Chapter 4

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4–1. Determine the internal normal force, shear force, and bending moment in the beam at points C and D. Assume the support at A is a pin and B is a roller.

6 kN 20 kN · m A

C

1m

Entire beam: + : a Fx = 0;

Ax = 0

a + a MA = 0;

By(6) - 20 - 6(2) = 0 By = 5.333 kN

+ c a Fy = 0;

Ay + 5.333 - 6 = 0 Ay = 0.6667 kN

Segment AC: + : a Fx = 0;

NC = 0

+ c a Fy = 0;

0.6667 - VC = 0

Ans.

VC = 0.667 kN a + a MC = 0;

Ans.

MC - 0.6667(1) = 0 MC = 0.667 kN # m

Ans.

+ : a Fx = 0;

ND = 0

Ans.

+ c a Fy = 0;

VD + 5.333 = 0

Segment DB:

VD = -5.33 kN a + a MD = 0;

Ans.

-MD + 5.333(2) - 20 = 0 MD = -9.33 kN # m

Ans.

80

1m

B

D

2m

2m


4–2. Determine the internal normal force, shear force, and bending moment in the beam at points C and D. Assume the support at B is a roller. Point D is located just to the right of the 10–k load.

10 k 25 k · ft

25 k · ft

A C 10 ft

Entire Beam: a + a MA = 0;

By (30) + 25 - 25 - 10(20) = 0 By = 6.667 k

+ c a Fy = 0;

Ay + 6.667 - 10 = 0 Ay = 3.333 k

+ : a Fx = 0;

Ax = 0

Segment AC: + : a Fx = 0;

NC = 0

+ c a Fy = 0;

-VC + 3.333 = 0

Ans.

VC = 3.33 k a + a MC = 0;

Ans.

MC - 25 - 3.333(10) = 0 MC = 58.3 k # ft

Ans.

+ : a Fx = 0;

ND = 0

Ans.

+ c a Fy = 0;

VD + 6.6667 = 0

Segment DB:

VD = -6.67 k a + a MD = 0;

Ans.

-MD + 25 + 6.667(10) = 0 MD = 91.7 k # ft

Ans.

81

D 10 ft

B 10 ft


4–3. 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 internal normal force, shear force, and bending moment in the crane at points A, B, and C.

D

2 ft

F

A

B 8 ft

3 ft

5 ft C 300 lb 7 ft

E

Equations of Equilibrium: For point A + ; a Fx = 0;

NA = 0

+ c a Fy = 0;

VA - 150 - 300 = 0

Ans.

VA = 450 lb a + a MA = 0;

Ans.

-MA - 150(1.5) - 300(3) = 0 MA = -1125 lb # ft = -1.125 kip # ft

Ans.

Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ; a Fx = 0;

NB = 0

+ c a Fy = 0;

VB - 550 - 300 = 0

Ans.

VB = 850 lb a + a MB = 0;

Ans.

-MB - 550(5.5) - 300(11) = 0 MB = -6325 lb # ft = -6.325 kip # ft

Ans.

Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ; a Fx = 0;

VC = 0

+ c a Fy = 0;

-NC - 250 - 650 - 300 = 0

Ans.

NC = -1200 lb = -1.20 kip a + a MC = 0;

Ans.

-MC - 650(6.5) - 300(13) = 0 MC = -8125 lb # ft = -8.125 kip # ft

Ans.

Negative sign indicate that NC and MC act in the opposite direction to that shown on FBD.

82


*4–4. Determine the internal normal force, shear force, and bending moment at point D. Take w = 150 N> m.

w B

A D 4m

3m

4m C

4m

a + a MA = 0;

-150(8)(4) +

3 F (8) = 0 5 BC

FBC = 1000 N + : a Fx = 0;

Ax -

4 (1000) = 0 5

Ax = 800 N

+ c a Fy = 0;

3 Ay - 150(8) + (1000) = 0 5 Ay = 600 N

+ : a Fx = 0; + c a Fy = 0;

ND = -800 N

Ans.

600 - 150(4) - VD = 0 VD = 0

a + a MD = 0;

Ans.

