331

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5–17.

Determine the force in members DC, HC, and HI of the truss and state if the members are in tension or compression.

A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m

Determine the force in members DC, HC, and HI of thetruss, and state if the members are in tension orcompression.

SOLUTIONSupport Reactions: Applying the moment equation of equilibrium about point A tothe free-body diagram of the truss, Fig. a,

a

Method of Sections: Using the bottom portion of the free-body diagram, Fig. b,

a

Ans.

a

Ans.

Ans.FDC = 125 kN (C)

+ c ©Fy = 0; 32.5 + 42.5 - FDC (35

) = 0

FHC = 100 kN (T)

+ ©MD = 0; 70(4.5) - 40(3) - 30(1.5) - FHC(1.5) = 0

FHI = 42.5 kN (T)

+ ©MC = 0; 70(3) - 32.5(2) - 40(1.5) - FHI(2) = 0

+ c ©Fy = 0; 57.5 - 40 - 50 + Ay = 0; Ay = 32.5 kN

:+ ©Fx = 0; Ax - 30 - 40 = 0; Ax = 70 kN

Fy = 57.5 kN

+ ©MA = 0; 40(1.5) + 30(3) + 40(2) - Fy(4) = 0A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m

Determine the force in members DC, HC, and HI of thetruss, and state if the members are in tension orcompression.

SOLUTIONSupport Reactions: Applying the moment equation of equilibrium about point A tothe free-body diagram of the truss, Fig. a,

a

Method of Sections: Using the bottom portion of the free-body diagram, Fig. b,

a

Ans.

a

Ans.

Ans.FDC = 125 kN (C)

+ c ©Fy = 0; 32.5 + 42.5 - FDC (35

) = 0

FHC = 100 kN (T)

+ ©MD = 0; 70(4.5) - 40(3) - 30(1.5) - FHC(1.5) = 0

FHI = 42.5 kN (T)

+ ©MC = 0; 70(3) - 32.5(2) - 40(1.5) - FHI(2) = 0

+ c ©Fy = 0; 57.5 - 40 - 50 + Ay = 0; Ay = 32.5 kN

:+ ©Fx = 0; Ax - 30 - 40 = 0; Ax = 70 kN

Fy = 57.5 kN

+ ©MA = 0; 40(1.5) + 30(3) + 40(2) - Fy(4) = 0A

C

G

E D

HF

IB

2 m 2 m 2 m

1.5 m

50 kN40 kN

40 kN

30 kN

1.5 m

1.5 m

Ans:FHI = 42.5 kN (T)FHC = 100 kN (T)FDC = 125 kN (C)

345

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SOLUTIONSupport Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A.

a+ΣMA = 0; ND(5.5) - 11(2) - 22(3.5) = 0 ND = 18.0 kN

Method of Sections. Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FBC and FFE can be determined directly by writing the moment equations of equilibrium about point E and B, respectively.

a+ΣME = 0; 18.0(2) - FBC(2) = 0 FBC = 18.0 kN (T) Ans.

a+ΣMB = 0; 18.0(3.5) - 22(1.5) - FFE(2) = 0 FFE = 15.0 kN (C) Ans.

Also, FEB can be obtained directly by writing force equation of equilibrium along the y axis

+ cΣFy = 0; FEB a45b + 18.0 - 22 = 0 FEB = 5.00 kN (C) Ans.

5–31.

Determine the force developed in members FE, EB, and BC of the truss and state if these members are in tension or compression.

11 kN

B

A D

C

F E

22 kN

2 m 1.5 m

2 m

2 m

Ans:FBC = 18.0 kN (T)FFE = 15.0 kN (C)FEB = 5.00 kN (C)

347

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5–33.

In each case, determine the force P required to maintain equilibrium of the 100-lb block.

P

(a) (b) (c)

P

P

SOLUTIONEquations of Equilibrium:

(a)

Ans.

(b)

Ans.

(c)

Ans.P = 11.1 lb

3P - 33.33 = 0+ c ©Fy = 0;

P¿ = 33.33 lb

3P¿ - 100 = 0+ c ©Fy = 0;

P = 33.3 lb

3P - 100 = 0+ c ©Fy = 0;

P = 25.0 lb

4P - 100 = 0+ c ©Fy = 0;

In each case, determine the force P required to maintainequilibrium. The block weighs 100 lb.

P

(a) (b) (c)

P

P

SOLUTIONEquations of Equilibrium:

(a)

Ans.

(b)

Ans.

(c)

Ans.P = 11.1 lb

3P - 33.33 = 0+ c ©Fy = 0;

P¿ = 33.33 lb

3P¿ - 100 = 0+ c ©Fy = 0;

P = 33.3 lb

3P - 100 = 0+ c ©Fy = 0;

P = 25.0 lb

4P - 100 = 0+ c ©Fy = 0;

In each case, determine the force P required to maintainequilibrium. The block weighs 100 lb.

P

(a) (b) (c)

P

P

SOLUTIONEquations of Equilibrium:

(a)

Ans.

(b)

Ans.

(c)

Ans.P = 11.1 lb

3P - 33.33 = 0+ c ©Fy = 0;

P¿ = 33.33 lb

3P¿ - 100 = 0+ c ©Fy = 0;

P = 33.3 lb

3P - 100 = 0+ c ©Fy = 0;

P = 25.0 lb

4P - 100 = 0+ c ©Fy = 0;

In each case, determine the force P required to maintainequilibrium. The block weighs 100 lb.

P

(a) (b) (c)

P

P

Ans:a) P = 25.0 lbb) P = 33.3 lbc) P = 11.1 lb