Solutions (8th ed structural analysis) chapter 4

Determine the internal normal force, shear force, and bending moment in the beam at points C and D... Determine the internal normal force, shear force, and bending moment in the beam at

VD = - 5.33 kN

VD + 5.333 = 0+ c a Fy = 0;

ND = 0:+ a Fx = 0;

MC = 0.667 kN#m

MC - 0.6667(1) = 0+a MC = 0;

VC = 0.667 kN

0.6667 - VC = 0+ c a Fy = 0;

NC = 0 :+ a Fx = 0;

Ay = 0.6667 kN

Ay + 5.333 - 6 = 0+ c a Fy = 0;

By = 5.333 kN

By(6) - 20 - 6(2) = 0+a MA = 0;

Ax = 0:+ a Fx = 0;

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.

ND = 0:+ a Fx = 0;

MC = 58.3 k#ft

MC - 25 - 3.333(10) = 0+a MC = 0;

VC = 3.33 k

-VC + 3.333 = 0+ c a Fy = 0;

NC = 0 :+ a Fx = 0;

Ax = 0:+ a Fx = 0;

Ay = 3.333 k

Ay + 6.667 - 10 = 0+ c a Fy = 0;

By = 6.667 k

By (30) + 25 - 25 - 10(20) = 0+a MA = 0;

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

Negative sign indicates that M Bacts in the opposite direction to that shown on FBD.

Equations of Equilibrium: For point C

NC = - 1200 lb = - 1.20 kip

-NC - 250 - 650 - 300 = 0+ c a Fy = 0;

VC = 0

;+ a Fx = 0;

MB = - 6325 lb#ft = - 6.325 kip#ft

-MB - 550(5.5) - 300(11) = 0+a MB = 0;

VB = 850 lb

VB - 550 - 300 = 0+ c a Fy = 0;

NB = 0

;+ a Fx = 0;

MA = - 1125 lb#ft = - 1.125 kip#ft

-MA - 150(1.5) - 300(3) = 0+a MA = 0;

VA = 450 lb

VA - 150 - 300 = 0+ c a Fy = 0;

NA = 0

;+ a Fx = 0;

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.

VD = 0

600 - 150(4) - VD = 0+ c a Fy = 0;

ND = - 800 N :+ a Fx = 0;

Ay = 600 N

Ay - 150(8) + 3

5(1000) = 0+ c a Fy = 0;

Ax = 800 N

Ax 4

-5(1000) = 0:+ a Fx = 0;

FBC = 1000 N

-150(8)(4) + 3

5 FBC(8) = 0 +a MA = 0;

*4–4. Determine the internal normal force, shear force,

and bending moment at point D Take w = 150 N m.>

w = 100 N>m

TBC = 666.7 N 6 1500 N

-800(4) + TBC(0.6)(8) = 0+a MA = 0;

w = 100 N>m

800 = 8 w

MD - 8w

2(4) + 4w (2) = 0+a MD = 0;

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

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.

1.5 m1.5 m

4 kN/m

1.5 m1.5 m

Support Reactions. Referring to the FBD of the entire beam in Fig a,

a

Internal Loadings. Referring to the FBD of the left segment of the beam

sectioned through point C, Fig b,

VD = 1.25 kN

8 - 1

2 (3)(4.5) - VD = 0 + c a Fy = 0;

ND = 0 :+ a Fx = 0;

MC = - 0.375 kN#m

MC + 1

2(1)(1.5)(0.5) = 0 +a MC = 0;

VC = - 0.75 kN -

1

2 (1)(1.5) - VC = 0 + c a Fy = 0;

NC = 0 :+ a Fx = 0;

Ay = 8 kN 1

2(4)(6)(2) - Ay(3) = 0+a MB = 0;

VC = 1.75 kN

VC + 0.5 + 1.5 - 3.75 = 0 + Ta Fy = 0;

NC = 0 :+ a Fx = 0;

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.

VD = - 1.25 kN

3.75 - 3 - 2 - VD = 0 + c a Fy = 0;

ND = 0 :+ a Fx = 0;

*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.

