Explanation Concept used: A hinge is placed at the center of each girder, since this is assumed to be a point of zero moment. A hinge is placed at center of each column, since this is assumed to be a point of zero moment. At the given floor level the shear at the interior column hinges is twice that at the exterior column hinges, since the frame is considered to be a superposition of portals. Calculation: The following figure shows the free body diagram of the frame with shear. Figure-(1) Consider the upper half of the frame. Write the Equation for sum of forces in horizontal direction. Σ F X = 0 9 kN − ( P + 2 P + 2 P + P ) = 0 6 P = 9 kN P = 1.5 kN Here, P is shear acting in frame as shown in Figure-(1). Consider the lower half of the frame. Write the Equation for sum of forces in horizontal direction. Σ F X = 0 9 kN + 12 kN − ( Q + 2 Q + 2 Q + Q ) = 0 6 Q = 21 kN Q = 3.5 kN Here, Q is the shear acting in frame as shown in Figure-(1). Calculate the horizontal reactions at A, B, C, and D. Calculate horizontal reaction at A. H A = Q Here, horizontal reaction at A is H A Substitute 3.5 kN for Q. H A = 3.5 kN Calculate horizontal reaction at B. H B = 2 Q Substitute 3.5 kN for Q. H B = 2 × 3.5 kN H B = 7 kN Here, horizontal reaction at B is H B . Calculate horizontal reaction at C. H C = 2 Q Substitute 3.5 kN for Q. H C = 2 × 3.5 kN H C = 7 kN Here, horizontal reaction at C is H C . Calculate horizontal reaction at A. H D = Q Here, horizontal reaction at D is H D . Substitute 3.5 kN for Q. H D = 3.5 kN Calculate the vertical reactions at support A, B, C, and D. Consider the joint I as shown below. Figure-(2) Write the Equation for moment about I. ( − R × 2.5 + P × 2 + 9 × 0 ) kN ⋅ m = 0 ...... (I). Substitute 1.5 k for P in Equation (I). ( − R × 2.5 + 1.5 kN × 2 + 9 × 0 ) kN ⋅ m = 0 R × 2.5 = 3 kN ⋅ m R = 1.2 kN Here, shear in member IJ is R. Consider joint J as shown below. Figure-(3) Write the Equation for moment about J. ( − S × 2.5 − 1.2 × 2.5 + 3 × 2 ) kN ⋅ m = 0 S × 2.5 = 3 kN ⋅ m S = 1.2 kN Here, shear in member JK is S. Consider joint K as shown below. Figure-(4) Write the Equation for moment about K. ( − T × 2.5 − 1.2 × 2.5 + 3 × 2 ) kN ⋅ m = 0 T × 2.5 = 3 kN ⋅ m T = 1.2 kN Here, shear in member KL is T. Consider joint E as shown below. Figure-(5) Write the Equation for moment about E. ( − U × 2.5 + 3.5 × 2 + 1.5 × 2 ) kN ⋅ m = 0 U × 2.5 = 10 kN ⋅ m U = 4 kN Here, shear in member EF is U. Consider joint F as shown below. Figure-(6) Write the Equation for moment about F. ( − V × 2.5 − 4 × 2.5 + 7 × 2 ) kN ⋅ m = 0 V × 2

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Author: HIBBELER

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The trusses and frames are the two most important structural members of civil engineering. Trusses are the interconnection of links and other structural members. The interconnection with other members makes the system behave as a single member that suppo…

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Chapter 12, Problem 12.37P

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The reactions at supports A, B, C, and D by using portal method.

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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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