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Home , Area, Moment (physics), Second moment of area

MOMENT OF Moments of

Example 1: on a dam The left figure shows a dam bounding a body of water. Acting at any is a water pressure which is a per unit of area Water pressure is a of both water specific (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 p x=⋅x

2 First of Area

Assume pressure p(x) is applied to elemental area dA. Element dA is located a 'x' below the z-axis. If we multiply this elemental area by distance 'x', we can define elemental moment: dM = x⋅dA or if expressed algebraically

M i = x⋅Ai Integrating, we find the statical or first M = x⋅dA moment of area as: ∫A expressed algebraically as

M = ∑ x⋅Ai i 3 Moment of Area

If we take an element of area dA and multiply it by pressure p(x)...... the elemental force acting on dA is: dF = p x⋅dA

If we multiply this elemental moment by its distance 'x' from the z-axis, we have: dF = p x⋅dA

The moment of dF about the z-axis is:

dM = x⋅dF dM = x⋅⋅x⋅dA = ⋅x 2⋅dA

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If we define γ as a constant and divide it out, we are left with: dM = x⋅ x⋅dA 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 of dM over the entire area.

M = dM = x2 dA ∫A ∫A Algebraically, we can replace elemental area dA with some

area Ai of finite . Total moment is then the sum of all individual moments.

2 M = ∑  x ⋅Ai i

5 Second Moment of Area

Thus, the second moment of area...... is often called area due to the to the mathematical form of the integral for the resultant moments of the inertia in the case of rotating bodies (i.e.: 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: I = distance 2 dA ∫A An algebraic can be expressed as:

2 I = ∑ distance Ai i

6 Area Moment of Inertia

Example 2: - A common application is to determine the internal bending stress in a beam. Such stresses are linear. of these stresses is a strength of materials topic and does require the moment of inertia of the beam cross section. y This will be addressed in detail in your strength of materials course M⋅y  = b I

7 Area Moment of Inertia

Example 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'? Study the equation for moment of inertia again.

I = distance 2 dA ∫A Notice that for a given cross section, it is not area that matters, it is the orientation of the section. 8 Area Moment of Inertia

As stated previously, the second moment of area, or area moment of inertia, is given by: I = distance2 dA ∫A 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 Y x distance 'x' from the y-axis and a distance 'y' from the x-axis. Then: dA I = y2 dA y x ∫A 2 I y = ∫ x dA A X The units of moment of inertia will be to the fourth such as in4, ft4, mm4, or m4 9 Centroidal Moment of Inertia

In most cases, we are interested in the moment of inertia about an x-y axis superimposed on the of the cross section. This is known as the centroidal moment of inertia. Centroidal moments of inertia for simple geometric 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.

10 Centroidal Moment of Inertia

Consider a with a width 'b' and a height 'h'. The application of will yield: y = ⋅ h/2 dA b dy 2 2 I xc =∫ y ⋅dA = ∫ y b⋅dy dy A −h / 2 3 3 y y3 h/ 2 b h −h b⋅h3 h c x = b⋅ ∣ = − = 3 −h/ 2 3 [ 2   2  ] 12 y x dx b dA = h⋅dx

b/ 2 2 2 h c I yc =∫ x ⋅dA = ∫ x h⋅dx x A −b/ 2 3 3 x3 b/2 h b −b h⋅b3 = h⋅ ∣ = − = 3 −b/ 2 3 [ 2   2  ] 12 b

11 Centroidal Moment of Inertia

We 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:

y h Ai = b⋅ 2 4 I xc =  y ⋅Ai 2 2 2 2 h/ 4 3h h h h h h 3h h 3h = ⋅b⋅  ⋅b⋅  ⋅b⋅  ⋅b⋅ 8 h/ 4  8  4 8  4 8  4  8  4 h [ ] [ ] [ ] [ ] h 8 h x 2 2 2 2 8 3h h/ 4 3h h h 3h h 8 =    ⋅b⋅ [ 8  8  8   8  ] 4 h/ 4 5bh3 b = 64

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 .

12 Centroidal Moment of Inertia

For example, by dividing the rectangle into 8 equal , we would calculate the moment of inertia to be:

2 I xc =  y ⋅Ai 2 2 2 2 2 2 2 2 7h 5h 3h h h 3h 5h 7h h =        ⋅b⋅ [ 16  16  16  16  16  16  16  16  ] 8 168 h2 h 21bh 3 = ⋅b⋅ =  256  8 256 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.

13 Centroidal Moment of Inertia

When 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.

y y' b⋅h3 b3⋅h I = I = x ' 12 y ' 12 h x' b⋅h3 b3⋅h c I = I = x 3 y 3 b x

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