7.61 kN # m
Ans.
Since Tc 6 (TY)c, the core is still linearly elastic. Thus, f t = f tc = ft =
gi L; ci
TcL = JcGst
7610.97(0.45) p 4 9 2 (0.03 )(75)(10 )
0.3589 =
= 0.03589 rad
gi (0.45) 0.03
gi = 0.002393 rad Since gi 7 (gY) ∞ = 0.002 rad, the tube is indeed fully plastic.
Ans: Tt = 7.39 kN # m, Tc = 7.61 kN # m 452
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R5–1. The shaft is made of A992 steel and has an allowable shear stress of tallow = 75 MPa. When the shaft is rotating at 300 rpm, the motor supplies 8 kW of power, while gears A and B withdraw 5 kW and 3 kW, respectively. Determine the required minimum diameter of the shaft to the nearest millimeter. Also, find the rotation of gear A relative to C.
300 mm
300 mm
C
B
Solution A
Applied Torque: The angular velocity of the shaft is v = a300
rev 1 min 2prad ba ba b = 10p rad>s min 60 s 1 rev
Thus, the torque at C and gear A are TC =
8(103) PC = = 254.65 N # m v 10p
TA =
5(103) PA = = 159.15 N # m v 10p
Internal Loading: The internal torque developed in segment BC and AB of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: By inspection, segment BC is critical. tallow =
254.65(d2 ) TBC c ; 75(106) = p d 4 J 2 (2 ) d = 0.02586 m Ans.
Use d = 26 mm Angle of Twist: Using d = 26 mm, f A>C = Σ =
TiLi TBCLBC TABLAB = + JiGi JG JG 0.3
(159.15 p 4 9 2 (0.013 )(75)(10 )
+ 254.65) Ans.
= 0.03689 rad = 2.11°
Ans: Use d = 26 mm, f A>C = 2.11° 453
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R5–2. The shaft is made of A992 steel and has an allowable shear stress of tallow = 75 MPa. When the shaft is rotating at 300 rpm, the motor supplies 8 kW of power, while gears A and B withdraw 5 kW and 3 kW, respectively. If the angle of twist of gear A relative to C is not allowed to exceed 0.03 rad, determine the required minimum diameter of the shaft to the nearest millimeter.
300 mm
300 mm
C
B
Solution A
Applied Torque: The angular velocity of the shaft is v = a300
rev 1 min 2prad ba ba b = 10p rad>s min 60 s 1 rev
Thus, the torque at C and gear A are TC =
8(103) PC = = 254.65 N # m v 10p
TA =
5(103) PA = = 159.15 N # m v 10p
Internal Loading: The internal torque developed in segment BC and AB of the shaft are shown in Figs. a and b, respectively. Allowable Shear Stress: By inspection, segment BC is critical. tallow =
254.65(d2 ) TBC c ; 75(103) = p d 4 J 2 (2 ) d = 0.02586
Angle of Twist: f A>C = Σ 0.03 =
TiLi TBCLBC TABLAB = + JiGi JG JG 0.3
p d 4 9 2 ( 2 ) (75)(10 )
(159.15 + 254.65)
d = 0.02738 m(controls) Ans.
Use d = 28 mm
Ans: Use d = 28 mm 454
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R5–3. The A36 steel circular tube is subjected to a torque of 10 kN # m. Determine the shear stress at the mean radius r = 60 mm and calculate the angle of twist of the tube if it is 4 m long and fixed at its far end. Solve the problem using Eqs. 5–7 and 5–15 and by using Eqs. 5–18 and 5–20.
r 60 mm 4m
t 5 mm 10 kNm
Solution We show that two different methods give similar results: Shear Stress: Applying Eq. 57, ro = 0.06 + tr = 0.06 m =
0.005 = 0.0625 m 2
Tr = J
ri = 0.06 
10(103)(0.06) p 4 2 (0.0625
 0.05754)
0.005 = 0.0575 m 2
= 88.27 MPa
Applying Eq. 518, tavg =
10(103) T = = 88.42 MPa 2 t Am 2(0.005)(p)(0.062)
Angle of Twist: Applying Eq. 515, f = =
TL JG 10(103)(4)
p 4 2 (0.0625
 0.05754)(75.0)(109)
= 0.0785 rad = 4.495° Applying Eq. 520, f = = =
=
TL ds 2 4AmG L t TL ds 4A2mG t L
Where
L
ds = 2pr
2pTLr 4A2mG t 2p(10)(103)(4)(0.06) 4[(p)(0.062)]2 (75.0)(109)(0.005)
= 0.0786 rad = 4.503° Rounding to three significant figures, we find t = 88.3 MPa
Ans.
f = 4.50°
Ans. Ans: t = 88.3 MPa, f = 4.50° 455
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*R5–4. A portion of an airplane fuselage can be approximated by the cross section shown. If the thickness of its 2014T6aluminum skin is 10 mm, determine the maximum wing torque T that can be applied if tallow = 4 MPa. Also, in a 4mlong section, determine the angle of twist.
0.75 m
T 2m
Solution tavg
0.75 m
T = 2tAm
4(106) =
T 2(0.01)[(p)(0.75)2 + 2(1.5)]
T = 381.37(103) = 381 kN # m f =
f =
Ans.
ds TL 4A2mG L t 381.37(103)(4) 2
2
9
4[(p(0.75) + 2(1.5)) 27(10 )]
f = 0.542(103) rad = 0.0310°
c
4 + 2p(0.75) 0.010
d
Ans.
Ans: T = 381 kN # m, f = 0.0310° 456
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R5–5. The material of which each of three shafts is made has a yield stress of tY and a shear modulus of G. Determine which shaft geometry will resist the largest torque without yielding. What percentage of this torque can be carried by the other two shafts? Assume that each shaft is made from the same amount of material and that it has the same crosssectional area A.
A
60 A
A 60
60
Solution For circular shaft: 1
A = p c 2; tmax =
T =
c = a
Tc , J
A 2 b p
tY = =
p c3 t = 2 Y
Tc a 4 c 2
3 A 2 p( p )
tY
2 1
Tcir = 0.2821 A2 tY For the square shaft: 1
A = a2; a = A2 tmax =
4.81 T 4.81 T ; tY = 3 a3 A2 3
T = 0.2079 A2tY For the triangular shaft: A =
1 (a)(a sin 60°); 2
tmax =
20 T ; a3
tY =
1
a = 1.5197A2 20 T 3
(1.5197)3A2
3
T = 0.1755A2 tY Ans.
The circular shaft will resist the largest torque. For the square shaft: , =
0.2079 (100,) = 73.7, 0.2821
Ans.
For the triangular shaft: , =
0.1755 (100,) = 62.2 , 0.2821
Ans.
Ans: The circular shaft will resist the largest torque. For the square shaft: 73.7%, For the triangular shaft: 62.2% 457
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R5–6. Segments AB and BC of the assembly are made from 6061T6 aluminum and A992 steel, respectively. If couple forces P = 15 kN are applied to the lever arm, determine the maximum shear stress developed in each segment. The assembly is fixed at A and C.
P
1.2 m A
0.8 m
0.8 m E
al
1.2 m
100 mm
Solution
B
Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a, ©Mx = 0;
(1)
TA + TC  15 (1.6) = 0
D
st 100 mm
C
P
Compatibility Equation: It is required that fB>A = fB>C TA LAB TC LBC = JGal JGst
TAL
J[26(109 )]
=
TC L
J[75(109 )]
TA = 0.3467 TC
(2)
Solving Eqs. (1) and (2),
TC = 17.82 kN ⋅ m
TA = 6.178 kN ⋅ m
Maximum Shear Stress: (tmax)AB =
TA c [6.178(10 3 )](0.05) 2 31.5 MPa = 31.47(106 ) N−m = = π (0.054 ) J 2
Ans.
(tmax)BC =
TC c [17.82(10 3 )](0.05) 2 90.8 MPa = 90.77(106 ) N−m = = π (0.054 ) J 2
Ans.
Ans. (τ max ) AB = 31.5 MPa
(τ max )BC = 90.8 MPa 458
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R5–7. Segments AB and BC of the assembly are made from 6061T6 aluminum and A992 steel, respectively. If the allowable shear stress for the aluminum is (tallow)al = 90 MPa and for the steel (tallow)st = 120 MPa, determine the maximum allowable couple forces P that can be applied to the lever arm. The assembly is fixed at A and C.
P
1.2 m A
0.8 m
0.8 m E 1.2 m
100 mm
Solution
B
Equilibrium: Referring to the freebody diagram of the assembly shown in Fig. a, ©Mx = 0;
TA + TC  P(1.6) = 0
(1)
D 100 mm
C
P
Compatibility Equation: It is required that fB>A = fB>C TCLBC TA LAB = JGal JGst
TAL
J[26(109 )]
=
TC L
J[75(109 )]
TA = 0.3467 TC
(2)
Solving Eqs. (1) and (2),
= TC 1.1881 = P TA 0.4119 P Allowable Shear Stress:
= (τ allow )al
TAc (0.4119 P )(0.05) = ; 90(106 ) π (0.054 ) J 2 3 = ) N 42.9 kN = P 42.90(10
(τ allow )st =
(1.1881P)(0.05) TC c ; 120(106 ) = π (0.054 ) J 2 3 = ) N 19.8 kN (contπols) P 19.83(10 =
Ans.
Ans: P = 19.8 kN 459
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*R5–8. The tapered shaft is made from 2014T6 aluminum alloy, and has a radius which can be described by the equation > r = 0.02(1 + x 3 2) m, where x is in meters. Determine the angle of twist of its end A if it is subjected to a torque of 450 N # m.
x
r = 0.02(1 + x3/2) m 4m
450 Nm x
Solution
A
T = 450 N # m fA =
Tdx = L0 L JG
4
4
450 dx p 4 2 (0.02) (1
3 2
4
9
+ x ) (27)(10 )
= 0.066315
L0 (1 + x2)4 dx
3
Evaluating the integral numerically, we have f A = 0.066315 [0.4179] rad Ans.
= 0.0277 rad = 1.59°
Ans: f A = 1.59° 460
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R5–9. The 60mmdiameter shaft rotates at 300 rev>min. This motion is caused by the unequal belt tensions on the pulley of 800 N and 450 N. Determine the power transmitted and the maximum shear stress developed in the shaft. 300 rev/min
100 mm
Solution
450 N
rev 2p rad 1 min v = 300 c d = 10 p rad>s min 1 rev 60 s
800 N
T + 450(0.1)  800(0.1) = 0 T = 35.0 N # m
Ans.
P = Tv = 35.0(10p) = 1100 W = 1.10 kW tmax =
35.0(0.03) Tc = p = 825 kPa 4 J 2 (0.03 )
Ans.
Ans: P = 1.10 kW, tmax = 825 kPa 461
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6–1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.
B
A
250 mm
800 mm
24 kN
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.
Ans: x = 0.25 , V = 24, M = 6
329 329
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6–2. The deadweight loading along the centerline of the airplane wing is shown. If the wing is fixed to the fuselage at A, determine the reactions at A, and then draw the shear and moment diagram for the wing.
3.75 250kN/m lb/ft
15 kN 3000 lb
6 kN/m 400 lb/ft
A
Support Reactions: + c ©Fy = 0;
0.9 3 ftm
kNlb 1575000
–4.5 – 5.625 – 1.6875 – Ay =A0y = 0 1.00– 15  3+ 75 + 15  1.25  0.375
Ay = 48.1875 kN = 48.2 kN a + ©MA = 0;
0.6 m 2 ft
2.4 8 ftm
5.625 kN
Ans.
4.5 kN 15 kN
1.6875 kN 0.75 m
4.5(2.3) + 15(1.5) – 75(0.9) + 1.6875(0.5) ++MM 0 0 + 5.625(0.75) 1.25(2.5) ++0.375(1.667) AA= =
1.6 m
MA = 18.583 29.5875kip kN# ft · m==18.6 29.6kip kN# ·ftm
Ans.
+ ©F = 0; : x
Ans.
Ax = 0
0.5 m 0.4 m
75 kN
V (kN)
Shear and Moment Diagram:
0.6 m 0.8 m
M (kN · m)
53.0
29.6
48.2 2.4 –4.5
3
x (m) 3.9
2.4
3
x (m) 3.9
–3.6
–19.5 – 22.0 –16.0
Ans: Ay = 48.2 kN, Ax = 0 351 351
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6–3. Draw the shear and moment diagrams for the overhang beam.
8 kN/m
C
A B 4m
2m
Ans: x = 1.5, V = 0, M = 9, x = 4 , V = 20, M = 16 331 331
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*6–4. Express the shear and moment in terms of x for 0 6 x 6 3 m and 3 m 6 x 6 4.5 m, and then draw the shear and moment diagrams for the simply supported beam.
300 N/m
A
B 3m
1.5 m
Solution Support Reactions: Referring to the freebody diagram of the entire beam shown in Fig. a, a+ ΣMA = 0;
By(4.5) 
1 1 (300)(3)(2)  (300)(1.5)(3.5) = 0 2 2
By = 375 N + c ΣFy = 0; Ay + 375 
1 1 (300)(3)  (300)(1.5) = 0 2 2
Ay = 300 N Shear and Moment Function: For 0 … x 6 3 m, we refer to the freebody diagram of the beam segment shown in Fig. b. + c ΣFy = 0; 300 
1 (100x)x  V = 0 2
V = {300  50x2} N a+ ΣM = 0;
M +
Ans.
1 x (100x)xa b  300x = 0 2 3
M = e 300x 
50 3 x fN#m 3
Ans.
When V = 0, from the shear function, 0 = 300  50x2
x = 26 m
Substituting this result into the moment equation, Mx = 26 m = 489.90 N # m
For 3 m 6 x … 4.5 m, we refer to the freebody diagram of the beam segment shown in Fig. c. + c ΣFy = 0; V + 375 
1 3200(4.5  x) 4(4.5  x) = 0 2
V = e 100(4.5  x)2  375 f N
a+ ΣM = 0; 375(4.5  x) 
Ans.
1 4.5  x [200(4.5  x)](4.5  x)a b  M = 0 2 3
M = e 375(4.5  x) 
100 (4.5  x)3 f N # m Ans. 3
Shear and Moment Diagrams: As shown in Figs. d and e.
Ans: For 0 … x 6 3 m, V = {300  50x2} N, M = e 300x 
50 3 x f N # m, 3
For 3 m 6 x … 4.5 m, V = e 100(4.5  x)2  375 f N, M = e 375(4.5  x) 
464
100 (4.5  x)3 f N # m 3
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6–5. Draw the shear and moment diagrams for the simply supported beam.
4 kN M � 2 kN�m A
B 2m
2m
2m
Ans: x = 2 , V = 1, M = 2, x = 4 , V = 1, M = 6 329 329
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6–6. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x within the region 125 mm 6 x 6 725 mm.
1500 N 800 N A
B
x 125 mm
600 mm 75 mm
Solution + c ΣFy = 0;
815.63  800  V = 0 Ans.
V = 15.6 N a+ ΣM = 0;
M + 800(x  0.125)  815.63 x = 0 M = (15.6x + 100) N # m
Ans.
Ans: V = 15.6 N, M = (15.6x + 100) N # m 466
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6–7. Express the internal shear and moment in terms of x for 0 … x 6 L>2, and L>2 6 x 6 L, and then draw the shear and moment diagrams.
w0
A
B x L 2
Solution Support Reactions: Referring to the freebody diagram of the entire beam shown in Fig. a, a+ ΣMA = 0; By(L) By = + c ΣFy = 0; Ay + Ay =
1 L 5 w a ba L b = 0 2 0 2 6
5 wL 24 0
L 5 1 w L  w0 a b = 0 24 0 2 2 w0L 24
Shear and Moment Function: For 0 … x 6
L , we refer to the freebody diagram of 2
the beam segment shown in Fig. b. + c ΣFy = 0;
w0L  V = 0 24 V =
M a+ ΣM = 0; M =
w0L 24
Ans.
w0L x = 0 24 w0L x 24
Ans.
L 6 x … L, we refer to the freebody diagram of the beam segment shown in 2 Fig. c.
For
w0L 1 w0 1 + c ΣFy = 0;  c (2x  L) d c (2x  L) d  V = 0 24 2 L 2 V = a+ ΣM = 0; M + M =
w0 c L2  6(2x  L)2 d 24L
Ans.
w0 1 w0 1 1 c (2x  L) d c (2x  L) d c (2x  L) d x = 0 2 L 2 6 24L w0 c L2x  (2x  L)3 d 24L
Ans.
467
L 2
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6–7. Continued
When V = 0, the shear function gives 0 = L2  6(2x  L)2 x = 0.7041L Substituting this result into the moment equation, Mx = 0.7041L = 0.0265w0L2 Shear and Moment Diagrams: As shown in Figs. d and e.
Ans:
w0L w0L L :V = ,M = x, 2 24 24 w0 L For 6 x … L: V = 3 L2  6(2x  L)2 4 , 2 24L w0 M = 3 L2x  (2x  L)3 4 24L For 0 … x 6
V
w0L
0.704 L
24 0
0.0208 w0L2
5 wL 0 24
0.0265 w0L2
0.5 L 0.704 L
468
x
0.5 L
M
L
L
x
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*6–8. Draw the shear and moment diagrams for the beam. Hint: The 100kN load must be replaced by equivalent loadings at point C on the axis of the beam.
15 kN kip 75 20 kip 100 kN
1 ftm 0.25
A 75 kN
ft 14 m
100 kN 0.25 m
1m
1m
C ft 14 m
B ft 14 m
1m
75 kN 25 kN · m
100 kN 100 kN
58.33 kN
16.67 kN
V (kN) 58.3 2
3
1
x (m)
–16.67 M (kNm)
58.3
41.7 16.7
1
2
3
x (m)
333 333
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6–9. The load binder is used to support a load. If the force applied to the handle is 225 N, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC.
T1
A
C B 225 N 300 mm 75 mm
a + ©MC = 0;
225(375)  T1 (75) = 0 Ans.
T1 = 1125 N + T ©Fy = 0;
T2
225  1125 + T2 = 0 T2 = 900 N
Ans.

Ans: T1 = 1125 N, T2 = 900 N V (N )
900 x ⫺225
M (N⭈m )
x
⫺67 . 5
463
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6–10. Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load.
P
Support Reactions:
P
A
D
B
C
From the FBD of segment BD a + ©MC = 0; + c ©Fy = 0; + ©F = 0; : x
By (a)  P(a) = 0 Cy  P  P = 0
By = P
a
a
a
a
Cy = 2P
Bx = 0
From the FBD of segment AB a + ©MA = 0; + c ©Fy = 0;
P(2a)  P(a)  MA = 0
MA = Pa
P  P = 0 (equilibrium is statisfied!)
Ans: x = 3a , V = P, M = Pa 335 335
CH 06.indd 335
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6–11. The engine crane is used to support the engine, which has a weightof 6 kN. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown.
A
3 ftm 0.9
5 ftm 1.5 B
C
ft 1.24 m
a + ©MA = 0;
4 6(2.4) = = 0 0; FB(0.9) (3) – 1200(8) 5 A
+ c ©Fy = 0;
4 Ay + (20) (4000) – 6= 1200 0; = 0; 5
+ ©F = 0; ; x
Ax 
3 (4000) = 0; (20) = 0; 5
F kN lb FBA==204000
0.9 m
1.5 m
V (kN) 6
Ayy==102000 A kN lb
FB
6 kN –10
0.9
x (m) 2.4
M (kN . m)
Ax = 2400 Ax =lb12 kN
0.9
2.4
x (m)
–9.00
Ans:
x = 0.9m, V = 10 kN # x = 0.9+m, V = 6 kN x = 0.9 m, M =  9.00 kN # m
330 330
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*6–12. Draw the shear and moment diagrams for the cantilevered beam.
1.5 300kN lb
3 kN/m 200 lb/ft
A A 26m ft
The freebody diagram of the beam’s left segment sectioned through an arbitrary point shown in Fig. b will be used to write the shear and moment equations. The intensity of the triangular distributed load at the point of sectioning is x x 3 a ab =b1.5x = 33.33x w = 200 2 6
1 (3)(2) kN 2
1.5 kN
4.5 kN
4m 3
2m 3
5 kN · m
Referring to Fig. b, + c ©Fy = 0;
11 2 300 (33.33x)(x) V = V0 = {–1.5 V = { 300  16.67x2} lb (1) –1.5 –  (1.5x)(x) – V =0 – 0.75x } kN 22
a+ © M = 0;
1 x {−1.5 x − 0.25 x 3 } kN ⋅ m M + (1.5 x)( x) + 1.5 x = 0 M= 2 3
1.5 kN
1 (1.5x) x 2
(2)
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1) and (2), respectively.
V (kN)
x (m)
–1.5 –4.5
M (kN · m) x (m)
–5
337 337
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6–13. Draw the shear and moment diagrams for the beam.
M0
M0
M0
A
B
L/2
L/2
Solution Support Reactions: Referring to the FBD of the beam, Fig. a, a+ ΣMA = 0;
NB(L) + M0  M0  M0 = 0 NB =
a+ ΣMB = 0;
Ay(L) + M0  M0  M0 = 0 Ay =
+ ΣFx = 0; d
Ax = 0.
M0 L M0 L
Shear and Moment Diagram: Using the results of the support reaction, the shear and moment diagrams shown in Fig. b and c respectively can be plotted.
Ans: V = 
M0 , L
M0 L , M = M0  a bx, L 2 M0 L For 6 x … L, M =  a bx L 2 For 0 … x 6
474
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6–14. Draw the shear and moment diagrams for the beam.
30 kN/m 2 kip/ft 45 kip�ft kNm 30
30 kN/m
B
45 kN · m
A 1.5 m V (kN)
41.25 kN 1.5
0
1.5 m
1.5 m 3
1.5 5 ftm
3.75 kN 4.5 x (m)
1.5 5 ftm
1.5 5 ftm
–3.75 –45
M (kN · m) 0
1.5
–33.75
3
5.625 x (m)
–39.375
Ans:
x = 1.5 m, V = 45 kN # x = 1.5+m, V =  3.75 kN x = 1.5 m, M = 33.75 kN # m # x = 3  m M =  39.375 kN # m x = 3 + m, M = 5.625 kN # m 338 338
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6–15. Members ABC and BD of the counter chair are rigidly connected at B and the smooth collar at D is allowed to move freely along the vertical slot. Draw the shear and moment diagrams for member ABC. Equations of Equilibrium: Referring to the freebody diagram of the frame shown in Fig. a, + c ©Fy = 0;
150 750 lb N P�
Ay  750 150 = 0 150 N lb Ay = 750
0.45 m 1.5 ft
0.45 m 1.5 ft D
ND(0.45) – 750(0.9) = 0
a + ©MA = 0;
B
A
N 300 N lb NDD ==1500
C 0.45 m 1.5 ft
Shear and Moment Diagram: The couple moment acting on B due to ND is
300(1.5) == 450 MB ==1500(0.45) . The 675 lb N #· ftm. Theloading loadingacting acting on on member member ABC ABC is is shown in Fig. b and the shear and moment diagrams are shown in Figs. c and d.
750 N 0.45 m
0.45 m
0.45 m
0.45 m
0.45 m V (N)
M(N · m) 337.5
750 x (m) 0.45
0.9
0.45 –337.5
x (m)
750 N
0.9 Ay = 750 N
1500 N
MB = 675 N· m
Ans: 0 … x … 0.9 m, V = 750 N # x = 0.45m, M =  337.5 N # m # x = 0.45 + m, M = 337.5 N # m
333 333
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*6–16. A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier. Assume the columns at A and B exert only vertical reactions on the pier.
60 kN 60 kN 35 kN 35 kN 35 kN 1 m 1 m 1.5 m 1.5 m 1 m 1 m
A
B
Solution
Ans: N/A 477
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6–17. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.
w
A
B a
wL2 wL  wx = 0 2a
+ c ©Fy = 0;
x = L 
L
L2 2a
wL2 x Mmax (+) + wx a b  a wL bx = 0 2 2a
a + ©M = 0;
Substitute x = L 
L2 ; 2a
Mmax (+) = a wL =
L2 w L2 2 wL2 b aL b aL b 2a 2a 2 2a
w L2 2 aL b 2 2a
Mmax ()  w(L  a)
©M = 0;
Mmax () =
(L  a) = 0 2
w(L  a)2 2
To get absolute minimum moment, Mmax (+) = Mmax () w L2 2 w (L ) = (L  a)2 2 2a 2 L a =
L2 = L  a 2a L 22
Ans.
‚
Ans: L 12 x = 0, V = 0.243 wL # x = 0.243 L, V = 0 # x = 0.707 L , V =  0.414 wL x = 0.707 L+ , V = 0.293 wL x = 0.243 L, M = 0.0429 wL2 # x = 0.707 L, M = 0.0429 wL2
a =
342 342
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6–18. The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam.
2 kN/m
Equations of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, a + ©MA = 0;
3 FBC a b (2)  2(3)(1.5) = 0 5
B
A 1.5 m
FBC = 7.5 kN + c ©Fy = 0;
C
3 Ay + 7.5 a b  2(3) = 0 5
2m
1m
Ay = 1.5 kN
3 Shear and Moment Diagram: The vertical component of FBC is A FBC B y = 7.5a b 5 = 4.5 kN. The shear and moment diagrams are shown in Figs. c and d.
Ans: x = 0.75, V = 0, M = 0.5625, FBC = 7.5 kN, Ay = 1.5 kN 340 340
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6–19. w0
Draw the shear and moment diagrams for the beam.
B
A L 3
L 3
L 3
Solution Support Reactions: As shown on FBD. Shear and Moment Diagram: Shear and moment at x = L>3 can be determined using the method of sections. w0 L w0 L w0 L + c ΣFy = 0;  V = 0 V = 3 6 6 w0 L L w0 L L a + ΣMNA = 0; M + a b a b = 0 6 9 3 3 M =
5w0 L2 54
Ans: VA = Mmax 489
w0 L , 3 23 w0 L2 = 216
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*6–20. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kN>m caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw the shear and moment diagram for the pin.
0.4 kN/m
C
A
B
w0
w0
20 mm 60 mm 20 mm
Solution + c ΣFy = 0;
1 2(w0)(20)a b  60(0.4) = 0 2
Ans.
w0 = 1.2 kN>m
Ans: w0 = 1.2 kN>m 493
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6–21. Draw the shear and moment diagrams for the compound beam.
5 kN 3 kN/m A
B 3m
3m
D
C 1.5 m
1.5 m
Ans: x = 3 , V = 11.5, M = 21 349 349
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6–22. Draw the shear and moment diagrams for the simply supported beam.
10 kN
10 kN
15 kN�m A
B 2m
2m
2m
Ans:
x = 0, V = 7.5 kN # x = 2m, V = 7.5 kN x = 2 + m, V = 2.5 kN x = 4 + m V =  12.5 kN x = 2 m, M = 15 kN # m # x = 4 m, M = 10 kN # m # x = 6 m, M =  15 kN # m 352 352
CH 06.indd 352
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6–23. Draw the shear and moment diagrams for the beam.
18 kN/ m 12 kN/ m
A
B 3m
Solution
490
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*6–24. The footing supports the load transmitted by the two columns. Draw the shear and moment diagrams for the footing if the reaction of soil pressure on the footing is assumed to be uniform.
60 kN
60 kN
2m
0
2
4
30 kN.m 0
2
6
8
4m
2m
x(m)
30 kN.m 4
6
8
x(m)
Ans: V (kN) 30
30 2
6
8
6
8
4 ⫺30
x (m)
⫺30
M (kzN⭈m) 30 0
484
30 2
4
x (m)
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6–25. Draw the shear and moment diagrams for the overhanging beam.
45 kN/m
A
Support Reactions: Referring to the freebody diagram of the beam shown in Fig. a, a + ©MA = 0;
+ c ©Fy = 0;
1 By(6) (45)(4)(2.67) = 0 2 By = 40.0 kN
1 0 Ay + 40.0 − (45)(4) = 2 Ay = 50.0 kN
Shear and Moment Functions: For 0 … x 6 4 m, we refer to the freebody diagram of the beam segment shown in Fig. b. + c ©Fy = 0;
1 50.0 − (11.25 x)( x) − V = 0 2 V {50.0 − 5.625 x 2 } kN =
a + ©M = 0;
Ans.
1 x M + (11.25 x)( x) − 50.0 x = 0 2 3 M= {50.0x − 1.875 x 3 } kN ⋅ m
Ans.
When V = 0, from the shear function,
= 0 50.0 − 5.625 x 2
x = 2.981 m
Substituting this result into the moment function,
M x= 99.38 kN ⋅ m =2.981 m For 4 m 6 x … 6 m, we refer to the freebody diagram of the beam segment shown in Fig. c. + c ©Fy = 0;
V + 40.0 kN = 0 V =
Ans.
− 40.0 kN
a + ©M = 0; 40.05(6  x)  M = 0
M = {40.0(6 − x)} kN ⋅ m
Ans.
Shear and Moment Diagrams: As shown in Figs. d and e.
481
B 4m
2m
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6–26. The support at A allows the beam to slide freely along the vertical guide so that it cannot support a vertical force. Draw the shear and moment diagrams for the beam.
w B
A
L
Solution Equations of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, L a+ ΣMB = 0; wL a b  MA = 0 2 MA =
wL2 2
+ c ΣFy = 0; By  wL = 0 By = wL Shear and Moment Diagram: As shown in Figs. b and c.
Ans: V L
x
wL M 2
wL 2
L
492
x
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6–27. Draw the shear and moment diagrams for the beam.
M0
M0
M0
A
B
a
Solution
481
a
a
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*6–28. Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition.
P
P
L – 2
L – 2
A
B a
Solution Support Reactions: As shown on FBD. Absolute Minimum Moment: In order to get the absolute minimum moment, the maximum positive and maximum negative moment must be equal that is Mmax( + ) = Mmax(  ). For the positive moment: a+ ΣMNA = 0; Mmax( + )  a2P 
3PL L ba b = 0 2a 2
Mmax( + ) = PL 
3PL2 4a
For the negative moment: a+ ΣMNA = 0; Mmax(  )  P(L  a) = 0 Mmax(  ) = P(L  a) Mmax( + ) = Mmax(  ) PL 
3PL2 = P(L  a) 4a
4a L  3L2 = 4a L  4a2 a =
23 L = 0.866L 2
Ans.
Shear and Moment Diagram:
Ans: a = 0.866L 480
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6–29. Draw the shear and moment diagrams for the beam.
w0
B L 3
A
2L 3
Solution + c ΣFy = 0;
w0L 1 w0x  a b(x) = 0 4 2 L x = 0.7071 L
a+ ΣMNA = 0;
M +
Substitute x = 0.7071L,
w0L 1 w0x x L a ax  b = 0 b(x)a b 2 L 3 4 3
M = 0.0345 w0L2
Ans: V 7w0L 36 x w0L 18
w0L 4
0.707 L M
0.0345w0L2
x 2
0.00617w0L
488
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6–30. The 700N man sits in the center of the boat, which has a uniform width and a weight per linear foot of 450 N/m. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat. 2.25 m
2.25 m
450 N/m 605.55 N/m
155.55 N/m 394 2.25
0
4.50 x(m)
394 0
x(m) 2.25
4.50
Ans.
Mmax = 39 4 N # m
Ans: Mmax = 39 4 N # m V (N ) 3 50 x ⫺3 50 M ( N m)
#
3394 9 3 . 0
482
x
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6–31. Draw the shear and moment diagrams for the overhang beam.
18 kN 6 kN
A B 2m
2m
M � 10 kN�m
2m
Ans:
x = 0, V = 3.5 kN # x = 2+ m, V = 14.5 kN # x = 4 + m, V = 6 kN x = 2 m, M = 7 kN # m x = 4 m, M = 22 kN # m # x = 6 m M = 10 kN # m 350 350
CH 06.indd 350
1/18/11 2:22:06 PM
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*6–32. Draw the shear and moment diagrams for the beam.
5 kN/m
5 kN/m
B
A 4.5 m
4.5 m
From FBD(a) + c ©Fy = 0; a + ©MNA = 0;
9.375  0.5556x2 = 0
x = 4.108 m
M + (0.5556) A 4.1082 B a M = 25.67 kN # m
4.108 b  9.375(4.108) = 0 3
From FBD(b) a + ©MNA = 0;
M + 11.25(1.5)  9.375(4.5) = 0 M = 25.31 kN # m
345 345
CH 06.indd 345
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6–33. The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed loading on the beam over its 0.6m length. Draw the shear and moment diagrams for the beam if it supports a uniform loading of 30 kN/m.
30 kN/m
B A
2.4 m
0.6 m
0.3 m
0 0.3
2.7
3.3
2.7
3.3
1.5
x(m)
32.4
0 0.3
1.5
x(m)
Ans: V (kN) 36 3 6 0
0.3
M (kN m) M
1.5
490
3.3
x (m)
36 6 32.4 5
10 . 10.8 0.3
2.7
10.8 1 0 . 1.5
2.7
3.3
x (m)
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6–34. Draw the shear and moment diagrams for the simply supported beam.
10 kN
10 kN
15 kNm A
B 2m
Solution
501
2m
2m
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6–35. A short link at B is used to connect beams AB and BC to form the compound beam. Draw the shear and moment diagrams for the beam if the supports at A and C are considered fixed and pinned, respectively.
15 kN 3 kN/m
B
A
4.5 m
C
1.5 m
1.5 m
Solution Support Reactions: Referring to the freebody diagram of segment BC shown in Fig. a, a+ΣMC = 0; 15(1.5)  FB(3) = 0 FB = 7.5 kN + c ΣFy = 0; Cy + 7.5  15 = 0 Cy = 7.5 kN Using the result of FB and referring to the freebody diagram of segment AB, Fig. b, 1 + c ΣFy = 0; Ay  (3)(4.5)  7.5 = 0 2 Ay = 14.25 kN 1 (3)(4.5)(3)  7.5(4.5) = 0 2 MA = 54 kN # m
a+ΣMA = 0; MA 
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans: V (kN) 14.25 0
7.5 6
7.5
4.5
x (m)
7.5
M (kNm) 11.25 0
54
506
4.5
6
7.5
x (m)
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*6–36. The compound beam is fixed at A, pin connected at B, and supported by a roller at C. Draw the shear and moment diagrams for the beam.
2 kN 3 kN/m
C
B
A 3m
3m
Solution Support Reactions: Referring to the freebody diagram of segment BC shown in Fig. a, a+ΣMB = 0; Cy(3)  3(3)(1.5) = 0 Cy = 4.5 kN
c ΣFy = 0;
By + 4.5  3(3) = 0 By = 4.5 kN
Using the result of By and referring to the freebody diagram of segment AB, Fig. b,
c ΣFy = 0;
Ay  2  4.5 = 0 Ay = 6.5 kN
a+ΣMA = 0; MA  2(3)  4.5(3) = 0 MA = 19.5 kN # m
Shear and Moment Diagrams: As shown in Figs. c and d.
Ans: V (kN) 6.5 0
4.5 6 3
4.5
x (m)
4.5 M (kNm) 3.375 0
19.5
504
3
4.5
6
x (m)
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6–37. Draw the shear compound beam.
and
moment
diagrams
for
the
5 kN/m
A
B 2m
D
C 1m
1m
Solution Support Reactions: From the FBD of segment AB a+ ΣMA = 0;
By (2)  10.0(1) = 0
By = 5.00 kN
+ c ΣFy = 0;
Ay  10.0 + 5.00 = 0
Ay = 5.00 kN
From the FBD of segment BD a+ ΣMC = 0;
5.00(1) + 10.0(0)  Dy (1) = 0 Dy = 5.00 kN
+ c ΣFy = 0;
Cy  5.00  5.00  10.0 = 0 Cy = 20.0 kN
+ ΣFx = 0; Bx = 0 S From the FBD of segment AB + ΣFx = 0; Ax = 0 S Shear and Moment Diagram:
Ans: V (kN) 5.00 0
2 1 5.00
3 10.0
503
M (kNm)
10.0 5.00 x (m) 4
2.50 0
1
3
4
2 7.50
x (m)
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6–38. The compound beam is fixed at A, pin connected at B, and supported by a roller at C. Draw the shear and moment diagrams for the beam.
600 N 400 N/m
A
C
B 2m
2m
2m
Solution Support Reactions: Referring to the freebody diagram of segment BC shown in Fig. a, a+ΣMB = 0; Cy(2)  400(2)(1) = 0 Cy = 400 N + c ΣFy = 0; By + 400  400(2) = 0 By = 400 N Using the result of By and referring to the freebody diagram of segment AB, Fig. b, + c ΣFy = 0; Ay  600  400 = 0 Ay = 1000 N a+ΣMA = 0; MA  600(2)  400(4) = 0 MA = 2800 N Shear and Moment Diagrams: As shown in Figs. c and d.
Ans: V (N) 1000 400 0
2
5
6
4
x (m)
400 M (Nm)
0
4 800
2800
502
200
2
5
6
x (m)
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6–39. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at B. Draw the shear and moment diagrams for the shaft.
400 Nm
B
A
1m
1m
1m
900 N
Solution Equations of Equilibrium: Referring to the freebody diagram of the shaft shown in Fig. a, a+ ΣMA = 0; By(2) + 400  900(1) = 0 By = 250 N Ay + 250  900 = 0 + c ΣFy = 0; Ay = 650 N Shear and Moment Diagram: As shown in Figs. b and c.
Ans: V (N) 650 1
0
2
3
x (m)
250
M (Nm) 650 0
494
1
400 2
3
x (m)
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*6–40. The beam is used to support a uniform load along CD due to the 6kN weight of the crate. Also, the reaction at the bearing support B can be assumed uniformly distributed along its width. Draw the shear and moment diagrams for the beam.
0.5 m
0.75 m
2.75 m
2m C
A
D
B
Solution Equations of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, a+ ΣMA = 0;
FB(3)  6(5) = 0 FB = 10 kN
+ c ΣFy = 0;
10  6  Ay = 0 Ay = 4 kN
Shear and Moment Diagram: The intensity of the distributed load at support B and FB 10 6 portion CD of the beam are wB = = = 20 kN>m and wCD = = 3 kN>m, 0.5 0.5 2 Fig. b. The shear and moment diagrams are shown in Figs. c and d.
Ans: V (kN)
0
2.95 2.75
6
3.25 4
6
3.25 4
6
x (m)
4
M (kNm)
0
2.95 2.75
6 11
499
10.5 11.4
x (m)
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6–41. Draw the shear and moment diagrams for the beam.
50 kN/m
50 kN/m
B A 4.5 m
4.5 m
Solution
Ans: V (kN) 112.5 x 112.5 M (kNm) 169 x
498
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6–42. Draw the shear cantilever beam.
and
moment
diagrams
for
the
2 kN
A
3 kNm 1.5 m
1.5 m
Solution Equations of Equilibrium: Referring to the freebody diagram of the beam shown in Fig. a, + c ΣFy = 0; Ay  2 = 0 Ay = 2 kN a+ ΣMA = 0;
MA  3  2(3) = 0 MA = 9 kN # m
Shear and Moment Diagram: As shown in Figs. b and c.
