Don’t be stumped when you have questions that seem out of the blue! Companies often like to disguise beam bending questions in the form of an application or hypothetical situation.
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This is a classic beam deflection question. Think of the general stress and strain equations:
Stress Equation
Stress: σ = \frac{force}{area} = \frac{P}{A}
Strain Equation
Strain: ε = \frac{displacement}{length} = \frac{ΔL}{L} = \frac{δ}{L}
Hooke’s Law
σ = Eε
- A is the cross-sectional area of the stool leg
- P is the load
- δ is the displacement in length
- L is the length of the stool leg
- E is the material’s Young’s Modulus
- ε is the strain
You can rearrange these equations to find a relationship using the variables we know from the question.
δ ∝ \frac{PL}{AE}
The trick here is to identify a constraint or any variable that has to remain constant by nature of the problem. Now that we’ve set up the problem and equations, see if you can figure out what that constraint or variable is before checking out the next reply for the solution!
following Ben’s thought, I want to ask if this question was asking about load path or asking a critical load. Still, where is the load place relative to the support location is important. In terms of critical load, we can already determine one leg will have a higher critical load to buckle.
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Did you figure it out?
Since the stool has legs with identical geometry, but different material, we need the deflection of each leg under an applied load to remain constant.
δ = \frac{PL}{AE}
Now we know that we want to keep the following parameters constant while we apply a vertical load to this stool seat:
- A – cross-sectional area of the stool leg
- δ – displacement length of each leg
- L – length of each stool leg
We can set the leg deflections equal to each other as follows:
δ_1 = δ_2
\frac{P_1L}{AE_1} = \frac{P_2L}{AE_2}
\frac{E_2}{E_1} = \frac{P_2}{P_1}
Since we know steel typically has a Young’s Modulus of 200 GPa compared to 70 GPa for aluminum, we can derive this relationship:
\frac{E_2}{E_1} = \frac{200}{70} > 1
∴\frac{P_2}{P_1} > 1
Solution
From the relationship above, we see that the steel legs will take a higher load than the aluminum legs in this situation. This makes intuitive sense since the stiffer material supports the larger load.
Extra Practice
If you thought this question was interesting, check out another one about stools!
Hey @jacker2011, those are some good points! We should’ve been more clear in the question, but the intent here is to see which legs would take more load if someone is sitting on the stool’s seat and placing their weight down into the stool. The legs should be assumed to be equally spaced apart and in the same orientation, size, geometry, etc. (but this should be clarified in an actual interview setting to avoid confusion!) Note that a 4-legged stool is an indeterminate structure.
The question about critical load to buckling is a good one, maybe that should be added as a new question!
That’s a detailed explanation. Thank you.
I pictured if load path is the same. Then both legs are going to experience same load from the person, the Al leg would deflect more
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How is constant deflection an acceptable answer? That will not occur in any situation, if calculate the reaction forces at T=0 (time 0), the force applied to both legs will be equal since the force is centered.
This will cause a stress in the legs, and we can use hookes law to determine the strain caused by this.
Strain = Stress/Modulus
Since the modulus of steel is typically 3x that of aluminum, the strain will be 3x higher for aluminu, meaning the legs will deflect different amounts.
Then the stool will tilt, at that point it will depend whether the force stays perpendicular to the ground or to the seat.
I would like to agree with @bmachado, that the answer is equal load placed on all 4 legs.
The question is also explicit about a load being applied to the seat, with the implication of it being a constant load, such as the weight of a person. As opposed to placing it in an Instron, programmed to apply constant displacement (in compression).
By analogy, to enhance the difference in materials, if you imagine sitting down on a stool with 2 legs made of steel and 2 legs made of rubber, the rubber would compress (low elastic modulus = flexible) and the steel won’t (high elastic modulus = stiff).