Explanation Given: The cross sectional area of the members, A = 0.0015 m 2 . The modulus of elasticity, E = 200 GPa . 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, 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 free body diagram of the frame is shown below. Figure (1) Joint 1 is the origin of the global coordinate system. Consider the member-1. Calculate direction cosine in x -direction. Substitute 4 m for x F , 4 m for x N and 3 m for L in Equation (I). λ x = 4 m − 4 m 3 m 0 Calculate direction cosine in y -direction. Substitute 0 m for y F , 3 m for y N and 3 m for L in Equation (II). λ y = 0 m − 3 m 3 m = − 1 Determine the stiffness matrix for member-1. Substitute 0 for λ x , − 1 for λ y and 3 m for L in Equation (III). k 1 = A E 3 [ ( 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 ( 0 ) ( − 1 ) ( − 1 ) 2 ] 1 2 3 4 = A E [ 0 0 0 0 0 0.333 0 − 0.333 0 0 0 0 0 − 0.333 0 0.333 ] 1 2 3 4 Consider the member- 2 . Calculate direction cosine in x -direction. Substitute 0 m for x F , 4 m for x N and 4 m for L in Equation (I). λ x = 0 m − 4 m 4 m = − 1 Calculate direction cosine in y direction. Substitute 3 m for y F , 3 m for y N and 4 m for L in Equation (II). λ y = 3 m − 3 m 4 m = 0 The stiffness matrix for member-2 is shown below. Substitute − 1 for λ x , 0 for λ y and 4 m for L . k 2 = A E 4 [ ( − 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 ( − 1 ) ( 0 ) ( 0 ) 2 ] 1 2 7 8 = A E [ ( 0.25 ) 0 − ( 0.25 ) 0 0 0 0 0 − ( 0.25 ) 0 ( 0.25 ) 0 0 0 0 0 ] 1 2 7 8 Consider the member- 3 . Calculate direction cosine in x -direction. Substitute 0 m for x F , 4 m for x N and 4 m for L in Equation (I). λ x = 0 m − 4 m 4 m = − 1 Calculate direction cosine in y -direction. Substitute 0 m for y F , 0 m for y N and 4 m for L in Equation (II). λ y = 0 m − 0 m 4 m = 0 The stiffness matrix for member-3 is shown below. Substitute − 1 for λ x , 0 for λ y , 4 m for L in Equation (III). k 3 = A E 4 [ ( − 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 ( − 1 ) ( 0 ) ( 0 ) 2 ] 3 4 6 5 = A E [ ( 0.25 ) 0 − ( 0.25 ) 0 0 0 0 0 − ( 0.25 ) 0 ( 0.25 ) 0 0 0 0 0 ] 3 4 6 5 Consider the member- 4 . Calculate direction cosine in x -direction. Substitute 0 m for x F , 0 m for x N and 3 m for L in Equation (I). λ x = 0 m − 0 m 3 m = 0 Calculate direction cosine in y -direction. Substitute 3 m for y F , 0 m for y N and 3 m for L in Equation (II). λ y = 3 m − 0 m 3 m = 1 The stiffness matrix for member-4 is shown below. Substitute 0 for λ x , 1 for λ y , 3 m for L in Equation (III). k 4 = A E 3 [ ( 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 ( 0 ) ( 1 ) ( 1 ) 2 ] 6 5 7 8 = A E [ 0 0 0 0 0 0.333 0 − 0.333 0 0 0 0 0 − 0

Structural Analysis, Student Value Edition

10th Edition

ISBN: 9780134622088

Author: HIBBELER, Russell C.

Publisher: PEARSON

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Chapter 14 Solutions

Structural Analysis, Student Value Edition

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

Ch. 14 - Prob. 14.11P Ch. 14 - Prob. 14.12P Ch. 14 - Prob. 14.13P

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