Subject: Structural Analysis 1

Topic: Deflection of Beams using Energy Method

Difficulty: Medium

$\begin{align} &\sum M_A=0(\circlearrowright+ve) \\& (-V_d\times4)+(15\times4\times2)(8\times5)=0 \\& \boxed{V_d=40\ kN} \\& \sum F_y=0(\uparrow+ve) \\&V_A+40=60=0 \\& \boxed{\therefore V_A=20\ kN} \\ \\& \sum M_A=0(\circlearrowright+ve) \\& -V_d\times4+(1\times2)=0 \\& \therefore 4V_d=2 \\& \boxed{V_d=0.5\ kN}(\uparrow) \\& \boxed{V_A=0.5\ kN}(\downarrow) \end{align}$

$\begin{align} E=200\times10^3 \ MPa \\& \phantom{E}=\frac{200\times10^3\times10^{-3}\ kN}{(10^{-6})m^2} \\& \underline{E=200\times10^6\ kN/m^2} \\& I=4\times10^8mm^4=4\times10^8\times\left(10^{-3}\right)^4m^4 \\& \underline{I=4\times10^{-4}m^4} \end{align}$

$\begin{align} &\therefore \Delta_{DH}=\int_0^L \frac{M_um_u}{EI} dx \\& \phantom{\therefore\Delta_{DH}}=\frac{1}{EI}\left[ \int_0^5(8x)(x)dx+\int_0^4(20x+40-7.5x^2)(5-0.5x)dx \right] \\& \phantom{\therefore \Delta_{DH}}=\frac{1}{EI}\left[ \int_0^5(8x^2)dx+\int_0^4(100x-10x+200-20x-37.5x^2+3.75x^3)dx \right] \\& \phantom{\therefore \Delta_{DH}}=\frac{1}{EI}\left[ \int_0^5(8x^2)dx+\int_0^4(70x-37.5x^2+3.75x^3+200)dx \right] \\& \phantom{\therefore \Delta_{ DH}}=\frac{1}{EI}\left[ \left(\frac{8x^3}{3}\right)_0^5+\left(\frac{70x^2}{2}-\frac{37.5x^3}{3}+\frac{3.75x^4}{4}+200x\right)_0^4 \right] \\& \therefore \Delta_{DH}=\frac{1133.33}{EI} \\& \phantom{\therefore \Delta_{DH}}=\frac{1133.33}{200\times10^6\times4\times10^{-4}} \\& \phantom{\therefore \Delta_{DH}}=0.01416m \\& \boxed{\phantom{\therefore}\Delta_{DH}=y_D=14.166\ m} \underline{Ans.} \end{align}$