# Module 3. Analysis of Statically Indeterminate Structures by the Displacement Method. Version 2 CE IIT, Kharagpur

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1 odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod

2 Lesson 20 The oment- Distribution ethod: rames without Sidesway

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 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 ig. 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,cd and CE according to their distribution factors. ew examples are solved to explain procedure. The moment-distribution method is carried out on a working diagram.

4 Example 20.1 Calculate reactions and beam end moments for the rigid frame shown in ig. 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. = 5.0 kn. m DB = 5.0 kn. m = 0.0 kn. m (1) = 0.0 kn. m Also, the fixed end moment acting at B on is clockwise. = 10.0 kn. m

5 In the next step calculate stiffness and distribution factors. At joint B : EI K = 25EI 4 = 0. and K = EI 25EI 4 = 0. K = 0. 50EI 0.25EI D = = 0.5 ; D = 0.5 (2) 0.5EI All the calculations are shown in ig. 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 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 θ 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 ig. 20.2c for easy reference. The bending moment diagram is shown in ig. 20.2d.

6

7 Example 20.2 Analyse the rigid frame shown in ig. 20.3a by moment-distribution method. oment of inertia of different members are shown in the diagram. Solution: Calculate fixed end moments by locking the joints AB = = 4.0 kn.m 20 A, B, C, D and E = kn.m = 7.5 kn.m = 7.5 kn.m = = = = 0 (1) DB CE The frame is restrained against sidesway. In the next step calculate stiffness and distribution factors. 2EI K = 0. 25EI and K = = EI 6 EC

8 K 3 EI = = EI ; K CE = 0. EI (2) At joint B : D K = K + K = EI + K = ; D = D = (3) At joint C : K = EI D = CD ; D = In ig. 20.3b, the complete procedure is shown on a working diagram. The moment-distribution method is started from joint C. When joint C is unlocked, it will rotate under the action of unbalanced moment of 7.5 kn.m. Hence the 7.5 kn.m is distributed among members and CE according to their distribution factors. Now joint C is balanced. To indicate that the joint C is balanced a horizontal line is drawn. This balancing moment in turn developed moments kn.m at and kn.m at EC. Now unlock joint B. The joint B is unbalanced and the unbalanced moment is ( ) = kn.m. This moment is distributed among three members meeting at B in proportion to their distribution factors. Also there is no carry over to joint D from beam end moment as it was left unlocked. or member, modified stiffness factor is used as the end D is hinged. Example 20.3 Analyse the rigid frame shown in ig. 20.4a by moment-distribution method. Draw bending moment diagram for the rigid frame. The flexural rigidities of the members are shown in the figure.

9 Solution: Assuming that the joints are locked, calculate fixed end moments. = kn.m ; = kn.m AB = kn.m; = kn.m = kn.m ; = kn.m CD = 0.0 kn.m ; = 0.0 kn.m BE DC EB = 5.0 kn.m ; = 5.0 kn.m (1) C C The frame is restrained against sidesway. Calculate stiffness and distribution factors. K = 0.5EI ; K = 0.333EI ; K BE = EI 3 2EI K = 0.333EI ; K CD = 0.5EI ; K C = = EI 4 4 K = 0.5EI ; K = 0. EI (2) DC DG 5 Joint B : K = 0.5EI EI EI = EI

10 D D BE ; D = = = Joint C : K = 0.333EI + 0.5EI EI = EI D D C = CD ; D = = 0.31 Joint D : K = 1. 0EI D 0.50 ; D = 0.50 (3) DC = DG

11 The complete moment-distribution method is shown in ig. 20.4b. The momentdistribution is stopped after three cycles. The moment-distribution is started by releasing and balancing joint D. This is repeated for joints C and B respectively in that order. After balancing joint, it is left unlocked throughout as it is a hinged joint. After balancing each joint a horizontal line is drawn to indicate that joint has been balanced and locked. When moment-distribution method is finally stopped all joints except fixed joints will be left unlocked. Summary In this lesson plane frames which are restrained against sidesway are analysed by the moment-distribution method. As many equilibrium equations are written as there are unknown displacements. The reactions of the frames are computed from equations of static equilibrium. The bending moment diagram is drawn for the frame. A few problems are solved to illustrate the procedure. ree-body diagrams are drawn wherever required.

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