Find the vertical and horizontal deflections of the joint E of the truss shown in figure. Area of each horizontal member = 1200 mm^2; Area of other members = 1500 mm². Modulus of elasticity= 200KN/mm^2

Given Data:-

• A_horizontal $=1200 \mathrm{~mm}^{2}$
• $A_{\text {other member }}=1500 \mathrm{~mm}^{2}$
• Modulus of Elasticity, $E=200 \mathrm{~kN} / \mathrm{mm}^{2}$
• $\sum V=0$

$\Rightarrow R_{A}+R_{D}=180 \mathrm{kN}$

• $\sum M_{A}=0$

$-\left(R_{D} \times 6\right)+(120 \times 1.5)+(60 \times 4.5)=0$

• $R_{D}=75 \mathrm{KN}$
• $R_{A}=105 \mathrm{KN}$
• $F_{A B}=R_{A} \operatorname{cosec} 60^{\circ}=105 \times \frac{2}{\sqrt{3}}=121.24 \mathrm{kN}$ (C)
• $F_{A E}=R_{A} \cot 60^{\circ}=105 \times \frac{1}{\sqrt{3}}=60.62 \mathrm{KN}(T)$
• $F_{C D}=R_{D} \operatorname{cosec} 60^{\circ}=75 \times \frac{2}{\sqrt{3}}=86.60 \mathrm{KN}(C)$
• $F_{E D}=R_{D} \cot 60^{\circ}=75 \times \frac{1}{\sqrt{3}}=43.30 \mathrm{KN}(T)$

Joint E

• $F_{B E} \sin 60+F_{E} \sin 60^{\circ}=0$

• $F_{B E}=-F_{E C}$

• $F_{B E} \cos 60-F_{C E} \cos 60+F_{E D}-F_{A E}=0$

$\Rightarrow \quad 2 F_{B E} \cos 60+43.30-60.62=0$ $\Rightarrow \quad F_{BE}=17.32 \mathrm{KN}$ (C)

• $\therefore \quad F_{C E}=17.32 \mathrm{kN}(T)$,

Joint B

• K-Value(Vertical)

• K-Value (Horizontal)

$\begin{equation} \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text { Member } & P(K N) & K_{V} & K_{H} & L & A E×10^7 & \delta V=\frac{P K_{V }L}{A E} & \delta H=\frac{P K_{H }L}{A E} \\ \hline A B & -121.24 & -0.583 & 0 & 3 & 30 & 0.706 & 0 \\ B C & -51.95 & -0.6 & 0 & 3 & 24 & 0.390 & 0 \\ C D & -86.60 & -0.583 & 0 & 3 & 30 & 0.505 & 0 \\ A E & 60.62 & 0.3 & 1 & 3 & 24 & 0.227 & 0.7 \\ E D & 43.30 & 0.3 & 0 & 3 & 24 & 0.162 & 0 \\ B E & -17.32 & 0.583 & 0 & 3 & 30 & -0.1 & 0 \\ C E & 17.32 & 0.583 & 0 & 3 & 30 & +0.1 & 0 \\ \hline \end{array} \end{equation}$ Hence,

• Vertical Deflection = 1.99mm
• Horizontal Deflection = 0.76mm