Page 1 curve " AB rotates by an angle AB ? , rotates by " "C B BC ? and rotates by DC CD ? as shown in figure. Due to symmetry, AB CD ? ? = . From the geometry of the figure, AB AB AB L L BB 1 " ? - = = ? But a cos 1 ? = ? Thus, 5 cos ? - = ? - = a ? AB AB L 5 ? - = CD ? 5 tan 2 tan 2 2 2 ? = ? = ? = ? = a a ? BC (1) We have three independent unknowns for this problem C B ? ? , and . The ends ? A and are fixed. Hence, D . 0 = = D A ? ? Fixed end moments are, . 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = = F DC F CD F CB F BC F BA F AB M M m kN M m kN M M M Now, writing the slope-deflection equations for the six beam end moments, [] AB A AB I E M ? ? 3 1 . 5 ) 2 ( 2 - = ? + = EI EI M B AB 471 . 0 784 . 0 ? ? + = EI EI M B BA 471 . 0 568 . 1 ? ? - + + = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? - + + - = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? + = EI EI M C CD 471 . 0 568 . 1 ? ? + = EI EI M C DC 471 . 0 784 . 0 ? (2) Now, considering the joint equilibrium of B andC , yields Page 2 curve " AB rotates by an angle AB ? , rotates by " "C B BC ? and rotates by DC CD ? as shown in figure. Due to symmetry, AB CD ? ? = . From the geometry of the figure, AB AB AB L L BB 1 " ? - = = ? But a cos 1 ? = ? Thus, 5 cos ? - = ? - = a ? AB AB L 5 ? - = CD ? 5 tan 2 tan 2 2 2 ? = ? = ? = ? = a a ? BC (1) We have three independent unknowns for this problem C B ? ? , and . The ends ? A and are fixed. Hence, D . 0 = = D A ? ? Fixed end moments are, . 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = = F DC F CD F CB F BC F BA F AB M M m kN M m kN M M M Now, writing the slope-deflection equations for the six beam end moments, [] AB A AB I E M ? ? 3 1 . 5 ) 2 ( 2 - = ? + = EI EI M B AB 471 . 0 784 . 0 ? ? + = EI EI M B BA 471 . 0 568 . 1 ? ? - + + = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? - + + - = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? + = EI EI M C CD 471 . 0 568 . 1 ? ? + = EI EI M C DC 471 . 0 784 . 0 ? (2) Now, considering the joint equilibrium of B andC , yields 0 0 = + ? = ? BC BA B M M M 5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI C B ? ? (3) 0 0 = + ? = ? CD CB C M M M 5 . 2 129 . 0 568 . 3 = ? - + EI EI EI B C ? ? (4) Shear equation for Column AB 0 ) 1 ( 5 1 1 = + - - V M M H BA AB (5) Column CD 0 ) 1 ( 5 2 2 = + - - V M M H DC CD (6) Beam BC 0 10 2 0 1 = - - - = ? CB BC C M M V M (7) Page 3 curve " AB rotates by an angle AB ? , rotates by " "C B BC ? and rotates by DC CD ? as shown in figure. Due to symmetry, AB CD ? ? = . From the geometry of the figure, AB AB AB L L BB 1 " ? - = = ? But a cos 1 ? = ? Thus, 5 cos ? - = ? - = a ? AB AB L 5 ? - = CD ? 5 tan 2 tan 2 2 2 ? = ? = ? = ? = a a ? BC (1) We have three independent unknowns for this problem C B ? ? , and . The ends ? A and are fixed. Hence, D . 0 = = D A ? ? Fixed end moments are, . 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = = F DC F CD F CB F BC F BA F AB M M m kN M m kN M M M Now, writing the slope-deflection equations for the six beam end moments, [] AB A AB I E M ? ? 3 1 . 5 ) 2 ( 2 - = ? + = EI EI M B AB 471 . 0 784 . 0 ? ? + = EI EI M B BA 471 . 0 568 . 1 ? ? - + + = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? - + + - = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? + = EI EI M C CD 471 . 0 568 . 1 ? ? + = EI EI M C DC 471 . 0 784 . 0 ? (2) Now, considering the joint equilibrium of B andC , yields 0 0 = + ? = ? BC BA B M M M 5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI C B ? ? (3) 0 0 = + ? = ? CD CB C M M M 5 . 2 129 . 0 568 . 3 = ? - + EI EI EI B C ? ? (4) Shear equation for Column AB 0 ) 1 ( 5 1 1 = + - - V M M H BA AB (5) Column CD 0 ) 1 ( 5 2 2 = + - - V M M H DC CD (6) Beam BC 0 10 2 0 1 = - - - = ? CB BC C M M V M (7) ? = + = 5 0 2 1 H H F X (8) ? = - - = 0 10 0 2 1 V V F Y (9) From equation (7), 2 10 1 + + = CB BC M M V From equation (8), 2 1 5 H H - = From equation (9), 10 2 10 10 1 2 - + + = - = CB BC M M V V Substituting the values of and in equations (5) and (6), 1 1 ,H V 2 V 0 2 2 10 60 2 = + + - - - CB BC BA AB M M M M H (10) 0 2 2 10 10 2 = + + - - + - CB BC DC CD M M M M H (11) Eliminating in equation (10) and (11), 2 H 25 = - - + + + CB BC DC CD BA AB M M M M M M (12) Substituting the values of in (12) we get the required third equation. Thus, DC CD BA AB M M M M , , , + ? + EI EI B 471 . 