Question

The moment of inertia of a rectangular section 3 cm wide and 4 cm deep about an axis passing through its centre of gravity and parallel to width is

Answer 1

Bd^3/12. So b = 3cm and d= 4 cm Then (3x 4x 4x4)/(12). Answer is 16 cm ^4

Answer 2

The moment of inertia of rectangular section passing through its center of gravity and parallel to width is 16 $cm^{2}$

Given : Width of rectangular section is 3 cm

Depth of rectangular section is 4 cm

To Find :  The moment of inertia of rectangular section passing through its center of gravity and parallel to width

Solution : moment of inertia of rectangular section passing through its center of gravity and parallel to width is 16 $cm^{2}$

Here we have  to find the moment of inertia of an area or second moment of area

The Moment of inertia of a body about any axis is defined as the summation of the second moment of all elementary areas about the axis

I = ∑( A × $d^{2}$ )

Moment of inertia of a rectangular section passing through its center of gravity and parallel to width  is given as

Ixx = $\frac{(b) d^{3} }{12}$

where b is width of the rectangle

and d is depth of the rectangle

Here it is given that

b = 3cm

d = 4 cm

Ixx = $\frac{(3)4^{3} }{12}$

Ixx = $\frac{192}{12}$

Ixx = 16 $cm^{2}$

So moment of inertia of rectangular section passing through its center of gravity and parallel to width is 16 $cm^{2}$

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