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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Hint: Aluminum vs. Steel Material Properties

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I’d use the steel just because it has a higher Young’s modulus, and probably use an I-beam cross section. Right?

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I think you’re pretty much right.

Deflection is a function of the applied load, the cantilever distance, the modulus of elasticity and the moment of inertia. The first two are not within our control, so it just comes down to the modulus of elasticity and the moment of inertia.

The optimization of the moment of inertia is not unique to any material because we have the same amount of mass for each. So whatever design has the most amount of mass away from the bending axis (I beam is a good candidate) will suffice.

Basically it comes down to the modulus of elasticity. Aluminum’s is 69 GPa, whereas steel’s is 180 GPa. So I’d pick steel.

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I think steel is the wrong choice here. It is true that the moment inertia is not unique to any material. However, because the density of steel and aluminum are different (steel is roughly 3 times denser than aluminum) and because both beams are 1m in length and weigh 5 kg, the cross sectional area of the steel beam must be 3 times smaller than that of the aluminum beam.

Going back to the deflection equation, we still are left with the bending stiffness (EI) as the factors we can choose between. The elastic modulus of steel is roughly 3 times greater than that of aluminum. So now, the question is can we come up with a cross sectional area that is 3 times greater in area but more than 3 times the moment of inertia?

The answer is clearly yes. We can look at a steel rectangular cross section with width x and height y. The aluminum cross section can have width x and height 3y (because of differing densities). Because the moment of inertia increases with the cube of the height, the moment of inertia for the aluminum cross section is 27 times greater than the steel one. This principle would also apply for an I beam.

In short, the specific stiffness (sqrt(E/rho)) of aluminum and steel are roughly the same. Because the mass and the length of the beams are the same, the cross sectional area is different, and we can design a much better cross section with more area.

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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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I’m quite confused here. The formula for deflection is PL^3/(3E*I), but the formula in this post is different. Can you please clarify?

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Hey! You’re correct in providing the specific equation for a cantilever beam with a point load applied. The answer above is providing the general equation, where you have four key variables to modify - Load (P) & Beam Length (L) on the numerator, and Modulus of Elasticity (E) and Area Moment of Inertia (I) on the denominator.

Depending on the supports & loads applied, the four variables will be adjusted by a scaling factor to your beam length (i.e L^3 or L4), and a constant on the denominator.

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@hardwarefyi , thanks for the clarification.
I feel it would remove any confusion if the proportional sign “∝” is used in place of “=” sign
Deflection ∝ PL/EI

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Fixed! Thanks for flagging the confusion. :slight_smile:

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