Solutions (8th ed structural analysis) chapter 7

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FBE = 20.0 kN (C)14.14 sin 45° + 14.14 sin 45° - FBE = 0

+a MD = 0;

FDE = 10.0 kN (C)14.14 cos 45°(3) - FDE(3) = 0

FBD = FCE = F2

FAB= 10.0 kN (T)

FAB(3) - 14.14 cos 45°(3) = 0+a MF = 0;

FEF = 10.0 kN (C)

FEF(3) - 14.14 cos 45°(3) = 0+a MA = 0;

FAE = 14.1 kN (C)

FBF = 14.1 kN (T)

F1 = 14.14 kN

70 - 50 - 2F1sin 45° = 0+ ca Fy = 0;

FBF = FAE = F1

Ax = 0+

: a Fx = 0;

Ay = 70 kN 40(3) + 50(6) - Ay(6) = 0

+a MC = 0;

Cy = 40 kN

Cy(6) - 40(3) - 20(6) = 0+a MA = 0;

7–1. Determine (approximately) the force in each

member of the truss Assume the diagonals can support

either a tensile or a compressive force

7–1 Contiuned

7–2. Solve Prob 7–1 assuming that the diagonals cannot

support a compressive force

Support Reactions Referring to Fig a,

: a Fx = 0;

Ay = 70 kN40(3) + 50(6) - Ay(6) = 0

+a MC = 0;

Cy = 40 kN

Cy(6) - 40(3) - 20(6) = 0+a MA = 0;

FDE = 20.0 kN (C)28.28 cos 45°(3) - FDE(3) = 0

FEF = 20.0 kN (C)

FEF(3) - 28.28 cos 45°(3) = 0+a MA = 0;

Assume is carried equally by and , so

FGF = 7.5 k (C)

4.17 + 5.89 cos 45° - 1.18 cos 45° - FGF = 0:+

a Fx = 0;

FBF =

1.6672 cos 45° = 1.18 k (T)

FGC =

1.6672 cos 45° = 1.18 k (C)

FHB =

8.332 cos 45° = 5.89 k (T)

7–3. Determine (approximately) the force in each member

of the truss Assume the diagonals can support either a tensile

or a compressive force

FFE = 0.833 k (C)

5 + FFE - 8.25 cos 45° = 0:+

a Fx = 0;

FED = 15.83 k (C)

21.667 - 8.25 sin45° - FED = 0+ ca Fy = 0;

FCD= 8.25 cos 45° = 5.83 k (T):+

a Fx = 0;

FDF =

11.5672 cos 45° = 8.25 k (C)

FEC =

11.6672 cos 45° = 8.25 k (T)

VPanel = 21.667 - 10 = 11.667 k

FBC = 12.5 k (T)

FBC+ 1.18 cos 45° - 9.17 - 5.89 cos 45° = 0:+

*7–4. Solve Prob 7–3 assuming that the diagonals cannot

support a compressive force

a Fx = 0;

+ ca Fy = 0; FED = 21.7 k (C)

FCD= 0:+

FGB = 10 k (C)

- FGB + 11.785 sin45° + 2.36 sin 45° = 0+ ca Fy = 0;

FBC = 11.7 k (T)

FBC+ 2.36 cos 45° - 11.785 cos 45° - 5 = 0:+

a Fx = 0;

+ ca Fy = 0; FAN = 18.3 k (C)

7–4 Continued

+a MH = 0; FAB(6) + 2(6) - 12.08a45b(6) = 0 FAB = 7.667 k (T) = 7.67 k (T)

FBH = 12.1 k (T) FAG = 12.1 k (C)

+ ca Fy = 0; 21.5 - 7 - 2F1a35b = 0 F1= 12.08 k

FBH= FAG = F1+a MD = 0; 14(8) + 14(16) + 7(24) + 2(6) - Ay(24) = 0 Ay = 21.5 k

+a MA = 0; Dy(24) + 2(6) - 7(24) - 14(16) - 14(8) = 0 Dy = 20.5 k

:+

a Fx = 0; Ax - 2 = 0 Ax = 2 k

7–5. Determine (approximately) the force in each member

of the truss Assume the diagonals can support either a tensile

7–6. Solve Prob 7–5 assuming that the diagonals cannot

support a compressive force

2 1 0

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7–6 Continued

7–7. Determine (approximately) the force in each member of the truss.

