A) \[\frac{d+D}{8}\]
B) \[\frac{{{d}^{2}}+{{D}^{2}}}{8d}\]
C) \[\frac{{{d}^{2}}+{{D}^{2}}}{8D}\]
D) \[\sqrt{\frac{{{d}^{2}}+{{D}^{2}}}{8}}\]
Correct Answer: B
Solution :
Tensile stress \[=-\frac{P}{A}+\frac{Pe}{l}+\frac{D}{2}=0\] \[\therefore \,\,\,\frac{1}{A}=\frac{eD}{2l}\] \[\frac{4}{\pi ({{D}^{2}}-{{d}^{2}})}=\frac{eD\times 64}{2\times \pi ({{D}^{4}}-{{d}^{4}})}\] \[e=\frac{8\pi ({{D}^{4}}-{{d}^{4}})}{64\times \pi \times D({{D}^{2}}-{{d}^{2}})}\] \[=\frac{{{D}^{2}}+{{d}^{2}}}{8D}\]
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