Subject: Structural Analysis II

Topic: Flexibility Method

Difficulty: Medium / High

1)Free Bending moment

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

For BC $\ \ \ M_{L}=-7.5\times2 =-15kN$

$M_{R}=7.5\times2$

$=15kN.M$

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

2) Free Bending moment diagram

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

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

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

$=20-40$

$=-20$

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

$=0-0$

$=0$

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

$=0-(-0.08)$

$=0.008$

$3(M_{A})+2M_{B}(3+4)+M_{c}(4)=\frac{6\times54\times1.5}{3}+\frac{6\times(-20)}{4}+0+\frac{6\times0.008\times1600}{4}=0$

$3M_{A}+14M_{B}+4M_{c}+162+(-30)+0+19.2=0$

$\underline{3M_{A}+14M_{B}+4M_{c}=-151.2}\tag{2}$

For , B-C-D

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

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

$=-20+40$ =20

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

$=-0.008-0$

$\int_{1}=-0.008$

$int_{2}=int_{c}-\int_{D}$

$=-0.008-0$

$\int_{2}=-0.008$

$4M_{B}+18M_{c}+5M_{0}+\frac{6\times20}{4}+\frac{6\times75\times2.67}{5}+\frac{6\times(-0.008)1600}{4}+\frac{6\times(-0.008)\times1600}{5}=0$

$4M_{B}+18M_{c}+5M_{D}+60+240.3+(-19.2)+(-15.36)=0$ $4M_{B}+18M_{c}+5M_{D}+300.3-19.2-15.36=0$ $4M_{B}+18M_{c}+5M_{D}+265.74=0$

$\underline{4M_{B}+18M_{c}+5M_{D}=-265.74}\tag{3}$

For, $C-D-D^\prime$

$\{l_{1}=5 \ l_{2}=0\}$

$M_{c}(l_{1})+2M_{D}(l_{1}+l_{2})+M_{D}^\prime(l_{2})+gA_{1}x_{1}+6A_{2}x_{2}+\frac{6\times0.008\times1600}{5}+0$

$5M_{c}+10m_{D}+209.7+15.36=0$

$\underline{5m_{c}+10m_{D}=-225.0}\tag{4}$

From equation $(1)$ we get

$6m_{A}+3M_{B}=-162$

$6M_{A}=-162-3m_{B}$

$M_{A}=\frac{-162-3M_{B}}{6}$

=$\frac{-162}{6}-\frac{3}{6}M_{B}$

$\underline{M_{A}=-27-0.5M_{B}}\tag{5}$

Putting the value of $M_{A}$ in equation $(2)$

$3(-27-0.5M_{B})+14M_{B}+4M_{c}=-151.25$

$-81-1.5m_{B}+14M_{B}+4M_{C}=-151.2$

$-81+12.5M_{B}+4M_{c}=-151.2$

$12.5M_{B}+4M_{c}=-151.2+81$

$\underline{12.5M_{B}+4M_{c}=-70.2}\tag{6}$

Solving 3,4 & 6 we get

$\underline{M_{B}=-26.7kN.m}$

$\underline{M_{c}=-9.19kN.m}$

$\underline{M_{D}=17.90kN.m}$

$M_{A}=-27-0.5M_{B}$

$=-27-0.5(-2.67)$

$\underline{25.67kN.m}$