-600(4) + 150(4)(2) + MD = 0 MD = 1200 N # m = 1.20 kN # m

Ans.

83


4–5. The beam AB will fail if the maximum internal moment at D reaches 800 N . m or the normal force in member BC becomes 1500 N. Determine the largest load w it can support.

w B

A D 4m

3m

4m C

4m

Assume maximum moment occurs at D; 8w MD a + a MD = 0; (4) + 4w (2) = 0 2 800 = 8 w w = 100 N>m a + a MA = 0;

-800(4) + TBC(0.6)(8) = 0 TBC = 666.7 N 6 1500 N

(O. K!)

w = 100 N>m

Ans.

84


4–6. Determine the internal normal force, shear force, and bending moment in the beam at points C and D. Assume the support at A is a roller and B is a pin.

4 kN/m

A

C 1.5 m

Support Reactions. Referring to the FBD of the entire beam in Fig. a, a + a MB = 0;

1 (4)(6)(2) - Ay(3) = 0 2

Ay = 8 kN

Internal Loadings. Referring to the FBD of the left segment of the beam sectioned through point C, Fig. b, + : a Fx = 0;

+ c a Fy = 0;

a + a MC = 0;

NC = 0

Ans.

1 (1)(1.5) - VC = 0 2

-

MC +

1 (1)(1.5)(0.5) = 0 2

VC = - 0.75 kN

MC = -0.375 kN # m

Ans.

Ans.

Referring to the FBD of the left segment of the beam sectioned through point D, Fig. c, + : a Fx = 0;

+ c a Fy = 0;

a + a MD = 0;

ND = 0

8-

Ans.

1 (3)(4.5) - VD = 0 2

MD +

VD = 1.25 kN

Ans.

1 (3)(-4.5)(1.5) - 8(1.5) = 0 2 MD = 1.875 kN # m

85

Ans.

1.5 m

B

D 1.5 m

1.5 m


4–7. Determine the internal normal force, shear force, and bending moment at point C. Assume the reactions at the supports A and B are vertical.

1.5 kN/m 0.5 kN/m

A

B C 6m

3m

+ : a Fx = 0;

NC = 0

+ T a Fy = 0;

VC + 0.5 + 1.5 - 3.75 = 0

Ans.

VC = 1.75 kN

a + a MC = 0;

Ans.

MC + 0.5(1) + 1.5(1.5) - 3.75(3) = 0 MC = 8.50 kN # m

Ans.

*4–8. Determine the internal normal force, shear force, and bending moment at point D. Assume the reactions at the supports A and B are vertical.

1.5 kN/m 0.5 kN/m

A

B D 6m

+ : a Fx = 0;

ND = 0

+ c a Fy = 0;

3.75 - 3 - 2 -VD = 0

Ans.

VD = -1.25 kN

a + a MD = 0;

Ans.

MD + 2(2) + 3(3) - 3.75(6) = 0 MD = 9.50 kN # m

Ans.

86

3m


4–9. Determine the internal normal force, shear force, and bending moment in the beam at point C. The support at A is a roller and B is pinned.

5 kN 3 kN/m

A

Support Reactions. Referring to the FBD of the entire beam in Fig a, a + a MA = 0; + : a Fx = 0;

By (4) + 5(1) - 3(4)(2) = 0

1m

By = 4.75 kN

Bx = 0

Internal Loadings. Referring to the FBD of the right segment of the beam sectioned through point c, Fig. b, + : a Fx = 0;

NC = 0

+ c a Fy = 0;

VC + 4.75 - 3(2) = 0

VC = 1.25 kN

Ans.

a + a MC = 0;

4.75(2) - 3(2)(1) - MC = 0

MC = 3.50 kN # m

Ans.

Ans.

87

B

C

2m

2m


4–10. Determine the internal normal force, shear force, and bending moment at point C. Assume the reactions at the supports A and B are vertical.

400 lb/ft

300 lb/ft

A

B

C 12 ft

Support Reactions: a + a MA = 0;

By(20) - 6(10) - 1.8(23) = 0 By = 5.07 kip

+ c a Fy = 0;

Ay + 5.07 - 6 - 1.8 = 0 Ay = 2.73 kip

Equations of Equilibrium: For point C + : a Fx = 0;

+ c a Fy = 0;

NC = 0

Ans.