Assume the diagonals can support either a tensile or compressive force

*7–8. Solve Prob 7–7 assuming that the diagonals cannot

support a compressive force

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Method of Sections It is required that Referring to Fig a,

7–9. Determine (approximately) the force in each member

of the truss Assume the diagonals can support both tensile

and compressive forces

15 ft

15 ft

2 k

2 k1.5 k

D

Method of Sections It is required that

7–10. Determine (approximately) the force in each member

of the truss Assume the diagonals DG and AC cannot

support a compressive force

15 ft

15 ft

2 k

2 k1.5 k

D

2 1 6

7–10 Continued

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Method of Sections It is required that Referring to Fig a,

7–11. Determine (approximately) the force in each

member of the truss Assume the diagonals can support

D

10 kN

2 m

2 m

2 1 8

7–11 Continued

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Method of Sections It is required that

*7–12. Determine (approximately) the force in each

member of the truss Assume the diagonals cannot support

D

10 kN

2 m

2 m

2 2 0

7–12 Continued

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The frame can be simplified to that shown in Fig a, referring to Fig b,

+a MA = 0; MA - 7.2(0.6) - 3(0.6)(0.3) = 0 MA = 4.86 kN#m

7–13. Determine (approximately) the internal moments

at joints A and B of the frame.

7–14. Determine (approximately) the internal moments

at joints F and D of the frame.

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The frame can be simplified to that shown in Fig a, Referring to Fig b,

a

Ans.

MA= 40.32 kN#m = 40.3 kN#m+a MA = 0; MA - 5(0.8)(0.4) - 16(0.8) - 9(0.8)(0.4) - 28.8(0.8) = 0

7–15. Determine (approximately) the internal moment at

A caused by the vertical loading.

8 m

A

C E

2 2 4

The frame can be simplified to that shown in Fig a The reactions of the

3 kN/m and 5 kN/m uniform distributed loads are shown in Fig b and c

respectively Referring to Fig d,

*7–16. Determine (approximately) the internal moments

at A and B caused by the vertical loading.

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E

F K

L G

ML = 20.25 k#ft

ML - 6.0(3) - 1.5(1.5) = 0+a ML = 0;

MI = 9.00 k#ft

MI - 1.0(1) - 4.0(2) = 0+a MI = 0;

7–17. Determine (approximately) the internal moments

at joints I and L Also, what is the internal moment at joint

H caused by member HG?

A H I

7–18. Determine (approximately) the support actions at

A, B, and C of the frame.

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A D F

C

H G

E

B

400 lb/ft

1200 lb/ft

7–19. Determine (approximately) the support reactions

at A and B of the portal frame Assume the supports are

(a) pinned, and (b) fixed

For printed base, referring to Fig a and b,

Gx = 6.00 kN6.00 - Gx = 0

:+

a Fx = 0;

Ey = 9.00 kN9.00 - Ey = 0

+ ca Fy = 0;

Ex = 6.00 kN

12 - 6.00 - Ex = 0:+

a Fx = 0;

Fx = 6.00 kN

Fy = 9.00 kN

Fy(2) - Fx(3) = 0+a MG = 0;

Fx(3) + Fy(2) - 12(3) = 0+a ME = 0;

By = 18.0 kN

By - 18.0 = 0+ ca Fy = 0;

Bx = 6.00 kN6.00 - Bx = 0

:+ a Fx = 0;

Ay = 18.0 kN18.0 - Ay = 0

+ ca Fy = 0;

Ax = 6.00 kN

12 - 6.00 - Ax = 0:+ a Fx = 0;

Ex = 6.00 kN

Ey = 18.0 kN

Ey(6) - Ex(6) = 0+a MB = 0;

Ex(6) + Ey(2) - 12(6) = 0+a MA = 0;

Bx = 6.00 kN6.00 - Bx = 0

:+

a Fx = 0;

MA = 18.0 kN#m

MA - 6.00(3) = 0+a MA = 0;

Ay = 9.00 kN9.00 - Ay = 0

+ ca Fy = 0;

Ax = 6.00 kN6.00 - Ax = 0

Ey =2Ph3b = 0+ ca Fy = 0;

Gy = Pa2h3bb

Gy(b) - Pa2h3 b = 0+a MB = 0;

*7–20. Determine (approximately) the internal moment

and shear at the ends of each member of the portal frame

Assume the supports at A and D are partially fixed, such

that an inflection point is located at h/3 from the bottom of

h

a Fx = 0;

FCF = 1.77 k(T)250(7.5) - FCF(sin 45°)(1.5) = 0;

+a ME = 0;

Ay = 375 lb

Ay(10) - 500(7.5) = 0;

+a MB = 0;

7–21. Draw (approximately) the moment diagram for

member ACE of the portal constructed with a rigid member

EG and knee braces CF and DH Assume that all points of

connection are pins Also determine the force in the knee

brace CF.

