Explanation Given information: The truss is given in the Figure. The value of E is 10,000 ksi. 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: Let apply a horizontal load P 1 at joint B in the horizontal direction to find the horizontal deflection. The value of load P 1 is 0. Let C x and C y be the horizontal and vertical reactions at the hinged support C . Let A x be the horizontal reaction at the roller support A . Sketch the resultant truss diagram for horizontal load P 1 at joint B as shown in Figure 1. Find the reactions at the supports using equilibrium equations: Summation of moments about C is equal to 0. ∑ M C = 0 − A x ( 10 ) + P 1 ( 15 ) = 0 A x = 3 P 1 2 Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 C y = 0 Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − A x + C x + P 1 = 0 − 3 P 1 2 + C x + P 1 = 0 C x = P 1 2 Find the member forces using method of joints: Apply equilibrium equation to the joint A : Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − 3 P 1 2 + F A B cos 14.04 ° = 0 F A B = 1.546 P 1 Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 F A C − F A B sin 14.04 ° = 0 F A C − ( 1.546 P 1 ) sin 14.04 ° = 0 F A C = 0.375 P 1 Apply equilibrium equation to the joint C : Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 P 1 2 + F B C cos 36.87 ° = 0 F B C = 0.625 P 1 Let apply a vertical load P 2 at joint B in the vertical direction to find the vertical deflection. The value of load P 2 is 45 k. Sketch the resultant truss diagram for vertical load P 2 at joint B as shown in Figure 2. Find the reactions at the supports using equilibrium equations: Summation of moments about C is equal to 0. ∑ M C = 0 − A x ( 10 ) + P 2 ( 20 ) = 0 A x = 2 P 2 Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − P 2 + C y = 0 C y = P 2 Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 A x − C x = 0 2 P 2 − C x = 0 C x = 2 P 2 Find the member forces using method of joints: Apply equilibrium equation to the joint A : Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 2 P 2 + F A B cos 14.04 ° = 0 F A B = − 2.062 P 2 Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 F A C − F A B sin 14.04 ° = 0 F A C − ( − 2.062 P 2 ) sin 14.04 ° = 0 F A C = 0.5 P 2 Apply equilibrium equation to the joint C : Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − 2 P 2 + F B C cos 36.87 ° = 0 F B C = 2.5 P 2 The total force F in a member is equal to the sum of the force in the member due to horizontal load P 1 and vertical load P 2 . Refer Figure 1 and Figure 2. The force in the member AB is, F A B = 1

6th Edition

ISBN: 9780357030967

Author: KASSIMALI

Publisher: CENGAGE L

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Equilibrium and Support Reactions

Civil engineers deal with structures that are composed of several members connected or with individual members such as beams and columns. To analyze such structures, the condition of statics must be applied. Under the static condition, the members are co…

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Chapter 7, Problem 54P

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Calculate the horizontal deflection and vertical deflection at joint B of the truss.

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Chapter 7 Solutions

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