Using unit load method or Castigliano's second theorem for rigid jointed frame as shown in fig. Find a horizontal displacement of roller support. Take E=210$\times10^3 N/mm^2$, I=2$\times10^8 mm^4$

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written 6.9 years ago by

teamques10 ★ 68k

• modified 6.7 years ago

$\begin{align} & \sum M_A=0 \\& 10\times3+\times4\times\frac{4}{2}-V_D\times4=0 \\& \boxed{V_D=27.5\ kN}\ (\uparrow) \\ \\ &\sum F_Y=0\ (\uparrow+ve) \\& V_A-(10\times4)+27.5=0 \\&\boxed{V_A=12.5\ kN}\ (\uparrow) \\ \\& H_A=-10\ kN \\& \boxed{H_A=10\ kN} \ (\leftarrow) \end{align}$

2. enter image description here

Region	origin	limit	$M_u$	$m_u$	EI
AE	A	0-3	10x	1x	EI
EB	E	0-2	10(x+3)-10x	(1x+3)	EI
BC	C	0-4	27.5x-$\frac{10x^2}{2}$	1$\times$5	EI
CD	D	0-5	0	1x	EI

$ y_D= \int_{0}^{L} \frac{M.m}{EI}\ dx $

$\begin{align} & y_D=\frac{1}{EI}\left[ \int_0^3 (10x)(1x)\ dx+\int_0^2 (10(x+3)-10x)(1x+3)\ dx+\\ \int_0^4 (27.5x-5x^2)(1\times5)\ dx+\int_0^5 (0)(1x)\ dx \right] \\& \phantom{y_D}=\frac{1}{EI}\left[ 90+240+566.66\right] \\& y_D=\frac{896.66}{EI} \dots \left[ E=210\times10^3\ N/mm^2;\ I=2\times10^8\ mm^4 \right] \\& y_D=\frac{896.66}{210\times10^3\times 10^{-3}\times10^6\times2\times10^8\times10^{-2}} \\& \phantom{y_D}=21.34\times10^{-3}\ m \\& \boxed{y_D=21.34\ mm\ (\rightarrow)}\ \underline{Ans.} \end{align}$

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