This time, we will explain how to actually calculate the second moment of area.
In actual design work, the second moment of area can be easily calculated using a command in CAD software.
However, when designing a new structure, if you were to create accurate cross-sectional drawings or 3D models using CAD for every shape you came up with during the planning stage, you wouldn't have enough time.
That's why it's important to be able to visualize, "For this cross-section, the second moment of area will be approximately this much."
At the very least, it is desirable to be able to instantly visualize whether the second moment of area is high or low in relation to the direction in which the external force is applied.
First, before discussing the second moment of area, let's briefly explain the mathematical properties of the first moment of area.
Properties of the first moment of area
First, let's recall the definition of the first moment of area. A is the area.
First moment of area = \int_{A}ydA
In the properties explained last time, if we take the center of the beam as the origin, the first moment of area becomes 0 (the sum of the bending stresses in the cross-section is 0).
First moment of area = \int_{A}ydA=0
Let's examine this characteristic mathematically.
If the cross-section is rectangular and the coordinate origin is the center G of the beam, the calculation of the first moment of area will look like the diagram below.
Since the quadrilateral is symmetrical in all directions around the origin G, the first moment of area obtained by integrating with respect to the area is 0.
When we actually calculate the first moment of area, as shown in the figure below, the areas of the zy-plane and the -zy-plane are dA, and the areas of the zy-plane and the -zy-plane are -dA, so they cancel each other out and become 0.
The origin coordinates where the first moment of area is zero are called the centroid (the centroid, neutral axis, and neutral plane are important concepts).
In other words, the first moment of area with the centroid G as the origin is 0.
As shown above, when the centroid is known, the first moment of area is 0 by definition. However, let's consider the case where the centroid is unknown.
Let's consider the first moment of area with an arbitrary coordinate system that does not pass through the centroid as the origin. Let O be the origin of the arbitrary coordinate system and let (x1, y1) be the new coordinate system.
Let's assume the centroid is G (yG, xG) and place a coordinate system (x, y) in the coordinate system shown in the diagram above.
In this case, the coordinate system (x, y) of the centroid can be expressed as follows.
$ x=x1-zG $
$ y = y1 - yG $
Substitute this into the first moment of area of the centroid coordinate system (x, y).
$ \int_{A}x=\int_{A}x1dA-\int_{A}xG dA$
$ \int_{A}ydA=\int_{A}y1dA-\int_{A}yGdA $
The first moment of area in the coordinate system of the centroid is 0, and $ \int_{a}yGdA$ about the centroid in the coordinate system (x1, y1) is yGA (where A is the area of the rectangle).
$ 0 = \int_{A} x 1 d A - x GA $
$xG=\frac{\int_{A}x1dA}{A} $
$ 0 = \int_{A} y 1 d A - y GA $
$ yG=\frac{\int_{A}y1dA}{A} $
Therefore, the coordinates of the centroid can be found by dividing the first moment of area by the area. In other words, if the centroid is unknown, it can be found if the first moment of area and the area of a suitable coordinate system are known.
Conversely, if the centroid is known, the first moment of area of a suitable coordinate system can be easily determined.
$ \int_{A}x1=xGA, \int_{A}y1=yGA $
I'll explain this in more detail later in terms of mechanical dynamics, but you can think of it as an area-based version of an object's center of mass.
The center of gravity of an object is the centroid of the cross-sectional area.
Next, let's look at how to calculate the second moment of area.
Calculate the second moment of area
Let's start by reviewing the second moment of area.
By definition
The second moment of area in the y-axis direction is Iy = \int_{A}z^2dA
You might be wondering, "How do you integrate an area here?" but there's a proper way to do it.
Let's actually calculate some of the second moment of area.
Calculate the second moment of area of a rectangular cross-section.
First, let's find the case where the cross-section is square.
As an example, let's consider the second moment of area in the z-axis direction for a cross section with width b, height h (half of which is $ \frac{h}{2}$), and centroid at the origin of the coordinate system.
hereHere's the key point: Define the area using the direction perpendicular to the axis you're looking for as a variable.
In this example, the axis we want to find is the Z-axis, so the variable is the y-axis. When we express the area using the y-axis, the width is b, so it becomes by. Therefore, the infinitesimal cross-sectional area dA that we integrate can be expressed by the following equation.
