Subject: Structural Analysis 1

Topic: Deflection of Beams using Energy Method

Difficulty: Medium

1. Reaction

$\begin{align} &\sum M_A=0\ (\circlearrowright+ve) \\& 12\times3+18\times4-V_B\times120 \\& \boxed{V_B=9\ kN}\ (\uparrow) \\ \\&\sum F_X=0\ (\rightarrow+ve) \\& H_A+12=0 \\& H_A=-12\ kN \\& \boxed{H_A=12\ kN}\ (\leftarrow) \\ \\& \sum F_y=0\ (\uparrow+ve) \\& V_A-18+9=0 \\& \boxed{V_A=9\ kN}\ (\uparrow) \end{align}$

2. P-Analysis:

Joint A:

$ \tan \theta=\frac{3}{4} ; \sin \theta=\frac{3}{5};\cos \theta=\frac{4}{5}$

$\begin{align} &\sum F_Y=0\ (\uparrow+ve) \\&9+P_{AC}\sin\theta=0 \\&P_{AC}=\frac{-9}{\frac{3}{5}}=\underline{-15\ kN} (C) \\ \\& \sum F_X=0\ (\rightarrow+ve) \\&-12+P_{AF}+(-15\cos\theta)=0\implies P_{AF}=24\ kN\ (T) \end{align}$

Joint C:

$ \tan \theta=\frac{4}{3} ; \cos \theta=\frac{3}{5};\sin \theta=\frac{4}{5}$

$\begin{align} &\sum F_X=0\ (\rightarrow+ve) \\& P_{CD}+12+15\sin\theta=0 \\& P_{CD}=\left(-12-15\times\frac{4}{5}\right)=-24\ kN=\underline{24\ kN\ \left(C\right)} \\ \\& \sum F_Y=0\ (\uparrow+ve) \\& 15\cos\theta-P_{CF}=0 \\& P_{CF}=15\times\frac{3}{5}=\underline{9\ kN\ (T)} \end{align}$

Joint F:

$ \cos \theta=\frac{4}{5};\sin \theta=\frac{3}{5}$

$\begin{align} &\sum F_Y=0\ (\uparrow+ve) \\& P_{FD}\sin\theta+9-18=0 \\& P_{FD}=\frac{18-9}{\frac{3}{5}}=\underline{15\ kN\ (T)} \\ \\&\sum F_X=0\ (\rightarrow+ve) \\& P_{FE}-24+P_{FD}\cos\theta=0 \\& P_{FE}=\left( 24-15\times\frac{4}{5} \right)=\underline{12\ kN\ (T)} \end{align}$

Joint E:

$obviously,\ P_{ED}=0;\ P_{EB}=\underline{12\ kN\ (T)}$

Joint B:

$ \tan \theta=\frac{3}{4} ; \sin \theta=\frac{3}{5};\cos \theta=\frac{4}{5}$

$\begin{align} &\sum F_Y=0\ (\uparrow+ve) \\& P_{BD}\sin\theta+9=0 \\& \therefore P_{BD}=\frac{-9}{\frac{3}{5}}=-15\ kN=\underline{15\ kN\ (C)} \end{align}$

3. K-Analysis:

$\begin{align} &\sum M_A=0\ (\circlearrowright+ve) \\&-1\times8+V_B\times12=0 \\& \boxed{V_B=0.67\ kN} \\ \\& \sum F_Y=9\ (\uparrow+ve) \\& V_A-1+0.67=0 \\& \boxed{V_A=0.33\ kN} \end{align}$

Joint A:

$\begin{align} &\sum F_Y=0\ (\uparrow+ve) \\& K_{AC}\sin\theta+0.33=0 \\& K_{AC}=\frac{-0.33}{\frac{3}{5}}=-0.55\ kN=\underline{0.55\ kN\ (C)} \\ \\& \sum F_X=0\ (\rightarrow+ve) \\& K_{AF}=-0.55\cos\theta=0 \\& K_{AF}=0.55\times\frac{4}{5}=\underline{0.44\ kN\ (T)} \end{align}$

Joint C:

$ \tan \theta=\frac{4}{3} ; \cos \theta=\frac{3}{5};\sin \theta=\frac{4}{5}$

$\begin{align} &\sum F_Y=0\ (\uparrow+ve) \\& -K_{CF}+0.55\cos\theta=0 \\& K_{CF}=\left(0.55\times\frac{3}{5}\right)=0.33\ kN=\underline{0.33\ kN\ (T)} \\ \\& \sum F_X=0\ (\rightarrow+ve) \\& K_{CD}+0.55\sin\theta=0 \\& K_{CD}=\left(-0.55\times\frac{4}{5}\right)=-0.44\ kN=\underline{0.44\ kN\ (C)} \end{align}$

Joint F:

$ \tan \theta=\frac{3}{4} ; \sin \theta=\frac{3}{5};\cos \theta=\frac{4}{5}$

$\begin{align} &\therefore \sum F_Y=0\ (\uparrow+ve) \\& K_{FD}+0.33\sin\theta=0 \\& K_{FD}=\left(-0.33\times\frac{5}{3}\right)=-0.55\ kN=\underline{0.55\ kN\ (C)} \\ \\& \sum F_X=0\ (\rightarrow+ve) \\& K_{FE}-0.55\cos\theta=0 \\&\therefore K_{FE}=\left(0.55\times\frac{4}{3}\right)+0.44=(0.44+0.44)=\underline{0.88\ kN\ (T)} \end{align}$

Joint E:

Joint B:

$ \tan \theta=\frac{3}{4} ; \sin \theta=\frac{3}{5};\cos \theta=\frac{4}{5}$

$\begin{align} &\sum F_Y=0\ (\uparrow+ve) \\& K_{BD}\sin\theta+0.67=0 \\& K_{BD}=\left(-0.67\times\frac{5}{3}\right)=-1.1166\ kN=\underline{1.1166\ kN\ (C)} \end{align}$

$\boxed{\therefore \Delta E_V=\frac{261.615}{AE}}$