0 784 . 0 ? + ? + EI EI B 471 . 0 568 . 1 ? + ? + EI EI C 471 . 0 568 . 1 ? ? + EI EI C 471 . 0 784 . 0 ? -( ? - + + EI EI EI C B 6 . 0 2 5 . 2 ? ? )- ( ? - + + - EI EI EI C B 6 . 0 2 5 . 2 ? ? ) 25 = Simplifying, 25 084 . 3 648 . 0 648 . 0 = ? + - - EI EI EI B C ? ? (13) Solving simultaneously equations (3) (4) and (13), yields 205 . 1 ; 741 . 0 = - = C B EI EI ? ? and 204 . 8 = ? EI . Substituting the values of C B EI EI ? ? , and ? EI in the slope-deflection equation (3), one could calculate beam end moments. Thus, 3.28 kN.m AB M = Page 4 curve " AB rotates by an angle AB ? , rotates by " "C B BC ? and rotates by DC CD ? as shown in figure. Due to symmetry, AB CD ? ? = . From the geometry of the figure, AB AB AB L L BB 1 " ? - = = ? But a cos 1 ? = ? Thus, 5 cos ? - = ? - = a ? AB AB L 5 ? - = CD ? 5 tan 2 tan 2 2 2 ? = ? = ? = ? = a a ? BC (1) We have three independent unknowns for this problem C B ? ? , and . The ends ? A and are fixed. Hence, D . 0 = = D A ? ? Fixed end moments are, . 0 ; 0 ; . 50 . 2 ; . 50 . 2 ; 0 ; 0 = = - = + = = = F DC F CD F CB F BC F BA F AB M M m kN M m kN M M M Now, writing the slope-deflection equations for the six beam end moments, [] AB A AB I E M ? ? 3 1 . 5 ) 2 ( 2 - = ? + = EI EI M B AB 471 . 0 784 . 0 ? ? + = EI EI M B BA 471 . 0 568 . 1 ? ? - + + = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? - + + - = EI EI EI M C B BC 6 . 0 2 5 . 2 ? ? ? + = EI EI M C CD 471 . 0 568 . 1 ? ? + = EI EI M C DC 471 . 0 784 . 0 ? (2) Now, considering the joint equilibrium of B andC , yields 0 0 = + ? = ? BC BA B M M M 5 . 2 129 . 0 568 . 3 - = ? - + EI EI EI C B ? ? (3) 0 0 = + ? = ? CD CB C M M M 5 . 2 129 . 0 568 . 3 = ? - + EI EI EI B C ? ? (4) Shear equation for Column AB 0 ) 1 ( 5 1 1 = + - - V M M H BA AB (5) Column CD 0 ) 1 ( 5 2 2 = + - - V M M H DC CD (6) Beam BC 0 10 2 0 1 = - - - = ? CB BC C M M V M (7) ? = + = 5 0 2 1 H H F X (8) ? = - - = 0 10 0 2 1 V V F Y (9) From equation (7), 2 10 1 + + = CB BC M M V From equation (8), 2 1 5 H H - = From equation (9), 10 2 10 10 1 2 - + + = - = CB BC M M V V Substituting the values of and in equations (5) and (6), 1 1 ,H V 2 V 0 2 2 10 60 2 = + + - - - CB BC BA AB M M M M H (10) 0 2 2 10 10 2 = + + - - + - CB BC DC CD M M M M H (11) Eliminating in equation (10) and (11), 2 H 25 = - - + + + CB BC DC CD BA AB M M M M M M (12) Substituting the values of in (12) we get the required third equation. Thus, DC CD BA AB M M M M , , , + ? + EI EI B 471 . 0 784 . 0 ? + ? + EI EI B 471 . 0 568 . 1 ? + ? + EI EI C 471 . 0 568 . 1 ? ? + EI EI C 471 . 0 784 . 0 ? -( ? - + + EI EI EI C B 6 . 0 2 5 . 2 ? ? )- ( ? - + + - EI EI EI C B 6 . 0 2 5 . 2 ? ? ) 25 = Simplifying, 25 084 . 3 648 . 0 648 . 0 = ? + - - EI EI EI B C ? ? (13) Solving simultaneously equations (3) (4) and (13), yields 205 . 1 ; 741 . 0 = - = C B EI EI ? ? and 204 . 8 = ? EI . Substituting the values of C B EI EI ? ? , and ? EI in the slope-deflection equation (3), one could calculate beam end moments. Thus, 3.28 kN.m AB M = 2.70 kN.m BA M = 2.70 kN.m BC M = - 5.75 kN.m CB M = - 5.75 kN.m CD M = 4.81 kN.m DC M = . (14) The bending moment diagram for the frame is shown in Fig. 17.5 d. Summary In this lesson, slope-deflection equations are derived for the plane frame undergoing sidesway. Using these equations, plane frames with sidesway are analysed. The reactions are calculated from static equilibrium equations. A couple of problems are solved to make things clear. In each numerical example, the bending moment diagram is drawn and deflected shape is sketched for the plane frame.Read More

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- Introduction: The Moment Distribution Method - 1
- Introduction: The Moment Distribution Method - 2
- The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 1
- Displacement Methods: Slope Deflection and Moment Distribution Methods
- The Moment Distribution Method: Statically Indeterminate Beams With Support Settlements - 2
- Slope Deflection Method