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

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$\begin{align} &\left\{ \tan Q=\frac{5}{3} ;\ Q=59.03 ^{\circ} \right\} \\&\sum M_A=0\ (\circlearrowright+ve) \\& 25\times5+30\times4\times\frac{4}{2}-V_D\times7=0 \\& \boxed{V_D=52.14\ kN} \\ \\&\sum F_Y=0\ (\uparrow+ve) \\& V_A-30\times4+52.14=0 \\& \boxed{V_A=67.86\ kN} \\ \\&\sum F_Y=0\ (\rightarrow+ve) \\& H_A+25=0 \implies H_A=-25\ kN \\& \boxed{H_A=25\ kN}\ (\leftarrow) \end{align}$

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$l=\sqrt{5^2+3^2} \\ l=\underline{5.83\ m}$

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Region	origin	limit	$M_u$	$m_u$	EI
AB	A	0-5	$25x$	$1x$	EI
BC	C	0-4	$52.14(x+3)-\frac{30x^2}{2} \\=52.14x+156.42-15x^2$	1$\times$5	EI
CD	D	0-5.83	$52.14\cos\theta.x=(26.83x)$	$1\sin\theta.x=(0.86x)$	EI

$\begin{align} &y_D=\int_0^L \frac{Mm}{EI} dx \\& \phantom{y_D}=\frac{1}{EI}\left[ \int_0^5 (25x)(1x)dx \int_0^4 (52.14x+156.42-15x^2)(1\times5)dx +\int_0^{5.83} (26.83x)(0.86x)dx\right] \\& \phantom{y_D}=\frac{1}{EI}\left[1041.67+3614+1524.07\right] \\&y_D=\frac{6179.74}{EI} \\& \left \{ 1MPa=1N1mm^2\ 1Pa=1N/m^2=10^9N/m^2 \right\} \\& y_D=\frac{6179.94}{200\times4\times10^8\times10^6\times10^{-12}} \\& \phantom{y_D}=0.077m\ (\rightarrow) \\& \boxed{y_D=77mm \ (\rightarrow)} \end{align}$

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