273 - 3.60 - VC = 0 VC = -0.870 kip

a + a MC = 0;

Ans.

MC + 3.60(6) - 2.73(12) = 0 MC = 11.2 kip # ft

Ans.

Negative sign indicates that VC acts in the opposite direction to that shown on FBD.

88

8 ft

9 ft


4–11. Determine the internal normal force, shear force, and bending moment at points D and E. Assume the reactions at the supports A and B are vertical.

400 lb/ft

300 lb/ft

A

B

C 12 ft


By (20) - 6(10) - 1.8(23) = 0 By = 5.07 kip

+ c a Fy = 0;

Ay + 5.07 - 6 - 1.8 = 0 Ay = 2.73 kip

Equations of Equilibrium: For point D + : a Fx = 0;

+ c a Fy = 0;

ND = 0

Ans.

2.73 - 1.8 - VD = 0 VD = 0.930 kip

a + a MD = 0;

Ans.

MD + 1.8(3) - 2.73(6) = 0 MD = 11.0 kip # ft

Ans.

Equations of Equilibrium: For point E + ; a Fx = 0;

NE = 0

+ c a Fy = 0;

VE - 0.45 = 0

Ans.

VE = 0.450 kip

a + a ME = 0;

Ans.

-ME - 0.45(1.5) = 0 ME = - 0.675 kip # ft

Ans.

Negative sign indicates that ME acts in the opposite direction to that shown on FBD.

89

8 ft

9 ft


*4–12. Determine the shear and moment throughout the beam as a function of x.

P a

b

A

B x L

Support Reactions: Referring to the FBD of the entire beam in Fig. a, a + a MA = 0;

NB (L) - Pa = 0

NB =

Pa L

a + a MB = 0;

Pb - Ay (L) = 0

Ay =

Pb L

+ : a Fx = 0;

Ax = 0

Internal Loading: For 0 … x 6 a, refer to the FBD of the left segment of the beam in Fig. b. + c a Fy = 0;

Pb - V=0 L

a + a MO = 0;

M -

V =

Pb L

Ans.

Pb Pb x=0 M = x L L

Ans.

For a 6 x … L, refer to the FBD of the right segment of the beam in Fig. c. + c a Fy = 0; a + a MO = 0;

V+

Pa =0 L

V= -

Pa L

Ans.

Pa (L - x) - M = 0 L M=

Pa (L - x) L

Ans.

90


4–13. Determine the shear and moment in the floor girder as a function of x. Assume the support at A is a pin and B is a roller.

6 kN 4 kN

A

B 1m x

Support Reactions: Referring to the FBD of the entire beam in Fig. a. a + a MA = 0;

By (4) - 4(1) - 6(3) = 0 By = 5.50 kN

a + a MB = 0;

6(1) + 4(3) - Ay(4) = 0 Ay = 4.50 kN

+ : a Fx = 0;

Ax = 0

Internal Loadings: For 0 … x 6 1 m, Referring to the FBD of the left segment of the beam in Fig. b, + c a Fy = 0;

4.50 - V = 0

a + a MO = 0;

M - 4.50 x =0

V = 4.50 kN

Ans.

M = 54.50 x6 kN # m

Ans.

For 1 m 6 x 6 3 m, referring to the FBD of the left segment of the beam in Fig. c, + c a Fy = 0;

4.50 - 4 - V = 0 V = 0.500 kN

a + a MO = 0;

Ans.

M + 4 (x - 1) - 4.50 x = 0 M = 50.5 x + 46 kN # m

Ans.

For 3 m 6 x … 4 m, referring to the FBD of the right segment of the beam in Fig. d, + c a Fy = 0; a + a MO = 0;

V + 5.50 = 0

V = -5.50 kN

Ans.

5.50(4 - x) - M = 0 M = 5-5.50x + 226 kN # m

Ans.

