Page 1 Instructional Objectives After reading this chapter the student will be able to 1. Derive slope-deflection equations for the frames undergoing sidesway. 2. Analyse plane frames undergoing sidesway. 3, Draw shear force and bending moment diagrams. 4. Sketch deflected shape of the plane frame not restrained against sidesway. 17.1 Introduction In this lesson, slope-deflection equations are applied to analyse statically indeterminate frames undergoing sidesway. As stated earlier, the axial deformation of beams and columns are small and are neglected in the analysis. In the previous lesson, it was observed that sidesway in a frame will not occur if 1. They are restrained against sidesway. 2. If the frame geometry and the loading are symmetrical. In general loading will never be symmetrical. Hence one could not avoid sidesway in frames. For example, consider the frame of Fig. 17.1. In this case the frame is symmetrical but not the loading. Due to unsymmetrical loading the beam end moments and are not equal. If is greater than , then . In BC M CB M b a CB BC M M > Page 2 Instructional Objectives After reading this chapter the student will be able to 1. Derive slope-deflection equations for the frames undergoing sidesway. 2. Analyse plane frames undergoing sidesway. 3, Draw shear force and bending moment diagrams. 4. Sketch deflected shape of the plane frame not restrained against sidesway. 17.1 Introduction In this lesson, slope-deflection equations are applied to analyse statically indeterminate frames undergoing sidesway. As stated earlier, the axial deformation of beams and columns are small and are neglected in the analysis. In the previous lesson, it was observed that sidesway in a frame will not occur if 1. They are restrained against sidesway. 2. If the frame geometry and the loading are symmetrical. In general loading will never be symmetrical. Hence one could not avoid sidesway in frames. For example, consider the frame of Fig. 17.1. In this case the frame is symmetrical but not the loading. Due to unsymmetrical loading the beam end moments and are not equal. If is greater than , then . In BC M CB M b a CB BC M M > such a case joint B andC are displaced toward right as shown in the figure by an unknown amount ? . Hence we have three unknown displacements in this frame: rotations C B ? ? , and the linear displacement ?. The unknown joint rotations B ? and C ? are related to joint moments by the moment equilibrium equations. Similarly, when unknown linear displacement occurs, one needs to consider force-equilibrium equations. While applying slope-deflection equation to columns in the above frame, one must consider the column rotation ? ? ? ? ? ? ? = h ? as unknowns. It is observed that in the columnAB , the end B undergoes a linear displacement with respect to end ? A. Hence the slope-deflection equation for column AB is similar to the one for beam undergoing support settlement. However, in this case is unknown. For each of the members we can write the following slope-deflection equations. ? [ AB B A F AB AB ] M M ? ? ? 3 2 2 - + + = h EI where h AB ? - = ? AB ? is assumed to be negative as the chord to the elastic curve rotates in the clockwise directions. [] AB A B F BA BA h EI M M ? ? ? 3 2 2 - + + = [] C B F BC BC h EI M M ? ? + + = 2 2 [] B C F CB CB h EI M M ? ? + + = 2 2 [] CD D C F CD CD h EI M M ? ? ? 3 2 2 - + + = h CD ? - = ? [ CD C D F DC DC h EI M M ? ? ? 3 2 2 - + + = ] (17.1) As there are three unknowns ( C B ? ? , and ?), three equations are required to evaluate them. Two equations are obtained by considering the moment equilibrium of joint B and C respectively. 0 0 = + ? = ? BC BA B M M M (17.2a) 0 0 = + ? = ? CD CB C M M M (17.2b) Now consider free body diagram of the frame as shown in Fig. 17.2. The horizontal shear force acting at A and B of the column AB is given by Page 3 Instructional Objectives After reading this chapter the student will be able to 1. Derive slope-deflection equations for the frames undergoing sidesway. 2. Analyse plane frames undergoing sidesway. 