Subject: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

As a support is fixed support. Therefore we have to use an imaginary span near the fixed end as shown in fig

1) Free bending moment

For Span AB BM at C=$\frac{wL^{2}}{8}=\frac{5\times(4)^{2}}{8}=10kN.m$

2) Free Bending Moment diagram

3) Applying theorem of three moments or elapeyron's theorem

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

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

As $A^{1}$ is an imaginary Span

${M_{A}}^{1}=0 \ \ \ \ l_{1}=0$

$0+2M_{A}(0+4)+0+0+\frac{6\times26.67\times2}{4}=0$

$8M_{A}=80.01$

$M_{A}=\frac{80.01}{8}=-10kN.m$

$M_{A}=-10kN.m$

4) Find Reaction at Support

$\sum m_{B}=0$

$-R_{B}\times4+5\times4\times\frac{4}{2}-10=0$

$R_{B}=7.5kN \ \ \ \sum F_{y}=0(+ve)$

$-(5\times4)+7.5+R_{B}=0$

$\underline{R_{A}=12.5kN}$

5)B.M.D 6) S.F.D