Second moment of area

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This article is about the moment of inertia as related to the bending of a plane. For the moment of inertia dealing with rotation of an object, see Moment of inertia.

The second moment of area, also known as the area moment of inertia or second moment of inertia, is a property of a shape that is used to predict its resistance to bending and deflection which are directly proportional. This is why beams with higher area moments of inertia, such as I-beams, are so often seen in building construction as opposed to other beams with the same area. It is analogous to the polar moment of inertia, which characterizes an object's ability to resist torsion.

The second moment of area is not the same thing as the moment of inertia, which is used to calculate angular acceleration. Many engineers refer to the second moment of area as the moment of inertia and use the same symbol $I$ for both, which may be confusing. Which inertia is meant (accelerational or bending) is usually clear from the context and obvious from the units.

[edit] Notes on notation

In some circumstances I_x and I_xx mean the same thing depending on what notation the person conducting the calculations is using. For example some people referring to a section across the x-axis will just use I_x, but someone who refers to a section across the x-axis as 'section x-x' will use $I x x$ for the same variable.

The same applies in the case of I_y and I_yy.

[edit] Definition

$I_x = \int y^2\,\mathrm dA$

I_x = the moment of inertia about the axis x
dA = an elemental area
y = the perpendicular distance from the axis x to the element dA

Intuitively, the second moment of area, measuring the resistance to bending, can be likened to a person's attempt to stop a force from turning a lever: the farther a person places a hand from the pivot the more leverage is obtained and the easier it is to resist the turning force. In the above formula, the hand's resisting turning are replaced with the sum of small sections of the object (infinitesimally small in the limit); the leverage is proportional to the square of the distance from the 'pivot'. Each small section adds its own contribution depending on its position and proportional to how big it is in cross section; each piece can be split into smaller pieces being summed up until the infinitesimal size is reached and the result is accurate (i.e. the limit of the integral).

The above can only be used on its own when sections are symmetrical about the x-axis. When this is not the case, the second moment of area about both the x- and the y-axis and the product moment of area, I_xy, are required.

[edit] Product moment of area

The product moment of area, I_xy is defined as

$I_{xy} = \int xy\,\mathrm dA$

dA = an elemental area
x = the perpendicular distance to the element dA from the axis y
y = the perpendicular distance to the element dA from the axis x

Note that I_xy is defined here with positive sign following common practice adopted in structural analysis (see e.g. Pilkey 2002, p. 15). The alternative approach to definition of moments of area could be to define the symmetric tensor for moment of area similarly to the tensor of mass moment of inertia; note that non-diagonal members of this tensor are equal to -I_xy:

$J_{ij} = \int ( r^2 \delta_{ij} - r_i r_j ) \mathrm dA$

is significant for determining the bending stress in an asymmetric cross section. Unlike the second moments of area, the product moment may give both negative and positive values. A coordinate system, in which the product moment is zero, is referred to as a set of principal axes, and the second moments of area calculated with respect to the principal axes will assume their maxima and minima. A coordinate system with origin in the centroid of the cross section and with both axes being axes of symmetry are always principal axes.

[edit] Unit

The unit for second moment of area is length to the fourth power (typically mm⁴, in⁴, etc.)

[edit] Co-ordinate transformations

When calculating moments of the section it is often practical to compute them in one co-ordinate system (typically bound to the section shape) and then transform to another one using co-ordinate transformations.

[edit] Parallel axis theorem

Main article: parallel axis theorem

The parallel axis theorem can be used to determine the moment of an object about any axis, given the moment of inertia of the object about the parallel axis through the object's center of mass (or centroid) and the perpendicular distance between the axes.

$I_x = I_{xCG}+A d^2\,$

I_x = the second moment of area with respect to the x-axis
I_xCG = the second moment of area with respect to an axis parallel to x and passing through the centroid of the shape (coincides with the neutral axis)
A = area of the shape
d = the distance between the x-axis and the centroidal axis

[edit] Axes rotation

The following formulae can be used to calculate moments of the section in a co-ordinate system rotated relative to the original co-ordinate system:

${I_x}^* = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} - I_{y}}{2} \cos(2 \phi) - I_{xy} \sin(2 \phi)$

${I_y}^* = \frac{I_{x} + I_{y}}{2} - \frac{I_{x} - I_{y}}{2} \cos(2 \phi) + I_{xy} \sin(2 \phi)$

${I_{xy}}^* = \frac{I_{x} - I_{y}}{2} \sin(2 \phi) + I_{xy} \cos(2 \phi)$

$φ$ = the angle of rotation:

x * = x cosφ + y sinφ

y * = - x sinφ + y cosφ

I_x, I_y and I_xy = the second moments and the product moment of area in the original coordinate system
I_x^*, I_y^* and I_xy^* = the second moments and the product moment of area in the rotated coordinate system.

The value of the angle $φ$ , which will give a product moment of zero, is equal to:

$\phi = -\frac{1}{2} \arctan \frac{ 2 I_{xy}} {I_x-I_y}$

This angle is the angle between the axes of the original coordinate system and the principal axes of the cross section.

[edit] Stress in a beam

The general form of the classic bending formula for a beam in co-ordinate system having origin located at the neutral axis of the beam is (Pilkey 2002, p. 17):

$\sigma=-\frac{M_y I_x + M_x I_{xy}}{I_x I_y - {I_{xy}}^2 } x + \frac{M_x I_y + M_y I_{xy}}{I_x I_y - {I_{xy}}^2} y$

$σ$ is the normal stress in the beam due to bending
x = the perpendicular distance to the centroidal y-axis
y = the perpendicular distance to the centroidal x-axis
M_y = the bending moment about the y-axis
M_x = the bending moment about the x-axis
I_x = the second moment of area about x-axis
I_y = the second moment of area about y-axis
I_xy = the product moment of area

If the coordinate system is chosen to give a product moment of area equal to zero, the formula simplifies to:

$\sigma=-\frac{M_y}{I_y} x + \frac{M_x}{I_x} y$

If additionally the beam is only subjected to bending about one axis, the formula simplifies further:

${\sigma}= \frac{M}{I_x} y$

[edit] Second moment of area for various cross sections

See list of area moments of inertia for other cross sections.

