Subject: Structural Analysis 1

Topic: Deflection of Beams using Energy Method

Difficulty: High

1. Support reaction calculation

$\begin{align} &\sum M_C=0\ (\circlearrowright +ve) \\& H_A\times2+10\times4=0 \\& H_A=-20\ kN \\& \boxed{H_A=20\ kN} \ (\leftarrow) \\ \\& \sum F_Y=0\ (\uparrow+ve) \\& V_C-10=0 \\& \boxed{V_C=10\ kN}\ \\ \\& \sum F_X=0\ (\rightarrow+ve) \\& -20+H_C=0 \\& \boxed{H_C=20\ kN} \end{align}$

Force in members due to given loads: [P-Analysis]

Joint F:

$\tan\theta=\frac{1}{2}\ \ \theta=26.56 ^\circ $

$\begin{align} &\sum F_Y=0\ (\uparrow+ve). \\& -10-P_{FD}\sin (26.56)=0 \\& P_{FD}=-22.36\ kN \\& \boxed{P_{FD}=22.36\ kN\ (C)} \\ \\& \sum F_X=0\ (\rightarrow+ve). \\& -P_{FE}+22.36\cos (26.56)=0 \\& P_{FE}=-20\ kN \\& \boxed{P_{FE}=20\ kN\ (T)} \end{align}$

Vertical deflection at F: [K-Analysis]

• Remove the given load system. Apply a vertical load of 1kN at F

$\begin{align} &\sum M_A=0\ (\circlearrowright+ve) \\& H_A\times2+1\times4=0 \\& \boxed{H_A=-2\ kN} (\leftarrow) \end{align}$

Joint F:

$\begin{align} &\theta=26.56^\circ \\& \sum F_Y=0(\uparrow+ve) \\&-1-P_{ED}\sin\theta=0 \\&P_{ED}=-2.23\ kN \\& \boxed{P_{ED}=2.23\ kN(C)} \\ \\& \sum F_X=0(\rightarrow+ve) \\&-P_{EF}+2.23\cos(26.56)=0 \\& \boxed{P_{EF}=2\ kN(T)} \end{align}$

$Vertical\ deflection\ of\ joint\ F=y_F=\frac{\sum Pkl}{AE}$

$\boxed{y_E=\frac{383}{AE}}\underline{Ans.}$