$ dA = b × dy $
Therefore, if we apply this to the formula for the second moment of area,
$ Iz=\int_{A}y^2dA=b\int_\frac{-h}{2}^\frac{h}{2}y^2dy $
And if you solve this then
$ Iz=\frac{bh^3}{12} $
Becomes
hereThe key point is that if the wall thickness is in the y-axis direction, the second moment of area increases by the cube of that thickness.
As the second moment of area increases with the cube of the beam, the bending stress generated within the beam's cross-section decreases drastically (as discussed previously, the bending stresses are $σ1=-\frac{M}{Iz}h1, σ2=\frac{M}{Iz}h2$).
Determining the second moment of area of an I-shaped section.
Next, let's consider an I-shaped cross-section.
The example setup is as shown in the diagram. Let's find the second moment of area Iz about the Z-axis.
Unlike the example above, the width of the cross-section changes midway through, so a little ingenuity is required.
Looking at the cross-section, there are two types, so we will consider each separately.
Next, we define the infinitesimal area in each y-axis direction and calculate the second moment of area.Since the shape is symmetrical with respect to the z-axis, we only need to calculate half of it and then multiply by 2 later.
The cross-sectional area of centroid G and centroid G1 in a small interval on the positive side in the y-axis direction can be calculated using the following formulas.
The cross-sectional area of G1 is tdy(0≦y≦\frac{h}{2}-c)
The cross-sectional area of G2 is bdy(\frac{h}{2}-c≦y≦\frac{h}{2}).
Applying this to the definition of the second moment of area,
$ Iz=\int_{A}y^2dA=2t\int_{0}^{\frac{h}{2}-c} y^2dy+2b\int_{\frac{h}{2}-c}^{\frac{h}{2}} y^2dy $
When you calculate
$ Iz=\frac{bh^3}{12}-\frac{(bt)(h-2c)^3}{12} $
Becomes
like thisEven with an I-shaped cross section, the second moment of area increases as the y-axis direction h is cubed.So, be aware that this will happen.
Polar moment of inertia of area
Up until now, calculating the infinitesimal area was easy because the value in the Z-axis direction was a constant (like b or t). However, when the Z-axis direction also changes, the equation becomes complex and the calculation becomes quite troublesome. Therefore, we will consider a method to simplify the process by utilizing the properties of the second moment of area.
This new property is called the polar moment of inertia of the area. It's not difficult at all; in fact, it's quite simple.
Up until now, we've been thinking about a cross-section of a beam and considering it using two coordinate systems, Z and Y. But now, we just need to add an x-axis perpendicular to the Z and Y coordinates and consider it using the distance r around that axis.
Anyone with mathematical knowledge would see this as simply a conversion from Cartesian coordinates to polar coordinates.
Now let's consider the second moment of area Ip with respect to the x-axis.
$ Ip=\int_{A}r^2dA $
And thisThis is called the polar moment of inertia of the area.
Moreover, since $r^2=y^2+z^2$ (considering r as a vector)$, substituting this into the above equation gives
$ Ip = Iy (second moment of area with respect to the y-axis) + Iz (second moment of area with respect to the z-axis) $
In other words, the polar moment of inertia of an area can be found by adding the second moments of area of two perpendicular axes of any cross-section.
This property allows us to easily determine the second moment of area of a circular shaft.
Second moment of area of a round bar
Let's set up an example problem.
Set the Z, Y, and X axes with the centroid of the cross-section of a round bar with diameter d as the origin.
Now let's consider the polar moment of inertia of the area.
From the diagram, the area of a small circle at a distance r from the origin can be defined as (circumference × width of the circle).
$ dA=2πrdr $
Substituting this into the definition of the polar moment of inertia of area,
$ Ip=\int_{A}r^2dA=2π\int_{0}^{\frac{d}{2}}r^3dr=\frac{πd^4}{32} $
Here, due to the symmetry of the cross-section, $Iy=Iz$
$ Iz=\frac{Ip}{2}=\frac{πd^4}{64} $
Becomes
What's important here is the case of a round bar,This means that the second moment of area increases with the fourth power of the diameter.
In other words, Increasing the diameter significantly reduces the bending stress (as discussed previously, the bending stresses are $σ1=-\frac{M}{Iz}r1, σ2=\frac{M}{Iz}r1$)..
Furthermore, even if it is hollowed out, the second moment of area does not decrease that much.
