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Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer» <html><body><p></p>Solution :Here, `Y=2.0xx10^(11)<a href="https://interviewquestions.tuteehub.com/tag/pa-1145246" style="font-weight:bold;" target="_blank" title="Click to know more about PA">PA</a>`, Mass (M) = 51,000 kg. <br/> inner radius, `r_(1)=<a href="https://interviewquestions.tuteehub.com/tag/30cm-306639" style="font-weight:bold;" target="_blank" title="Click to know more about 30CM">30CM</a>= 0.3 m` <br/> Outer radius, `r_(2)=60cm = 0.6m` <br/> Force acting on each columns, <br/> `F=(Mg)/(4)=(50,000 xx 9.8)/(4)=122500N` <br/> Area of corss-section of each column, <br/> i.e., `A=pi r_(2)^(2)-pi r_(1)^(2)` <br/> `=pi (r_(2)^(2)-r_(1)^(2))=3.14(0.36-0.09)=0.85m^(2)`<br/> <a href="https://interviewquestions.tuteehub.com/tag/compressional-7673081" style="font-weight:bold;" target="_blank" title="Click to know more about COMPRESSIONAL">COMPRESSIONAL</a> stress on each column, <br/> `(F)/(A)=(122500)/(0.85)=1.44xx10^(5)Pa "As Y"=("stress")/("strain")`, <br/> strain (compressional) `=("stress(compressional)")/(Y)(or) (1.44xx10^(5))/(2.0xx10^(11))=<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>.2xx10^(-7)`</body></html>


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