Explanation Calculation: The diagram of the beam is shown below. Figure (1) Consider the beam A B . Calculate moment about point A . ∑ M A = 0 ( R B 1 × 2 m ) − ( 12 kN/m × 2 m × 2 m 2 ) = 0 R B 1 × 2 m = 24 kN-m R B 1 = 12 kN Calculate all forces in y-direction. ∑ F y = 0 R A + R B 1 = 12 kN/m × 2 m R A + R B 1 = 24 kN ...... (I) Substitute 12 kN for R B 1 in the Equation (I). R A + 12 kN = 24 kN R A = 12 kN Here, the vertical component of force at point A is R A and the vertical component of force at point B in beam A B is R B 1 . Consider the beam B C . Calculate moment about point C . ∑ M C = 0 ( R B 2 × 2 m ) − ( 12 kN/m × 2 m × 2 m 2 ) = 0 R B 2 × 2 m = 24 kN-m R B 2 = 12 kN Calculate all forces in y-direction. ∑ F y = 0 R C + R B 2 = 12 kN/m × 2 m R C + R B 2 = 24 kN ...... (II) Substitute 12 kN for R B 2 in the Equation (II). R C + 12 kN = 24 kN R C = 12 kN Here, the vertical component of force at point C is R C and the vertical component of force at point B in beam B C is R B 2 . Calculate total reaction at point B . R B = R B 1 + R B 2 ...... (III) Substitute 12 kN for R B 1 and 12 kN for R B 2 in the Equation (III). R B = 12 kN + 12 kN = 24 kN The free body diagram of the floor deck and truss is shown below. Figure (2) The free body diagram of the truss is shown below. Figure (3) Here, the vertical component of force at point K is V K , the vertical component of force at point A is V A , the horizontal component of force at point K is H K and the horizontal component of force at point A is H A . Calculate all forces in y-direction. ∑ F y = 0 V A + V K = ( 12 kN × 2 ) + ( 24 kN × 4 ) = 120 kN ...... (IV) Calculate all forces in x-direction. ∑ F x = 0 H A + H K = 0 ...... (V) Calculate moment about point A . ∑ M A = 0 [ ( H K × 3 m ) − ( 24 kN × 2 m ) − ( 24 kN × 4 m ) − ( 24 kN × 6 m ) − ( 24 kN × 8 m ) − ( 12 kN × 10 m ) ] = 0 3 H K = 600 kN H K = 200 kN Substitute 12 kN for H K in Equation (V). H A + 12 kN = 0 H A = − 12 kN The free body diagram of the joint A is shown below. Figure (4) Here, force in member A B is F A B . Calculate all forces in y-direction. ∑ F y = 0 V A − 12 kN = 0 V A = 12 kN Calculate all forces in x-direction. ∑ F x = 0 F A B − 200 kN = 0 F A B = 200 kN ( T ) Substitute 12 kN for V a in Equation (IV). 12 kN + V K = 120 kN V K = 108 kN The free body diagram of the joint K is shown below. Figure (5) Here, force in member K B is F K B and force in member K J is F K J . Consider from Figure (5).to calculate the angle tan θ 1 = 2 3 sin θ 1 = 2 13 cos θ 1 = 3 13 tan θ = 3 10 sin θ = 3 109 cos θ = 10 109 Calculate all forces in y-direction. ∑ F y = 0 F K B cos θ 1 + F K J sin θ + 108 kN = 0 F K B ( 3 13 ) + F K J ( 3 109 ) + 108 = 0 0.287 F K J + 0.832 F K B + 108 = 0 ...... (VI) Calculate all forces in x-direction. ∑ F x = 0 F K J cos θ + 200 kN + F K J sin θ 1 = 0 F K B ( 10 109 ) + 200 + F K J ( 2 13 ) = 0 0.958 F K J + 0.555 F K B + 200 = 0 ...... (VII) Solve the equation (VI) and (VII). F K J = − 167.045 kN ( C ) F K B = − 72.111 kN ( C ) The free body diagram of the joint J is shown below. Figure (6) Here, force in member J I is F J I and force in member B J is F B J . Calculate all forces in y-direction. ∑ F y = 0 F B J = 0 ...... (VIII) Calculate all forces in x-direction. ∑ F x = 0 F K J cos θ + F J I cos θ = 0 − F J I = F K J ...... (IX) Substitute − 167.045 kN for F K J in Equation (IX) − F J I = − 167

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ISBN: 8220106821596

Author: HIBBELER

Publisher: PEARSON

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Chapter 3, Problem 3.27P

To determine

The force in member BC, CI, HI, and DI, and to state if the members are in tension or compression.

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Chapter 3, Problem 3.26P

Chapter 3, Problem 3.28P

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Chapter 3 Solutions

EBK STRUCTURAL ANALYSIS

Ch. 3 - Classify each of the following trusses as...Ch. 3 - Prob. 3.2P Ch. 3 - Prob. 3.3P Ch. 3 - Prob. 3.4P Ch. 3 - Prob. 3.5P Ch. 3 - Prob. 3.6P Ch. 3 - Prob. 3.7P Ch. 3 - Prob. 3.8P Ch. 3 - Prob. 3.9P Ch. 3 - Prob. 3.10P

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