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 Write the load displacement relationship. Q = K × D ...... (III) Here, global displacement loading is Q and global displacement matrix is K . 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 12 ft for y F , 0 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 = [ 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 ] ...... (IV) Calculate the value of A E L in stiffness matrix. Substitute 8 in 2 for A , 29 × 10 3 ksi for E and 12 ft for L . A E L = ( 8 in 2 ) × ( 29 × 10 3 ksi ) ( 12 × 12 ) in = 1611.11 kips/in Calculate the value of 6 E I L 2 in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 12 ft for L . 6 E I L 2 = 6 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 12 × 12 ) 2 in 2 = 67.12 kips Calculate the value of 12 E I L 3 in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 12 ft for L . 12 E I L 3 = 12 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 12 × 12 ) 3 in 3 = 0.9323 kips/in Calculate the value of 4 E I L in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 12 ft for L . 4 E I L = 4 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 12 × 12 ) in = 6444.44 kip-in Calculate the value of 2 E I L in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 12 ft for L . 2 E I L = 2 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 12 × 12 ) in = 3222.22 kip-in Substitute 1611.11 kips/in for A E L , 67.12 kips for 6 E I L 2 , 0.9323 kips/in for 12 E I L 3 , 6444.44 kip-in for 4 E I L and 3222.22 kip-in for 2 E I L in matrix (IV). k 1 = [ 1611.11 0 0 − 1611.11 0 0 0 0.9323 67.12 0 − 0.9323 67.12 0 67.12 6444.44 0 − 67.12 3222.22 − 1611.11 0 0 1611.11 0 0 0 − 0.9323 − 67.12 0 0.9323 − 67.12 0 67.12 3222.22 0 − 67.12 6444.44 ] ...... (V) Consider the member-2. Calculate direction cosine in x-direction. Substitute 0 ft for x F , 8 ft for x N and 8 ft for L in Equation (I). λ x = 0 ft − 8 ft 8 ft = − 1 Calculate direction cosine in y-direction. Substitute 0 ft for y F , 0 ft for y N and 8 ft for L in Equation (II). λ y = 0 ft − 0 ft 8 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 ] ...... (VI) Calculate the value of A E L in stiffness matrix. Substitute 8 in 2 for A , 29 × 10 3 ksi for E and 8 ft for L . A E L = ( 8 in 2 ) × ( 29 × 10 3 ksi ) ( 8 × 12 ) in = 2416.667 kips/in Calculate the value of 6 E I L 2 in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 8 ft for L . 6 E I L 2 = 6 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 8 × 12 ) 2 in 2 = 151.041 kips Calculate the value of 12 E I L 3 in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 8 ft for L . 12 E I L 3 = 12 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 8 × 12 ) 3 in 3 = 3.146 kips/in Calculate the value of 4 E I L in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 8 ft for L . 4 E I L = 4 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 12 × 12 ) in = 9666.668 kip-in Calculate the value of 2 E I L in stiffness matrix. Substitute 29 × 10 3 ksi for E , 450 in 4 for I and 8 ft for L . 2 E I L = 2 × ( 29 × 10 3 ksi ) × ( 450 in 4 ) ( 8 × 12 ) in = 4833.334 kip-in Substitute 2416.667 kips/in for A E L , 151.041 kips for 6 E I L 2 , 3.146 kips/in for 12 E I L 3 , 9666.668 kip-in for 4 E I L and 4833.334 kip-in for 2 E I L in matrix (VI). k 2 = [ 2416.667 0 0 − 2416.667 0 0 0 3.146 151.041 0 − 3.146 151.041 5454.28 151.041 9666.668 0 − 151.041 4833.334 − 2416.667 0 0 2416.667 0 0 0 − 3.146 − 151.041 0 3.146 − 151.041 5454.28 151.041 4833.334 0 − 151.041 9666.668 ] ...... (VII) Consider the member-3. 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 12 ft for y F , 0 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 3 = [ 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 ]

10th Edition

ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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