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Moment Of Inertia Of A Rectangular Plate

In the case of a rectangular plate, we usually find the mass moment of inertia when the axis is passing through the centre perpendicular to the plane.

Moment Of Inertia Of A Rectangular Plate

We use the following expressions to calculate the moment of inertia of a rectangular plate;

For x-axis;

Ix = (1/12) mb2

For y-axis;

Ix = (1/12) ma2

And,

IZ = (1/12) m (a2 + b2)

Moment Of Inertia Of A Rectangular Plate Derivation

1. Line Passing Through The Base

Moment Of Inertia Of A Rectangular Plate - Line Passing Through The Base

For the derivation of the moment of inertia formula for a rectangular plate, we will consider a rectangular section and cut out an elemental part at a distance (y) from the x-axis. Let its thickness be dy and s be the mass per unit volume of the plate.

Mass of the rectangular body = density x volume

M = ρ x bdt

Mass of the elemental section;

dM =ρ  x btdy

Now we will find the mass moment of inertia of the elemental strip about the x-axis.

I = ∫ y2 dM

I = ∫ (ρ b dy t)y2

In the next step, we will find the mass moment of inertia of the rectangular plate.

I = od ρ b dy t y2

I = ρ bt od y2dy

I = ρ bt d3 / 3

(since ρ  = M/bdt)

I = Md2 / 3

At this point, we apply the parallel axis theorem which gives us;

Ibase = ICG + M x h2

ICG = Ibase – Mh2

ICG = Md2 / 3 – M x (d / 2)2

ICG = Md2 / 12

This is how we determine the mass moment of inertia of a rectangular section about a line passing through the base.

2. Line Passing Through The Centre of Gravity

Now if we want to derive the expression for a line passing through the centre of gravity then the steps for derivation are almost the same. We only have to make a few changes.

Moment Of Inertia Of A Rectangular Plate - Line Passing Through The Centre Of Gravity

We will take one rectangular elementary strip and consider the thickness to be (dY) and it will be at a distance (Y) from the X-X axis.

The rectangular elementary strip area will be dA = dY.B

Mass of the rectangular elemental part is given by;

dm = ρ x T x dA

dm = ρ x T x dY. B

dm =ρBT. dY

Now we can say that the mass moment of inertia of the rectangular elemental section about the X-X axis ,

(Im)xx = dm.Y2

If we substitute the values for dm we get;

(Im)xx = ρBT. Y2dY

Next step involves determining the mass moment of inertia of the entire rectangular section. We have to integrate the above equation between limit (-D/2) to (D/2). We will write it as;

(Im)xx = -D/2D/2 ρBT. Y2dY

(Im)xx = ρBT -D/2D/2 Y2dY

(Im)xx = ρBT [y3 / 3]-D/2D/2

(Im)xx = ρBT / 3 [(D/2)3 – (-D/2)3]

Im) XX = ρBT.D3/12

(Im) XX = ρBTD.D2/12

(Im) XX = ρV.D2/12

(Im) XX = M.D2/12

Likewise, we can also determine the mass moment of inertia of the rectangular section about the Y-Y axis. We will obtain;

(Im) YY = M.B2/12

⇒ Check Other Object’s Moment of Inertia:

Video lesson – Parallel Axis Theorem

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