Ans: V (kN) 2 0
1.5
3
1.5
3
x (m)
M (kNm)
0
3 6 9
495
x (m)
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6–43. Draw the shear and moment diagrams for the beam.
400 N/m 200 N/ m A
B x 3m
3m
Solution Support Reactions: As shown on FBD. Shear and Moment Functions: For 0 … x 6 3 m: + c ΣFy = 0;
V = 200 N
Ans.
M = {200 x} N # m
Ans.
200  V = 0
a+ ΣMNA = 0;
M  200 x = 0
For 3 m 6 x … 6 m:
+ c ΣFy = 0;
200  200(x  3) V = e
Set V = 0, x = 3.873 m a+ ΣMNA = 0;
M +
1 200 c (x  3) d (x  3)  V = 0 2 3
100 2 x + 500 f N 3
Ans.
x  3 1 200 c (x  3) d (x  3)a b 2 3 3
+ 200(x  3)a M = e
x  3 b  200x = 0 2
100 3 x + 500x  600 f N # m 9
Ans.
Substitute x = 3.87 m, M = 691 N # m
Ans: For 0 … x 6 3 m: V = 200 N, M = {200x} N # m, For 3 m 6 x … 6 m: V = e 
M = e
100 2 x + 500 f N, 3
100 3 x + 500x  600 f N # m 9 M (Nm)
V (N)
600 691 200 0
3.87
6
3 700
496
x (m)
0
3 3.87
6
x (m)
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*6–44. Draw the shear and moment diagrams for the beam.
w
w0
p w � w0 sin – x L
A
B L – 2
FR =
LA
dA = w0
L0
L
sin a
x
L – 2
2w0 L p x bdx = p L
355 355
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6–45. Draw the shear and moment diagrams for the beam.
w
w�
L
A
L
w0 L dA = wdx = 2 x2 dx = FR = 3 L L0 LA L0 w0
w0
w0 2 x L2
w0
B
x
L
L
x3dx L2 L0 3L x = = = w0 L 4 dA 3 LA LA
xdA
+ c ©Fy = 0;
w0L w0x3 = 0 12 3L2
1 1>3 x = a b L = 0.630 L 4 w0L w0x3 1 a + ©M = 0; (x) a xb  M = 0 12 3L2 4 M =
w0Lx w0x4 12 12L2
Substitute x = 0.630L M = 0.0394 w0L2
Ans: x = 0.630 L, V = 0, M = 0.0394w0L2, M =
w0Lx w0x4 12 12L2
354 354
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6–46. The truck is to be used to transport the concrete column. If the column has a uniform weight of w (force/length), determine the equal placement a of the supports from the ends so that the absolute maximum bending moment in the column is as small as possible. Also, draw the shear and moment diagrams for the column.
L a
a
Solution Support Reactions: As shown on FBD. Absolute Minimum Moment: In order to get the absolute minimum moment, the maximum positive and maximum negative moment must be equal that is Mmax(+) = Mmin(). For the positive moment: a+ΣMNA = 0; Mmax(+) +
wL L wL L a b a  ab = 0 2 4 2 2
Mmax(+) =
wL2 waL 8 2
For the negative moment: a a+ΣMNA = 0; wa a b  Mmax() = 0 2 Mmax(  ) =
wa2 2
Mmax(+) = Mmax() wL2 wL wa2 a = 8 2 2 4a2 + 4La  L2 = 0 a =
 4L { 216L2  4(4)(  L2) 2(4)
Ans.
a = 0.207L Shear and Moment Diagram:
Ans: a = 0.207L 507
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6–47. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the maximum bending stress in the beam. Sketch a threedimensional view of the stress distribution and cover the cross section.
25 mm
150 mm 20 mm 200 mm M 600 Nm 20 mm
Solution y = I =
(0.0125)(0.24)(0.025) + 2(0.1)(0.15)(0.02) 0.24(0.025) + 2(0.15)(0.02)
= 0.05625 m
1 (0.24) ( 0.0253 ) + (0.24)(0.025) ( 0.043752 ) 12 +2a
1 b(0.02) ( 0.153 ) + 2(0.15)(0.02) ( 0.043752 ) 12
= 34.53125 ( 106 ) m4 smax = sB = =
Mc I 600(0.175  0.05625) 34.53125 ( 106 ) Ans.
= 2.06 MPa sc =
600(0.05625) My = = 0.977 MPa I 34.53125 ( 106 )
Ans: smax = 2.06 MPa 513
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*6–48. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N # m, determine the resultant force the bending stress produces on the top board.
25 mm
150 mm 20 mm 200 mm M 600 Nm 20 mm
Solution y = I =
(0.0125)(0.24)(0.025) + 2(0.15)(0.1)(0.02) 0.24(0.025) + 2(0.15)(0.02)
= 0.05625 m
1 (0.24) ( 0.0253 ) + (0.24)(0.025) ( 0.043752 ) 12 + 2a
1 b(0.02) ( 0.153 ) + 2(0.15)(0.02) ( 0.043752 ) 12
= 34.53125 ( 106 ) m4 st =
600(0.05625) My = = 0.9774 MPa I 34.53125 ( 106 )
sb =
600(0.05625  0.025) My = = 0.5430 MPa I 34.53125 ( 106 )
F =
1 (0.025)(0.9774 + 0.5430) ( 106 ) (0.240) = 4.56 kN 2
Ans.
Ans: F = 4.56 kN 514
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6–49. Determine the moment M that will produce a 70 ksi MPa the crosssection. maximum stress of 10 onon the cross section.
0.5mm in. 12 A 12 0.5mm in.
753 mm in.
12 mm 0.5 in. B
C
753 mm in. M 25010mm in.
D 12 mm 0.5 in.
Section Properties: y =
=
©yA ©A
0.006(0.099)(0.012) + 2[0.0495(0.012)(0.075)] + 0.137(0.012)(0.25) = 0.08471 m 0.099(0.012) + 2(0.012)(0.075) + 0.012(0.25)
1 I NA =(0.099)(0.012 3 ) + (0.099)(0.012)(0.08471 − 0.006)2 12 1 + 2 (0.012)(0.0753 ) + (0.012)(0.075)(0.08471 − 0.0495)2 12 +
1 (0.012)(0.253 ) + (0.012)(0.25)(0.137 − 0.08471)2 12
= 34.2773(10 −6 ) m 4 Maximum Bending Stress: Applying the flexure formula smax =
70(106 ) =
Mc I
M (0.262 − 0.08471) 34.2773(10 −6 )
⋅ m 13.5 kN ⋅ m = M 13.53(10 3 ) N=
Ans.
Ans:
M = 13.5 kN # m 356 356
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6–50. Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of # ft. M ==64kN kip· m.
0.5mm in. 12 A 12 0.5mm in.
753mm in.
12 0.5mm in. B
C
753 mm in. M 25010 mm in.
D 12 0.5mm in.
Section Properties: y =
=
©yA ©A
0.006(0.099)(0.012) + 2[0.0495(0.012)(0.075)] + 0.137(0.012)(0.25) = 0.08471 m 0.099(0.012) + 2(0.012)(0.075) + 0.012(0.25)
1 I NA =(0.099)(0.012 3 ) + (0.099)(0.012)(0.08471 − 0.006)2 12 1 + 2 (0.012)(0.0753 ) + (0.012)(0.075)(0.08471 − 0.0495)2 12 +
1 (0.012)(0.253 ) + (0.012)(0.25)(0.137 − 0.08471)2 12
= 34.2773(10 −6 ) m 4
Maximum Bending Stress: Applying the flexure formula smax =
Mc I
= (σ t )max
[6(10 3 )](0.262 − 0.08471) 6 = 31.03(10 = ) N−m 2 31.0 MPa 34.2773(10 −6 )
Ans.
= (σ c )max
[6(10 3 )](0.08471) 6 = 14.83(10 = ) N−m 2 14.8 MPa 34.2773(10 −6 )
Ans.
Ans: y = 0.08471 m, INA = 34.2773 (10  6) m4, (st) max = 31.0 MPa, (sc) max = 14.8 MPa 357 357
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6–51. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards, A and B, of the beam.
A 25 mm
I =
1 1 (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10  6 B m4 12 12
150 mm 25 mm 25 mm
Bending Stress: Applying the flexure formula s =
M
D
Section Property:
B 150 mm
25 mm
My I
sE =
sD =
M(0.1) 91.14583(10  6) M(0.075) 91.14583(10  6)
= 1097.143 M
= 822.857 M
Resultant Force and Moment: For board A or B F = 822.857M(0.025)(0.2) +
1 (1097.143M  822.857M)(0.025)(0.2) 2
= 4.800 M M¿ = F(0.17619) = 4.80M(0.17619) = 0.8457 M sc a
M¿ b = 0.8457(100%) = 84.6 % M
Ans.
Ans: 84.6, 359 359
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*6–52. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of sD = 30 MPa . Also sketch the stress distribution acting over the cross section and compute the maximum stress developed in the beam.
A 25 mm M
D 150 mm 25 mm 25 mm
B 150 mm
25 mm
Section Property: I =
1 1 (0.2) A 0.23 B (0.15) A 0.153 B = 91.14583 A 10  6 B m4 12 12
Bending Stress: Applying the flexure formula s = 30 A 106 B =
My I M(0.075) 91.14583(10  6)
M = 36458 N # m = 36.5 kN # m
smax =
Ans.
36458(0.1) Mc = 40.0 MPa = I 91.14583(10  6)
Ans.
Ans: M = 36.5 kN # m, smax = 40.0 MPa 359 359
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6–53. An A36 steel strip has an allowable bending stress of 165 MPa. If it is rolled up, determine the smallest radius r of the spool if the strip has a width of 10 mm and a thickness of 1.5 mm. Also, find the corresponding maximum internal moment developed in the strip.
r
Solution Bending StressCurvature Relation: sallow =
200(109) 30.75(10  3) 4 Ec ; 165(106) = r r Ans.
r = 0.9091 m = 909 mm Moment Curvature Relation: 1 M = ; r EI
1 = 0.9091
M 200(109) c
1 (1)(0.00153) d 12
M = 61.875 N # m = 61.9 N # m
Ans.
Ans: r = 909 mm, M = 61.9 N # m 508
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6–54. If the beam is subjected to an internal moment of M = 30 kN # m, determine the maximum bending stress in the beam. The beam is made from A992 steel. Sketch the bending stress distribution on the cross section.
50 mm 50 mm 15 mm A
10 mm
M
150 mm
15 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is I =
1 1 (0.1)(0.153) (0.09)(0.123) = 15.165(10  6) m4 12 12
Maximum Bending Stress: The maximum bending stress occurs at the top and bottom surfaces of the beam since they are located at the furthest distance from the neutral axis. Thus, c = 75 mm = 0.075 m. smax =
30(103)(0.075) Mc = 148 MPa = I 15.165(10  6)
Ans.
At y = 60 mm = 0.06 m, sy = 0.06 m =
My 30(103)(0.06) = 119 MPa = I 15.165(10  6)
The bending stress distribution across the cross section is shown in Fig. a.
Ans: s max = 148 MPa 513
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6–55. If the beam is subjected to an internal moment of M = 30 kN . m, determine the resultant force caused by the bending stress distribution acting on the top flange A.
50 mm 50 mm 15 mm A
10 mm
M
150 mm
15 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is 1 1 (0.1)(0.153) (0.09)(0.123) = 15.165(10  6) m4 12 12
I =
Bending Stress: The distance from the neutral axis to the top and bottom surfaces of flange A is yt = 75 mm = 0.075 m and yb = 60 mm = 0.06 m. st =
Myt 30(103)(0.075) = 148.37 = 148 MPa = I 15.165(10  6)
sb =
Myb 30(103)(0.06) = 118.69 = 119 MPa = I 15.165(10  6)
Resultant Force: The resultant force acting on flange A is equal to the volume of the trapezoidal stress block shown in Fig. a. Thus, FR =
1 (148.37 + 118.69)(106)(0.1)(0.015) 2 Ans.
= 200 296.74 N = 200 kN
Ans: FR = 200 kN 514
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*6–56. If the builtup beam is subjected to an internal moment of M = 75 kN # m, determine the maximum tensile and compressive stress acting in the beam.
150 mm 20 mm 150 mm 10 mm
150 mm
M
10 mm
Solution Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is y =
300 mm A
0.15(0.3)(0.02) + 23 0.225(0.15)(0.01)] + 230.295(0.01)(0.14) 4 Σy~A = = 0.2035 m ΣA 0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
Thus, the moment of inertia of the cross section about the neutral axis is I = ΣI + Ad 2 =
1 (0.02)(0.33) + 0.02(0.3)(0.2035  0.15)2 12 + 2c + 2c
1 (0.01)(0.153) + 0.01(0.15)(0.225  0.2035)2 d 12 1 (0.14)(0.013) + 0.14(0.01)(0.295  0.2035)2 d 12
= 92.6509(10  6) m4
Maximum Bending Stress: The maximum compressive and tensile stress occurs at the top and bottommost fiber of the cross section. (smax)c =
75(103)(0.3  0.2035) My = = 78.1 MPa I 92.6509(10  6)
Ans.
(smax)t =
75(103)(0.2035) Mc = = 165 MPa I 92.6509(10  6)
Ans.
Ans: (smax)c = 78.1 MPa, (smax)t = 165 MPa 515
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6–57. If the builtup beam is subjected to an internal moment of M = 75 kN # m, determine the amount of this internal moment resisted by plate A.
150 mm 20 mm 150 mm 10 mm
150 mm
M
10 mm
Solution
300 mm
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
y =
A
~A 0.15(0.3)(0.02) + 23 0.225(0.15)(0.01)] + 230.295(0.01)(0.14) 4 Σy = = 0.2035 m ΣA 0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
Thus, the moment of inertia of the cross section about the neutral axis is I = I + Ad 2 =
1 (0.02)(0.33) + 0.02(0.3)(0.2035  0.15)2 12
+ 2c + 2c
1 (0.01)(0.153) + 0.01(0.15)(0.225  0.2035)2 d 12 1 (0.14)(0.013) + 0.14(0.01)(0.295  0.2035)2 d 12
= 92.6509(10  6) m4
Bending Stress: The distance from the neutral axis to the top and bottom of plate A is yt = 0.3  0.2035 = 0.0965 m and yb = 0.2035 m. st =
75(103)(0.0965) Myt = = 78.14 MPa (C) I 92.6509(10  6)
sb =
75(103)(0.2035) Myb = = 164.71 MPa (T) I 92.6509(10  6)
The bending stress distribution across the cross section of plate A is shown in Fig. b. The resultant forces of the tensile and compressive triangular stress blocks are
(FR)t =
1 (164.71)(106)(0.2035)(0.02) = 335 144.46 N 2
(FR)c =
1 (78.14)(106)(0.0965)(0.02) = 75 421.50 N 2
Thus, the amount of internal moment resisted by plate A is 2 2 M = 335144.46c (0.2035) d + 75421.50c (0.0965) d 3 3 = 50315.65 N # m = 50.3 kN # m
Ans.
Ans: M = 50.3 kN # m 516
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6–58. The beam is subjected to a moment M. Determine the percentage of this moment that is resisted by the stresses acting on both the top and bottom boards of the beam.
90 mm 20 mm 20 mm 20 mm 20 mm
D
90 mm 20 mm 20 mm
M
100 mm
Solution
100 mm
Section Properties: The moment of inertia of the beam’s crosssection about the neutral axis is 1 1 I = (0.24) ( 0.243 ) (0.18) ( 0.23 ) = 0.15648 ( 10  3 ) m4 12 12
20 mm
My Bending Stress: Applying the flexure formula, s = , the bending stress on points I D and E, Fig. a, is sD = sE =
M(0.1) 0.15648 ( 10  3 ) M(0.12) 0.15648 ( 10  3 )
= 639.06M = 766.87M
Resultant Force and Moment: The resultant of the stress block acting on boards A and B, Fig. a, is F =
1 (639.06M + 766.87M)(0.02)(0.24) = 3.3742M 2
The location of line of action of F is a =
1 2(766.87M) + 639.06M c d (0.02) = 0.010303 m 3 639.06M + 766.87M
Thus, the moment arm of F is
a′ = 2(0.1 + 0.010303) = 0.22061 m Then M′ = Fd = 3.3742M(0.22061) = 0.74438M %a
M′ 0.74438M b = a b(100) = 74.4, M M
Ans.
Ans: M′ = 74.4, M 517
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6–59. Determine the moment M that should be applied to the beam in order to create a compressive stress at point D of sD = 10 MPa. Also sketch the stress distribution acting over the cross section and calculate the maximum stress developed in the beam.
90 mm 20 mm 20 mm 20 mm 20 mm
D
90 mm 20 mm 20 mm
M
100 mm
Solution
100 mm
Section Properties: The moment of inertia of the beam’s crosssection about the neutral axis is I =
20 mm
1 1 (0.24) ( 0.243 ) (0.18) ( 0.23 ) = 0.15648 ( 10  3 ) M4 12 12
Bending Stress: Applying the flexure formula, sD = 10 ( 106 ) =
MyD I M(0.1) 0.15648 ( 10  3 )
M = 15.648 ( 10  3 ) N # m = 15.6 kN # m
Ans.
The maximum bending stress is smax =
[15.648 ( 103 ) ](0.12) MC = = 12.0 ( 106 ) Pa = 12.0 MPa I 0.15648 ( 10  3 )
Ans.
The sketch of Bending stress distribution on the beam’s crosssection is shown in Fig. a.
Ans: M = 15.6 kN # m, smax = 12.0 MPa 518
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*6–60. If the beam is subjected to an internal moment of of M = 150 kN · m, determine the maximum tensile and compressive bending stress in the beam.
75 mm 3 in. 75 mm 3 in. in. 1506mm MM 2 in. 50 mm in. 37.1.5 5 mm
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
y =
©yA 0.1(0.15)(0.2) − 0.05[π (0.03752 )] 0.10863 m = = ©A 0.15(0.2) − π (0.03752 )
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
1 π = (0.15)(0.2 3 ) + 0.15(0.2)(0.10863 − 0.1)2 − (0.03754 ) + π (0.03752 )(0.10863 − 0.05)2 12 4 = 85.4948(10 −6 ) m 4 Maximum Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section.
(σ max= )T
MC [150(10 3 )](0.10863) 6 190.60(10 = ) N−m 2 191 MPa = = I 85.4948(10 −6 )
Ans.
(σ max= )C
MC [150(10 3 )](0.2 − 0.10863) 2 = = 160.30(106 ) N−m = 160 MPa I 85.4948(10 −6 )
Ans.
Ans:
(σ max )T = 191 MPa, (σ max )C = 160 MPa 362 362
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6–61. If the beam is made of material having an allowable tensile tensileand and compressive of )(s = allowable compressive stressstress of (sallow 24)tksi allow t = 168 MPa = 154 MPa,determine respectively, and (sallowand the determine maximum )c = (s 22allow ksi,)crespectively, the maximum allowable moment M to that be allowable internal moment internal M that can be applied thecan beam. applied to the beam.
75 mm 3 in. 75 mm 3 in. 1506mm in. MM 50 mm 2 in. 37.1.5 5 mm in.
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
y =
©yA 0.1(0.15)(0.2) − 0.05[π (0.03752 )] 0.10863 m = = ©A 0.15(0.2) − π (0.03752 )
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
1 π = (0.15)(0.2 3 ) + 0.15(0.2)(0.10863 − 0.1)2 − (0.03754 ) + π (0.03752 )(0.10863 − 0.05)2 12 4 = 85.4948(10 −6 ) m 4 Allowable Bending Stress: The maximum compressive and tensile bending stress occurs at the top and bottom edges of the cross section. For the top edge, (sallow)c =
My ; I
154(106 ) =
M (0.2 − 0.10863) 85.4948(10 −6 )
M 144.11(10 3 ) N= = ⋅ m 144 kN ⋅ m For the bottom edge,
A smax B t =
Mc ; I
154(106 ) =
M (0.10863) 85.4948(10 −6 )
M 132.22(10 3 ) N= = ⋅ m 132 kN ⋅ m (controls!)
Ans.
Ans:
M = 132 kN # m 363 363
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6–62. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. If d = 90 mm, determine the absolute maximum bending stress in the beam, and sketch the stress distribution acting over the cross section.
12 kN/m d A
B 3m
1.5 m
Solution
Absolute Maximum Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax = =
Mmax c I 11.34(103)(0.045) p 4
(0.0454) Ans.
= 158 MPa
Ans: smax = 158 MPa 528
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6–63. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Determine its smallest diameter d if the allowable bending stress is sallow = 180 MPa.
12 kN/m d A
B 3m
1.5 m
Solution
Allowable Bending Stress: The maximum moment is Mmax = 11.34 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 180 ( 106 ) =
Mmax c I 11.34(103) 1 d2 2 p 4
1 d2 24
Ans.
d = 0.08626 m = 86.3 mm
Ans: d = 86.3 mm 529
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*6–64. The pin is used to connect the three links together. Due to wear, the load is distributed over the top and bottom of the pin as shown on the freebody diagram. If the diameter of the pin is 10 mm, determine the maximum bending stress on the crosssectional area at the center section a–a. For the solution it is first necessary to determine the load intensities w1 and w2.
3600 N
25 mm
w2
a
w2
w1
25 mm a 10 mm 40 mm
1800 N
1800 N
1 = = w2 (0.25) 1800; w2 144(10 3 ) N−m 12 = = w1(0.04) 3600; w1 90(10 3 ) N−m
0.01833 m
= = 33.0 N ⋅ m M 1800(0.01833) = I
1 π (0.0054 ) 0.49087(10 −9 ) m 4 = 4
σ max =
Mc 33.0(0.005) 2 = = 336.14(106 ) N−m = 336 MPa I 0.49087(10 −9 )
Ans.
Ans: smax = 324.45 MPa 535
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6–65. The shaft is supported by a thrust bearing at A and journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress in the shaft.
40 mm A
B 0.75 m
1.5 m 3 kN
D
C
25 mm
0.75 m 3 kN
Solution Shear and Moment Diagrams: As shown in Fig. a. Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the freebody diagram of the segment shown in Fig. b. Section Properties: The moment of inertia of the cross section about the neutral axis is p I = (0.044  0.0254) = 1.7038(10  6) m4 4 Absolute Maximum Bending Stress: smax =
2.25(103)(0.04) Mmaxc = = 52.8 MPa I 1.7038(10  6)
Ans.
Ans: smax = 52.8 MPa 536
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6–66. Determine the absolute maximum bending stress in the 40mmdiameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.
1800 N A
B
300 mm
1350 N
450 mm 375 mm
Mmax 506.25 N ⋅ m = = σ
Mc 506.25(0.02) 2 = = 80.57(106 ) N−m = 80.6 MPa π (0.02 4 ) I 4
V (N)
Ans.
1350
x
M (N.m)
x
Ans: s max = 52.8 MPa 533
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6–63. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is sallow = 154 MPa.
1800 N A 1350 N
B
300 mm 450 mm
375 mm
Mmax 506.25 N ⋅ m = 10.5(106 ) =
(506.25)c
V (N)
π c4 4
c = 0.01612 m
1350
x
= d 2= c 2(0.01612) = 0.03223 m = 32.2 mm
Ans. M (N.m)
x
Ans: d = 32.2 mm 534
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*6–68. A shaft is made of a polymer having an elliptical cross section. If it resists an internal moment of M = 50 N # m, determine the maximum bending stress in the material (a) using the flexure formula, where Iz = 14 p(0.08 m)(0.04 m)3, (b) using integration. Sketch a threedimensional view of the stress distribution acting over the crosssectional area. Here Ix = 14 p(0.08 m)(0.04 m)3.
y
y2 z2 ——— ———2 1 2 (40) (80) 80 mm M 50 Nm
z
160 mm
x
Solution (a) I =
1 1 p ab3 = p(0.08)(0.04)3 = 4.021238(10  6) m4 4 4 50(0.04) Mc = = 497 kPa I 4.021238(10  6)
smax =
Ans.
(b) M = =
smax y2dA c LA smax y22zdy c L
z = 20.0064  4y2 = 22(0.04)2  y2 0.04
2
L 0.04
0.04
2
y zdy = 4
L 0.04
= 4c
=
2
y 2(0.04)2  y2 dy
(0.04)4 8
(0.04)4 2
sin  1a
sin  1a
0.04 y 1 b  y2(0.04)2  y2(0.042  2y2) d ` 0.04 8 0.04
0.04 y b2 0.04  0.04
= 4.021238(10  6) m4 smax =
50(0.04) 4.021238(10  6)
Ans.
= 497 kPa
Ans: (a) smax = 497 kPa, (b) smax = 497 kPa 526
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6–69. Solve Prob. 6–65 if the moment M = 50 N # m is applied about the y axis instead of the x axis. Here Iy = 14 p (0.04 m)(0.08 m)3.
y
y2 z2 ——— ———2 1 2 (40) (80) 80 mm M 50 Nm
z
160 mm
x
Solution (a) I =
1 1 p ab3 = p(0.04)(0.08)3 = 16.085(10  6) m4 4 4
smax =
50(0.08) Mc = 249 kPa = I 16.085(10  6)
Ans.
(b) M =
LA
50 = 2 a
z(s dA) =
LA
za
smax b(z)(2y)dz 0.08
0.08 1>2 smax z2 b z2 a1 b (0.04)dz 0.04 L0 (0.08)2
50 = 201.06(10  6)smax
Ans.
smax = 249 kPa
Ans: (a) smax = 249 kPa, (b) smax = 249 kPa 527
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6–70. The beam is subjected to a moment of M = 40 kN # m. Determine the bending stress at points A and B. Sketch the results on a volume element acting at each of these points.
A
B
50 mm
M = 40 kNm
50 mm 50 mm
Solution I =
50 mm 50 mm 50 mm
1 1 (0.150)(0.05)3 + 2c (0.05)(0.05)3 + (0.05)(0.05)(0.05)2 d = 15.1042 ( 10  6 ) m4 12 12
sA =
40 ( 103 ) (0.075) Mc = 199 MPa = I 15.1042 ( 10  6 )
Ans.
sB =
40 ( 103 ) (0.025) My = = 66.2 MPa I 15.1042 ( 10  6 )
Ans.
Ans: sA = 199 MPa, sB = 66.2 MPa 523
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6–71. Determine the dimension a of a beam having a square cross section in terms of the radius r of a beam with a circular cross section if both beams are subjected to the same internal moment which results in the same maximum bending stress.
a
r
a
Section Properties: The moments of inertia of the square and circular cross sections about the neutral axis are IS =
1 a4 a A a3 B = 12 12
IC =
1 4 pr 4
Maximum Bending Stress: For the square cross section, c = a>2.
A smax B S =
M(a>2) 6M Mc = 3 = 4 IS a >12 a
For the circular cross section, c = r.
A smax B c =
Mc Mr 4M = Ic 1 4 pr3 pr 4
It is required that
A smax B S = A smax B C
4M 6M = 3 a pr3
Ans.
a = 1.677r
Ans: a = 1.677r 366 366
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*6–72.
A portion of the femur can be modeled as a tube diameter of of 0.375 9.5 mm having an inner diameter in. and an outer diameter 32 mm. of 1.25 in. Determine the maximum elastic static force P that can be applied to its center. Assume the bone to be roller supported at its ends. The s– P diagram for the bone mass is shown and is the same in tension as in compression.
P
ss(MPa) (ksi) 16.10 2.30 8.75 1.25 100 mm 4 in.
0.02
I=
π 4
(0.016 4 − 0.004754 ) = 51.0720(10 −9 ) m 4
= Mmax
0.05
100 mm 4 in.
P (mm/mm) (in./ in.)
0.1 m
P = (0.1) 0.05P 2
8.75 ksi MPa Require smax = 1.25 smax =
Mc I
8.75(106 ) =
(0.05P )(0.016) 51.0720(10 −9 )
= = N 559 N P 558.6
Ans.
Ans:
= P 559 N 389 389
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xle ofofthethe freight car is to wheel freight carsubjected is subjected to 6–73. The axle loadings of 20 kip. it iskN. supported by two journal bearings at wheel loadings of If100 If it is supported by two journal C and D,at determine maximum stress developed bearings C and D, the determine the bending maximum bending stress at the center thecenter axle, where diameter 5.5diameter in. developed atofthe of thethe axle, where is the is 137.5 mm.
A
C
10mm in. 250 20kN kip 100
σ= max
MC [25(10 3 )](0.06875) 2 = = 97.96(106 ) N−m = 98 MPa π (0.068754 ) I 4
B
60 in. 1500 m
D
10 250in. mm 20kN kip 100
Ans.
100 kN Mmax = 25 kN · m
0.25 m
100 kN
Ans: s max = 98.0 MPa 370 370
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6–74. The chair is supported by an arm that is hinged so it rotates about the vertical axis at A. If the load on the chair is 900 hollow tube section having the 180 N lb and and the the arm arm isis aa hollow dimensions shown, determine the maximum bending stress at section a–a.
180 900 lb N
1 in. 25 mm a A 8 in. 200 mm
c + ©M = 0;
M − 900(0.9) = 0
in. 75 3mm
a
2.5mm in. 63 0.5mm in. 13
900 N
= M 810 N ⋅ m 0.9 m
1 1 I x = (0.025)(0.0753 ) − (0.013)(0.0633 ) = 0.608022(10 −6 ) m 4 12 12
σ= max
MC 810(0.0375) 2 = 50.0 MPa = = 49.96(10 −6 ) N−m I 0.608022(10 −6 )
Ans.
Ans: s max = 11.1 MPa 389 389
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6–75. The boat has a weight of 11.5 kN and a center of gravity at G. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolute maximum bending stress developed in the main strut of the trailer. Consider the strut to be a boxbeam having the dimensions shown and pinned at C.
B 10.3 ft m
G
C
A D 3 ftm 0.9
5 ftm 1.5
4 ftm 1.2
1.75 in. 45 mm
1 ft 0.3 m
1.75 in. 45 mm
in. 75 3mm 1.5 mm in. 37.5
Boat:
11.5 kN
+ ©F = 0; : x
Bx = 0 –N 11.5(1.5)== 00 NAA(2.7) (9) ++ 2300(5)
a + ©MB = 0;
1.2 m
1277.78 NAA ==6.3889 kNlb + c ©Fy = 0;
1.5 m
6.3889 – 11.5 + B+y =B0y = 0 1277.78 2300 By == 5.1111 1022.22 kNlb
Assembly: –N NDD(3.0) (10) ++ 11.5(2.7) 2300(9) = = 00
a + ©MC = 0;
N 2070kN lb NDD ==10.35 + c ©Fy = 0;
2070–11.5 2300 Cyy ++10.35 = 0= 0 Cy == 1.15 230 kN lb
1 1 I = (0.045)(0.0753 ) − (0.0375)(0.0453 ) =1.2973(10 −6 ) m 4 12 12
σ= max
MC [5.75(10 3 )](0.0375) = = 166.21(106 ) N− = m 2 166 MPa I 1.2973(10 −6 )
Ans.
11.5 kN 2.7 m 3.0 m
5.1111 kN
6.3889 kN
1.8 m 0.9 m 10.35 kN V (kN)
1.2 m
1.15 kN
3.9611 –1.15
–6.3889 M (kN · m)
1.38
Ans: s max = 166 MPa
–5.75
372 372
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*6–76. The steel beam has the crosssectional area shown. Determine the largest intensity of the distributed load w0 that it can support so that the maximum bending MPa. stress in the beam does not exceed sallow = 160 22 ksi.
w0
39 m ft
39 m ft
200 mm 9 in. 6 mmin. 0.25 300in. mm 12 6 mmin. 0.25
6 mm 0.25 in.
Support Reactions. The FBD of the beam is shown in Fig. a. The shear and moment diagrams are shown in Fig. a and b, respectively. As 3wo.o. indicated on the moment diagram, Mmax = 27w The moment of inertia of the crosssection about the neutral axis is
I=
1 1 (0.2)(0.312 3 ) − (0.194)(0.33 ) = 69.6888(10 −6 ) m 4 12 12
Hence, c/ = 0.156 m. Thus, sallow =
Mmax c ; I
160(106 ) =
(3wo− )(0.156)
69.6888(10 −6 )
3 = ) N−m 23.8 kN−m = wo 23.83(10
Ans.
1 w0(6) 2
V
1.5w0 6 x (m) 3 3m Ay = 1.5w0
–1.5w0
3m By = 1.5w0 M 3w0
x (m) 3
6
Ans:
= wo 23.8 kN−m 391 391
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6–77. The steel beam has the crosssectional area kN/m, determine the shown. If w w00 ==30 the maximum maximum bending 2 kip>ft, stress in the beam.
w0
ft 39 m
ft 39 m
200 mm 9 in. 60.25 mmin. 12 in. 300 mm
0.25 in. 6 mm
60.25 mmin.
The FBD of the beam is shown in Fig. a The shear and moment diagrams are shown in Fig. b and c, respectively. As 54 kN kip ·# m. ft. indicated on the moment diagram, Mmax = 90 The moment of inertia of the I crosssection about the bending axis is
= I
1 1 (0.2)(0.312 3 ) − (0.194)(0.33 ) 12 12
= 69.6888(10 −6 ) m 4 Hence, c/ = 0.156 m. Thus, smax =
=
Mmax c I
[90(0.33 )](0.156) 69.6888(10 −6 )
6 = 201.47(10 = ) N−m 2 201 MPa
Ans.
1 (30)(6) kN 2
V (kN)
45 6 x (m) 3 3m Ay = 45 kN
–45
3m By = 45 kN M (kN · m) 90
x (m) 3
6
Ans: I = 69.6888 (10 6) m4, s max = 201 MPa 392 392
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6–78. If the beam is subjected to a moment of M = 100 kN # m, determine the bending stress at points A, B, and C. Sketch the bending stress distribution on the cross section.
A M
300 mm
30 mm 30 mm B 150 mm
C
Solution Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ~ 0.015(0.03)(0.3) + 0.18(0.3)(0.03) Σ yA y = = = 0.0975 m ΣA 0.03(0.3) + 0.3(0.03)
150 mm
Thus, the moment of inertia of the cross section about the neutral axis is I =
1 1 (0.3)(0.033) + 0.3(0.03)(0.0975  0.015)2 + (0.03)(0.33) 12 12 + 0.03(0.3)(0.18  0.0975)2
= 0.1907(10  3) m4 Bending Stress: The distance from the neutral axis to points A, B, and C is yA = 0.33  0.0975 = 0.2325 m, yB = 0.0975 m, and yC = 0.0975  0.03 = 0.0675 m. sA =
100(103)(0.2325) MyA = = 122 MPa (C) I 0.1907(10  3)
Ans.
sB =
100(103)(0.0975) MyB = = 51.1 MPa (T) I 0.1907(10  3)
Ans.
sC =
100(103)(0.0675) MyC = = 35.4 MPa (T) I 0.1907(10  3)
Ans.
Using these results, the bending stress distribution across the cross section is shown in Fig. b.
Ans: sA = 122 MPa (C), sB = 51.1 MPa (T), sC = 35.4 MPa (T) 542
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6–79. If the beam is made of material having an allowable tensile and compressive stress of (sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively, determine the maximum moment M that can be applied to the beam.
A M
300 mm
30 mm 30 mm B 150 mm
C
Solution Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is ~ 0.015(0.03)(0.3) + 0.18(0.3)(0.03) Σ yA y = = = 0.0975 m ΣA 0.03(0.3) + 0.3(0.03)
150 mm
Thus, the moment of inertia of the cross section about the neutral axis is I =
1 1 (0.3)(0.033) + 0.3(0.03)(0.0975  0.015)2 + (0.03)(0.33) 12 12 + 0.03(0.3)(0.18  0.0975)2
= 0.1907(10  3) m4 Allowable Bending Stress: The maximum compressive and tensile stress occurs at the top and bottommost fibers of the cross section. For the topmost fiber, (sallow)c =
Mc I
150(106) =
M(0.33  0.0975) 0.1907(10  3)
M = 123024.19 N # m = 123 kN # m (controls)
Ans.
For the bottommost fiber, (sallow)t =
My I
125(106) =
M(0.0975) 0.1907(10  3)
M = 244 471.15 N # m = 244 kN # m
Ans: M = 123 kN # m 543
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*6–80. The two solid steel rods are bolted together along their length and support the loading shown. Assume the support at A is a pin and B is a roller. Determine the required diameter d of each of the rods if the allowable bending stress is sallow = 130 MPa.
20 kN/m
80 kN
A B
2m 2m
Section Property: I = 2B
p d 4 p d 2 5p 4 a b + d2 a b R = d 4 2 4 2 32
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 130 A 106 B =
Mmax c I 100(103)(d) 5p 32
d4 Ans.
d = 0.1162 m = 116 mm
Ans: d = 116 mm 388 388
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6–81. Solve Prob. 6–94 if the rods are rotated 90° so that both rods rest on the supports at A (pin) and B (roller).
20 kN/m
80 kN
A B
2m 2m
Section Property: I = 2B
p d 4 p 4 a b R = d 4 2 32
Allowable Bending Stress: The maximum moment is Mmax = 100 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 130 A 106 B =
Mmax c I
100(103)(d) p 32
d4 Ans.
d = 0.1986 m = 199 mm
Ans: d = 199 mm 388 388
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6–82. If the compound beam in Prob. 6–42 has a square cross section of side length a, determine the minimum value of a if the allowable bending stress is sallow = 150 MPa.
Solution
Allowable Bending Stress: The maximum moment is Mmax = 7.50 kN # m as indicated on moment diagram. Applying the flexure formula smax = sallow = 150 1 106 2 =
Mmax c I 7.50(103) 1 a2 2 1 12
a4
Ans.
a = 0.06694 m = 66.9 mm
Ans: a = 66.9 mm 551
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6–83. If the beam in Prob. 6–28 has a rectangular cross section with a width b and a height h, determine the absolute maximum bending stress in the beam.
Solution Absolute Maximum Bending Stress: The maximum moment is Mmax = indicated on the moment diagram. Applying the flexure formula smax
Mmax c = = I
23w0 L2 h 2 216 1 3 bh 12
1 2
=
23w0 L2
23w0 L2 as 216
Ans.
36bh2
Ans: smax = 552
23w0 L2 36 bh2
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*6–84. Determine the absolute maximum bending stress in the 80mmdiameter shaft which is subjected to the concentrated forces. There is a journal bearing at A and a thrust bearing at B.
A
0.5 m 12 kN
B
0.4 m
0.6 m 20 kN
Solution The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 6 kN # m .
