Develop the flexibility matrix from the frame A B C D from the co-ordinate shown:

1views

written 6.3 years ago by

teamques10 ★ 68k

enter image description here

M1 Analysis

enter image description here

M2 Analysis

enter image description here

M3 Analysis

2) Table:- [+ve -ve]

Past	Origin	Limit	M.I	$M_1 \| M_2 \| M_3$
DC	D	0-5	I	1x	0	-1
CB	C	0-10	4I	1*5	1x	-1
BA	B	0-10	4I	-1(x-5)	1*10	-1

3) Flexibility coefficient:-

$f_{11} = $$\int_{0}^{l} \frac{(M_1)^2}{EI} dx$ =$$\int_{0}^{5} \frac{(1x)^2}{EI} dx$$ + $$\int_{0}^{10} \frac{(5)^2}{4EI} dx$$ + $$\int_{0}^{10} \frac{-(x-5)^2}{4EI} dx$$

=41.67 + 62.5 + 20.83

=$\frac{125}{EI}$

$f_{11} = \frac{125}{EI}$

$f_{22} = $$\int_{0}^{l} \frac{(M_2)^2}{EI} dx$ =$$\int_{0}^{10} \frac{(1x)^2}{4EI} dx$$ + $$\int_{0}^{10} \frac{(10)^2}{4EI} dx$$

=83.33 + 250

$f_{22} = \frac{333.33}{EI}$

$f_{33} = $$\int_{0}^{l} \frac{(M_3)^2}{EI} dx$ =$$\int_{0}^{5} \frac{(-1)^2}{EI} dx$$ + $$\int_{0}^{10} \frac{(-1)^2}{4EI} dx$$ + $$\int_{0}^{10} \frac{(-1)^2}{4EI} dx$$

$f_{33} = \frac{10}{EI}$

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

= $\int_{0}^{10} \frac{(5)(1x)}{4EI}dx + \int_{0}^{10} \frac{-(x-5)(10)}{4EI}$dx

=$\frac{62.5}{EI}$ + 0

$f_{12} = f_{21} = \frac{62.5}{EI}$

$f_{31} = f_{13} = \int_{0}^{l} \frac{M_3.M_1}{EI}$dx

=$$\int_{0}^{5} \frac{(-1)(1x)}{EI} dx$$ + $$\int_{0}^{10} \frac{(-1)(5)}{4EI} dx$$ + $$\int_{0}^{10} \frac{(-1)(-(x-5))}{4EI} dx$$

=-12.5 + (-12.5) + 0

$f_{31} = f_{13} = \frac{-25}{EI}$

$f_{32} = f_{23} = \int{0}^{l} \frac{M_3.M_2}{EI}$dx

= $\int{0}^{10} \frac{(-1)(1x)}{4EI}dx + \int{0}^{10} \frac{(-1)(10)}{4EI}$dx

$f_{32} = f_{23} = \frac{-37.5}{EI}$

ADD COMMENT EDIT