# How to calculate deflection of a simple beam with a load in the center?

Given this type of arrangement:

Where the beam is of rectangular cross-section.

I am specifically thinking of a wooden beam where I can look up some material characteristics (e.g. modulus of elasticity) from places such as here

If I only know

• Force
• Length between supports
• Width and thickness(height) of beam
• Modulus of elasticity of beam material

Wikipedia suggests I also need to know "Area moment of inertia of cross section" but I have no idea how to obtain this.

Do I have enough information to calculate deflection?

Footnote: I ultimately want to know how thick the beam should be so that it won't break if I sit on it - but to avoid asking unacceptably many questions at once, I have broken my problem up into simpler and more general questions.

• It is worth noting that the question you ask here and what you actually want to know are not necessarily the same. A beam can deflect significantly and not break. Or it can barely budge and shatter. Checking for deflections is considered a "service limit state", since if the beam deflects too much you won't feel safe sitting on it. Checking to see if the beam will hold your weight is an "ultimate limit state". This is the really important one for safety and the methods for each limit state are different. If you wish for me to expound on the subject, please edit your question accordingly. – Wasabi Oct 16 '15 at 17:14

I am specifically thinking of a wooden beam... Wikipedia suggests I also need to know "Area moment of inertia of cross section" but I have no idea how to obtain this.

If you're using standard wood dimensional lumber sizes, the American Wood Council (AWC) publishes a table of different sizes vs. section properties in the 2015 NDS Supplement, Section 3.1. For example, consider the screenshot below If you were using a 6x12 beam, its moment of inertia, $I_x$, is 697.1 in4 about its major axis. This is one part of the measurement of the beam's stiffness, or its resistance to deflection - the other part being its modulus of elasticity, as you've already identified.

If you don't already have the formulation, the deflection of a simply-supported beam with a point load at midspan is given by the following (per the AISC 14th Ed. Steel Construction Manual): You are solving for the variable $\Delta_{max}$. As you've stated, you have all of the parts needed to calculate deflection except for $I$. @GlenH7's answer already gives you the formulation to determine $I$ if you have a shape not covered by the NDS Supplement, so I won't repeat that information here.

Note that the NDS contains a code for designing wood structures. This, or another similar code, is usually the code of record for wooden structures designed in the United States. There are additional provisions to consider if you are doing a "proper" deflection analysis per the code that are beyond the scope of this Question, but I get the feeling that you don't really need to go that in-depth.

For what it's worth, this type of scenario is very easily worked out by trial and error if all you're trying to do is see if a wooden beam is OK to sit on. And given how cheap dimensional lumber is (in the United States, at least), I would just err on the side of caution and get a bigger size than you think you need and it will probably work.

• If wood is to be used, as RedGrittyBrick has indicated, and if this is to be a real project with a real purpose, then the grade and orientation of the grain becomes critical too. The modulus of elasticity, as the only material property given, would not be enough. – AsymLabs Oct 22 '15 at 9:04
• @AsymLabs, There are additional provisions to consider if you are doing a "proper" deflection analysis per the code that are beyond the scope of this Question This was meant to be my catch-all for that type of consideration. This isn't really the format to get into that specific level of detail for wood design. – grfrazee Oct 22 '15 at 13:25

Yes, you likely have enough information in order to solve your equation.

Have a look at this article listing various area moment of inertias

I would use the rectangular area equation for your bench.

a filled rectangular area with a base width of b and height h Which has an area moment of inertia of: $I = b*h^3/12$

In order to apply that, take a reasonable amount of deflection (say $1/4^"$ or so) and work the equations backwards in order to determine the minimum area moment of inertia that's required.

From there, one of the two dimensions of your material is likely fixed. So you can determine the remaining dimension based upon the area moment of inertia equation.