The moment of inertia of the cross section about the neutral axis is I =
p (0.044) = 0.64(10  6)p m4 4
Here, c = 0.04 m. Thus smax =
6(103)(0.04) Mmax c = I 0.64(10  6)p = 119.37(106) Pa Ans.
= 119 MPa
Ans: smax = 119 MPa 553
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6–85. Determine, to the nearest millimeter, the smallest allowable diameter of the shaft which is subjected to the concentrated forces. There is a journal bearing at A and a thrust bearing at B. The allowable bending stress is sallow = 150 MPa.
A
0.5 m 12 kN
B
0.4 m
0.6 m 20 kN
Solution The FBD of the shaft is shown in Fig. a. The shear and moment diagrams are shown in Fig. b and c, respectively. As indicated on the moment diagram, Mmax = 6 kN # m .
The moment of inertia of the cross section about the neutral axis is I =
p d 4 pd 4 a b = 4 2 64
Here, c = d>2. Thus sallow =
Mmax c ; I
150(106) =
6(103)(d > 2) pd 4 >64
d = 0.07413 m = 74.13 mm = 75 mm
Ans.
Ans: d = 75 mm 554
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6–86. If the beam is subjected to an internal moment of M = 3 kN # m, determine the maximum tensile and compressive stress in the beam. Also, sketch the bending stress distribution on the cross section.
A
100 mm 25 mm 25 mm
Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a. The location of C is
M
©y~A 2[0.0375(0.025)(0.075)] + 0.0875(0.2)(0.025) + 0.15(0.025)(0.1) = y = ©A 2(0.025)(0.075) + 0.2(0.025) + 0.025(0.1) = 0.08472 m
75 mm
25 mm 75 mm 75 mm 25 mm
Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
1 1 (0.2)(0.0253 ) + 0.2(0.025)(0.0875 − 0.08472)2 = 2 (0.025)(0.0753 ) + 0.025(0.075)(0.08472 − 0.0375)2 + 12 12 +
1 (0.025)(0.13 ) + 0.025(0.1)(0.15 − 0.08472)2 12
= 23.1554(10 −6 ) m 4 Maximum Bending Stress: The maximum compressive and tensile stress occurs at the top and bottommost fibers of the cross section.
(σ max= )c
Mc [3(10 3 )](0.2 − 0.08472) 2 = = 14.94(106 ) N−m = 14.9 MPa I 23.1554(10 −6 )
Ans.
(σ max= )t
My [3(10 3 )](0.08472) = = 10.98(106 ) N−m 2 = 11.0 MPa I 23.1554(10 −6 )
Ans.
The bending stresses at y = 0.01528 m and y = −0.009722 m are
sy = 0.1528 m =
My =
I
sy =  0.009722 m =
[3(10 3 )](0.01528) 6 = 1.979(10 = ) N−m 2 1.98 MPa (C) 23.1554(10 −6 )
My I
=
[3(10 3 )](0.009722) 6 = 1.260(10 = ) N−m 2 1.26 MPa (T) 23.1554(10 −6 )
The bending stress distribution across the cross section is shown in Fig. b.
14.9 MPa 1.98 MPa
1.26 MPa
11.0 MPa
Ans: (smax)c = 14.9 MPa, (smax)t = 11.0 MPa 538
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6–87. If the allowable tensile and compressive stress for the beam are (sallow)t = 14 MPa and (sallow)c = 21 MPa, respectively, determine the maximum allowable internal moment M that can be applied on the cross section.
A
100 mm 25 mm 25 mm M
Section Properties: The neutral axis passes through the centroid C of the cross section as shown in Fig. a. The location of C is given by y=
©y~A = ©A
2[0.0375(0.025)(0.075)] + 0.0875(0.2)(0.025) + 0.15(0.025)(0.1) 2(0.025)(0.075) + 0.2(0.025) + 0.025(0.1)
= 0.08472 m Thus, the moment of inertia of the cross section about the neutral axis is
75 mm
25 mm 75 mm 75 mm 25 mm
I = ©I + Ad2
1 1 (0.2)(0.0253 ) + 0.2(0.025)(0.0875 − 0.08472)2 = 2 (0.025)(0.0753 ) + 0.025(0.075)(0.08472 − 0.0375)2 + 12 12 +
1 (0.025)(0.13 ) + 0.025(0.1)(0.15 − 0.08472)2 12
= 23.1554(10 −6 ) m 4 Allowable Bending Stress: The maximum compressive and tensile stress occurs at the top and bottommost fibers of the cross section. (sallow)c =
Mc ; I
21(106 ) =
M (0.2 − 0.08472) 23.1554(10 −6 )
M 4.218(10 3 ) N= = ⋅ m 4.22 kN ⋅ m For the bottommost fiber, (sallow)t =
My ; I
21(106 ) =
M (0.08472) 23.1554(10 −6 )
M 3.826(10 3 ) N= ⋅ m 3.83 kN ⋅ m (Controls) =
Ans.
Ans:
M = 3.83 kN # m
539
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*6–88. If the beam is subjected to an internal moment of M = 3 kN . m, determine the resultant force of the bending stress distribution acting on the top vertical board A.
A
100 mm 25 mm 25 mm M
Section Properties: The neutral axis passes through the centroid C of the cross section as shown in Fig. a. The location of C is given by
75 mm
25 mm 75 mm 75 mm 25 mm
~ © yA 2[0.0375(0.025)(0.075)] + 0.0875(0.2)(0.025) + 0.15(0.025)(0.1) y = = 2(0.025)(0.075) + 0.2(0.025) + 0.025(0.1) ©A = 0.08472 m Thus, the moment of inertia of the cross section about the neutral axis is I = ©I + Ad2
1 1 (0.2)(0.0253 ) + 0.2(0.025)(0.0875 − 0.08472)2 = 2 (0.025)(0.0753 ) + 0.025(0.075)(0.08472 − 0.0375)2 + 12 12 +
1 (0.025)(0.13 ) + 0.025(0.1)(0.15 − 0.08472)2 12
= 23.1554(10 −6 ) m 4 Bending Stress: The distance from the neutral axis to the top and bottom of board A is yt = 0.2 − 0.08472 = 0.11528 m and yb = 0.1 − 0.08472 = 0.01528 m. We have
σt =
Myt [3(10 3 )](0.11528) 2 = = 14.94(106 ) N−m = 14.94 MPa I 23.1554(10 −6 )
σb =
Myb [3(10 3 )](0.01528) 2 = = 1.979(106 ) N−m = 1.979 MPa I 23.1554(10 −6 )
Resultant Force: The resultant force acting on board A is equal to the volume of the trapezoidal stress block shown in Fig. b. Thus,
1 21.14(10 3 ) N = 21.1 kN FR = [(14.94 + 1.979)(106 )](0.025)(0.1) = 2
Ans. 0.025 m
14.94 MPa
0.1 m
1.979 MPa
Ans: FR = 21.1 kN 540
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6–89. A timber beam has a cross section which is originally square. If it is oriented as shown, determine the dimension h′ so that it can resist the maximum moment possible. By what factor is this moment greater than that of the beam without its top or bottom flattened?
h
Solution x 2h = ; h  h′ h y = h′ + I = 2e = = = =
x = 2(h  h′)
h  h′ 2h′ + h = 3 3
1 1 1 2h′ + h 2 (2h) ( h3 )  c (2)(h  h′)(h  h′)3 + (2)(h  h′)(h  h′)a b df 12 36 2 3
1 4 1 2 h  (h  h′)4  (h  h′)2(2h′ + h)2 3 9 9 1 4 1 h  (h  h′)2 3 3h2 + 9h′2 + 6hh′ 4 3 9 1 4 1 h  ( 3h4 + 9h′4  12hh′3 ) 3 9 4 hh′3  h′4 3
smax = M = =
Mc I
I s c max 4 3 3 hh′
(1)  h′4
h′
4 smax = a hh′2  h′3 bsmax 3
dM 8 = a hh′  3h′2 bsmax dh′ 3
Inorder to have maximum moment, dM 8 = 0 = hh′  3h′2 dh′ 3 h′ =
8 h 9
Ans.
For the square beam, I = I + Ad 2 I = 2c
1 h4 1 h 2 (2h)(h)3 + (2h)(h)a b d = 36 12 3 3
537
h¿
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6–89. Continued
From Eq. (1) M =
h4 3
h
smax =
h3 s = 0.3333h3 smax 3 max
For the flattened beam: I =
4 8 3 8 4 ha hb  a hb = 0.312147 h4 3 9 9
From Eq. (1) M′ =
0.312147 h4
Factor =
8 9h
smax = 0.35117 h3 smax
0.35117 h3 smax M′ = = 1.05 M′ 0.3333 h3 smax
Ans.
Ans:
8 h, 9 Factor = 1.05 h′ =
538
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6–90. The wood beam has a rectangular cross section in the proportion shown. Determine its required dimension b if the allowable bending stress is sallow = 10 MPa.
500 N/m
A
B 2m
1.5b b
2m
Allowable Bending Stress: The maximum moment is Mmax = 562.5 N # m as indicated on the moment diagram. Applying the flexure formula smax = sallow = 10 A 106 B =
Mmax c I 562.5(0.75b) 1 12
(b)(1.5b)3 Ans.
b = 0.05313 m = 53.1 mm
Ans: b = 53.1 mm 380 380
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6–91. Determine the absolute maximum bending stress in the tubular shaft if di = 160 mm and do = 200 mm.
15 kN/m 60 kN � m d i do A
B 3m
1m
Section Property: I =
p A 0.14  0.084 B = 46.370 A 10  6 B m4 4
Absolute Maximum Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax = =
Mmaxc I
60.0(103)(0.1) 46.370(10  6) Ans.
= 129 MPa
Ans: s max = 129 MPa 378 378
CH 06.indd 378
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*6–92. The tubular shaft is to have a cross section such that its inner diameter and outer diameter are related by di = 0.8do. Determine these required dimensions if the allowable bending stress is sallow = 155 MPa.
15 kN/m 60 kN � m d i do A
B 3m
1m
Section Property: I =
0.8do 4 do 4 dl 4 p do 4 p  a b R = 0.009225pd4o Ba b  a b R = B 4 2 2 4 16 2
Allowable Bending Stress: The maximum moment is Mmax = 60.0 kN # m as indicated on the moment diagram. Applying the flexure formula smax = sallow =
Mmax c I d
155 A 10
6
Thus,
B =
60.0(103) A 2o B
0.009225pd4o
do = 0.1883 m = 188 mm
Ans.
dl = 0.8do = 151 mm
Ans.
Ans: do = 0188 mm, dl = 151 mm 379 379
CH 06.indd 379
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6–93. If the intensity of the load w = 15 kN>m, determine the absolute maximum tensile and compressive stress in the beam.
w A
B 6m 300 mm
150 mm
Solution Support Reactions: Shown on the freebody diagram of the beam, Fig. a. Maximum Moment: The maximum moment occurs when V = 0. Referring to the freebody diagram of the beam segment shown in Fig. b, + c ΣFy = 0;
45  15x = 0
x = 3m
3 Mmax + 15(3)a b  45(3) = 0 Mmax = 67.5 kN # m 2 Section Properties: The moment of inertia of the cross section about the neutral axis is a+ ΣM = 0;
I =
1 (0.15)(0.33) = 0.1125(10  3) m4 36
Absolute Maximum Bending Stress: The maximum compressive and tensile stresses occur at the top and bottommost fibers of the cross section. (smax)c =
67.5(103)(0.2) Mc = = 120 MPa (C) I 0.1125(10  3)
Ans.
(smax)t =
67.5(103)(0.1) My = = 60 MPa (T) I 0.1125(10  3)
Ans.
Ans: (smax)c = 120 MPa (C), (smax)t = 60 MPa (T) 546
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6–94. If the allowable bending stress is sallow = 150 MPa, determine the maximum intensity w of the uniform distributed load.
w A
B 6m 300 mm
150 mm
Solution Support Reactions: As shown on the freebody diagram of the beam, Fig. a. Maximum Moment: The maximum moment occurs when V = 0. Referring to the freebody diagram of the beam segment shown in Fig. b, + c ΣFy = 0;
3w  wx = 0
x = 3m
3 9 Mmax + w(3)a b  3w(3) = 0 Mmax = w 2 2 Section Properties: The moment of inertia of the cross section about the neutral axis is a+ ΣM = 0;
I =
1 (0.15)(0.33) = 0.1125(10  3) m4 36
Absolute Maximum Bending Stress: Here, c =
sallow
Mc = ; I
150(106) =
2 (0.3) = 0.2 m. 3
9 w(0.2) 2 0.1125 ( 10  3 ) Ans.
w = 18750 N>m = 18.75 kN>m
Ans: w = 18.75 kN>m 547
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6–95. The beam has a rectangular cross section as shown. Determine the largest intensity w of the uniform distributed load so that the bending stress in the beam does not exceed smax = 10 MPa.
w
50 mm 150 mm
2m
2m
2m
Solution Absolute Maximum Bending Stress: The support reactions, shear diagram, and moment diagrams are shown in Figs. a, b, and c, respectively. From the moment diagram, the maximum moment is Mmax = 2.00 w, which occurs at the pin support and roller support. Applying the flexure formula, sabs = max
(2.00 w)(0.075) MmaxC ; 10 ( 106 ) = I 1 (0.05) ( 0.153 ) 12 w = 937.5 N>m
Ans.
Ans: w = 937.5 N>m 564
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*6–96. The beam has the rectangular cross section shown. If w = 1 kN>m, determine the maximum bending stress in the beam. Sketch the stress distribution acting over the cross section.
w
50 mm 150 mm
2m
2m
2m
Solution Absolute Maximum Bending Stress: The support reactions, shear diagram, and moment diagram are shown in Figs. a, b, and c, respectively. From the moment diagram, the maximum moment is Mmax = 2000 N # m, which occurs at the supports. Applying the flexure formula, sabs = max
MmaxC = I
2000(0.075) 1 12
(0.05) ( 0.153 )
= 10.67 ( 106 ) MPa = 10.7 MPa
Ans.
Using this result, the bending stress distribution on the beam’s crosssection shown in Fig. d can be sketched.
Ans: sabs = 10.7 MPa max 565
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6–97. The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. top member memberisisa pipe a pipe having an outer diameter of The top having an outer diameter of 1 in. 3 25 mm and thickness 5 mm, and themember bottom is member is a and thickness of 16 the bottom a solid rod in.,ofand 1 solid rod having a of diameter of 12 mm. having a diameter 2 in.
1.8 6 ftm
∑ yA 0 + 0.16[π (0.01252 − 0.00752 )] = = 0.11765 m ∑ A π (0.006 2 ) + π (0.01252 − 0.00752 )
= y I=
1.5 100kN/m lb/ft
π 4
(0.006 4 ) + π (0.006 2 )(0.117652 ) +
π 4
1.8 6 ftm
141.5 5.75mm in.
1.8 6 ftm
0.16 m
y
(0.01254 − 0.00754 )
2.1466(10 −6 ) m 4 + π (0.01252 − 0.00752 )(0.16 − 0.11765)2 = Mmax =1.35(2.7) − 1.35(0.45) =3.0375 kN ⋅ m
σ= max
MC [3.0375(10 3 )](0.11765 + 0.006) = I 2.1466(10 −6 )
6 = 174.96(10 = ) N−m 2 175 MPa
1.35 kN
Ans.
0.45 m
2.7 m 1.35 kN
Ans: (s max )c = 120 MPa (C), (s max )t = 60 MPa (T) 370 370
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6–98. If d = 450 mm, determine the absolute maximum bending stress in the overhanging beam.
12 kN 8 kN/m
125 mm 25 mm 25 mm 75 mm d
A B 4m
75 mm 2m
Solution Support Reactions: Shown on the freebody diagram of the beam, Fig. a. Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, Mmax = 24 kN # m. Section Properties: The moment of inertia of the cross section about the neutral axis is I =
1 1 (0.175)(0.453) (0.125)(0.33) 12 12
= 1.0477(10  3) m4 Absolute Maximum Bending Stress: Here, c = smax =
0.45 = 0.225 m. 2
24(103)(0.225) Mmax c = = 5.15 MPa I 1.0477(10  3)
Ans.
Ans: smax = 5.15 MPa 562
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6–99. If the allowable bending stress is sallow = 6 MPa, determine the minimum dimension d of the beam’s cross‑sectional area to the nearest mm.
12 kN 8 kN/m
125 mm 25 mm 25 mm 75 mm d
A B 4m
75 mm 2m
Solution Support Reactions: Shown on the freebody diagram of the beam, Fig. a. Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram, Mmax = 24 kN # m. Section Properties: The moment of inertia of the cross section about the neutral axis is I =
1 1 (0.175)d 3 (0.125)(d  0.15)3 12 12
= 4.1667(10  3)d 3 + 4.6875(10  3)d 2  0.703125(10  3)d + 35.15625(10  6) Absolute Maximum Bending Stress: Here, c = sallow =
d . 2
Mc ; I 24(103)
6
6(10 ) =
d 2
4.1667(10  3)d 3 + 4.6875(10  3)d 2  0.703125(10  3)d + 35.15625(10  6)
4.1667(10  3)d 3 + 4.6875(10  3)d 2  2.703125(10  3)d + 35.15625(10  6) = 0
Solving, Ans.
d = 0.4094 m = 410 mm
Ans: d = 410 mm 563
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*6–100. If the reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown, determine the maximum bending stress developed in the tie. The tie has the rectangular cross section with thickness tt ==150 6 in. mm.
15 75 kip kN 0.45 1.5 ftm
15 75 kip kN 0.45 m 1.5 ft
5 ftm 1.5
300 12 mm in. t
w
Support Reactions: Referring to the free  body diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
w(2.4)–2(15) 2(75) = = 00 w(8) w == 62.5 3.75kN/m kip>ft
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As
# ft·. m 7.5 kipkN indicated on the moment diagram, the maximum moment is � Mmax � = 11.25 Absolute Maximum Bending Stress: = σ max
Mmax c [11.25(10 3 )](0.075) 6 = 10.0(10 ) N−m 2 10.0 MPa = 1= I (0.3)(0.153 ) 12
75 kN
75 kN 0.45 m
0.75 m
0.75 m
Ans.
V (kN)
0.45 m 46.875
28.125
1.95 0.45
1.2
x (m) 2.4
–28.125 –46.875 w(2.4)
M (kN · m)
6.328
6.328 1.2
x (m) 1.95
0.45
–11.25
Ans:
σ max = 10.0 MPa 376 376
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6–101. The reaction of the ballast on the railway tie can be assumed uniformly distributed over its length as shown. If the wood has an allowable bending stress of sallow = 1.5 determine the the required minimum thickness t of tthe 10.5ksi, MPa, determine required minimum thickness of 1 rectangular crosscross sectional areaarea of the the in. the rectangular sectional of tie thetotie tonearest the nearest 8 multiples of 5 mm.
15 kN kip 75 0.45 1.5 ftm
15 kN kip 75 0.45 m 1.5 ft
5 ftm 1.5
300 12 mm in. t
w
Support Reactions: Referring to the freebody diagram of the tie shown in Fig. a, we have + c ©Fy = 0;
75 kN 0.45 m
w(8)  2(15) = 0= 0 w = (2.4) – 2(75)
75 kN 0.75 m
0.75 m
0.45 m
w == 62.5 3.75kN/m kip>ft Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As
# ft·. m. 7.5 kipkN indicated on the moment diagram, the maximum moment is � Mmax � = 11.25 w(2.4)
Absolute Maximum Bending Stress:
smax =
Mc ; I
10.5(106 ) =
[11.25(10 3 )]
( 2t )
1 (0.3)(t 3 ) 12
m 146.38 mm = = t 0.14638
Ans.
Use t = 150 mm t =
V (kN) M (kN · m)
0.45
6.328
6.328
46.875
28.125
1.2
1.2
1.95
x (m) 2.4
x (m) 0.45
1.95
2.4
–28.125 –46.875
–11.25
Ans: Use t = 150 mm 377 377
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6–102. A log that is 0.6 m in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is sallow = 56 MPa, determine the required width b and height h of the beam that will support the largest load possible. What is this load?
h b 0.6 m P
2 b2 + h 2 0.6=
h2 0.36 − b2 =
2.4 m
P Mmax = (2.4) 1.2 P = 2 sallow =
2.4 m
Mmax(h2 ) Mc = 1 3 I 12 (b)(h) 0.6 m
sallow =
6 Mmax
56(106 ) =
bh2
6(1.2P) b(0.36 − b2 )
P 7.7778(106 )(0.36b − b3 ) = Set
dP = 0 gives db
dP )(0.36 − b2 ) 0 = 7.7778(106= db b = 0.3464 m Thus, from the above equations,
Ans.
b = 346.41 mm = 346 mm
Ans.
h = 0.4899 m = 490 mm
Ans.
3 P 646.63(10 = ) N 647 kN =
Ans: b = 346 mm, h = 490 mm, P = 647 kN 557
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6–103. A log that is 0.6 m in diameter is to be cut into a rectangular section for use as a simply supported beam. If the wood is sallow = 56 MPa, the allowable bending stress for determine the largest load P that can be supported if the width of the beam is b = 200 mm.
h b 0.6 m P
2 0.6= h2 + 0.2 2
2.4 m
2.4 m
P = (2.4) 1.2 P 2
sallow =
Mmax c I
56(106 ) =
0. 6
Mmax =
m
h = 0.5657 m
(1.2 P )(0.5657 2)
0.2 m
1 (0.2)(0.5657 3 ) 12
3 P 497.78(10 = ) 498 kN =
Ans.
2.4 m
Ans: P = 498 kN 558
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*6–104. The member has a square cross section and is subjected to the moment M = 850 N # m. Determine the stress at each corner and sketch the stress distribution. Set u = 45°.
B 250 mm
z
125 mm
125 mm E
A
C
M 850 Nm
u D y
Solution
My = 850 cos 45° = 601.04 N # m Mz = 850 sin 45° = 601.04 N # m
1 (0.25)(0.25)3 = 0.3255208(10  3) m4 12
Iz = Iy = s = 
Mzy
sA = sB = sD = sE = 
Iz
+
Myz Iy
601.04(  0.125) 3
0.3255208(10 ) 601.04(  0.125) 0.3255208(10  3) 601.04(0.125) 3
0.3255208(10 ) 601.04(0.125) 3
0.3255208(10 )
+ + + +
601.04(  0.125) 0.3255208(10  3) 601.04(0.125) 0.3255208(10  3) 601.04(  0.125) 0.3255208(10  3) 601.04(0.125) 0.3255208(10  3)
= 0
Ans.
= 462 kPa
Ans.
=  462 kPa
Ans.
= 0
Ans.
The negative sign indicates compressive stress.
Ans: sA = sB = sD = sE = 566
0, 462 kPa,  462 kPa, 0
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6–105. The member has a square cross section and is subjected to the moment M = 850 N # m as shown. Determine the stress at each corner and sketch the stress distribution. Set u = 30°.
B 250 mm
z
125 mm
125 mm E
A
C
M 850 Nm
u D y
Solution
My = 850 cos 30° = 736.12 N # m Mz = 850 sin 30° = 425 N # m Iz = Iy = s = 
1 (0.25)(0.25)3 = 0.3255208(10  3) m4 12
Mzy
sA = sB = sD = sE = 
Iz
+
Myz Iy
425( 0.125) 3
0.3255208(10 ) 425( 0.125) 0.3255208(10  3) 425(0.125)
+ +
3
+
3
+
0.3255208(10 ) 425(0.125) 0.3255208(10 )
736.12( 0.125) 0.3255208(10  3) 736.12(0.125) 0.3255208(10  3) 736.12( 0.125) 0.3255208(10  3) 736.12(0.125) 0.3255208(10  3)
=  119 kPa
Ans.
= 446 kPa
Ans.
=  446 kPa
Ans.
= 119 kPa
Ans.
The negative signs indicate compressive stress.
Ans: sA =  119 kPa, sB = 446 kPa, sD =  446 kPa, sE = 119 kPa 567
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6–106. Consider the general case of a prismatic beam subjected to bendingmoment components My and Mz when the x, y, z axes pass through the centroid of the cross section. If the material is linear elastic, the normal stress in the beam is a linear function of position such that s = a + by + cz. Using the equilibrium conditions 0 = 1A s dA, My = 1Azs dA, Mz = 1A  ys dA, determine the constants a, b, and c, and show that the normal stress can be determined from the equation s = [  (Mz Iy + My Iyz)y + (MyIz + MzIyz)z]>(Iy Iz  I yz2), where the moments and products of inertia are defined in Appendix A.
y z My dA sC y Mz
x
z
Solution Equilibrium Condition: sx = a + by + cz 0 =
LA
sx dA
0 =
LA
(a + by + cz) dA
0 = a
LA
dA + b
My =
LA
z sx dA
=
LA
z(a + by + cz) dA
= a
LA
LA
z dA + b
y dA + c
LA
LA
LA
y sx dA
=
LA
 y(a + by + cz) dA LA
ydA  b
(1)
LA
(2)
yz dA + c
Mz =
= a
z dA
LA
y2 dA  c
z2 dA
LA
(3)
yz dA
Section Properties: The integrals are defined in Appendix A. Note that LA
y dA =
LA
z dA = 0. Thus,
From Eq. (1)
Aa = 0
From Eq. (2)
My = bIyz + cIy
From Eq. (3)
Mz =  bIz  cIyz
Solving for a, b, c: a = 0 (Since A ≠ 0) b = ¢
MzIy + My Iyz Iy Iz 
Thus, sx =  ¢
I 2yz
Ans. ≤
Mz Iy + My Iyz Iy Iz 
I 2yz
c =
My Iz + Mz Iyz
≤y + ¢
Ans.
Iy Iz  I 2yz My Iy + MzIyz Iy Iz  I 2yz
(Q.E.D.)
≤z
Ans: a = 0, b =  ¢
568
MzIy + MyIyz IyIz  I 2yz
≤, c =
MyIz + MzIyz IyIz  I 2yz
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6–107. If the resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown, determine the bending stress at points A and B. The location y of the centroid C of the strut’s crosssectional area must be determined. Also, specify the orientation of the neutral axis.
y M � 520 N�m 20 mm z
–y
12 5
13
B
200 mm
C 20 mm 200 mm
A
20 mm
200 mm
Internal Moment Components: Mz = 
12 (520) =  480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
Ans.
= 0.057368 m = 57.4 mm Iz =
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368  0.01)2 12
+
1 (0.04) A 0.183 B + 0.04(0.18)(0.110  0.057368)2 12
= 57.6014 A 10  6 B m4
Iy =
1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10  3 B m4 12 12
Maximum Bending Stress: Applying the flexure formula for biaxial at points A and B s = 
Mzy
sA = 
Iz
+
Myz Iy
480( 0.142632) 6
57.6014(10 )
+
200(  0.2) 0.366827(10  3) Ans.
= 1.298 MPa = 1.30 MPa (C) sB = 
480(0.057368)
+
6
57.6014(10 )
200(0.2) 0.366827(10  3) Ans.
= 0.587 MPa (T) Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10  6) 0.366827(10  3)
tan ( 22.62°) Ans.
a = 3.74°
Ans: y = 57.4 mm, sA = 1.30 MPa (C), sB = 0.587 MPa (T), a = 3.74°
402 402
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*6–108. The resultant internal moment acting on the cross section of the aluminum strut has a magnitude of M = 520 N # m and is directed as shown. Determine maximum bending stress in the strut. The location y of the centroid C of the strut’s crosssectional area must be determined. Also, specify the orientation of the neutral axis.
y M � 520 N�m 20 mm z
–y
12 5
13
B
200 mm
C 20 mm 200 mm
A
20 mm
200 mm
Internal Moment Components: Mz = 
12 (520) =  480 N # m 13
My =
5 (520) = 200 N # m 13
Section Properties: 0.01(0.4)(0.02) + 2[(0.110)(0.18)(0.02)] ©yA = ©A 0.4(0.02) + 2(0.18)(0.02)
y =
Ans.
= 0.057368 m = 57.4 mm Iz =
1 (0.4) A 0.023 B + (0.4)(0.02)(0.057368  0.01)2 12
1 (0.04) A 0.183 B + 0.04(0.18)(0.110  0.057368)2 12
+
= 57.6014 A 10  6 B m4
Iy =
1 1 (0.2) A 0.43 B (0.18) A 0.363 B = 0.366827 A 10  3 B m4 12 12
Maximum Bending Stress: By inspection, the maximum bending stress can occur at either point A or B. Applying the flexure formula for biaxial bending at points A and B s = 
Mz y
sA = 
Iz
+
My z Iy
480(  0.142632) 6
57.6014(10 )
+
200(  0.2) 0.366827(10  3) Ans.
=  1.298 MPa = 1.30 MPa (C) (Max) sB = 
 480(0.057368) 6
57.6014(10 )
+
200(0.2) 0.366827(10  3)
= 0.587 MPa (T) Orientation of Neutral Axis: tan a =
tan a =
Iz Iy
tan u
57.6014(10  6) 0.366827(10  3)
tan ( 22.62°) Ans.
a = 3.74°
403 403
CH 06.indd 403
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6–109. The steel shaft is subjected to the two loads. If the journal bearings at A and B do not exert an axial force on the shaft, determine the required diameter of the shaft if the allowable bending stress is sallow = 180 MPa.
4 kN
30 30
4 kN
B 1.25 m
1m A
1.25 m
Solution Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. The moment diagram for each component is drawn. Allowable Bending Stress: Since all the axes through the circle’s center for a circular shaft are principal axes, then the resultant moment M = 2M2y + M2z can be used for the design. The maximum resultant moment is Mmax = 24.3302 + 0.71432 = 4.389 kN # m. Applying the flexure formula, smax = sallow = 180 ( 106 ) =
Mmaxc I
4.389 ( 103 ) 1 d2 2 p 4
1 d22 4
Ans.
d = 0.06286 m = 62.9 mm
Ans: d = 62.9 mm 571
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6–110. The 65mmdiameter steel shaft is subjected to the two loads. If the journal bearings at A and B do not exert an axial force on the shaft, determine the absolute maximum bending stress developed in the shaft.
4 kN
30 30
4 kN
B 1.25 m
1m A
1.25 m
Solution Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. The moment diagram for each component is drawn. Maximum Bending Stress: Since all the axes through the circle’s center for a circular shaft are principal axes, then the resultant moment M = 2M2y + M2z can be used to determine the maximum bending stress. The maximum resultant moment is Mmax = 24.3302 + 0.71432 = 4.389 kN # m. Applying the flexure formula, smax = =
Mmaxc I
4.389 ( 103 ) (0.0325) p 4
( 0.03254 ) Ans.
= 163 MPa
Ans: smax = 163 MPa 572
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6–111. For the section, Iz′ = 31.7(10−6) m4, Iy′ = 114(10−6) m4, Iy′z′ = −15.8(10−6) m4. Using the techniques outlined in Appendix A, the member’s crosssectional area has principal moments of inertia of Iz = 28.8(10−6) m4 and Iy = 117(10−6) m4, calculated about the principal axes of inertia y and z, respectively. If the section is subjected to the moment M = 15 kN # m, determine the stress at point A using Eq. 6–17.
y
y¿
10.5 60 mm M 15 kNm z¿
C
z 140 mm
Solution
A
60 mm
60 mm
80 mm 60 mm
Internal Moment Components: The y and z components are My = 15 sin 10.5° = 2.7335 kN # m Mz = 15 cos 10.5° = 14.7488 kN # m Section Properties: Given that Iy = 117 ( 106 ) m4 and Iz = 28.8 ( 106 ) m4, the coordinates of point A with respect to the y and z axes are y = 0.06 cos 10.5°  0.14 sin 10.5° = 0.03348 m z =  (0.06 sin 10.5° + 0.14 cos 10.5°) = 0.14859 m Bending Stress: Applying the flexure formula for biaxial bending, Myz Mzy s = + Iz Iy sA = 
14.7488 ( 103 ) (0.03348) 28.8 ( 106 )
+
2.7335 ( 103 ) (  0.14869) 117 ( 106 )
=  20.62 ( 106 ) Pa Ans.
= 20.6 MPa (C)
Ans: sA = 20.6 MPa (C) 573
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*6–112. Solve Prob. 6–111 using the equation developed in Prob. 6–106.
y
y¿
10.5 60 mm M 15 kNm z¿
C
z 140 mm
Solution
A
60 mm
60 mm
80 mm 60 mm
Internal Moment Components: The y′and z′ components are Mz′ = 15 kN # m My′ = 0 Section Properties: Given that Iy′ = 114 ( 106 ) m4, Iz′ = 31.7 ( 106 ) m4 and Iy′z′ =  15.8 ( 106 ) m4, the coordinates of point A with respect to the y′ and z axes are y′ = 0.06 m z′ =  0.14 m Bending Stress: Using the formula, s =
 (Mz′Iy′ + My′Iy′z′)y′ + (My′Iz′ + Mz′Iy′z′)z′ Iy′Iz′  I 2y′z′
 3 15 ( 10 ) (114) ( 10 3
sA =
) + 0 4 (0.06) + 3 0 + 15 ( 103 ) (  15.8) ( 106 ) 4 (  0.14) 3 114 ( 106 ) 4 3 31.7 ( 106 ) 4  3  15.8 ( 106 ) 4 2 6
= 20.64 ( 106 ) Pa
Ans.
= 20.6 MPa (C)
Ans: sA = 20.6 MPa (C) 574
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6–113. If the beam is subjected to the internal moment of M = 1200 kN # m, determine the maximum bending stress acting on the beam and the orientation of the neutral axis.
y 150 mm 150 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus,
M 300 mm 30�
My = 1200 sin 30° = 600 kN # m
150 mm
Mz =  1200 cos 30° =  1039.23 kN # m
x 150 mm
z
Section Properties: The location of the centroid of the crosssection is given by ©yA 0.3(0.6)(0.3)  0.375(0.15)(0.15) = = 0.2893 m ©A 0.6(0.3)  0.15(0.15)
y =
150 mm
The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = Iz =
1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10  3 B m4 12 12
1 (0.3) A 0.63 B + 0.3(0.6)(0.3  0.2893)2 12  c
1 (0.15) A 0.153 B + 0.15(0.15)(0.375  0.2893)2 d 12
= 5.2132 A 10  3 B m4
Bending Stress: By inspection, the maximum bending stress occurs at either corner A or B. s = 
Mzy
sA = 
+
Iz
Myz Iy
c  1039.23 A 103 B d (0.2893) 5.2132 A 10  3 B
+
600 A 103 B (0.15) 1.3078 A 10  3 B
= 126 MPa (T)
sB = 
c  1039.23 A 103 B d ( 0.3107) 5.2132 A 10  3 B
+
600 A 103 B ( 0.15)
=  131 MPa = 131 MPa (C)(Max.)
1.3078 A 10  3 B
Ans.
Orientation of Neutral Axis: Here, u =  30°. tan a =
tan a =
Iz Iy
tan u
5.2132 A 10  3 B
1.3078 A 10  3 B
tan( 30°) Ans.
a = 66.5°
The orientation of the neutral axis is shown in Fig. b.
Ans: sB = 131 MPa (C), a = 66.5° 407 407
CH 06.indd 407
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6–114. If the beam is made from a material having an allowable tensile and compressive stress of (sallow)t = 125 MPa and (sallow)c = 150 MPa, respectively, determine the maximum allowable internal moment M that can be applied to the beam.
y 150 mm 150 mm M 300 mm
Internal Moment Components: The y component of M is positive since it is directed towards the positive sense of the y axis, whereas the z component of M, which is directed towards the negative sense of the z axis, is negative, Fig. a. Thus, My = M sin 30° = 0.5M
30� 150 mm x 150 mm
z
Mz = M cos 30° =  0.8660M Section Properties: The location of the centroid of the cross section is 150 mm
0.3(0.6)(0.3)  0.375(0.15)(0.15) ©yA y = = = 0.2893 m ©A 0.6(0.3)  0.15(0.15) The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = Iz =
1 1 (0.6) A 0.33 B (0.15) A 0.153 B = 1.3078 A 10  3 B m4 12 12
1 (0.3) A 0.63 B + 0.3(0.6)(0.3  0.2893)2 12  c
1 (0.15) A 0.153 B + 0.15(0.15)(0.375  0.2893)2 d 12
= 5.2132 A 10  3 B m4
Bending Stress: By inspection, the maximum bending stress can occur at either corner A or B. For corner A which is in tension, sA = (sallow)t = 125 A 106 B = 
Mz yA Iz
+
My zA Iy
(  0.8660M)(0.2893) 5.2132 A 10
3
B
+
0.5M(0.15) 1.3078 A 10  3 B
M = 1185 906.82 N # m = 1186 kN # m (controls)
Ans.
For corner B which is in compression, sB = (sallow)c =  150 A 106 B = 
Mz yB Iz
+
My zB Iy
( 0.8660M)(0.3107) 5.2132 A 10
3
B
M = 1376 597.12 N # m = 1377 kN # m
+
0.5M( 0.15) 1.3078 A 10  3 B
Ans:
M = 1186 kN # m 408 408
CH 06.indd 408
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6–115. The shaft is subjected to the vertical and horizontal loadings of two pulleys D and E as shown. It is supported on two journal bearings at A and B which offer no resistance to axial loading. Furthermore, the coupling to the motor at C can be assumed not to offer any support to the shaft. Determine the required diameter d of the shaft if the allowable bending stress is sallow = 180 MPa.
y
z 1m 1m 1m 1m A D x
Solution
E
B
C 400 N
100 mm 400 N 60 mm
150 N 150 N
Support Reactions: As shown on FBD. Internal Moment Components: The shaft is subjected to two bending moment components My and Mz. Allowable Bending Stress: Since all the axes through the circle’s center for a circular shaft are principal axes, then the resultant moment M = 2M2y + M2z can be used for the design.The maximum resultant moment is Mmax = 24002 + 1502 = 427.2 N # m.
Applying the flexure formula smax = sallow = 180 ( 106 ) =
Mmaxc I 427.2 ( d2 )
( )
p d 4 4 2
Ans.
d = 0.02891 m = 28.9 mm
Ans: d = 28.9 mm 577
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*6–116. For the section, Iy′ = 31.7(106) m4, Iz′ = 114(106) m4, 6 4 Iy′z′ = 15.8(10 ) m . Using the techniques outlined in Appendix A, the member’s crosssectional area has principal moments of inertia of Iy = 28.8(106) m4 and Iz = 117(106) m4, calculated about the principal axes of inertia y and z, respectively. If the section is subjected to a moment of M = 2500 N # m, determine the stress produced at point A, using Eq. 6–17.
y 60 mm
y¿ 60 mm 60 mm
M 2500 Nm z¿ 10.5 z
80 mm C 140 mm
60 mm
Solution Iz = 117(10  6) m4
A
Iy = 28.8(10  6) m4
My = 2500 sin 10.5° = 455.59 N # m
Mz = 2500 cos 10.5° = 2458.14 N # m y =  0.06 sin 10.5°  0.14 cos 10.5° =  0.14859 m z = 0.14 sin 10.5°  0.06 cos 10.5° =  0.03348 m sA = =
 Mzy Iz
+
My z Iy
2458.14( 0.14859) 6
117(10 )
+
455.59(  0.03348) 28.8(10  6)
= 2.59 MPa (T)
Ans.
