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## Q. 1.19

Three rods are hinged together to support a vertical load Was shown in Fig. 1.29. Find the force carried by each rod if A, L and E denote their respective cross-sectional area, length and modulus of elasticity. Outer rods AD and CD are identical.

## Verified Solution

Equilibrium condition: For static eq-uilibrium of joint D,

$\sum F_x=0: F_{A D} \sin \theta=F_{C D} \sin \theta$

or,  $F_{A D}=F_{C D}=F_1$

$\sum F_y=0: 2 F_1 \cos \theta + F_2=W$ (i)

Compatibility condition:The deformed shape of three members is shown by dotted line. It is assumed that the angle $\theta$ does not change appreciably, then the compatibility equation,

$\Delta L_1= \Delta L_2 \cos \theta$ (ii)

or, $\frac{F_1 L_1}{A_1 E_1}=\frac{F_2 L_2}{A_2 E_2} \cos \theta$

But, $L_2= L_1 \cos \theta$

Therefore, $F_1 = F_2 \frac{A_1 E_1}{A_2 E_2} \cos^{2} \theta$ (iii)

From Eqs (i) and (ii),
$F_2=\frac{W}{1+\frac{2 A_1 E_1}{A_2 E_2} \cos ^3 \theta}$

and $F_1=\frac{W \cos ^2 \theta}{\frac{A_2 E_2}{A_1 E_1}+2 \cos ^3 \theta}$