Explanation Concept Used: Write the expression for direction cosine in x-direction. λ x = x F − x N L ...... (I) Here, coordinate of x at near end of the member is x F and coordinate of x at far end of the member is x N . Write the expression for direction cosine in x-direction. λ y = y F − y N L ...... (II) Here, coordinate of y at near end of the member is y F and coordinate of y at far end of the member is y N . Write the stiffness matrix for the member. k = A E L [ λ x 2 λ x λ y − λ x 2 − λ x λ y λ x λ y λ y 2 − λ x λ y − λ y 2 − λ x 2 − λ x λ y λ x 2 λ x λ y − λ x λ y − λ y 2 λ x λ y λ y 2 ] ...... (III) Here, the stiffness matrix for the member is k , cross-sectional area of the member is A , length of the member is L , modulus of elasticity is E and moment of inertia is I . Write the expression for total structural stiffness matrix. K = k 1 + k 2 + k 3 + k 4 + k 5 + k 6 ...... (IV) Here, the structural stiffness matrix is K . Calculation: The given truss is shown below. Figure-(1) Consider the member-1. Consider the node 5 is as the origin. Calculate the length of member 1 . L = ( 10 ft ) 2 + ( 10 ft ) 2 = 200 ft =14 .142 ft Calculate the direction cosine in x -direction. Substitute 10 ft for x F , 0 ft for x N and 14.142 ft for L in Equation (I). λ x = 10 ft − 0 ft 14.142 ft = 0.707 Calculate direction cosine in y -direction. Substitute 10 ft for y F , 20 ft for y N and 14.142 ft for L in Equation (II). λ y = 10 ft − 20 ft 14.142 ft = − 0.707 Determine the stiffness matrix for member-1. Substitute 0.707 for λ x , − 0.707 for λ y and 14.142 ft for L in Equation (III). k 1 = A E 14.142 [ ( 0.707 ) 2 ( 0.707 ) ( − 0.707 ) − ( 0.707 ) 2 − ( 0.707 ) ( − 0.707 ) ( 0.707 ) ( − 0.707 ) ( − 0.707 ) 2 − ( 0.707 ) ( − 0.707 ) − ( − 0.707 ) 2 − ( 0.707 ) 2 − ( 0.707 ) ( − 0.707 ) ( − 0.707 ) 2 ( 0.707 ) ( − 0.707 ) − ( 0.707 ) ( − 0.707 ) − ( − 0.707 ) 2 ( − 0.707 ) ( 0.707 ) ( − 0.707 ) 2 ] = A E [ 0.0353 − 0.0353 − 0.0353 0.0353 − 0.0353 0.0353 0.0353 − 0.0353 − 0.0353 0.0353 0.0353 − 0.0353 0.0353 − 0.0353 − 0.0353 0.0353 ] 1 2 5 6 Consider the member- 2 . Calculate direction cosine in x -direction. Substitute 0 ft for x F , 0 ft for x N and 10 ft for L in Equation (I). λ x = 0 ft − 0 ft 10 ft = 0 Calculate direction cosine in y direction. Substitute 10 ft for y F , 20 ft for y N and 10 ft for L in Equation (II). λ y = 10 ft − 20 ft 10 ft = − 1 Determine the stiffness matrix for member-2. Substitute 0 for λ x , − 1 for λ y and 10 ft for L in Equation (III). k 2 = A E 10 [ ( 0 ) 2 ( 0 ) ( − 1 ) − ( 0 ) 2 − ( 0 ) ( − 1 ) ( 0 ) ( − 1 ) ( − 1 ) 2 − ( 0 ) ( − 1 ) − ( − 1 ) 2 − ( 0 ) 2 − ( 0 ) ( − 1 ) ( 0 ) 2 ( 0 ) ( − 1 ) − ( 0 ) ( − 1 ) − ( − 1 ) 2 ( − 1 ) ( 0 ) ( − 1 ) 2 ] = A E [ 0 0 0 0 0 0.1 0 − 0.1 0 0 0 0 0 − 0.1 0 0.1 ] 1 2 3 4 Consider the member- 3 . Calculate direction cosine in x -direction. Substitute 0 ft for x F , 0 ft for x N and 10 ft for L in Equation (I). λ x = 0 ft − 0 ft 10 ft = 0 Calculate direction cosine in y direction. Substitute 0 ft for y F , 10 ft for y N and 10 ft for L in Equation (II). λ y = 0 ft − 10 ft 10 ft = − 1 Determine the stiffness matrix for member-3. Substitute 0 for λ x , − 1 for λ y and 10 ft for L in Equation (III). k 3 = A E 10 [ ( 0 ) 2 ( 0 ) ( − 1 ) − ( 0 ) 2 − ( 0 ) ( − 1 ) ( 0 ) ( − 1 ) ( − 1 ) 2 − ( 0 ) ( − 1 ) − ( − 1 ) 2 − ( 0 ) 2 − ( 0 ) ( − 1 ) ( − 1 ) 2 ( 0 ) ( − 1 ) − ( 0 ) ( − 1 ) − ( − 1 ) 2 ( − 1 ) ( 0 ) ( − 1 ) 2 ] = A E [ 0 0 0 0 0 0.1 0 − 0.1 0 0 0 0 0 − 0.1 0 0.1 ] 3 4 9 10 Consider the member- 4 Calculate direction cosine in x -direction. Substitute 10 ft for x F , 0 ft for x N and 10 ft for L in Equation (I). λ x = 10 ft − 0 ft 10 ft = 1 Calculate direction cosine in y direction. Substitute 10 ft for y F , 10 ft for y N and 10 ft for L in Equation (II). λ y = 0 ft − 10 ft 10 ft = 0 Determine the stiffness matrix for member-4. Substitute 1 for λ x , 0 for λ y and 10 ft for L in Equation (III). k 4 = A E 10 [ ( 1 ) 2 ( 1 ) ( 0 ) − ( 1 ) 2 − ( 1 ) ( 0 ) ( 1 ) ( 0 ) ( 0 ) 2 − ( 1 ) ( 0 ) − ( 0 ) 2 − ( 1 ) 2 − ( 1 ) ( 0 ) ( 1 ) 2 ( 1 ) ( 0 ) − ( 1 ) ( 0 ) − ( 0 ) 2 ( 0 ) ( 1 ) ( 0 ) 2 ] = A E [ 0

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ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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A force is an external influence or agent that changes the state of a system. According to Newton's second law, the force applied to a body is proportional to the change in momentum of a body. Hence, whenever a force is applied to a body, its momentum ch…

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Chapter 14, Problem 14.13P

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The displacement at joint 3 and force in the member 1.

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Ch. 14 - Prob. 14.1P Ch. 14 - Prob. 14.2P Ch. 14 - Prob. 14.3P Ch. 14 - Prob. 14.4P Ch. 14 - Prob. 14.5P Ch. 14 - Prob. 14.6P Ch. 14 - Prob. 14.7P Ch. 14 - Prob. 14.8P Ch. 14 - Prob. 14.9P Ch. 14 - Prob. 14.10P

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