m3l21

m3l21 - Module 3 Analysis of Statically Indeterminate...

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Version 2 CE IIT, Kharagpur Module 3 Analysis of Statically Indeterminate Structures by the Displacement Method
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Version 2 CE IIT, Kharagpur Lesson 21 The Moment- Distribution Method: Frames with Sidesway
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Instructional Objectives After reading this chapter the student will be able to 1. Extend moment-distribution method for frames undergoing sidesway. 2. Draw free-body diagrams of plane frame. 3. Analyse plane frames undergoing sidesway by the moment-distribution method. 4. Draw shear force and bending moment diagrams. 5. Sketch deflected shape of the plane frame not restrained against sidesway. 21.1 Introduction In the previous lesson, rigid frames restrained against sidesway are analyzed using moment-distribution method. It has been pointed in lesson 17, that frames which are unsymmetrical or frames which are loaded unsymmetrically usually get displaced either to the right or to the left. In other words, in such frames apart from evaluating joint rotations, one also needs to evaluate joint translations (sidesway). For example in frame shown in Fig 21.1, the loading is symmetrical but the geometry of frame is unsymmetrical and hence sidesway needs to be considered in the analysis. The number of unknowns is this case are: joint rotations B θ and C and member rotation ψ . Joint B and C get translated by the same amount as axial deformations are not considered and hence only one independent member rotation need to be considered. The procedure to analyze rigid frames undergoing lateral displacement using moment-distribution method is explained in section 21.2 using an example. Version 2 CE IIT, Kharagpur
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21.2 Procedure A special procedure is required to analyze frames with sidesway using moment- distribution method. In the first step, identify the number of independent rotations ( ψ ) in the structure. The procedure to calculate independent rotations is explained in lesson 22. For analyzing frames with sidesway, the method of superposition is used. The structure shown in Fig. 21.2a is expressed as the sum of two systems: Fig. 21.2b and Fig. 21.2c. The systems shown in figures 21.2b and 21.2c are analyzed separately and superposed to obtain the final answer. In system 21.2b, sidesway is prevented by artificial support at C . Apply all the external loads on frame shown in Fig. 21.2b. Since for the frame, sidesway is prevented, moment-distribution method as discussed in the previous lesson is applied and beam end moments are calculated. Let and be the balanced moments obtained by distributing fixed end moments due to applied loads while allowing only joint rotations ( ' ' ' ' ' , , , , CD CB BC BA AB M M M M M ' DC M B θ and C ) and preventing sidesway. Now, calculate reactions and (ref. Fig 21.3a).they are , 1 A H 1 D H Version 2 CE IIT, Kharagpur
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2 2 ' ' 1 h Pa h M M H BA AB A + + = 1 ' ' 1 h M M H DC CD D + = ( 2 1 . 1 ) again, ( 2 1 . 2 ) ) ( 1 1 D A H H P R + = Version 2 CE IIT, Kharagpur
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In Fig 21.2c apply a horizontal force in the opposite direction of F R . Now , then the superposition of beam end moments of system (b) and times (c) gives the results for the original structure. However, there is no way one could
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This note was uploaded on 06/30/2009 for the course CE 358 taught by Professor Trifunac during the Fall '07 term at USC.

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m3l21 - Module 3 Analysis of Statically Indeterminate...

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