Using unit load method (virtual-work method) for rigid jointed frame shown in fig. calculate a horizontal displacement ($\Delta D

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teamques10 ★ 68k

• modified 6.6 years ago

1. Reaction

$\begin{align} &\sum M_A=0\ (\circlearrowright+ve) \\& 10\times3\times\frac{3}{2}+15\times3-V_D\times5=0 \\& \boxed{V_D=18\ kN\ (\uparrow)} \\ \\& \sum F_Y=0\ (\uparrow+ve) \\&V_A-15+18=0 \\& V_A=-3\ kN \\& \boxed{V_A=-3\ kN\ (\downarrow)} \\ \\& \sum F_X=0\ (\rightarrow+ve) \\&H_A=10\times3=0 \\&H_A=-30\ kN \\& \boxed{H_A=-30\ kN\ (\leftarrow)} \end{align}$

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2. Table: [$\circlearrowright|\circlearrowleft+ve$]

Region	origin	Limit	Mu	Mu(H)	EI
AB	A	0-3	$30x-10\times x\times\frac{x}{2}=30x-5x^2$	1.x	EI
BE	B	0-3	$30\times3-10\times\frac{3^2}{2}-3x$	1$\times$3	EI
EC	C	0-2	18x	1$\times$2	EI
CD	D	0-2	0	1x	EI

$ \left\{\Delta D_H= \int_{0}^{l}\frac{Mu.Mu(H)}{EI} dx \right\} \ \underline{formula}\\ \Delta D_H=\frac{1}{EI}[ \int_{0}^{3}(30x-5x^3).x\ dx+\int_{0}^{3}(45-3x).3\ dx+\int_{0}^{3}(18x).2\ dx] \\ \Delta D_H= \frac{1}{EI}\left[\left( \frac{30x^3}{3}-\frac{5x^4}{4}\right)_0^3+\left(135x-\frac{9x^2}{2}\right)_0^3+\left(36\frac{x^2}{2}\right)_0^2\right]\\=\frac{605.25}{EI} \ [EI \ constant] \\ \boxed{\Delta D_H=\frac{605.25}{EI}} \ \underline{Ans.}$

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