91

2m

1m


4–14. Determine the shear and moment throughout the beam as a function of x.

a

b M0

A

B x L

Support Reactions: Referring to the FBD of the entire beam in Fig. a + : a Fx = 0;

Ax = 0

a + a MA = 0;

MO - NB (L) = 0

a + a MB = 0;

MO - Ay (L) = 0

MO L MO Ay = L By =

Internal Loadings: For 0 … x 6 a, refer to the FBD of the left segment of the beam is Fig. b. + c a Fy = 0; a + a Mo = 0;

MO - V=0 L Mo M x=0 L

MO L MO M= x L V=

Ans. Ans.

For a 6 x … L, refer to the FBD of the right segment of the beam in Fig. c + c a Fy = 0; a + a Mo = 0;

MO =0 L Mo -M (L - x) = 0 L

V -

M= -

V=

MO L

Ans. Ans.

Mo (L - x) L

Ans.

92



7 kN 12 kN⭈ m A

B x 2m

Reaction at A: + : a Fx = 0;

Ax = 0

c + a MB = 0;

Ay(8) - 7(6) + 12 = 0; Ay = 3.75 kN

Ans.

0…x62m + c a Fy = 0;

3.75 - V = 0;

V = 3.75 kN

Ans.

c + a M = 0;

3.75x - M = 0;

M = 3.75x kN

Ans.

+ c a Fy = 0;

- V + 3.75 - 7 = 0;

V = -3.25

Ans.

c + a M = 0;

-M + 3.75x - 7(x - 2) = 0;

2m6x64m Segment:

M = -3.25x + 14

Ans.

V = -3.25 kN

Ans.

4m6x…8m + c a Fy = 0;

3.75 - 7 - V = 0;

a + a M = 0;

- 3.75x + 7(x - 2) - 12 + M = 0; M = 26 - 3.25x

93

Ans.

2m

4m


*4–16. Determine the shear and moment throughout the beam as a function of x.

8 kN/m

A

B x 3m

Support Reactions. Referring to the FBD of the entire beam in Fig. a, a + a MA = 0;

By (6) - 8(3)(4.5) -

1 (8)(3)(2) = 0 2

By = 22 kN a + a MB = 0;

8(3)(1.5) +

1 (8)(3)(4) - Ay (6) = 0 2

Ay = 14 kN + : a Fx = 0;

Ax = 0

Internal Loadings: For 0 … x 6 3 m, refer to the FBD of the left segment of the beam in Fig. b, + c a Fy = 0;

a + a MO = 0;

14 -

1 8 a xbx - V = 0 2 3

V=

E -1.33x 2 + 14 F kN

M+

x 1 8 a xb(x)a b - 14x = 0 2 3 3

M=

Ans.

E -0.444 x 3 + 14 x F kN # m

Ans.

For 3 m 6 x … 6 m, refer to the FBD of the right segment of the beam in Fig. c + c a Fy = 0;

V + 22 - 8(6 - x) = 0 V=

a + a MO = 0;

E -8x + 26 F kN

22(6 - x) - 8(6 - x)a

Ans. 6-x b -M=0 2

M = 5 -4 x2 + 26x - 126 kN # m

Ans.

94

3m



8 kN

8 kN

4 kN

A

x 1m

Internal Loadings. beam in Fig. a,

For 0 … x … 1 m, referring to the FBD of the left segment of the

+ c a Fy = 0;

-V-4=0

V = -4 kN

Ans.

a + a MO = 0;

M + 4x =0

M = 5-4 x6 kN # m

Ans.

For 1 m 6 x 6 2 m, referring to the FBD of the left segment of the beam in Fig. b, V = 5-126 kN # m

+ c a Fy = 0;

-4-8 - V=0

a + a MO = 0;

M + 8 (x - 1) + 4x = 0

Ans.

M = 5 -12 x + 86 kN # m

Ans.

For 2 m 6 x … 3 m, referring to the FBD of the left segment of the beam in Fig. c, V = 5-206 kN

+ c a Fy = 0;

-4 - 8 - 8 - V = 0

a + a MO = 0;

M + 4x + 8 (x - 1) + 8(x - 2) = 0

Ans.

M = 5-20x + 24 6 kN # m

Ans.

