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Analyse the contineous beam by using three moment theorem

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Subject: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

1 Answer
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1)Free Bending moment

  • For Span BC at $F=\frac{wl^{2}}{8}=\frac{2\times(6)^{2}}{8}=9kN.m$

  • For Span CD at $G=\frac{wl^{2}}{8}=\frac{2\times(8)^{2}}{8}=16kN.m$

2) Free Bending moment diagram

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3) Find Reaction at Support:-

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Reaction at $B=4+4.94=8.94kN$

Reaction at $C=7.06+9.16=16.23kN$

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