Elastic Bending Properties

Area and Centroid

The area and centroid are fundamental properties of a beam. The centroid of the section is very important in elastic bending as this is where the neutral axis of the section passes through, this is the point from which the section will bend about. For a given moment about an axis through the centroid, all the section above the centroid will have the same sign stress (say, in compression), while below the centroid all the stress will be the opposite sign (in tension in this example). This is discussed in more detail in the article on "Euler-Bernoulli Beam bending equations".

Neutral axis for symmetrical and asymmetrical sections
Figure 1: Neutral axis for bending lies at a sections centroids and effects the stress distribution within the section

Centroid and Area are discussed in more detail in the article on "Area and Centroid" and won't be recalcualted here.

Second Moment of Area

The second moment of area of a section is a measure of how far area from an arbitary axis the area of a section is arranged. A low second moment of area indicates the area of a section is located close to the axis in question, a high second moment of area indicates the area is located further away from the axis in question.

Figure 2: The same total area of a section can be arranged in different ways to give a higher or lower second moment of area.

The second moment of area is important when examining the elastic behaviour of a beam as it allows:

  • The flexural stiffness of an element to be calculated, the higher the second moment of area, the higher the stiffness
  • It allows the Section Modulii to be calculated, which is important in establishing the bending stresses within a section
  • It allows the calculation of the Radius of Gyration of the section, this is important in the calculating the buckling resistance of the section and is discussed in more detail in a seperate article.

The second moment of area is mathematically defined as:

$$I_{xx} = \int y^2 \; dA$$

I.e the integral of the square of the perpendicular distance from the axis under consideration to a point on the section, with respect to area. As the integral term is squared it means that the perpendicular offset is always positive, meaning the value of \(I_{xx}\) is always positive The second moment of area can be calculated about any axis, usually perpendicular axes x and y are chosen, the expressions for second moment of area for each of these axes would be:

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

$$I_{xx} = \int y^2 \; dA$$ $$I_{yy} = \int x^2 \; dA$$

Note that as the axis under consideration changes (from \(I_{xx}\) to \(I_{yy}\)) the distance perpendicular also changes (from \(y^2\) to \(x^2\))

Second Moment of Area of a Rectangle

To demonstrate the above, the second moment of area of a rectangle about its centroid can be found as:

Second moment of area, rectangle
Figure 3: Key dimensions for finding the second moment of area of a rectangle

$$I_{xx} = \int y^2 \; dA$$ $$I_{xx} = \int_{y = -d/2}^{y = d/2} \int_{x=-b/2}^{x=b/2} y^2 \; dx \;dy$$ $$I_{xx} = \int_{-d/2}^{d/2} y^2\Big[x\Big]_{-b/2}^{b/2}\;dy = \int_{-d/2}^{d/2} y^2\left[\frac{b}{2}- \frac{-b}{2}\right]\;dy = \int_{-d/2}^{d/2} b y^2 \;dy$$ $$I_{xx} = b\left[\frac{y^3}{3}\right]_{-d/2}^{d/2} = b\left[\frac{d^3}{24}- \frac{-d^3}{24}\right]\;dy$$ $$I_{xx} = \frac{bd^3}{12} $$

Second Moment of Area of a Circle

The above example uses cartesian coordinates x and y. Second moment of area can also be calculated using the polar coordinate system, \(r\) and \(\theta\). To demonstrate this above, the second moment of area of a circle about its centroid can be found as:

Second moment of area, circle
Figure 4: Key dimensions for finding the second moment of area of a circle

$$I_{xx} = \int y^2 dA$$ $$\text{The value of y can be related to the radius of the circle, r, and the angle from the horizontal, }\theta \text{ using the equation }y = rsin\theta $$ $$I_{xx} = \int (rsin\theta)^2 \;dA = \int r^2sin^2\theta \; dA $$ $$\text{Substituting the area, }dA \text{ for an area defined in polar coordinate system, }r \;dr \;d\theta \text{ and adding in the bounds of the integral gives:}$$ $$I_{xx} = \int_{\theta = 0}^{\theta = 2\pi} \int_{r=0}^{r=d/2} r^2sin^2\theta \; r \;dr \;d\theta$$ $$\text{Using the double cosine rule } sin^2\theta \text{ can be written as } \frac{1}{2}(1 - cos2\theta) \text{, hence}$$ $$I_{xx} = \int_{0}^{2\pi} \int_{0}^{d/2} r^2sin^2\theta \; r \;dr\; d\theta = \int_{0}^{2\pi} \int_{0}^{d/2} \frac{r^2}{2}(1 - cos2\theta) \; r\; dr\; d\theta$$ $$\text{Using standard solutions } \frac{1}{2}(1 - cos2\theta) \text{ integrates to } \frac{1}{2}(\theta - \frac{1}{2}sin2\theta) \text{ hence}$$ $$I_{xx} = \int_{0}^{2\pi} \int_{0}^{d/2} \frac{r^2}{2}(1 - cos2\theta) \; r \;dr \;d\theta = \int_{0}^{d/2} \frac{r^3}{2}\left[\theta - \frac{1}{2}sin2\theta\right]_{0}^{2\pi} \; dr $$ $$I_{xx} = \int_{0}^{d/2} \frac{r^3}{2}\left[(2\pi - \frac{1}{2}sin4\pi)-(0-\frac{1}{2}sin0)\right] \; dr = \int_{0}^{d/2} \frac{r^3}{2}\left[2\pi\right] \; dr $$ $$I_{xx} = 2\pi\left[\frac{r^4}{8}\right]_{0}^{d/2}=2\pi\frac{d^4}{8\times16}$$ $$I_{xx} = \frac{\pi d^4}{64} $$

