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Analysis the Continuous beam loaded and supported as shown in fig 1)Using clapeyron's theorem of three moments method Note that support C settles by 8mm during loading Take $EI=1600kN/M^{2}$

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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 AB, BM at $E=\frac{wl^{2}}{8}=\frac{24\times3^{2}}{8}=27kN.m$

For BC enter image description here $\ \ \ M_{L}=-7.5\times2 =-15kN$

$M_{R}=7.5\times2$

$=15kN.M$

For CD BM at G=$\frac{wab}{l}=\frac{15\times2\times3}{5}=30kN.m$

2) Free Bending moment diagram

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3)Three moment theorem

For $A^{1}-A-B$

$MA^{1}(l_{1})+2M_{A}(l_{1}+l_{2})+M_{B}(l_{2})+\frac{6A_{1}x_{1}}{l_{1}}+\frac{6A_{2}x_{2}}{l_{2}}=0$

$0+2M_{A}(0+3)+M_{B}(3)+\frac{6\times0\times0}{0}\frac{6\times54\times1.5}{3}=0$

$\underline{6M_{A}+3M_{B}=-162} \tag{1}$

For A-B-C

$M_{A}(l_{1})+2M_{B}(l_{1}+l_{2})+M_{c}(l_{2})+\frac{5A_{1}x_{1}}{l_{1}}+\frac{6A_{2}x_{2}}{l_{2}}+\frac{6 \int_{1}E_{1}x_{1}}{l_{1}}+\frac{6\int_{2}E_{2}x_{2}}{l_{2}}=0$

$A_{2}x_{2}=\{\frac{1}{2}\times15\times2\times(\frac{2}{3}\times2)\}-\{\frac{1}{2}\times15\times2\times(2+\frac{1}{3}\times2)\}$

$=20-40$

$=-20$

$\int_{1}=\int_{B}-\int_{A}$

$=0-0$

$=0$

$\int_{2}=\int_{B} - \int_{C}$

$ …

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