Ans: sA = 2.59 MPa (T) 578
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6–117. Solve Prob. 6–116 using the equation developed in Prob. 6–106.
y 60 mm
y¿ 60 mm 60 mm
M 2500 Nm z¿ 10.5 z
=
140 mm
60 mm
Solution sA =
80 mm C
A
 (Mz′Iy′ + My′Iy′z′)y′ + (My′Iz′ + Mz′Iy′z′)z′ Iy′Iz′  Iy′z′2  32500(31.7)(10  6) + 04( 0.14) + 30 + 2500(15.8)(10  6) 4(  0.06) 31.7(10  6)(114)(10  6)  3(15.8)(10  6) 4 2
Ans.
= 2.59 MPa (T)
Ans: sA = 2.59 MPa (T) 579
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6–118. If the applied distributed loading of w = 4 kN>m can be assumed to pass through the centroid of the beam’s crosssectional area, determine the absolute maximum bending stress in the joist and the orientation of the neutral axis. The beam can be considered simply supported at A and B.
w
A
15 6m 15
10 mm 15 mm
Solution Internal Moment Components: The uniform distributed load w can be resolved into its y and z components as shown in Fig. a.
w(6 m)
B
15 mm 100 mm 100 mm
15
15 100 mm
wy = 4 cos 15° = 3.864 kN>m wz = 4 sin 15° = 1.035 kN>m wy and wz produce internal moments in the beam about the z and y axes, respectively. For the simply supported beam subjected to the uniform distributed wL2 load, the maximum moment in the beam is Mmax = . Thus, 8 (Mz)max = (My)max =
wyL2
=
8 wzL2
=
8
3.864(62) 8 1.035(62) 8
= 17.387 kN # m = 4.659 kN # m
As shown in Fig. b, (Mz)max and (My)max are positive since they are directed towards the positive sense of their respective axes. Section Properties: The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = 2c
1 1 (0.015)(0.13) d + (0.17)(0.013) = 2.5142(10  6) m4 12 12
1 1 (0.1)(0.23) (0.09)(0.17 3) = 29.8192(10  6) m4 12 12 Bending Stress: By inspection, the maximum bending stress occurs at points A and B. Iz =
s = 
(Mz)max y Iz
smax = sA = 
+
(My)max z Iy
17.387(103)( 0.1) 29.8192 (10  6)
+
4.659(103)(0.05) 2.5142(10  6) Ans.
= 150.96 MPa = 151 MPa (T) 3
smax = sB = 
17.387(10 )(0.1) 6
29.8192 (10 )
3
+
4.659(10 )( 0.05) 2.5142(10  6) Ans.
=  150.96 MPa = 151 MPa (C) Orientation of Neutral Axis: Here, u = tan  1 c tan a = tan a =
Iz Iy
(My)max (Mz)max
tan u
29.8192(10  6) 2.5142(10  6)
d = tan  1 a
4.659 b = 15°. 17.387
tan 15° Ans.
a = 72.5° The orientation of the neutral axis is shown in Fig. c. 580
Ans: smax = 151 MPa, a = 72.5°
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6–119. Determine the maximum allowable intensity w of the uniform distributed load that can be applied to the beam. Assume w passes through the centroid of the beam’s crosssectional area, and the beam is simply supported at A and B. The allowable bending stress is sallow = 165 MPa.
w
A
15 6m 15
10 mm 15 mm
Solution Internal Moment Components: The uniform distributed load w can be resolved into its y and z components as shown in Fig. a.
w(6 m)
B
15 mm 100 mm 100 mm
15
15 100 mm
wy = w cos 15° = 0.9659w wz = w sin 15° = 0.2588w wy and wz produce internal moments in the beam about the z and y axes, respectively. For the simply supported beam subjected to the a uniform distributed load, the wL2 maximum moment in the beam is Mmax = . Thus, 8 (Mz)max = (My)max =
wyL2 8 wzL2 8
= =
0.9659w(62) 8 0.2588w(62) 8
= 4.3476w = 1.1647w
As shown in Fig. b, (Mz)max and (My)max are positive since they are directed towards the positive sense of their respective axes. Section Properties: The moments of inertia of the cross section about the principal centroidal y and z axes are Iy = 2c Iz =
1 1 (0.015)(0.13) d + (0.17)(0.013) = 2.5142(10  6) m4 12 12
1 1 (0.1)(0.23) (0.09)(0.173) = 29.8192(10  6) m4 12 12
Bending Stress: By inspection, the maximum bending stress occurs at points A and B. We will consider point A. sA = sallow = 165(106) = 
(Mz)maxyA Iz
+
4.3467w(  0.1) 6
29.8192(10 )
(My)maxzA Iy +
1.1647w(0.05) 2.5142(10  6) Ans.
w = 4372.11 N >m = 4.37 kN>m
Ans: w = 4.37 kN>m 581
Ltd. All rights This is protected all copyright laws as they currently laws exist.asNo portion 2018 Pearson Education, ©© 2010 Pearson Education, Inc., Upper Saddlereserved. River, NJ. Allmaterial rights reserved. Thisunder material is protected under all copyright they currently this material be reproduced, in any form any or means, without in writingin from the publisher. exist. Noofportion of thismay material may be reproduced, in or anybyform by any means,permission without permission writing from the publisher.
*6–120. The composite beam is made of 6061T6 aluminum (A) and C83400 red brass (B). Determine the dimension h of the brass strip so that the neutral axis of the beam is located at the seam of the two metals. What maximum moment will this beam support if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa?
h B
50 mm
A 150 mm
Section Properties: n =
68.9(109) Eal = 0.68218 = Ebr 101(109)
bbr = nbal = 0.68218(0.15) = 0.10233 m y = 0.05 =
©yA ©A 0.025(0.10233)(0.05) + (0.05 + 0.5h)(0.15)h 0.10233(0.05) + (0.15)h Ans.
h = 0.04130 m = 41.3 mm INA =
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.05  0.025)2 12
+
1 (0.15) A 0.041303 B + 0.15(0.04130)(0.070649  0.05)2 12
= 7.7851 A 10  6 B m4
Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =
Mc INA M(0.04130) 7.7851(10  6)
M = 6598 N # m = 6.60 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow)al = n
Mc INA
128 A 106 B = 0.68218 c
M(0.05) 7.7851(10  6)
d
M = 29215 N # m = 29.2 kN # m
Ans: h = 41.3 mm, M = 6.60 kN # m (controls!) 415 415
CH 06.indd 415
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6–121. The composite beam is made of 6061T6 aluminum (A) and C83400 red brass (B). If the height h = 40 mm, determine the maximum moment that can be applied to the beam if the allowable bending stress for the aluminum is 1sallow2al = 128 MPa and for the brass 1sallow2br = 35 MPa.
h B
50 mm
A
Section Properties: For transformed section.
150 mm
68.9(109) Eal = 0.68218 = n = Ebr 101.0(109) bbr = nbal = 0.68218(0.15) = 0.10233 m y = =
©yA ©A 0.025(0.10233)(0.05) + (0.07)(0.15)(0.04) 0.10233(0.05) + 0.15(0.04)
= 0.049289 m INA =
1 (0.10233) A 0.053 B + 0.10233(0.05)(0.049289  0.025)2 12
+
1 (0.15) A 0.043 B + 0.15(0.04)(0.07  0.049289)2 12
= 7.45799 A 10  6 B m4
Allowable Bending Stress: Applying the flexure formula Assume failure of red brass (sallow)br = 35 A 106 B =
Mc INA M(0.09  0.049289) 7.45799(10  6)
M = 6412 N # m = 6.41 kN # m (controls!)
Ans.
Assume failure of aluminium (sallow)al = n
Mc INA
128 A 106 B = 0.68218 c
M(0.049289) 7.45799(10  6)
d
M = 28391 N # m = 28.4 kN # m
Ans:
M = 6.41 kN # m 416 416
CH 06.indd 416
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6–122. The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown. If it is subjected to a moment of M = 6.5 kN # m, determine the maximum bending stress in the brass and steel. Also, what is the stress in each material at the seam where they are bonded together? Ebr = 100 GPa, Est = 200 GPa.
A
y 50 mm 200 mm M
B z
Solution n = y = I =
x 175 mm
200(109) Est = = 2 Ebr 100(109) (350)(50)(25) + (175)(200)(150) 350(50) + 175(200)
= 108.33 mm
1 1 (0.35)(0.053) + (0.35)(0.05)(0.083332) + (0.175)(0.23) + 12 12 (0.175)(0.2)(0.041672) = 0.3026042(10  3) m4
Maximum stress in brass: (sbr)max =
6.5(103)(0.14167) Mc1 = = 3.04 MPa I 0.3026042(10  3)
Ans.
Maximum stress in steel: (sst)max =
(2)(6.5)(103)(0.10833) nMc2 = = 4.65 MPa I 0.3026042(10  3)
Ans.
Stress at the junction: sbr =
6.5(103)(0.05833) Mr = = 1.25 MPa I 0.3026042(10  3)
Ans. Ans.
sst = nsbr = 2(1.25) = 2.51 MPa
Ans: (sbr)max = 3.04 MPa, (sst)max = 4.65 MPa, sbr = 1.25 MPa, sst = 2.51 MPa 582
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6–123. The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown. If the allowable bending stress for the steel is (sallow)st = 180 MPa, and for the brass (sallow)br = 60 MPa, determine the maximum moment M that can be applied to the beam. Ebr = 100 GPa, Est = 200 GPa.
A
y 50 mm 200 mm M
B z
Solution n = y = I =
x 175 mm
200(109) Est = = 2 Ebr 100(109) (350)(50)(25) + (175)(200)(150) 350(50) + 175(200)
= 108.33 mm
1 1 (0.35)(0.053) + (0.35)(0.05)(0.083332) + (0.175)(0.23) + 12 12 (0.175)(0.2)(0.041672) = 0.3026042(10  3) m4
(sst)allow =
nMc2 ; I
180(106) =
(2) M(0.10833) 0.3026042(10  3)
M = 251 kN # m (sbr)allow =
Mc1 ; I
60(106) =
M(0.14167) 0.3026042(10  3)
M = 128 kN # m (controls)
Ans.
Ans: M = 128 kN # m 583
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*6–124. The reinforced concrete beam is made using two steel reinforcing rods. If the allowable tensile stress for for the steel is (sst)allow = 280 MPa and the allowable compressive stress for the concrete is (sconc)allow = 21 MPa, determine the maximum moment M that can be applied to the section. Assume the concrete cannot support a tensile stress. Est = 200 GPa, Econc = 26.5 GPa.
8 in. 200 mm 6 150 in. mm 8 in. 200 mm
4 in. 100 mm
M M 18 mm in. 450 in. 50 2mm 1in. diameter 25mm diameterrods rods
π (0.01252 )] 0.3125π (10 −3 ) m 2 = Ast 2[=
550 mm
200 −3 −3 2 [0.3125π (10 )] = 7.4094(10 ) m 26.5
©yA = 0;
150 mm
400 mm
h′ (h′ + 0.05)(0.55)(0.1) + (0.15)h′ − 7.4094(10 −3 )(0.4 − h′) = 0 2 0 0.075h′2 + 0.062409 h′ − 0.21377(10 −3 ) =
Solving for the positive root: h′ = 0.0034112 m I=
1 (0.55)(0.13 ) + (0.55)(0.1)(0.0034112 + 0.05)2 12 +
1 (0.15)(0.0034112 3 ) + 0.15(0.0034112)(0.0034112 2)2 12
+ 7.4094(10 −3 )(0.4 − 0.0034112)2 = 1.36811(10 −3 ) m 4
Assume concrete fails: (scon)allow =
My ; I
200 M (0.4 − 0.0034112) 21(106 ) = 26.5 1.36811(10 −3 ) ⋅ m 278 kN ⋅ m M = 277.83(10 3 ) N=
Assume steel fails: (sst)allow = na
My b; I
280(106 ) =
M (0.4 + 0.0034112) 1.36811(10 −3 )
= M 127.98(10 3 ) N= ⋅ m 128 kN ⋅ m (Controls)
Ans.
Ans: M = 128 kN ⋅ m (Controls) 429 429
CH 06.indd 429
1/18/11 2:24:38 PM
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6–125. The low strength concrete floor slab is integrated with a wideflange A36 steel beam using shear studs (not shown) to form the composite beam. If the allowable bending stress for the concrete is (sallow)con = 10 MPa, and allowable bending stress for steel is (sallow)st = 165 MPa, determine the maximum allowable internal moment M that can be applied to the beam.
1m
100 mm
15 mm 400 mm M 15 mm 15 mm
Section Properties: The beam cross section will be transformed into Econ 22.1 that of steel. Here, Thus, = = 0.1105. n = Est 200 bst = nbcon = 0.1105(1) = 0.1105 m. The location of the transformed section is y = =
200 mm
©yA ©A 0.0075(0.015)(0.2) + 0.2(0.37)(0.015) + 0.3925(0.015)(0.2) + 0.45(0.1)(0.1105) 0.015(0.2) + 0.37(0.015) + 0.015(0.2) + 0.1(0.1105)
= 0.3222 m The moment of inertia of the transformed section about the neutral axis is I = ©I + Ad2 =
1 (0.2) A 0.0153 B 12
+ 0.2(0.015)(0.3222  0.0075)2 + + +
1 (0.015) A 0.373 B + 0.015(0.37)(0.3222  0.2)2 12
1 (0.2) A 0.0153 B + 0.2(0.015)(0.3925  0.3222)2 12 1 (0.1105) A 0.13 B + 0.1105(0.1)(0.45  0.3222)2 12
= 647.93 A 10  6 B m4
Bending Stress: Assuming failure of steel, (sallow)st =
M(0.3222) Mcst ; 165 A 106 B = I 647.93 A 10  6 B
M = 331 770.52 N # m = 332 kN # m
Assuming failure of concrete,
(sallow)con = n
Mccon ; I
10 A 106 B = 0.1105C
M(0.5  0.3222) 647.93 A 10  6 B
S
M = 329 849.77 N # m = 330 kN # m (controls) Ans. Ans:
M = 330 kN # m 427 427
CH 06.indd 427
1/18/11 2:24:31 PM
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6–126. The wooden section of the beam is reinforced with two steel plates as shown. Determine the maximum moment M that the beam can support if the allowable stresses for the wood and steel are (sallow)w = 6 MPa, and (sallow)st = 150 MPa, respectively. Take Ew = 10 GPa and Est = 200 GPa.
15 mm
150 mm M
15 mm
Solution Section Properties: The cross section will be transformed into that of steel as shown Ew 10 in Fig. a. Here, n = = = 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m. The Est 200 moment of inertia of the transformed section about the neutral axis is I =
100 mm
1 1 (0.1)(0.183) (0.095)(0.153) = 21.88125(10  6) m4 12 12
Bending Stress: Assuming failure of the steel, (sallow)st =
Mcst ; I
150(106) =
M(0.09) 21.88125(10  6)
M = 36 468.75 N # m = 36.5 kN # m Assuming failure of wood, (sallow)w = n
Mcw ; I
6(106) = 0.05c
M(0.075) 21.88125(10  6)
d
M = 35 010 N # m = 35.0 kN # m (controls)
Ans.
Ans: M = 35.0 kN # m 587
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6–127. The wooden section of the beam is reinforced with two steel plates as shown. If the beam is subjected to a moment of M = 30 kN # m, determine the maximum bending stresses in the steel and wood. Sketch the stress distribution over the cross section. Take Ew = 10 GPa and Est = 200 GPa.
15 mm
150 mm M
15 mm
Solution Section Properties: The cross section will be transformed into that of steel as shown Ew 10 in Fig. a. Here, n = = = 0.05. Thus, bst = nbw = 0.05(0.1) = 0.005 m. The Est 200 moment of inertia of the transformed section about the neutral axis is I =
100 mm
1 1 (0.1)(0.183) (0.095)(0.153) = 21.88125(10  6) m4 12 12
Maximum Bending Stress: For the steel, (smax)st =
30(103)(0.09) Mcst = = 123 MPa I 21.88125(10  6)
sst y = 0.075 m =
Ans.
30(103)(0.075) My = = 103 MPa I 21.88125(10  6)
For the wood, (smax)w = n
30(103)(0.075) Mcw = 0.05c d = 5.14 MPa I 21.88125(10  6)
Ans.
The bending stress distribution across the cross section is shown in Fig. b.
Ans: (smax)st = 123 MPa, (smax)w = 5.14 MPa 588
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*6–128. y
A wood beam is reinforced with steel straps at its top and bottom as shown. Determine the maximum bending stress developed in the wood and steel if the beam is subjected to a moment of M = 150 kN # m. Sketch the stress distribution acting over the cross section. Take Ew = 10 GPa, Est = 200 GPa.
40 mm
400 mm M 150 kNm 40 mm x
Solution Ew 10 GPa = = 0.05. Then Est 200 GPa bst = nbw = 0.05(0.2) = 0.01 m. For the transformed section shown in Fig. a,
z
Section Properties: Here, n =
INA =
200 mm
1 1 (0.2) ( 0.483 ) (0.19) ( 0.43 ) = 0.82987 ( 103 ) m4 12 12
Maximum Bending Stress: Applying the flexure formula, (sst)max = (sw)max = n
150 ( 103 ) (0.24) Mcst = = 43.38 ( 106 ) Pa = 43.4 MPa I 0.82987 ( 103 )
150 ( 103 ) (0.2) Mcw = 0.05 c d = 1.8075 ( 106 ) Pa = 1.81 MPa I 0.82987 ( 103 )
Ans. Ans.
The bending stress on the steel at y = 0.2 m is sst =
150 ( 103 ) (0.2) My = = 36.15 ( 106 ) Pa = 36.2 MPa I 0.82987 ( 103 )
Using these results, the stress distribution on the beam’s crosssection shown in Fig. b can be sketched.
Ans: (sst)max = 43.4 MPa, (sw)max = 1.81 MPa 591
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6–129. The Douglas Fir beam is reinforced with A992 steel straps at its sides. Determine the maximum stress in the wood and steel if the beam is subjected to a moment of Mz = 80 kN # m. Sketch the stress distribution acting over the cross section.
y 20 mm
400 mm
200 mm
20 mm
z
Solution Ew 13.1 GPa = = 0.0655. Then Est 200 GPa bst = nbw = 0.0655(0.2) = 0.0131 m. For the transformed section shown in Fig. a,
Section Properties: Here, n =
INA =
1 (0.04 + 0.0131) ( 0.43 ) = 0.2832 ( 103 ) m4 12
Maximum Bending Stress: Applying the flexure formula, 80 ( 103 ) (0.2) Mc = = 56.497 ( 106 ) Pa = 56.5 MPa Ans. I 0.2832 ( 103 ) 80 ( 103 ) (0.2) Mc = n = 0.0655 c d = 3.7006 ( 106 ) Pa = 3.70 MPa Ans. I 0.2832 ( 103 )
(sst)max = (sw)max
Using these results, the stress distribution on the beam’s crosssection shown in Fig. b can be sketched.
Ans: (sst)max = 56.5 MPa, (sw)max = 3.70 MPa 589
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6–130. If P = 3 kN, determine the bending stress at points A, B, and C of the cross section at section a–a. Using these results, sketch the stress distribution on section a–a.
D
600 mm
E
a 300 mm A
Solution Internal Moment: The internal moment developed at section a–a can be determined by writing the moment equation of equilibrium about the neutral axis of the cross section at a–a. Using the freebody diagram shown in Fig. a, a+ ΣMNA = 0; 3(0.6)  Ma  a = 0 Ma  a = 1.8 kN # m
B C a P
20 mm 50 mm 25 mm 25 mm 25 mm Section a – a
Here, M a  a is considered negative since it tends to reduce the curvature of the curved segment of the beam. Section Properties: Referring to Fig. b, the location of the centroid of the cross section from the center of the beam’s curvature is r =
0.31(0.02)(0.075) + 0.345(0.05)(0.025) ΣrA = = 0.325909 m ΣA 0.02(0.075) + 0.05(0.025)
The location of the neutral surface from the center of the beam’s curvature can be determined from R =
A dA ΣLA r
where A = 0.02(0.075) + 0.05(0.025) = 2.75(10  3) m2 dA 0.32 0.37 Σ = 0.075 ln + 0.025 ln = 8.46994(10  3) m r 0.3 0.32 LA Thus, R =
2.75(10  3) 8.46994(10  3)
= 0.324678 m
then e = r  R = 0.325909  0.324678 = 1.23144(10  3) m Normal Stress: sA = sB = sC =
M(R  rA) AerA M(R  rB) AerB M(R  rC) AerC
= = =
1.8(103)(0.324678  0.3) 2.75(10  3)(1.23144)(10  3)(0.3) 1.8(103)(0.324678  0.32) 2.75(10  3)(1.23144)(10  3)(0.32) 1.8(103)(0.324678  0.37) 2.75(10  3)(1.23144)(10  3)(0.37)
= 43.7 MPa (T)
Ans.
= 7.77 MPa (T)
Ans.
= 65.1 MPa (C)
Ans.
Ans: sA = 43.7 MPa (T), sB = 7.77 MPa (T), sC = 65.1 MPa (C) 604
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6–131. If the maximum bending stress at section a–a is not allowed to exceed sallow = 150 MPa, determine the maximum allowable force P that can be applied to the end E.
D
600 mm
E
a 300 mm A
Solution Internal Moment: The internal Moment developed at section a–a can be determined by writing the moment equation of equilibrium about the neutral axis of the cross section at a–a.
B C a P
20 mm 50 mm 25 mm 25 mm 25 mm Section a – a
a+ ΣMNA = 0; P(0.6)  Ma  a = 0 Ma  a = 0.6P Here, Ma–a is considered positive since it tends to decrease the curvature of the curved segment of the beam. Section Properties: Referring to Fig. b, the location of the centroid of the cross section from the center of the beam’s curvature is r =
0.31(0.02)(0.075) + 0.345(0.05)(0.025) Σr~A = = 0.325909 m ΣA 0.02(0.075) + 0.05(0.025)
The location of the neutral surface from the center of the beam’s curvature can be determined from R =
A dA r
A L
Σ
where A = 0.02(0.075) + 0.05(0.025) = 2.75(10  3) m2 Σ
dA 0.32 0.37 = 0.075 ln + 0.025 ln = 8.46994(10  3) m r 0.3 0.32
A L
Thus, R =
2.75(10  3) 8.46994(10  3)
= 0.324678 m
then e = r  R = 0.325909  0.324678 = 1.23144(10  3) m Allowable Normal Stress: The maximum normal stress occurs at either points A or C. For point A, which is in tension, sallow =
M(R  rA) AerA
; 150(106) =
0.6P(0.324678  0.3) 2.75(10  3)(1.23144)(10  3)(0.3)
P = 10 292.09 N = 10.3 kN For point C, which is in compression, sallow =
M(R  rC) AerC
;  150(106) =
0.6P(0.324678  0.37) 2.75(10  3)(1.23144)(10  3)(0.37)
P = 6911.55 N = 6.91 kN (controls)
Ans.
Ans: P = 6.91 kN 605
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*6–132. If the beam is subjected to a moment of M = 45 kN # m, determine the maximum bending stress in the A36 steel section A and the 2014T6 aluminum alloy section B.
A 50 mm
M 15 mm
Solution
150 mm
B
Section Properties: The cross section will be transformed into that of steel as shown 73.1 1 109 2 Eal in Fig. a. Here, n = = = 0.3655. Thus, bst = nbal = 0.3655(0.015) = Est 200 1 109 2 0.0054825 m. The location of the transformed section is ΣyA y = = ΣA
0.075(0.15)(0.0054825) + 0.2c p 1 0.052 2 d 0.15(0.0054825) + p 1 0.052 2
= 0.1882 m
The moment of inertia of the transformed section about the neutral axis is I = ΣI + Ad 2 =
1 (0.0054825) ( 0.153 ) + 0.0054825(0.15)(0.1882  0.075)2 12 +
1 p 10.0542 + p 10.0522 (0.2  0.1882)2 4
= 18.08 110  62 m4
Maximum Bending Stress: For the steel, (smax)st =
45 11032 (0.06185) Mcst = = 154 MPa I 18.08 110  62
Ans.
For the aluminum alloy, (smax)al = n
45 11032 (0.1882) Mcal = 0.3655 C S = 171 MPa I 18.08 110  62
Ans.
Ans: (smax)st = 154 MPa, (smax)al = 171 MPa 596
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6–133. For the curved beam in Fig. 6–40a, show that when the radius of curvature approaches infinity, the curvedbeam formula, Eq. 6–24, reduces to the flexure formula, Eq. 6–13.
Solution Normal Stress: Curvedbeam formula s = s =
M(R  r)
dA LA r
where A′ =
Ar(r  R)
and R =
A
dA 1A r
=
A A′
M(A  rA′)
(1)
Ar(rA′  A)
(2)
r = r + y rA′ = r
r dA = a  1 + 1bdA LA r LA r + y =
LA
a
= A 
r  r  y r + y y
LA r + y
+ 1b dA (3)
dA
Denominator of Eq. (1) becomes, y
LA r + y
Ar(rA′  A) = Ar ¢A 
dA  A≤ = Ar
y
LA r + y
dA
Using Eq. (2), Ar(rA′  A) =  A = A =
LA
¢
ry r + y
+ y  y≤dA  Ay
y dA  A 1A y dA  Ay dA LA r + y LA r + y
y2 y Ay A ¢ ¢ A y dA y ≤dA  A 1 y ≤ dA r LA 1 + r LA 1 + r r as
y r
S 0
Ar(rA′  A) S
Then, Eq. (1) becomes Using Eq. (2),
A I r
Mr (A  rA′) AI Mr s = (A  rA′  yA′) AI s =
Using Eq. (3), s =
=
dA
y2
1A y dA = 0,
But,
y
LA r + y
y Mr dA C A  ¢A S dA≤  y AI LA r + y LA r + y y dA Mr C S dA  y AI LA r + y r LA + y 598
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6–133. (Continued)
y
y dA Mr r = C ¢ ≤dA ¢ y≤S AI LA 1 + y r LA 1 + r r As
y r
S 0 y r
LA 1 + ¢
y≤
dA = 0 and
r
Therefore, s =
y
r LA 1 + ¢
dA
y≤ r
=
yA A dA = 1 r r
y
yA My Mr ab = AI I r
(Q.E.D)
Ans: N/A 599
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6–134. The curved member is subjected to the moment of M = 50 kN # m. Determine the percentage error introduced in the calculation of maximum bending stress using the flexure formula for straight members.
M 100 mm
M 200 mm
200 mm
Solution Straight Member: The maximum bending stress developed in the straight member smax =
50(103)(0.1) Mc = 75 MPa = I 1 (0.1)(0.23) 12
Curved Member: When r = 0.2 m, r = 0.3 m, rA = 0.2 m and rB = 0.4 m, Fig. a. The location of the neutral surface from the center of curvature of the curve member is R =
0.1(0.2) A = = 0.288539 m 0.4 dA 0.1ln 0.2 LA r
Then e = r  R = 0.011461 m
Here, M = 50 kN # m. Since it tends to decrease the curvature of the curved member, sB =
M(R  rB) AerB
=
50(103)(0.288539  0.4) 0.1(0.2)(0.011461)(0.4)
=  60.78 MPa = 60.78 MPa (C) sA =
M(R  rA) AerA
=
50(103)(0.288539  0.2) 0.1(0.2)(0.011461)(0.2)
= 96.57 MPa (T) (Max.) Thus, % of error = a
96.57  75 b100 = 22.3% 96.57
Ans.
Ans: , of error = 22.3, 600
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6–135. The curved member is made from material having an allowable bending stress of sallow = 100 MPa. Determine the maximum allowable moment M that can be applied to the member.
M 100 mm
M 200 mm
200 mm
Solution Internal Moment: M is negative since it tends to decrease the curvature of the curved member. Section Properties: Referring to Fig. a, the location of the neutral surface from the center of curvature of the curve beam is R =
0.1(0.2) A = = 0.288539 m 0.4 dA 0.1ln 0.2 LA r
Then e = r  R = 0.3  0.288539 = 0.011461 m Allowable Bending Stress: The maximum stress occurs at either point A or B. For point A, which is in tension, sallow =
M(R  rA) AerA
; 100(106) =
M(0.288539  0.2) 0.1(0.2)(0.011461)(0.2)
M = 51 778.27 N # m = 51.8 kN # m (controls)
Ans.
For point B, which is in compression, sallow =
M(R  rB) AerB
;  100(106) =
M(0.288539  0.4) 0.1(0.2)(0.011461)(0.4)
M = 82 260.10 N # m = 82.3 kN # m
Ans: M = 51.8 kN # m 601
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*6–136. The curved beam is subjected to a bending moment of M = 900 N # m as shown. Determine the stress at points A and B, and show the stress on a volume element located at each of these points.
A C B
100 mm C
A
30�
20 mm
15 mm 400 mm
150 mm B
M
Internal Moment: M =  900 N # m is negative since it tends to decrease the beam’s radius curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10  3) m3 r = ©
2.18875 (10  3) ©rA = = 0.5150 m ©A 0.00425
dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10  3) m 0.4 0.55 LA r
R =
A
©1A
dA r
=
0.00425 = 0.509067 m 8.348614(10  3)
r  R = 0.515  0.509067 = 5.933479(10  3) m Normal Stress: Applying the curvedbeam formula sA =
M(R  rA) ArA (r  R)
=
900(0.509067  0.57) 0.00425(0.57)(5.933479)(10  3) Ans.
= 3.82 MPa (T) sB =
M(R  rB)  900(0.509067  0.4) = ArB (r  R) 0.00425(0.4)(5.933479)(10  3) Ans.
=  9.73 MPa = 9.73 MPa (C)
Ans: sA = .1B 5
sB = 9.73 MPa (C) 435 435
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6–137. The curved beam is subjected to a bending moment of M = 900 N # m. Determine the stress at point C. A C B
100 mm C
A
30�
20 mm
15 mm 400 mm
150 mm B
M
Internal Moment: M =  900 N # m is negative since it tends to decrease the beam’s radius of curvature. Section Properties: ©A = 0.15(0.015) + 0.1(0.02) = 0.00425 m2 ©rA = 0.475(0.15)(0.015) + 0.56(0.1)(0.02) = 2.18875(10  3) m r = ©
2.18875 (10  3) ©rA = = 0.5150 m ©A 0.00425
dA 0.57 0.55 = 0.015 ln + 0.1 ln = 8.348614(10  3) m 0.4 0.55 LA r
R =
A
©1A
dA r
=
0.00425 = 0.509067 m 8.348614(10  3)
r  R = 0.515  0.509067 = 5.933479(10  3) m Normal Stress: Applying the curvedbeam formula sC =
M(R  rC) ArC(r  R)
=
900(0.509067  0.55) 0.00425(0.55)(5.933479)(10  3) Ans.
= 2.66 MPa (T)
Ans: ΣA = 0.00425 m2, r = 0.5150 m, dA = 8.348614(10  3) m, Σ LA r sC = 2.66 MPa (T) (s max )pvc = 12.3 MPa 436 436
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6–138. The beam is made from three types of plastic that are identified and have the moduli of elasticity shown in the figure. Determine the maximum bending stress in the PVC.
2.5 500 kN lb
2.5 lb kN 500 PVC � 450 ksiGPa PVCEE 3.15 PVC PVC � 160 ksiGPa Escon EsconEE 1.12 EE 800 Bakelite BakeliteEBE� 5.6ksi GPa B
3 ftm 0.9
1.12 (b= = 0.015 m bk )1 n= 1bEs (0.075) 5.6
4 ftm 1.2
3 ft m 0.9
in. 251mm in. 502mm 2 in. 50 mm
3.15 n= = (b = 0.0421875 m 2 bpvc bk )2 (0.075) 5.6
3 in. 75 mm
∑ yA 0.025(0.075)(0.05) + 0.075(0.015)(0.05) + 0.1125(0.0421875)(0.025) = y = ∑A 0.075(0.05) + 0.015(0.05) + 0.0421875(0.025)
2.5 kN
= 0.048365 m
0.9 m
1 I =2 (0.075)(0.053 ) + 0.075(0.05)(0.048365 − 0.025)2 12
2.5 kN
2.5 kN 0.9 m
1 + (0.015)(0.053 ) + 0.015(0.05)(0.075 − 0.048365)2 12
2.5 kN
1.2 m
0.9 m
2.5 kN
Mmax = 2.25 kN · m
1 + (0.0421875)(0.0253 ) + 0.0421875(0.025)(0.1125 − 0.048365)2 12
2.5 kN
0.025 m 0.05 m
= 7.90996(10 −6 ) m 4 (σ max ) pvc
0.05 m
Mc 3.15 [2.25(10 3 )](0.1125 − 0.048365) = = n2 I 7.90996(10 −6 ) 5.6
0.075 m
6 = 12.26(10 = ) N−m 2 12.3 MPa
Ans.
Ans: (s max )pvc = 12.3 MPa 426 426
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6–139. The composite beam is made of A36 steel (A) bonded to C83400 red brass (B) and has the cross section shown. If it is subjected to a moment of M = 6.5 kN # m, determine the maximum stress in the brass and steel. Also, what is the stress in each material at the seam where they are bonded together?
A
y
100 mm
M
100 mm
B z
Solution
x 125 mm
Section Properties: For the transformed section. n =
101 ( 109 ) Ebr = = 0.505 Est 200 ( 109 )
bst = nbbr = 0.505(0.125) = 0.063125 m y = =
ΣyA ΣA 0.05(0.125)(0.1) + 0.15(0.1)(0.063125) 0.125(0.1) + 0.1(0.063125)
= 0.08355 m INA =
1 (0.125) ( 0.13 ) + 0.125(0.1)(0.08355  0.05)2 12 +
1 (0.063125) ( 0.13 ) + 0.063125(0.1)(0.15  0.08355)2 12
= 57.62060 ( 10  6 ) m4 Maximum Bending Stress: Applying the flexure formula (smax)st =
6.5 ( 103 ) (0.08355) My = I 57.62060 ( 106 ) Ans.
= 9.42 MPa (smax)br = n
6.5 ( 103 ) (0.2  0.08355) Mc = 0.505c d I 57.62060 ( 106 )
Ans.
= 6.63 MPa
Bending Stress: At seam sst = =
My I 6.5 ( 103 ) (0.1  0.08355) 57.62060 ( 106 ) Ans.
= 1.855 MPa = 1.86 MPa sbr = n
My = 0.505(1.855) = 0.937 MPa I
Ans. Ans: (smax)st = 9.42 MPa, (smax)br = 6.63 MPa, sst = 1.86 MPa, sbr = 0.937 MPa 594
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*6–140. The composite beam is made of A36 steel (A) bonded to C83400 red brass (B) and has the cross section shown. If the allowable bending stress for the steel is (sallow)st = 180 MPa and for the brass (sallow)br = 60 MPa, determine the maximum moment M that can be applied to the beam.
A
y
100 mm
M
100 mm
B z
Solution
x 125 mm
Section Properties: For the transformed section. 101 ( 109 ) Ebr = = 0.505 Est 200 ( 109 )
n =
bst = nbbr = 0.505(0.125) = 0.063125 m y = =
ΣyA ΣA 0.05(0.125)(0.1) + 0.15(0.1)(0.063125) 0.125(0.1) + 0.1(0.063125)
= 0.08355 m INA =
1 (0.125) ( 0.13 ) + 0.125(0.1)(0.08355  0.05)2 12 +
1 (0.063125) ( 0.13 ) + 0.063125(0.1)(0.15  0.08355)2 12
= 57.62060 ( 10  6 ) m4 Allowable Bending Stress: Applying the flexure formula Assume failure of steel (smax)st = (sallow)st = 180 ( 106 ) =
My I M(0.08355) 57.62060 ( 106 )
M = 124130 N # m = 124 kN # m Assume failure of brass (smax)br = (sallow)br = n
Mc I
60 ( 106 ) = 0.505c
M(0.2  0.08355) 57.62060 ( 106 )
M = 58792 N # m
d
= 58.8 kN # m (Controls!)
Ans.
Ans: M = 58.8 kN # m 595
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6–141. The member has a brass core bonded to a steel casing. If a couple moment of 8 kN # m is applied at its end, determine the maximum bending stress in the member. Ebr = 100 GPa, Est = 200 GPa.
8 kN�m
3m 20 mm 100 mm 20 mm 20 mm
n =
Ebr 100 = = 0.5 Est 200
I =
1 1 (0.14)(0.14)3 (0.05)(0.1)3 = 27.84667(10  6)m4 12 12
100 mm
20 mm
Maximum stress in steel: (sst)max =
8(103)(0.07) Mc1 = 20.1 MPa = I 27.84667(10  6)
Ans.
(max)
Maximum stress in brass: (sbr)max =
0.5(8)(103)(0.05) nMc2 = 7.18 MPa = I 27.84667(10  6)
Ans: (sst) max = 20.1 MPa 422 422
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6–142. y
The Douglas Fir beam is reinforced with A36 steel straps at its sides. Determine the maximum stress in the wood and steel if the beam is subjected to a bending moment of Mz = 4 kN # m. Sketch the stress distribution acting over the cross section.
15 mm
350 mm
200 mm
15 mm
z
Solution Section Properties: For the transformed section, n =
13.1 ( 109 ) Ew = = 0.0655 Est 200 ( 109 )
bst = nbw = 0.0655(0.2) = 0.0131 m 1 (0.03 + 0.0131) ( 0.353 ) = 153.99 ( 106 ) m4 12 Maximum Bending Stress: Applying the flexure formula INA =
(smax)st =
4 ( 103 ) (0.175) Mc = = 4.55 MPa I 153.99 ( 106 )
(smax)w = n
Ans.
4 ( 103 ) (0.175) Mc = 0.0655c d = 0.298 MPa I 153.99 ( 106 )
Ans.
Ans: (smax)st = 4.55 MPa, (smax)w = 0.298 MPa 597
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6–143. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stress acting at section a–a. Sketch the stress distribution on the section in three dimensions.
