Using unit load method or castigliano's theorem for rigid jointed frame shown in fig.

Calculate a horizontal displacement of roller support at D.

Take $E=GPa, I=4\times 10^8mm^4$

Solution:

$\sum M_A=$

$25\times 5+30\times 4\times \frac{4}{2}-V_D\times 7=0$

$V_D= 52.14 kN$

$\sum F_y=0$

$V_A-30\times 30\times 4+52.14=0$

$V_A=67.86kN$

$\sum F_y=0(\rightarrow +ve)$

$H_A+25=0$

$H_A=-25kN$

$H_A=25kN(\leftarrow)$

$y_D=\int _0^L\frac{Mm}{EI}dx$

$\frac{1}{EI}[\int _0^5(25x)(1x)dx \int _0^4(52.14x+156.42-15x^2)(1\times 5)dx+\int _0^{5.83}(26.83x)(0.86x)dx]$

$\frac{1}{EI}[1041.67+3614+1524.07]$

$y_D=\frac{6179.74}{EI}$

$1MPa=1N/mm^2$

$1Pa=1N/m^2=10^9N/m^2$

$y_D=\frac{6179.74}{200\times 4\times 10^8\times 10^6\times 10^{-12}}$

$y_D=0.077m(\rightarrow)$

$y_D=77mm(\rightarrow)$