# Calculating deflection of a cantilever beam using Young's modulus

I have looked at a lot of deflection questions on here and I didn't see anything that quite matched what I needed to know.

I am working through some curriculum for students and I am working through the equations given for beam deflection given Young's modulus and the moment of inertia. This is a class for kids who have had algebra at the very most, so these things are going to be very (over)simplified -- no calculus. The point is just to show how it is possible get some first approximations.

That out of the way, I see this formula for beam deflection:

$$\delta = \frac{FL^3}{3IE}\\\\$$

Well and good, as $$\delta$$ is the deflection, F is the force, L is the length of the beam. I is the moment of inertia. E is Young's modulus.

So I tried a sample problem. I will use Young's modulus of pine wood, given as 8500 MPa with a quick google search. (Yes I know that there's all kinds of ways it isn't constant and is affected by moisture or whatever, that isn't the point of the exercise).

I assumed a beam length of 2 meters. And that I put a 10 kg mass on the end of it. We assume it is supported only at one end. The 10 kg mass will be 98 N of force.

To get the moment of inertia I used $$I= \frac{bh^3}{12}$$ and the size of a 2x4 (converted to meters). This gets me

$$b = 1.5" = 3.81 \text{cm} = 0.0381 \text{m}\\ h = 3.5" = 8.89 \text{cm} = 0.089 \text{m} \\ I = \frac{(0.0381 \text{m})(0.089 \text{m})^3}{12}=\frac{2.69 \times 10^{-5} \text{m}^4}{12}=2.23 \times 10^{-6} \text{m}^4$$

From here I simply plug in the numbers: $$\delta =\frac{FL^3}{3IE} = \frac{(98 \text{N})((2\text{m})^3)}{3(2.23 \times 10^{-6}\text{m}^4)(8500 \text{MPa})}$$

Since our modulus is in megapascals I will convert that to Pascals and then to N per meter squared, though it won't be in "proper" scientific notation for convenience's sake

$$\delta =\frac{FL^3}{3IE} = \frac{(98 \text{N})((2\text{m})^3)}{3(2.23 \times 10^{-6} \text{m}^4)(8500 \times 10^6 \text{N/m}^2)}=\frac{(98 \text{N})(8\text{m}^3)}{(6.69 \times 10^{-6}\text{m}^4)(8500 \times 10^6 \text{N/m}^2)}=\frac{(784 \text{Nm}^3)}{(56865 \text{m}^4 \cdot \text{N/m}^2)} \\ =0.0138 \text{m}$$

This all seems to make perfect sense, and I wanted to make sure I wasn't doing something terribly wrong; 0.014 meters (about 1.4 cm) seems a perfectly reasonable amount of deflection for a 10kg weight on the end of a 2 meter 2x4 oriented so that the short side is the base.

So really, what I am asking is if this all looks kosher to a first approximation, and that I haven't done anything wildly wrong. (As a class we will do some comparisons of say, steel, and the different MoI of differing shapes).

• It seems Ok to me. May 14 at 20:37