100 mm 50 mm
200 mm
Solution
a
Internal Moment: Referring to the FBD of the lower segment of the curve beam sectioned through a–a, Fig. a, 4 a+ΣMo = 0; 5 a b(0.2)  M = 0 M = 0.8 kN # m = 800 N # m 5 Section Properties: Referring to Fig. a, 0.2 + 0.3 r = = 0.25 m A = (0.05)(0.1) = 0.005 m2 2
5 kN
a
200 mm
5 4 3
100 mm 5
5 kN
4 3
r2 dA 0.3 = b>n = 0.05>n a b = 0.020273255 m r1 0.2 LA r R =
A 0.005 = = 0.2466303 m 0.020273255 dA LA r
r  R = 0.25  0.2466303 = 0.003369654 m Normal Stress: Here M = 800 N # m is positive, since it tends to increase the beam’s radius of curvature. Applying the curved beam formula, (sT)max =
= (sC)max =
=
M(R  r1) Ar1(r  R) 800(0.2466303  0.2) 0.005(0.2)(0.003369654)
= 11.07 ( 106 ) Pa = 11.1 MPa (T)
Ans.
=  8.447 ( 106 ) Pa = 8.45 MPa (C)
Ans.
M(R  r2) Ar2(r  R) 800(0.2466303  0.3) 0.005(0.3)(0.003369654)
Using these results, the normal stress distribution on the beam’s crosssection shown in Fig. b can be sketched.
Ans: (sT)max = 11.1 MPa (T), (sC)max = 8.45 MPa (C) 612
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*6–144. The curved member is symmetric and is subjected to a moment of M = 900 N # m. Determine the bending stress in the member at points A and B. Show the stress acting on volume elements located at these points.
0.5mm in. 12 B 2 in. 50 mm A 1.5mm in. 37 8 in.mm 200
M
M
1 A= 0.012(0.05) + (0.025)(0.05) = 0.001225 m 2 2
r =
0.012 m
0.225(0.012)(0.05) + 0.21667 21 (0.025)(0.05) ©rA = 0.220748 m = ©A 0.001225
0.05 m 0.25 m
dA 0.25 0.025(0.25) 0.25 = 0.012 ln 0.005570667 m + − 0.025 = ln LA r 0.2 0.25 − 0.2 0.2 R =
A
dA 1A r
=
0.037 m 0.2 m
0.001225 = 0.2199018 0.005570667
86.4 MPa
r  R = 0.220748 − 0.2199018 = 0.846427(10 −3 ) m s =
σA =
M(R  r)
TA = 86.4 MPa
Ar(r  R)
105 MPa
900(0.2199018 − 0.2) 6 = 86.37(10 = ) N−m 2 86.4 MPa (T) 0.001225(0.2)[0.846427(10 −3 )] 900(0.2199018 − 0.25)
σB =
−3
0.001225(0.25)[0.846427(10 )]
Ans.
= −104.50(106 ) N−m 2 = 105 MPa (C) Ans.
TB = 105 MPa
Ans: sA = 86.4 MPa (T), sB = 105 MPa (C) 437 437
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6–145. The ceilingsuspended Carm is used to support the Xray camera used in medical diagnoses. If the camera has a mass of 150 kg, with center of mass at G, determine the maximum bending stress at section A.
G 1.2 m A
Section Properties: r =
1.22(0.1)(0.04) + 1.25(0.2)(0.02) ©rA = = 1.235 m ©A 0.1(0.04) + 0.2(0.02)
200 mm
1.24 1.26 dA © = 0.1 ln + 0.2 ln = 6.479051 A 10  3 B m 1.20 1.24 LA r
40 mm
A = 0.1(0.04) + 0.2(0.02) = 0.008 m2 R =
A
dA 1A r
=
100 mm
20 mm
0.008 = 1.234749 m 6.479051 (10  3)
r  R = 1.235  1.234749 = 0.251183 A 10  3 B m
Internal Moment: The internal moment must be computed about the neutral axis as shown on FBD. M =  1816.93 N # m is negative since it tends to decrease the beam’s radius of curvature. Maximum Normal Stress: Applying the curvedbeam formula sA = =
M(R  rA) ArA (r  R)  1816.93(1.234749  1.26) 0.008(1.26)(0.251183)(10  3)
= 18.1 MPa (T) sB = =
M(R  rB) ArB (r  R) 1816.93(1.234749  1.20) 0.008(1.20)(0.251183)(10  3)
=  26.2 MPa = 26.2 MPa (C)
Ans.
(Max)
Ans: LA A = 0.008 m2, sB = 26.2 MPa (C)
r = 1.235 m, Σ
dA 3 r = 6.479051(10 ) m,
439 439
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6–146 The member has a circular cross section. If it is subjected to a moment of M = 5 kN # m, determine the stress at points A and B. Is the stress at point A′, which is located on the member near the wall, the same as that at A? Explain.
200 mm
A¿ M
A 200 mm
Solution Section Properties: Referring to Fig. a B
0.2 + 0.4 r = = 0.3 m A = pC 2 = p ( 0.12 ) = 0.01p m2 2 dA 2 = 2p 1 r  2r 2  C 2 LA r = 2p 1 0.3  20.32  0.12 2 = 0.1078024 m
R =
A 0.012r = = 0.291421 m 0.1078024 dA LA r
r  R = 0.3  0.291421 = 0.00857864 m Bending Stress: Here, M = 5 kN # m is negative since it tends to decrease the beam’s radius of curvature. Applying the curved beam formula, sA =
sB =
M(R  rA) ArA(r  R)
M(R  rB) ArB(r  R)
=
 5 ( 103 ) (0.291421  0.2) (0.01p)(0.2)(0.00857864)
= 8.4805 ( 106 ) Pa = 8.48 MPa (C) Ans.
=
 5 ( 103 ) (0.291421  0.4) (0.01p)(0.4)(0.00857864)
= 5.0360 ( 106 ) Pa = 5.04 MPa (T) Ans.
No! Point A′ is at the fixed support where stress concentration occurs. The application of the curved beam formula is only valid for sections sufficiently removed from the support in accordance to SaintVenant’s principle.
Ans: sA = 8.48 MPa (C), sB = 5.04 MPa (T) No, it is not the same. 610
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6–147. The member has a circular cross section. If the allowable bending stress is sallow = 100 MPa, determine the maximum moment M that can be applied to the member.
200 mm
A¿ M
A 200 mm
Solution Section Properties: Referring to Fig. a, B
0.2 + 0.4 r = = 0.3 m 2
A = pC 2 = p ( 0.12 ) = 0.01p m2
dA = 2p 1 r  2r 2  C 2 2 LA r
= 2p 1 0.3  20.32  0.12 2 = 0.1078024 m
R =
A 0.01p = = 0.291421 m 0.1078024 dA LA r
r  R = 0.3  0.291421 = 0.00857864 m Bending Stress: Here, M is negative since it tends to decrease the beam’s radius of curvature. Applying the curve beam’s formula by assuming tension failure, sB = sallow =
M(R  rB) ArB(r  R)
; 100 ( 106 ) =
 M(0.291421  0.4) (0.01p)(0.4)(0.00857864)
M = 99.29 ( 103 ) N # m = 99.3 kN # m Assuming compression failure, sA = sallow =
M(R  rA) ArA(r  R)
;  100 ( 106 ) =
 M(0.291421  0.2) (0.01p)(0.2)(0.00857864)
M = 58.96 ( 103 ) N # m = 59.0 kN # m (control!) Ans.
Ans: M = 59.0 kN # m 611
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*6–148. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stresses acting at section a–a. Sketch the stress distribution on the section in three dimensions.
10 kN 30 150 mm 30 10 kN
150 mm a
200 mm
Solution
a
100 mm
Internal Moment: Referring to the FBD of the upper segment of the curved beam sectioned through a– a, Fig. a a+ ΣM0 = 0 (10 sin 30°)(0.2) + 10 cos 30°(0.15)  M = 0 M = 2.2990 kN # m
50 mm
Section Properties: Referring to Fig. a
100 mm
0.15 + 0.25 = 0.2 m A = 0.05(0.1) = 0.005 m2 2 r2 dA 0.25 = b>n = 0.05 m a b = 0.025541281 m r1 0.15 LA r r =
R =
A
rA
dA
=
0.005 = 0.1957615 m 0.025541281
r
r  R = 0.2  0.1957615 = 0.00423848 m
Normal Stress: Here, M =  2.2990 kN # m is negative since it tends to decrease the beam’s radius of curvature. Applying curvedbeam formula, (sC)max =
M(R  r1) Ar1(r  R)
=
 2.2990 ( 103 ) (0.1957615  0.15)
= 33.10 ( 106 ) Pa
0.005(0.15)(0.00423848)
=  33.10 ( 106 ) Pa = 33.1 MPa (C) (sT)max =
M(R  r2) Ar2(r  R)
=
 2.2990 ( 103 ) (0.1957615  0.25) 0.005(0.25)(0.00423848)
Ans.
=
= 23.54 ( 106 ) Pa = 23.5 MPa (T)
Ans.
Using these results, the normal stress distribution on the beam’s crosssection shown in Fig. b can be sketched.
Ans: (sC)max = 33.1 MPa (C), (sT)max = 23.5 MPa (T) 607
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6–149. The curved bar used on a machine has a rectangular cross section. If the bar is subjected to a couple as shown, determine the maximum tensile and compressive stresses acting at section a–a. Sketch the stress distribution on the section in three dimensions.
a 75 mm a
50 mm
100 mm 50 mm
250 N
150 mm
Solution
250 N
Internal Moment: M = 37.5 N # m is positive since it tends to increase the beam’s 75 mm radius of curvature. Section Properties: r =
0.1 + 0.175 = 0.1375 m 2
A = 0.075(0.05) = 0.00375 m2 dA 0.175 = 0.05 ln = 0.027981 m 0.1 LA r R =
A
dA 1A r
=
0.00375 = 0.134021 m 0.027981
r  R = 0.1375  0.134021 = 3.479478 ( 103 ) m Normal Stress: Applying the curvedbeam formula (smax)t = =
M(R  r1) Ar1(r  R) 37.5(0.134021  0.1) 0.00375(0.1)(3.479478) ( 103 ) Ans.
= 0.978 MPa (T) (smax)c = =
M(R  r2) Ar2(r  R) 37.5(0.134021  0.175) 0.00375(0.175)(3.479478) ( 103 ) Ans.
=  0.673 MPa = 0.673 MPa (C)
Ans: (smax)t = 0.978 MPa (T), (smax)c = 0.673 MPa (C) 608
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6–150. If the radius of each notch on the plate is rdetermine = 12.5 mm, the largest that canThe be the determine largest moment that moment can be applied. applied. allowable bending stressisfor the=material allowableThe bending stress for the material sallow 18 ksi. is stress of sallow = 125 MPa.
362.5 m 14.5 in.
M
b5 =
25 mm 1 in.
312.5 mm 12.5 in.
M
14.5 – 312.5 12.5 362.5 25 in. mm =51.0 22
1 b 25 2.0 = =52.0 r 0.5 12.5
r 0.5 12.5 0.04 = =50.04 h 12.5 312.5
From Fig. 644: K = 2.60 smax = K
Mc I
M (0.15625) 125(106 ) = 2.60 1 (0.025)(0.31253 ) 12 M 19.56(10 3 ) N= ⋅ m 19.6 kN ⋅ m =
Ans.
Ans:
K = 2.60, M = 19.6 kN # m 442 442
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6–151. The symmetric notched plate is subjected to bending. If the radius of each notch is r = 12.5 mm and the applied moment is M = 15 kN · m, determine the maximum bending stress in the plate.
= b
14.5 in. 362.5 m
M
in. 251 mm
12.5 mm in. 312.5
M
362.5 − 312.5 = 25 mm 2
1 b 25 2.0 =52.0 = r 0.5 12.5
r 0.5 12.5 0.04 = =50.04 h 12.5 312.5
From Fig. 644: K = 2.60 smax = K
[15(10 3 )](0.15625) Mc = 2.60 1 (0.025)(0.31253 ) I 12
=
6 36.864(10 = ) N−m 2 36.9 MPa
Ans.
Ans: s max = 36.9 MPa 443 443
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*6–152. The bar is subjected to a moment of M = 100 N # m. Determine the maximum bending stress in the bar and sketch, approximately, how the stress varies over the critical section.
80 mm
10 mm
20 mm
10 mm M
M 10 mm
Solution
r 10 w 80 Stress Concentration Factor: Entering = = 4.0 and = = 0.5 into h 20 h 20 Fig. 6–43, we obtain K = 1.25. Maximum Bending Stress: smax = K
Mc I
= 1.25 £
100(0.01) 1 12 (0.01)
( 0.023 )
§
= 187.5 ( 106 ) Pa = 187.5 MPa
Ans.
The bending stress distribution across the critical section is shown in Fig. a.
Ans: smax = 187.5 MPa 613
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6–153. The allowable bending stress for the bar is sallow = 200 MPa. Determine the maximum moment M that can be applied to the bar.
80 mm
10 mm
20 mm
10 mm M
M 10 mm
Solution
r 10 w 80 Stress Concentration Factor: Entering = = 4.0 and = = 0.5 into h 20 h 20 Fig. 6–43, we obtain K = 1.25. Maximum Bending Stress: smax = sallow = K
Mc I
200(106) = 1.25 £
M(0.01) 1 12
M = 106.67 N # m
(0.01)(0.023)
§
= 107 N # m
Ans.
Ans: M = 107 N # m 614
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6–15 4. The simply supported notched bar is subjected to two forces P. Determine the largest magnitude of P that can be applied without causing the material to yield.The material mm. in. is A36 steel. Each notch has a radius of r == 30.125
P
0.5mm in. 12 1.75 in. 42 mm
1.25 in. 30 mm
20 mm in. 500
b5 =
P
20mm in. 500
20 in. 500 mm
20 in. 500 mm
1.75  1.25 42 – 30 = 0.25 5 6 mm 2 2
b 60.25 = 5 2; = 2; r 0.125 3
M = 0.5 P
r 0.125 3 5 0.1 = = 0.1 h 1.25 30
0.5 m 0.5 m
From Fig. 644. K = 1.92 sY = K
Mc ; I
1m
0.5 m
0.5P(0.015) 250(106 ) = 1.92 1 (0.012)(0.33 ) 12 Ans.
= = P 468.75 N 469 N
Ans: P 469 N 617
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6–155. The stepped bar has a thickness of 10 mm. Determine the maximum moment that can be applied to its ends if the allowable bending stress is sallow = 150 MPa.
90 mm
M
60 mm 7.5 mm
20 mm 15 mm M
Solution Stress Concentration Factor: For the smaller section, entering
w 60 = = 3.0 h 20
r 15 = = 0.75 into Fig. 6–43, we obtain K = 1.15. For the larger section, h 20 w 90 r 7.5 = = 1.5 and = = 0.125 gives K = 1.65. h 60 h 60
and
Maximum Bending Stress: For the smaller section, Mc smax = sallow = K I 150 ( 106 ) = 1.15£
M(0.01) 1 12 (0.01)
( 0.023 )
§
M = 86.96 N # m = 87.0 N # m (controls!) For the larger section, Mc smax = sallow = K I 150(106) = 1.65£
M(0.03) 1 3 12 (0.01)(0.06 )
M = 545.45 N # m
Ans.
§
Ans: M = 87.0 N # m 619
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*6–156. If the radius of each notch on the plate is r = 10 mm, determine the largest moment M that can be applied. The allowable bending stress is sallow = 180 MPa.
M 20 mm
125 mm
Solution
b 20 Stress Concentration Factor: From the graph in the text with = = 2 and r 10 r 10 = = 0.08, then K = 2.1. h 125 Allowable Bending Stress: smax = sallow = K
165 mm
M
Mc I
180 ( 106 ) = 2.1 c
M(0.0625) 1 12 (0.02)
( 0.1253 )
d
M = 4464 N # m = 4.46 kN # m
Ans.
Ans: M = 4.46 kN # m 618
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6–157. Determine the length L of the center portion of the bar so that the maximum bending stress at A, B, and C is the same. The bar has a thickness of 10 mm.
7 mm
350 N 60 mm
A 200 mm
60 w = = 1.5 h 40
40 mm 7 mm
C L 2
B L 2
200 mm
r 7 = = 0.175 h 40
From Fig. 643, K = 1.5 (sA)max = K
(35)(0.02) MAc d = 19.6875 MPa = 1.5c 1 3 I (0.01)(0.04 ) 12
(sB)max = (sA)max = 19.6875(106) =
MB c I
175(0.2 + L2 )(0.03) 1 3 12 (0.01)(0.06 )
Ans.
L = 0.95 m = 950 mm
Ans:
M = 87.0 kN # m 445 445
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6–158. Determine the shape factor for the cross section of the Hbeam.
200 mm
20 mm
20 mm
200 mm
1 1 (0.2)(0.023) + 2 a b (0.02)(0.23) = 26.8(10  6)m4 12 12
Ix =
Mp
20 mm
C1 = T1 = sY(2)(0.09)(0.02) = 0.0036sy C2 = T2 = sY(0.01)(0.24) = 0.0024sy
Mp = 0.0036sY(0.11) + 0.0024sY(0.01) = 0.00042sY sY = MY = k =
MYc I sY(26.8)(10  6) = 0.000268sY 0.1
Mp MY
=
0.00042sY = 1.57 0.000268sY
Ans.
Ans: Ix = 26.8 (10 6) m4, Mp = 0.00042sY, MY = 0.000268sY, k = 1.57 451 451
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6–159. Determine the shape factor for the wideflange beam. 15 mm
20 mm 200 mm Mp 15 mm
Solution
200 mm
1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10  6) m4 12 12
Ix =
C1 = T1 = sY(0.2)(0.015) = 0.003 sY C2 = T2 = sY(0.1)(0.02) = 0.002 sY Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY MYc I
sY = MY = k =
sY(82.78333)(10  6)
Mp MY
0.115 =
= 0.000719855 sY
0.000845 sY = 1.17 0.000719855 sY
Ans.
Ans: k = 1.17 621
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*6–160. Determine the plastic moment Mp that can be supported by a beam having the cross section shown. sY = 210 MPa.
50 mm 25 mm
Mp
250 mm
1 s dA = 0
25 mm
C1 + C2  T1 = 0
[π (0.052 − 0.0252 )][210(106 )] + [(0.25 − d)(0.025)][210(106 )] − [d(0.025)][210(106 )] = 0 d = 0.24281 m < 0.25 m (o.k.!) = M p [π (0.052 − 0.0252 )][210(106 )](0.0571903) + [0.0071903(0.025)][210(106 )](0.0035951) + [0.242810(0.025)][210(106 )](0.121405) = 225.64(10 3 ) N= ⋅ m 226 kN ⋅ m
Ans.
Ans: Mp = 226 kN # m. 626
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The beam is made of an elastic perfectly plastic material for which sY = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment Mp is applied and then released.
t h t t
Solution
b
Plastic analysis: T1 = C1 = sYbt ; T2 = C2 = sY a Mp = sYbt(h  t) + sY a = sY 3 bt(h  t) +
h  2t bt 2
h  2t h  2t b(t)a b 2 2
t (h  2t)2 4 4
Elastic analysis: 1 1 I = bh3 (b  t)(h  2t)3 12 12 =
1 3 bh3  (b  t)(h  2t)3 4 12 MY =
syI c
=
=
1 ) [bh3  (b  t)(h  2t)3] sY ( 12 h 2
bh3  (b  t)(h  2t)3 6h
sY
Shape factor: k =
[bt(h  t) + 4t (h  2t)2]sY MP = bh3  (b  t)(h  2t)3 MY s 6h
=
Y
2 3h 4bt(h  t) + t(h  2t) c 3 d 3 2 bh  (b  t)(h  2t)
Ans.
Ans: k = 622
2 3h 4bt(h  t) + t(h  2t) c 3 d 3 2 bh  (b  t)(h  2t)
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6–162. The rod has a circular cross section. If it is made of an elastic perfectly plastic material, determine the shape factor.
100 mm
Solution Maximum Elastic Moment: Applying the flexure formula with s = sY and p p ( 0.14 ) = 25 ( 106 ) p m4, I = r4 = 4 4 s =
MY(0.1) Mc ; sY = I 25 ( 106 ) p MY = 0.25 ( 10  3 ) psY
Plastic Moment: Referring to Fig. a, p ( 0.1 ) pr 2 b = sY c d = 5 ( 103 ) psY 2 2 2
C = T = sY a
Thus,
d = 2a
4(0.1) 4r 0.26667 b = 2c = m d = p 3p 3p
Mp = Cd = ( 5 ( 103 ) p sY ) a
Shape Factor: k =
Mp MY
=
1.3333 ( 103 ) sY 0.25 ( 103 ) psY
0.26667 b = 1.3333 ( 103 ) sY p Ans.
= 1.6976 = 1.70
Ans: k = 1.70 623
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6–163. The rod has a circular cross section. If it is made of an elastic perfectly plastic material where sY = 345 MPa, determine the maximum elastic moment and plastic moment that can be applied to the cross section.
100 mm
Solution Maximum Elastic Moment: Applying the flexure formula with s = sy = 345 MPa p p ( 0.14 ) = 25 ( 106 ) p m4, and I = r 4 = 4 4 s =
MY(0.1) Mc ; 345 ( 106 ) = I 25 ( 106 ) p
MY = 270.96 ( 103 ) N # m = 271 kN # m
Ans.
Plastic Moment: Referring to Fig. a, p ( 0.1 ) pr 2 b = 345 ( 106 ) c d = 1.725 ( 106 ) p N 2 2 2
C = T = sy a Thus,
d = 2a
4(0.1) 4r 0.26667 b = 2c m d = p 3p 3p
MP = Cd =
3 1.725 ( 106 ) p 4 c
0.26667 d p
= 460 ( 103 ) N # m = 460 kN # m
Ans.
Ans: MY = 271 kN # m, MP = 460 kN # m 624
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*6–164. The beam is made of an elastic perfectly plastic material for which sY = 200 MPa. If the largest moment in the beam occurs within the center section a–a, determine the magnitude of each force P that causes this moment to be (a) the largest elastic moment and (b) the largest plastic moment.
P
P a
a 2m
2m
2m
2m
200 mm
Solution (1)
M = 2P
100 mm
a) Elastic moment I =
1 (0.1)(0.23) = 66.667(10  6) m4 12
sY = MY =
MYc I 200(106)(66.667)(10  6) 0.1
= 133.33 kN # m From Eq. (1) 133.33 = 2 P Ans.
P = 66.7 kN b) Plastic moment Mp = =
b h2 s 4 Y 0.1(0.22) 4
(200)(106)
= 200 kN # m From Eq. (1) 200 = 2 P Ans.
P = 100 kN
Ans: Elastic: P = 66.7 kN, Plastic: P = 100 kN 630
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6–165. Determine the shape factor of the beam’s cross section. in. 503 mm
Referring to Fig. a, the location of centroid of the crosssection is y =
∑ yA 0.125(0.1)(0.05) + 0.05(0.05)(0.1) = = 0.0875 m ∑A 0.1(0.05) + 0.05(0.1)
in. 1006mm
The moment of inertia of the crosssection about the neutral axis is 1 I = (0.05)(0.13 ) + 0.05(0.1)(0.0875 − 0.05)2 12
25 mm 1.5 in. 503 mm in. 25 mm 1.5 in.
1 (0.1)(0.053 ) + 0.1(0.05)(0.125 − 0.0875)2 + 12 = 19.27083(10 −6 ) m 4 Here σ max= σ Y and c= y= 0.0875 m. Thus
smax =
Mc ; I
σY =
MY (0.0875)
19.27083(10 −6 )
MY = 0.22024(10 −3 )σ Y
Referring to the stress block shown in Fig. b, sdA = 0; LA
T  C1  C2 = 0
[d(0.05)]σ Y − [(0.1 − d)(0.05)]σ Y − [0.05(0.1)]σ Y = 0 d = 0.1 m = Since d 0.1 = m, c1 0, Fig. c. Here T= C= [0.05(0.1)]σ Y= 0.005σ Y
Thus, = (0.075) 0.005σ Y= (0.075) 0.375(10 −3 )σ Y M p T=
Thus,
Ans.
Mp 0.375(10 −3 )σ Y k == = 1.7027 = 1.70 MY 0.22024(10 −3 )σ Y
0.05 m
0.05 m
0.075 m
0.1 – d 0.1 m
Ans: k = 1.70 449 449
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6–166. The beam is made of elasticperfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take sYY ==250 36 MPa. ksi.
503 mm in.
Referring to Fig. a, the location of centroid of the crosssection is y =
1006mm in.
∑ yA 0.125(0.1)(0.05) + 0.05(0.05)(0.1) = = 0.0875 m ∑A 0.1(0.05) + 0.05(0.1)
The moment of inertia of the crosssection about the neutral axis is
25 mm 1.5 in. 503 mm in.
1 I = (0.05)(0.13 ) + 0.05(0.1)(0.0875 − 0.05)2 12 +
25 mm 1.5 in.
1 (0.1)(0.053 ) + 0.1(0.05)(0.125 − 0.0875)2 12
= 19.27083(10 −6 ) m 4 Here σ max= σ Y =250 MPa and c/= y= 0.0875 m. Then
smax =
Mc ; I
250(106 ) =
MY (0.0875)
19.27083(10 −6 )
Ans.
MY = 55.06(10 3 ) N ⋅ m = 55.1 kN ⋅ m
Referring to the stress block shown in Fig. b, sdA = 0; LA
T  C1  C2 = 0
[d(0.05)][250(106 )] − [(0.1 − d)(0.05)][250(106 )] − [0.05(0.1)][250(106 )] = 0 d = 0.1 m = Since d 0.1 m, c1 0, =
Here, T= C= [0.05(0.1)][250(106 )]= 1250(10 3 ) N= 1250 kN
Thus, (0.075) (1250)(0.075) = 93.75 kN = M p T=
Ans.
0.05 m
0.05 m
0.1 – d
0.075 m 0.1 m
Ans:
MY = 55.1 kN # m, MP = 93.75 kN # m 450 450
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6–167. Determine the shape factor for the beam. 10 mm
15 mm 200 mm Mp 10 mm
Solution I =
200 mm
1 1 (0.2)(0.22)3 (0.185)(0.2)3 = 54.133(10  6) m4 12 12
C1 = sY(0.01)(0.2) = (0.002) sY C2 = sY(0.1)(0.015) = (0.0015) sY Mp = 0.002 sY(0.21) + 0.0015 sY(0.1) = 0.0005 sY MYc I
sY = MY = k =
sY(54.133)(10  6)
Mp MY
0.11 =
= 0.000492 sY
0.00057 sY = 1.16 0.000492 sY
Ans.
Ans: k = 1.16 629
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*6–168. The beam is made of an elastic perfectly plastic material for which sY = 250 MPa. Determine the residual stress in the beam at its top and bottom after the plastic moment Mp is applied and then released.
15 mm
20 mm 200 mm Mp 15 mm
Solution Ix =
200 mm
1 1 (0.2)(0.23)3 (0.18)(0.2)3 = 82.78333(10  6) m4 12 12
C1 = T1 = sY(0.2)(0.015) = 0.003 sY C2 = T2 = sY(0.1)(0.02) = 0.002 sY Mp = 0.003 sY(0.215) + 0.002 sY(0.1) = 0.000845 sY = 0.000845(250)(106) = 211.25 kN # m
s ′ =
Mpc I
=
y 0.115 = ; 250 293.5
211.25(103)(0.115) 82.78333(10  6)
= 293.5 MPa
y = 0.09796 m = 98.0 mm Ans.
stop = sbottom = 293.5  250 = 43.5 MPa
Ans: stop = sbottom = 43.5 MPa 640
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6–169. The box beam is made of an elastic perfectly plastic material for which sY = 250 MPa. Determine the residual stress in the top and bottom of the beam after the plastic moment Mp is applied and then released.
25 mm
Solution
150 mm 25 mm
Plastic Moment:
25 mm 150 mm 25 mm
MP = 250 11062 (0.2)(0.025)(0.175) + 250 11062 (0.075)(0.05)(0.075) = 289062.5 N # m
Modulus of Rupture: The modulus of rupture sr can be determined using the flexure formula with the application of reverse, plastic moment MP = 289062.5 N # m. I =
1 1 (0.2) 10.232 (0.15) 10.1532 12 12
= 91.14583 ( 10  6 ) m4 sr =
289062.5 (0.1) MP c = = 317.41 MPa I 91.14583 110  62
Residual Bending Stress: As shown on the diagram. = = stop = sbot = sr  sY
Ans.
= 317.14  250 = 67.1 MPa
Ans: stop = sbottom = 67.1 MPa 639
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6–170. Determine the shape factor of the cross section. a a a
Solution Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first. 1 1 INA = (a)(3a)3 + (2a) ( a3 ) = 2.41667 a4 12 12
a
a
a
Applying the flexure formula with s = sY , we have sY =
MY c I
MY =
sY ( 2.41667a4 ) sYI = = 1.6111a3sY c 1.5a
Plastic Moment: Mp = sY(a)(a)(2a) + sY(0.5a)(3a)(0.5a) = 2.75a3sY Shape Factor: k =
Mp MY
=
2.75a3sY 1.6111a3sY
Ans.
= 1.71
Ans: k = 1.71 638
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6–171. Determine the shape factor for the member having the tubular cross section. 2d
d
Solution Maximum Elastic Moment: The moment of inertia about neutral axis must be determined first. INA =
15p 4 p 4 p d 4 d  a b = d 4 4 2 64
Applying the flexure formula with s = sY , we have sY = MY =
MY c I 4 sY ( 15p sY I 15p 3 64 d ) = = d sY c d 64
Plastic Moment:
y =
ΣyA = ΣA
MP = sY a
4d pd 2 2 3p
(
pd 2 2
2
MP = MY
4 ( d2 ) 3p
pd 2 p 2 4d
2
Shape Factor: k =
) 
7 3 6 d sY 15p 3 64 d sY
b
a
p 2 4d
p 2 4d
2
b
=
14 d 9p
2
28 7 d = d 3 sY 9p 6
Ans.
= 1.58
Ans: k = 1.58 637
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*6–172.
Determine the shape factor for the member. –h 2
–h 2
Plastic analysis: T = C = MP =
h 1 bh (b) a b sY = s 2 2 4 Y
b
b h2 bh h sY a b = s 4 3 12 Y
Elastic analysis: I = 2c
1 h 3 b h3 (b)a b d = 12 2 48 3
sY A bh sYI 48 B b h2 MY = = = s h c 24 Y 2
Shape factor: k =
Mp MY
=
bh2 12
sY
bh2 24
sY
Ans.
= 2
Ans: k =
Mp MY
=
bh2 12
sY
bh2 24
sY
= 2
456 456
CH 06.indd 456
1/18/11 2:25:33 PM
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6–173. The member is made from an elasticplastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. mm, h =6 150 250 MPa. Take b == 100 sY =sY36= ksi. 4 in., h = in., mm,
–h 2
–h 2
Elastic analysis: 1 3 4 −6 I 2= = 12 (0.1)(0.075 ) 7.03125(10 ) m MY=
σY I c
=
b
[250(106 )][7.03125(10 −6 )] = 23.4375(10 3 ) N ⋅ m= 23.4 kN ⋅ m 0.075
Ans.
Plastic analysis:
h 3
1 T= C= (0.1)(0.075) [250(106 )]= 937.5(10 3 ) N 2 h 3 0.15 3 ⋅ m 46.9 kN ⋅ m = M p T= [937.5(10 )] = 46.875(10 ) N= 3 3
Ans.
Ans:
MY = 23.4 kN # m, MP = 46.9 kN # m 456 456
CH 06.indd 456
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6–174. Determine the shape factor of the cross section. a
2a
Solution
a
Maximum Elastic Moment: The centroid and the moment of inertia about neutral axis must be determined first. y = INA =
a 2
a 2
0.5a(a) (2a) + 2a(2a) (a) ΣyA = = 1.25a ΣA a(2a) + 2a(a) 1 (2a) ( a3 ) + 2a(a) (1.25a  0.5a)2 12 +
1 (a) (2a)3 + a(2a) (2a  1.25a)2 12
= 3.0833a4 Applying the flexure formula with s = sY , we have sY = MY =
MY c I sY ( 3.0833a4 ) sY I = c (3a  1.25a) = 1.7619a3 sY
Plastic Moment: LA
sdA = 0; T  C1  C2 = 0 sY (d) (a)  sY (2a  d) (a)  sY (a) (2a) = 0 d = 2a MP = sY (2a) (a) (1.5a) = 3.00a3 sY
Shape Factor: k =
MP 3.00a3 sY = = 1.70 MY 1.7619a3 sY
Ans.
Ans: k = 1.70 635
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6–175. The beam is made of elastic perfectly plastic material. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section. Take a = 50 mm and sY = 230 MPa.
a
2a
Solution
a
Maximum Elastic Moment: The centroid and the moment of inertia about neutral axis must be determined first. y = INA =
a 2
a 2
0.025(0.05) (0.1) + 0.1 (0.1) (0.05) ΣyA = = 0.0625 m ΣA 0.05(0.1) + 0.1(0.05) 1 (0.1) ( 0.053 ) + 0.1(0.05) (0.0625  0.025)2 12 +
1 (0.05) ( 0.13 ) + 0.05(0.1) (0.1  0.0625)2 12
= 19.2709 ( 106 ) m4 Applying the flexure formula with s = sY , we have sY = MY =
MY c I 230 ( 106 ) (19.2709) ( 106 ) sY I = c (0.15  0.0625)
= 50654.8 N # m = 50.7 kN # m
Ans.
Plastic Moment: LA
sdA = 0; T  C1  C2 = 0 sY (d) (0.05)  sY (0.1  d)(0.05)  sY (0.05)(0.1) = 0 d = 0.100 m MP = 230 ( 10
6
) (0.100) (0.05) (0.075)
= 86250 N # m = 86.25 kN # m
Ans.
Ans: MY = 50.7 kN # m, MP = 86.25 kN # m 636
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*6–176. The beam is made of elastic perfectly plastic material for which sY = 345 MPa. Determine the maximum elastic moment and the plastic moment that can be applied to the cross section.
100 mm 100 mm 100 mm
Solution Maximum Elastic Moment: Applying the flexure formula with s = sy = 345 MPa 1 1 and I = (0.1) ( 0.33 ) + (0.2) ( 0.13 ) = 0.24167 ( 103 ) m4, 12 12 s =
100 mm 100 mm 100 mm
MY(0.15) Mc ; 345 ( 106 ) = I 0.24167 ( 103 )
MY = 555.83 ( 103 ) N # m = 556 kN # m
Ans.
Plastic Moment: Referring to Fig. a, C1 = T1 = sy(0.1)(0.1) = 0.01sy = 0.01 3 345 ( 106 )4 = 3.45 ( 106 ) N
C2 = T2 = sy(0.3)(0.05) = 0.015sy = 0.015 3 345 ( 106 ) 4 = 5.175 ( 106 ) N d 1 = 4(0.05) = 0.2 m d 2 = 2(0.025) = 0.05 m Thus, MP = C1d 1 + C2d 2 =
3 3.45 ( 106 ) 4 (0.2)
+
3 5.175 ( 106 ) 4 (0.05)
= 948.75 ( 103 ) N # m = 949 kN # m
Ans.
Ans: MY = 556 kN # m, MP = 949 kN # m 631
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6–177. Determine the shape factor of the cross section. 100 mm 100 mm 100 mm
Solution Maximum Elastic Moment: Applying the flexure formula with s = sy and 1 1 I = (0.1) ( 0.33 ) + (0.2) ( 0.13 ) = 0.24167 ( 103 ) m4, 12 12 My(0.15) Mc s = ; sy = My = 1.6111 ( 10  3 ) sy I 0.24167 ( 103 ) Plastic Moment: Referring to Fig. a,
100 mm 100 mm 100 mm
C1 = T1 = sy(0.1)(0.1) = 0.01 sy C2 = T2 = sy(0.3)(0.05) = 0.015 sy d 1 = 4(0.05) = 0.2 m d 2 = 2(0.025) = 0.05 m Thus, Mp = C1d 1 + C2d 2 = (0.01 sy)(0.2) + (0.015 sy)(0.05) = 2.75 ( 103 ) sy Shape Factor:
k =
Mp My
=
2.75 ( 103 ) sy 1.6111 ( 103 ) sy
Ans.
= 1.7069 = 1.71
Ans: k = 1.71 632
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6–178. The plexiglass bar has a stress–strain curve that can be approximated by the straightline segments shown. Determine the largest moment M that can be applied to the bar before it fails.
s (MPa)
20 mm M 20 mm
failure
60 40
tension
0.06 0.04 0.02
Solution
P (mm/mm)
compression
Ultimate Moment: LA
0.04
80 100
s dA = 0; C  T2  T1 = 0
1 1 d 1 d sc (0.02  d)(0.02) d  40 ( 106 ) c a b(0.02) d  (60 + 40) ( 106 ) c (0.02) d = 0 2 2 2 2 2 Since
0.04 P = , then 0.02  d = 25Pd. d 0.02 d
And since
40 ( 106 ) s s = = 2 ( 109 ) , then P = . P 0.02 2 ( 109 )
So then 0.02  d =
23sd 2 ( 109 )
= 1.25 ( 10  8 ) sd.
Substituting for 0.02  d, then solving for s, yields s = 74.833 MPa. Then P = 0.037417 mm>mm and d = 0.010334 m. Therefore, 1 C = 74.833 ( 106 ) c (0.02  0.010334)(0.02) d = 7233.59 N 2 T1 =
1 0.010334 b d = 5166.85 N (60 + 40) ( 106 ) c (0.02)a 2 2
1 0.010334 T2 = 40 ( 106 ) c (0.02)a b d = 2066.74 N 2 2 y1 = y2 = y3 =
2 (0.02  0.010334) = 0.0064442 m 3 2 0.010334 a b = 0.0034445 m 3 2
0.010334 1 2(40) + 60 0.010334 + c1  a b da b = 0.0079225 m 2 3 40 + 60 2
M = 7233.59(0.0064442) + 2066.74(0.0034445) + 5166.85(0.0079255) = 94.7 N # m
Ans.
Ans: M = 94.7 N # m 641
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6–179. The beam is made of phenolic, a structural plastic, that has the stress–strain curve shown. If a portion of the curve can be represented by the equation s = (5(106)P)1/2 MPa, determine the magnitude w of the distributed load that can be applied to the beam without causing the maximum strain in its fibers at the critical section to exceed Pmax = 0.005 mm>mm.
w
150 mm 150 mm
2m
2m
s(MPa)
Solution
s2 5(106)P
Resultant Internal Forces: The resultant internal forces T and C can be evaluated from the volume of the stress block which is a paraboloid. When P = 0.005 mm>mm, then
P(mm/mm)
s = 25 ( 106 ) (0.005) = 158.11 MPa T = C =
2 3 158.11 ( 106 ) (0.075) 4 (0.150) = 1.1859 MN 3
3 d = 2 c (0.075) d = 0.090 m 5
Maximum Internal Moment: The maximum internal moment M = 2w occurs at the overhang support as shown on FBD. Mmax = Td 2w = 1.1859 ( 106 ) (0.090) Ans.
w = 53363 N>m = 53.4 kN>m
Ans: w = 53.4 kN>m 645
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*6–180. The stress–strain diagram for a titanium alloy can be approximated by the two straight lines. If a strut made of this material is subjected to bending, determine the moment resisted by the strut if the maximum stress reaches in. 75 3mm
a value of (a) sA and (b) sB.