*7–22. Solve Prob 7–21 if the supports at A and B are

fixed instead of pinned

Inflection points are as mid-points of columns

H F

D C

6 ft 1.5 ft

7 ft

500 lb

B A

H F

D C

6 ft 1.5 ft 1.5 ft 1.5 ft

7–23. Determine (approximately) the force in each truss

member of the portal frame Also find the reactions at the

fixed column supports A and B Assume all members of the

truss to be pin connected at their ends

Assume that the horizontal reactive force component at fixed supports A and B are equal.

FFG = 1.00 k (C)

FFG(6) - 2(6) + 1.5(6) + 1.875(8) = 0+a MD= 0;

FCD = 2.00 k (C)

FCD(6) + 1(6) - 1.50(12) = 0+a MG = 0;

FDG = 3.125 k (C)

FDGa35b - 1.875 = 0+ ca Fy = 0;

MB = 9.00 k#ft

MB - 1.50(6) = 0+a MB= 0;

By = 1.875 k

By - 1.875 = 0+ ca Fy = 0;

Bx = 1.50 k1.50 - Bx = 0

:+

a Fx = 0;

MA= 9.00 k#ft

MA - 1.50(6) = 0+a MA= 0;

Ay = 1.875 k1.875 - Ay = 0

+ ca Fy = 0;

Hx = 1.50 k

Hx - 1.50 = 0:+

a Fx = 0;

Iy = 1.875 k

Iy - 1.875 = 0+ ca Fy = 0;

Hy = 1.875 k

Hy(16) - 1(6) - 2(12) = 0+a MI= 0;

1 k

2 3 2

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7–23 Continued

*7–24. Solve Prob 7–23 if the supports at A and B are

pinned instead of fixed

Assume that the horizontal reactive force component at pinal supports

A and B are equal Thus,

Ay = 3.00 k

Ay(16) - 1(12) - 2(18) = 0+a MB = 0;

1 k

Using the method of sections and referring to Fig b,

:+

a Fx = 0;

FDF = 5.00 k (T)

FDFa35b - 5.00a35b = 0+ ca Fy = 0;

FCD = 3.50 k (T)

FCD(6) + 1(6) - 1.50(18) = 0+a MG = 0;

FGF = 1.00 k (C)

FGF(6) - 2(6) - 1.5(12) + 3(8) = 0+a MD = 0;

FDG = 5.00 k (C)

FDGa35b - 3.00 = 0+ ca Fy = 0;

7–24 Continued

FEG = 27.5 kN (T)

FEGa35b - 16.5 = 0+ ca Fy = 0;

Ay = 16.5 kN16.5 - Ay = 0

+ ca Fy = 0;

By = 16.5 kN

By(4) - 8(5) - 4(6.5) = 0+a MA= 0;

Ax = Bx = 4 + 8

2 = 6.00 kN

7–25. Draw (approximately) the moment diagram for

column AGF of the portal Assume all truss members and

the columns to be pin connected at their ends Also

determine the force in all the truss members

4 kN

8 kN

7–26. Draw (approximately) the moment diagram for

column AGF of the portal Assume all the members of the

truss to be pin connected at their ends The columns are

fixed at A and B Also determine the force in all the truss

Hy = 9.00 kN

Hy(4) - 8(2.5) - 4(4) = 0+a MI= 0;

a Fx = 0;

FCE = 27.5 kN (C)

FCEa35b - 27.5a35b = 0+ ca Fy = 0;

4 kN

8 kN

FEG= 15.0 kN(T)

FEGa35b - 9.00 = 0+ ca Fy = 0;

MA = 15.0 kN#m

MA - 6.00(2.5) = 0+a MA = 0;

Ay = 9.00 kN9.00 - Ay = 0

+ ca Fy = 0;

Hx = 6.00 kN

Hx - 6.00 = 0:+ a Fx = 0;

7–26 Continued

7–27. Determine (approximately) the force in each truss

member of the portal frame Also find the reactions at the

fixed column supports A and B Assume all members of the

truss to be pin connected at their ends

a Fx = 0;