3, Draw shear force and bending moment diagrams. 4. Sketch deflected shape of the plane frame not restrained against sidesway. 17.1 Introduction In this lesson, slope-deflection equations are applied to analyse statically indeterminate frames undergoing sidesway. As stated earlier, the axial deformation of beams and columns are small and are neglected in the analysis. In the previous lesson, it was observed that sidesway in a frame will not occur if 1. They are restrained against sidesway. 2. If the frame geometry and the loading are symmetrical. In general loading will never be symmetrical. Hence one could not avoid sidesway in frames. For example, consider the frame of Fig. 17.1. In this case the frame is symmetrical but not the loading. Due to unsymmetrical loading the beam end moments and are not equal. If is greater than , then . In BC M CB M b a CB BC M M > such a case joint B andC are displaced toward right as shown in the figure by an unknown amount ? . Hence we have three unknown displacements in this frame: rotations C B ? ? , and the linear displacement ?. The unknown joint rotations B ? and C ? are related to joint moments by the moment equilibrium equations. Similarly, when unknown linear displacement occurs, one needs to consider force-equilibrium equations. While applying slope-deflection equation to columns in the above frame, one must consider the column rotation ? ? ? ? ? ? ? = h ? as unknowns. It is observed that in the columnAB , the end B undergoes a linear displacement with respect to end ? A. Hence the slope-deflection equation for column AB is similar to the one for beam undergoing support settlement. However, in this case is unknown. For each of the members we can write the following slope-deflection equations. ? [ AB B A F AB AB ] M M ? ? ? 3 2 2 - + + = h EI where h AB ? - = ? AB ? is assumed to be negative as the chord to the elastic curve rotates in the clockwise directions. [] AB A B F BA BA h EI M M ? ? ? 3 2 2 - + + = [] C B F BC BC h EI M M ? ? + + = 2 2 [] B C F CB CB h EI M M ? ? + + = 2 2 [] CD D C F CD CD h EI M M ? ? ? 3 2 2 - + + = h CD ? - = ? [ CD C D F DC DC h EI M M ? ? ? 3 2 2 - + + = ] (17.1) As there are three unknowns ( C B ? ? , and ?), three equations are required to evaluate them. Two equations are obtained by considering the moment equilibrium of joint B and C respectively. 0 0 = + ? = ? BC BA B M M M (17.2a) 0 0 = + ? = ? CD CB C M M M (17.2b) Now consider free body diagram of the frame as shown in Fig. 17.2. The horizontal shear force acting at A and B of the column AB is given by h M M H AB BA + = 1 (17.3a) Similarly for memberCD , the shear force is given by 3 H 3 CD DC MM H h + = (17.3b) Now, the required third equation is obtained by considering the equilibrium of member , BC 13 00 X FH =? + ? H= 0 = + + + h M M h M M DC CD AB BA (17.4) Substituting the values of beam end moments from equation (17.1) in equations (17.2a), (17.2b) and (17.4), we get three simultaneous equations in three unknowns C B ? ? , and , solving which joint rotations and translations are evaluated. ? Page 4 Instructional Objectives After reading this chapter the student will be able to 1. Derive slope-deflection equations for the frames undergoing sidesway. 2. Analyse plane frames undergoing sidesway. 3, Draw shear force and bending moment diagrams. 4. Sketch deflected shape of the plane frame not restrained against sidesway. 17.1 Introduction In this lesson, slope-deflection equations are applied to analyse statically indeterminate frames undergoing sidesway. As stated earlier, the axial deformation of beams and columns are small and are neglected in the analysis. In the previous lesson, it was observed that sidesway in a frame will not occur if 1. They are restrained against sidesway. 