M
in. 502 mm �s(MPa) (ksi) s
B
180 ss �1260 BB � 980 140 sA
A
0.01
0.04
(in./in.) P (mm/mm)
a) Maximum Elastic Moment : Since the stress is linearly related to strain up to point A, the flexure formula can be applied. sA =
Mc I
–1260 9.375 mm
=
b)
1 (0.05)(0.075 ) [980(10 )] 12 0.0375 3
1260
⋅ m 45.9 kN ⋅ m 45.9375(10 3 ) N=
Ans.
The Ultimate Moment :
28.125 mm 9.375 mm
12.5 mm
48.05 mm
28.125 mm
1 [(1260 + 980)(106 )](0.028125)(0.05) = 1575(10 3 ) N C1 = T1 = 2 C= 2 T= 2
s (MPa) 980
6
=
9.375 mm
e (mm/mm)–980
sA I M = c
1 [980(106 )](0.009375)(0.05) = 229.6875(10 3 ) N 2
= M [1575(10 3 )](0.04805) + [229.6875(10 3 )](0.0125) = 78.54(10 3 ) N= ⋅ m 78.5 kN ⋅ m
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment.
461 461
CH 06.indd 461
1/18/11 2:25:44 PM
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6–181. The bar is made of an aluminum alloy having a stress–strain diagram that can be approximated by the straight line segments shown. Assuming that this diagram is the same for both tension and compression, determine the moment the bar will support if the maximum strain at the top and bottom fibers of the beam is P max = 0.03.
�s (ksi) s(MPa) 90 630 80 560 60 420
in. M 1004mm
MPa s == 574 82 ksi
0.05
P (in./ in.) (mm/mm)
y1 = 10 mm
1 C1 = T1 = [(560 + 574)(106 )](0.008333)(0.075) = 354.375(10 3 ) N 2 yz = 41.667 mm
1 [(560 + 420)(106 )](0.031667)(0.075) = 1163.75(10 3 ) N C2 = T2 = 2 C= 3 T= 3
in. 753 mm
53.17 mm 13.33 mm
80  80 ss –560 63090– 560 5= ; ; 0.03 –0.025 0.025 0.05 0.05– 0.025  0.025
0.025
91.70 mm
0.006
420 MPa
560 MPa 574 MPa
1 [420(106 )](0.01)(0.075) = 157.5(10 3 ) N 2
M = [354.375(10 3 )](0.09170) + [1163.75(10 3 )](0.05317) + [157.5(10 3 )](0.01333) = 96.48(10 3 ) N= ⋅ m 96.5 kN ⋅ m
Ans.
Note: The centroid of a trapezodial area was used in calculation of moment areas.
Ans:
s = 574 MPa, M = 96.5 kN # m 463 463
CH 06.indd 463
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6–182. A beam is made from polypropylene plastic and has a stress–strain diagram that can be approximated by the curve shown. If the beam is subjected to a maximum tensile and compressive strain of P = 0.02 mm>mm, determine the moment M.
M
s (Pa)
s 10(106)P1/ 4
100 mm
M 30 mm
P (mm/ mm)
Solution Pmax = 0.02 smax = 10 1 106 2 (0.02)1>4 = 3.761 MPa 0.02 P = 0.05 y
P = 0.4 y
s = 10 1 106 2 (0.4)1>4y1>4 M =
LA
y s dA = 2
M = 0.47716 1 106 2 M =
251 N # m
L0
L0
0.05
0.05
y(7.9527) 1 106 2 y1>4(0.03)dy
4 y5>4dy = 0.47716 1 106 2 a b(0.05)9>4 9
Ans.
Ans: M = 251 N # m 643
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R6–1. Determine the shape factor for the wideflange beam. 20 mm
30 mm 180 mm Mp 20 mm
Solution I =
180 mm
1 1 (0.18)(0.223) (0.15)(0.183) 12 12
= 86.82(10  6) m4 Plastic moment: Mp = sY(0.18)(0.02)(0.2) + sY(0.09)(0.03)(0.09) = 0.963(10  3)sY Shape Factor: MY = k =
sY(86.82)(10  6) sYI = = 0.789273(10  3)sY c 0.11
Mp MY
=
0.963(10  3)sY 0.789273(10  3)sY
= 1.22
Ans.
Ans: k = 1.22 646
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R6–2. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment diagrams if it supports the distributed loading shown.
w
A 2/3 L
Solution + c ΣFy = 0;
2wL 1 w 2 x = 0 27 2 L x =
a+ ΣM = 0;
M +
4 L = 0.385 L A 27
1w 1 2wL (0.385L)2a b(0.385L) (0.385L) = 0 2L 3 27
M = 0.0190 wL2
647
C
B 1/3 L
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R6–3. y
The composite beam consists of a wood core and two plates of steel. If the allowable bending stress for the wood is (sallow)w = 20 MPa, and for the steel (sallow)st = 130 MPa, determine the maximum moment that can be applied to the beam. Ew = 11 GPa, Est = 200 GPa.
z 125 mm
M
Solution
x 9
200(10 ) Est = = 18.182 Ew 11(109) 1 I = (0.80227)(0.1253) = 0.130578(10  3)m4 12
20 mm 75 mm 20 mm
n =
Failure of wood: Mc I
(sw)max =
M(0.0625)
20(106) =
0.130578(10  3)
;
M = 41.8 kN # m
Failure of steel: (sst)max = 130(106) =
nMc I 18.182(M)(0.0625) 0.130578(10  3)
M = 14.9 kN # m (controls)
Ans.
Ans: M = 14.9 kN # m 648
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*R6–4. A shaft is made of a polymer having a parabolic upper and lower cross section. If it resists a moment of M = 125 N # m, determine the maximum bending stress in the material (a) using the flexure formula and (b) using integration. Sketch a threedimensional view of the stress distribution acting over the crosssectional area. Hint: The moment of inertia is determined using Eq. A–3 of Appendix A.
y
100 mm y 100 – z 2/ 25 M 125 N· m z
Solution Maximum Bending Stress: The moment of inertia about y axis must be determined first in order to use flexure formula I = = 2
LA
x
y2 dA 100 mm
L0
= 20
50 mm 50 mm
y2 (2z) dy
L0
100 mm
y2 2100  y dy
100 3 5 7 3 8 16 = 20J  y2 (100  y)2 y (100  y)2 (100  y)2 R ` mm 2 15 105 0
Thus,
= 30.4762 1 106 2 mm4 = 30.4762 1 10  6 2 m4
smax =
125(0.1) Mc = = 0.410 MPa I 30.4762(10  6)
Ans.
Maximum Bending Stress: Using integration dM = 2[y(s dA)] = 2b yc a M = 125 1 103 2 = 125 1 103 2 =
smax by d (2z dy) r 100
smax 100 mm 2 y 2100  y dy 5 L0
100 smax 3 5 7 3 8 16 J  y2(100  y)2 y(100  y)2 (100  y)2 R ` mm 5 2 15 105 0
smax (1.5238) 1 106 2 5
smax = 0.410 N>mm2 = 0.410 MPa
Ans.
Ans: smax = 0.410 MPa 649
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20 20�
Determine the maximum bending stress in the R6–5. N is handle of the cable cutter cutter at at section section a–a. a–a.A Aforce forceof of225 45 lb applied to the handles. The crosssectional area is shown in the figure.
a 100 mm 4 in.
225 45 lbN 125 mm 5 in.
75 mm 3 in. A
a
18 mm 0.75 in. 12 mm 0.50 in.
225 45 lbN
a + ©M = 0;
M − 225(0.125 + 0.1cos 20°) =0 = M 49.27 N ⋅ m
smax =
Mc = I
49.27(0.009) =
1 (0.012)(0.018 3 ) 12
0.1 m
6 76.03(10 ) N−m 2 76.0 MPa =
225 N 0.125 m
Ans.
Ans: s max = 76.0 MPa 468 468
CH 06.indd 468
1/18/11 2:25:55 PM
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R6–6. The curved beam is subjected to a bending m oment of M = 85 N # m as shown. Determine the stress at points A and B and show the stress on a volume element located at these points.
M 85 Nm
100 mm
A 400 mm
A
20 mm
15 mm
B
150 mm
30 B
Solution
20 mm
r2 dA 0.42 0.57 0.59 = b ln = 0.1 ln + 0.015 ln + 0.1 ln r1 0.40 0.42 0.57 LA r = 0.012908358 m A = 2(0.1)(0.02) + (0.15)(0.015) = 6.25(10  3) m2 R =
A LA
dA r
=
6.25(10  3) 0.012908358
= 0.484182418 m
r  R = 0.495  0.484182418 = 0.010817581 m sA =
M(R  rA) ArA(r  R)
=
85(0.484182418  0.59) 6.25(10  3)(0.59)(0.010817581)
= 225.48 kPa Ans.
sA = 225 kPa (C) sB =
M(R  rB ) ArB (r  R)
=
85(0.484182418  0.40) 6.25(10  3)(0.40)(0.010817581)
= 265 kPa (T)
Ans.
Ans: sA = 225 kPa (C), sB = 265 kPa (T) 651
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R6–7. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as 8 .m. functions of x, where 0 … x 6 1. 6 ft
40 kN 8 kip
30 kN/m 2 kip/ft 75 50 kNm kip�ft
x 1.8 6 ftm
+ c ©Fy = 0;
20 –30x 2x –VV= = 94 0 0
243.6 kN · m
Ans.
= V {94 − 30 x} kN c + ©MNA = 0;
1.2 4 ftm 40 kN
30 kN/m
75 kN · m
1.8 m
1.2 m
94 kN
x 94 x − 243.6 − 30 x − M = 0 2
V (kN) 94
{−15 x 2 + 94x − 243.6} kN ⋅ m M=
40
Ans. 0 M (kN · m) 0
x (m)
1.8
3
1.8
3
x (m)
–48 –123 30x
–243.6 243.6 (kN · m) 94 kN
Ans: V 94 30x, M 15x 2 94x 243.6 654
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*R6–8. A wooden beam has a square cross section as shown. Determine which orientation of the beam provides the greatest strength at resisting the moment M. What is the difference in the resulting maximum stress in both cases?
a
a
M
a
M
a (a)
(b)
Solution Case (a): smax =
M(a>2) Mc 6M = 1 4 = 3 I a 12 (a)
Case (b): I = 2c
3 1 2 1 1 1 2 1 1 2 aba bd d = 0.08333 a4 ab c a aba ab + a aba a 3 2 22 36 22 22 22 22
smax
1 ab Ma 8.4853 M Mc 22 = = = 4 I 0.08333 a a3
Case (a) provides higher strength, since the resulting maximum stress is less for a given M and a. ∆smax =
M 8.4853 M 6M  3 = 2.49 a 3 b 3 a a a
Ans.
Ans: Case (a), ∆smax = 2.49 a 653
M b a3
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R6–9. Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings. The bearings at A and B exert only vertical reactions on the shaft.
300 N 450 N A
B
200 mm
400 mm
300 mm
200 mm 150 N
Solution
654
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R6–10. The strut has a square cross section a by a and is subjected to the bending moment M applied at an angle u as shown. Determine the maximum bending stress in terms of a, M, and u. What angle u will give the largest bending stress in the strut? Specify the orientation of the neutral axis for this case.
y
a
z
x
a M
Solution Internal Moment Components: Mz = M cos u
My =  M sin u
Section Property: Iy = Iz =
1 4 a 12
Maximum Bending Stress: By inspection, maximum bending stress occurs at A and B. Applying the flexure formula for biaxial bending at point A smax = = =
Mzy Iz
+
My z Iy
 M cos u (a2) 1 12
a4
+
 M sin u (  a2) 1 12
a4
6M (cos u + sin u) a3
Ans.
ds 6M = 3 (  sin u + cos u) = 0 du a cos u  sin u = 0 Ans.
u = 45° Orientation of Neutral Axis: tan a =
Iz Iy
tan u
tan a = (1) tan (45°) Ans.
a = 45°
Ans: 6M (cos u + sin u), a3 u = 45°, a = 45° smax =
655
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7–1. If the wideflange beam is subjected to a shear of V = 20 kN, determine the shear stress on the web at A. Indicate the shearstress components on a volume element located at this point.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
Solution
20 mm
The moment of inertia of the crosssection about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10  3) m4 12 12
From Fig. a, QA = y′A′ = 0.16 (0.02)(0.2) = 0.64(10  3) m3 Applying the shear formula, tA =
20(103)[0.64(10  3)] VQA = It 0.2501(10  3)(0.02) = 2.559(106) Pa = 2.56 MPa
Ans.
The shear stress component at A is represented by the volume element shown in Fig. b.
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.
656
Ans: tA = 2.56 MPa
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7–2. If the wideflange beam is subjected to a shear of V = 20 kN, determine the maximum shear stress in the beam.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
Solution
20 mm
The moment of inertia of the crosssection about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10  3) m4 12 12
From Fig. a. Qmax = Σy′A′ = 0.16 (0.02)(0.2) + 0.075 (0.15)(0.02) = 0.865(10  3) m3 The maximum shear stress occurs at the points along neutral axis since Q is maximum and thickness t is the smallest. tmax =
20(103) [0.865(10  3)] VQmax = It 0.2501(10  3) (0.02) = 3.459(106) Pa = 3.46 MPa
Ans.
Ans: tmax = 3.46 MPa 657
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7–3. If the wideflange beam is subjected to a shear of V = 20 kN, determine the shear force resisted by the web of the beam.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
Solution
20 mm
The moment of inertia of the crosssection about the neutral axis is I =
1 1 (0.2)(0.343) (0.18)(0.33) = 0.2501(10  3) m4 12 12
For 0 … y 6 0.15 m, Fig. a, Q as a function of y is Q = Σy′A′ = 0.16 (0.02)(0.2) +
1 (y + 0.15)(0.15  y)(0.02) 2
= 0.865(10  3)  0.01y2 For 0 … y 6 0.15 m, t = 0.02 m. Thus, t =
20(103) 3 0.865(10  3)  0.01y2 4 VQ = It 0.2501(10  3) (0.02) =
5 3.459(106)
The sheer force resisted by the web is Vw = 2
L0
0.15 m
tdA = 2
L0
0.15 m
 39.99(106) y2 6 Pa
3 3.459(106)
6
2
 39.99(10 ) y
= 18.95 (103) N = 19.0 kN
4 (0.02 dy)
Ans.
Ans: Vw = 19.0 kN 658
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*7–4. If the beam is subjected to a shear of V = 30 kN, determine the web’s shear stress at A and B. Indicate the shearstress components on a volume element located at these points. Set w = 200 mm. Show that the neutral axis is located at y = 0.2433 m from the bottom and I = 0.5382(10−3) m4.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
Solution
20 mm
Section Properties: The location of the centroid measured from the bottom is y =
(0.01)(0.2)(0.02) + 0.22 (0.02)(0.4) + 0.430 (0.3)(0.02) 0.2 (0.02) + 0.02 (0.4) + 0.3 (0.02)
= 0.2433 m The moment of inertia of the crosssection about the neutral axis is 1 I = (0.2) ( 0.023 ) + 0.2 (0.02)(0.2433  0.01)2 12 1 + (0.02) ( 0.43 ) + 0.02 (0.4)(0.2433  0.22)2 12 1 + (0.3) ( 0.023 ) + 0.3(0.02)(0.43  0.2433)2 12 = 0.5382 ( 103 ) m4 Referring to Fig. a, QA = y′A A′A = 0.1867 [0.3(0.02)] = 1.12 ( 103 ) m3 QB = y′B A′B = 0.2333 [0.2(0.02)] = 0.9333 ( 103 ) m3 Shear Stress: Applying the shear formula, 30 ( 103 ) 3 1.12 ( 103 ) 4 VQA tA = = = 3.122 ( 106 ) Pa = 3.12 MPa I tA 0.5382 ( 103 ) (0.02) tB =
30 ( 103 ) 3 0.9333 ( 103 ) 4 VQB = = 2.601 ( 106 ) Pa = 2.60 MPa I tB 0.5382 ( 103 ) (0.02)
Ans.
Ans.
These shear stresses on the volume element at points A and B are shown in Fig. b.
Ans: tA = 3.12 MPa, tB = 2.60 MPa 659
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7–5. If the wideflange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam. Set w = 300 mm.
200 mm
A
20 mm
20 mm B
V 300 mm 200 mm
Solution
20 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is I =
1 1 (0.3) ( 0.443 ) (0.28) ( 0.43 ) = 0.63627 ( 103 ) m4 12 12
Maximum shear stress occurs at the neutral axis. Referring to Fig. a, Qmax = Σy′A′ = 0.21[0.3(0.02)] + 0.1[0.2(0.02)] = 1.66 ( 103 ) m3 Maximum Shear Stress: Applying the shear formula, tmax =
30 ( 103 ) 3 1.66 ( 103 ) 4 VQmax = It 0.63627 ( 103 ) (0.02)
= 3.913 ( 106 ) Pa = 3.91 MPa
Ans.
Ans: tmax = 3.91 MPa 660
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7–6. The wood beam has an allowable shear stress of tallow = 7 MPa. Determine the maximum shear force V that can be applied to the cross section.
50 mm
100 mm
50 mm
50 mm
200 mm V 50 mm
Solution I =
1 1 (0.2)(0.2)3 (0.1)(0.1)3 = 125(10  6) m4 12 12
tallow =
VQmax It
7(106) =
V[(0.075)(0.1)(0.05) + 2(0.05)(0.1)(0.05)] 125(10  6)(0.1) Ans.
V = 100 kN
Ans: Vmax = 100 kN 661
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7–7. The shaft is supported by a thrust bearing at A and a journal bearing at B. If P = 20 kN, determine the absolute maximum shear stress in the shaft.
A
C
1m
Solution
B
1m
D
1m P
P 30 mm
Support Reactions: As shown on the freebody diagram of the beam, Fig. a. Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = 20 kN.
40 mm
Section Properties: The moment of inertia of the hollow circular shaft about the neutral axis is p I = (0.044  0.034) = 0.4375(106)p m4 4 Qmax can be computed by taking the first moment of the shaded area in Fig. c about the neutral axis. Here, y′1 =
4(0.04) 3p
=
4(0.03) 4 1 m and y′2 = = m. Thus, 75p 3p 25p
Qmax = y′1A′1  y′2A′2
=
4 p 1 p c (0.042) d c (0.032) d = 24.667(106) m3 75p 2 25p 2
Shear Stress: The maximum shear stress occurs at points on the neutral axis since Q is maximum and the thickness t = 2(0.04  0.03) = 0.02 m is the smallest. tmax =
20(103)(24.667)(106) Vmax Qmax = = 17.9 MPa It 0.4375(106)p (0.02)
Ans.
Ans: tmax = 17.9 MPa 662
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*7–8. The shaft is supported by a thrust bearing at A and a journal bearing at B. If the shaft is made from a material having an allowable shear stress of tallow = 75 MPa, determine the maximum value for P.
A
C
1m
B
1m
1m P
P
Solution
D
30 mm
Support Reactions: As shown on the freebody diagram of the shaft, Fig. a. Maximum Shear: The shear diagram is shown in Fig. b. As indicated, Vmax = P.
40 mm
Section Properties: The moment of inertia of the hollow circular shaft about the neutral axis is p I = (0.044  0.034) = 0.4375(106)p m4 4 Qmax can be computed by taking the first moment of the shaded area in Fig. c about the neutral axis. Here, y′1 =
4(0.04) 3p
=
4(0.03) 4 1 m and y′2 = = m. Thus, 75p 3p 25p
Qmax = y′1A′1  y′2A′2
=
4 p 1 p c (0.042) d c (0.032) d = 24.667(106) m3 75p 2 25p 2
Shear Stress: The maximum shear stress occurs at points on the neutral axis since Q is maximum and the thickness t = 2(0.04  0.03) = 0.02 m. tallow =
Vmax Qmax ; It
75(106) =
P(24.667)(106) 0.4375(106)p(0.02)
P = 83 581.22 N = 83.6 kN
Ans.
Ans: P = 83.6 kN 663
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7–9. Determine the largest shear force V that the member can sustain if the allowable shear stress is tallow = 856ksi. MPa.
in. 753 mm 251mm in. V in.mm 753 mm in. 1 25
Solution
125in. mm
y =
(0.0125)(0.125)(0.025) 2[0.05(0.025)(0.05)] 0.029167 m 0.125(0.025) 2(0.025)(0.05)
I =
1 (0.125)(0.0253 ) + 0.125(0.025)(0.029167 − 0.0125)2 12
1 + 2 (0.025)(0.053 ) + 0.025(0.05)(0.05 − 0.029167)2 = 2.6367(10 −6 ) m 4 12
0.04583 m 0.022917 m
6 3 Qmax = ©y¿A¿ = 2[(0.022917)(0.025)(0.04583)] 52.5174(10 ) m
tmax = tallow =
56(106 )
VQmax It
V [52.5174(10 6 )] [2.6367(10 6 )][2(0.025)]
V = 140.58(10 3 ) N 141 kN
Ans.
Ans: V = 141 kN 666
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applied shear shearforce forceVV == 18 90 kip, kN, determine the 7–10. If the applied maximum shear stress in the member. 753mm in. 251mm in. V in.mm 753mm in. 125
Solution
25 mm 1 in.
y =
(0.0125)(0.125)(0.025) 2[0.05(0.025)(0.05)] 0.029167 m 0.125(0.025) 2(0.025)(0.05)
I =
1 (0.125)(0.0253 ) + 0.125(0.025)(0.029167 − 0.0125)2 12
1 + 2 (0.025)(0.053 ) + 0.025(0.05)(0.05 − 0.029167)2 = 2.6367(10 −6 ) m 4 12 6 3 Qmax = ©y¿A¿ = 2[(0.022917)(0.025)(0.04583)] 52.5174(10 ) m
tmax =
VQmax [90(10 3 )][52.5174(10 −6 ) m 3 ] 6 = 35.85(10 = ) N/m 3 35.9 MPa Ans. = It [2.6367(10 −6 )][2(0.025)]
0.04583 m 0.022917 m
Ans: tmax = 35.9 MPa 667
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7–11. w
The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kN>m. Determine the maximum shear stress in the beam.
A B 3m
3m 50 mm
100 mm
Solution tmax =
150(103) N (0.025 m)(0.05 m)(0.05 m) VQ = 1 3 It 12 (0.05 m)(0.1 m) (0.05 m) Ans.
tmax = 45.0 MPa Because the cross section is a rectangle, then also, tmax = 1.5
150(103) N V = 1.5 = 45.0 MPa A (0.05 m)(0.1 m)
Ans.
Ans: tmax = 45.0 MPa 666
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*7–12. A member has a cross section in the form of an equilateral triangle. If it is subjected to a shear force V, determine the maximum average shear stress in the member using the shear formula. Should the shear formula actually be used to predict this value? Explain.
a V
h
Solution I =
1 (a)(h)3 36
y h ; = x a>2 Q =
LA¿
Q = a
y =
y dA = 2 c a
2h x a
1 2 2 b (x)(y) a h  y b d 2 3 3
4h2 2x b (x2)a 1 b a 3a
t = 2x t = t =
V(4h2>3a)(x2)(1  2x VQ a) = 3 It ((1>36)(a)(h ))(2x) 24V(x  a2 x2) a2h
24V 4 dt = 2 2 a1  xb = 0 a dx ah At x = y =
a 4 h 2h a a b = a 4 2
tmax =
24V a 2 a a b a1  a b b 2 a 4 ah 4
tmax =
3V ah
Ans.
No, because the shear stress is not perpendicular to the boundary. See Sec. 73.
Ans. tmax = 483
3V ah
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7–13. Determine the shear stress at point B on the web of the cantilevered strut at section a–a.
2 kN 250 mm
a
250 mm
4 kN 300 mm
a
20 mm 70 mm
Solution (0.01)(0.05)(0.02) + (0.055)(0.07)(0.02)
y = I =
(0.05)(0.02) + (0.07)(0.02)
20 mm
B
50 mm
= 0.03625 m
1 (0.05) ( 0.023 ) + (0.05)(0.02)(0.03625  0.01)2 12 +
1 (0.02) ( 0.073 ) + (0.02)(0.07)(0.055  0.03625)2 = 1.78625 ( 106 ) m4 12
y′B = 0.03625  0.01 = 0.02625 m QB = (0.02)(0.05)(0.02625) = 26.25 ( 106 ) m3 tB =
6 ( 103 ) (26.25) ( 106 ) VQB = It 1.78622 ( 106 ) (0.02) Ans.
= 4.41 MPa
Ans: tB = 4.41 MPa 679
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7–14. Determine the maximum shear stress acting at section a–a of the cantilevered strut.
2 kN 250 mm
a
250 mm
4 kN 300 mm
a
20 mm 70 mm
Solution y = I =
(0.01)(0.05)(0.02) + (0.055)(0.07)(0.02) (0.05)(0.02) + (0.07)(0.02)
20 mm
B
50 mm
= 0.03625 m
1 (0.05) ( 0.023 ) + (0.05)(0.02)(0.03625  0.01)2 12 +
1 (0.02) ( 0.073 ) + (0.02)(0.07)(0.055  0.03625)2 = 1.78625 ( 106 ) m4 12
Qmax = y′A′ = (0.026875)(0.05375)(0.02) = 28.8906 ( 106 ) m3 tmax =
6 ( 103 ) (28.8906) ( 106 ) VQmax = It 1.78625 ( 106 ) (0.02) Ans.
= 4.85 MPa
Ans: tmax = 4.85 MPa 680
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7–15. Determine the maximum shear stress in the Tbeam at the critical section where the internal shear force is maximum.
10 kN/m
A
The FBD of the beam is shown in Fig. a,
1.5 m
3m
The shear diagram is shown in Fig. b. As indicated, Vmax = 27.5 kN
B
C
1.5 m
150 mm
The neutral axis passes through centroid c of the crosssection, Fig. c. ' 0.075(0.15)(0.03) + 0.165(0.03)(0.15) © y A = y = ©A 0.15(0.03) + 0.03(0.15)
150 mm
30 mm 30 mm
= 0.12 m 1 I = (0.03)(0.153) + 0.03(0.15)(0.12  0.075)2 12 +
1 (0.15)(0.033) + 0.15(0.03)(0.165  0.12)2 12
= 27.0 (10  6) m4 From Fig. d, Qmax = y¿A¿ = 0.06(0.12)(0.03) = 0.216 (10  3) m3 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =
27.5(103) C 0.216(10  3) D Vmax Qmax = It 27.0(10  6)(0.03) = 7.333(106) Pa = 7.33 MPa
Ans.
Ans. t max = 7.33 MPa 490
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*7–16.
10 kN/m
Determine the maximum shear stress in the Tbeam at point C. Show the result on a volume element at this point. A
B
C
1.5 m
3m
1.5 m
150 mm
Solution
150 mm
30 mm 30 mm
using the method of sections, + c ©Fy = 0;
VC + 17.5 
1 (5)(1.5) = 0 2
VC =  13.75 kN The neutral axis passes through centroid C of the crosssection, ©yA 0.075 (0.15)(0.03) + 0.165(0.03)(0.15) = ©A 0.15(0.03) + 0.03(0.15)
y =
= 0.12 m I =
1 (0.03)(0.15) + 0.03(0.15)(0.12  0.075)2 12
+
1 (0.15)(0.033) + 0.15(0.03)(0.165  0.12)2 12
= 27.0 (10  6) m4 Qmax = y¿A¿ = 0.06 (0.12)(0.03) = 0.216 (10  3) m3 490 The maximum shear stress occurs at points on the neutral axis since Q is maximum and thickness t = 0.03 m is the smallest. tmax =
13.75(103) C 0.216(10  3) D VC Qmax = It 27.0(10  6) (0.03)
= 3.667(106) Pa = 3.67 MPa
Ans.
Ans. t max = 3.67 MPa 491
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7–17. The strut is subjected to a vertical shear of V = 130 kN. Plot the intensity of the shearstress distribution acting over the crosssectional area, and compute the resultant shear force developed in the vertical segment AB.
B
150 mm 50 mm
A
Solution I =
V ⫽ 130 kN
150 mm
1 1 (0.05)(0.353) + (0.3)(0.053) = 0.18177083(103) m4 12 12
QC = y ¿A ¿ = 10.1210.05210.152 = 0.7511032 m3
150 mm 50 mm 150 mm
QD = ©y ¿A¿ = 10.1210.05210.152 + 10.0125210.35210.0252 = 0.85937511032 m3 t =
VQ It
1tC2t = 0.05 m =
= 10.7 MPa
1301103210.75211032
= 1.53 MPa
0.181770831103210.052
1tC2t = 0.35 m = tD =
1301103210.75211032
0.181770831103210.352
1301103210.859375211032 0.181770831103210.352
= 1.76 MPa
A¿ = 10.05210.175  y2 y¿ = y +
10.175  y2 2
=
1 10.175 + y2 2
Q = y¿A¿ = 0.02510.030625  y22 t =
=
VQ It 13010.025210.030625  y22 0.181770831103210.052
= 10951.3  357593.1 y2 VAB =
L
t dA
dA = 0.05 dy
0.175
=
L0.025
110951.3  357593.1y2210.05 dy2
0.175
=
L0.025
1547.565  17879.66y 2 dy 2
= 50.3 kN
Ans.
Ans: VAB = 50.3 kN 672
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7–18. Plot the shearstress distribution over the cross section of a rod that has a radius c. By what factor is the maximum shear stress greater than the average shear stress acting over the cross section? c y V
Solution x = 2c 2  y2 ;
I =
p 4 c 4
t = 2x = 22c 2  y2 dA = 2x dy = 22c 2  y2 dy dQ = ydA = 2y2c 2  y2 dy c
c
3 3 2 2 Q = 2y2c  y dy =  ( c 2  y2 ) 2 ` = ( c 2  y2 ) 2 3 3 Ly y
2
VQ t = = It
2
V 3 23 ( c 2  y2 ) 2 4 3
(
p 4 4c
) 1 22c  y 2 2
2
=
4V 2 ( c  y2 ) 3pc 4
The maximum shear stress occur when y = 0 tmax =
4V 3pc 2
tavg =
V V = A pc 2
The factor =
tmax = tavg
4V 3pc 2 V pc 2
=
4 3
Ans.
Ans: The factor = 671
4 3
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7–19. Determine the maximum shear stress in the strut if it is subjected to a shear force of V = 20 kN.
12 mm
60 mm V 12 mm 80 mm
20 mm
20 mm
Solution Section Properties: INA =
1 1 (0.12) ( 0.0843 ) (0.04) ( 0.063 ) 12 12
= 5.20704 ( 10  6 ) m4 Qmax = Σy′A′ = 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 ( 10  6 ) m3 Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax = =
VQmax It 20(103)(87.84)(10  6) 5.20704(10  6)(0.08) Ans.
= 4. 22 MPa
Ans: tmax = 4.22 MPa 668
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Determine the maximum shear force V that the strut can support if the allowable shear stress for the material is tallow = 40 MPa.
12 mm
60 mm V 12 mm 80 mm
20 mm
20 mm
Solution Section Properties: INA =
1 1 (0.12) ( 0.0843 ) (0.04) ( 0.063 ) 12 12
= 5.20704 ( 10  6 ) m4 Qmax = Σy′A′ = 0.015(0.08)(0.03) + 0.036(0.012)(0.12) = 87.84 ( 10  6 ) m3 Allowable Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section. Applying the shear formula tmax = tallow = 40 ( 106 ) =
VQmax It V(87.84)(10  6) 5.20704(10  6)(0.08) Ans.
V = 189 692 N = 190 kN
Ans: Vmax = 190 kN 669
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7–21. Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum.
3 kN/m 200 lb/ft
2.5 150kN/m lb/ft
D D
A A 26 m ft
26 m ft
0.6 2 ftm
100 mm 4 in.
18 mm 0.75 in.
150 mm 6 in.
120.5 mm in.
18 mm 0.75 in.
4 in. 100 mm
Solution Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, Vmax = 4.783 kN. 2.5(2) = 5 kN
1 2 (3)(2) = 3 kN
Section Properties: INA =
1 1 (0.1)(0.186 3 ) − (0.088)(0.153 ) = 28.8738(10 −6 ) m 4 12 12
1m
Qmax = ©y¿A¿
1.333 m
2.267 m
3.217 kN
4.783 kN
−3
3
= 0.084(0.1)(0.018) + 0.0375(0.012)(0.075) = 0.18495(10 ) m Maximum Shear Stress: Maximum shear stress occurs at the point where the neutral axis passes through the section.
V (kN) 4.783 2
Applying the shear formula
4.6
2.6
–0.217
tmax
VQmax = It =
–3.217
[4.783(10 3 )][0.18495(10 −3 )] 6 = 2.553(10 = ) N/m 3 2.55 MPa [28.8738(10 −6 )](0.012)
Ans.
0.1 m 0.018 m y2 = 0.084 m
0.075 m
y1 = 0.0375 m
0.012 m
0.075 m 0.018 m
Ans. tmax = 2.55 MPa 492
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If the beam is subjected to a shear of V = 15 kN, determine the web’s shear stress at A and B. Indicate the shearstress components on a volume element located at these points. Set w = 125 mm. Show that the neutral axis is located at y = 0.1747 m from the bottom and INA = 0.2182(10 −3) m4.
200 mm A
30 mm
25 mm V B
250 mm 30 mm
w
Solution (0.015)(0.125)(0.03) + (0.155)(0.025)(0.25) + (0.295)(0.2)(0.03)
y = I =
0.125(0.03) + (0.025)(0.25) + (0.2)(0.03)
= 0.1747 m
1 (0.125)(0.033) + 0.125(0.03)(0.1747  0.015)2 12
+
1 (0.025)(0.253) + 0.25(0.025)(0.1747  0.155)2 12
+
1 (0.2)(0.033) + 0.2(0.03)(0.295  0.1747)2 = 0.218182(103) m4 12
QA = yA′A = (0.310  0.015  0.1747)(0.2)(0.03) = 0.7219(103) m3 QB = yA′B = (0.1747  0.015)(0.125)(0.03) = 0.59883(103) m3 tA =
15(103)(0.7219)(103) VQA = = 1.99 MPa It 0.218182(103)0.025
Ans.
tB =
15(103)(0.59883)(103) VQB = = 1.65 MPa It 0.218182(103)0.025
Ans.
Ans: tA = 1.99 MPa, tB = 1.65 MPa 672
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If the wideflange beam is subjected to a shear of V = 30 kN, determine the maximum shear stress in the beam. Set w = 200 mm.
200 mm A
30 mm
25 mm V B
250 mm 30 mm
w
Solution Section Properties: I =
1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10)6 m4 12 12
Qmax = ΣyA = 0.0625(0.125)(0.025) + 0.140(0.2)(0.030) = 1.0353(10)3 m3 tmax =
30(10)3 (1.0353)(10)3 VQ = = 4.62 MPa It 268.652(10)6(0.025)
Ans.
Ans: tmax = 4.62 MPa 673
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If the wideflange beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam. Set w = 200 mm.
200 mm A
30 mm
25 mm V B
250 mm 30 mm
w
Solution I =
1 1 (0.2)(0.310)3 (0.175)(0.250)3 = 268.652(10)  6 m4 12 12
Q = a tf = Vf =
0.155 + y b(0.155  y)(0.2) = 0.1(0.024025  y2) 2
30(10)3(0.1)(0.024025  y2) 268.652(10)6(0.2) L
0.155
tf dA = 55.8343(10)6
L0.125
(0.024025  y2)(0.2 dy) 0.155
1 = 11.1669(10) c 0.024025y  y3 d 3 6
0.125
Vf = 1.457 kN
Ans.
Vw = 30  2(1.457) = 27.1 kN
Ans: Vw = 27.1 kN 674
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7–25. Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent to the maximum shear stress.
P
h L
b
Solution Vmax = P Mmax = PL smax =
PL ( h>2 ) Mc PLh = = I I 2I
tmax =
P ( h>2 ) (b) ( h>4 ) VQ Ph2 = = It Ib 8I
Require, smax = tmax PLh Ph2 = 2I 8I L =
h 4
Ans.
Ans: L = 675
h 4
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7–26. The steel rod is subjected to a shear of 150 kN. Determine the shear stress at point A. Show the result on a volume element at this point. 25 mm
The moment of inertia of the circular crosssection about the neutral axis (x axis) is I =
A
50 mm
p 4 p 44 r = (2 in4 p(106) mm4 (50) )== 4p 1.5625 4 4
150 k N
Q for the differential area shown in Fig. a is dQ = ydA = y (2xdy) = 2xy dy 1
1
2 , Then (4  y–2)y22, )Then However, from the equation of the circle, x = (2500 1 2
1
dQ = 2y (2500 (4  y–2y)22)dydy Thus, Q for the area above y is 2 in. 50 mm
Q =
Ly
= 
1
1
2 2 2y2y(2500 (4  y2–)2ydy ) dy
3 23 in.50 mm 3 3 2 2 2 (2500 (4  y–2)y22)` 2 0 = =(4 (2500  y2)2– y2) 2 y 3 3 3 y
1
1
2 (4  y–2)y22. )Thus, Here t = 2x = 2 (2500 . Thus
30 150(10 )2 D C 23 (4 3)[y232(2500 VQ – y2) 2 ] = t = 1 1 2 2 It 4p )D C 2(4  y6)[2(2500 1.5625p(10 – y2) 2 ] 54 2 t = (4 (2500  y2) – yksi ) Ksi 2p 125p 3
3
Thus For point A, y = 125inmm. . Thus tA =
54 2 (4 (2500  12)–=252.39 ksi MPa )= 19.1 2p 125p
Ans.
The state of shear stress at point A can be represented by the volume element shown in Fig. b.
19.1 MPa
50 mm
x = (2500 – y2)1/2
Ans: P = 5.9988 kN 678
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7–27. The beam is slit longitudinally along both sides. If it is subjected to a shear of V = 250 kN, compare the maximum shear stress in the beam before and after the cuts were made.
25 mm 200 mm V 100 mm
25 mm 25 mm
Solution Section Properties: The moment of inertia of the cross section about the neutral axis is
25 mm
25 mm 200 mm
1 1 I = (0.2)(0.23) (0.125)(0.153) = 98.1771(106) m4 12 12 Qmax is the first moment of the shaded area shown in Fig. a about the neutral axis. Thus, Qmax = 3y′1A′1 + y′2 A′2 = 3 3 0.0375(0.075)(0.025) 4 + 0.0875(0.025)(0.2) = 0.6484375(103) m3
Maximum Shear Stress: The maximum shear stress occurs at the points on the neutral axis since Q is maximum and t is minimum. Before the cross section is slit, t = 3(0.025) = 0.075 m. tmax =
250(103)(0.6484375)(103) VQmax = = 22.0 MPa It 98.1771(106)(0.075)
Ans.
