Page 1 ? ? = 0 B M 0 = + + BD BC BA M M M (3) ? = 0 D M ? 0 = DB M (4) ? ? = 0 C M 0 = + CE CB M M (5) Substituting the values of and in the equations (3), (4), and (5) CB DB BD BC BA M M M M M , , , , CE M 3.333 0.667 0.5 4.833 BC D EI EI EI ? ?? + +=- 0.5 0 BD EI EI ? ? += 2.333 0.667 7.5 CB EI EI ? ? + = (6) Solving the above set of simultaneous equations, C B ? ? , and D ? are evaluated. 2.4125 B EI ? = - Page 2 ? ? = 0 B M 0 = + + BD BC BA M M M (3) ? = 0 D M ? 0 = DB M (4) ? ? = 0 C M 0 = + CE CB M M (5) Substituting the values of and in the equations (3), (4), and (5) CB DB BD BC BA M M M M M , , , , CE M 3.333 0.667 0.5 4.833 BC D EI EI EI ? ?? + +=- 0.5 0 BD EI EI ? ? += 2.333 0.667 7.5 CB EI EI ? ? + = (6) Solving the above set of simultaneous equations, C B ? ? , and D ? are evaluated. 2.4125 B EI ? = - 3.9057 C EI ? = 1.2063 D EI ? = (7) Substituting the values of C B ? ? , and D ? in (2), beam end moments are computed. 2.794 kN.m AB M = 5.080 kN.m BA M = - 1.8094 kN.m BD M = - 0 DB M = 6.859 kN.m BC M = 3.9028 kN.m CB M = - 3.9057 kN.m CE M = 1.953 kN.m EC M = (8) The reactions are computed in Fig 16.5(d), using equilibrium equations known beam-end moments and given loading. Page 3 ? ? = 0 B M 0 = + + BD BC BA M M M (3) ? = 0 D M ? 0 = DB M (4) ? ? = 0 C M 0 = + CE CB M M (5) Substituting the values of and in the equations (3), (4), and (5) CB DB BD BC BA M M M M M , , , , CE M 3.333 0.667 0.5 4.833 BC D EI EI EI ? ?? + +=- 0.5 0 BD EI EI ? ? += 2.333 0.667 7.5 CB EI EI ? ? + = (6) Solving the above set of simultaneous equations, C B ? ? , and D ? are evaluated. 2.4125 B EI ? = - 3.9057 C EI ? = 1.2063 D EI ? = (7) Substituting the values of C B ? ? , and D ? in (2), beam end moments are computed. 2.794 kN.m AB M = 5.080 kN.m BA M = - 1.8094 kN.m BD M = - 0 DB M = 6.859 kN.m BC M = 3.9028 kN.m CB M = - 3.9057 kN.m CE M = 1.953 kN.m EC M = (8) The reactions are computed in Fig 16.5(d), using equilibrium equations known beam-end moments and given loading. ( ) 6.095 kN Ay R = ? ( ) 9.403 kN Dy R = ? ( ) 4.502 kN Ey R = ? ( ) 1.013 kN Ax R = ? ( ) 0.542 kN Dx R = ? ( ) 1.465 kN Ex R =- ? (9) The bending moment diagram is shown in Fig 16.5.(e) and the elastic curve is shown in Fig 16.5(f). Page 4 ? ? = 0 B M 0 = + + BD BC BA M M M (3) ? = 0 D M ? 0 = DB M (4) ? ? = 0 C M 0 = + CE CB M M (5) Substituting the values of and in the equations (3), (4), and (5) CB DB BD BC BA M M M M M , , , , CE M 3.333 0.667 0.5 4.833 BC D EI EI EI ? ?? + +=- 0.5 0 BD EI EI ? ? += 2.333 0.667 7.5 CB EI EI ? ? + = (6) Solving the above set of simultaneous equations, C B ? ? , and D ? are evaluated. 2.4125 B EI ? = - 3.9057 C EI ? = 1.2063 D EI ? = (7) Substituting the values of C B ? ? , and D ? in (2), beam end moments are computed. 2.794 kN.m AB M = 5.080 kN.m BA M = - 1.8094 kN.m BD M = - 0 DB M = 6.859 kN.m BC M = 3.9028 kN.m CB M = - 3.9057 kN.m CE M = 1.953 kN.m EC M = (8) The reactions are computed in Fig 16.5(d), using equilibrium equations known beam-end moments and given loading. ( ) 6.095 kN Ay R = ? ( ) 9.403 kN Dy R = ? ( ) 4.502 kN Ey R = ? ( ) 1.013 kN Ax R = ? ( ) 0.542 kN Dx R = ? ( ) 1.465 kN Ex R =- ? (9) The bending moment diagram is shown in Fig 16.5.(e) and the elastic curve is shown in Fig 16.5(f). Summary In this lesson plane frames restrained against sidesway are analysed using slope-deflection equations. Equilibrium equations are written at each rigid joint of the frame and also at the support. Few problems are solved to illustrate the procedure. The shear force and bending moment diagrams are drawn for the plane frames.Read More

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