# Elastic Bending Properties about a Rotated Axis

#### The Product Moment of Area

Calculating the section properties about a rotated axis is often required for elements that are non-orthogonal or that are non-symmetric about both axes. To calculate the second moment of area about a rotated axis, the product moment of area, $$I_{xy}$$, must first be calculated for the section This is expressed as:

In the metric system, the typical units for Product Moment of Area are m⁴ or mm⁴. However this can give rise to a large number of digits, values of the order ×10⁹ mm⁴ are not unknown, so some section catalogues will use the more concise cm⁴ or dm⁴.

$$I_{xy} = \int\int x y \; dx \; dy \text{ in cartesian coordinates}$$ $$I_{xy} = \int\int (rcos\theta) (rsin\theta) \; r \; dr \; d\theta\text{ in polar coordinates}$$ $$\qquad\text{Using the equation }x = rcos\theta \text{ and the equation }y = rsin\theta \text{ and substituting the the area, }dA \text{ for an area defined in polar coordinate system, }r \;dr \;d\theta$$

It is useful to note that as both the $$x$$ and $$y$$ terms in the product moment of inertia are not squared and so can be both negative and positive. This means that if there is full symmetry in either the $$x$$ and $$y$$ directions, then the product moment of inertia is zero.

#### Product Moment of Area of a Rectangle

The product moment of area of a rectangle about its centroid can be found as:

$$I_{xy} = \int x y \; dA$$ $$I_{xy} = \int_{y = -d/2}^{y = d/2} \int_{x=-b/2}^{x=b/2} xy \; dx \;dy$$ $$I_{xy} = \int_{-d/2}^{d/2} y\Big[\frac{x^2}{2}\Big]_{-b/2}^{b/2}\;dy = \int_{-d/2}^{d/2} y\left[\frac{b^2}{8}- \frac{b^2}{8}\right]\;dy$$ $$I_{xy} = \int_{-d/2}^{d/2} y\left[0\right]\;dy$$ $$I_{xy} = 0$$ $$\text{The rectangular section is symmetric about at least one axis so the above demonstrates that the product moment of inertia is zero if the section is symmetric about at least one axis}$$

#### Product Moment of Area of a Circle

The product moment of area of a circle about its centroid can be found as:

$$I_{xy} = \int xy dA$$ $$\text{The value of x and y can be related to the radius of the circle, r, and the angle from the horizontal, }\theta \text{ using the equations }x = rcos\theta\text{ and }y = rsin\theta$$ $$I_{xy} = \int (rsin\theta)(rcos\theta) \; dA$$ $$\text{Substituting the area, }dA \text{ for an area defined in polar coordinate system, }r \;dr \;d\theta \text{ and adding in the bounds of the integral gives:}$$ $$I_{xy} = \int_{\theta = 0}^{\theta = 2\pi} \int_{r=0}^{r=d/2} r^2(sin\theta)(cos\theta) \; r \;dr \;d\theta$$ $$\text{Using the double sin rule } (sin\theta)(cos\theta) \text{ can be written as } \frac{1}{2}(sin2\theta) \text{, hence}$$ $$I_{xy} = \int_{0}^{2\pi} \int_{0}^{d/2} r^2(sin\theta)(cos\theta) \; r \;dr \;d\theta = \int_{0}^{2\pi} \int_{0}^{d/2} \frac{r^2}{2}(sin2\theta) \; r\; dr\; d\theta$$ $$\text{Using standard solutions } \frac{1}{2}(sin2\theta) \text{ integrates to } \frac{-1}{4}cos2\theta \text{ hence}$$ $$I_{xy} = \int_{0}^{2\pi} \int_{0}^{d/2} \frac{r^2}{2}(sin2\theta) \; r \;dr \;d\theta = \int_{0}^{d/2} \frac{r^3}{2}\left[\frac{-1}{4}cos2\theta\right]_{0}^{2\pi} \; dr$$ $$I_{xy} = \int_{0}^{d/2} \frac{r^3}{2}\left[\frac{-1}{4}-\frac{-1}{4}\right] \; dr$$ $$I_{xy} = 0$$ $$\text{Again, the circular section is symmetric about at least one axis so the above demonstrates that the product moment of inertia is zero if the section is symmetric about at least one axis}$$

#### Parallel Axis Theorum

The product moment of inertia can be found about a pair of parallel axis not passing through the centroid of the section using the parallel axis theorum:

