Explanation Given information: Force at point H is F H = 5 kN . Force at point G is F G = 5 kN . Force at point F is F F = 5 kN . Free body diagram of the truss is Reaction at A is vertical only because the support is a roller and both horizontal and vertical reactions are there at E due to hinge support. For the truss to be in equilibrium ∑ F H = 0 ∑ F V = 0 ∑ M = 0 There are two methods to find force in members of the truss Methods of joint Method of section The reactions at supports are given as Force equilibrium in the horizontal direction ∑ F H = 0 R E x = 0 Force equilibrium in the vertical direction ∑ F V = 0 R A − F H − F G − F F + R E y = 0 R A − 5 kN − 5 kN − 5 kN + R E y = 0 R A + R E y = 15 kN ............................. 1 Moment equilibrium about A . ∑ M A = 0 F H × A H + F G × A G + F F × A F − R E y × A F + D E − R E x × B H = 0 5 kN × 3 m + 5 kN × 6 m + 5 kN × 9 m − R E y × 12 m − 0 kN × 3 m = 0 R E y = 7.5 kN ..................................... 2 From 1 and 2 . R A + R E y = 15 kN R A + 7.5 kN = 15 kN R A = 7.5 kN Calculation: Consider the joint A . Angle θ is tan θ = B H A H = 3 m 3 m = 1 θ = tan − 1 1 = 45 ° Force equilibrium in the vertical direction ∑ F V = 0 F A B sin θ + R A = 0 F A B = − 7.5 kN sin 45 ° F A B = − 10.61 kN F A B = 10.61 kN C Force equilibrium in the horizontal direction ∑ F H = 0 F A H + F A B cos θ = 0 F A H = − F A B cos 45 ° F A H = − − 10.61 kN cos 45 ° F A H = 7.5 kN T Consider the joint B . Force equilibrium in the vertical direction ∑ F V = 0 F A B sin θ + F B H = 0 F B H = − F A B sin θ F B H = − − 10.61 kN sin 45 ° F B H = 7.5 kN T Force equilibrium in the horizontal direction ∑ F H = 0 F B C − F A B cos θ = 0 F B C = F A B cos 45 ° F B C = − 10.61 kN cos 45 ° F B C = − 7.5 kN F B C = 7.5 kN C Consider the joint H . Force equilibrium in the vertical direction ∑ F V = 0 F H C sin θ + F B H − 5 kN = 0 F H C = − F B H + 5 kN sin θ F H C = − 7

15th Edition

ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Question

Chapter 6, Problem 22P

To determine

The Force in each member of the truss.

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Chapter 6, Problem 21P

Chapter 6, Problem 23P

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