95

1m

1m


4–18. Determine the shear and moment throughout the beam as functions of x.

2 k/ft

10 k

8k 40 k⭈ft

x 6 ft

Support Reactions: As shown on FBD. Shear and Moment Functions: For 0 … x 6 6 ft + c a Fy = 0;

30.0 - 2x - V = 0 V = 530.0 - 2x6 k

Ans.

x M + 216 + 2xa b - 30.0x = 0 2

a + a MNA = 0;

M = 5-x2 + 30.0x - 2166 k # ft

Ans.

For 6 ft 6 x … 10 ft + : a Fy = 0; a + a MNA = 0;

V-8=0

V = 8.00 k

Ans.

-M - 8 (10 -x) - 40 = 0 M = 58.00x - 1206 k # ft

Ans.

96

4 ft


4–19. Determine the shear and moment throughout the beam as functions of x.

250 lb

250 lb 150 lb/ ft

A

B

x 4 ft

Support Reactions: As shown on FBD. Shear and moment Functions: For 0 … x 6 4 ft + c a Fy = 0;

-250 - V = 0

a + a MNA = 0;

M + 250x = 0

V = -250 lb

Ans.

M = 5-250x F lb # ft

Ans.

For 4 ft 6 x 6 10 ft + c a Fy = 0;

-250 + 700 - 150(x - 4) - V = 0 V = 51050 - 150x F lb

a + a MNA = 0;

M + 150(x - 4)a

Ans.

x-4 b + 250x - 700(x - 4) = 0 2

M = 5- 75x2 + 1050x - 40006lb # ft

Ans.

For 10 ft 6 x … 14 ft + c a Fy = 0; a + a MNA = 0;

V - 250 = 0

V = 250 lb

Ans.

-M - 250(14 - x) = 0 M = 5250x - 35006lb # ft

Ans.

97

6 ft

4 ft


*4–20. Determine the shear and moment in the beam as functions of x.

w0

A

B x L __ 2

Support Reactions: As shown on FBD. Shear and Moment Functions: For 0 … x 6 L> 2 + c a Fy = 0; a + a MNA = 0;

3woL - wo x - V = 0 4

V=

wo (3L-4x) 4

Ans.

7woL2 3woL x x + woxa b + M = 0 24 4 2 wo 2 (-12x + 18Lx - 7L2) M= 24

Ans.

For L> 2 6 x … L + c a Fy = 0;

a + a MNA = 0;

V-

1 2wo c (L-x) d (L-x) = 0 2 L

V=

wo (L -x)2 L

Ans.

1 2wo L -x c (L -x) d(L -x) a b =0 2 L 3 wo (L -x)2 M= 3L

-M -

Ans.

98

L __ 2


4–21. Determine the shear and moment in the beam as a function of x.

200 lb/ft

1200 lb ⭈ ft A

B x 10 ft

800 lb

Internal Loadings: Referring to the FBD of the left segment of the beam in Fig. a,

+ c a Fy = 0;

-800 -

1 200 a xb (x) - V = 0 2 10

V = 5 -10x2 - 8006lb

a + a Mo = 0;

M +

Ans.

1 200 x a xb(x)a b + 800x + 1200 = 0 2 10 3

M = 5 -3.33x3 - 800x - 12006 lb # ft

Ans.

4–22 Determine the shear and moment throughout the tapered beam as a function of x.

8 kN/m

B A

x 9m

1 8 8 8 + : a Fy = 0; 36 - a xb (x) - a 8 - xb x - V = 0 2 9 9 9 V = 36 -

4 2 8 x - 8x + x2 9 9

V = 0.444x2 - 8x + 36 a + a M = 0; 108 +

1 8 2 8 8 x a xb (x)a xb + a 8 - xb(x)a b - 36x + M = 0 2 9 3 9 9 2

M = -108 -

8 3 8 3 x - 4x2 + x + 36x 27 18

M = 0.148x3 - 4x2 + 36x - 108

Ans.

99


4–23.

Draw the shear and moment diagrams for the beam.

500 lb

300 lb

200 lb B

A

C

6 ft

*4–24. beam.

