Explanation Given information: Each A-36 steel member has a cross-sectional area of A = 2 in . 2 . Method adopted: Castigliano’s theorem to determine displacement and Method of Joints to determine the force in the member. Assumptions: Young’s modulus or modulus of elasticity of A-36 steel is E = 29 ( 10 3 ) ksi . Explanation: Castigliano’s theorem for the entire truss is as follows: Δ = ∑ N ( ∂ N ∂ P ) L A E (1) Here, Δ is displacement of the truss joint, P an external force of variable magnitude applied to the truss joint in the direction of Δ , N is internal axial force in a member caused by both force P and the actual loads of the truss, L is length of a member, A is cross-sectional area of a member, and E is modulus of elasticity of a member. Internal axial force ( N ) in the member calculation: Apply a vertical external force P in the direction of displacement to be determined. Determine the reactions: Entire truss: Show the free-body diagram of the entire truss as on Figure 1. Moment about point C : Determine the vertical reaction at point A by taking moment about the point C . ∑ M C = 0 A y ( 12 ) − P ( 8 ) = 0 (2) Along the vertical direction: Determine the vertical reaction at point C by resolving the vertical component of forces. ∑ F y = 0 A y − C y = 0 (3) Along the horizontal direction: Determine the horizontal reaction at point C by resolving the horizontal component of forces. ∑ F x = 0 P − C x = 0 (4) Show the calculation of values as follows: Solve Equation (2). 12 A y − 8 P = 0 A y = 0 .667 P Substitute 0 .667 P for A y in Equation (3). 0 .667 P − C y = 0 C y = 0 .667 P Solve Equation (4). C x = P 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 (5) Along the vertical direction: Determine the internal force in the member AB by resolving the vertical component of forces. ∑ F y = 0 A y − N A B sin θ = 0 (6) Along the horizontal direction: Determine the internal force in the member AC by resolving the horizontal component of forces. ∑ F x = 0 N A B cos θ − N A C = 0 (7) Show the calculation of internal force in the members as follows: Substitute 8 ft for opposite side and 6 ft for adjacent side in Equation (5). tan θ = 8 6 θ = 53.13 ° Substitute 0 .667 P for A y and 53.13 ° for θ in Equation (6). 0 .667 P − N A B sin 53

10th Edition

ISBN: 9780134319650

Author: Russell C. Hibbeler

Publisher: PEARSON

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 14.9, Problem 14.127P

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The horizontal displacement of joint B.

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Chapter 14.9, Problem 14.126P

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The horizontal displacement of joint B .

Solution Summary: The author explains Castigliano's theorem for the entire truss and Method of Joints to determine the force in the member.

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The horizontal displacement of joint B.

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