Page 1 ? = 0 B M ? 0 = + BA BC M M (4) Substituting the value of and in the above equation, BC M BA M 5 . 12 667 . 0 333 . 2 = + A B EI EI ? ? (5) Or, EI A B 741 . 18 498 . 3 = + ? ? Solving equation (3) and (4) ) ( 245 . 16 ) ( 002 . 10 clockwise EI ckwise counterclo EI B B - = = ? ? (6) Substituting the value of A ? and B ? in equation (2) beam end moments are evaluated. 0 002 . 10 667 . 0 245 . 16 333 . 1 15 = ? ? ? ? ? ? + ? ? ? ? ? ? - + = EI EI EI EI M AB 1 002 . 10 33 . 1 . 245 . 16 667 . 0 15 - = ? ? ? ? ? ? + ? ? ? ? ? ? - + - = EI EI EI EI M BA kN.m 5 . 12 002 . 10 5 . 2 = ? ? ? ? ? ? + = EI EI M BC kN.m 5 . 2 002 . 10 5 . 0 5 . 2 = ? ? ? ? ? ? + - = EI EI M CB (7) Page 2 ? = 0 B M ? 0 = + BA BC M M (4) Substituting the value of and in the above equation, BC M BA M 5 . 12 667 . 0 333 . 2 = + A B EI EI ? ? (5) Or, EI A B 741 . 18 498 . 3 = + ? ? Solving equation (3) and (4) ) ( 245 . 16 ) ( 002 . 10 clockwise EI ckwise counterclo EI B B - = = ? ? (6) Substituting the value of A ? and B ? in equation (2) beam end moments are evaluated. 0 002 . 10 667 . 0 245 . 16 333 . 1 15 = ? ? ? ? ? ? + ? ? ? ? ? ? - + = EI EI EI EI M AB 1 002 . 10 33 . 1 . 245 . 16 667 . 0 15 - = ? ? ? ? ? ? + ? ? ? ? ? ? - + - = EI EI EI EI M BA kN.m 5 . 12 002 . 10 5 . 2 = ? ? ? ? ? ? + = EI EI M BC kN.m 5 . 2 002 . 10 5 . 0 5 . 2 = ? ? ? ? ? ? + - = EI EI M CB (7) Using these results, reactions are evaluated from equilibrium equations as shown in Fig 16.4 (e) The shear force and bending moment diagrams are shown in Fig 16.4(g) and 16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). Page 3 ? = 0 B M ? 0 = + BA BC M M (4) Substituting the value of and in the above equation, BC M BA M 5 . 12 667 . 0 333 . 2 = + A B EI EI ? ? (5) Or, EI A B 741 . 18 498 . 3 = + ? ? Solving equation (3) and (4) ) ( 245 . 16 ) ( 002 . 10 clockwise EI ckwise counterclo EI B B - = = ? ? (6) Substituting the value of A ? and B ? in equation (2) beam end moments are evaluated. 0 002 . 10 667 . 0 245 . 16 333 . 1 15 = ? ? ? ? ? ? + ? ? ? ? ? ? - + = EI EI EI EI M AB 1 002 . 10 33 . 1 . 245 . 16 667 . 0 15 - = ? ? ? ? ? ? + ? ? ? ? ? ? - + - = EI EI EI EI M BA kN.m 5 . 12 002 . 10 5 . 2 = ? ? ? ? ? ? + = EI EI M BC kN.m 5 . 2 002 . 10 5 . 0 5 . 2 = ? ? ? ? ? ? + - = EI EI M CB (7) Using these results, reactions are evaluated from equilibrium equations as shown in Fig 16.4 (e) The shear force and bending moment diagrams are shown in Fig 16.4(g) and 16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). Page 4 ? = 0 B M ? 0 = + BA BC M M (4) Substituting the value of and in the above equation, BC M BA M 5 . 12 667 . 0 333 . 2 = + A B EI EI ? ? (5) Or, EI A B 741 . 18 498 . 3 = + ? ? Solving equation (3) and (4) ) ( 245 . 16 ) ( 002 . 10 clockwise EI ckwise counterclo EI B B - = = ? ? (6) Substituting the value of A ? and B ? in equation (2) beam end moments are evaluated. 0 002 . 10 667 . 0 245 . 16 333 . 1 15 = ? ? ? ? ? ? + ? ? ? ? ? ? - + = EI EI EI EI M AB 1 002 . 10 33 . 1 . 245 . 16 667 . 0 15 - = ? ? ? ? ? ? + ? ? ? ? ? ? - + - = EI EI EI EI M BA kN.m 5 . 12 002 . 10 5 . 2 = ? ? ? ? ? ? + = EI EI M BC kN.m 5 . 2 002 . 10 5 . 0 5 . 2 = ? ? ? ? ? ? + - = EI EI M CB (7) Using these results, reactions are evaluated from equilibrium equations as shown in Fig 16.4 (e) The shear force and bending moment diagrams are shown in Fig 16.4(g) and 16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). Example 16.3 Compute reactions and beam end moments for the rigid frame shown in Fig 16.5(a). Draw bending moment diagram and sketch the elastic curve for the frame. Solution Page 5 ? = 0 B M ? 0 = + BA BC M M (4) Substituting the value of and in the above equation, BC M BA M 5 . 12 667 . 0 333 . 2 = + A B EI EI ? ? (5) Or, EI A B 741 . 18 498 . 3 = + ? ? Solving equation (3) and (4) ) ( 245 . 16 ) ( 002 . 10 clockwise EI ckwise counterclo EI B B - = = ? ? (6) Substituting the value of A ? and B ? in equation (2) beam end moments are evaluated. 0 002 . 10 667 . 0 245 . 16 333 . 1 15 = ? ? ? ? ? ? + ? ? ? ? ? ? - + = EI EI EI EI M AB 1 002 . 10 33 . 1 . 245 . 16 667 . 0 15 - = ? ? ? ? ? ? + ? ? ? ? ? ? - + - = EI EI EI EI M BA kN.m 5 . 12 002 . 10 5 . 2 = ? ? ? ? ? ? + = EI EI M BC kN.m 5 . 2 002 . 10 5 . 0 5 . 2 = ? ? ? ? ? ? + - = EI EI M CB (7) Using these results, reactions are evaluated from equilibrium equations as shown in Fig 16.4 (e) The shear force and bending moment diagrams are shown in Fig 16.4(g) and 16.4 h respectively. The qualitative elastic curve is shown in Fig 16.4 (h). Example 16.3 Compute reactions and beam end moments for the rigid frame shown in Fig 16.5(a). Draw bending moment diagram and sketch the elastic curve for the frame. Solution The given frame is kinematically indeterminate to third degree so three rotations are to be calculated, C B ? ? , and D ? . First calculate the fixed end moments (see Fig 16.5 b). 2 54 4 kN.m 20 F AB M × == 2 54 2.667 kN.m 30 F BA M -× ==- 2 2 10 3 3 7.5 kN.m 6 F BC M ×× == 2 2 10 3 3 7.5 kN.m 6 F CB M -× × ==- (1) 0 = = = = F EC F CE F DB F BD M M M M The frame is restrained against sidesway. Four spans must be considered for rotating slope – deflection equation: AB, BD, BC and CE. The beam endRead More

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