Parallel Axis Theorum

The above examples both calculate the second moment of area about the centroid of the section, however it is often useful to calculate the second moment of area about an axes that doesn't pass through the centroid of a section. What is needed as a method of calculated the second moment of area about an axes parallel to but offset from the centroidal axis, this is known as "The Parallel Axis Theorum". A governing equation of the parallel axis theorum can be calculated by finding the second moment of area of a rectangle about an offset parallel axis, instead of about its centroid:

Second moment of area using parallel axis theorem rectangle
Figure 5: Dimensions defining a parallel axis of bending

$$I_{xx} = \int y^2 \; dA$$ $$\text{Let the offset from the parallel axis to the centroid of the rectangle be } \overline{y}$$ $$I_{xx} = \int_{y = -d/2}^{y = d/2} \int_{x=-b/2}^{x=b/2} (y + \overline{y})^2 \; dx \;dy$$ $$I_{xx} = \int_{-d/2}^{d/2} \int_{b/2}^{b/2} (y^2 + 2y\overline{y}+\overline{y}^2) \; dx \;dy = \int_{-d/2}^{d/2} b\left[y^2 + 2y\overline{y}+\overline{y}^2\right] \;dy $$ $$I_{xx} = b\left[\frac{y^3}{3} + y^2\overline{y}+\overline{y}^2y\right]_{-d/2}^{d/2} = b\left[\left(\frac{d^3}{24} + \cancel{\frac{\overline{y}d^2}{4}}+\frac{d\overline{y}^2}{2}\right)-\left(\frac{-d^3}{24} + \cancel{\frac{\overline{y}d^2}{4}}+\frac{-d\overline{y}^2}{2}\right)\right] $$ $$I_{xx} = b\left[\left(\frac{d^3}{24} + \frac{d\overline{y}^2}{2}\right) + \left(\frac{d^3}{24} +\frac{d\overline{y}^2}{2}\right)\right] = \frac{bd^3}{12} + bd\overline{y}^2$$ $$I_{xx} = \frac{bd^3}{12}+A\overline{y}^2 $$ $$I_{xx} = I_{xx,local}+A\overline{y}^2 $$

The parallel axis theorum is useful as it allows complex sections to be broken down into simple elements that are offset from the centroid of the whole complex section by a known distance. Calculating the individual values of \(I_{xx}\) and \(A\overline{y}^2\) terms for each simple element and adding all these values up allows the \(I_{xx}\) of the whole complex section to be calculated much more easily.

Adding and subtracting Second Moments of Area

Values of second moment of area can be added up or subtracted from one another, allowing simple standard solutions to be combined to find the second moment of area of a more complex shape.

Where the centroids of each shape being superimposed both lie along the axis under consideration then the values of \(I_{xx}\) can simply be added or subtracted. This is particularly useful when calculating the second moment of area of Circular Hollow Sections (CHSs), Square Hollow Sections (SHSs) or Rectangular Hollow Sections (RHSs). An example is given below of calculating \(I_{xx}\) using both first principals and also the subtraction method.

