A Guide To Section Properties

Table of contents

Introduction

In this post I want to walk through the principles behind different common section properties that are required in structural design.

Some properties, like the area of a shape, are naturally intuitive to understand whereas others are more complex and require a bit more thought. If you can walk through all the properties in the order described you will be rewarded with a deeper understanding for section properities for complex shapes and an ability to derive them.

I will often refer to one of the most complex shapes you would see in structural assessments - that of a fillet radius. With I-Sections this part of the section is often ignored due to the complexity added to the shape for minimal gain in the section properties. However if you can understand calculations for a corner you certainly can understand the common simplification of an I-section as 3 combined rectangles.

In this post we cover the area, the centroid and the second moment of area or area moment of inertia of a shape.

Area of a Shape

Formula Summary

$A = \int_{}^{} b(y) \, dy$

$A = \sum A_i$

Area of a Shape Through Integration

$A = \int_{}^{} b(y) \, dy$

The area of a shape is quite intuitive to understand. We draw it on our 2D graph and find the unit area. For something like a rectangle or circle we already know formula to calculate the area but how could we mathematically calculate the area for a complex shape ?

One way could be to draw small strips inside the shape and add up the area of those strips. As we make the strips smaller and smaller our approximation would get better and better and if we made our strips infintesimally small we would get the correct value.

This is similar to the process of integration and we could express the area of the shape mathematically as:

$A = \int_{}^{} b(y) \, dy$

where:

• b(y) is the width as a function of y
• dy represents the infinitesimal height of a strip.

We will come back to this understanding of area later.

Area of a Shape through Superposition

$A = \sum A_i$

A more intuitive way we might calculate an area is to break it up into simpler shapes and add the area of those shapes together.

We can chop it in any way we like, and so long as we add up all the areas of the components of the shape we will get the correct area.

We can divide this shape into 4 squares. If we take each square as being dimensions 1 x 1 we can calculate the area as 4 by adding the area of the 4 shapes together.

The fact that you can break up a shape into smaller shapes, calculate their individual areas, and add them together to get the total area is an example of the principle of superposition. This principle states that a system’s total result is the sum of its individual results, and applies to the calculation of an area.

For the example corner we can use superposition to calculate the area by taking the area of the rectangle and subtracting the area of a quarter-circle.

$A_{rect} = b \times d = 100 \times 100 = 10000$

$A_{qrt,circ} = \pi r^2 / 4 = \pi \times {100}^2 / 4 = 7854$

$A = A_{rect} - A_{qrt,circ} = 10000 - 7854 = 2146$

Centroid

Formula Summary

$\bar{y} = \frac{ \sum A_i\times y_i}{A}$

$\bar{x} = \frac{ \sum A_i\times x_i}{A}$

Concept

The centroid is, as the name implies, the center of our shape.

Of course it is slightly more complicated than that because how would we categorise the center of our green shape above? We could potentially use the red dot which divides the areas based on distance or the blue dot which divides the areas based on, well, area.

The centroid is actually the geometric center of our shape and is similar to the centre of gravity. With the centre of gravity we find a balance point by multiplying the distance to a weight with the centre of that weight to get the spinning action that it causes. The centroid (denoted $\text{ }\bar{x} , \bar{y}\text{ }$ ) is similar except instead of using weight we use area.

Just like with the area we can use the principle of superposition to break up our shape into simpler shapes. From this we then have that for the centroid in the y direction:

$A\times \bar{y} = \sum A_i\times y_i$

If we rearrange this we can then get the centroid for the group. Note that we can take the distance with y with respect to any datum and we will get the same position (just with respect to whatever datum we define).

$\bar{y} = \frac{ \sum A_i\times y_i}{A}$

We can use the same result to determine the centroid with respect to the x direction:

$\bar{x} = \frac{ \sum A_i\times x_i}{A}$

The centroid will also lie on an axis of symmetry for a shape if it has one. For an I Section which is double symmetric the centroid can be entirely determined upon this principle. For a shape which is symmetrical about one axis only we would need to perform the calculation for the axis that is not symmetrical.