After the cross section is slit, t = 0.025 m. (tmax)s =
250(103)(0.6484375)(103) VQmax = = 66.0 MPa It 98.1771(106)(0.025)
Ans.
Ans: tmax = 22.0 MPa, (tmax)s = 66.0 MPa 682
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*7–28. The beam is to be cut longitudinally along both sides as shown. If it is made from a material having an allowable shear stress of tallow = 75 MPa, determine the maximum allowable shear force V that can be applied before and after the cut is made.
25 mm 200 mm V 100 mm
25 mm 25 mm
Solution Section Properties: The moment of inertia of the cross section about the neutral axis is I =
25 mm
25 mm 200 mm
1 1 (0.2)(0.23) (0.125)(0.153) = 98.1771(106) m4 12 12
Qmax is the first moment of the shaded area shown in Fig. a about the neutral axis. Thus, Qmax = 3y′1A′1 + y′2 A′2 = 3(0.0375)(0.075)(0.025) + 0.0875(0.025)(0.2) = 0.6484375(103) m3 Shear Stress: The maximum shear stress occurs at the points on the neutral axis since Q is maximum and thickness t is minimum. Before the cross section is slit, t = 3(0.025) = 0.075 m. tallow =
VQmax ; It
75(106) =
V(0.6484375)(103) 98.1771(106)(0.075) Ans.
V = 851 656.63 N = 852 kN After the cross section is slit, t = 0.025 m. tallow =
VQmax ; It
75(106) =
Vs (0.6484375)(103) 98.1771(106)(0.025) Ans.
Vs = 283 885.54 N = 284 kN
Ans: V = 852 kN, Vs = 284 kN 683
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7–29. The composite beam is constructed from wood and reinforced with a steel strap. Use the method of Sec. 6.6 and calculate the maximum shear stress in the beam when it is subjected to a shear of V = 50 kN. Take Est = 200 GPa, Ew = 15 GPa.
10 mm V= 50 kN 300 mm
10 mm 175 mm
Solution bst = nbw = I =
15 (0.175) = 0.013125 m 200
1 1 (0.175) ( 0.323 ) (0.175  0.013125) ( 0.33 ) = 0.113648 ( 103 ) m4 12 12
Qmax = Σy′A′ = 0.075(0.013125)(0.15) + 0.155(0.175)(0.01) = 0.4189 ( 103 ) m3 tmax = n
50 ( 103 ) (0.4189) ( 103 ) VQmax 15 = a b It 200 0.113648 ( 103 ) (0.013125)
Ans.
= 1.05 MPa
Ans: tmax = 1.05 MPa 684
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7–30. The beam has a rectangular cross section and is subjected to a load P that is just large enough to develop a fully plastic moment Mp = PL at the fixed support. If the material is elastic perfectly plastic, then at a distance x 6 L the moment M = Px creates a region of plastic yielding with an associated elastic core having a height 2y′. This situation has been described by Eq. 6–30 and the moment M is distributed over the cross section as shown in Fig. 6–48e. Prove that the maximum shear stress in the beam is given by tmax = 32(P>A′), where A′ = 2y′b, the crosssectional area of the elastic core.
P x Plastic region 2y¿
h
b Elastic region
L
Solution Force Equilibrium: The shaded area indicates the plastic zone. Isolate an element in the plastic zone and write the equation of equilibrium. { ΣFx = 0;
tlong A2 + sg A1  sg A1 = 0 tlong = 0
This proves that the longitudinal shear stress. tlong, is equal to zero. Hence the corresponding transverse stress, tmax, is also equal to zero in the plastic zone. Therefore, the shear force V = P is carried by the material only in the elastic zone. Section Properties: INA =
1 2 (b)(2y′)3 = b y′3 12 3 y′ y′2b (y′)(b) = 2 2
Qmax = y′ A′ =
Maximum Shear Stress: Applying the shear formula tmax However,
VQmax = = It
1
V1
2 3
y′2b 2
A′ = 2by′ hence tmax =
2
by′ 2 (b) 3
=
3P 4by′
3P ‚ (Q.E.D.) 2A′
Ans: N/A 685
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7–31. The beam in Fig. 6–48f is subjected to a fully plastic moment Mp . Prove that the longitudinal and transverse shear stresses in the beam are zero. Hint: Consider an element of the beam shown in Fig. 7–4d.
Solution Force Equilibrium: If a fully plastic moment acts on the cross section, then an element of the material taken from the top or bottom of the cross section is subjected to the loading shown. For equilibrium { ΣFx = 0;
sg A1 + tlong A2  sg A1 = 0 tlong = 0
Thus no shear stress is developed on the longitudinal or transverse plane of the element. (Q. E. D.)
Ans: N/A 686
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*7–32. 20 mm
The double Tbeam is fabricated by welding the three plates together as shown. Determine the shear stress in the weld necessary to support a shear force of V = 80 kN.
150 mm V 50 mm
Solution
20 mm
y =
0.01(0.215)(0.02) + 2[0.095(0.15)(0.02)] ΣyA = = 0.059515 m ΣA 0.215(0.02) + 2(0.15)(0.02)
I =
1 (0.215) ( 0.023 ) + 0.215(0.02)(0.059515  0.01)2 12 + 2c
75 mm
50 mm 20 mm
1 (0.02) ( 0.153 ) + 0.02(0.15)(0.095  0.059515)2 d = 29.4909 ( 106 ) m4 12
y′ = 0.059515  0.01 = 0.049515 m
Q = y′A′ = 0.049515(0.215)(0.02) = 0.2129 ( 103 ) m3 Shear stress: t =
80 ( 103 ) (0.2129) ( 103 ) VQ = It 29.4909 ( 106 ) (2)(0.02) Ans.
= 14.4 MPa
Ans: t = 14.4 MPa 691
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The double Tbeam is fabricated by welding the three plates together as shown. If the weld can resist a shear stress tallow = 90 MPa, determine the maximum shear V that can be applied to the beam.
20 mm
150 mm V 50 mm
Solution
20 mm
y =
0.01(0.215)(0.02) + 2[0.095(0.15)(0.02)] ΣyA = = 0.059515 m ΣA 0.215(0.02) + 2(0.15)(0.02)
I =
1 (0.215) ( 0.023 ) + 0.215(0.02)(0.059515  0.01)2 12 + 2c
75 mm
50 mm 20 mm
1 (0.02) ( 0.153 ) + 0.02(0.15)(0.095  0.059515)2 d = 29.4909 ( 106 ) m4 12
y′ = 0.059515  0.01 = 0.049515 m
Q = y′A′ = 0.049515(0.215)(0.02) = 0.2129 ( 103 ) m3 t = 90 ( 106 ) =
VQ It V(0.2129) ( 103 ) 29.491 ( 106 ) (2)(0.02) Ans.
V = 499 kN
Ans: V = 499 kN 692
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7–34. The beam is constructed from two boards fastened in. apart. three rows rows of ofnails nailsspaced spaceds s= =502mm together with three apart. If nail can can support support aa2.25kN 450lb shear each nail shear force, force, determine the maximum shear force V that can be applied to the beam. The allowable shear stress for the wood is ttallow == 2.1 300MPa. psi.
s s 1.5 in. 40 mm 1.5 in. 40 mm
V
Solution
6150 in.mm
The moment of inertia of the crosssection about the neutral axis is I =
1 (0.15)(0.08 3 ) = 6.40(10 −6 ) m 4 12
Refering to Fig. a,
0.15 m −3
QA = Qmax = y¿A¿ = 0.02(0.15)(0.04) = 0.12(10 ) m
3
The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 0.15 m. tallow =
VQmax ; It
2.1(106 ) =
0.04 m
V [0.12(10 −3 )] [6.40(10 −6 )](0.15)
y = 0.02 m
3 = V 16.8(10 = ) N 16.8 kN
Shear Flow: Since there are three rows of nails, 2.25(10 3 ) F 3 qallow = 3 a b = 3 = 135(10 ) N/m s 0.05 qallow =
VQA ; I
135(10 3 ) =
V [0.12(10 −3 )] 6.40(10 −6 )
V = 7.20(10 3 ) N = 7.20 kN
Ans.
Ans. V = 7.20 kN 497
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7–35. The beam is constructed from two boards fastened together with three rows of nails. If the allowable shear stress for the the wood wood isis ttallow = 1150MPa. psi, determine the allow = maximum shear force V that can be applied to the beam. Also, find the maximum spacing s of the nails if each nail can resist 3.25 kNininshear. shear. 650 lb
s s 1.5 in. 40 mm V
1.5 in. 40 mm
6150 in.mm
Solution The moment of inertia of the crosssection about the neutral axis is I =
1 (0.15)(0.08 3 ) = 6.40(10 −6 ) m 4 12
Refering to Fig. a, QA = Qmax = y¿A¿ = 0.02(0.15)(0.04) = 0.12(10 −3 ) m 3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 0.15 m. tallow =
VQmax ; It
1(106 ) =
V [0.12(10 −3 )] [6.40(10 −6 )](0.15)
3 = V 8.00(10 = ) N 8.00 kN
Ans.
2.25(10 3 ) 9750 Since there are three rows of nails, qallow = 3 = N/m s s qallow =
VQA ; I
9750 [8(10 3 )][0.12(10 −3 )] = s 6.40(10 −6 )
s = 0.0650 m = 65.0 mm
Ans.
0.15 m
0.04 m
y = 0.02 m
Ans. V = 8.00 kN, s = 65.0 mm 498
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*7–36. The beam is constructed from four boards which are nailed together. If the nails are on both sides of the beam and each can resist a shear of 3 kN, determine the maximum load P that can be applied to the end of the beam.
3 kN
A
P
B
C
2m
2m
100 mm
30 mm 150 mm
30 mm
250 mm 30 mm
30 mm
Solution Support Reactions: As shown on FBD. Internal Shear Force: As shown on shear diagram, VAB = (P + 3) kN. Section Properties: INA =
1 1 (0.31) A 0.153 B (0.25) A 0.093 B 12 12
= 72.0 A 10  6 B m4
Q = y¿A¿ = 0.06(0.25)(0.03) = 0.450 A 10  3 B m3 Shear Flow: There are two rows of nails. Hence the allowable shear flow is 3(2) = 60.0 kN>m. q = 0.1 VQ q = I (P + 3)(103)0.450(10  3) 60.0 A 103 B = 72.0(10  6) P = 6.60 kN
Ans.
Ans. P = 6.60 kN 507
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7–37. The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 150 a thickness 12in. mm. shearofofV V= kN is 6 in.mm andand a thickness ofof0.5 If If a ashear = 250 50 kip applied to the cross section, determine the maximum spacing 75 kip. kN. of the bolts. Each bolt can resist a shear force of 15
0.5 in. 12 mm ss
75 mm 3 in. 125in.mm A A V V
1506mm in.
Solution
0.5 in. 12 mm
N N
Section Properties: INA
375in.mm
1 1 = (0.075)(0.224 3 ) − (0.063)(0.2 3 ) 12 12 −
75 mm
1 1 (0.012)(0.053 ) + (0.024)(0.153 ) 12 12
12 mm 12 mm
= 34.8714(10 −6 ) m 4
106 mm
−3
75 mm 75 mm
62.5 mm
25 mm 25 mm 75 mm 75 mm
3
0.15165(10 ) m Q = ©y¿A¿ = 0.0625(0.012)(0.075) + 0.106(0.075)(0.012) =
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(75)(10 3 ) 150(10 3 ) = q = . s s
12 mm 12 mm
q =
VQ I
150(10 3 ) [250(10 3 )][0.15165(10 −3 )] = s 34.8714(10 −6 ) = s 0.1380 = m 138 mm
Ans.
Ans. s = 138 mm 499
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7–38. The beam is fabricated from two equivalent structural tees and two plates. Each plate has a height of 6 in.mm andand a thickness ofof0.5 If Ifthe 150 a thickness 12 in. mm. thebolts boltsare are spaced spaced at 8 in., determine the s ==200 mm, determine themaximum maximumshear shearforce force V V that that can appliedtotothe the cross section. Each can resist be applied cross section. Each bolt bolt can resist a sheara shear of force of 15 kip. force 75 kN.
12 mm 0.5 in. s
753 mm in. mm 125in. A
Solution
1506mm in.
V
N
12 mm 0.5 in. mm 375in.
Section Properties: INA =
1 1 (0.075)(0.224 3 ) − (0.063)(0.2 3 ) 12 12
75 mm
1 1 − (0.012)(0.053 ) + (0.024)(0.153 ) 12 12
12 mm 12 mm
106 mm
= 34.8714(10 −6 ) m 4 −3
75 mm 75 mm
62.5 mm
25 mm 25 mm 75 mm 75 mm
3
0.15165(10 ) m Q = ©y¿A¿ = 0.0625(0.012)(0.075) + 0.106(0.075)(0.012) =
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is = q
3
12 mm 12 mm
2(75)(10 ) = 750(10 3 ) N/m 0.2
q = 750(10 3 ) =
VQ I V [0.15165(10 −3 )] 34.8714(10 −6 )
3 = V 172.46(10 ) N 172 kN =
Ans.
Ans. INA = 34.8714(10 6) m4, Q = 0.15165 (10 3) m3 V =172 kN 499
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7–39. The doubleweb girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. If each lb in single each fastener fastener can can support support 600 3 kN shear, determine the required spacing s of the fasteners needed to support support the the loading loading PP == 3000 15 kN. lb. Assume Assume A is pinned and B is a roller.
2 in. 50 mm 2 in. 50 mm
P s
10 in. 250 mm
A
4 ftm 1.2
B
4 ftm 1.2
2 in. 50 mm 2 in. 50 mm 6 in. 150 mm 0.5mm in. 0.5mm in. 12 12
Solution Support Reactions: As shown on FBD.
15 kN
7.5 kN. lb. Internal Shear Force: As shown on shear diagram, Vmax = 1500 Section Properties: INA =
7.5 kN
1 1 (0.174)(0.453 ) − (0.15)(0.253 ) = 1.126(10 −3 ) m 4 12 12
7.5 kN
V (kN) 7.5
−3 3 Q = y¿A¿ = 0.175(0.15)(0.1) = 2.625(10 ) m
x (m) –7.5
Shear Flow: Since there are two shear planes on the bolt, the allowable shear flow is 2(600) 3) 1200 2(3)(10 6000 = . . q = ss s s VQ q = I 3
12 mm
150 mm
12 mm
100 mm 175 mm
125 mm
−3
6000 7.5(10 )[2.625(10 )] = s 1.126(10 −3 )
125 mm
= s 0.3432 = m 343 mm
Ans.
100 mm
Ans. s = 343 mm 501
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*7–40. The doubleweb girder is constructed from two plywood sheets that are secured to wood members at its top and bottom. The allowable bending stress for the wood is MPa the allowable = andand the allowable shearshear stress stress is tallowis =tallow sallow = 56 8 ksi 3 ksi. 21the MPa. If the fasteners are s = 150 and each If fasteners are spaced s spaced = 6 in. and eachmm fastener can fastener600 canlbsupport kN indetermine single shear, determineload the support in single3 shear, the maximum maximum load P thattocan applied to the beam. P that can be applied thebebeam.
mm 250in. mm 250in.
P s
250 mm 10 in.
A
1.2 4 ftm
B
1.2 4 ftm
mm 250in. mm 250in. 6 in. 150 mm 12 12 0.5mm in. 0.5mm in.
Solution Support Reactions: As shown on FBD.
Internal Shear Force and Moment: As shown on shear and moment diagram, Vmax = 0.500P and Mmax = 2.00P. Section Properties: 1.2 m
INA
1.2 m
1 1 (0.174)(0.453 ) − (0.15)(0.253 ) = 1.126(10 −3 ) m 4 = 12 12
Q = yœ2 A¿ = 0.175(0.15)(0.1) = 2.625(10 −3 ) m 3 2.4
3.2325(10 −3 ) m 3 Qmax = ©y¿A¿ = 0.175(0.15)(0.1) + 0.1125(0.024)(0.225) =
Shear Flow: Assume bolt failure. Since there are two shear planes on the bolt, the 2(3)(10 3 ) allowable shear = flow is q = 40(10 3 ) N/m 0.15 VQ q = I 40(10 3 ) =
x (m) 1.2
0.012 m
Ans.
(Controls!)
2.4
tmax = tallow =
VQmax It
0.15 m
0.012 m
0.1 m 0.175 m
Shear Stress: Assume failure due to shear stress.
21(10 ) =
0.6P
0.5P[2.625(10 −3 )]
1.126(10 −3 ) P = 34.32(10 3 ) N = 34.3 kN
6
x (m)
1.2
0.125 m 0.125 m 0.1 m
0.5P[3.2325(10 −3 )] [1.126(10 −3 )](0.024)
3 P 351.12(10 ) N 351 kN = =
Bending Stress: Assume failure due to bending stress. smax = sallow = 56(106 ) =
Mc I 0.6P(0.225) 1.126(10 −3 )
3 P 467.08(10 = = ) N 467 kN
Ans. P = 34.3 kN 502
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A beam is constructed from three boards bolted together as shown. Determine the shear force in each bolt if the bolts are spaced s = 250 mm apart and the shear is V = 35 kN.
25 mm 25 mm 100 mm 250 mm
V
Solution y =
2 (0.125)(0.25)(0.025) + 0.275 (0.35)(0.025) 2 (0.25)(0.025) + 0.35 (0.025)
I = (2) a +
350 mm
s 250 mm
= 0.18676 m
25 mm
1 b(0.025)(0.253) + 2 (0.025)(0.25)(0.18676  0.125)2 12
1 (0.025)(0.35)3 + (0.025)(0.35)(0.275  0.18676)2 12
= 0.270236 (10  3) m4 Q = y′A′ = 0.06176(0.025)(0.25) = 0.386(10  3) m3 q =
35(0.386)(103) VQ = = 49.997 kN>m I 0.270236(103) Ans.
F = q(s) = 49.997 (0.25) = 12.5 kN
Ans: F = 12.5 kN 694
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7–42. The simply supported beam is built up from three boards by nailing them together as shown. The wood has an allowable shear stress of tallow = 1.5 MPa, and an allowable bending stress of sallow = 9 MPa. The nails are spaced at s = 75 mm, and each has a shear strength of 1.5 kN. Determine the maximum allowable force P that can be applied to the beam.
P
s
A
B 1m
1m 100 mm
Solution
25 mm
Support Reactions: As shown on the freebody diagram of the beam shown in Fig. a. Maximum Shear and Moment: The shear diagram is shown in Fig. b. As indicated, P Vmax = . 2 Section Properties: The moment of inertia of the cross section about the neutral axis is 1 1 I = (0.1)(0.253) (0.075)(0.23) 12 12 = 80.2083(10  6) m4 Referring to Fig. d, QB = y ′2 A′2 = 0.1125(0.025)(0.1) = 0.28125(10  3) m3 Shear Flow: Since there is only one row of nails, qallow =
qallow =
VmaxQB ; I
20 ( 103 ) =
1.5(103) F = = 20 (103) N>m. s 0.075
p c 0.28125 ( 103 ) d 2
80.2083 ( 106 ) P = 11417.41 N = 11.4 kN (controls)
Bending, smax = s(106) N>m2 =
Mc I
1 P2 2(0.125 m)
80.2083(106) m4
P = 11.550 N = 11.55 kN Shear, tmax =
VQ It
Q = (0.1125)(0.025)(0.1) + (0.05)(0.1)(0.025) = 0.40625(103) m3 1.5 1 106 2 =
1 P2 2 (0.40625)(103) m3
80.2083(106) m4(0.025 m)
P = 14.808 N = 14.8 kN
695
Ans.
25 mm
200 mm
25 mm
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7–42.
Continued
Ans: P = 11.4 kN 696
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7–43. The simply supported beam is built up from three boards by nailing them together as shown. If P = 12 kN, determine the maximum allowable spacing s of the nails to support that load, if each nail can resist a shear force of 1.5 kN.
P
s
A
B 1m
1m 100 mm
Solution
25 mm
Support Reactions: As shown on the freebody diagram of the beam shown in Fig. a.
25 mm
200 mm
Maximum Shear and Moment: The shear diagram is shown in Fig. b. As indicated, Vmax =
P 12 = = 6 kN. 2 2
25 mm
Section Properties: The moment of inertia of the cross section about the neutral axis is 1 1 I = (0.1)(0.253) (0.075)(0.23) 12 12 = 80.2083(106) m4 Referring to Fig. d, QB = y′2 A′2 = 0.1125(0.025)(0.1) = 0.28125(103) m3 Shear Flow: Since there is only one row of nails, qallow =
qallow =
VmaxQB ; I
1.5(1031) s
=
1.5(1031) F = . s s
6000 3 0.28125(103) 4 80.2083(1061)
Ans.
s = 0.07130 m = 71.3 mm
Ans: s = 71.3 mm 697
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*7–44. The Tbeam is nailed together as shown. If the nails can each each support determine the the can support aa shear shear force force of of 4.5 950 kN, lb, determine maximum shear shear force forceVVthat thatthethe beam support maximum beam cancan support and and the 1 the corresponding maximum spacing s tonearest the nearest corresponding maximum nail nail spacing s to the 8 in. multiples of 5 mm. The allowable shear stress for the wood The allowable shear stress for the wood is tallow = 450 psi. is tallow = 3 MPa.
502mm in.
ss
300 mm 12 in. ss
30012mm in.
Solution
V V
The neutral axis passes through the centroid c of the crosssection as shown in Fig. a. ' © y A 0.325(0.3)(0.05) + 0.15(0.05)(0.3) = 0.2375 m = y = ©A 0.3(0.05) + 0.05(0.3) I =
50 mm 2 in.
1 (0.05)(0.33 ) + 0.05(0.3)(0.2375 − 0.15)2 12 +
1 (0.3)(0.053 ) + 0.3(0.05)(0.325 − 0.2375)2 12
= 0.3453125(10 −3 ) m 4
Refering to Fig. a, Qmax and QA are Qmax = y1œ A1œ = 0.11875(0.05)(0.2375) = 1.41016(10 −3 ) m 3 QA = y2œ A2œ = 0.0875(0.3)(0.05) = 1.3125(10 −3 ) m 3 The maximum shear stress occurs at the points on the neutral axis where Q is maximum and t = 0.05 m. tallow =
VQmax ; It
3(106 ) =
V [1.41016(10 −3 )] [0.3453125(10 −3 )](0.05)
3 V 36.73(10 = = ) N 36.7 kN
Here, qallow =
Ans.
F 4.5(10 3 ) = N/m . Then s s −3 3 3 VQA ; 4.5(10 ) = [36.73(10 )][1.3125(10 )] qallow = I s 0.3453125(10 −3 ) = s 0.03223 = m 32.23 mm
Ans.
Use s 35 mm 0.3 m
0.3 m 0.05 m
0.05 m
0.875 m
= 0.11875 m 0.3 m
0.2375 m
0.05 m
0.05 m
Ans. V = 36.7 kN, s = 32.23 mm 503
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7–45. The nails are on both sides of the beam and each can resist a shear of 2 kN. In addition to the distributed loading, determine the maximum load P that can be applied to the end of the beam. The nails are spaced 100 mm apart and the allowable shear stress for the wood is tallow = 3 MPa.
P
2 kN/m
A
B
C
1.5 m
Solution
1.5 m
100 mm
The FBD is shown in Fig. a. 40 mm
As indicated the shear diagram, Fig. b, the maximum shear occurs in region AB of Constant value, Vmax = (P + 3) kN. The neutral axis passes through Centroid C of the crosssection as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04) = 0.14 m I =
1 (0.04)(0.23) + 0.04(0.2)(0.14  0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18  0.142) 12
Refering to Fig. d, Qmax = y1œ A1œ = 0.07(0.14)(0.04) = 0.392(10  3) m3 QA = y2œ A2œ = 0.04(0.04)(0.2) = 0.32(10  3) m3 The maximum shear stress occurs at the points on Neutral axis where Q is maximum and t = 0.04 m. Vmax Qmax ; It
3(106) =
(P + 3)(103) C 0.392(10  3) D 53.333(10  6)(0.04)
P = 13.33 kN Since there are two rows of nails qallow = 2 a qallow =
Vmax QA ; I
40 000 =
200 mm 20 mm
20 mm
= 53.333(10  6) m4
tallow =
200 mm
2(103) F b = 2c d = 40 000 N>m. s 0.1
(P + 3)(103) C 0.32(10  3) D 53.333(10  6)
P = 3.67 kN (Controls!)
Ans.
505
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7–45. Continued
Ans. P = 3.67 kN 506
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7–46. Determine the average shear stress developed in the nails within region AB of the beam. The nails are located on each side of the beam and are spaced 100 mm apart. Each nail has a diameter of 4 mm. Take P = 2 kN.
P
2 kN/m
A
B
C
1.5 m
The FBD is shown in Fig. a. As indicated in Fig. b, the internal shear force on the crosssection within region AB is constant that is VAB = 5 kN.
1.5 m
100 mm
The neutral axis passes through centroid C of the cross section as shown in Fig. c. ' 0.18(0.04)(0.2) + 0.1(0.2)(0.04) © y A = y = ©A 0.04(0.2) + 0.2(0.04)
40 mm
= 0.14 m
200 mm
1 (0.04)(0.23) + 0.04(0.2)(0.14  0.1)2 12 1 + (0.2)(0.043) + 0.2(0.04)(0.18  0.14)2 12
I =
200 mm 20 mm
20 mm
= 53.333(10  6) m4 Q for the shaded area shown in Fig. d is Q = y¿A¿ = 0.04(0.04)(0.2) = 0.32(10  3) m3 Since there are two rows of nail, q = 2 a q =
VAB Q ; I
20F =
F F b = 2a b = 20F N>m. s 0.1
5(103) C 0.32(10  3) D 53.333(10  6)
F = 1500 N Thus, the average shear stress developed in each nail is
A tnail B avg =
F 1500 = = 119.37(106)Pa = 119 MPa p Anail 2 (0.004 ) 4
504
Ans.
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7–47. The beam is made from four boards nailed together as shown. If the nails can each support a shear force of 500 N, determine determinetheir theirrequired requiredspacing spacing ssand andss if the beam 100 lb., is subjected to a shear of V V ==3.5 700kN. lb.
D
s¿ s¿ s
A
C s
1 in. 25 mm
Solution
10 in. 250 mm V B
Section Properties: y =
10 in. 250 mm
1 in. 25 mm 1 in. 25 mm 2 in. 50 mm
1.5mm in. 40
©yA 0.0125(0.25)(0.025) + 0.0375(0.05)(0.75) + 0.15(0.04)(0.25) = ©A 0.25(0.025) + 0.05(0.075) + 0.04(0.25) = 0.0859375 m
INA =
1 (0.25)(0.0253 ) + 0.25(0.025)(0.0859375 − 0.0125)2 12 +
1 (0.05)(0.0753 ) + 0.05(0.075)(0.0859375 − 0.0375)2 12
+
1 (0.04)(0.253 ) + 0.04(0.25)(0.15 − 0.0859375)2 12
= 0.137712(10 −3 ) m 4 QC = y1¿A¿ = (0.0859375 − 0.0375)(0.25)(0.075) = 90.8203(10 −6 ) m 3 0.640625(10 −3 ) m 3 QD = y2¿A¿ = (0.0859375 − 0.0125)(0.25)(0.025) + [(0.0859375 − 0.0375)](0.25)(0.075) =
Shear Flow: The allowable shear flow at points C and D is qC = 500 100 , respectively. qB = s¿ VQC qC = I 500 [3.5(10 3 )][90.8203(10 −6 )] = s 0.137712(10 −3 )
500 100 and s
Ans.
= m 216 mm = s 0.2166 VQD qD = I 500 [3.5(10 3 )][0.640625(10 −3 )] = s′ 0.137712(10 −3 ) = s′ 0.03071 = m 30 mm
Ans.
Ans. s = 216 mm, s′ = 30 mm 508
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*7–48. The beam is made from three polystyrene strips that are glued together as shown. If the glue has a shear strength of 80 kPa, determine the maximum load P that can be applied without causing the glue to lose its bond.
30 mm
P 1 —P 4
40 mm 60 mm 40 mm
1 —P 4
20 mm A
B 0.8 m
1m
1m
0.8 m
Solution Maximum shear is at supports. Vmax = I = t =
3P 4 1 1 (0.02)(0.06)3 + 2c (0.03)(0.04)3 + (0.03)(0.04)(0.05)2 d = 6.68(106) m4 12 12
VQ ; It
80(103) =
(3P>4)(0.05)(0.04)(0.03) 6.68(106)(0.02)
Ans.
P = 238 N
Ans: P = 238 N 704
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7–50. The beam is subjected to a shear force of V ow at V ==25 5 kN. Determinethe theshear shearflflow atpoints pointsAAand andB. B. kip.Determine
0.5mm in. 12 C C
5 in. 125 mm 1255 mm in. 0.5 in. 12 mm
Solution y =
0.5mm in. 12 in. 502mm
©yA 0.006(0.274)(0.012) + 2[0.112(0.012)(0.2)] + 0.156(0.25)(0.012) = ©A 0.274(0.012) + 2(0.012)(0.2) + 0.25(0.012)
8 in. 200 mm V V
1 (0.274)(0.012 3 ) + 0.274(0.012)(0.092472 − 0.006)2 12 1 + 2 (0.012)(0.2 3 ) + (0.012)(0.2)(0.112 − 0.092472)2 12 +
0.5 in. 12 mm
D D
= 0.092472 m I =
A A
1 (0.25)(0.012 3 ) + 0.25(0.012)(0.156 − 0.092472)2 12
12 mm
= 54.5990(10 −6 ) m 4
B B
250 mm
12 mm 12 mm 138 mm 12 mm 50 mm
œ 0.086472 m yA = 0.092472 − 0.006 =
0.063528 m yBœ = 0.156 − 0.092472 = −3 œ QA = yA A¿ = 0.086472(0.274)(0.012) = 0.28432(10 ) −3 QB = yBœ A¿ = 0.063528(0.25)(0.012) = 0.19058(10 )
qA =
1 VQA 1 [25(10 3 )][0.28432(10 −3 )] 3 a b = = 65.09(10 ) N/m 2 I 2 54.5990(10 −6 ) Ans.
= 65.1 kN/m
qB =
1 VQB 1 [25(10 3 )][0.19058(10 −3 )] 3 a b = = 43.63(10 ) N/m 2 I 2 54.5990(10 −6 ) = 43.6 kN/m
Ans.
Ans. qA = 65.1 kN>m, qB = 43.6 kN>m 515
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7–51. The beam is constructed from four plates and is subjected to to aa shear shear force force of of VV == 25 5 kip kN.. Determine the maximum shear flow in the cross section.
0.5mm in. 12 C
5 in. 125 mm 1255 mm in. 0.5 in. 12 mm
0.5mm in. 12 in. 50 2mm
Solution y =
A
0.5 in. 12 mm
D 8 in. 200 mm
©yA 0.006(0.274)(0.012) + 2[0.112(0.012)(0.2)] + 0.156(0.25)(0.012) = ©A 0.274(0.012) + 2(0.012)(0.2) + 0.25(0.012)
V
B
= 0.092472 m I =
1 (0.274)(0.012 3 ) + 0.274(0.012)(0.092472 − 0.006)2 12 1 + 2 (0.012)(0.2 3 ) + (0.012)(0.2)(0.112 − 0.092472)2 12 +
1 (0.25)(0.012 3 ) + 0.25(0.012)(0.156 − 0.092472)2 12 −6
= 54.5990(10 ) m
86.47 mm
40.235 mm
4
0.36203(10 −3 ) m 3 Qmax = 0.086472(0.274)(0.012) + 0.040236(0.012)(0.086472) =
qmax =
1 VQmax 1 [25(10 3 )][0.36203(10 −3 )] 3 a b = = 82.88(10 ) N/m 2 I 2 54.5990(10 −6 )
= 82.9 kN/m
Ans.
Ans. y = 0.092472 m, I = 54.5990(106) m4, qmax = 82.9 kN>m 515
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*7–52. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V = 150 N, determine the shear flow at points A and B. 10 mm 40 mm
B A
10 mm 30 mm
Solution y =
10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
I = 2c
30 mm 10 mm
= 0.027727 m
1 (0.03)(0.01)3 + 0.03(0.01)(0.027727  0.005)2 d 12
+ 2c +
V 40 mm
1 (0.01)(0.06)3 + 0.01(0.06)(0.03  0.027727)2 d 12
1 (0.04)(0.01)3 + 0.04(0.01)(0.055  0.027727)2 = 0.98197(10  6) m4 12
yB ′ = 0.055  0.027727 = 0.027272 m yA ′ = 0.027727  0.005 = 0.022727 m QA = yA ′A′ = 0.022727(0.04)(0.01) = 9.0909(10  6) m3 QB = yB ′A′ = 0.027272(0.03)(0.01) = 8.1818(10  6) m3 qA =
150(9.0909)(10  6) VQA = = 1.39 kN>m I 0.98197(10  6)
Ans.
qB =
150(8.1818)(10  6) VQB = = 1.25 kN>m I 0.98197(10  6)
Ans.
Ans: qA = 1.39 kN>m, qB = 1.25 kN>m 709
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7–53. The aluminum strut is 10 mm thick and has the cross section shown. If it is subjected to a shear of V = 150 N, determine the maximum shear flow in the strut. 10 mm 40 mm 10 mm 30 mm
Solution y =
B A V 40 mm 10 mm
30 mm 10 mm
2[0.005(0.03)(0.01)] + 2[0.03(0.06)(0.01)] + 0.055(0.04)(0.01) 2(0.03)(0.01) + 2(0.06)(0.01) + 0.04(0.01)
= 0.027727 m I = 2c
1 (0.03)(0.01)3 + 0.03(0.01)(0.027727  0.005)2 d 12
+ 2c +
1 (0.01)(0.06)3 + 0.01(0.06)(0.03  0.027727)2 d 12
1 (0.04)(0.01)3 + 0.04(0.01)(0.055  0.027727)2 12
= 0.98197(10  6) m4 Qmax = (0.055  0.027727)(0.04)(0.01) + 2[(0.06  0.027727)(0.01)]a = 21.3(10  6) m3 qmax =
6 1 VQmax 1 150(21.3(10 )) a b = a b = 1.63 kN>m 2 I 2 0.98197(10  6)
0.06  0.0277 b 2
Ans.
Ans: qmax = 1.63 kN>m 710
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7–54. A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at points A and B.
10 mm 30 mm 10 mm
A
100 mm C
B
100 mm 10 mm 30 mm 10 mm
Solution
150 mm
V
Section Properties:
150 mm
INA =
10 mm 125 mm
1 1 (0.145) ( 0.33 ) (0.125) ( 0.283 ) 12 12
10 mm
1 + 2c (0.125) ( 0.013 ) + 0.125(0.01) ( 0.1052 ) d 12
= 125.17 ( 10  6 ) m4
QA = y=2 A′ = 0.145(0.125)(0.01) = 0.18125 ( 10  3 ) m3 QB = y=1 A′ = 0.105(0.125)(0.01) = 0.13125 ( 10  3 ) m3 Shear Flow: qA = =
1 VQA c d 2 I
3 3 1 18(10 )(0.18125)(10 ) c d 6 2 125.17(10 )
Ans.
= 13033 N>m = 13.0 kN>m qB = =
1 VQB d c 2 I
3 3 1 18(10 )(0.13125)(10 ) c d 2 125.17(10  6)
Ans.
= 9437 N>m = 9.44 kN>m
Ans: qA = 13.0 kN>m, qB = 9.44 kN>m 707
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7–55. A shear force of V = 18 kN is applied to the box girder. Determine the shear flow at point C.
10 mm 30 mm 10 mm
A
100 mm C
B
100 mm 10 mm 30 mm 10 mm
Solution
150 mm
V
Section Properties:
150 mm
INA =
10 mm 125 mm
1 1 (0.145) ( 0.33 ) (0.125) ( 0.283 ) 12 12
10 mm
1 + 2c (0.125) ( 0.013 ) + 0.125(0.01) ( 0.1052 ) d 12
= 125.17 ( 10  6 ) m4 QC = Σy′A′
= 0.145(0.125)(0.01) + 0.105(0.125)(0.01) + 0.075(0.15)(0.02) = 0.5375 ( 10  3 ) m3 Shear Flow: qC = =
1 VQC c d 2 I
3 3 1 18(10 )(0.5375)(10 ) d c 2 125.17(10  4)
Ans.
= 38648 N>m = 38.6 kN>m
Ans: qC = 38.6 kN>m 708
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*7–56. A shear force of V = 300 kN is applied to the box girder. Determine the shear flow at points A and B. 90 mm A
90 mm C D
200 mm B
V
100 mm
190 mm
Solution 200 mm
The moment of inertia of the cross section about the neutral axis is I =
10 mm 180 mm
1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10  3) m4 12 12
10 mm
Referring to Fig. a, Fig. b, QA = y=1A=1 = 0.195(0.01)(0.19) = 0.3705(10  3) m3 QB = 2y=2A=2 + y=3A=3 = 2 [0.1(0.2)(0.01)] + 0.195(0.01)(0.18) = 0.751(10  3) m3 Due to symmetry, the shear flow at points A and A′, Fig. a, and at points B and B′, Fig. b, are the same. Thus qA =
3 3 1 VQA 1 300(10 ) 3 0.3705(10 ) 4 a b = c s 2 I 2 0.24359(10  3)
= 228.15(103) N>m = 228 kN>m
qB =
Ans.
3 3 1 VQB 1 300(10 ) 3 0.751(10 ) 4 a b = c s 2 I 2 0.24359(10  3)
= 462.46(103) N>m = 462 kN>m
Ans.
Ans: qA = 228 kN>m, qB = 462 kN>m 705
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7–57. A shear force of V = 450 kN is applied to the box girder. Determine the shear flow at points C and D. 90 mm A
90 mm C D
200 mm B
V
100 mm
190 mm
Solution 200 mm
The moment of inertia of the cross section about the neutral axis is I =
10 mm 180 mm
1 1 (0.2)(0.43) (0.18)(0.383) = 0.24359(10  3) m4 12 12
10 mm
Referring to Fig. a, due to symmetry AC= = 0. Thus QC = 0 Then referring to Fig. b, QD = y1= A1= + y2= A2= = 0.195(0.01)(0.09) + 0.15(0.1)(0.01) = 0.3255(10  3) m3 Thus, qC =
VQC = 0 I
qD =
450(103) 3 0.3255(10  3) 4 VQD = I 0.24359(10  3)
Ans.
= 601.33(103) N>m = 601 kN>m
Ans.
