Explanation Concept Used: Write the compatibility equation for the truss member E D . Δ E D + F E D × f E D E D = 0 ..... (I) Here, the deflection in truss member E D due to real loads is Δ E D , the force in redundant truss member E D is F E D and the flexibility coefficient of deflection in truss member E D is f E D E D . Write the expression for the deflection in truss member E D due to real loads. Δ D E = ∑ n N L A E = [ n A F N A F L A F A E + n B F N B F L B F A E + n E F N E F L E F A E + n E C N E C L E C A E ] ....(II) Here, the force in member in redundant structure is n , the force in member in primary structure is N , length of the member of the truss is L , the cross-sectional area for the truss member is A and modulus of elasticity of the truss member is E . Write the expression for the flexibility coefficient of deflection in truss member E D . f E D E D = ∑ n n L A E = [ ∑ ( n A F ) ( n A F ) L A F A E + ∑ ( n B F ) ( n B F ) L B F A E + ∑ ( n E F ) ( n E F ) L E F A E + ∑ ( n E D ) ( n E D ) L E D A E + ∑ ( n E C ) ( n E C ) L E C A E ] ..... (III) Calculation: Consider truss member D E as redundant member. The free body diagram of deflection of truss member D E by applying superposition is shown below. Figure (1) Consider the member of the truss D E as redundant having zero force. The free body diagram of the primary truss is shown below. Figure (2) Calculate the forces in the members of the truss. The free body diagram of the joint E is shown below. Figure (3) Here, the force in member E F in primary truss is F E F , the force in member E C in primary truss is F E C and the force in member E D in primary truss is F E D . Calculate all forces in x-direction. ∑ F x = 0 ( F E D × 4 m 5 m ) + 8 kN − F E F = 0 ..... (IV) Substitute 0 for F E D in Equation (IV). ( 0 × 4 m 5 m ) + 8 kN − F E F = 0 F E F = 8 kN ( tension ) Calculate all forces in y-direction. ∑ F y = 0 ( F E D × 3 m 5 m ) + F E C = 0 ..... (V) Substitute 0 for F E D in Equation (V). ( 0 × 3 m 5 m ) + F E C = 0 F E C = 0 The free body diagram of the joint F is shown below. Figure (4) Here, the force in member F A in primary truss is F F A and the force in member F B in primary truss is F F B . Calculate all forces in x-direction. ∑ F x = 0 F E F − ( F F A × 4 m 5 m ) = 0 ..... (VI) Substitute 8 kN for F E F in Equation (VI). 8 kN − ( F F A × 4 m 5 m ) = 0 F F A = 10 kN ( tension ) Calculate all forces in y-direction. ∑ F y = 0 ( F F A × 3 m 5 m ) + F F B = 0 ..... (VII) Substitute 10 kN for F F A in Equation (VII). ( 10 kN × 3 m 5 m ) + F F B = 0 F F B = − 6 kN ( compression ) Use virtual method and consider the redundant truss member D E having unit load. The free body diagram of the redundant truss is shown below. Figure (5) Calculate the forces in the member of the truss. The free body diagram of the joint E is shown below. Figure (6) Here, the force in member E F in redundant truss is F E F , the force in member E C in redundant truss is F E C and the force in member E D in redundant truss is F E D . Calculate all forces in x-direction. ∑ F x = 0 ( F E D × 4 m 5 m ) − F E F = 0 ..... (VIII) Substitute 1 kN for F E D in Equation (VIII). ( 1 kN × 4 m 5 m ) − F E F = 0 F E F = 0.8 kN ( tension ) Calculate all forces in y-direction. ∑ F y = 0 ( F E D × 3 m 5 m ) + F E C = 0 ..... (IX) Substitute 1 kN for F E D in Equation (IX). ( 1 kN × 3 m 5 m ) + F E C = 0 F E C = 0

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ISBN: 9780134713649

Author: HIBBELER

Publisher: PEARSON

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Chapter 9, Problem 9.32P

To determine

The force in each member of the truss.

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Chapter 9, Problem 9.31P

Chapter 9, Problem 9.33P

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Problem 5 (A, B, C and D are fixed). Find the reactions at A and D 8 k B 15 ft A -20 ft C 10 ft D

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Problem 1 (A, C and D are pins) Find the reactions and A, C and D. D 6 m B 12 kN/m 8 m A C 6 m

Chapter 9 Solutions

STRUCTURAL ANAL.W/MOD MASTERING ACCE

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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