Subject: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

Free Bending Moment:

For Span AB BM at $E=\frac{wab}{l}=\frac{80\times2\times4}{6}=106.67KN.m$

For Span BC BM at $F=\frac{wl^{2}}{8}=\frac{20\times5^{2}}{8}=62.5kN.m$

For Span CD BM at $G=\frac{wab}{l}=\frac{60\times3\times2}{5}=72kN.m$

Support B sinks by 5mm =0.005m=-0.005(Downward)

$E=2.1\times10^{5}N/mm^{2}$

=$2.1\times10^{5}\frac{N}{(10^{-3})^{2}m^{2}}$

$=2.1\times10^{5}\times10^{6}N/m^{2}$

$I=9300cm^{4}$

$=9300\times{(10^{-2}})^{4}m^{4}$

$9300\times10^{-8}m^{4}$

$E\times I=2.1\times{10^{5}}\times{10}^{6}\times9300\times10^{-8}$

$(N/m^{2}) \times (m^{4})$

$=19830\times10^{3}N/m^{2}$

$=19530kN/m^{2}$

2) Free bending moment diagram

3)Applying Three moment theorem

For A-B-C

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

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

$-0.005-0$

$\int_{1}=-0.005$

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

$=-0.05-0$

$\int_{2}=-0.005$

$0+2MB(6+5)+M_{c}5+\frac{6\times320.01\times2.67}{6}+\frac{6\times208.33\times2.5}{5}+\frac{6\times19530\times(-0.005)}{6}+\frac{6\times19530\times(-0.005)}{5}=0$

$22M_{B}+5M_{c}+1479.42-214.83=0$

$\underline{22M_{B}+rM_{c}=-1264.59}\tag{1}$

For B-C-D

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

$\int_{1}=y_{c}-y_{B}$

$=0-(-0.005)$

$\int_{1}=0.005$

$\int_{2}=y_{C}-y_{D}$

=0-0

=0

$\underline{\int_{2}=0}$

$5M_{B}+2M_{c}(5+5)+0+\frac{6\times208.33\times2.5}{5}+\frac{6\times180\times2.67}{5}+\frac{6\times0.005\times19530}{5}+0=0$

$5M_{B}+20M_{c}+624.99+576.72+117.18=0$

$\underline{5M_{B}+20M_{C}=-1318.89}$\tag{2}

solving 1 & 2 we get

$M_{B}=-45kN.m$

$M_{C}=-54.68kN.m$

Reaction at $A=\underline{45.83kN}$

Reaction at $B=34.17+48.2=\underline{82.37kN}$

Reaction at $C=51.8+46.8=\underline{98.6}kN$

Reaction at $D=\underline{13.2kN}$