Explanation Calculation: The following figure shows the free body diagram of the frame. Figure-(1) Consider P and Q as inflection points. Write the equation for the forces acting in the horizontal direction. ∑ F X = 0 H A + H B − 10 kN / m − 18 kN / m = 0 H A + H B = 28 kN / m 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 the forces acting in the horizontal direction for column A P . ∑ F x = 0 H P − H A = 0 H P = H A ...... (I) Here, summation of all horizontal forces is ∑ F X , horizontal reaction at point P is R P . Substitute, 14 kN for H A in Equation (I) H P = 14 kN Write the equilibrium equation for the forces acting in the horizontal direction for column B Q . ∑ F x = 0 H Q − H B = 0 H Q = H B ...... (II) Here, summation of all horizontal forces is ∑ F X , horizontal reaction at point Q is R Q . Substitute, 14 kN for H B in Equation (II) H Q = 14 kN Write the equilibrium equation for the moment about joint Q . ∑ M Q = 0 10 kN × 4 m + 18 kN × 2 m − R P × 6 m = 0 40 + 36 − 6 R P = 0 R P = 12.667 kN Here, moment about the point Q is ∑ M Q , vertical reaction at point P is R P . Write the equilibrium equation for the forces acting in the vertical direction for the frame. ∑ F Y = 0 R P − R Q = 0 R Q = R P ...... (III) Here, summation of all vertical forces is ∑ F Y , vertical reaction at point Q is R Q . Substitute, 12.667 kN for R P in Equation (III) R A = 12.667 kN Write the equilibrium equation for the forces acting in the vertical direction for column A P . ∑ F Y = 0 R P − R A = 0 R A = R P ...... (IV) Here, summation of all vertical forces is ∑ F Y , vertical reaction at point A is R A . Substitute, 12.667 kN for R P in Equation (IV) R A = 12.667 kN Write the equilibrium equation for the forces acting in the vertical direction for column B Q . ∑ F Y = 0 R Q − R B = 0 R B = R Q ...... (V) Here, summation of all vertical forces is ∑ F Y , vertical reaction at point B is R B . Substitute, 12.667 kN for R Q in Equation (V) R B = 12.667 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 P 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 P = 0 F CG = R P cos θ ...... (VI) Here, summation of all vertical forces is ∑ F Y , and the force in the member C G is F CG . Substitute, 12.667 kN for R P , and 0.5547 for cos θ in Equation (VI). F C G = 12

10th Edition

ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 12, Problem 12.22P

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The force in each truss member of the portal frame

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Chapter 12 Solutions

Structural Analysis (10th Edition)

Ch. 12 - Prob. 12.1P Ch. 12 - Prob. 12.2P Ch. 12 - Prob. 12.3P Ch. 12 - Prob. 12.4P Ch. 12 - Prob. 12.5P Ch. 12 - Prob. 12.6P Ch. 12 - Prob. 12.7P Ch. 12 - Prob. 12.8P Ch. 12 - Prob. 12.9P Ch. 12 - Prob. 12.10P

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