Support Reactions. Referring to the FBD of the entire beam in Fig a,

a

Internal Loadings. Referring to the FBD of the right segment of the beam

sectioned through point c, Fig b,

NC = 0 :+ a Fx = 0;

Bx = 0:+ a Fx = 0;

By = 4.75 kN

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

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

VC = - 0.870 kip

273 - 3.60 - VC = 0+ c a Fy = 0;

NC = 0 :+ a Fx = 0;

Ay = 2.73 kip

Ay + 5.07 - 6 - 1.8 = 0+ c a Fy = 0;

By = 5.07 kip

By(20) - 6(10) - 1.8(23) = 0 +a MA = 0;

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.

VE = 0.450 kip

VE - 0.45 = 0+ c a Fy = 0;

NE = 0

;+ a Fx = 0;

MD = 11.0 kip#ft

MD + 1.8(3) - 2.73(6) = 0+a MD = 0;

VD = 0.930 kip

2.73 - 1.8 - VD = 0+ c a Fy = 0;

ND = 0 :+ a Fx = 0;

Ay = 2.73 kip

Ay + 5.07 - 6 - 1.8 = 0+ c a Fy = 0;

By = 5.07 kip

By (20) - 6(10) - 1.8(23) = 0 +a MA = 0;

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.

L = 0

M = Pb

L x+ a MO = 0; M - PbL x = 0

V = PbL+ c a Fy = 0; Pb

L - V = 0

Ax = 0 :+ a Fx = 0;

Ay=PbL

Pb - Ay (L) = 0 + a MB = 0;

NB = PaL

NB (L) - Pa = 0 +a MA = 0;

B A

x

L

P

Support Reactions: Referring to the FBD of the entire beam in Fig a.

a

a

Internal Loadings: For 0 … x 6 1 m, Referring to the FBD of the left segment of

the beam in Fig b,

V = - 5.50 kN+ c a Fy = 0; V + 5.50 = 0

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

V = 0.500 kN+ c a Fy = 0; 4.50 - 4 - V = 0

M = 54.50x6 kN#m

M - 4.50 x = 0+a MO = 0;

V = 4.50 kN4.50 - V = 0

+ c a Fy = 0;

:+ a Fx = 0; Ax = 0

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

By = 5.50 kN+a MA = 0; By (4) - 4(1) - 6(3) = 0

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

V = MOL

V - MO

L = 0+ c a Fy = 0;

M = MO

L x

M - Mo

L x = 0+a Mo = 0;

V = MOL

MO

L - V = 0+ c a Fy = 0;

Ay = MOL

MO - Ay (L) = 0+a MB = 0;

By = MOL

MO - NB (L) = 0+a MA = 0;

:+ a Fx = 0; Ax = 0

4–14. Determine the shear and moment throughout the

beam as a function of x.

B A

x

M0

L

+a M = 0;

V = 3.75 kN3.75 - V = 0;

4–15. Determine the shear and moment throughout the

beam as a function of x.

B A

V = E-8x + 26F kN+ c a Fy = 0; V + 22 - 8(6 - x) = 0

M = E-0.444 x3 +14 xF kN#m

M + 1

2a83 xb(x)ax3b - 14x = 0+a MO = 0;

V = E-1.33x2 +14F kN

14 - 1

2a83xbx - V = 0+ c a Fy = 0;

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

V = 5-206 kN+ c a Fy = 0; -4 - 8 - 8 - V = 0

M = 5-12x + 86 kN#m+a MO = 0; M + 8 (x - 1) + 4x = 0

V = 5-126 kN#m+ c a Fy = 0; - 4 - 8 - V = 0

M = 5-4x6 kN#m

M + 4x = 0+a MO = 0;

V = - 4 kN

- V - 4 = 0+ c a Fy = 0;

4–17. Determine the shear and moment throughout the

9 6

Support Reactions: As shown on FBD

Shear and Moment Functions:

V = 8.00 k:+ a Fy = 0; V - 8 = 0

M = 5-x2 + 30.0x - 2166 k#ft+a MNA = 0; M + 216 + 2xax2b - 30.0x = 0

V = 530.0 - 2x6 k+ ca Fy = 0; 30.0 - 2x - V = 0

4–18. Determine the shear and moment throughout the

Support Reactions: As shown on FBD.

