Explanation Figure 1 is the free body diagram of the second pipe. Write the equation to find the weight of the pipes. W = ( 30 lb / ft ) ( 7.5 ft ) = 225 lb Write the equation to find the force at points F and D. F 8 = D 17 = 225 lb 15 F = 120 lb D = 255 lb Figure 2 shows the geometry of the second pipe. The radius of the pipe is 4.5 in . Figure 2 shows that CF and CF are symmetric. C F = C D (I) Equate the horizontal distance of CF and CD. r + 8 17 r = C D ( 15 17 ) 25 17 r = C D ( 15 17 ) C D = 25 15 r = 5 3 r Write the equation to find the value of C F from equation (I). C F = 5 3 r Here, r is the radius of the pipe. Substitute 4.5 in for r to get C F . C F = 5 3 ( 4.5 in ) = 7.5 in Consider the free body diagram of member CFG. Write the equation to find the sum of moments about the point C. ∑ M C = F × C F − G x ( 16 in ) Here, F is the force, C F is the length of the pipe CF, G x is the x component of reaction at point G. F × C F − G x ( 16 in ) = 0 (II) Figure 4 is the free body diagram of the system consisting of frames and pipes

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Chapter 6.3, Problem 6.93P

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The components of reactions at point A and G.

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The components of reactions at point A and G.

Solution Summary: The author shows the components of reactions at points A and G. The x component of reaction force at point A is underset_176.3 lb, and the y component

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