Page 1 Instructional Objectives After reading this chapter the student will be able to 1. Solve plane frame restrained against sidesway by the moment-distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the plane frame. 20.1 Introduction In this lesson, the statically indeterminate rigid frames properly restrained against sidesway are analysed using moment-distribution method. Analysis of rigid frames by moment-distribution method is very similar to that of continuous beams described in lesson 18. As pointed out earlier, in the case of continuous beams, at a joint only two members meet, where as in case of rigid frames two or more than two members meet at a joint. At such joints (for example joint C in Fig. 20.1) where more than two members meet, the unbalanced moment at the beginning of each cycle is the algebraic sum of fixed end beam moments (in the first cycle) or the carry over moments (in the subsequent cycles) of the beam meeting at C. The unbalanced moment is distributed to members and according to their distribution factors. Few examples are solved to explain procedure. The moment-distribution method is carried out on a working diagram. CD CB, CE Page 2 Instructional Objectives After reading this chapter the student will be able to 1. Solve plane frame restrained against sidesway by the moment-distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the plane frame. 20.1 Introduction In this lesson, the statically indeterminate rigid frames properly restrained against sidesway are analysed using moment-distribution method. Analysis of rigid frames by moment-distribution method is very similar to that of continuous beams described in lesson 18. As pointed out earlier, in the case of continuous beams, at a joint only two members meet, where as in case of rigid frames two or more than two members meet at a joint. At such joints (for example joint C in Fig. 20.1) where more than two members meet, the unbalanced moment at the beginning of each cycle is the algebraic sum of fixed end beam moments (in the first cycle) or the carry over moments (in the subsequent cycles) of the beam meeting at C. The unbalanced moment is distributed to members and according to their distribution factors. Few examples are solved to explain procedure. The moment-distribution method is carried out on a working diagram. CD CB, CE Example 20.1 Calculate reactions and beam end moments for the rigid frame shown in Fig. 20.2a. Draw bending moment diagram for the frame. Assume EI to be constant for all the members. Solution In the first step, calculate fixed end moments. m kN M m kN M m kN M m kN M F CB F BC F DB F BD . 0 . 0 . 0 . 0 . 0 . 5 . 0 . 5 = = - = = (1) Also, the fixed end moment acting at B on BA is clockwise. m kN M F BA . 0 . 10 - = Page 3 Instructional Objectives After reading this chapter the student will be able to 1. Solve plane frame restrained against sidesway by the moment-distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the plane frame. 20.1 Introduction In this lesson, the statically indeterminate rigid frames properly restrained against sidesway are analysed using moment-distribution method. Analysis of rigid frames by moment-distribution method is very similar to that of continuous beams described in lesson 18. As pointed out earlier, in the case of continuous beams, at a joint only two members meet, where as in case of rigid frames two or more than two members meet at a joint. At such joints (for example joint C in Fig. 20.1) where more than two members meet, the unbalanced moment at the beginning of each cycle is the algebraic sum of fixed end beam moments (in the first cycle) or the carry over moments (in the subsequent cycles) of the beam meeting at C. The unbalanced moment is distributed to members and according to their distribution factors. Few examples are solved to explain procedure. The moment-distribution method is carried out on a working diagram. CD CB, CE Example 20.1 Calculate reactions and beam end moments for the rigid frame shown in Fig. 20.2a. Draw bending moment diagram for the frame. Assume EI to be constant for all the members. Solution In the first step, calculate fixed end moments. m kN M m kN M m kN M m kN M F CB F BC F DB F BD . 0 . 0 . 0 . 0 . 0 . 5 . 0 . 5 = = - = = (1) Also, the fixed end moment acting at B on BA is clockwise. m kN M F BA . 0 . 10 - = In the next step calculate stiffness and distribution factors. EI EI K BD 25 . 0 4 = = and EI EI K BC 25 . 0 4 = = At jointB : EI K 50 . 0 = ? 5 . 0 ; 5 . 0 5 . 0 25 . 0 = = = BC BD DF EI EI DF (2) All the calculations are shown in Fig. 20.2b. Please note that cantilever member does not have any restraining effect on the joint B from rotation. In addition its stiffness factor is zero. Hence unbalanced moment is distributed between members and BC BD only. In this problem the moment-distribution method is completed in only one cycle, as equilibrium of only one joint needs to be considered. In other words, there is only one equation that needs to be solved for the unknown B ? in this problem. This problem has already been solved by slop- deflection method wherein reactions are computed from equations of statics. The free body diagram of each member of the frame with external load and beam end moments are again reproduced here in Fig. 20.2c for easy reference. The bending moment diagram is shown in Fig. 20.2d. Page 4 Instructional Objectives After reading this chapter the student will be able to 1. Solve plane frame restrained against sidesway by the moment-distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the plane frame. 