Develop the flexibility matrix for the beam width respect to the coordinates shown in fig.

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written 7.3 years ago by

teamques10 ★ 70k

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M-3 Analysis

Table :-

Part	origin	limit	$M_1 \| M_2 \| M_3$
AB	A	0-4	-0.25x+1	-1x	-0.25x
BC	C	0-4	0	-1x	-1

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

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

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

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

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

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

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

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

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