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## Q. 1.18

Determine the central transverse load W which will produce central deflection $\delta$ in the initially straight bar AB of area A and length 2L hinged at both ends as shown in Fig. 1.28. ## Verified Solution

Solution Equilibrium condition: The forces in section AC and CB will be equal due to symmetry. Therefore, from free body diagram,

$W = 2F \sin \alpha= 2F \times \frac{\delta }{L^\prime}$ (i)

Compatibility condition: Elongation of section AC is given by,

$\Delta L = L^\prime — L=\frac{FL}{AE}$ (ii)

From Eqs (i) and (ii),

$W =\frac{2\left(L^\prime-L\right) A E}{L} \times \frac{\delta}{L^\prime}=\frac{2 \delta A E}{L}\left(1-\frac{L}{L^\prime}\right)$

$=\frac{2 \delta A E}{L}\left(1-\frac{1}{\sqrt{L^2+\delta^2}}\right)$