## Elastic Section Modulus Of a Composite Beam System

The concept of a “section modulus” is sometimes not very well understood. Although, most people are familiar with the terms “mass moment of inertia”, “second moment of area” or “area moment of inertia” etc. The “section modulus” of a structural member or a built up beam system is a geometric indicator of how efficiently the part or system was designed.

Basically, the section modulus is the ratio of the overall area moment of inertia to the most extreme fiber distance (y or c) from the overall bending neutral axis of a part or beam system.

Zxx = Ixx / ymax_y

Zyy = Iyy / ymax_x

The bending stress is also expressed in terms of the section modulus:

That is, Bending Stress Sigma = My/I = M / (I/y) = M / Z

For a particular material or set of materials:

The higher the section modulus for the same total cross sectional area, the more efficient and optimized the design is.

Examples of built up beam systems:

• Built up shear web and beam systems with various bays and stiffeners
• Frames with channels and doublers
• Box beam systems
• Spars with caps
• Floor beams
• Seat Track Beams etc.

In this post, we will learn how to use classical hand calculation methods to calculate the section modulus of a sample shear web system. Specifically, we will look at a doubly symmetric composite beam system for simplicity. Hence, the overall section CG would be at the mid height of the system.

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But first of all, let us look at what our beam system is composed of.

## Section Modulus – Composite Beam System Example

The figure below illustrates the various components in our beam system. The upper and lower caps comprise a doubler and two L angles, and the beam center web (shear web) is essentially a sheet sandwiched between the upper and lower cap components.

It is noteworthy that this type of a construction is typical in light weight aerospace structures. In other words, the main idea is to use simple parts and the concept of “separation” to achieve a high section modulus relative to the weight of the net section. Also note that the phantom lines in the figure indicate where the fasteners (or rivets) join the parts together. In this example however, we will focus on the x’-x’ bending neutral axis calculations.

## Section Modulus – Calculation Steps

So, the basic sequence of calculation steps is as follows:

1. First, break up the parts into rectangular (or near) segments
2. Then label each segment
3. Next, choose a local coordinate system that is convenient and define the datum (x’-x’ Vs y’)
• Typically, this is at a bottom corner
4. Following that, for each segment, list the width and height dimensions
5. Similarly, for each segment, list the segment CG (centers of the rectangles) height w.r.t the datum zero along vertical axis, y’
6. Next, calculate the areas of each segment, A
7. Then Calculate the product values of A*y’ and A*y’^2
8. Then, calculate the area moment of inertia of each rectangular segment about its own CG horizontal bending neutral axis, Ix’x’
9. After that, using the parallel axis theorem, account for the separation of each area w.r.t to the local datum zero by adding Ix’x’ with A*y’^2
10. Next, sum all the calculated values in the bottom row
11. In order to determine the overall section CG location along vertical axis y’, divide the sum of A*y’ with the sum of A.
• This is distance y (or ‘c’, commonly used in the Bending Stress = Mc/I equation, the extreme fiber distance) of the overall system CG from the datum zero
12. Then, to determine the overall system’s area moment of inertia, use (Sum Ix’x’-Sum A*y^2), about the overall section’s CG horizontal bending neutral axis, Ixx
13. Next, calculate the section radius of gyration using sqrt(Ixx/Sum A)
14. Finally, calculate the x-x section modulus of the composite beam system using Zxx = Ixx/y

## Conclusion:

This is a basic example of a doubly symmetric composite beam system. However, if the components are larger in terms of area to the top side, then the 'y' we calculated will be the largest fiber distance from the bottom datum vertical zero (the CG shifts up in this case). On the contrary, if it is bottom heavy, then the largest 'y' or 'c' would be the total system height minus the calculated 'y'.

Furthermore, if the system is unsymmetrical, then other variable come into play, such as:

• Calculate Iyy
• You need (both x and y) for the overall system centroid
• Principal axes (about which the product of inertias is zero) may very well be rotated with the horizontal
• The angle of the principal axes with horizontal
• Shear center offset
• Polar moment of inertia
• etc.

We will look into the above additional terms in the next blog post considering an open unsymmetrical composite beam system.

So that is all for now folks, what did you think about this post? Comment below the post....

Cheers...

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Surya Batchu

Surya Batchu is the founder of Stress Ebook LLC. A senior stress engineer specializing in aerospace stress analysis and finite element analysis, Surya has close to two decades of real world aerospace industry experience. He shares his expertise with you on this blog and the website via paid courses, so you can benefit from it and get ahead in your own career.