Simple Example Calculation

For the example shown above we can take our datum at the left of the image and calculate the x centroid (assuming 1 x 1 squares) as:

$\bar{x} = \frac{ 1 \times 0.5 + 1 \times 0.5 + 1 \times 1.5 + 1 \times 2.5}{4}$ $= \frac{ 0.5 + 0.5 + 1.5 + 2.5}{4}$ $= \frac{5}{4} = 1.25$

This happens to be the position halfway between the blue and red dot that we previously considered.

We can also try the same calculation with a different datum. Lets set our datum to be the position that we calculated as the centroid and see what result we get. We now take our x distances with reference to position 1.25 from the left of the shape.

$\bar{x} = \frac{ 1 \times (0.5 - 1.25) + 1 \times (0.5 - 1.25) + 1 \times (1.5 - 1.25) + 1 \times (2.5 - 1.25)}{4}$ $= \frac{ -0.75 + -0.75 + 0.25 + 1.25}{4}$ $= \frac{ 0}{4} = 0$

Unsurprisingly the result we get is 0 since our centroid is at the position we defined as the datum.

In the example shown we do not need to calculate the y centroid since the shape is symetric about the x axis. If we were to calculate the centroid position from the bottom of the shape we would get a value of 1 as indicated by the shapes plane of symmetry.

Corner Example Calculation

We can calculate the centroid of the corner using superposition with a positive rectangular area and a negative quarter circle area. (Take the radius as being dimension 100).

$\bar{x} = (A_{rect} \times \bar{x}_{rect} - A_{qrt,circ} \times \bar{x}_{qrt,circ}) / A_{tot}$

$A_{rect} = b \times d = 100 \times 100 = 10000$

$A_{qrt,circ} = \pi r^2 / 4 = \pi \times {100}^2 / 4 = 7854$

$\bar{x}_{rect} = b / 2 = 50$

$\bar{x}_{qrt,circ} = 4 r / 3 \pi = 4 \times 100 / 3 \times \pi = 42.44$

$\bar{x} = (10,000 \times 50 - 7854 \times 42.44) / (10,000 - 7854) = 77.66$

Where the centroid of a quarter circle is at $4r / 3 \pi$ (taken from a wikipedia).

Since the shape is only dependent on the variable r we can make the generalisation based on our result that the centroid will always be at position 77.668% of r towards the end with more material.

Second Moment of Area

Formula Summary

General:

$I_x = \int y^2 dA$

$I_y = \int x^2 dA$

Rectangle:

$I_{x} = \frac{b d^3}{12}$

$I_{y} = \frac{b^3d}{12}$

Parallel Axes thereom:

$I_{x} = \bar{I_{x}} + A \times y^2$

$I_{y} = \bar{I_{y}} + A \times x^2$

Concept

The second moment of area is one of the most common engineering properties since it is linearly proportional to the bending deflection of a beam and is required to solve for indeterminate structures.

The second moment of area is different depending on the cardinal direction considered (just like it was with the centroid). Typically we care about the x and y axis directions.

It also depends on where we take our datum. Unlike the centroid where we could take our datum anywhere and get the same position, with the second moment of area our datum MUST be the centroid position.

$I_x = \int y^2 dA$

Just like we did with the area we can break our shape up into infinitesimally small areas to calculate the second moment of area. We multiply each area with the distance from the centroid (purple dot) squared in the axes we are considering. For I_x since we are multiplying by y squared we can group areas at the same y distance together in strips to make things easier to see.

It would of course be much better if we could derive an equation for this shape.

Derivation for a Rectangle

We can derive the most common second moment of area formula for a rectangle of width b and depth d as follows. First we split the integral into its x and y component.

$I_x = \int y^2 dA = \int \int y^2 dx dy$

We should remember that we always take our calculations about the centroid which for a rectangle is at position b/2, d/2. We can add in the limits to the integral based on this datum.