[edit] Rectangular cross section

$I_{x}=\frac{bh^3}{12}$

b = width (x-dimension),
h = height (y-dimension)

$I_{y}=\frac{hb^3}{12}$

b = width (x-dimension),
h = height (y-dimension)

[edit] Circular cross section

$I_0 = \frac{\pi}{4} r ^4$

r = radius

[edit] Hollow Cylindrical Cross Section

$I_0 = \frac{\pi}{64} (D_O ^4 - D_I ^4)$

$D O$ = Outside Diameter
$D I$ = Inside Diameter

[edit] Composite cross section

When it is easier to compute the moment for an item as a combination of pieces, the second moment of area is calculated by applying the parallel axis theorem to each piece and adding the terms:

$I_{x}= \sum \left(I_\mathrm{local}+y^{2}A\right)$

$I_{y}= \sum \left(I_\mathrm{local}+x^{2}A\right)$

y = distance from x-axis
x = distance from y-axis
A = surface area of part
I_local is the second moment of area for that part of the composite, in the appropriate direction (i.e. I_x or I_y respectively).

[edit] "I-beam" cross section

I-beam

The I-beam can be analyzed as either three pieces added together or as a large piece with two pieces removed from it. Either of these methods will require use of the formula for composite cross section. This section only covers doubly symmetric I-beams, meaning the shape has two planes of symmetry.

b = width (x-dimension),
h = height (y-dimension)
t_w = width of central webbing
h₁ = inside distance between flanges

This formula uses the method of a block with two pieces removed. (While this may not be the easiest way to do this calculation, it is instructive in demonstrating how to subtract moments).

I-beam diagram, moment by subtraction

Since the I-beam is symmetrical with respect to the y-axis the $I x$ has no component for the centroid of the blocks removed being offset above or below the x axis.

$I_{x}=\frac{{bh^3}-2{{\frac{b-t_{w}}{2}}{h_{1}}^3}}{12}$

When computing I_y it is necessary to allow for the fact that the pieces being removed are offset from the X axis, this results in the Ax² term.

$I_{y}=\frac{hb^3}{12}-2\left({\frac{h_{1}\left({\frac{b-t_{w}}{2}}\right)^3}{12}+Ax^2}\right)$

A = Area contained with in the middle of one of the 'C' shapes of created by two flanges and the webbing on one side of the cross section = $h_{1}{\frac{b-t_{w}}{2}}$
x = distance of the centroid of the area contained in the 'C' shape from the y-axis of the beam = ${\frac{b+t_{w}}{4}}$

Doing the same calculation by combining three pieces, the center webbing plus identical contributions for the top and bottom piece:

I-beam diagram, moment by addition

Since the centroids of all three pieces are on the y-axis I_y can be computed just by adding the moments together.

$I_{y}=\frac{h_{1} {t_{w}}^3}{12} + 2 \frac{\frac{h-h_{1}}{2} b^3}{12}$

However, this time the law for composition with offsets must be used for I_x because the centroids of the top and bottom are offset from the centroid of the whole I-beam.

A = Area of the top or bottom piece= $b \frac{h-h_{1}}{2}$
y = offset of the centroid of the top or bottom piece from the centroid of the whole I-beam= $\frac{h+h_{1}}{4}$

$I_{x}=\frac{t_{w} {h_{1}}^3}{12} + 2 \left( \frac{b \left(\frac{h-h_{1}}{2}\right)^3}{12} + Ay^2\right) =\frac{t_{w} {h_{1}}^3}{12} + 2 \left( \frac{b \left(\frac{h-h_{1}}{2}\right)^3}{12} + b \frac{h-h_{1}}{2} \left(\frac{h+h_{1}}{4}\right)^2\right)$

[edit] Any cross section defined as polygon

The second moments of area for any cross section defined as a simple polygon on XY plane can be computed in a generic way by summing contributions from each segment of a polygon.

For each segment defined by two consecutive points of the polygon, consider a triangle with two corners at these points and third corner at the origin of the co-ordinates. Integration by the area of that triangle and summing by the polygon segments yields:

$I_x = \frac{1}{12} \sum_{i = 1}^{n} ( y_i^2 + y_i y_{i+1} + y_{i+1}^2 ) a_i \,$

$I_y = \frac{1}{12} \sum_{i = 1}^{n} ( x_i^2 + x_i x_{i+1} + x_{i+1}^2 ) a_i \,$

$I_{xy} = \frac{1}{24} \sum_{i = 1}^{n} ( x_i y_{i+1} + 2 x_i y_i + 2 x_{i+1} y_{i+1} + x_{i+1} y_i ) a_i \,$

$a i = x i y i + 1 - x i + 1 y i$ is twice the (signed) area of the elementary triangle,
index i passes over all n points in the polygon, which is considered closed, i.e. point n+1 is point 1

These formulae imply that points defining the polygon are ordered in anticlockwise manner; for clockwisely defined polygons it will give negative values. See polygon area for calculating area and centroid of the section using similar formulae.

[edit] References

Pilkey, Walter D. (2002). Analysis and Design of Elastic Beams. John Wiley & Sons, Inc.. ISBN 0-471-38152-7.

[edit] External links

http://mywebsite.bigpond.com/npajkic/solid_mechanics/second_moment_of_area/index.html

Categories: Structural analysis | Physical quantity