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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.
Ans.
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)
Ans.
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.
Ans.
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.
Ans.
a
Ans.
Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
Ans.
Ans.
a
Ans.
Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
Ans.
Ans.
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,
1–11. The forearm and biceps support the 2-kg load at A. If C can be assumed as a pin support, determine the resultant internal loadings acting on the cross section of the bone of the forearm at E .The biceps pulls on the bone along BD.
Support Reactions: In this case, all the support reactions will be completed. Referring to the free-body diagram of the forearm, Fig. a ,
a
Internal Loadings: Using the results of C x and C y , section CE of the forearm will be considered. Referring to the free-body diagram of this part shown in Fig. b ,
Ans.
Ans.
a Ans.
The negative signs indicate that N E , V E and M E act in the opposite sense to that shown on the free-body diagram.
+ ©ME = 0; ME + 64.47(0.035) = 0 ME = - 2.26 N #^ m
- c©Fy = 0; -VE - 64.47 = 0 VE = - 64.5 N
: NE + 22.53 = 0 N E = -22.5 N
©Fx = 0;
- c ©Fy = 0; 87.05 sin 75° - 2(9.81) - Cy = 0 C (^) y = 64.47 N
: C (^) x - 87.05 cos 75° = 0 Cx = 22.53 N
©Fx = 0;
- ©MC = 0; FBD sin 75°(0.07) - 2(9.81)(0.3) = 0 FBD = 87.05 N
75
230 mm 35 mm35 mm
C (^) E B
D
A
Ans:
NE = -22.5 N , VE = - 64.5 N, M E = - 2.26 N #^ m
Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a ,
Ans.
Ans.
a Ans.
The negative sign indicates that N a–a and M a–a act in the opposite sense to that shown on the free-body diagram.
+ ©MD = 0; - M a - a - 100(0.15) = 0 Ma - a = -15 N #^ m
; Na - a + 100 = 0 N (^) a - a = -100 N
©Fx = 0;
1–13. The blade of the hacksaw is subjected to a pretension force of Determine the resultant internal loadings acting on section a–a that passes through point D.
F = 100 N.
A B
C
D
F F
a
b
b a
30
225 mm
150 mm
Ans:
Na - a = -100 N , Va - a = 0 , Ma - a = -15 N #^ m
1–14. The blade of the hacksaw is subjected to a pretension force of. Determine the resultant internal loadings acting on section b–b that passes through point D.
F = 100 N
Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a ,
Ans.
Ans.
a Ans.
The negative sign indicates that N b–b and M b–b act in the opposite sense to that shown on the free-body diagram.
+ ©M D = 0; - Mb - b - 100(0.15) = 0 Mb - b = -15 N #^ m
©Fy¿ = 0; Vb - b - 100 sin 30° = 0 Vb - b = 50 N
©Fx¿ = 0; Nb - b + 100 cos 30° = 0 Nb - b = - 86.6 N
A B
C
D
F F
a
b
b a
30
225 mm
150 mm
Ans: , ,
Mb - b = -15 N #^ m
N (^) b - b = - 86.6 N Vb - b = 50 N
*1–16. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings acting on the cross section at E.
Support Reactions: We will only need to compute A x , A y , and F BI. Referring to the free-body diagram of the frame, Fig. a ,
a
Internal Loadings: Using the results of A x and A y , section AE of member AB will be considered. Referring to the free-body diagram of this part shown in Fig. b ,
Ans.
Ans.
a Ans.
The negative sign indicates that N E acts in the opposite sense to that shown on the free-body diagram.
+ ©MD = 0; 100(3) - ME = 0 ME = 300 lb # ft
- c ©Fy = 0; 100 - VE = 0 VE = 100 lb
: NE + 323.21 = 0 NE = - 323 lb
©Fx = 0;
- c ©Fy = 0; A (^) y - 200 cos 30° - 150 = 0 A (^) y = 323.21 lb
:^ + ©Fx = 0; A^ x -^ 200 sin 30°^ =^0 A^ x =^ 100 lb
- ©MA = 0; F (^) BI sin 30°(6) - 150(4) = 0 FBI = 200 lb
2 ft
2 ft
3 ft
1 ft 1 ft
1 ft E
D (^) C
B
I A
30 G
H
45
1.5 m
1.5 m
3 m
45
A
C
B
b a
a
b
5 kN
Referring to the FBD of the entire beam, Fig. a ,
a
Referring to the FBD of this segment (section a–a ), Fig. b ,
Ans.
Ans.
a Ans.
Referring to the FBD (section b–b ) in Fig. c ,
Ans.
Ans.
a
Mb - b = 3.75 kN #^ m Ans.
= -1.77 kN
;^ + ©Fx = 0; Nb - b - 5 cos 45° + 5.303 = 0 Nb - b = -1.768 kN
+ ©MC = 0; 5.303 sin 45°(3) - 5(1.5) - Ma - a = 0 Ma - a = 3.75 kN #^ m
+a ©Fy¿ = 0; Va - a + 5.303 sin 45° - 5 = 0 Va - a = 1.25 kN
+b©Fx¿ = 0; Na - a + 5.303 cos 45° = 0 Na - a = -3.75 kN
- ©MA = 0; NB sin 45°(6) - 5(4.5) = 0 NB = 5.303 kN
1–17. Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C.
Ans:
V b - b = 3.54 kN #^ m, Mb - b = 3.75 kN #^ m
Ma - a = 3.75 kN #^ m, Nb - b = -1.77 kN,
Na - a = -3.75 kN, V (^) a - a = 1.25 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 = 31.5 kip #^ ft Ans.
+ ©MC = 0; MC + (3)(3)(1.5) +
(3)(3) - (3)(3) - VC = 0 VC = 4.50 kip
:^ + ©Fx = 0; NC = 0
+ ©MB = 0;
(6)(6)(10) - A (^) y(12) = 0 Ay = 18.0 kip
1–19. Determine the resultant internal loadings acting on the cross section through point C. Assume the reactions at the supports A and B are vertical.
3 ft 3 ft
C D A B
6 ft
6 kip/ft 6 kip/ft
Ans:
NC = 0 , V C = 4.50 kip, MC = 31.5 kip #^ ft
Referring to the FBD of the entire beam, Fig. a ,
a
Referring to the FBD of this segment, Fig. b ,
Ans.
Ans.
a + ©MA = 0; MD - 18.0 (2) = 0 MD = 36.0 kip #^ ft Ans.
(6)(6) - VD = 0 VD = 0
:^ + ©Fx = 0; ND = 0
+ ©MB = 0;
(6)(6)(10) - A (^) y(12) = 0 Ay = 18.0 kip
*1–20. Determine the resultant internal loadings acting on the cross section through point D. Assume the reactions at the supports A and B are vertical.
3 ft 3 ft
C D A B
6 ft
6 kip/ft 6 kip/ft