4 ft

E

D

4 ft

6 ft

2 ft

Draw the shear and moment diagrams for the

2k

2k

2k

2k

A 4 ft

100

4 ft

4 ft

4 ft

4 ft


4–25.


6 kN 0.4 m

5

4

20 kN · m

3

0.6 m A C 1m

Vmax = -4.89 kN

Ans.

Mmax = -20 kN # m

Ans.

4–26.

D 1m

2m

A

6 ft

Vmax = -10.1 k

Ans.

Mmax = -60 k # ft

Ans.

101

2m

0.8 k/ft

10 k

Draw the shear and moment diagrams of the beam.

B

8k

C

12 ft

B

12 ft

6 ft


4–27. Draw the shear and moment diagrams for the beam.

400 lb/ft 600 lb ⭈ ft A

B x 15 ft

*4–28. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set MO = 500 N . m, L = 8 m.

L/3

(a)

For

0 … x …

L 3

+ c a Fy = 0; V = 0

Ans.

a + a M = 0; M = 0

Ans.

For

2L L 6 x 6 3 3

+ c a Fy = 0; V = 0

Ans.

a + a M = 0; M = MO

Ans.

For

M0

M0

2L 6 x … L 3

+ c a Fy = 0; V = 0

Ans.

a + a M = 0; M = 0

Ans.

102

L/3

L/3


4–28. (b)

Continued Set MO = 500 N # m, L = 8 m

For

For

For

4–29.

0 … x 6

8 m 3

+ c a Fy = 0; V = 0

Ans.

c + a M = 0; M = 0

Ans.

8 16 m 6 x 6 m 3 3 + c a Fy = 0; V = 0

Ans.

c + a M = 0; M = 500 N # m

Ans.

16 m 6 x … 8m 3 + c a Fy = 0; V = 0

Ans.

c + a M = 0; M = 0

Ans.


1.5 kN/m

A

Support Reactions:

B 2m

a + a MA = 0; Cx(3) - 1.5(2.5) = 0 Cx = 1.25 kN

3m

+ c a Fy = 0; Ay - 1.5 + 1.25 = 0 Ay = 0.250 kN

Shear and Moment Functions: For 0 … x < 2 m [FBD (a)], + c a Fy = 0; 0.250 - V = 0 V = 0.250 kN

Ans.

a + a M = 0; M - 0.250x = 0 M = (0.250x) kN # m

Ans.

For 2 m < x … 3 m [FBD (b)]. + c a Fy = 0; 0.25 - 1.5(x - 2) - V = 0 V = (3.25 - 1.50x) kN a + a M = 0;

-0.25x + 1.5(x + 2) a

Ans. x-2 b + M = 0 2

M = ( -0.750x2 + 3.25x - 3.00) kN # m

Ans.

103

C


4–30. Draw the shear and bending-moment diagrams for the beam.

50 lb/ft 200 lb⭈ft A

C

B 20 ft

Support Reactions: a + a MB = 0; 1000(10) - 200 - Ay(20) = 0 Ay = 490 lb

Shear and Moment Functions: For 0 … x < 20 ft [FBD (a)]. + c a Fy = 0; 490 - 50x - V = 0 V =

E 490 - 50.0x F lb

Ans.

x a + a M = 0; M + 50xa b - 490x = 0 2 M = (490x - 25.0x2) lb # ft

Ans.

For 20 ft < x … 30 ft [FBD (b)], + c a Fy = 0; V = 0 a + a M = 0;

Ans.

-200 - M = 0 M = -200 lb # ft

Ans.

104

10 ft


4–31.


w A

B L –– 2 L

Support Reactions: From FBD(a), a + a MA = 0; Cy(L) + c a Fy = 0; Ay +

wL 3L a b = 0 2 4

3wL wL - a b = 0 8 2

Shear and Moment Functions: For 0 … x < + c a Fy = 0;

wL - V = 0 8

a + a M = 0; M -

Cy =

V =

3wL 8

Ay =

wL 8

L [FBD (b)], 2

wL 8

Ans.

wL wL (x) = 0 M = (x) 8 8

Ans.

