Calculate the vertical & horizontal displacements at point C in the simple frame with uniform flexural rigidity as shown in fig. Use unit load method. Also draw the deflected profile of the frame.

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written 6.8 years ago by

teamques10 ★ 68k

• modified 6.6 years ago

1. Reaction

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2. For horizontal deflection at C

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3. For vertical deflection at C

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Region	origin	limits	Mu	Mu Hor.	Mu Vert.
AB	A	0-4	$$8x-\frac{2x^2}{2}-20$$	$x-4$	$-2$
BC	C	0-2	$-2x$	$-x$	$-x$

$\begin{align} &\Delta H_C= \int_0^4 (8x-x^2-20)(x-4)\frac{dx}{EI}+\int_0^2 (-2x)(0)\frac{dx}{EI} \\&\phantom{\Delta H_C}=\frac{1}{EI} \int_0^4 \left[8x^2-32x-x^3+4x^2-20x+80\right]dx \\&\phantom{\Delta H_C}=\frac{1}{EI} \int_0^4 \left[-x^3+12x^2-\frac{52x^2}{2}+80\right]dx \\& \Delta H_C=\frac{1}{EI} \left[ \frac{-x^4}{4}+\frac{12x^3}{3}-\frac{52x^2}{2}+80x \right]_0^4 \\& \boxed{ \Delta H_C=\frac{96}{EI}}\ (\rightarrow) \\ \\& \Delta V_C= \int_0^4 (8x-x^2-20)(-2) \frac{dx}{EI}+ \int_0^2 (-2x)(-x) \frac{dx}{EI} \\& \phantom{\Delta V_C}=\frac{1}{EI} \left[ \int_0^4 (-16x+2x^2+40)dx+\int_0^2 2x^2 dx \right] \\& \phantom{\Delta V_C}=\frac{1}{EI} \left[-8x^2-\frac{2x^3}{3}+40x \right]_0^4+\left[\frac{2x^3}{3} \right]_0^2 \\& \boxed{ \Delta V_C=\frac{80}{EI}}\ (\downarrow) \end{align}$

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