FCE = 15.0 kN (C)

FCEa35b - 15.0a35b = 0+ ca Fy = 0;

B

2 m1.5 m

D E

12 kN

7–27 Continued

*7–28. Solve Prob 7–27 if the supports at A and B are

pinned instead of fixed

B

2 m1.5 m

D E

12 kN

7–28 Continued

7–29. Determine (approximately) the force in members

GF, GK, and JK of the portal frame Also find the reactions

at the fixed column supports A and B Assume all members

of the truss to be connected at their ends

L J

Assume that the horizontal force components at pin supports A and B are equal.

7–30. Solve Prob 7–29 if the supports at A and B are pin

connected instead of fixed

+a MD= 0; FFHa35b(16) - 2.00(15) = 0 FFH = 3.125 k (C)

FEH = 0.500 k (T)+a MF = 0; 4(6) + 1.875(8) - 2.00(21) + FEH(6) = 0

+a MH = 0; FFGa35b(16) + 1.875(16) - 2.00(15) = 0 FFG = 0

+a MB = 0; Ay(32) - 4(15) = 0 Ay = 1.875 k

Ax = Bx = 4

2 = 2.00 k

7–31. Draw (approximately) the moment diagram for

column ACD of the portal Assume all truss members and

the columns to be pin connected at their ends Also

determine the force in members FG, FH, and EH.

D

C

K F

L

J G

*7–32. Solve Prob 7–31 if the supports at A and B are

fixed instead of pinned

L

J G

+a MD = 0; FFHa35b(16) - 2.00(9) = 0 FFH = 1.875 k (C)

+a MF = 0; -FEH(6) + 4(6) + 1.125(8) - 2.00(15) = 0 FEH = 0.500 k (C)

+a MH = 0; FFGa35b(16) + 1.125(16) - 2.00(9) = 0 FFG = 0

7–32 Continued

Assume the horizontal force components at pin supports A and B to be

FHG = 4.025 kN (C) = 4.02 kN (C)+a ML = 0; FHG cos 6.340° (1.167) + FHG sin 6.340° (1.5) + 2.889(3) - 2(1) - 3.00(4) = 0

+a MB = 0; Ay(9) - 4(4) - 2(5) = 0 Ay = 2.889 kN

Ax = Bx =

2 + 4

2 = 3.00 kN

7–33. Draw (approximately) the moment diagram for

column AJI of the portal Assume all truss members and the

columns to be pin connected at their ends Also determine

the force in members HG, HL, and KL.

FIH = 14.09 kN (C)+a MJ = 0; FIH cos 6.340° (1) - 2(1) - 3.00(4) = 0

FJK = 5.286 kN (T)+a MH = 0; FJK(1.167) + 2(0.167) + 4(1.167) + 2.889(1.5) - 3.00(5.167) = 0

7–33 Continued

Assume that the horizontal force components at fixed supports A and B

are equal Therefore,

Also, the reflection points P and R are located 2 m above A and B

respectively Referring to Fig a

FKL = 1.857 kN (T) = 1.86 kN (T)+a MH = 0; FKL(1.167) + 2(0.167) + 4(1.167) + 1.556(1.5) - 3.00(3.167) = 0

FHG = 2.515 kN (C) = 2.52 kN (C)+a ML = 0; FHG cos 6.340° (1.167) + FHG sin 6.340° (1.5) + 1.556(3) - 3.00(2) - 2(1) = 0

FIH = 8.049 kN (C)+a MJ = 0; FIH cos 6.340° (1) - 2(1) - 3.00(2) = 0

FJK = 1.857 kN (T)+a MH = 0; FJK(1.167) + 4(1.167) + 2(0.167) + 1.556(1.5) - 3.00(3.167) = 0

7–34 Continued

7–35. Use the portal method of analysis and draw the

moment diagram for girder FED.

2 5 2

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*7–36. Use the portal method of analysis and draw the

moment diagram for girder JIHGF.

7–37. Use the portal method and determine (approximately)

the reactions at supports A, B, C, and D. 9 kN

D C

B A

I

12 kN

2 5 4

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7–37 Continued

7–37 Continued

2 5 6

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7–38. Use the cantilever method and determine

(approximately) the reactions at supports A, B, C, and D.

All columns have the same cross-sectional area

D C

B A

I

12 kN