MOMENT OF INERTIA

Moments of AreaExample 1: Pressure on a dam

The left figure shows a dam bounding a body of water. Acting at any point is a water pressure which is a force per unit of area

Water pressure is a function of both water specific weight (γ ) and water depth (x). The pressure is constant across an area of infinitely small height dx and width 'L' but will vary linearly with height 'x' as

2

p x =⋅x

Assume pressure p(x) is applied to elemental area dA.

Element dA is located a distance 'x' below the z-axis.

If we multiply this elemental area by distance 'x', we can define elemental moment:

Integrating, we find the statical or first moment of area as:

3

First Moment of Area

M =∫Ax⋅dA

expressed algebraically as

M =∑i

x⋅Ai

dM = x⋅dAor if expressed algebraicallyM i = x⋅Ai

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Second Moment of Area

dF= p x⋅dA

dF= p x⋅dA

dM = x⋅dFdM = x⋅⋅x⋅dA = ⋅x2⋅dA

If we take an element of area dA and multiply it by pressure p(x)...

...the elemental force acting on dA is:

If we multiply this elemental moment by its distance 'x' from the z-axis, we have:

The moment of dF about the z-axis is:

If we define γ as a constant and divide it out, we are left with:

Mathematically, this is a moment of the first moment of area, thus we refer to this as the second moment of area.

The moment of the entire distribution is equal to the integral of dM over the entire area.

Algebraically, we can replace elemental area dA with some area Ai of finite size. Total moment is then the sum of all individual moments.

M =∫AdM =∫A

x2 dA

M =∑i

x2⋅Ai

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Second Moment of Area

dM = x⋅ x⋅dA

Thus, the second moment of area...

...is often called area moment of inertia due to the similarity to the mathematical form of the integral for the resultant moments of the inertia forces in the case of rotating bodies (i.e.: mass moment of inertia).

For our purposes, we will hereafter refer to the second moment of area simply as the moment of inertia

As such, the moment of interia is defined as:

An algebraic approximation can be expressed as:

I =∫Adistance2dA

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Second Moment of Area

I =∑i

distance2Ai

Area Moment of InertiaExample 2: Beam Bending - Stress

A common application is to determine the internal bending stress in a beam. Such stresses are linear.

Analysis of these stresses is a strength of materials topic and does require the moment of inertia of the beam cross section.

This will be addressed in detail in your strength of materials course

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b =M⋅yI

y

Area Moment of InertiaExample 3:Beam Bending – Deflection

Consider bending the same beam in two planes...for example, a 2 x 4 laid across a pair of carpenter's horses.

Now try with your own ruler. About which axis is the ruler 'stiffer'?

I =∫Adistance2dA

8

Study the equation for moment of inertia again.

Notice that for a given cross section, it is not area that matters, it is the orientation of the section.

Area Moment of InertiaAs stated previously, the second moment of area, or area moment of inertia, is given by:

Let's consider finding the moment of inertia about some arbitrary 'x' or 'y' axis. Consider cross section below with area 'A'. Elemental area dA is located a distance 'x' from the y-axis and a distance 'y' from the x-axis. Then:

The units of moment of inertia will be length to the fourth power such as in4, ft4, mm4, or m4

I =∫Adistance2dA

X

Y

y

x

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I x =∫Ay2dA

I y =∫Ax2dA

dA

Centroidal Moment of InertiaIn most cases, we are interested in the moment of inertia about an x-y axis superimposed on the centroid of the cross section. This is known as the centroidal moment of inertia. Centroidal moments of inertia for simple geometric shapes can be found in many engineering handbooks, on the Internet, and in the appendix of this online text.

But before we use such handbook data, let's apply the general equation for moment of inertia to a rectangular cross section.

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Centroidal Moment of InertiaConsider a rectangle with a base width 'b' and a height 'h'. The application of calculus will yield:

11

x

ydA= b⋅dy

dy

y

b

h c

x

y

dA= h⋅dxdxx

b

h cI yc =∫A

x2⋅dA= ∫−b/2

b/2

x 2h⋅dx

= h⋅x3

3∣−b /2

b /2

=h3 [ b2

3

−−b2 3

] = h⋅b3

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I xc =∫A

y2⋅dA= ∫−h /2

h /2

y2 b⋅dy

= b⋅y3

3∣−h/2

h/2

=b3 [h2

3

−−h2 3

] = b⋅h3

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Centroidal Moment of InertiaWe can also estimate the moment of inertia using algebraic methods. By breaking the area into finite sections (lets try four), then the moment of inertia (Ixc only) calculates as:

12

x

yAi = b⋅

h4

h / 4

b

h

h / 4

h / 4

h / 4

h8

3h8

3h8

h8

Note this does not equal the moment of inertia calculated using calculus methods. We can get closer by breaking the area into more and smaller rectangles.

I xc= y2⋅A i

= [ 3h8 2

⋅b⋅h4 ] [h8

2

⋅b⋅h4 ] [h8

2

⋅b⋅h4 ] [ 3h8

2

⋅b⋅h4 ]

= [ 3h8 2

h8 2

h8 2

3h8 2

]⋅b⋅h4=5bh3

64

Centroidal Moment of InertiaFor example, by dividing the rectangle into 8 equal areas, we would calculate the moment of inertia to be:

13

Although much closer to the result obtained through integration, it is still not equal. Although we could continue to divide this rectangle into even smaller areas to refine the solution, it is unneccessary. Integrations have been accomplished for common geometric shapes and are tabulated in the appendix to this online text.

I xc = y2⋅Ai

= [ 7h16 2

5h16 2

3h16 2

h16 2

h16 2

3h16 2

5h16 2

7h16 2

]⋅b⋅h8= 168h

2

256 ⋅b⋅h8 =21bh 3

256

Centroidal Moment of InertiaWhen applying the relationships listed in the appendix, take care to note the location of the x-y coordinate axis. Moments of inertia can be found about any axis one wishes.

For example, the moment of inertia of a rectangle about an x-axis depends on the x-axis in which you are interested, the centroidal axis or the base axis.

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I x ' =b⋅h3

12I y ' =

b3⋅h12

I x =b⋅h3

3I y =

b3⋅h3

x

x'

y'y

b

h c