Explanation Method adopted: Method of section. Assumptions: Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating the right side as positive and left side as negative in horizontal components of forces. Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces. Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. Entire truss: Show the free body diagram of entire truss as in Figure 1. Using Figure 1, Reactions: Moment about point A : Determine the vertical reaction at point G by taking moment about point A . ∑ M A = 0 G y ( 24 ) − 2 , 000 ( 20 ) − 2 , 000 ( 16 ) − 4 , 000 ( 12 ) − 3 , 000 ( 8 ) − 3 , 000 ( 4 ) = 0 (I) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − 3 , 000 − 3 , 000 − 4 , 000 − 2 , 000 − 2 , 000 + G y = 0 (II) Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 A x = 0 Conclusion: Solve Equation (I). 24 G y − 40 , 000 − 32 , 000 − 48 , 000 − 24 , 000 − 12 , 000 = 0 24 G y = 156 , 000 G y = 6 , 500 lb Substitute 6,500 lb for G y in Equation (II). A y − 14 , 000 + 6 , 500 = 0 A y = 7 , 500 lb Section a-a : Pass a section a-a as shown in Figure 1 and consider left side of the section. Show the free body diagram of left side of the section a-a as in Figure 2. Determine the value of θ : tan θ = Opposite side Adjacent side (III) Force in the member LK : Moment about point C : Take the moment about joint C , it will eliminate the force in the members CL and CB . ∑ M C = 0 3 , 000 ( 4 ) − A y ( 8 ) + F L K sin θ ( 8 ) = 0 (IV) Force in the member LC : Along the vertical direction: Determine the force in the member LC by resolving the vertical component of forces. ∑ F y = 0 A y − 3 , 000 − F L K sin θ + F L C sin θ = 0 (V) Force in the member BC : Along the horizontal direction: Determine the force in the member BC by resolving the horizontal component of forces. ∑ F x = 0 F B C − F L K cos θ − F L C cos θ = 0 (VI) Conclusion: Substitute 6 ft for opposite side and 8 ft for adjacent side in Equation (III)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 6.4, Problem 42P

To determine

The force in members LK, LC, and BC of the truss and the state of members are in tension or compression.

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Chapter 6.4, Problem 41P

Chapter 6.4, Problem 43P

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Chapter 6 Solutions

Engineering Mechanics: Statics

Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - Prob. 3FP Ch. 6.3 - Determine the greatest load P that can be applied...Ch. 6.3 - Identify the zero-force members in the truss....Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - Prob. 1P Ch. 6.3 - Prob. 2P Ch. 6.3 - Determine the force in each member of the truss,...Ch. 6.3 - Determine the force in each member of the truss,...

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The force in members LK , LC , and BC of the truss and the state of members are in tension or compression.

The force in members LK, LC, and BC of the truss and the state of members are in tension or compression.

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Chapter 6 Solutions