The moments of inertia of flat figures – Steiner theorem (example)

In this article, I will present the calculations of the parameters of a composite cross-section. In the example, we will determine the positions of the centroids of simple shapes. Knowledge of these will be necessary for calculating the moments of inertia using Steiner’s theorem.

The following steps:

1. we will begin by analyzing the simple cross-sections,
2. determine the position of the centroid of the cross-section,
3. using Steiner’s theorem, we will calculate the moments of inertia with respect to the central axis,
4. we will conclude with the calculations of the principal moments and axes.

In the previous article, I presented the procedure for calculating moments of inertia for symmetrical shapes. If you haven’t read it yet, I encourage you to catch up.

In this example we will deal with the calculations of a much more complex cross-section:

Just like before, we will start with a simple classification of the cross-section.

In this case, to put it briefly — in terms of calculations, we are in a bit of a tough spot. The cross-section is definitely not a symmetrical figure — we will have to calculate everything from scratch. Additionally, it includes students’ favorite simple shapes — a semicircle and a quarter circle — determining the centroids and central moments of inertia for these shapes will not be pleasant ????. Nevertheless, we will tackle this cross-section, and I am confident we will emerge victorious from this battle.

Let’s start by summarizing the data for the simple shapes.

Dividing into simple shapes

The division of the cross-section is visible in the first diagram — we have a right triangle, a quarter circle, and a cut-out semicircle. Let’s now gather the data on their dimensions and determine the locations of their centroids and other parameters in the given coordinate system.

For the right triangle b=2a, h=a, the centroid along the z-axis is 1/3 of the width, and along the y-axis, it is the sum of the quarter circle’s radius and 1/3 of the height:

c_{z1}=\frac 1 3 b=\frac 23 a,c_{y1}=2a+\frac1 3⋅h=2\frac 13 a

Other parameters include the area and central moments of inertia and deviation relative to its centroid (note the negative sign for the moment of deviation, which results from the triangle’s geometric configuration):

A_1=\frac 12 bh=\frac12⋅2a⋅a=a^2,I_{zc1}=\frac {bh^3} {36}=\frac{2a⋅a^3}{36}=\frac1 {18} a^4

I_{yc1}=\frac {b^3h} {36}=\frac{(2a)^3⋅a}{36}=\frac2 9 a^4, D_{zy1}=\frac {b^2h^2} {72}=\frac{(2a)^2⋅a^2}{72}=\frac1 {18} a^4

For the quarter circle R=2a, the centroid along the z-axis will be determined using the general formula (below), and along the y-axis, we subtract the value from the position of the quarter circle’s centroid (2a):

c_{z2}=\frac {4R} {3π}=\frac{4⋅2a}{3π}=\frac 8{3π}  a≈0.84883a,c_{y2}=2a-\frac{4R}{3π}=\big(2-\frac 8{3π}\big)a≈1.15117a

The area is, of course, 1/4 of the area of a circle. As for the central moments — the formulas and descriptions can be found in our article Introduction to Cross Section Analysis (1). Also, be mindful of the sign for the moment of deviation.

A_2=\frac14 πR^2=\frac 14 π(2a)^2=πa^2≈3.1416a^2

I_{zc2}=I_{yc2}=r^4 \Big(\frac π{16}-\frac 4{9π}\Big)=(2a)^4⋅\Big(\frac π{16}-\frac 4{9π}\Big)≈0.87806a^4

D_{zy2}=-r^4 \Big(\frac 1 8-\frac 4{9π}\Big)=-(2a)^4⋅\Big(\frac π{16}-\frac 4{9π}\Big)≈0.263537a^4

For the semicircle r=a, the centroid along the z-axis will be determined using the general formula (below), and along the y-axis, we will read it from the diagram:

c_{z3}=\frac {4r}{3π}=\frac {4⋅a}{3π}=\frac 4{3π}a≈0.42441a,c_{y3}=1.5a

The area is, of course, 1/2 of the area of a circle. The remaining formulas can also be found in the article above. The moment of deviation for a symmetrical figure in the given coordinate system will be 0.

A_3=\frac 12 πr^2=\frac12 πa^2≈1.5708a^2,D_{zy3}=0

 I_{zc3}=\frac {πr^4}8=\frac {πa^4}8=0.3927 a^4,I_{yc3}=r^4 \Big(\frac π8 -\frac 8{9π}\Big)=a^4⋅\Big(\frac π8-\frac 8{9π}\Big)≈0.10976a^4

Phew… We’ve made it through, and the worst is behind us. If we have determined all the parameters for the simple shapes, all that’s left is the tedious process of plugging numbers into formulas. Sure, it’s not the most pleasant part and it’s easy to make mistakes — which is why we created the EquiBeam app ????. Now, let’s calculate the centroid’s position.

Determining the centroid

We will use the same formulas we always use for these calculations — calculating the area, first moments, etc. The key thing to remember is that the cut-out semicircle will always be subtracted in the calculations.

A=a^2+3.1416a^2-1.5708a^2=2.5708a^2

S_z=A_1 c_{y1}+A_2 c_{y2}-A_3 c_{y3}=a^2⋅2 \frac13 a+3.1416a^2⋅1.15117a-1.5708a^2⋅1.5a=

=2 \frac13 a^3+3.616522a^3-2.3562a^3=3.59366a^3

S_y=A_1 c_{z1}+A_2 c_{z2}-A_3 c_{z3}=a^2⋅\frac 23 a+3.1416a^2⋅0.84883a-1.5708a^2⋅0.42441a=

=\frac2 3 a^3+2.66667a^3-0.666667a^3=2 \frac23 a^3

Finally, we calculate the centroid’s location:

c_y=\frac {S_z}A=\frac {3.59366a^3}{2.5708a^2}=1.39788 a,c_z=\frac {S_y} A=\frac {2.66667a^3}{2.5708a^2}=1.03729a

The position of the centroid and the central coordinate system aligned with the initial system (as well as the principal system, which we will determine shortly) is shown below:

Now we are left with the final calculations related to the moments of inertia and deviation.

