Explanation Check the determinacy of the truss as shown below. m + r − 2 j = 0 Here, number of members are m , number of reactions are r and number of joints are j . Substitute 16 for m , 3 for r and 8 for j . 16 + 3 − 2 × 8 > 0 3 > 0 Hence, the given truss is statically indeterminate of 3 degree. The following diagram shows the free body diagram of the truss: Figure-(1) Write the Equation for moment about A. Σ M A = 0 ( D y × 60 ft ) − ( 4 × 60 + 8 × 40 + 8 × 20 + 3 × 20 ) k ⋅ ft = 0 D y × 60 ft = 780 k ⋅ ft D y = 13 k Here, vertical reaction at D is D y . Write the Equation for sum of vertical forces. Σ F y = 0 A y + D y − ( 4 + 8 + 8 + 4 ) k = 0 A y + D y = 24 k .....(I) Here, vertical reaction at A is A y . Substitute 13 k for D y in Equation (I). 13 k + A y = 24 k A y = ( 24 − 13 ) k A y = 11 k Write the Equation for sum of horizontal forces. Σ F x = 0 A x − 3 k = 0 A x = 3 k Here, horizontal reaction at A is A x . Consider the forces in the diagonal member HB are in tension and AG is in compression as Zero member forces. F A G = 0 k Use the method of section and cut the section as shown in figure below and calculate the forces in the members. Figure-(2) Write the Equation for sum of vertical forces. Σ F y = 0 − 4 k + 11 k − F H B sin 45 ° − 0 × sin 45 ° = 0 F H B sin 45 ° − 7 k = 0 F H B = 9.9 k ( Tension ) Here, force in member HB and AG are F H B and F A G . Consider the joint A. Figure-(3) Write the Equation for sum of horizontal forces. Σ F x = 0 − 3 k + F A B − F A G cos 45 ° = 0 Substitute 0 k for F A G . − 3 k + F A B − 0 k × cos 45 ° = 0 F A B = 3 k ( Tension ) Here, force in the member AB is F A B . Write the Equation for sum of vertical forces. Σ F y = 0 11 k − F A H − 0 × sin 45 ° = 0 F A H = 11 k F A H = 11 k ( Compression ) Here, force in the member AH is F A H . Consider the joint H. Figure-(4) Write the Equation for sum of horizontal forces. Σ F x = 0 − F H G + 9.9 cos 45 ° + 3 k = 0 F H G = 10 k ( Compression ) Here, force in the member HG is F H G . Consider the forces in the diagonal member BF are in tension and GC is in compression as zero member forces. F G C = 0 k Use the method of section and cut the section as shown in figure below and calculate the forces in the members. Figure-(5) Write the Equation for sum of vertical forces. Σ F y = 0 11 k + 0 × sin 45 ° + F B F sin 45 ° − 8 k − 4 k = 0 F B F sin 45 ° − 1 = 0 F B F = 1.414 k ( Tension ) Here, force in the member BF is F B F . Write the Equation for moment about B. Σ M B = 0 ( F G F × 20 ) + ( F G C cos 45 ° × 20 ) + ( 4 × 20 ) k ⋅ ft − ( 3 × 20 ) k ⋅ ft − ( 11 × 20 ) k ⋅ ft = 0 Substitute 0 k for F G C . F G F × 20 ft + 0 × cos 45 ° × 20 + ( 4 × 20 ) k ⋅ ft − ( 11 × 20 ) k ⋅ ft − ( 3 × 20 ) k ⋅ ft = 0 F G F × 20 ft = − 200 k ⋅ ft F G F = 10 k ( Compression ) Here, force in the member GF is F G F

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ISBN: 9780134840857

Author: 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 12.8P

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The forces in each member of the truss.

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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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