Chapter 1
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.
Step-by-Step
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)