Explanation Given information: The cross-sectional area of each A992 steel member has A = 4.5 in . 2 Method adopted: Castigliano’s theorem to determine displacement and Method of Joints to determine the force in the member. Assumptions: Consider that the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider that the force indicating right side as positive and left side as negative in horizontal components of forces. Consider that the force indicating upward 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. Consideration: Young’s modulus or modulus of elasticity of A992 steel is E = 29 ( 10 3 ) ksi . Show Castigliano’s theorem 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. Determine the reactions: 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 ) − 5 ( 8 ) − P ( 8 ) = 0 C y = 2.5 + P 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 − P − 5 = 0 (2) Substitute ( 2.5 + P 2 ) for C y in Equation (2). A y + 2.5 + P 2 − P − 5 = 0 A y = 2.5 + P 2 Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of forces. ∑ F x = 0 A x = 0 Internal force ( N ) calculation: Joint F : Show the free-body diagram of the joint F as in Figure 2. Along the horizontal direction: Determine the internal force in the member EF by resolving the horizontal component of forces. ∑ F x = 0 N E F = 0 Along the vertical direction: Determine the internal force in the member AF by resolving the vertical component of forces. ∑ F y = 0 N A F = 0 Joint D : Show the free-body diagram of the joint D as in Figure 3. 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 = 0 Along the vertical direction: Determine the internal force in the member CD by resolving the vertical component of forces. ∑ F y = 0 N C D = 0 Joint C : Show the free-body diagram of the joint C as in Figure 4. Determine the value of θ using the relation. tan θ = Opposite side ( L B E ) Adjacent side ( L B C ) (3) Substitute 6 ft for L B E and 8 ft for L B C in Equation (3). tan θ = 6 8 θ = 36.87 ° Along the vertical direction: Determine the internal force in the member CE by resolving the vertical component of forces. ∑ F y = 0 C y + N C D − N C E sin θ = 0 (4) Substitute ( 2.5 + P 2 ) for C y , 0 for N C D , and 36.87 ° for θ in Equation (4). 2.5 + P 2 + 0 − N C E sin 36.87 ° = 0 N C E = 2.5 sin 36.87 ° + P 2 sin 36.87 ° N C E = ( 4.167 + 0.833 P ) ( C ) Along the horizontal direction: Determine the internal force in the member BC by resolving the horizontal component of forces. ∑ F x = 0 N C E cos θ − N B C = 0 (5) Substitute ( 4

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 14.9, Problem 14.128P

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The vertical displacement of point E.

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Chapter 14.9, Problem 14.127P

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