2. If the frame geometry and the loading are symmetrical. In general loading will never be symmetrical. Hence one could not avoid sidesway in frames. For example, consider the frame of Fig. 17.1. In this case the frame is symmetrical but not the loading. Due to unsymmetrical loading the beam end moments and are not equal. If is greater than , then . In BC M CB M b a CB BC M M > such a case joint B andC are displaced toward right as shown in the figure by an unknown amount ? . Hence we have three unknown displacements in this frame: rotations C B ? ? , and the linear displacement ?. The unknown joint rotations B ? and C ? are related to joint moments by the moment equilibrium equations. Similarly, when unknown linear displacement occurs, one needs to consider force-equilibrium equations. While applying slope-deflection equation to columns in the above frame, one must consider the column rotation ? ? ? ? ? ? ? = h ? as unknowns. It is observed that in the columnAB , the end B undergoes a linear displacement with respect to end ? A. Hence the slope-deflection equation for column AB is similar to the one for beam undergoing support settlement. However, in this case is unknown. For each of the members we can write the following slope-deflection equations. ? [ AB B A F AB AB ] M M ? ? ? 3 2 2 - + + = h EI where h AB ? - = ? AB ? is assumed to be negative as the chord to the elastic curve rotates in the clockwise directions. [] AB A B F BA BA h EI M M ? ? ? 3 2 2 - + + = [] C B F BC BC h EI M M ? ? + + = 2 2 [] B C F CB CB h EI M M ? ? + + = 2 2 [] CD D C F CD CD h EI M M ? ? ? 3 2 2 - + + = h CD ? - = ? [ CD C D F DC DC h EI M M ? ? ? 3 2 2 - + + = ] (17.1) As there are three unknowns ( C B ? ? , and ?), three equations are required to evaluate them. Two equations are obtained by considering the moment equilibrium of joint B and C respectively. 0 0 = + ? = ? BC BA B M M M (17.2a) 0 0 = + ? = ? CD CB C M M M (17.2b) Now consider free body diagram of the frame as shown in Fig. 17.2. The horizontal shear force acting at A and B of the column AB is given by h M M H AB BA + = 1 (17.3a) Similarly for memberCD , the shear force is given by 3 H 3 CD DC MM H h + = (17.3b) Now, the required third equation is obtained by considering the equilibrium of member , BC 13 00 X FH =? + ? H= 0 = + + + h M M h M M DC CD AB BA (17.4) Substituting the values of beam end moments from equation (17.1) in equations (17.2a), (17.2b) and (17.4), we get three simultaneous equations in three unknowns C B ? ? , and , solving which joint rotations and translations are evaluated. ? Knowing joint rotations and translations, beam end moments are calculated from slope-deflection equations. The complete procedure is explained with a few numerical examples. Example 17.1 Analyse the rigid frame as shown in Fig. 17.3a. Assume EI to be constant for all members. Draw bending moment diagram and sketch qualitative elastic curve. Solution In the given problem, joints B and rotate and also translate by an amount C ? . Hence, in this problem we have three unknown displacements (two rotations and one translation) to be evaluated. Considering the kinematically determinate structure, fixed end moments are evaluated. Thus, . 0 ; 0 ; . 10 ; . 10 ; 0 ; 0 = = - = + = = = F DC F CD F CB F BC F BA F AB M M m kN M m kN M M M (1) The ends A and are fixed. Hence, D . 0 = = D A ? ? Joints B and translate by the same amount . Hence, chord to the elastic curve C ? ' AB and ' rotates by an amount (see Fig. 17.3b) DC 3 ? - = = CD AB ? ? (2) Chords of the elastic curve ' AB and ' rotate in the clockwise direction; hence DC AB ? and CD ? are taken as negative. Page 5 Instructional Objectives After reading this chapter the student will be able to 1. Derive slope-deflection equations for the frames undergoing sidesway. 2. Analyse plane frames undergoing sidesway. 3, Draw shear force and bending moment diagrams. 4. Sketch deflected shape of the plane frame not restrained against sidesway. 17.1 Introduction In this lesson, slope-deflection equations are applied to analyse statically indeterminate frames undergoing sidesway. As stated earlier, the axial deformation of beams and columns are small and are neglected in the analysis. In the previous lesson, it was observed that sidesway in a frame will not occur if 1. They are restrained against sidesway. 