$$I_{xy} = \int xy \; dA$$ $$\text{Let the offsets from the parallel axes (NOTE: plural, can be offset on both x and y axes) to the centroid of the rectangle be } \overline{x} \text{ and }\overline{y}$$ $$\overline{I_{xy}} = \int_{y = -d/2}^{y = d/2} \int_{x=-b/2}^{x=b/2} (x + \overline{x})(y + \overline{y}) \; dx \;dy$$ $$\overline{I_{xy}} = \int_{-d/2}^{d/2} \int_{-b/2}^{b/2} (xy + x\overline{y}+y\overline{x}+\overline{x}\overline{y}) \; dx \;dy$$ $$\text{Both } \overline{x} \text{ and }\overline{y}\text{ are constants, so can be taken outside of the integrals}$$ $$\overline{I_{xy}} = \int_{-d/2}^{d/2} \int_{-b/2}^{b/2} (xy + x\overline{y}+y\overline{x}+\overline{x}\overline{y}) \; dx \;dy = \int_{-d/2}^{d/2} \int_{-b/2}^{b/2} xy \; dx \;dy + \overline{y} \int_{-d/2}^{d/2} \int_{-b/2}^{b/2}x\; dx \;dy+\overline{x} \int_{-d/2}^{d/2} \int_{-b/2}^{b/2}y \; dx \;dy+\overline{x}\overline{y}\int_{-d/2}^{d/2} \int_{-b/2}^{b/2} 1 \; dx \;dy$$ $$\text{We already know that } \int_{-d/2}^{d/2} \int_{-b/2}^{b/2} xy \; dx \;dy \text{ is the definition of local Ixy}$$ $$\text{As the offsets } \overline{x} \text{ and }\overline{y} \text{ are measured to the centroid of the section, the the terms } \overline{y} \int_{-d/2}^{d/2} \int_{-b/2}^{b/2}x\; dx \;dy+\overline{x} \int_{-d/2}^{d/2} \int_{-b/2}^{b/2}y \; dx \;dy \text{ become zero}$$ $$\text{Hence,}$$ $$\overline{I_{xy}} = \int_{-d/2}^{d/2} \int_{-b/2}^{b/2} (xy + x\overline{y}+y\overline{x}+\overline{x}\overline{y}) \; dx \;dy = I_{xy} + \overline{x}\overline{y}\int_{-d/2}^{d/2} \int_{-b/2}^{b/2} 1 \; dx \;dy$$ $$\overline{I_{xy}} = I_{xy} + \overline{x}\overline{y}\int_{-d/2}^{d/2} \Big[x\Big]_{-b/2}^{b/2} = I_{xy} + \overline{x}\overline{y}\int_{-d/2}^{d/2} b \;dy$$ $$\overline{I_{xy}} = I_{xy} + \overline{x}\overline{y}b\Big[y\Big]_{-d/2}^{d/2} = I_{xy} + \overline{x}\overline{y}bd$$ $$\overline{I_{xy}} = I_{xy} + A\overline{x}\overline{y}$$

This shows that the product of inertia of a section around a pair of parallel axes can be found by establishing the local product of inertia of the section about its centroid and adding the area of section times the perpendicular offsets to both axes.

# Principal Axes

## Principal Second Moments of Area and the Principal Axes

For doubly asymmetric, such as angles, the second moment of area calculated about the x-x or y-y axis doesn't actually give rise to either the highest or lowest value of second moment of area of the section. To find the highest and lowest value of second moment of area for these sections the axes need to be rotated. The highest value of second moment of area for a section is maximum principal second moment of area, often denoted $$I_{uu}$$, which occurs about an axis u-u. The lowest value of second moment of area for a section is minimum principal second moment of area, often denoted $$I_{vv}$$, which occurs about an axis v-v. The two principal axes, u-u and v-v will always be at 90° to each other, but can be at any angle to the x-x and y-y axes, depending on the section dimensions.

The easiest way to display and calculate the relationships between the second moments of area x-x, y-y, u-u and v-v is to draw the known values in a manner similar to a Mohr's circle, which is used to calculate pricipal stresses (see the section on Mohr's circles in the articles area). This requires us to know three of the following:

• $$I_{xx}$$
• $$I_{yy}$$
• $$I_{uu}$$
• $$I_{vv}$$
• $$I_{xy}$$
• $$\theta$$, the angle between the x-x and u-u axes (which is the same as the angle between the y-y and v-v axes)

The standard form of the principal axes circle is shown in the diagram below.

Useful derivations that can found using the principal axes circle are summarised below:

• The centre of the circle, C, is at $$0.5(I_{uu}+I_{vv})=0.5(I_{xx}+I_{yy})$$
• The radius of the circle, R, has a value of$$\sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$
• The angle is measured as $$2\theta$$, as the angle between u-u and v-v is always 90°, but needs to be drawn on the circle as 180°

The circle be used in a number of different ways, depending on what values are known for a section and what information needs to be found.

• If the user knows the values of $$I_{xx}$$ and $$I_{yy}$$ and needs to find the values of the principal second moments of area, then $$I_{xy}$$ must be found. This then allows the users to determine $$I_{uu}$$, $$I_{vv}$$ and $$\theta$$. This is typically the case for angle sections, where the values of $$I_{xx}$$ and $$I_{yy}$$ are easiest to calculate. However, a shortcut can be made for equal angle sections, by inspection the u-u and v-v axes will be at 45° to the x-x and y-y axes. This means that $$\theta$$ is known, so $$I_{xy}$$ doesn't need to be calculated to $$I_{uu}$$ and $$I_{vv}$$.
• If the user knows the values of the principal second moments of area, $$I_{uu}$$ and $$I_{vv}$$ and needs to find the values of $$I_{xx}$$ and $$I_{yy}$$, then again $$I_{xy}$$ must be found. This then allows the users to determine $$I_{xx}$$, $$I_{yy}$$ and $$\theta$$. This is typically the case for symmetrical sections that are not being bent about there principal axes, this can happen for troughing, sheet piling section, corrugated sections or inclined box sections, where the webs of the elements are not perpendicular to the flanges of the section.