Question

Find the moment of inertia of a hollow rectangular section about its centre of gravity, if the internal dimension are 40mm deep and 30 mm wide and internal dimension are 25 mm deep and 15 mm wide?​

Answer 1

Explanation:

plss detaills solution

Answer 2

The moment of inertia of the hollow rectangular section with given dimensions is 140468.75 mm⁴.

Given:

For a hollow rectangular section,

The external dimensions are; length (L) is = 40 mm, breadth (B) is = 30 mm

The internal dimensions are; length (l) is = 25 mm, breadth (b) is = 15 mm

To Find:

The moment of inertia of a hollow rectangular section about its center of gravity =?

Solution:

Here, to calculate the moment of inertia of the hollow rectangular frame, we will subtract the moment of inertia of the smaller rectangle from the moment of inertia of the bigger rectangle.

We know that the moment of inertia of a rectangular section about an axis passing through its center of gravity is given as;

I = Moment of inertia  =  $\frac{bh^{3} }{12}$

Let, the moment of inertia of the external rectangle be 'I₁' and that of the internal rectangle be 'I₂'.

Now,

I₁ = $\frac{BH^{3} }{12}$

∴ I₁ =  $\frac{(30)(40)^{3} }{12}$

∴ I₁ =  $\frac{1920000}{12}$  mm⁴.

Similarly,

I₂ =  $\frac{bh^{3} }{12}$

∴ I₂ =  $\frac{(15)(25)^{3} }{12}$

∴ I₂ =  $\frac{234375}{12}$  mm⁴.

Now, The required moment of Inertia of the hollow rectangular section I is;

∴ I =  I₁ - I₂

∴ I =  $\frac{1920000}{12}$ -  $\frac{234375}{12}$

∴ I =  $\frac{1920000-234375}{12}$

∴ I =  $\frac{1685625}{12}$

∴ I =  140468.75 mm⁴.

Therefore, The moment of inertia of a hollow rectangular section about its center of gravity is 140468.75 mm⁴.

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