Explanation Given information: The truss is given in the Figure. The value of E is 29,000 ksi. Procedure to find the deflection of truss by virtual work method is shown below. For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces ( F ) in all the members of the truss. For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces ( F v ) in all the member of the truss. Finally use the desired deflection equation. Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below. For summation of forces along x -direction is equal to zero ( ∑ F x = 0 ) , consider the forces acting towards right side as positive ( → + ) and the forces acting towards left side as negative ( ← − ) . For summation of forces along y -direction is equal to zero ( ∑ F y = 0 ) , consider the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For summation of moment about a point is equal to zero ( ∑ M at a point = 0 ) , consider the clockwise moment as negative and the counter clockwise moment as positive. Method of joints: The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T). Condition for zero force members: 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero. 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero. Calculation: Consider the real system. Find the member axial force ( F ) for the real system using method of joints: Let D x and D y be the horizontal and vertical reactions at the hinged support D . Let A x be the horizontal reaction at the roller support A . Sketch the resultant diagram of the truss as shown in Figure 1. Find the reactions at the supports using equilibrium equations: Summation of moments about D is equal to 0. ∑ M D = 0 A x ( 15 ) − 25 ( 20 ) − 25 ( 40 ) = 0 A x = 100 k → Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 D y − 25 − 25 = 0 D y = 50 k ↑ Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 A x − D x = 0 100 − D x = 0 D x = 100 k ← Find the member forces using method of joints: Apply equilibrium equation to the joint D : Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − 100 + F D E = 0 F D E = 100 k Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 50 − F A D = 0 F A D = 50 k Apply equilibrium equation to the joint A : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 F A D + F A E sin 36.87 ° = 0 50 + F A E sin 36.87 ° = 0 F A E = − 83.333 k Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 100 + F A B + F A E cos 36.87 ° = 0 100 + F A B + ( − 83.333 ) cos 36.87 ° = 0 F A B = − 33.333 k Apply equilibrium equation to the joint B : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − 25 + F B E = 0 F B E = 25 k Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − F A B + F B C = 0 − ( − 33.333 ) + F B C = 0 F B C = − 33.333 k Apply equilibrium equation to the joint C : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − 25 + F C E sin 36.87 ° = 0 F C E = 41.667 k Consider the virtual system: For vertical deflection apply l k at joint C in the vertical direction. Find the member axial force ( F v ) due to virtual vertical load using method of joints: Sketch the resultant diagram of the virtual system ( F v ) as shown in Figure 2. Find the reactions at the supports using equilibrium equations: Summation of moments about D is equal to 0. ∑ M D = 0 A x ( 15 ) − 1 ( 40 ) = 0 A x = 2.667 k → Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 D y − 1 = 0 D y = 1 k ↑ Summation of forces along x -direction is equal to 0

6th Edition

ISBN: 9780357030967

Author: KASSIMALI

Publisher: CENGAGE L

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Matrix Algebra for Structural analysis

Structural analysis is a process in which the effects of different loads and internal forces, acting on structures and their elements are determined. This method of analysis is effective in detecting the failure loads of the structural elements and the d…

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Chapter 7, Problem 6P

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Calculate the vertical deflection at joint C of the truss.

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Chapter 7 Solutions

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