You have 5kg of Aluminum and 5kg of Steel. You can make whatever cross section beam you want with these materials. If you have to make a cantilever beam 1m long with a load at the end and you want to minimize the deflection. What material would you use?

Mechanical Engineering Cantilever Beams
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Starting from the basics, let’s go over how to solve this problem! This is a tricky problem since it will challenge initial assumptions we make about material properties.

Our standard deflection equation for a beam is:

Deflection ∝ \frac{PL}{EI}

  1. Load = (P)
  2. Length: Length of the beam (L)
  3. Material: Represents the modulus of elasticity (E)
  4. Cross-Sectional Shape: Represents the moment of inertia (I)

Here in the problem statement we have fixed constraints. We know that load (P) is constant and that the length of the beam (L) is fixed at 1 meter.

Now we can see the variables of interest are Modulus of Elasticity (E) and Moment of Inertia (I). You have to ask here, which variable to tweak here would have more of an effect on minimizing the total deflection?

We know that steel is roughly 3x stiffer than aluminum, so it’s a 3x factor difference for modulus of elasticity (E).

Let’s look at the moment of inertia term, and say that it has a rectangular cross-section:

MomentofInertia = \frac{bh^3}{12}

Since we have a fixed constraint of material (5kg), the important thing to note now is the material density. Steel is around 7800 kg/m^3, while aluminum is 2700 kg/m^3 (i.e smaller cube of steel will weigh the same as a larger cube of aluminum), meaning steel is ~3x denser than aluminum.

If we keep b constant, we can set h_aluminum = 3y and h_steel = y. Once we factor in the cubic term, we’ll see that the momentum of inertia for aluminum is 27 times greater.

Hence, aluminum is the material to choose to reduce deflection because it’s 3x less dense, and this carries over to the cubic factor for height in the moment of inertia term

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