Explanation Calculation: The following figure shows the free body diagram of the frame. Figure-(1) Write the equation for the forces acting in the horizontal direction. ∑ F X = 0 H A + H B − 10 kN − 18 kN = 0 H A + H B = 28 kN Due to symmetry, H A = H B H A = 14 kN H B = 14 kN Here, summation of all horizontal forces is ∑ F X , the horizontal reaction at point A is H A , and the horizontal reaction at point B is H B . Write the equilibrium equation for net moment about joint B . ∑ M B = 0 18 kN × 4 ft + 10 kN × 6 ft − 6 ft × R A = 0 72 + 60 − 6 R A = 0 R A = 22 kN Here, net moment about joint B is ∑ M B , and the vertical reaction at point A is R A . Write the equilibrium equation for the forces acting in the vertical direction. ∑ F Y = 0 R A − R B = 0 R B = R A ...... (I) Substitute. 22 kN for R A in Equation (I) R B = 22 kN Consider the joint H of the truss as shown in Figure-(2) Figure-(2) Three members are joining at point H . Force in member C H and H G are collinear forces. Force in the member E H is non collinear. Therefore force in the member E H is zero. F EH = 0 kN Since the truss is symmetrical about vertical axis. Therefore force in the member F I is also zero. F FI = 0 kN Consider the triangle E C G Apply Pythagoras theorem C G = E G 2 + E C 2 = 3 2 + 2 2 = 9 + 4 = 3.605 cos θ = 2 3.605 = 0.5547 sin θ = 3 3.605 = 0.8321 Consider the section including the column A C E and members E G , G C , and C D . Figure-(3) Write the equilibrium equation for the forces acting in the vertical direction. ∑ F Y = 0 F CG cos θ − R A = 0 F CG = R A cos θ ...... (II) Here, summation of all vertical forces is ∑ F Y , and the force in the member C G is F CG . Substitute, 22 kN for R A , and 0.5547 for cos θ in Equation (II). F C G = 22 kN 0.5547 = 39.66 kN Consider the joint H as shown in Figure-(2) Three members are joining at point H . Force in member C H and H G are collinear forces. Therefore, F CH = 39.66 kN F HG = 39.66 kN Write the equilibrium equation for net moment about joint C . ∑ M C = 0 10 kN × 2 ft + 4 ft × H A − 2 ft × F EG = 0 20 + 4 H A − 2 F EG = 0 F EG = − 4 H A + 20 2 ...... (III) Here, net moment about joint C is ∑ M C , force in the member E G is F EG . Substitute, 14 kN for H A in Equation (III) F EG = − 4 × 14 kN + 20 2 = − 76 2 = − 38 kN Write the equilibrium equation for the forces acting in the horizontal direction

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ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Plane and Space Trusses

In civil engineering, a truss is the most important structure that forms the most fundamental member of any building structure and commercial complexes. Trusses are composed of individual members or links that are interconnected by suitable joints, such …

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Chapter 12, Problem 12.21P

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The force in each truss member of the portal frame

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