Question
Download Solution PDFThe polar moment of inertia of a hollow circular shaft of outer diameter (D) and inner diameter (d) is
This question was previously asked in
WBPSC JE Civil 2018 Official Paper
Answer (Detailed Solution Below)
Option 3 : \(\frac{\pi}{32}\times (D^4-d^4)\)
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Detailed Solution
Download Solution PDFExplanation:
Polar moment of inertia of Solid cylinder:
\({I_Z} = J = \frac{{\pi {D^4}}}{{32}}\)
Polar moment of inertia of Hollow Cylinder:
\({I_Z} = J = \frac{{\pi \left( {{D^4} - {d^4}} \right)}}{{32}}\)
where outer diameter = D and inner diameter = d
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