2. If the frame geometry and the loading are symmetrical. In general loading will never be symmetrical. Hence one could not avoid sidesway in frames. For example, consider the frame of Fig. 17.1. In this case the frame is symmetrical but not the loading. Due to unsymmetrical loading the beam end moments and are not equal. If is greater than , then . In BC M CB M b a CB BC M M > such a case joint B andC are displaced toward right as shown in the figure by an unknown amount ? . Hence we have three unknown displacements in this frame: rotations C B ? ? , and the linear displacement ?. The unknown joint rotations B ? and C ? are related to joint moments by the moment equilibrium equations. Similarly, when unknown linear displacement occurs, one needs to consider force-equilibrium equations. While applying slope-deflection equation to columns in the above frame, one must consider the column rotation ? ? ? ? ? ? ? = h ? as unknowns. It is observed that in the columnAB , the end B undergoes a linear displacement with respect to end ? A. Hence the slope-deflection equation for column AB is similar to the one for beam undergoing support settlement. However, in this case is unknown. For each of the members we can write the following slope-deflection equations. ? [ AB B A F AB AB ] M M ? ? ? 3 2 2 - + + = h EI where h AB ? - = ? AB ? is assumed to be negative as the chord to the elastic curve rotates in the clockwise directions. [] AB A B F BA BA h EI M M ? ? ? 3 2 2 - + + = [] C B F BC BC h EI M M ? ? + + = 2 2 [] B C F CB CB h EI M M ? ? + + = 2 2 [] CD D C F CD CD h EI M M ? ? ? 3 2 2 - + + = h CD ? - = ? [ CD C D F DC DC h EI M M ? ? ? 3 2 2 - + + = ] (17.1) As there are three unknowns ( C B ? ? , and ?), three equations are required to evaluate them. Two equations are obtained by considering the moment equilibrium of joint B and C respectively. 0 0 = + ? = ? BC BA B M M M (17.2a) 0 0 = + ? = ? CD CB C M M M (17.2b) Now consider free body diagram of the frame as shown in Fig. 17.2. The horizontal shear force acting at A and B of the column AB is given by h M M H AB BA + = 1 (17.3a) Similarly for memberCD , the shear force is given by 3 H 3 CD DC MM H h + = (17.3b) Now, the required third equation is obtained by considering the equilibrium of member , BC 13 00 X FH =? + ? H= 0 = + + + h M M h M M DC CD AB BA (17.4) Substituting the values of beam end moments from equation (17.1) in equations (17.2a), (17.2b) and (17.4), we get three simultaneous equations in three unknowns C B ? ? , and , solving which joint rotations and translations are evaluated. ? Knowing joint rotations and translations, beam end moments are calculated from slope-deflection equations. The complete procedure is explained with a few numerical examples. Example 17.1 Analyse the rigid frame as shown in Fig. 17.3a. Assume EI to be constant for all members. Draw bending moment diagram and sketch qualitative elastic curve. Solution In the given problem, joints B and rotate and also translate by an amount C ? . Hence, in this problem we have three unknown displacements (two rotations and one translation) to be evaluated. Considering the kinematically determinate structure, fixed end moments are evaluated. Thus, . 0 ; 0 ; . 10 ; . 10 ; 0 ; 0 = = - = + = = = F DC F CD F CB F BC F BA F AB M M m kN M m kN M M M (1) The ends A and are fixed. Hence, D . 0 = = D A ? ? Joints B and translate by the same amount . Hence, chord to the elastic curve C ? ' AB and ' rotates by an amount (see Fig. 17.3b) DC 3 ? - = = CD AB ? ? (2) Chords of the elastic curve ' AB and ' rotate in the clockwise direction; hence DC AB ? and CD ? are taken as negative. Now, writing the slope-deflection equations for the six beam end moments, [] AB B A F AB AB EI M M ? ? ? 3 2 3 2 - + + = . 3 ; 0 ; 0 ? - = = = AB A F AB M ? ? ? + = EI EI M B AB 3 2 3 2 ? ? + = EI EI M B BA 3 2 3 4 ? C B BC EI EI M ? ? 2 1 10 + + = C B CB EI EI M ? ? + + - = 2 1 10 ? + = EI EI M C CD 3 2 3 4 ?Read More

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