Subject:- Structural Analysis II

Title:- Flexibility Method

Difficulty:- Hard

M-3 Analysis

Table :-

2) Flexibility Coefficient:-

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

= $\int_{0}^{4} \frac{(-0.25x+1)^2}{EI}dx = \frac{1.33}{EI}$

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

= $\frac{1}{EI} \int_{0}^{4} (-1x)^2dx + \frac{1}{EI} \int_{0}^{4} (-1x)^2dx = \frac{42.67}{EI}$

$f_{33} = \int_{0}^{l} \frac{(M_3)^2}{EI}$dx

= $\frac{1}{EI} \int_{0}^{4} (-0.25x)^2dx + \frac{1}{EI} \int_{0}^{4} (-1)^2dx = \frac{5.33}{EI}$

$f_{12} = f_{21} = \int_{0}^{4} \frac{(-0.25x+1)(-1x)}{EI}dx$

$f_{12} = f_{21} = \frac{-2.67}{EI}$

$f_{13} = f_{31} = \int_{0}^{4} \frac{(-0.25x+1)(-0.25x)}{EI}dx$

$f_{13} = f_{31} = \frac{-0.67}{EI}$

$f_{23} = f_{32} = \int_{0}^{4} \frac{(-0.25x)(-1x)}{EI}dx + \int_{0}^{4} \frac{(-1x)(-1)}{EI}$

$f_{23} = f_{32} = \frac{13.33}{EI}$