Second moment of area, adding or subtracting
Figure 6: Calculating the second moment of area through subtraction. Note: centroids of both sections lie on the same axis

$$I_{xx} = \int y^2 \; dA$$ $$\text{Let, } D \text{ represent the outside diameter of a circular section and } d \text{ represent the inside diameter of a circular section.}$$ $$I_{xx} = \int_{\theta = 0}^{\theta = 2\pi} \int_{r=d/2}^{r=D/2} r^2sin^2\theta \; r \;dr \;d\theta$$ $$\text{Using the double cosine rule } sin^2\theta \text{ can be written as } \frac{1}{2}(1 - cos2\theta) \text{, hence}$$ $$I_{xx} = \int_{0}^{2\pi} \int_{d/2}^{D/2} r^2sin^2\theta \; r \;dr\; d\theta = \int_{0}^{2\pi} \int_{d/2}^{D/2} \frac{r^2}{2}(1 - cos2\theta) \; r\; dr\; d\theta$$ $$\text{Using standard solutions } \frac{1}{2}(1 - cos2\theta) \text{ integrates to } \frac{1}{2}(\theta - \frac{1}{2}sin2\theta) \text{ hence}$$ $$I_{xx} = \int_{0}^{2\pi} \int_{d/2}^{D/2} \frac{r^2}{2}(1 - cos2\theta) \; r \;dr \;d\theta = \int_{d/2}^{D/2} \frac{r^3}{2}\left[\theta - \frac{1}{2}sin2\theta\right]_{0}^{2\pi} \; dr $$ $$I_{xx} = \int_{d/2}^{D/2} \frac{r^3}{2}\left[(2\pi - \frac{1}{2}sin4\pi)-(0-\frac{1}{2}sin0)\right] \; dr = \int_{d/2}^{D/2} \frac{r^3}{2}\left[2\pi\right] \; dr $$ $$I_{xx} = 2\pi\left[\frac{r^4}{8}\right]_{d/2}^{D/2}=2\pi\left[\frac{D^4}{8\times16} - \frac{d^4}{8\times16}\right]$$ $$I_{xx} = \frac{\pi D^4}{64} - \frac{\pi d^4}{64} $$ $$\text{This is the same solution as } I_{xx,D} - I_{xx,d} $$

Where the centroids of each shape being superimposed do not lie along the axis under consideration then the parallel axis theorum must be used, in order to calculate an \(I_{xx}\) value about the axis under consideration. This is useful in establishing the second moment of area of I-shaped sections:

Second moment of area, adding sections that line on separate axes
Figure 7: Calculating the second moment of area by adding sections together. Note: centroids of both sections lie on the different axes, therefore the parallel axis theorum is needed.

$$I_{xx} = \int y^2 \; dA$$ $$I_{xx} = \sum I_{xx,local} + A\overline{y}^2 = (I_{xx,top} + A_{top}\overline{y}_{top}^2) + (I_{xx,mid} + A_{mid}\overline{y}_{mid}^2) + (I_{xx,bottom} + A_{bottom}\overline{y}_{bottom}^2) $$

It is useful to note that for doubly symmetric I-shaped sections, the same result can also be found by taking the outer dimensions of the section and finding the value of \(I_{xx}\) of the outer rectangle and subtracting the \(I_{xx}\) of an inner rectangle with dimensions equal to the inner height and breadth of the section. This only works for doubly symmetric I-sections as the centroids of both the outer rectangle and inner rectangle lie on the same axis.

Section Modulus of an Area

The final property that is often used for analysing a element under bending is the section modulus. This is defined as:

$$Z_{x} = \frac{I_{xx}}{y} $$ $$\text{Where }I_{xx} \text{ is the second moment of area of a section and } y \text{ is the distance from the elastic neutral axis (passes through the centroid)}$$ $$\text{to the part of the section being considered}$$

The section modulus has it's origins in Euler-Bernoulli Beam bending theory, which is discussed in more details in the articles section Which states :

In the metric system, the typical units for Section Modulus are the mm³. However this can give rise to a large number of digits, values of the order ×10⁶mm³ are not unknown, so some section catalogues will use the more concise cm³ or dm³.

$$\frac{M}{I} = \frac{\sigma}{y} \text{ which can be rearranged as } \sigma = \frac{My}{I}$$ $$\frac{I}{y}\text{ can be grouped together to give a more compact expression } \sigma = \frac{M}{Z}$$

Figure 8: Using section modulus to relate applied moment, M, fibre distance, y, to second moment of area and stress.

When establishing the elastic limits of a section, a limiting stress is applied to any point on the section. Euler-Bernoulli Beam bending theory states that under an applied moment the stress on a section varies linearly from zero at the neutral axis to a peak negative or positive value at the outer fibres (maximum value of y in either direction on a section). Hence the capacity of the section will be limited by either of the outer fibres reaching a given stress. Calculating the value of Z for each outer fibre therefore allows either the applied stress on a section to be established if the applied moment is known (\(\sigma = \frac{M}{Z_{min}}\)) or the elastic moment capacity of the section to be calculated if a limiting stress is known (\(M = \sigma Z_{min}\)).



To Top of Page