Explanation Given information: The truss is given in the Figure. The value of E is 200 GPa and the value of A is 1 , 100 mm 2 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 50 kN. Let A x and A y be the horizontal and vertical reactions at the hinged support A . Let C x and C y be the horizontal and vertical reactions at the hinged support C . 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 y ( 6 ) − P 1 ( 4 ) = 0 A y = 0.667 P 1 Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − A y + C y = 0 − 0.667 P 1 + C y = 0 C y = 0.667 P 1 Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − A x + P 1 = 0 A x = P 1 Apply equilibrium equation to the joint A : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − 0.667 P 1 + F A B sin 53.13 ° = 0 F A B = 0.833 P 1 Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 − P 1 + F A B cos 53.13 ° + F A C = 0 − P 1 + ( 0.833 P 1 ) cos 53.13 ° + F A C = 0 F A C = 0.5 P 1 Apply equilibrium equation to the joint C : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 0.667 P 1 + F B C sin 53.13 ° = 0 F B C = − 0.833 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 210 kN. 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 y ( 6 ) + P 2 ( 3 ) = 0 A y = 0.5 P 2 Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 A y + C y − P 2 = 0 0.5 P 2 + C y − P 2 = 0 C y = 0.5 P 2 Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 A x = 0 Apply equilibrium equation to the joint A : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 0.5 P 2 + F A B sin 53.13 ° = 0 F A B = − 0.625 P 2 Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 F A B cos 53.13 ° + F A C = 0 ( − 0.625 P 2 ) cos 53.13 ° + F A C = 0 F A C = 0.375 P 2 Apply equilibrium equation to the joint C : Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 0.5 P 2 + F B C sin 53.13 ° = 0 F B C = − 0.625 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 = 0.833 P 1 − 0

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 53P

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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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