m2l13

m2l13 - Module 2 Analysis of Statically Indeterminate...

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Module 2 Analysis of Statically Indeterminate Structures by the Matrix Force Method Version 2 CE IIT, Kharagpur
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Lesson 13 The Three-Moment Equations-Ii Version 2 CE IIT, Kharagpur
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Instructional Objectives After reading this chapter the student will be able to 1. Derive three-moment equations for a continuous beam with yielding supports. 2. Write compatibility equations of a continuous beam in terms of three moments. 3. Compute reactions in statically indeterminate beams using three-moment equations. 4. Analyse continuous beams having different moments of inertia in different spans and undergoing support settlements using three-moment equations. 13.1 Introduction In the last lesson, three-moment equations were developed for continuous beams with unyielding supports. As discussed earlier, the support may settle by unequal amount during the lifetime of the structure. Such future unequal settlement induces extra stresses in statically indeterminate beams. Hence, one needs to consider these settlements in the analysis. The three-moment equations developed in the pervious lesson could be easily extended to account for the support yielding. In the next section three-moment equations are derived considering the support settlements. In the end, few problems are solved to illustrate the method. 13.2 Derivation of Three-Moment Equation Consider a two span of a continuous beam loaded as shown in Fig.13.1. Let , and be the support moments at left, center and right supports respectively. As stated in the previous lesson, the moments are taken to be positive when they cause tension at the bottom fibers. and denote moment of inertia of left and right span respectively and and denote left and right spans respectively. Let L M C M R M L I R I L l R l C L δ , and R be the support settlements of left, centre and right supports respectively. C L , and R are taken as negative if the settlement is downwards. The tangent to the elastic curve at support makes an angle C CL θ at left support and CR at the right support as shown in Fig. 13.1. From the figure it is observed that, Version 2 CE IIT, Kharagpur
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CR CL θ = (13.1) The rotations CL β and CR due to external loads and support moments are calculated from the EI M diagram .They are (see lesson 12) L L C L L L L L L L CL EI l M EI l M l EI x A 3 6 + + = (13.2a) R R C R R R R R R R CR EI l M EI l M l EI x A 3 6 + + = (13.2b) The rotations of the chord and ' from the original position is given by ' ' C L ' R C L C L CL l δδ α = (13.3a) R C R CR l = (13.3b) From Fig. 13.1, one could write, Version 2 CE IIT, Kharagpur
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CL CL CL β α θ = (13.4a) CR CR CR = (13.4b) Thus, from equations (13.1) and (13.4), one could write, CR CR CL CL = (13.5)
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m2l13 - Module 2 Analysis of Statically Indeterminate...

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