Find the moment of inertia of a solid rectangular section 40 mm wide and 60 mm deep about its smaller side.

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written 5.6 years ago by

teamques10 ★ 69k

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written 5.6 years ago by

teamques10 ★ 69k

Given: For solid rectangular section, b = 40 mm, d = 60 mm

Solution:

For solid rectangular section:

$I_{base} = I_{G + (A_{Y})^{2}}$

$= \frac{bd^{3}}{12} + b.d (d/2)^{2}$

$= \frac{40 \times 60}{12}^{3} + 40 \times 60 \times \big(\frac{60}{2}\big)^{2}$

$= 7.20 \times 10^{5} mm^{4}$

$I_{base} = 28.8 \times 10^{5} mm^{4}$

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