L < x … L [FBD (c)], 2 3wL + c a Fy = 0; V + - w(L - x) = 0 8 w V = (5L - 8x) 8

For

a + a MB = 0;

3wL L - x (L - x) - w(L - x)a b - M = 0 8 2 M =

w (-L2 + 5Lx - 4x2) 8

Ans.

105

C


*4–32. beam.

Draw the shear and moment diagrams for the

250 lb/ft

A

B

150 lb ⭈ ft

a + a MA = 0;

150 lb ⭈ ft

20 ft

-5000(10) - 150 + By (20) = 0 By = 2500 lb

+ : a Fx = 0; Ax = 0 + c a Fy = 0; Ay = 5000 + 2500 = 0 Ay = 2500 lb For 0 … x … 20 ft + c a Fy = 0; 2500 - 250x - V = 0 V = 250(10 - x) a + a M = 0;

Ans.

x -2500(x) + 150 + 250xa b + M = 0 2 M = 25(100x - 5x2 - 6)

4–33.

Ans.


20 kN 40 kN/m

A B

0 … x < 8 8m

+ c a Fy = 0; 133.75 - 40x - V = 0 V = 133.75 - 40x

Ans.

x a + a M = 0; M + 40xa b - 133.75x = 0 2 M = 133.75x - 20x2

Ans.

8 < x … 11 + c a Fy = 0; V - 20 = 0 V = 20

Ans.

c + a M = 0; M + 20(11 - x) + 150 = 0 M = 20x - 370

Ans.

106

3m

150 kN⭈m


4–34.


200 lb/ft

C

D

E

F

G

A

B x 4 ft

Vmax = ;1200 lb

Ans.

Mmax = 6400 lb # ft

Ans.

4–35.

4 ft

4 ft


4 ft

200 lb/ft

1200 lb ⭈ ft A

B

30 ft 800 lb

Vmax = -3.80 k

Ans.

Mmax = -55.2 k # ft

Ans.

107


*4–36. Draw the shear and moment diagrams of the beam. Assume the support at B is a pin and A is a roller.

100 lb/ft 800 lb · ft A

4 ft

Vmax = 850 lb

Ans.

Mmax = -2.81 K # ft

Ans.

4–37. Draw the shear and moment diagrams for the beam. Assume the support at B is a pin.

B

16 ft

8 kN/m

A 1.5 m

6m

Vmax = 24.5 kN

Ans.

Mmax = 34.5 kN # m

Ans.

108

B


4–38. Draw the shear and moment diagrams for each of the three members of the frame. Assume the frame is pin connected at A, C, and D and there is fixed joint at B.

50 kN 1.5 m

40 kN 2m

1.5 m

B

C

4m 6m A

D

109

15 kN/m


4–39. Draw the shear and moment diagrams for each member of the frame. Assume the support at A is a pin and D is a roller.

0.8 k/ft B

C

0.6 k/ft 16 ft

A

D

20 ft

Vmax = -11.8 k

Ans.

Mmax = -87.6 k # ft

Ans.

110


*4–40. Draw the shear and moment diagrams for each member of the frame. Assume A is a rocker, and D is pinned.

4k 2 k/ ft B

C 3k

8 ft

4 ft

15 ft

A

111

D


4–41. Draw the shear and moment diagrams for each member of the frame. Assume the frame is pin connected at B, C, and D and A is fixed.

6k

3k

6k 8 ft

8 ft

3k 8 ft

0.8 k/ft B

C

15 ft

D

A

4–42. Draw the shear and moment diagrams for each member of the frame. Assume A is fixed, the joint at B is a pin, and support C is a roller.

20 k 5

3 4

0.5 k/ft B C

8 ft

Vmax = 20.0 k

Ans.

Mmax = -144 k # ft

Ans.

A 6 ft

112

6 ft


4–43. Draw the shear and moment diagrams for each member of the frame. Assume the frame is pin connected at A, and C is a roller.

4 k/ft B 15 k

C 4 ft

10 k 4 ft A 10 ft

*4–44. Draw the shear and moment diagrams for each member of the frame. Assume the frame is roller supported at A and pin supported at C.