20.1 Introduction In this lesson, the statically indeterminate rigid frames properly restrained against sidesway are analysed using moment-distribution method. Analysis of rigid frames by moment-distribution method is very similar to that of continuous beams described in lesson 18. As pointed out earlier, in the case of continuous beams, at a joint only two members meet, where as in case of rigid frames two or more than two members meet at a joint. At such joints (for example joint C in Fig. 20.1) where more than two members meet, the unbalanced moment at the beginning of each cycle is the algebraic sum of fixed end beam moments (in the first cycle) or the carry over moments (in the subsequent cycles) of the beam meeting at C. The unbalanced moment is distributed to members and according to their distribution factors. Few examples are solved to explain procedure. The moment-distribution method is carried out on a working diagram. CD CB, CE Example 20.1 Calculate reactions and beam end moments for the rigid frame shown in Fig. 20.2a. Draw bending moment diagram for the frame. Assume EI to be constant for all the members. Solution In the first step, calculate fixed end moments. m kN M m kN M m kN M m kN M F CB F BC F DB F BD . 0 . 0 . 0 . 0 . 0 . 5 . 0 . 5 = = - = = (1) Also, the fixed end moment acting at B on BA is clockwise. m kN M F BA . 0 . 10 - = In the next step calculate stiffness and distribution factors. EI EI K BD 25 . 0 4 = = and EI EI K BC 25 . 0 4 = = At jointB : EI K 50 . 0 = ? 5 . 0 ; 5 . 0 5 . 0 25 . 0 = = = BC BD DF EI EI DF (2) All the calculations are shown in Fig. 20.2b. Please note that cantilever member does not have any restraining effect on the joint B from rotation. In addition its stiffness factor is zero. Hence unbalanced moment is distributed between members and BC BD only. In this problem the moment-distribution method is completed in only one cycle, as equilibrium of only one joint needs to be considered. In other words, there is only one equation that needs to be solved for the unknown B ? in this problem. This problem has already been solved by slop- deflection method wherein reactions are computed from equations of statics. The free body diagram of each member of the frame with external load and beam end moments are again reproduced here in Fig. 20.2c for easy reference. The bending moment diagram is shown in Fig. 20.2d. Page 5 Instructional Objectives After reading this chapter the student will be able to 1. Solve plane frame restrained against sidesway by the moment-distribution method. 2. Compute reactions at the supports. 3. Draw bending moment and shear force diagrams. 4. Draw the deflected shape of the plane frame. 20.1 Introduction In this lesson, the statically indeterminate rigid frames properly restrained against sidesway are analysed using moment-distribution method. Analysis of rigid frames by moment-distribution method is very similar to that of continuous beams described in lesson 18. As pointed out earlier, in the case of continuous beams, at a joint only two members meet, where as in case of rigid frames two or more than two members meet at a joint. At such joints (for example joint C in Fig. 20.1) where more than two members meet, the unbalanced moment at the beginning of each cycle is the algebraic sum of fixed end beam moments (in the first cycle) or the carry over moments (in the subsequent cycles) of the beam meeting at C. The unbalanced moment is distributed to members and according to their distribution factors. Few examples are solved to explain procedure. The moment-distribution method is carried out on a working diagram. CD CB, CE Example 20.1 Calculate reactions and beam end moments for the rigid frame shown in Fig. 20.2a. Draw bending moment diagram for the frame. Assume EI to be constant for all the members. Solution In the first step, calculate fixed end moments. m kN M m kN M m kN M m kN M F CB F BC F DB F BD . 0 . 0 . 0 . 0 . 0 . 5 . 0 . 5 = = - = = (1) Also, the fixed end moment acting at B on BA is clockwise. m kN M F BA . 0 . 10 - = In the next step calculate stiffness and distribution factors. EI EI K BD 25 . 0 4 = = and EI EI K BC 25 . 0 4 = = At jointB : EI K 50 . 0 = ? 5 . 0 ; 5 . 0 5 . 0 25 . 0 = = = BC BD DF EI EI DF (2) All the calculations are shown in Fig. 20.2b. Please note that cantilever member does not have any restraining effect on the joint B from rotation. In addition its stiffness factor is zero. Hence unbalanced moment is distributed between members and BC BD only. In this problem the moment-distribution method is completed in only one cycle, as equilibrium of only one joint needs to be considered. In other words, there is only one equation that needs to be solved for the unknown B ? in this problem. This problem has already been solved by slop- deflection method wherein reactions are computed from equations of statics. The free body diagram of each member of the frame with external load and beam end moments are again reproduced here in Fig. 20.2c for easy reference. The bending moment diagram is shown in Fig. 20.2d. Example 20.2 Analyse the rigid frame shown in Fig. 20.3a by moment-distribution method. Moment of inertia of different members are shown in the diagram. Solution: Calculate fixed end moments by locking the joints and D C B A , , , E 2 54 4.0 kN.m 20 F AB M × == kN.m 667 . 2 - = F BA M kN.m 5 . 7 = F BC M kN.m 5 . 7 - = F CB M 0 = = = = F EC F CE F DB F BD M M M M (1) The frame is restrained against sidesway. In the next step calculate stiffness and distribution factors. EI K BA 25 . 0 = and EI EI K BC 333 . 0 6 2 = =Read More

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