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 global stiffness matrix for the member. 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 Calculation: The free body diagram of the frame is shown below. Figure (1) Consider the member-1. Calculate direction cosine in x-direction. Substitute 0 ft for x F , 0 ft for x N and 12 ft for L in Equation (I). λ x = 0 ft − 0 ft 12 ft = 0 Calculate direction cosine in y-direction. Substitute 0 ft for y F , 12 ft for y N and 12 ft for L in Equation (II). λ y = 12 ft − 0 ft 12 ft = 1 The stiffness matrix for member-1 is shown below. Substitute 0 for λ x and 1 for λ y in global stiffness matrix. k 1 = [ 12 E I L 3 0 − 6 E I L 2 − 12 E I L 3 0 − 6 E I L 2 0 A E L 0 0 − A E L 0 − 6 E I L 2 0 4 E I L 6 E I L 2 0 2 E I L − 12 E I L 3 0 6 E I L 2 12 E I L 3 0 6 E I L 2 0 − A E L 0 0 A E L 0 − 6 E I L 2 0 2 E I L 6 E I L 2 0 4 E I L ] ...... (III) Calculate the value of A E L in stiffness matrix. Substitute 30 in 2 for A , 29 × 10 3 ksi for E and 12 ft for L . A E L = ( 30 in 2 ) × ( 29 × 10 3 ksi ) ( 12 × 12 ) in = 6041.67 kips/in Calculate the value of 6 E I L 2 in stiffness matrix. Substitute 29 × 10 3 ksi for E , 700 in 4 for I and 12 ft for L . 6 E I L 2 = 6 × ( 29 × 10 3 ksi ) × ( 700 in 4 ) ( 12 × 12 ) 2 in 2 = 5873.84 kips Calculate the value of 12 E I L 3 in stiffness matrix. Substitute 29 × 10 3 ksi for E , 700 in 4 for I and 12 ft for L . 12 E I L 3 = 12 × ( 29 × 10 3 ksi ) × ( 700 in 4 ) ( 12 × 12 ) 3 in 3 = 81.58 kips/in Calculate the value of 4 E I L in stiffness matrix. Substitute 29 × 10 3 ksi for E , 700 in 4 for I and 12 ft for L . 4 E I L = 4 × ( 29 × 10 3 ksi ) × ( 700 in 4 ) ( 12 × 12 ) in = 563888.89 kip-in Calculate the value of 2 E I L in stiffness matrix. Substitute 29 × 10 3 ksi for E , 700 in 4 for I and 12 ft for L . 281944.4 2 E I L = 2 × ( 29 × 10 3 ksi ) × ( 700 in 4 ) ( 12 × 12 ) in = 281944.4 kip-in Substitute 6041.67 kips/in for A E L , 5873.84 kips for 6 E I L 2 , 81.58 kips/in for 12 E I L 3 , 563888.89 kip-in for 4 E I L and 281944.4 kip-in for 2 E I L in matrix (III). k 1 = [ 81.58 0 − 5873.84 − 81.58 0 − 5873.84 0 6041.67 0 0 − 6041.67 0 − 5873.84 0 563888.89 5873.84 0 281944.4 − 81.58 0 5873.84 81.58 0 5873.84 0 − 6041.67 0 0 6041.67 0 − 5873.84 0 281944.4 5873.84 0 563888.89 ] ...... (IV) Consider the member-2. Calculate direction cosine in x-direction. Substitute 0 ft for x F , 12 ft for x N and 12 ft for L in Equation (I). λ x = 0 ft − 12 ft 12 ft = − 1 Calculate direction cosine in y-direction. Substitute 12 ft for y F , 12 ft for y N and 12 ft for L in Equation (II). λ y = 12 ft − 12 ft 12 ft = 0 The stiffness matrix for member-1 is shown below. Substitute − 1 for λ x and 0 for λ y in global stiffness matrix. k 2 = [ A E L 0 0 − A E L 0 0 0 12 E I L 3 6 E I L 2 0 − 12 E I L 3 6 E I L 2 0 6 E I L 2 4 E I L 0 − 6 E I L 2 2 E I L − A E L 0 0 A E L 0 0 0 − 12 E I L 3 − 6 E I L 2 0 12 E I L 3 − 6 E I L 2 0 6 E I L 2 2 E I L 0 − 6 E I L 2 4 E I L ]

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ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Load on Structures

Civil engineers primarily deal with beams in their day-to-day applications. A beam is a structural member that resists shear loadings. There are various types of beams such as cantilever beam, simply supported beam, and over the hanging beam. All these b…

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Chapter 16, Problem 16.5P

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The structure stiffness matrix for the frame.

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Structural Analysis (10th Edition)

Ch. 16 - Prob. 16.1P Ch. 16 - Prob. 16.2P Ch. 16 - Prob. 16.3P Ch. 16 - Prob. 16.4P Ch. 16 - Prob. 16.5P Ch. 16 - Prob. 16.6P Ch. 16 - Prob. 16.7P Ch. 16 - Prob. 16.8P Ch. 16 - Prob. 16.9P Ch. 16 - Prob. 16.10P

Ch. 16 - Prob. 16.11P Ch. 16 - Prob. 16.12P Ch. 16 - Prob. 16.13P Ch. 16 - Prob. 16.14P Ch. 16 - Prob. 16.15P Ch. 16 - Prob. 16.16P Ch. 16 - Prob. 16.17P Ch. 16 - Prob. 16.18P

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