To determine The horizontal displacement of joint B . Explanation Given information: The members of the truss are A992 steel bars, each with a cross-sectional area of 2 , 500 mm 2 . Consideration: The modulus of elasticity or Young’s modulus of the A992 steel is E = 200 × 10 9 N/m 2 . Assumptions: Consider that the right side force as positive and left side as negative in horizontal components of forces. Consider that the upward force as positive and downward as negative in vertical components of forces. Consider that the clockwise moment as negative and anticlockwise moment as positive. Express the equation for strain energy of a bar of constant cross-sectional area A and the constant internal axial load N as follows: U i = ∑ N 2 L 2 A E (1) Here, N is internal axial force, L is the length of the member, A is the cross sectional area of the member, and E is Young’s modulus or modulus of elasticity. Joint B : Show the free-body diagram of the joint B as in Figure 1. Along the horizontal direction: Determine the axial force in the member AB by resolving the horizontal component of forces. ∑ F x = 0 N A B − 15 = 0 N A B = 15 kN ( C ) Along the vertical direction: Determine the axial force in the member BD by resolving the vertical component of forces. ∑ F y = 0 N B D = 0 Joint A : Show the free-body diagram of the joint A as in Figure 2. Determine the value of θ : tan θ = Opposite side Adjacent side (2) Substitute 1.5 m for opposite side and 2 m for adjacent side in Equation (2). tan θ = 1.5 2 θ = 36.87 ° Along the horizontal direction: Determine the axial force in the member AD by resolving the horizontal component of forces. ∑ F x = 0 N A D cos θ − N A B = 0 (3) Substitute 15 kN for N A B and 36.87 ° for θ in Equation (3). N A D cos 36.87 ° − 15 = 0 N A D = 18.75 kN ( T ) Along the vertical direction: Determine the axial force in the member AC by resolving the vertical component of forces. ∑ F y = 0 N A C − N A D sin θ = 0 (4) Substitute 18.75kN for N A D , and 36.87 ° for θ in Equation (4). N A C − 18.75 sin 36.87 ° = 0 N A C = 11.25 kip ( C ) Joint C : Show the free body diagram of the joint C as in Figure 3. Along the horizontal direction: Determine the axial force in the member CD by resolving the horizontal component of forces. ∑ F x = 0 N C D = 0 Resultant axial strain energy calculation: Member AB : Calculate the strain energy for the member AB using Equation (1)

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 14.3, Problem 14.25P

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The horizontal displacement of joint B.

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Chapter 14.2, Problem 14.24P

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