Find the vertical deflecttions of the Joint C of the loaded truss shown in fig. The cls area of members CD & DE are each $250mm^2$ & those of the other members are each $1250mm^2.$Take$E=200\ kN/mm^2$

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written 6.8 years ago by

teamques10 ★ 68k

• modified 6.6 years ago

Analysis of truss due to given loading:

We have,

$AC=2\cos(30^\circ)=2\sqrt3\ m \\ \sum M_A=0\ gives(\circlearrowright+ve) \\ H_e\times2=40\times2\sqrt3 \\ \therefore \boxed{H_e=40\sqrt3\ kN\rightarrow} \\ \boxed{V_A=40\ kN\uparrow} \\ \boxed{H_a=40\sqrt3\ kN\rightarrow}$

P-Analysis

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1. Joint C:

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$\begin{align} &\sum F_Y=0(\uparrow+ve) \\& -40-P_{CD}\sin30=0 \\& P_{CD}=-80\ kN \\&\boxed{ P_{CD}= 80\ kN\ (C)} \\ \\&\sum F_X=0(\rightarrow+ve) \\& -P_{CD}\cos30-P_{CB}=0 \\& -(-80\cos30)-P_{CB}=-0 \\&\boxed{ P_{CB}= 40\sqrt3\ kN\ (T)} \end{align}$

2. Joint B:

$P_{BD}=0;\ P_{BA}=40\sqrt3\ kN\ (T)$

3. Joint D:

$P_{DA}=0;\ P_{DE}=80\ kN\ (C)$

4. Joint E

Resolving vertically

$P_{CA}=80\sin30^\circ \\ \boxed{P_{CA}=40\ kN\ (T)}$

To find the vertical deflection at C

Remove the given load system & apply a vertical load of 1 kN at C

K-Analysis:

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Solve directly.

Member	P(kN)	K(kN)	l(kN)	A($mm^2$)	$\frac{Pkl}{A}$
AB	$-40\sqrt3$	$-\sqrt3$	$1000\sqrt3$	1250	166.3
BC	$-40\sqrt3$	$-\sqrt3$	$1000\sqrt3$	1250	166.3
CD	80	2	2000	2500	128
DE	80	2	2000	2500	128
AE	-40	-1	2000	1250	64
AD	0	0	2000	1250	0
BD	0	0	1000	1250	0
				$\sum =$	652.6

$\therefore$ Vertical deflection of $C=y=\sum\frac{Pkl}{AE}=\frac{652.6}{200}=\underline{3.263\ mm}\ \underline{Ans.}$

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