$I_x = \int_{-\frac{d}{2}}^{\frac{d}{2}} \int_{-\frac{b}{2}}^{\frac{b}{2}} y^2 dx dy$

Next we can integrate with respect to x. Since y and x are independent we can treat y as a constant and integrate with respect to x. When we substitute in the limits b/2 and -b/2 we get the constant b which we can move to the front of the equation.

$I_x = \int_{-\frac{d}{2}}^{\frac{d}{2}} [x]_{-\frac{b}{2}}^{\frac{b}{2}} y^2 dy$

$I_x = b \int_{-\frac{d}{2}}^{\frac{d}{2}} y^2 dy$

We can see that actually our position in the x direction is not important for the second moment of area calculation about the x axis it is our position in the y direction that is important for the I_x calculation. If we had moved c_x to be at the left of the rectangle we would have calculated the same result.

Finally we integrate with respect to y.

$I_x = b \times [\frac{y^3}{3}]_{-\frac{d}{2}}^{\frac{d}{2}}$

$I_x = b [\frac{(d/2)^3}{3} - \frac{(-d/2)^3}{3}]$

$I_x = b [\frac{(d^3)}{8 \times 3} - - \frac{d^3}{8 \times 3}]$

$I_x = \frac{b d^3}{12}$

We can use the same process to derive the second moment of area about the y axis.

$I_y = \frac{b^3 d}{12}$

Second Moment of Area for Composite Shapes

In our calculation of the centroid the y that we referenced was any datum of our choosing and the result we got was the distance to the centroid from our datum. For the calculation of the second moment of area we MUST use the centroid as our datum to get the correct value. If we take our datum as the corner of the rectangle instead the result we would get is instead:

$I_x = \frac{b d^3}{3}$

which is not the value we are after and not a fundamentally important value in engineering.

This presents us with a problem. Previously we were able to break up a shape to calculate an area using the principle of superposition. When we now go to break up a composite shape into multiple smaller shapes we are left with different shapes that all have different centroids that do not necessarily line up with the centroid of the entire shape. This means superposition wont work.

Instead we can use the parallel axis thereom to achieve this approach of breaking a shape up into smaller pieces.

The parallel axis theorem states that the I_x contribution of a shape about an axis a distance y from the centroid is:

$I_{x} = \bar{I_{x}} + A \times y^2$

where:

• $\bar{I_{x}}$ is the second moment of area about the centroid of the shape
• A is the area of the shape
• y is the perpendicular distance from the centroid to the axis we are considering

Similarly we also have:

$I_{y} = \bar{I_{y}} + A \times x^2$

We can therefore calculate the second moment of area for a composite shape by using the parallel axis theorem on each individual shape and shifting the values to be at the centroid of the entire shape.

$I_{x} = \sum{\bar{I_{xi}} + A_i \times y_i^2}$

$I_{y} = \sum{\bar{I_{yi}} + A_i \times x_i^2}$

From here onwards we will use the $\bar{I_{x}}$ notation to represent the second moment of area of a shape about its centroid, and $I_{x}$ to represent the second moment of area of a shape about the centroid of the entire shape.

Corner Example Calculation

We know that the centroid for our corner shape is at position 77.66. Lets calculate the second moment of area at this position.

First we get the contribution of the external rectangle.

$\bar{I_{x}} = \frac{b d^3}{12} = \frac{100 \times 100^3}{12} = 8.333 \times 10^6$

$I_{x} = \bar{I_{x}} + A \times y^2 = 8.333 \times 10^6 + 10000 \times (77.66 - 50)^2 = 15.984 \times 10^6$

Next we find the (negative) contribution of the quarter circle. We will use the second moment of area formula for a quarter circle about its centroid.

$\bar{I_{x}} = 0.0549 \times r^4 = 5.490 \times 10^6$

$I_{x} = \bar{I_{x}} + A \times y^2 = 5.490 \times 10^6 + 7854 \times (77.66 - 42.44)^2 = 15.232 \times 10^6$

Finally we combine the two results to get the total second moment of area.