Ans: qC = 0, qD = 601 kN>m 706
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7–58. The Hbeam is subjected to a shear of V = 80 kN. Determine the shear flow at point A. 250 mm
50 mm
Solution
V A 30 mm 300 mm
25 mm
25 mm
1 1 I = 2c (0.025) ( 0.253 ) + (0.3) ( 0.033 ) d = 65.7792 ( 106 ) m4 12 12 QA = y′A′ = 0.0875(0.075)(0.025) = 0.1641(10  3) m3 qA =
80 ( 103 ) (0.1641) ( 10  3 ) VQA = 200 kN>m = I 65.7792 ( 10  6 )
Ans.
Ans: qA = 200 kN>m 713
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7–59. The Hbeam is subjected to a shear of V = 80 kN. Sketch the shearstress distribution acting along one of its side segments. Indicate all peak values. 250 mm
50 mm
Solution
V A 30 mm 300 mm
25 mm
25 mm
1 1 I = 2c (0.025) ( 0.253 ) d + (0.3) ( 0.033 ) = 65.7792 ( 106 ) m4 12 12 QB = (0.070)(0.025)(0.110) = 0.1925(10  3) m3 tB =
80 ( 103 ) (0.1925) ( 103 ) VQ = 9.36 MPa = It 65.7792 ( 106 ) (0.025)
tB =
80 ( 103 ) 3 2(0.1925) ( 103 ) 4 VQ = = 1.3378 MPa It 65.7792 ( 106 ) (0.35)
Qmax = 2(0.07)(0.025)(0.110) + (0.0075)(0.35)(0.015) = 0.4244 ( 103 ) m3 tC =
80 ( 103 ) (0.4244) ( 103 ) VQ = = 1.47 MPa It 65.7792 ( 106 ) (0.35)
Ans: tmax = 9.36 MPa 714
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*7–60. The builtup beam is formed by welding together the thin plates of thickness 5 mm. Determine the location of the shear center O.
5 mm 200 mm O e
100 mm 100 mm
200 mm
Solution
300 mm
Shear Center: Referring to Fig. a and summing moments about point A, we have a+ Σ(MR)A = ΣMA;  Pe =  (Fw)1(0.3) e =
0.3(Fw)1
(1)
P
Section Properties: The moment of inertia of the cross section about the axis of symmetry is I =
1 1 (0.005)(0.43) + (0.005)(0.23) = 30(10  6) m4 12 12
Referring to Fig. b, y′ = (0.1  s) +
s = (0.1  0.5s) m. Thus, Q as a function of s is 2
Q = y′A′ = (0.1  0.5s)(0.005s) = [0.5(10  3) s  2.5(10  3) s2] m3 Shear Flow: q =
P[0.5(10  3) s  2.5(10  3) s2] VQ = = P(16.6667s  83.3333s2) I 30(10  6)
Resultant Shear Force: The shear force resisted by the shorter web is (Fw)1 = 2
L0
0.1 m
qds = 2
L0
0.1 m
2
P(16.6667s  83.3333s )ds = 0.1111P
Substituting this result into Eq. (1), Ans.
e = 0.03333 m = 33.3 m
Ans: e = 33.3 m 715
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7–61. The assembly is subjected to a vertical shear of V ow at V ==35 7 kN. Determine the the shear shear flflow at points points A and B and kip.Determine the maximum shear flow in the cross section.
A A
0.5mm in. 12
B B
V V
Solution in. 50 2mm
©yA 0.006(0.274)(0.012) + 2[0.081(0.012)(0.138)] + 0.156(0.174)(0.012) = ©A 0.274(0.012) + 2(0.012)(0.138) + 0.174(0.012) = 0.070641 m
y =
6 in.mm 150
1506 mm in. 120.5 mm in.
0.5mm in. 12
2 in. 50 mm
0.5mm in. 12
0.5mm in. 12
1 (0.274)(0.012 3 ) + 0.274(0.012)(0.070641 − 0.006)2 12 1 + 2 (0.012)(0.138 3 ) + (0.012)(0.138)(0.081 − 0.070641)2 12
I =
+
1 (0.174)(0.012 3 ) + 0.174(0.012)(0.156 − 0.070641)2 12
= 34.6283(10 −6 ) m 4 QA = y1¿A1¿ = 0.06464(0.05)(0.012) = 0.28432(10 −3 ) m 3 QB = y2¿A2¿ = 0.08536(0.174)(0.012) = 0.17823(10 −3 ) m 3 Qmax = ©y¿A¿ = 0.08536(0.174)(0.012) + 2[0.039680(0.012)(0.079359)] = 0.25380(10 −3 ) m 3 q =
VQ I
qA =
[35(10 3 )][38.7845(10 −6 )] 3 = ) N/m 39.2 kN/m = 39.20(10 34.6283(10 −6 )
Ans.
qB =
1 [35(10 3 )][0.17823(10 −3 )] 3 = 90.07(10 ) N/m = 90.1 kN/m 2 34.6283(10 −6 )
Ans.
qmax =
1 [35(10 3 )][0.25380(10 −3 )] 3 = 128.26(10 ) N/m = 128 kN/m 2 34.6283(10 −6 )
Ans.
0.06464 m 0.039680 m
0.85359 m 0.079359 m
Ans. y = 0.070641 m, I = 34.6283(10 6) m4, qA = 39.2 kN>m, qB = 90.1 kN>m, qmax = 128 kN>m 518
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7–62. The box girder is subjected to a shear of V = 15 kN. Determine the shear flow at point B and the maximum shear flow in the girder’s web AB.
A
15 mm
250 mm 15 mm V B 150 mm
Solution 1 1 I = (0.375) ( 0.283 ) (0.3) ( 0.253 ) = 0.295375 ( 103 ) m4 12 12
150 mm
QB = y′B A′ = 0.1325(0.375)(0.015) = 0.7453125 ( 103 ) m3
25 mm
25 mm
25 mm
Qmax = Σy′A′ = 0.1325(0.375)(0.015) + 3[(0.0625)(0.125)(0.025)] = 1.33125 ( 103 ) m3 qB =
3 3 1 VQB 1 15 ( 10 ) (0.7453125) ( 10 ) c d = c d = 12.6 kN>m 3 3 I 3 0.295375 ( 10 )
qmax =
3 3 1 VQmax 1 15 ( 10 ) (1.33125) ( 10 ) c d = c d = 22.5 kN>m 3 I 3 0.295375 ( 103 )
Ans.
Ans.
Ans: qB = 12.6 kN>m, qmax = 22.5 kN>m 717
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7–63. Determine the location e of the shear center, point O, for the thinwalled member having a slit along its section.
100 mm 100 mm O
e 100 mm
Solution Suming moments about A, Pe = 2V1 (100) + F(200)
(1)
1 I = 2 c t ( 0.23 ) d + 2 3 (0.1)(t) ( 0.12 ) 4 = 3.3333 ( 103 ) t m4 12 y Q1 = y1 ′A′ = (y)t = 0.5y2 t 2 Q2 = Σy′A = 0.05(0.1)(t) + 0.1(x)(t) = 0.005t + 0.1xt q1 =
P(0.5y2 t) VQ1 = 150P y2 = I 3.3333(103)t
q2 =
P(0.005t + 0.1xt) VQ2 = 300P(0.005 + 0.1x) = I 3.3333(103)t 0.1
0.1
0.1
y3 V1 = q1 dy = 150P y dy = 150P c d 3 L0 L0 2
= 0.05P
0
F =
L0
0.1
q2 dx = 300P
L0
0.1
(0.005 + 0.1x) dx 0.1
0.1x2 = 300P c 0.005x + d` 2
= 0.3P
0
From Eq. (1);
Pe = 2(0.05P)(100) + 0.3P(200) Ans.
e = 70 mm
Ans: e = 70 mm 718
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*7–64. Determine the location e of the shear center, point O, for the thinwalled member. The member segments have the same thickness t.
b
h1
h2
O e
Solution Summing moments about A, (1)
eP = bF1 I =
1 1 1 (t) (h1 ) 3 + (t) (h2 ) 3 = t (h31 + h32 ) 12 12 12
q1 =
P(h1 >2) (t) (h1 >4)
F1 =
Ph31t 2 q1 (h1 ) = 3 12I
I
=
Ph21t 8I
From Eq. (1), e = = =
b Ph31t a b P 12I h31b
(h31 + h32 ) b 1 + (h2 >h1 ) 3
Ans.
Ans: e =
719
b 1 + (h2 >h1 ) 3
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7–65. The angle is subjected to a shear of V = 10 kN. Sketch the distribution of shear flow along the leg AB. Indicate numerical values at all peaks.
A in. 1255 mm
125 mm 5 in. 45 45
0.25 in. 6 mm
Solution
B
V
Section Properties: = b
0.006 = 0.00848528 m sin 45°
= = cos 45° 0.08839 m h 0.125 1 3 4 −6 INA = 2 (0.00848528)(0.08839 ) = 0.97656(10 ) m 12 y Q = y¿A¿ = [0.25(0.08839) + 0.5y] 0.625 − (0.006) sin 45° = 8.2864(10 −6 ) − 4.2426(10 −3 ) y2
Shear Flow: q = =
VQ I [10(10 3 )][8.2864(10 −6 ) − 4.2426(10 −3 )y2 ] 0.97656(10 −6 )
3 6 2 = [84.8528(10 ) − 43.4446(10 )y ] N/m
Ans.
= [84.9 − 43.4y2 ] kN/m At y = 0, q = qmax = 84.9 kN/m
Ans.
0.125 m
6 mm
0.0625
84.9 kN/m
Ans. At y = 0, q = qmax = 84.9 kN>m 517
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7–66. Determine the shearstress variation over the cross section of the thinwalled tube as a function of elevation y and show that t max = 2V>A, where A = 2prt. Hint: Choose a differential area element dA = Rt du. Using dQ = y dA, formulate Q for a circular section from u to (p  u) and show that Q = 2R2t cos u, where cos u = 2R2  y2>R.
ds du y
u t
R
Solution dA = R t du dQ = y dA = yR t du Here y = R sin u Therefore dQ = R2 t sin u du pu
Q =
pu
R2 t sin u du = R2 t( cos u) 
Lu
u
= R2 t [ cos (p  u)  (  cos u)] = 2R2 t cos u dI = y2 dA = y2 R t du = R3 t sin2 u du 2p
I =
L0
2p
R3 t sin2 u du = R3 t 2p
=
t =
sin 2u R3 t [u ]冷 2 2 0
R3 t [2p  0] = pR3 t 2
VQ V(2R2t cos u) V cos u = = It pR t pR3 t(2t)
Here cos u = t =
=
L0
(1  cos 2u) du 2
2R2  y2 R
V 2R2  y2 pR2t
Ans.
tmax occurs at y = 0; therefore tmax =
V pR t
A = 2pRt; therefore tmax =
2V A
QED
Ans. t = 519
V 2R 2  y2 pR 2t
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7–67. Determine the location e of the shear center, point O, for the beam having the cross section shown. The thickness is t. 1 — r 2
e r
O
1 — r 2
Solution I = (2)c
1 r 2 (t)(r>2)3 + (r>2)(t) a r + b d + Isemicircle 12 4
= 1.583333t r3 + Isemicircle p>2
Isemicircle =
p>2 2
Lp>2
(r sin u) t r du = t r3
Lp>2
sin2 u du
p Isemicircle = t r3 a b 2 Thus, p I = 1.583333t r3 + t r3 a b = 3.15413t r3 2 r r Q = a bt a + rb + 2 4 Lu
p>2
r sin u (t r du)
Q = 0.625 t r2 + t r2 cos u q =
VQ P(0.625 + cos u)t r2 = I 3.15413 t r3
Summing moments about A: p>2
Pe =
Lp>2
(q r du)r p>2
Pe = e =
Pr (0.625 + cos u)du 3.15413 Lp>2
r (1.9634 + 2) 3.15413
e = 1.26 r
Ans.
Ans. e = 1.26 r 525
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*7–68. Determine the location e of the shear center, point O, for the thinwalled member. The member segments have the same thickness t.
d d — 2 e
d — 2
O
d
Solution d
Summing moments about point A: (1)
Pe = F2d + F1(2d) I = 2[dt(d)2] + 2[dt(d>2)2] = q1 = F1 =
q2 = F2 =
P(dt)(d) 19 3 6 td
=
1 19 3 t(2d)3 = td 12 6
6P 19d
1 6P 3 a b(d) = P 2 19d 19 P(dt)(d>2) 19 3 6 td
=
3P 19d
1 3P 1.5P bd = a 2 19d 19
From Eq. (1): Pe = 2d a e =
3 1.5P Pb + da b 19 19
7.5 15 d = d 19 38
Ans.
Ans: e = 723
15 d 38
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7–69. t
A thin plate of thickness t is bent to form the beam having the cross section shown. Determine the location of the shear center O.
r O e
Solution Section Properties: For the arc segment, Fig. a, y = r cos u and the area of the differential element shown shaded is dA = t ds = tr du. Then, the moment of inertia of the entire cross section about the axis of symmetry is I =
1 (t)(2r)3 + y2 dA 12 L p
=
2 3 r t + (r cos u)2 trdu 3 L0
=
2 3 r 3t r t + (cos 2u + 1) du 3 2 L0 p
= =
p 2 3 r 3t 1 r t + a sin 2u + u b ` 3 2 2 0
r 3t (4 + 3p) 6
r . Thus, Q as a function of s is 2 u r Q = y′A′ + ydA = (rt) + r cos u(trdu) 2 L L0
Referring to Fig. a, y′ =
u
=
1 2 r t + r 2t cos udu 2 L0
=
r2 t (1 + 2 sin u) 2
Shear Flow: P 3 r2 t(1 + 2 sin u) 4 VQ 3P = = (1 + 2 sin u) r 3t I (4 + 3p)r (4 + 3p) 2
q =
6
Resultant Shear Force: The shear force resisted by the arc segment is p p 3P F = qds = qrdu = (1 + 2 sin u)rdu (4 + 3p)r L L0 L0 = =
p 3P (u  2 cos u) ` 4 + 3p 0
3P(p + 4) 4 + 3p
Shear Center: Referring to Fig. b and summing the moments about point A, a+ Σ(MR)A = ΣMA;
Pe = r
L
Pe = r c e = c
dF
3P(p + 4) 4 + 3p
3(p + 4) 4 + 3p
d
Ans: Ans.
d r 724
e = c
3(p + 4) 4 + 3p
dr
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7–70. t
Determine the location e of the shear center, point O, for the tube having a slit along its length.
r
e
O
Solution Section Properties: dA = t ds = t r du y = r sin u dI = y2 dA = r 2 sin2 u(t r du) = r 3t sin2 u du I = r 3t
L0
2p
2
sin u du 2p
1  cos 2u a b du = pr 3t 2 L0 dQ = y′A′ = ydA = r sin u(t r du) = r 2 t sin u du = r 3t
Q = r2 t
L0
u
sin u du = r 2 t (1  cos u)
Shear Flow Resultant: q = F =
P r 2t (1  cos u) VQ P = = (1  cos u) I pr pr 3t L0
2p
2p
qds =
P (1  cos u) rdu L0 pr 2p
=
P (1  cosu) du p L0
= 2P Shear Center: Summing moments about point A. Pe = Fr Pe = 2Pr Ans.
e = 2r
Ans: e = 2r 725
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R7–1. Sketch the intensity of the shearstress distribution acting over the beam’s crosssectional area, and determine the resultant shear force acting on the segment AB. The shear acting at the section is V = 175 kN. Show that INA = 0.3048(10−3) m4.
C
200 mm 8 in.
Solution
V B A
6150 in.mm
Section Properties: ©yA 0.1(0.2)(0.2) + 0.275(0.05)(0.15)] = 0.12763 m = ©A 0.2(0.2) + 0.05(0.15)
y =
INA =
mm 250in.
1 (0.2)(0.2 3 ) + 0.2(0.2)(0.12763 − 0.1)2 12 0.12763
1 + (0.05)(0.153 ) + (0.05)(0.15)(0.275 − 0.12763)2 12 −3
= 0.3408(10 ) m
4
375 in.mm mm 375in.
0.2 m
0.063816 + 0.5y1
0.12763 m 0.2 m
(Q.E.D)
Q1 = yœ1A¿ = (0.063816 + 0.5 y1 )(0.2)(0.12763 − y1 )
0.15 m
0.22237
2 −3 = 1.62898(10 ) − 0.1y1
0.05 m
0.11118 + 0.5y2
Q2 = y2œ A¿ = (0.11118 + 0.5 y2 )(0.05)(0.22237 − y2 ) = 1.23619(10 −3 ) − 0.25y22 Shear Stress: Applying the shear formula t = tCB
4.18 MPa 2.84 MPa
VQ , It
11.4 MPa
VQ1 [175(10 3 )][1.62898(10 −3 ) − 0.1y12 ] = = It [0.3408(10 −3 )](0.2)
(
)
= 4.18218 − 256.74y12 (106 ) At y1 = 0,
tCB = 4.18 523 psi MPa
At y1 = −0.07237 m
tCB = 2.84 355 psi MPa
tAB =
VQ2 [175(10 3 )][1.23619(10 −3 ) − 0.25y22 ] = It [0.3408(10 −3 )](0.05)
)
(
= 12.6950 − 256.74y22 (106 ) At y2 = 0.07237 m
tAB = 11.4 MPa
Resultant Shear Force: For segment AB. VAB =
L
tAB dA 0.22237 m
=
(
)
12.6950 − 256.74y2 (106 ) (0.05dy) 2 L0.07237 m
= 49.78(10 3 ) N = 49.8 kN
Ans.
Ans. VAB = 49.78 kN 528
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R7–2. The Tbeam is subjected to a shear of V = 150 kN. Determine the amount of this force that is supported by the web B.
200 mm
40 mm V = 150 kN 200 mm
40 mm
Solution y = I =
B
(0.02)(0.2)(0.04) + (0.14)(0.2)(0.04) 0.2(0.04) + 0.2(0.04)
= 0.08 m
1 (0.2)(0.043) + 0.2(0.04)(0.08  0.02)2 12 1 + (0.04)(0.23) + 0.2(0.04)(0.14  0.08)2 = 85.3333(106) m4 12
A′ = 0.04(0.16  y) y′ = y +
(0.16  y) 2
=
(0.16 + y) 2
Q = y ′A′ = 0.02(0.0256  y2) t = V =
150(103)(0.02)(0.0256  y2) VQ = = 22.5(106)  878.9(106)y2 It 85.3333(106)(0.04) L
t dA,
dA = 0.04 dy
0.16
V =
(22.5(106)  878.9(106)y2) 0.04 dy L0.04 0.16
=
(900(103)  35.156(106)y2)dy L0.04 Ans.
= 131 250 N = 131 kN
Ans: N/A 727
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R7–3. The member is subject to a shear force of V = 2 kN. Determine the shear flow at points A, B, and C. The thickness of each thinwalled segment is 15 mm.
200 mm B 100 mm A C
300 mm
V 2 kN
Solution Section Properties: y = =
INA
ΣyA ΣA 0.0075(0.2)(0.015) + 0.0575(0.115)(0.03) + 0.165(0.3)(0.015)
0.2(0.015) + 0.115(0.03) + 0.3(0.015) = 0.08798 m 1 = (0.2)(0.0153) + 0.2(0.015)(0.08798  0.0075)2 12 1 + (0.03)(0.1153) + 0.03(0.115)(0.08798  0.0575)2 12 1 + (0.015)(0.33) + 0.015(0.3)(0.165  0.08798)2 12 = 86.93913(10  6) m4 Ans.
QA = 0 QB = y1′A′ = 0.03048(0.115)(0.015) = 52.57705(10  6) m  3 QC = Σy1′A′ = 0.03048(0.115)(0.015) + 0.08048(0.0925)(0.015) = 0.16424(10  3) m3 Shear Flow: qA =
VQA = 0 I
qB =
2(103)(52.57705)(10  6) VQB = = 1.21 kN>m I 86.93913(10  6)
Ans.
qC =
2(103)(0.16424)(10  3) VQC = = 3.78 kN>m I 86.93913(10  6)
Ans.
Ans: qA = 0, qB = 1.21 kN>m, qC = 3.78 kN>m 728
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*R7–4. The beam is constructed from four boards glued together at their seams. If the glue can withstand 15 75 kN/m, lb>in., what is the maximum vertical shear V that the beam can support?
753mm in.
753mm in.
Solution Section Properties:
= 35.4467(10 −6 ) m 4 Q = y¿A¿ = 0.0435(0.1)(0.012) = 52.2(10 ) m
3
12 0.5mm in.
0.1 m
0.012 m
0.0435 m
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2[15(10 3 )] = 30(10 3 ) N/m .
30(10 3 ) =
4 in. 100 mm
0.012 m
−6
q =
12 0.5mm in.
12 0.5mm in.
753mm in.
V V
1 1 INA = 2 (0.012)(0.249 3 ) + 2 (0.1)(0.012 3 ) + 0.1(0.012)(0.04352 ) 12 12
12 0.5mm in.
VQ I
0.012 m
0.012 m
0.1245 m
0.1245 m
V [52.2(10 −6 )] 35.4467(10 −6 )
3 = V 20.37(10 = ) N 20.4 kN
Ans.
Ans. V = 20.4 kN 531
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R7–5. Solve Prob. R7–4 if the beam is rotated 90° from the position shown. 753 mm in.
753 mm in.
12 0.5mm in.
12 0.5mm in.
753 mm in.
V
Solution
12 0.5mm in.
4 in. 100 mm
12 0.5mm in.
Section Properties: INA =
1 1 (0.249)(0.124 3 ) − (0.225)(0.13 ) = 20.8124(10 −6 ) m 4 12 12
Q = y¿A¿ = 0.056(0.249)(0.012) = 1.167328(10 −3 ) m 3
0.012 m
Shear Flow: There are two glue joints in this case, hence the allowable shear flow is 2[15(10 3 )] = 30(10 3 ) N/m . q = 30(10 3 ) =
0.012 m 0.05 m
0.056 m
0.05 m 0.012 m
0.249 m
VQ I V [1.167328(10 −3 )] 20.8124(10 −6 )
3 V 3.731(10 = ) N 3.73 kN =
Ans.
Ans. V = 3.73 kN 531
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8–1. A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p = 300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.
Solution sallow =
pr ; 2t
12(106) =
300(103)(1.5) 2t Ans.
t = 0.0188 m = 18.8 mm
These solutions represent a preliminary version of the Instructors' Solutions Manual (ISM). It is possible and even likely that at this preliminary stage of preparing the ISM there are some omissions and errors in the draft solutions. These will be corrected and this manual will be republished.
731
Ans: t = 18.8 mm
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8–2. A pressurized spherical tank is to be made of 12mmthick steel. If If itit is is subjected subjected to an internal pressure 0.5in.thick steel. of pp ==1.4 determine its outer radius if the maximum 200MPa, psi, determine normal stress is not to exceed 105 MPa. 15 ksi.
Solution sallow =
pr 2t
;
105(106 ) =
[1.4(106 )]ri 2(0.012)
ri = 1.80 m ro =1.80 + 0.012 =1.812 m
Ans.
Ans. ro = 1.812 m 532
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8–3. The thinwalled cylinder can be supported in one of two ways as shown. Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the Thewall wallhas has aa thickness thickness of internal pressure to to be be0.5 65 MPa. psi. The 6 mmin.and mm. 0.25 andthe theinner innerdiameter diameterofofthe thecylinder cylinderisis200 8 in.
P
Solution
P
8 in. 200 mm
200 mm 8 in.
(a)
(b)
Case (a): s1 =
pr ; t
s1 =
65(4) 0.5(100) 8.33 = = 1.04 ksiMPa 0.25 6
Ans.
s2 = 0
Ans.
Case (b): s1 =
pr ; t
s1 =
65(4) 0.5(100) = 1.04 ksiMPa = 8.33 0.256
Ans.
s2 =
pr ; 2t
s2 =
65(4) 0.5(100) 4.17psi MPa == 520 2(0.25) 2(6)
Ans.
Ans. Case (a): s1 = 8.33 MPa; s2 = 0 Case (b): s1 = 8.33 MPa; s2 = 4.17 MPa 532
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*8–4. The tank of the air compressor is subjected to an MPa. If the internal diameter internal pressure of 0.63 90 psi. If the internal diameter of of the tank thickness 6 mm, the tank is is22550 in.,mm, and and the the wallwall thickness is is0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element.
A
Solution Hoop Stress for Cylindrical Vessels: Since
11 r 275 ==45.8 thenthin thinwall wall = 44 > 7 10, 10,then t 0.25 6
28.9 MPa
analysis can be used. Applying Eq. 8–1 14.4 MPa
pr 90(11) 0.63(275) = 28.875 MPa = 28.9 MPa = s1 = t 0.25 6
Ans.
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2 s2 =
0.63(275) pr = = 14.4375 MPa = 14.4 MPa 2t 2(6 )
Ans.
Ans. s1 = 28.9 MPa, s2 = 14.4 MPa 533
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8–5. P
Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall of the cylinder.
47 mm
P
Solution p =
2(103) P = 314 380.13 Pa = A p(0.0452)
s1 =
314 380.13(0.045) pr = = 7.07 MPa t 0.002
Ans. Ans.
s2 = 0
The pressure P is supported by the surface of the pistons in the longitudinal direction.
Ans: s1 = 7.07 MPa, s2 = 0 735
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8–6. Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress in the cylinder does not exceed 3 MPa. Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm.
P 47 mm
P
Solution s =
pr ; t
3(106) =
p(0.045) 0.002
p = 133.3 kPa P = pA = 133.3 1 103 2 (p)(0.045)2 = 848 N
Ans.
Ans: P = 848 N 736
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8–7. A boiler is constructed of 8mmthick steel plates that are fastened together at their ends using a butt joint consisting of two 8mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress in the boiler’s plate away from the seam, (b) the circumferential stress in the outer cover plate along the rivet line a–a, and (c) the shear stress in the rivets.
a
8 mm
50 mm
0.75 m a
Solution a)
s1 =
1.35(106)(0.75) pr = = 126.56(106) = 127 MPa t 0.008
Ans.
126.56 (106)(0.05)(0.008) = s1 ′(2)(0.04)(0.008)
b)
s1 ′ = 79.1 MPa
Ans.
c) From FBD(a) + c ΣFy = 0;
Fb  79.1(106)[(0.008)(0.04)] = 0 Fb = 25.3 kN
(tavg)b =
Fb 25312.5  p = 322 MPa 2 A 4 (0.01)
Ans.
Ans: (a) s1 = 127 MPa, (b) s1 ′ = 79.1 MPa, (c) (tavg)b = 322 MPa 737
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*8–8. The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of longitudinal bolts per meter length at each side of the cylindrical shell. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m. Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =
pr ; t
150 A 106 B =
3 A 106 B (2) tc
tc = 0.04 m = 40 mm For the hemispherical cap, sallow =
pr ; t
150 A 106 B =
Ans. 3 A 106 B (2) 2ts
ts = 0.02 m = 20 mm Since
Ans.
r 6 10, thinwall analysis is valid. t
Referring to the freebody diagram of the per meter length of the cylindrical portion, Fig. a, where P = pA = 3 A 106 B [4(1)] = 12 A 106 B N, we have + c ©Fy = 0;
12 A 106 B  nc(Pb)allow  nc(Pb)allow = 0 nc =
6 A 106 B
(1)
(Pb)allow
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 122.72 A 103 B N 4
Substituting this result into Eq. (1), nc = 48.89 = 49 bolts>meter
Ans.
Ans. tc = 40 mm, ts = 20 mm, nc = 49 bolts>meter 536
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8–9. The gas storage tank is fabricated by bolting together two half cylindrical thin shells and two hemispherical shells as shown. If the tank is designed to withstand a pressure of 3 MPa, determine the required minimum thickness of the cylindrical and hemispherical shells and the minimum required number of bolts for each hemispherical cap. The tank and the 25 mm diameter bolts are made from material having an allowable normal stress of 150 MPa and 250 MPa, respectively. The tank has an inner diameter of 4 m.
Normal Stress: For the cylindrical portion of the tank, the hoop stress is twice as large as the longitudinal stress. sallow =
pr ; t
150 A 106 B =
3 A 106 B (2) tc
tc = 0.04 m = 40 mm For the hemispherical cap, sallow =
pr ; t
150 A 106 B =
Ans. 3 A 106 B (2) 2ts
ts = 0.02 m = 20 mm Since
Ans.
r 6 10, thinwall analysis is valid. t
The allowable tensile force for each bolt is (Pb)allow = sallowAb = 250 A 106 B c
p A 0.0252 B d = 122.72 A 103 B N 4
Referring to the freebody diagram of the hemispherical cap, Fig. b, where p P = pA = 3 A 106 B c A 42 B d = 12p A 106 B N, 4 + ©F = 0; : x
12p A 106 B ns =
ns ns (Pb)allow (Pb)allow = 0 2 2
12p A 106 B
(1)
(Pb)allow
Substituting this result into Eq. (1), ns = 307.2 = 308 bolts
Ans.
Ans. tc = 40 mm, ts = 20 mm, (Pb)allow = 122.72(103) N, ns = 308 bolts 537
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s
8–10. A wood wood pipe pipe having havingan aninner innerdiameter diameterofof0.93 m ft is bound together each having a crosssectogether using usingsteel steelhoops hoops each having a crosstional area of 125 mm sectional area of 0.2 in22..If If the the allowable allowable stress stress for the hoops is sallow ==8412MPa, determine their their maximum spacing s along ksi, determine the section of pipe so that the pipe can resist an internal gauge pressure kPa.Assume Assumeeach each hoop hoop supports the pressure of of 28 4 psi. pressure loading acting along the length s of the pipe.
4 psi 28 kPa
428psi kPa
s
s
Solution Equilibrium for the steel Hoop: From the FBD + : ©Fx = 0;
3
[28(103)](0.95)
3
2 P − [28(10 )](0.93) = 0 P= 12.6(10 )s 0.9 m
Hoop Stress for the Steel Hoop: s1 = sallow = 84(106 ) =
P A 12.6(10 3 )s 125(10 −6 )
ss = 833.33 33.3 in.mm = 0.833 m
Ans.
Ans. s = 0.833 m 538
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8–11. The staves or vertical members of the wooden tank are held together using semicircular hoops having a and a width of 2ofin.50 Determine the normal thickness of of 0.5 12 in. mm and a width mm. Determine the stress instress hoop AB ifAB theif the tank is subjected normal in hoop tank is subjectedtotoan an internal pressure of of 14 2 psi gauge pressure kPa and and this this loading loading is transmitted bolts are used directly to the the hoops. hoops.Also, Also,ifif0.25in.diameter 6mmdiameter bolts to connect each hoop together, determine the tensile stress in each bolt at A and B. Assume hoop AB supports the pressure loading within withinaa300mm 12in. length lengthofofthe thetank tankas as shown.
450 mm 18 in.
150 mm 6 in. 150 mm 6 in.
300 mm 12 in. A
B 300 mm 12 in.
Solution 0.9 m
3
3
3
FR = [14(10 )][(0.9)(0.3 )] = 3.78(10 ) N ©F = 0;
3.78(10 3 ) − 2 F = 0;
F = 1.89(10 3 ) N
1.89(10 3 ) F 6 3.15(10 = ) N/m 3 3.15 MPa = = Ah 0.012(0.05)
Ans.
1.89(10 3 ) F 6 66.84(10 = ) N/m 3 66.8 MPa sb = = = π Ab 2 (0.006 ) 4
Ans.
sh =
0.3 m
Ans. sh = 3.15 MPa, sb = 66.8 MPa 538
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*8–12. A pressurevessel head is fabricated by welding the circular plate to the end of the vessel as shown. If the vessel sustains an internal pressure of 450 kPa, determine the average shear stress in the weld and the state of stress in the wall of the vessel.
450 mm 10 mm 20 mm
Solution + c ΣFy = 0; p(0.225)2450(103)  tavg (2p)(0.225)(0.01) = 0; Ans.
tavg = 5.06 MPa 3
s1 =
450(10 )(0.225) pr = = 5.06 MPa t 0.02
Ans.
s2 =
450(103)(0.225) pr = = 2.53 MPa 2t 2(0.02)
Ans.
Ans: tavg = 5.06 MPa, s1 = 5.06 MPa, s2 = 2.53 MPa 742
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8–13. The 304 stainless steel band initially fits snugly around the smooth rigid cylinder. If the band is then subjected to a nonlinear temperature temperaturedrop dropofof¢T ΔT = = 12 nonlinear 20 sin22 u °C, °F, where u is in radians, determine the circumferential stress in the band.
1
250 10mm in.
in. 0.4 64 mm 25 mm 1 in.
u
Solution Compatibility: Since the band is fixed to a rigid cylinder (it does not deform under load), then dF  dT = 0 2p P(2pr) a¢Trdu = 0 AE L0
2pr P a b = 12 20ar E A L0 2p 6a s = 10a E c L0
2p
sin2 udu
however,
P = sc A
2p
(1  cos 2u)du
sc = 10aE 6aE 6 −6 = 6[17(10 = )][193(109 )] 19.686(10 = ) N/m 2 19.7 MPa
Ans.
Ans. dF  dT = 0, sc = 19.7 MPa 539
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8–14. The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a pressure p. Determine the change in the inner radius of the ring after this pressure is applied. The modulus of elasticity for the ring is E.
ro ri w p
Solution Equilibrium for the Ring: From the FBD + ΣFx = 0; 2P  2pri w = 0 S
P = pri w
Hoop Stress and Strain for the Ring: s1 =
pri w pri P = = A (ro  ri)w ro  ri
Using Hooke’s Law P1 = However, P1 =
pri s1 = E E(ro  ri)
2p(ri)1  2pri 2pri
=
(ri)1  ri ri
(1) =
dri . ri
Then, from Eq. (1) pri dri = ri E(ro  ri) dri =
pr 2i E(ro  ri)
Ans.
Ans: dri = 744
pr 2i E(ro  ri)
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8–15. The inner ring A has an inner radius r1 and outer radius r2. The outer ring B has an inner radius r3 and an outer radius r4, and r2 7 r3. If the outer ring is heated and then fitted over the inner ring, determine the pressure between the two rings when ring B reaches the temperature of the inner ring. The material has a modulus of elasticity of E and a coefficient of thermal expansion of a.
r4
r2 r1 A
r3 B
Solution Equilibrium for the Ring: From the FBD + ΣFx = 0;2P  2pri w = 0 S
P = pri w
Hoop Stress and Strain for the Ring: s1 =
pri w pri P = = A (ro  ri)w ro  ri
Using Hooke’s Law pri s1 = E E(ro  ri)
P1 = However, P1 =
2p(ri)1  2pri 2pri
=
(ri)1  ri ri
(1)
=
dri . ri
Then, from Eq. (1) pri dri = ri E(ro  ri) dri =
pr 2i E(ro  ri)
Compatibility: The pressure between the rings requires (2)
dr2 + dr3 = r2  r3 From the result obtained above dr2 =
pr 22 E(r2  r1)
dr3 =
pr 23 E(r4  r3)
Substitute into Eq. (2) pr 22 pr 23 + = r2  r3 E(r2  r1) E(r4  r3) p =
E(r2  r3)
Ans.
r 23 r 22 + r2  r1 r4  r3
Ans: p =
745
E(r2  r3) r 23 r 22 + r2  r1 r4  r3
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*8–16. The cylindrical tank is fabricated by welding a strip of thin plate helically, making an angle u with the longitudinal axis of the tank. If the strip has a width w and thickness t, and the gas within the tank of diameter d is pressured to p, show that the normal stress developed along the strip is given by su = (pd>8t)(3  cos 2u).
w u
Normal Stress: sh = s1 =
pr p(d>2) pd = = t t 2t
sl = s2 =
p(d>2) pd pr = = 2t 2t 4t
Equilibrium: We will consider the triangular element cut from the strip shown in Fig. a. Here, Ah = (w sin u)t and Thus, Al = (w cos u)t. pd pwd and (w sin u)t = sin u Fh = shAh = 2t 2 pd pwd (w cos u)t = cos u. 4t 4
Fl = slAl =
Writing the force equation of equilibrium along the x¿ axis, ©Fx¿ = 0;
c
pwd pwd sin u d sin u + c cos u d cos u  Nu = u 2 4 Nu =
pwd A 2 sin2 u + cos2 u B 4
However, sin2 u + cos2 u = 1. This equation becomes Nu = Also, sin2 u =
pwd A sin2 u + 1 B 4
1 (1  cos 2u), so that 2 pwd Nu = (3  cos 2u) 8
Since Au = wt, then pwd (3  cos 2u) Nu 8 = su = Au wt su =
pd (3  cos 2u) 8t
(Q.E.D.)
Ans. su = 542
pd (3  cos 2u) 8t
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8–17. In order to increase the strength of the pressure vessel, filament winding of the same material is wrapped around the circumference of the vessel as shown. If the pretension in the filament is T and the vessel is subjected to an internal pressure p, determine the hoop stresses in the filament and in the wall of the vessel. Use the freebody diagram shown, and assume the filament winding has a thickness t′ and width w for a corresponding length L of the vessel.
L w t¿
p
s1 T
t s1 T
Solution Normal Stress in the Wall and Filament Before the Internal Pressure is Applied: The entire length L of wall is subjected to pretension filament force T. Hence, from equilibrium, the normal stress in the wall at this state is 2T  (s′)w (2Lt) = 0
(s′)w =
T Lt
and for the filament the normal stress is (s′)fil =
T wt′
Normal Stress in the Wall and Filament After the Internal Pressure is Applied: In order to use s1 = pr>t, developed for a vessel of uniform thickness, we redistribute the filament’s crosssection as if it were thinner and wider, to cover the vessel with no gaps. The modified filament has width L and thickness t>w>L, still with crosssectional area wt> subjected to tension T. Then the stress in the filament becomes sfil = s + (s′)fil =
pr T + (t + t′w>L) wt′
Ans.
pr T (t + t′w>L) Lt
Ans.
And for the wall, sw = s  (s′)w = Check: 2wt′sfil + 2Ltsw = 2rLp
OK
Ans:
pr T + , t + t′w>L wt′ pr T sw = t + t′w>L Lt sfil =
747
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8–18. Determine the shortest distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses in the plate at section a–a. The plate has a thickness of 10 mm and P acts along the centerline of this thickness.
300 mm a
a
200 mm
500 mm
Solution
d
sA = 0 = sa  sb 0 =
P Mc A I
0 =
P(0.1  d)(0.1) P  1 3 (0.2)(0.01) 12 (0.01)(0.2 )
P
P(  1000 + 15000 d) = 0 Ans.
d = 0.0667 m = 66.7 mm
Ans: d = 66.7 mm 748
© 2018 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–19. Determine the maximum distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses on the plate at section a–a. The plate has a thickness of 20 mm and P acts along the centerline of this thickness.
200 mm
a
P d
a
Solution Internal Loadings: C