The extension of a circular bar tapering uniformly from diameter d1 at one end to diameter d2 at the other end and subjected to an axial pull of ‘P’ is given by
A. $$\delta l = \frac{{4{\text{PE}}}}{{\pi {l^2}}}$$
B. $$\delta l = \frac{{4\pi l{{\text{d}}^2}}}{{{\text{PE}}}}$$
C. $$\delta l = \frac{{4{\text{P}}l}}{{\pi {\text{E}}{{\text{d}}_1}{{\text{d}}_2}}}$$
D. $$\delta l = \frac{{4{\text{P}}l{\text{E}}}}{{\pi {{\text{d}}_1}{{\text{d}}_2}}}$$
Answer: Option C
This Question Belongs to Mechanical Engineering >> Strength Of Materials In ME
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