Using virtual work method, for rigid jointed frame as shown in fig. Find horizontal displacement of roller. Take $E=200\times 10^3MPa,I=4\times 10^6mm^4$

$\sum M_A=0$

$(-V_d\times 4)+(15\times 4\times 2)(8\times 5)=0$

$\therefore V_d=40kN$

$\sum F_y=0$

$V_A+40-60=0$

$V_a=20kN$

$\sum M_A=0$

$-V_d\times+(1\times 2)=0$

$4V_d=2$

$V_d=0.5kN(\uparrow)$

$V_a=0.5kN(\downarrow)$

$E=200\times 10^3MPa$

$=\frac{200\times 10^3\times 10^{-3}kN}{(10^{-6})m^2}$

$E=200\times 10^6kN/m^2$

$I=4\times 10^8mm^4=4\times 10^8\times (10^{-3})^4m^4$

$I=4\times 10^{-4}m^4$

$\therefore \Delta _{DH}=\int _0^L \frac{M_u.m_u}{EI}dx$

$=\frac{1}{EI}[\int _0^5(8x)(x)dx+ \int _0^4(20x+40-7.5x^2)(5-0.5x)dx]$

$=\frac{1}{EI}=[\int _0^5 8x^2dx+\int +0^4 (100x-10x+200-20x-37.5x^2+3.75x^3dx)]$

$=\frac{1}{EI}[\int _0^5 8x^2dx + \int _0^4 (70x-37.5x^2+3.75x^3+200)dx]$

$=\frac{1}{EI}[(\frac{8x^3}{3})_0^5+(\frac{70x^2}{2}-\frac{37.5x^3}{3}+\frac{3.75x^4}{4}+200x)_0^4]$

$\therefore \Delta _{DH}=\frac{1133.33}{EI}$

$=\frac{1133.33}{200\times 10^6\times 4\times 10^-4}$

$=0.01416 m$

$\Delta _{DH}=y_D=14.166mm$