A truss consists of horizontal members AC- CD- DB- and EF and vertical members CE and DF having length l each- The members AE- DE and BF are inclined at 45- to the horizontal- For the uniformly distributed load p per unit length on the members EF of the truss shown in figure given below- the force in the member CD is

R_A+R_B=pl Taking moment about A pl(l+l/2)=R_B× 3l ∴ R_B=pl2 ∴ R_A=pl2 Joint A Now at joint A Σ F_v= 0 F_EAsin 45^∘=pl2 ∴ F_EA=pl√(2) F_AC=F_EAcos 45 ...

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Engineering Mechanics

Method of Joints-2

A truss consi...

Question

A truss consists of horizontal members (AC, CD, DB, and EF) and vertical members (CE and DF) having length l each. The members AE, DE and BF are inclined at $45^{\circ}$ to the horizontal. For the uniformly distributed load p per unit length on the members EF of the truss shown in figure given below, the force in the member CD is

\frac{p l}{2}

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p l

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\frac{2 p l}{3}

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Solution

The correct option is A $\frac{p l}{2}$

$R_{A} + R_{B} = p l$

Taking moment about A

$p l (l + l / 2) = R_{B} \times 3 l$

$∴ R_{B} = \frac{p l}{2}$

$∴ R_{A} = \frac{p l}{2}$

Joint A

Now at joint A

$Σ F_{v} = 0$

$F_{E A} sin 45^{\circ} = \frac{p l}{2}$

$∴ F_{E A} = \frac{p l}{\sqrt{2}}$

$F_{A C} = F_{E A} cos 45^{\circ}$

$F_{A C} = \frac{p l}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

$A_{A C} = \frac{p l}{2}$

At joint C
$F_{C A} = F_{C D} = \frac{p l}{2} [∵ F_{C E} = 0]$

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The correct option is A pl2 RA+RB=pl Taking moment about A pl(l+l/2)=RB×3l ∴RB=pl2 ∴RA=pl2 Joint A Now at joint A ΣFv=0 FEAsin45∘=pl2 ∴FEA=pl√2 FAC=FEAcos45∘ FAC=pl√2×1√2 AAC=pl2 At joint C FCA=FCD=pl2[∵FCE=0]