Question
Download Solution PDFA semicircular lamina has radius R and diameter D. Determine the moment of inertia about the diametrical axis as shown below?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The moment of inertia of a circular lamina is
\(I = \frac{π }{{64}}{D^4}\) = π(2R)4/64 = πR4/4
The moment of inertia about the diametrical axis semicircular lamina is
= I/2 = (πR4/4)/2 = πR4/8
Additional Information
The formula of the moment of inertia for various other figures is given below.
Moment of inertia of different section:
|
Shape of cross-section |
INA |
Ymax |
Z |
|
Rectangle |
\(I = \frac{{b{d^3}}}{{12}}\) |
\({Y_{max}} = \frac{d}{2}\) |
\(Z = \frac{{b{d^2}}}{6}\) |
|
Circular |
\(I = \frac{π }{{64}}{D^4}\) |
\({Y_{max}} = \frac{d}{2}\) |
\(Z = \frac{π }{{32}}{D^3}\) |
|
Triangular |
\(I = \frac{{B{h^3}}}{{36}}\) |
\({Y_{max}} = \frac{{2h}}{3}\) |
\(Z = \frac{{B{h^2}}}{{24}}\) |
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