$I_{x} = 15.984 \times 10^6 - 15.232 \times 10^6 = 7.52 \times 10^5$

This is the same result as what we got when we used a slices method but with some differences due to rounding. Since the shape is symmetrical we can see the I_y will be the same value as I_x.

Generalisation for corners

We only are really depending upon one variable which is r so we can expect that we could generalise the formula to help us in future.

If we combine the full formulae we had before we get:

$I_{x,rect} = \frac{r^4}{12} + r^2 \times (0.77688 \times r - 0.5 \times r)^2$

$I_{x,rect} = \frac{r^4}{12} + 0.07666 \times r^4 = 0.1600 r^4$

$I_{x,circ} = 0.0549 \times r^4 + (\pi \times r^2 / 4) \times (0.77688 \times r - 0.4244 \times r)^2$

$I_{x,circ} = 0.0549 \times r^4 + 0.0975 \times r^4 = 0.1524 r^4$

$I_{x} = I_{x,rect} - I_{x,circ} = 0.1600 r^4 - 0.1525 r^4 = 0.0075 r^4$

Actually we can instantly see that this result is correct when we compare to our previous result of 7.52 x 10⁵ was for r = 100.

Second Moment of Area Discussion

The units of the second moment of area are in mm⁴ or in⁴ as we have an area (mm² or in²) multiplied by a distance twice.

If we had a square section I_x would equal I_y however if we increase the depth of the section to be twice its width then we would have I_x = 4 I_y. This goes to show the importance if the orientation of our beam to help in resisting bending and shows why we have separate calculations for I_x and I_y.

Think of a piece of cardboard. It is extremely easy to bend along its flatside but not very easy to bend in its deep direction, in fact the difference in stiffness is so drastic its pretty impossible to bend the cardboard without it buckling.

Putting it all together (150UB18 Example)

Lets now put together what we have learnt in this blog by calculating the properties of a 150 UB 18.0 section.

The section has the following dimensions:

• Depth (d): 155 mm
• Breadth (b): 75 mm
• Flange Thickness (tf): 9.5 mm
• Web Thickness (tw): 6.0 mm
• Fillet Radius (r) : 8.0 mm

Calculating Area

We can add up the components of the shape to get the total area.

$A_{flange} = b \times tf = 75 \times 9.5 = 712.5$

$A_{web} = (d - 2 \times tf) \times tw = (155 - 2 \times 9.5) \times 6 = 816$

$A_{corner} = r^2 - \pi r^2 /4 = 8^2 - \pi \times 8^2 / 4 = 13.7$

$A = 2 \times A_{flange} + A_{web} + 4 \times A_{corner} = 2 \times 712.5 + 816 + 4 \times 13.7 = 2296$

Calculating Centroid

Calculating the centroid of an I Section is extremely easy because it is doubly symmetric so the centroid will be at the center of the shape.

$\bar{x} = b / 2 = 75 / 2 = 37.5$

$\bar{y} = d / 2 = 155 / 2 = 77.5$

We will need the centroid location of our individual shapes however for our second moment of area calculation so we can write those values at here. We can use the generalisation for a fillet radius that we found earlier.

$\bar{x}_{flange,top} = d - t_f / 2 = 155 - 9.5 / 2 = 150.25$

$\bar{x}_{flange,bot} = t_f / 2 = 9.5 / 2 = 4.75$

$\bar{x}_{web} = d / 2 = 155 / 2 = 77.5$

$\bar{x}_{corner,bot} = t_f + (r - 0.77688 \times r) = 9.5 + (8 - 0.77688 \times 8) = 11.28$

$\bar{x}_{corner,top} = d - t_f - (r - 0.77688 \times r) = 155 - 11.28 = 143.72$

Calculating Second Moment of Area

We can calculate the individual contributions to the second moment of area and then sum them up. Since the shape is symmetrical the contribution for the top and bottom flange will be the same.