1.5 k/ ft B

A 10 ft

6 ft

2k

6 ft C

113


4–45. Draw the shear and moment diagrams for each member of the frame. The members are pin connected at A, B, and C.

15 kN 2m

A

10 kN

2m

2m

B

6m


-15(2) - 10(4) + By (6) = 0 By = 11.667 kN

+ c a Fy = 0; Ay - 25 + 11.667 = 0 Ay = 13.3 kN a + a Mc = 0; 12(2) - By¿ (6) = 0 By¿ = 4 kN +Q a Fy¿ = 0; 4 - 12 + Cy¿ = 0 Cy¿ = 8 kN

114

45⬚

4 kN/m C


4–46. Draw the shear and moment diagrams for each member of the frame.

5 kN

10 kN

B

5 kN

2 kN/m

C

4m D

A

a + a MD = 0; 10(2.5) + 5(3) + 10(5) + 5(7) - Ay(10) = 0 Ay = 12.5 kN + : a Fx = 0;

4 -10 a b + Dx = 0 5 Dx = 8 kN

3 + c a Fy = 0; 12.5 - 5 - 10 - 5 - 10a b + Dy = 0 5 Dy = 13.5 kN

115

3m

2m

2m

3m


4–47. Draw the shear and moment diagrams for each member of the frame. Assume the joint at A is a pin and support C is a roller. The joint at B is fixed. The wind load is transferred to the members at the girts and purlins from the simply supported wall and roof segments.

300 lb/ft

C

3.5 ft

30⬚ 500 lb/ft

B 7 ft


-3.5(7) - 1.75(14) - (4.20)(sin 30°)(7 cos 30°) 7 ft

-4.20( sin 30°)(14 + 3.5) + (21) = 0 A

Cx = 5.133 kN + : a Fx = 0; 1.75 + 3.5 + 1.75 + 4.20 sin 30° - 5.133 - Ax = 0 Ax = 3.967 kN + c a Fy = 0; Ay - 4.20 cos 30° = 0 Ay = 3.64 kN

116

3.5 ft


*4–48. Draw the shear and moment diagrams for each member of the frame. The joints at A, B and C are pin connected.

250 lb/ft B 120 lb/ft

A 6 ft

6 ft 8 ft 60 C

6k

4–49. Draw the shear and moment diagrams for each of the three members of the frame. Assume the frame is pin connected at B, C and D and A is fixed.

8 ft

3k 8 ft

8 ft

0.8 k/ ft B

C

15 ft

A

117

D


4–50. Draw the moment diagrams for the beam using the method of superposition. The beam is cantilevered from A.

600 lb

600 lb

600 lb

A

1200 lb⭈ft

3 ft

118

3 ft

3 ft


4–51. Draw the moment diagrams for the beam using the method of superposition.

80 lb/ft

12 ft

12 ft

600 lb

*4-52. Draw the moment diagrams for the beam using the method of superposition. Consider the beam to be cantilevered from end A.

250 lb/ft 150 lb ⭈ ft

150 lb⭈ ft

A

B 20 ft

119


4–53. Draw the moment diagrams for the beam using the method of superposition. Consider the beam to be simply supported at A and B as shown.

250 lb/ft 150 lb ⭈ ft

150 lb⭈ ft

A

B 20 ft

4–54. Draw the moment diagrams for the beam using the method of superposition. Consider the beam to be cantilevered from the pin support at A.

30 kN

4 kN/ m 80 kN⭈ m

C

A B

8m

120

4m


4–54.

Continued

121


4–55. Draw the moment diagrams for the beam using the method of superposition. Consider the beam to be cantilevered from the rocker at B.

30 kN

4 kN/ m 80 kN⭈ m

C

A B

8m

122

4m


*4–56. Draw the moment diagrams for the beam using the method of superposition. Consider the beam to be cantilevered from end C.

30 kN

4 kN/ m 80 kN⭈ m

C

A B

8m

123

4m


4–57. Draw the moment diagrams for the beam using the method of superposition. Consider the beam to be simply supported at A and B as shown.

200 lb/ft 100 lb⭈ft

100 lb⭈ft A

B 20 ft

124

Chapter 4

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