Shear and moment Functions:

V = 250 lb+ ca Fy = 0; V - 250 = 0

M = 5- 75x2 + 1050x - 40006lb#ft+a MNA = 0; M + 150(x - 4)ax - 42 b + 250x - 700(x - 4) = 0

V = 51050 - 150xF lb

+ ca Fy = 0; -250 + 700 - 150(x - 4) - V = 0

M = 5-250xF lb#ft+a MNA = 0; M + 250x = 0

V = - 250 lb+ ca Fy = 0; -250 - V = 0

9 8

Support Reactions: As shown on FBD

Shear and Moment Functions:

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

function of x.

:+ a Fy = 0; 36 - 1

2a89xb(x) - 89a8 - 89xbx - V = 0

4–22 Determine the shear and moment throughout the

tapered beam as a function of x.

4–25. Draw the shear and moment diagrams for the beam.

4–26. Draw the shear and moment diagrams of the beam

B D

+a M = 0; M = 0

+ c a Fy = 0; V = 0

0 … x … L

3

*4–28. Draw the shear and moment diagrams for the

beam (a) in terms of the parameters shown; (b) set M O=

500 N m, L = 8 m.

M0

M0

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

4–29. Draw the shear and moment diagrams for the beam

C A

0 … x < 20 ft+a MB = 0; 1000(10) - 200 - Ay(20) = 0 Ay = 490 lb

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

Support Reactions: From FBD(a),

0 … x … 20 ft

Ay = 2500 lb+ c a Fy = 0; Ay = 5000 + 2500 = 0

:+ a Fx = 0; Ax = 0

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

*4–32. Draw the shear and moment diagrams for the

V = 20+ c a Fy = 0; V - 20 = 0

8 < x … 11

M = 133.75x - 20x2

+a M = 0; M + 40xax2b - 133.75x = 0

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

4–34. Draw the shear and moment diagrams for the beam.

4–35. Draw the shear and moment diagrams for the beam

*4–36. Draw the shear and moment diagrams of the

beam Assume the support at B is a pin and A is a roller.

Ans.

Ans.

Mmax = 34.5 kN#m

Vmax = 24.5 kN

4–37. Draw the shear and moment diagrams for the beam

Assume the support at B is a pin.

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.

4–39. Draw the shear and moment diagrams for each

member of the frame Assume the support at A is a pin and

*4–40. Draw the shear and moment diagrams for each

member of the frame Assume A is a rocker, and D is

4 k

3 k

1 1 2

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.

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.

D A

15 ft0.8 k/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.

*4–44. 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.

2 k

By¿ = 4 kN+a Mc = 0; 12(2) - By¿ (6) = 0

Ay = 13.3 kN+ c a Fy = 0; Ay - 25 + 11.667 = 0

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

4–45. Draw the shear and moment diagrams for each

member of the frame The members are pin connected at A,

Dy = 13.5 kN+ c a Fy = 0; 12.5 - 5 - 10 - 5 - 10a35b + Dy = 0

Dx = 8 kN:+ a Fx = 0; -10a45b + Dx = 0

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

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

1 1 6

Support Reactions:

a

Ay = 3.64 kN+ c a Fy = 0; Ay - 4.20 cos 30° = 0

Ax = 3.967 kN:+ a Fx = 0; 1.75 + 3.5 + 1.75 + 4.20 sin 30° - 5.133 - Ax = 0

Cx = 5.133 kN

-4.20(sin 30°)(14 + 3.5) + (21) = 0+a MA = 0; -3.5(7) - 1.75(14) - (4.20)(sin 30°)(7cos30°)

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

7 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

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.

1 1 8

4–50. Draw the moment diagrams for the beam using the

method of superposition The beam is cantilevered from A.

4–51. Draw the moment diagrams for the beam using the

method of superposition

*4-52. Draw the moment diagrams for the beam using the

method of superposition Consider the beam to be

cantilevered from end A.

1 2 0

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.

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.

4–54 Continued

1 2 2

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.

*4–56. Draw the moment diagrams for the beam using

the method of superposition Consider the beam to be

cantilevered from end C.

1 2 4

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.