Explanation Given Information: Set P 1 = 6 k N , P 2 = 9 k N Set P 1 = 3 k N , P 2 = 6 k N Method adopted: Method of Joints. Assumptions: Consider the state of members as in Figure (1). Consider the force indicating the right side as positive and the left side as negative in horizontal components of forces. Consider the force indicating the upward as positive and the downward as negative in vertical components of forces. Consider clockwise moment as positive and anti-clock wise moment as negative, wherever applicable. Show the state of members as in Figure (1). Find the reactions. Show the free body diagram of the truss as in Figure (2). Take the moment at joint A and equate it to zero. ∑ M A = 0 D y ( 12 ) − 3 ( 4 ) − 6 ( 8 ) = 0 D y = 5 kN Along the vertical direction: Determine the vertical reaction at joint A by resolving the vertical component of forces. ∑ F y = 0 A y + D y − 3 − 6 = 0 A y + D y = 9 Substitute 5 kN for D y . A y + 5 = 9 A y = 4 kN Along the horizontal direction: Determine the horizontal reaction at joint A by resolving the horizontal component of forces. ∑ F x = 0 A x = 0 Joint A : Show the free body diagram of the Joint A as in Figure (3). Refer to Figure (3). Determine the value of tan θ , tan θ = Opposite side Adjacent side (I) Substitute 6 m for the opposite side and 6 m for the adjacent side in Equation (I). tan θ 1 = 6 6 θ 1 = 45 ° Along the vertical direction: Determine the force in the member AE by resolving the vertical component of forces. ∑ F y = 0 A y − F A E sin θ 1 = 0 (II) Along the horizontal direction: Determine the force in the member AB by resolving the horizontal component of forces. ∑ F x = 0 F A B − F A E cos θ 1 = 0 (III) Show the calculation of force in the members as follows: Conclusion: Substitute 4 kN for A y and 45 ° for θ 1 in Equation (II). 4 − F A E sin 45 ° = 0 F A E = 5.66 kN ( C ) Thus, the magnitude of force in the member AE is 5.66 kN ( C ) _ . Substitute 5.66 kN for F A E and 45 ° for θ 1 in Equation (III). F A B − 5.66 cos 45 ° = 0 F A B = 4 kN ( T ) Thus, the magnitude of force in the member AB is 4 kN ( T ) _ . Joint B : Show the free body diagram of the Joint B as in Figure (4). Refer to Figure (4). Substitute 6 m for opposite side and 2 m for adjacent side in Equation (I). tan θ 2 = 6 2 θ 2 = 71.57 ° Along the vertical direction: Determine the force in the member BE by resolving the vertical component of forces, ∑ F y = 0 F B E sin θ 2 − P 1 = 0 (IV) 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 A B + F B E cos θ 2 = 0 (V) Show the calculation of force in the members as follows: Conclusion: Substitute 71

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 6.3, Problem 9P

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The force in each member of truss and the state of members are in tension or compression.

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Chapter 6 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 6.3 - In each case, calculate the support reactions and...Ch. 6.3 - Identify the zero-force members in each truss....Ch. 6.3 - Determine the force in each member of the truss....Ch. 6.3 - Determine the force in each member of the truss....Ch. 6.3 - Determine the force in each member of the truss....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 - Determine the force in each member of the truss....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 each member of truss and the state of members are in tension or compression.

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The force in each member of truss and the state of members are in tension or compression.

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Chapter 6 Solutions