Subject:- Structural Analysis II

Title:- Flexibility Method

Difficulty:- Hard

Point B is act like a point load as shown in fig.

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

Treat $V_B and V_C$ as redundants,

$R_1 = V_B$

$R_2 = V_c$

*Table :-

3) ^$R_1 = \int_{0}^{l} \frac{M.M_1}{EI}dx$ =$\int_{0}^{4} \frac{(-10x^2)(1x)}{EI}dx$ ^$R_1 = \frac{-640}{EI}$ ^$R_2 = \int_{0}^{l} \frac{M.M_2}{EI}dx$ =$\int_{0}^{4} \frac{(-10x^2)(1(3+x))}{EI}dx$ ^$R_2 = \frac{-1280}{EI}$ $f_{11} = \int_{0}^{l} \frac{(M_1)^2}{EI} dx$ =$\int_{0}^{4} \frac{(1x)^2}{EI}$ [$f_{11} = \frac{21.33}{EI}$] $f_{22} = \int_{0}^{l} \frac{(M_2)^2}{EI} dx$ =$\int_{0}^{3} \frac{(1x)^2}{EI}dx + \int_{0}^{4} \frac{(1(3+x))^2}{EI}dx$ [$f_{22} = \frac{114.33}{EI}$] $f_{12} = f_{21} = \int_{0}^{l} \frac{M_1.M_2}{EI}dx$ = $\int_{0}^{4} (1x)(1(3+x))dx$ [$f_{12} = f_{21} = \frac{45.33}{EI}$] *Flexibility:-* $F_B = \frac{1}{K_B}$ $F_B = \frac{10}{EI}$ = flexibility of spring Because of the flexibility $\frac{10}{EI}$ support B will settled down by = $\frac{10}{EI}$ -10 = -640 + 21.33$V_B + 45.22 V_C$ 0 = -1280 + 45.33$V_B + 114.33 V_C$ 630 = 21.33$V_B + 45.33 V_C$ ------(1) 1280 = 45.33 $V_B + 114.33 V_C$ --------(2) solving 1 and 2 we get, $V_B = 36.48KN and V_C = -3.27KN$ ![enter image description here][3] **B.M calculation :-** B$M_C$ = 0 B.$M_B$ = -3.2*3 = -9.81 KN B.$M_A$ = -3.27*7 + 36.84*4 + -20*4*$\frac{4}{2}$

= -36.9 KN.m