Page 1 ) 001 . 0 2 ( 2 . 0 667 . 41 - + + = D C CD EI M ? ? ) 001 . 0 2 ( 2 . 0 667 . 41 - + + - = C D DC EI M ? ? (3) For the present problem, four joint equilibrium equations can be written, one each for each of the supports. They are 0 0 = ? = ? AB A M M 0 0 = + ? = ? BC BA B M M M 0 0 = + ? = ? CD CB C M M M (4) 0 0 = ? = ? DC D M M Substituting the values of beam end moments from equations (3) in equation (4), four equations are obtained in four unknown rotations A ? , B ? , C ? and D ? .They are, ( ) 2 6 3 kN.m 000 , 270 10 35 . 1 10 200 = × × × = - EI 3 10 2716 . 1 2 - × - = + B A ? ? 001 . 0 4 - = + + C B A ? ? ? 0005 . 0 4 = + + D C B ? ? ? (5) 3 10 7716 . 1 2 - × = + D C ? ? Solving the above sets of simultaneous equations, values of A ? , B ? , C ? and D ? are evaluated. radians 4 10 9629 . 5 - × - = A ? radians 5 10 9013 . 7 - × - = B ? radians 5 10 7653 . 8 - × - = C ? radians (6) 4 10 2963 . 9 - × = D ? Page 2 ) 001 . 0 2 ( 2 . 0 667 . 41 - + + = D C CD EI M ? ? ) 001 . 0 2 ( 2 . 0 667 . 41 - + + - = C D DC EI M ? ? (3) For the present problem, four joint equilibrium equations can be written, one each for each of the supports. They are 0 0 = ? = ? AB A M M 0 0 = + ? = ? BC BA B M M M 0 0 = + ? = ? CD CB C M M M (4) 0 0 = ? = ? DC D M M Substituting the values of beam end moments from equations (3) in equation (4), four equations are obtained in four unknown rotations A ? , B ? , C ? and D ? .They are, ( ) 2 6 3 kN.m 000 , 270 10 35 . 1 10 200 = × × × = - EI 3 10 2716 . 1 2 - × - = + B A ? ? 001 . 0 4 - = + + C B A ? ? ? 0005 . 0 4 = + + D C B ? ? ? (5) 3 10 7716 . 1 2 - × = + D C ? ? Solving the above sets of simultaneous equations, values of A ? , B ? , C ? and D ? are evaluated. radians 4 10 9629 . 5 - × - = A ? radians 5 10 9013 . 7 - × - = B ? radians 5 10 7653 . 8 - × - = C ? radians (6) 4 10 2963 . 9 - × = D ? Substituting the values in slope-deflection equations the beam end moments are evaluated. 0 )} 0005 . 0 ) 10 9013 . 7 ( ) 10 9629 . 5 ( 2 { 000 , 270 2 . 0 667 . 41 5 4 = + × - + × - × + = - - AB M kN.m 40 . 55 } 0005 . 0 10 9629 . 5 ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41 4 5 - = + × - × - × + - = - - BA M kN.m 40 . 55 } 0005 . 0 ) 10 7653 . 8 ( ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41 5 5 = + × - + × - × + = - - BC M kN.m 40 . 28 } 0005 . 0 10 9013 . 7 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41 5 5 - = + × - × - × + - = - - CB M kN.m 40 . 28 } 001 . 0 10 2963 . 9 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41 4 5 = - × + × - × × + = - - CD M kN.m 0 } 001 . 0 10 7653 . 8 10 2963 . 9 2 { 000 , 270 2 . 0 667 . 41 5 4 = - × - × × × + - = - - DC M (7) Reactions are obtained from equilibrium equations. Now consider the free body diagram of the beam with end moments and external loads as shown in Fig.15.3d. ) ( kN 46 . 19 ? = A R Page 3 ) 001 . 0 2 ( 2 . 0 667 . 41 - + + = D C CD EI M ? ? ) 001 . 0 2 ( 2 . 0 667 . 41 - + + - = C D DC EI M ? ? (3) For the present problem, four joint equilibrium equations can be written, one each for each of the supports. They are 0 0 = ? = ? AB A M M 0 0 = + ? = ? BC BA B M M M 0 0 = + ? = ? CD CB C M M M (4) 0 0 = ? = ? DC D M M Substituting the values of beam end moments from equations (3) in equation (4), four equations are obtained in four unknown rotations A ? , B ? , C ? and D ? .They are, ( ) 2 6 3 kN.m 000 , 270 10 35 . 1 10 200 = × × × = - EI 3 10 2716 . 1 2 - × - = + B A ? ? 001 . 0 4 - = + + C B A ? ? ? 0005 . 0 4 = + + D C B ? ? ? (5) 3 10 7716 . 1 2 - × = + D C ? ? Solving the above sets of simultaneous equations, values of A ? , B ? , C ? and D ? are evaluated. radians 4 10 9629 . 5 - × - = A ? radians 5 10 9013 . 7 - × - = B ? radians 5 10 7653 . 8 - × - = C ? radians (6) 4 10 2963 . 9 - × = D ? Substituting the values in slope-deflection equations the beam end moments are evaluated. 0 )} 0005 . 0 ) 10 9013 . 7 ( ) 10 9629 . 5 ( 2 { 000 , 270 2 . 0 667 . 41 5 4 = + × - + × - × + = - - AB M kN.m 40 . 55 } 0005 . 0 10 9629 . 5 ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41 4 5 - = + × - × - × + - = - - BA M kN.m 40 . 55 } 0005 . 0 ) 10 7653 . 8 ( ) 10 9013 . 7 ( 2 { 000 , 270 2 . 0 667 . 41 5 5 = + × - + × - × + = - - BC M kN.m 40 . 28 } 0005 . 0 10 9013 . 7 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41 5 5 - = + × - × - × + - = - - CB M kN.m 40 . 28 } 001 . 0 10 2963 . 9 ) 10 765 . 8 ( 2 { 000 , 270 2 . 0 667 . 41 4 5 = - × + × - × × + = - - CD M kN.m 0 } 001 . 0 10 7653 . 8 10 2963 . 9 2 { 000 , 270 2 . 0 667 . 41 5 4 = - × - × × × + - = - - DC M (7) Reactions are obtained from equilibrium equations. Now consider the free body diagram of the beam with end moments and external loads as shown in Fig.15.3d. ) ( kN 46 . 19 ? = A R ) ( kN 54 . 30 ? = BL R ) ( kN 7 . 27 ? = BR R ) ( kN 3 . 22 ? = CL R ) ( kN 84 . 27 ? = CR R ) ( kN 16 . 22 ? = D R The shear force and bending moment diagrams are shown in Fig.15.5e. Summary In this lesson, slope-deflection equations are derived for the case of beam with yielding supports. Moments developed at the ends are related to rotations and support settlements. The equilibrium equations are written at each support. The continuous beam is solved using slope-deflection equations. The deflected shape of the beam is sketched. The bending moment and shear force diagrams are drawn for the examples solved in this lesson.Read More

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- The Slope Deflection Method: Frames with Sidesway - 2