Subject: Structural Analysis 1

Topic: Deflection of Beams using Energy Method

Difficulty: Medium

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

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