written 5.8 years ago by | • modified 5.8 years ago |
Subject: Structural Analysis II
Topic: Flexibility Method
Difficulty: Medium / High
written 5.8 years ago by | • modified 5.8 years ago |
Subject: Structural Analysis II
Topic: Flexibility Method
Difficulty: Medium / High
written 5.8 years ago by | • modified 5.8 years ago |
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
$M_{A}(l_{1})+2M_{B}(l_{1}+l_{2})+M_{c}(l_{2})+\frac{5A_{1}x_{1}}{l_{1}}+\frac{6A_{2}x_{2}}{l_{2}}+\frac{6 \int_{1}E_{1}x_{1}}{l_{1}}+\frac{6\int_{2}E_{2}x_{2}}{l_{2}}=0$
$A_{2}x_{2}=\{\frac{1}{2}\times15\times2\times(\frac{2}{3}\times2)\}-\{\frac{1}{2}\times15\times2\times(2+\frac{1}{3}\times2)\}$
$=20-40$
$=-20$
$\int_{1}=\int_{B}-\int_{A}$
$=0-0$
$=0$
$\int_{2}=\int_{B} - \int_{C}$
$ …