Explanation Given information: Each A992 steel member has a cross-sectional area of A = 1.5 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 A992 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 A : Determine the vertical reaction at point C by taking moment about the point A . ∑ M A = 0 C y ( 16 ) − 2 ( 16 ) − P ( 8 ) = 0 (2) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + C y − 2 − 2 − P = 0 (3) Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 A x = 0 Show the calculation of values as follows: Solve Equation (2). C y ( 16 ) − 32 − 8 P = 0 C y = 2 + 0.5 P Substitute ( 2 + 0.5 P ) for C y in Equation (3). A y + 2 + 0.5 P − 4 − P = 0 A y = 2 + 0.5 P Joint C : Show the free-body diagram of the joint C as in Figure 2. Along the vertical direction: Determine the internal force in the member CD by resolving the vertical component of forces. ∑ F y = 0 C y − N C D = 0 (4) Along the horizontal direction: Determine the internal force in the member BC by resolving the horizontal component of forces. ∑ F x = 0 N B C = 0 Show the calculation of internal force in the members as follows: Substitute ( 2 + 0.5 P ) for C y in Equation (4). 3 + 0.5 P − N C D = 0 N C D = ( 2 + 0.5 P ) ( C ) Joint D : Show the free-body diagram of the joint D as in Figure 3. Determine the value of θ : tan θ = Opposite side Adjacent side (5) Along the vertical direction: Determine the internal force in the member BD by resolving the vertical component of forces. ∑ F y = 0 N C D − N B D sin θ − 2 = 0 (6) Along the horizontal direction: Determine the internal force in the member ED by resolving the horizontal component of forces. ∑ F x = 0 N E D − N B D cos θ = 0 (7) Show the calculation of internal force in the member as follows: Substitute 6 ft for opposite side and 8 ft for adjacent side in Equation (5). tan θ = 6 8 θ = 36.87 ° Substitute ( 2 + 0.5 P ) for N C D and 36.87 ° for θ in Equation (6). 2 + 0.5 P − N B D sin 36.87 ° − 2 = 0 N B D = 0.5 P sin 36.87 ° N B D = ( 0.833 P ) ( T ) Substitute 0.833 P for N B D and 36.87 ° for θ in Equation (7). N E D − ( 0.833 P ) cos 36.87 ° = 0 N E D = ( 0.667 P ) ( C ) Joint B : Show the free-body diagram of the joint B as in Figure 4. Along the vertical direction: Determine the internalforce in the member BE by resolving the vertical component of forces. ∑ F y = 0 N B D sin θ − N B E = 0 (8) Along the horizontal direction: Determine the internal force in the member AB by resolving the horizontal component of forces

11th Edition

ISBN: 9780137605521

Author: HIBBELER

Publisher: RENT PEARS

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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 126P

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The vertical displacement of joint E.

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Chapter 14.9, Problem 125P

Chapter 14.9, Problem 127P

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The vertical displacement of joint E .

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 vertical displacement of joint E.

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