Explanation Concept Used: Write the expression for the force in the member G B using the principle of superposition. Δ ′ G B + F G B f G B G B = Δ G B .......(I) Here, total displacement of member G B is Δ G B , the relative displacement of member G B is Δ ′ G B , the flexibility coefficient is f G B G B and the force in member G B is F G B . Write the expression for relative displacement of member G B . Δ ′ G B = ∑ n N L A E = [ ( n 1 N 1 L 1 A E ) + ( n 2 N 2 L 2 A E ) + ( n 3 N 3 L 3 A E ) + ( n 4 N 4 L 4 A E ) + ( n 5 N 5 L 5 A E ) + ( n 6 N 6 L 6 A E ) + ( n 7 N 7 L 7 A E ) + ( n 8 N 8 L 8 A E ) ] .......(II) Here, length of member is L , the area of member is A , the modulus of elasticity is E , the real load is N and the virtual load is n Write the expression for the flexibility coefficient of member G B . f G B G B = 3 ( n 1 2 L 1 A E ) + 3 ( n 2 2 L 2 A E ) + 2 ( n 3 2 L 3 A E ) + 2 ( n 4 2 L 4 A E ) .......(III) Calculation: The free body diagram of truss is shown below. Figure (1) Calculate moment about point A ∑ M A = 0 ( E y × 40 ) = ( 5 k × 30 ft ) + ( 10 k × 10 ft ) E y = 550 40 E y = 13.75 k Calculate all forces in vertical directions. ∑ F y = 0 A y + E y = 10 k + 15 k + 5 k A y = 30 k − 13.75 k A y = 16.25 k Calculate all forces in the horizontal direction. ∑ F x = 0 A x = 0 k The free body diagram of joint A is shown below. Figure (2) Calculate all the forces in the vertical direction. ∑ F y = 0 F A H ( sin 45 ° ) = A y F A H = 16.25 k ( sin 45 ° ) F A H = 22.98 k = A y Calculate all the forces in horizontal direction. ∑ F x = 0 F A B = − F A H ( cos 45 ° ) F A B = − 22.98 k × ( cos 45 ° ) F A B = − 16.25 k The free body diagram of joint B is shown below. Figure (3) Calculate all the forces in the vertical direction. ∑ F y = 0 F B H = 10 k Calculate all the forces in the horizontal direction. ∑ F x = 0 F B C = − F A B = 16.25 k The free body diagram of joint H is shown below. Figure (4) Calculate all the forces in the vertical direction. ∑ F y = 0 F H C ( cos 45 ° ) + ( F H A ) ( cos 45 ° ) + F B H = 0 F H C ( cos 45 ° ) + ( − 22.98 k ) ( cos 45 ° ) + 10 k = 0 F H C = 6.249 k cos ( 45 ° ) F H C = 8.837 k Calculate all the forces in the horizontal direction. ∑ F x = 0 F H C ( sin 45 ° ) − ( F A H ) ( sin 45 ° ) + F H G = 0 ( 8.837 k ) ( sin 45 ° ) − ( − 22.98 k ) ( cos 45 ° ) + F H G = 0 F H C = − 22.50 k The free body diagram of joint C is shown below. Figure (5) Calculate all the forces in the vertical direction. ∑ F y = 0 F C H ( sin 45 ° ) + F C F ( sin 45 ° ) = 15 k 8.837 k × ( sin 45 ° ) + F C F ( sin 45 ° ) = 15 k F C F ( sin 45 ° ) = 15 k − 6.248 k F C F = 8 .751 ( sin 45 ° ) F C F = 12.37 k Calculate all the forces in the horizontal direction. ∑ F x = 0 F H C ( cos 45 ° ) − ( F A H ) ( cos 45 ° ) + F B C − F C D = 0 ( 8.837 k ) ( cos 45 ° ) − ( 12 .37 k ) ( cos 45 ° ) + 16.25 k − F C D = 0 6.248 k − 8.746 k+16 .25 k − F C D = 0 F C D = − 13.75 k The free body diagram of joint G is shown below. Figure (6) Calculate all the forces in the vertical direction. ∑ F y = 0 F G B ( cos 45 ° ) + F G D ( cos 45 ° ) = 0 ( 0 k ) × ( cos 45 ° ) + F G D ( cos 45 ° ) = 0 F G D = 0 k Calculate all the forces in the horizontal direction. ∑ F x = 0 F G H − F G F + F G B ( cos 45 ° ) − F G D ( cos 45 ° ) = 0 F G H − F G F − 0 − 0 = 0 F G H = F G F F G F = − 22.5 k The free body diagram of joint D is shown below. Figure (7) Calculate all the forces in the vertical direction. ∑ F y = 0 F D F + F D G ( sin 45 ° ) = 5 k F D F + ( 0 k ) ( sin 45 ° ) = 5 k F D F = 5 k Calculate all the forces in the horizontal direction. ∑ F x = 0 F D C = F D E F D C = − 13.75 k The free body diagram of joint E is shown below. Figure (8) Calculate all the forces in the vertical direction. ∑ F y = 0 F E F ( sin 45 ° ) + E y = 0 F E F ( sin 45 ° ) + 13.75 k = 0 F D F = − 19.45 k Calculate the forces in the members with unit load at member G using method of joints. Assume the member G B as redundant and calculate the force in member of the truss. The free body diagram of joint A is shown below. Figure (9) Calculate all the forces in the vertical direction. ∑ F y = 0 F A H ( sin 45 ° ) = A y F A H ( sin 45 ° ) = 0 k F A H = 0 k Calculate all the forces in the horizontal direction. ∑ F x = 0 F A B + ( 0 k ) ( cos 45 ° ) = 0 F A z b = 0 k The free body diagram of joint B is shown below. Figure (10) Calculate all the forces in the vertical direction. ∑ F y = 0 F B H = − 1 ( sin 45 ° ) F B H = − 0

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ISBN: 8220106821596

Author: HIBBELER

Publisher: PEARSON

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Chapter 9, Problem 9.33P

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The force in the member GB of the truss.

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Chapter 9, Problem 9.32P

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Chapter 9 Solutions

EBK STRUCTURAL ANALYSIS

Ch. 9 - Determine the reactions at the suooorts, then draw...Ch. 9 - Prob. 9.2P Ch. 9 - Prob. 9.3P Ch. 9 - Prob. 9.4P Ch. 9 - Prob. 9.5P Ch. 9 - Prob. 9.6P Ch. 9 - Prob. 9.7P Ch. 9 - Prob. 9.8P Ch. 9 - Prob. 9.9P Ch. 9 - Prob. 9.10P

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The force in the member G B of the truss.

The force in the member GB of the truss.

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Chapter 9 Solutions