$I_{flange,x} = \frac{b \times t_f^3}{12} + A_{flange} \times (\bar{x} - \bar{x}_{bot,f})^2 = \frac{75 \times 9.5^3}{12} + 712.5 \times (77.5 - 4.75)^2 = 5358 + 3770950.8 = 3776308.8$

$I_{web,x} = \frac{t_w \times (d - 2 \times t_f)^3}{12} + A_{web} \times (\bar{x} - \bar{x}_{web}) = \frac{6 \times (155 - 2 \times 9.5)^3}{12} + 816 \times (77.5 - 77.5)^2 = 1257728 + 0$

We will calculate the second moment of area for one corner and by symmetry all the other corners will have the same value. We will use the generalisation for I_corner,x that we derived earlier.

$I_{corner,x} = 0.0075 \times r^4 + A_{corner} \times y^2 = 0.0075 \times 8^4 + 13.7 \times (77.5 - 11.28)^2$

$I_{corner,x} = 30 + 60075 = 60105$

Finally we can sum up the contributions to get the total second moment of area.

$I_{x} = 2 \times I_{flange,x} + I_{web,x} + 4 \times I_{corner,x} = 2 \times 3776309 + 1257728 + 4 \times 60105 \approx 9050000$

We can see the contributions and Percentage contribution of each component to the second moment of area in the table below.

Based on the above table in future we may choose to conservatively ignore some parts of the calculation to simplify the process. For example we could ignore the $\bar{I_{x}}$ corner contributions since they are so small and just add in the $Ay^2$ contributions.

We can see that the values are the same as the property values which are given on beamdimensions with minor differences due to rounding.

I Sections (I_x) A General Formula

Based on the example we can see a general formula for the second moment of area for an I Section where we consider the more significant component of the root radius in the calculation that might typically be ignored in simplifications.

To get the contributions of the flange we can just use the $Ay^2$ term:

$I_{flange,x} = b \times t_f \times (d/2 - t_f/2)^2$

For the web we can use the $\bar{I_{x}}$ term since the $Ay^2$ term is always 0 due to symmetry.

$I_{web,x} = t_w \times (d - 2 \times t_f)^3 / 12$

For the corners we can completely ignore OR we can add in the contribution of the $Ay^2$ terms (the other term was negligible). We can consider the centroid for the corner as being at 0.75 x r which is approximately equal to 0.77688 x r but is easier to remember.

$I_{x,corner} = (4r^2 - \pi * r^2) * (d/2 - t_f - r/4)^2$

This gives us the general (improved) second moment of area for a rectangle bending about its major axis as:

$I_{x} = 2 \times b \times t_f \times (d/2 - t_f/2)^2 + t_w \times (d - 2 \times t_f)^3 / 12 + (4r^2 - \pi * r^2) * (d/2 - t_f - r/4)^2$

For the 150UB18 this formula would yield:

$I_{x} = 2 \times 75 \times 9.5 \times (155/2 - 9.5/2)^2 + 6 \times (155 - 2 \times 9.5)^3 / 12 + (4 \times 8^2 - \pi \times 8^2) \times (155/2 - 9.5 - 8/4)^2 = 7541900 + 1257700 + 239300 = 9040000$

which is within 0.1% of the actual value.

Conclusion

We have learnt about the princple of superposition and its application to the area and centroid calculations to derive complex shapes.

We have learnt about how we can use smaller and smaller slices to approximate values like the area and second moment of area. We have also seen how this relates to integration and this is essentially what integration is doing for us.

We have seen how the principle of superposition can not be used for the second moment of area calculation but the parallel axis theorem gives us a simple formula that still allows us to break up complex shapes into simpler shapes that we can process.

We have looked at some simple and some complex examples and how to correctly calculate values for these shapes. By addressing one of the most complex shapes (that of a fillet radius or corner) we were able to undertake the calculations for an I Section in full without any simplifications required. We were able to investigate the contributions to the second moment of area (and thereby bending strength) for the different components of an I section to see what impact ignoring a particular part of the shape might have. We can now see that ignoring the fillet radius may lose us around 3% of the bending strength of the section which is conservative but may result in a less efficient design than if we considered it.