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

$\sum M_A=0$

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

$\therefore V_d=40kN$

$\sum F_y=0$

$V_A+40-60=0$

$V_a=20kN$

$\sum M_A=0$

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

$4V_d=2$

$V_d=0.5kN(\uparrow)$

$V_a=0.5kN(\downarrow)$

$E=200\times 10^3MPa$

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

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

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

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