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1–1. 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.
Support Reactions: We will only need to compute C y by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a.
a
Internal Loadings: Using the result for C y , section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b ,
Ans.
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a
Ans.
The negative signs indicates that V E and M E act in the opposite sense to that shown on the free-body diagram.
M E = - 2400 lb #^ ft = - 2.40 kip #^ ft
+ © ME = 0;1000(4) - 800(8) - ME = 0
- c © Fy = 0; VE + 1000 - 800 = 0 V (^) E = -200 lb
:^ + ©F (^) x = 0; N (^) E = 0
- © MB = 0; C y(8) + 400(4) - 800(12) = 0 Cy = 1000 lb
A B^ E C D
4 ft 400 lb 800 lb
4 ft 4 ft 4 ft
Ans:
NE = 0 , VE = -200 lb, M E = - 2.40 kip #^ ft
1–2. 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 500-lb load is applied along the centroidal axis of the member.
(a)
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Ans.
(b)
Ans.
V b = 250 lb Ans.
+Q© Fy = 0; Vb - 500 sin 30° = 0
Nb = 433 lb
R+ © F x = 0; N b - 500 cos 30° = 0
Na = 500 lb
:^ + © Fx = 0; N (^) a - 500 = 0
30
A
a b
b (^) a
500 lb 500 lb
Ans: , , Nb = 433 lb, V b = 250 lb
N (^) a = 500 lb Va = 0
*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.
Support Reactions: We will only need to compute B y by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a.
a
Internal Loadings: Using the result of B y , section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b ,
Ans.
Ans.
a
Ans.
The negative sign indicates that V C act in the opposite sense to that shown on the free-body diagram.
M C = 433 N #^ m
+ ©M C = 0; 1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0
- c©Fy = 0; VC - 600(1) + 1733.33 - 900 = 0 VC = -233 N
; NC = 0
©F (^) x = 0;
- ©MA = 0; By(4.5) - 600(2)(2) - 900(6) = 0 By = 1733.33 N
A (^) B D
C
900 N
1.5 m
600 N/m
1 m 1 m 1 m 1.5 m
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 3-kip load.
Support Reactions: For member AB
a
Equations of Equilibrium: For point D
Ans.
Ans.
a
Ans.
Equations of Equilibrium: For point E
Ans.
Ans.
a
Ans.
Negative signs indicate that M (^) E and VE act in the opposite direction to that shown on FBD.
ME = -24.0 kip #^ ft
+ ©ME = 0; ME + 6.00(4) = 0
VE = -9.00 kip
- c ©Fy = 0; - 6.00 - 3 - VE = 0
:^ + ©Fx = 0; NE = 0
MD = 13.5 kip #^ ft
+ ©MD = 0; MD + 2.25(2) - 3.00(6) = 0
VD = 0.750 kip
- c ©Fy = 0; 3.00 - 2.25 - VD = 0
:^ +^ ©Fx = 0; ND = 0
- c ©Fy = 0; By + 3.00 - 9.00 = 0 By = 6.00 kip
:^ + ©Fx = 0; Bx = 0
- ©MB = 0; 9.00(4) - Ay(12) = 0 Ay = 3.00 kip
6 ft 4 ft
A
4 ft
D B E C
6 ft
3 kip 1.5 kip/ ft
Ans: , ,
NE = 0 , VE = -9.00 kip, ME = -24.0 kip #^ ft
N D = 0 VD = 0.750 kip MD = 13.5 kip #^ ft,
Support Reactions:
a
Ans.
Equations of Equilibrium: For point C
Ans.
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a
Ans.
Negative signs indicate that NC and VC act in the opposite direction to that shown on FBD.
MC = 0.400 kN #^ m
+ ©MC = 0; 0.5333(0.75) - MC = 0
VC = -0.533 kN
- c ©Fy = 0; VC + 0.5333 = 0
NC = -2.00 kN
:^ + ©Fx = 0; - NC - 2.00 = 0
- c ©Fy = 0; A (^) y - 0.5333 = 0 A (^) y = 0.5333 kN
:^ + ©Fx = 0; 2 - A (^) x = 0 A (^) x = 2.00 kN
P = 0.5333 kN = 0.533 kN
+ ©MA = 0; P(2.25) - 2(0.6) = 0
1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading.
0.75 m
C
P
A
B
0.1 m 0.5 m
0.75 m 0.75 m
Ans: , , ,
MC = 0.400 kN #^ m
P = 0.533 kNNC = -2.00 kN VC = -0.533 kN
Referring to the FBD of the entire beam, Fig. a ,
a
Referring to the FBD of this segment, Fig. b ,
Ans.
Ans.
a + ©MC = 0; MC + 6(0.5) - 7.5(1) = 0 MC = 4.50 kN #^ m Ans.
- c ©Fy = 0; 7.50 - 6 - VC = 0 VC = 1.50 kN
:^ + ©Fx = 0; NC = 0
- ©MB = 0; - Ay(4) + 6(3.5) +
(3)(3)(2) = 0 Ay = 7.50 kN
*1–8. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical.
0.5 m 0.5 m 1.5 m 1.5 m
C
A B
3 kN/m
6 kN
D
Equations of Equilibrium: For point A
Ans.
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a
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
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a
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Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
Ans.
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a
Ans.
Negative signs indicate that N (^) C and MC act in the opposite direction to that shown on FBD.
MC = -8125 lb #^ ft = -8.125 kip #^ ft
+ ©MC = 0; - MC - 650(6.5) - 300(13) = 0
NC = -1200 lb = -1.20 kip
- c © Fy = 0; - NC - 250 - 650 - 300 = 0
;^ + © Fx = 0; VC = 0
MB = -6325 lb #^ ft = -6.325 kip #^ ft
+ © MB = 0; - MB - 550(5.5) - 300(11) = 0
VB = 850 lb
- c © Fy = 0; VB - 550 - 300 = 0
;^ + © Fx = 0; NB = 0
MA = -1125 lb #^ ft = -1.125 kip #^ ft
+ ©MA = 0; - MA - 150(1.5) - 300(3) = 0
VA = 450 lb
- c © Fy = 0; VA - 150 - 300 = 0
;^ + © Fx = 0; NA = 0
1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A , B , and C.
5 ft
7 ft
C
D (^) F
E
B A
300 lb
2 ft 8 ft 3 ft
Ans: , , , ,
VC = 0 , NC = -1.20 kip,MC = -8.125 kip #^ ft
NB = 0 VB = 850 lb MB = -6.325 kip #^ ft,
NA = 0 VA = 450 lb MA = -1.125 kip #^ ft,