For example, the second moment of area of a hollow shaft with outer diameter d1 and inner diameter d2 is
$ Iz=\frac{πd1^4}{32}-\frac{πd2^4}{64} $
Since $d2^4≪d1^4$, a little weight reduction won't affect it at all.
Just like with the torsion of a round bar, the bending of a round bar's cross-section is highly dependent on its diameter, and the stress generated hardly changes even if it is hollow.
In other words, if cost allows, making the shaft hollow will result in a lighter yet reasonably strong structure.
Next, and finally, I will explain the mathematical properties of the second moment of area.
Parallel axis theorem
Let's set up an example to explain the parallel axis theorem.
A cross-section has a coordinate system centered on its centroid G, and an axis z' that passes through centroid G and is parallel to the z-axis, at a distance e from it. The area of the cross-section is A.
Now let's find the second moment of area Iz' with respect to the axis z'.
$ Iz'=\int_{A}y^2dA=\int_{A}(y+e)^2dA=\int_{A}y^2dA+2e\int_{A}ydA+e^2\int_{A}dA $
The expanded $ \int_{A}y^2dA$ is the second moment of area with respect to the z-axis, and $ \int_{A}ydA$ is the first moment of area at the centroid, which is zero. The remaining $ \int_{A}dA$ is simply the area, so it becomes as follows.
$ Iz'=Iz1+e^2A $
This is called the parallel axis theorem.
Basically, to find the second moment of area at a parallel axis separated by a certain distance e, you just need to add $e^2A$ to the original area.
Incidentally, although explained using mechanical dynamics, the moment of inertia has exactly the same properties.
Now, let me show you how to actually use it.
Second moment of area of an I-shaped section, parallel axis version
Now let's consider the second moment of area of the I-shaped cross section, which is exactly the same as the solution explained above.
By applying the concept of the parallel axis theorem, we can arrive at the idea shown in the diagram below.
First, the second moment of area Iz of the green bar in the middle diagram is obtained from the second moment of area of the bar $ \frac{bh^3}{12}$.
$ Iz=\frac{t(h-2c)^3}{12} $
If we consider the second moment of area Iz' of one of the blue squares in the figure around the z' axis, it is the same as above.
$ Iz'=\frac{bc^3}{12} $
Becomes
Here, we use the parallel axis theorem to transform the green area moment of inertia around the z-axis.
The area of the blue bar is bc, and the distance between the axes is $\frac{hc}{2}$, so by the parallel axis theorem...
$ Iz=Iz’+e^2A=\frac{bc^3}{12}+(\frac{(h-c)}{2})^2bc=\frac{bc(3h^2-6hc+4c^2)}{12} $
Therefore, the total area is obtained by adding the second moment of area of the blue section and the second moment of area of the two green sections (calculation omitted).
$ Iz=\frac{bh^3}{12}-\frac{(bc)(h-2c)^3}{12} $
It can be obtained relatively easily.
Summary
Up to this point, we have explained how to calculate the first moment of area, how to find the centroid and calculate the second moment of area, the two-axis polar moment of area, and the parallel axis theorem.
- The first moment of area is 0 when the coordinate axis is set with the centroid (neutral plane) as the origin.
The centroid can be found by dividing the first moment of area by the area (and vice versa).
The second moment of area can be found by defining an infinitesimal area with respect to the axis and integrating over that area.
The second moment of area of a rectangular cross-section can be found using $ \frac{bh^3}{12}$ (with respect to the h direction).
The second moment of area of a circular cross-section can be calculated as $ \frac{πd^4}{64}$ (where d is the diameter).
The cross-section of most structures is made up of a combination of circles and squares.No matter how complex it is, if you remember the second moment of area for circles and squares, you can find the answer using the parallel axis theorem.
The second moment of area for typical cross-sections will be introduced later on a separate page.
These days, calculations can be done using CAD software, so you rarely need to do them yourself.
However, the second moment of area increases exponentially with increasing axial dimensions (by the cube of the cube, or by the fourth power for a circular axis), so be sure to check the load direction and the dimensions of the cross-section of the designed structure.
No matter how wide it is, it doesn't really make much difference, so let's focus on making it taller.
That's all for now. Next time, we'll calculate the deflection of the beam.
To those who found this article helpful in understanding design:
While there is basically no textbook covering this content, and it is my own original work, I will introduce the textbook that I have been using since I was a student.
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