Subject:- Structural Analysis II

Title:- Flexibility Method

Difficulty:- Hard

Σ$M_A = 0$ ----(M-1)

1 - $V_B$*5 = 0

$V_B$ = 0.2KN

Σ$F_Y$ = 0

-$V_A$ + 0.2 = 0

$V_A$ = 0.2KN

Σ$M_A = 0$ -----(M-2)

12 - $V_B$5 = 0

$V_B$ = 0.4KN

Σ$F_Y$ = 0

$V_A$ - 1 + 0.4 = 0

$V_A$ = 0.6KN

$f_{11} = \frac{1}{EI} \int_{0}^{l} (M_1)^2$dx

= $\frac{1}{EI} \int_{0}^{2} (-0.2x)^2 dx + \frac{1}{EI} \int_{0}^{3}(0.2x)^2dx$

$f_{11} = \frac{0.466}{EI}$

$f_{22} = \frac{1}{EI} \int_{0}^{l} (M_2)^2$dx

= $\frac{1}{EI} \int_{0}^{2} (0.6x)^2 dx + \frac{1}{EI} \int_{0}^{3}(0.4x)^2dx$

$f_{22} = \frac{2.4}{EI}$

$f_{12} = f_{21} = \frac{1}{EI} \int_{0}^{l} M_1.M_2 dx$

=$\frac{1}{EI} [\int_{0}^{2} (-0.2x)(0.6x)dx + \int_{0}^{3} (0.2x)(0.4x) dx]$

=$\frac{-0.32}{EI} + \frac{0.72}{EI} = \frac{0.4}{EI}$

= [$f_{12} = f_{21} = \frac{0.4}{EI}$]