Calculating central moments of inertia -Steiner theorem

How should I put this — no matter how unpleasant the input data may be, the formulas for calculating the central moments of inertia remain the same (and always are).

We remember the moments of inertia for all simple shapes, add the effects of the Steiner theorem, and account for the signs associated with whether a shape is added or subtracted from the cross-section:

I_zc=(I_zc1+A_1⋅(c_y1-c_y )^2 )+(I_zc2+A_2⋅(c_y2-c_y )^2 )-(I_zc3+A_3⋅(c_y3-c_y )^2 )=

=(\frac1{18} a^4+a^2⋅(2\frac13 a-1.39788 a)^2 )+

+(0.87806a^4+3.1416a^2⋅(1.15117a-1.39788 a)^2 )-

-(0.3927 a^4+1.5708a^2⋅(1.5a-1.39788 a)^2 )=1.59081a^4

I_{yc}=(I_{yc1}+A_1⋅(c_z1-c_z )^2 )+(I_{yc2}+A_2⋅(c_z2-c_z )^2 )-(I_{yc3}+A_3⋅(c_z3-c_z )^2 )=

(\frac 29 a^4+a^2⋅(\frac23 a-1.03729a)^2 )+

+(0.87806a^4+3.1416a^2⋅(0.84883a-1.03729a)^2 )-

-(0.10976a^4+1.5708a^2⋅(0.42441a-1.03729a)^2 )=0.64945 a^4

One topic that wasn’t discussed in the previous article is calculating the moment of deviation.

These calculations are done similarly to the moments of inertia — we add the values from the simple shapes and apply the Steiner theorem. However, unlike the moments of inertia, moments of deviation can be negative, so we need to be particularly careful with the signs.

If the centroid of a simple shape is in the 1st or 3rd quadrant of the central coordinate system, the Steiner deviation moment must be positive; if it’s in the 2nd or 4th quadrant, it will be negative.

For this reason, it’s most convenient to use the following formula:

D_{zcyc}=(D_{zy1}+A_1⋅(c_{z1}-c_z )(c_{y1}-c_y ))+(D_{zy2}+A_2⋅(c_{z2}-c_z )(c_{y2}-c_y ))-(D_{zy3}+A_3⋅(c_{z3}-c_z )(c_{y1}-c_y ))=

\Bigg(-\frac1{18} a^4+a^2⋅\Big(\frac23 a-1.03729a\Big)\Big(2 \frac13 a-1.39788 a\Big)\Bigg)+

+(0.263537a^4+3.1416a^2⋅(0.84883a-1.03729a)(1.15117a-1.39788 a))-

-(0+1.5708a^2⋅(0.42441a-1.03729a)(1.5a-1.39788 a))=0.10566a^4

Having calculated all the unpleasant formulas that needed to be covered in this article, we observe that the moment of deviation is not zero, which means that besides the chosen central system for the calculations, we can also determine the principal central system, relative to which strength calculations for the cross-section should be performed (if we want to use standard formulas for stress).

Calculating principal central moments of inertia

Fortunately, calculating the principal moments of inertia and locating the principal axes is very simple. We use the formulas associated with Mohr’s circle:

I_{1,2}=\frac{I_{zc}+I_{yc}}2±\sqrt{\Big(\frac {I_zc-I_{yc}}2\Big)^2+D_{zy}^2}

We simply substitute the data:

I_1=\frac {1.59081a^4+0.64945 a^4}2+\sqrt{\Big(\frac {1.59081 a^4-0.64945 a^4}2\Big)^2+(0.10566a^4 )^2}

I_1=1.60253a^4

I_2=\frac {1.59081a^4+0.64945 a^4}2-\sqrt{\Big(\frac {1.59081 a^4-0.64945 a^4}2\Big)^2+(0.10566a^4 )^2}

I_2=0.63773a^4

We can observe that the principal moment I₁  is slightly larger than the moment I_zc, while the moment I₂  is slightly smaller than the moment I_yc.

As for the principal axes, we determine one angle and interpret it:

2φ=arctg\Bigg(\frac{-2D_{zc}}{I_{zc}-I_{yc} }\Bigg)=arctg⁡\Bigg(\frac{-2⋅0.10566a^4}{1.59081 a^4-0.64945 a^4}\Bigg)→φ=-6.32613°

The result means that one of the principal axes will be rotated by an angle φ relative to the current central coordinate system. It’s worth noting that if I_zc>I_yc the angle points to axis I₁, whereas if I_yc>I_zc the angle points to axis I₂. In our case, we are dealing with the first situation. You could have seen the diagram with the principal system above ????. That’s all for now.

For more examples, check out the EquiBeam app!

The moments of inertia for flat figures (cross-sections) are, along with internal forces, the most important parameters in strength analysis. The examples in this article showed how to correctly calculate central and principal moments of inertia for different types of cross-sections, allowing for a better understanding of a structure’s resistance to bending.

Mastering this skill is a crucial step in structural strength analysis and design. With the presented calculations and examples in the EquiBeam app, you can confidently proceed to more advanced calculations.