Subject:- Structural Analysis II

Title:- Flexibility Method

Difficulty:- Hard

1) $D_{SE}$ = r-3 = 5-3 = 2

$V_C$ and $V_B$ are redundant

$V_C = R_1$

$V_B = R_2$

{Note :- In fixed beam type problem no need to find reaction always take the section at farther end}

Table :-

3) ^$R_1 = \int_{0}^{l} \frac{M.M_1}{EI}dx$ = $\int_{0}^{l} (-8(2+x)-12)(1x)dx + \int_{0}^{1} (-8(3+x)+2-10x)(1(1+x))dx + \int_{0}^{3}(-8(4+x)-12-10(1+x)-\frac{5x^2}{2})(1(2+x))dx$ =-16.67 - 69 - 986.62 [^$R_1 = \frac{-1072.29}{EI}$] ^$R_2 = \int_{0}^{l} \frac{M.M_2}{EI}dx$ =$\int_{0}^{3} (-8(4+x)-12-10(1+x)-2.5x^2)(1x)dx$ [^$R_2 = \frac{-455.625}{EI}$] *4) Flexibility Coefficient :-* $f_{11} = \int_{0}^{l} \frac{(M_1)^2}{EI}dx$ =$\int_{0}^{1} \frac{(1x)^2}{EI}dx + \int_{0}^{1} \frac{(1(1+x))^2}{EI}dx + \int_{0}^{3} 1(2+x)^2dx$ = 0.333 + 2.333 + 39 $f_{11} = \frac{41.666}{EI}$ $f_{22} = \int_{0}^{l} \frac{(M_2)^2}{EI}dx$ =$\int_{0}^{2} \frac{(1x)^2}{EI}dx$ $f_{22} = \frac{9}{EI}$ $f_{12} = f_{21} = \int_{0}^{l} \frac{M_1.M_2}{EI}dx$ =$\int_{0}^{2} \frac{(1(2+x)(1x))}{EI}dx$ [$f_{12} = f_{21} = \frac{18}{EI}$] *5) Applying Compatibility Condition:-* 0=-1072.29 + 41.76$V_C + 18V_B$ ------(1) 0= -455.62 + 18$V_C + 9V_B$ -------(2) solving 1 and 2 we get, $V_C$ = 28.40 KN $V_B$ = -6.183 KN