written 2.5 years ago by
Chandan15
• 300
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•
modified 2.5 years ago
|
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$
$\Rightarrow R_{A}+R_{D}=180 \mathrm{kN}$
$-\left(R_{D} \times 6\right)+(120 \times 1.5)+(60 \times 4.5)=0$
- $R_{D}=75 \mathrm{KN}$
- $R_{A}=105 \mathrm{KN}$
- $F_{A B}=R_{A} \operatorname{cosec} 60^{\circ}=105 \times
\frac{2}{\sqrt{3}}=121.24 \mathrm{kN}$ (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
$\Rightarrow \quad 2 F_{B E} \cos 60+43.30-60.62=0$
$\Rightarrow \quad F_{BE}=17.32 \mathrm{KN}$ (C)
- $\therefore \quad F_{C E}=17.32 \mathrm{kN}(T)$,
Joint B
$$
\begin{array}{c}
\Sigma H=0 \\
F_{B A} \cos 60+F_{B E} \cos 60+F_{B C}=0\\
F_{B C}=51.95 \mathrm{kN} \text { (C) }
\end{array}
$$
$\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