Basic Stress and Strain question -- Rock on top of a vertical column

[Hamiltonian]

Homework Statement

A $300$ kg rock puts on an one-meter long aluminium column ($E = 7 * 10^{10} N/m^2$) with radius $20$ cm. (show in figure)

(a) Calculate the compressive strain of the column if a displacement of $10$mm
occurred.

(b) What is the compressive force, $G_{rock}$ applied to the column

Relevant Equations

$\sigma = E\epsilon$ and $\epsilon = \frac{\Delta L}{L}$

a) I can find the compressive strain on the aluminium column using the formula $\sigma = E\epsilon$ as we know $\sigma = F/A$. The area of the column is $A = \pi r^2 = 0.126m^2$ and the force on the column is $F = 300*(9.8)N = 2940N$. The stress therefore is $\sigma = \frac{2940N}{0.126m^2} = 23333.333N/m^2$

hence plugging the stress($\sigma$) and the Modulus of Elacticity(E) into our original equation we find the strain, $\epsilon = \frac{\sigma}{E} = \frac{23333.33}{7*10^{10}} = 3333.332857*10^{-10}$.

But if we use the definition of strain $\epsilon = \frac{\Delta L}{L}$ we get $\epsilon = \frac{0.01m}{1m} = 0.01$. The compressive strain obviously can't have two different values, so where have I gone wrong in my reasoning?

b) Also Just to be safe the compressive force $G_{rock}$ applied to the column would be the weight of the rock i.e., $300g N = 2940N$?

 

[erobz]

Seems like they over constrained the problem to me.

 

[Hamiltonian]

Seems like they over constrained the problem to me.

are you saying there's a mistake in the question?

 

[erobz]

are you saying there's a mistake in the question?

If you are given a modulus of elasticity and a strain, then the stress is fixed.

If they were intending for you to calculate $G_{rock}$ they bungled it when they gave its weight, and the deflection (and other relevant properties) it caused. Deflection is given by

$$ \epsilon = \frac{ \delta}{L} = \frac{P\cancel{L}}{\cancel{L}AE}$$

You have everything on both sides, and if you did it the math correctly (which I didn't check as a yet) they don't match...

 

[TonyStewart]

a) the compressive strain = the displacement under force.
This was given and does not need to be computed !

 

[erobz]

a) the compressive strain = the displacement under force.
This was given and does not need to be computed !

Well it needs to be calculated as has been done in the following:

But if we use the definition of strain $\epsilon = \frac{\Delta L}{L}$ we get $\epsilon = \frac{0.01m}{1m} = 0.01$. The compressive strain obviously can't have two different values, so where have I gone wrong in my reasoning?

 

[haruspex]

a) the compressive strain = the displacement under force.
This was given and does not need to be computed !

No, it is the displacement divided by the total length.

 

[haruspex]

are you saying there's a mistake in the question?

Not only is it overconstrained but the numbers are in violent disagreement.
A 10mm displacement is huge. Maybe it should be 10μm? That still leaves a factor of 30 to be found somewhere.

 

[Hamiltonian]

so what do you reckon I do now? This problem is on my homework and I need to put one of the answers which one should I put.

 

[TonyStewartEE75]

The modulus method should equal the compression displacement method which is actual for low levels , which this is.

I cannot imagine 300 kg doing 10mm compression on a 40 mm diameter aluminum column and this is faulty data.

——-
To compute the compression of the aluminum column, we need to use the following formula:

δ = (F * L) / (A * E)

where:

• δ is the compression of the column
• F is the compressive force applied to the column
• L is the original length of the column
• A is the cross-sectional area of the column
• E is the modulus of elasticity of aluminum

From the problem statement, we have:

• Original length of the column, L = 1 m
• Diameter of the column, d = 40 mm = 0.04 m, so the radius r = d/2 = 0.02 m
• Cross-sectional area of the column, A = π * r^2 = 3.14 * (0.02)^2 = 0.001256 m^2
• Modulus of elasticity of aluminum, E = 70 GPa = 70 * 10^9 N/m^2
• Compressive force applied to the column, F = 400 kg = 400 * 9.8 N = 3920 N

Now we can substitute the values into the formula to obtain:

δ = (3920 N * 1 m) / (0.001256 m^2 * 70 * 10^9 N/m^2) = 0.000005788 m = 5.788 μm

Therefore, the compression of the aluminum column is 5.788 μm.

 

[haruspex]

so what do you reckon I do now? This problem is on my homework and I need to put one of the answers which one should I put.

Are you sure it gives the displacement as 10mm, not 10μm?
Are you saying it is multiple choice? If so, what are the options?

 

[Hamiltonian]

Are you sure it gives the displacement as 10mm, not 10μm?
Are you saying it is multiple choice? If so, what are the options

its 10mm.
It's not multiple choice.

 

[scottdave]

Maybe we are not on Earth anymore.

 

[haruspex]

View attachment 325359
its 10mm.
It's not multiple choice.

Then I give up. Even the grammar is faulty. What book is this?

 

[erobz]

Diameter of the column, d = 40 mm = 0.04 m, so the radius r = d/2 = 0.02 m
Cross-sectional area of the column, A = π * r^2 = 3.14 * (0.02)^2 = 0.001256 m^2

$20~\rm{cm} = 0.2 ~\rm{m}$

Homework Statement: A $300$ kg rock puts on an one-meter long aluminium column ($E = 7 * 10^{10} N/m^2$) with radius $20$ cm. (show in figure)

Also, the mass of the rock is $300 ~\rm{kg}$, not $400 ~\rm{kg}$.

Compressive force applied to the column, F = 400 kg = 400 * 9.8 N = 3920 N

 

[Hamiltonian]

Then I give up. Even the grammar is faulty. What book is this?

It's actually on my homework problem set for a physics class meant for cs majors. This is what happens when the computer science department tries to do physics:/

 

[TonyStewart]

It's your attempt to find contradictions that counts for points.

Everyone makes errors, even String Theorists and ChatGPT.

e.g. Diameter of the column, d = 40 mm = 0.04 m, so the radius r = d/2 = 0.02 m

But look at who makes the big bucks? What's wrong with this picture?

 

[TonyStewart]

 

[haruspex]

Diameter of the column, d = 40 mm

Why are you repeating that claim?

Homework Statement: …. with radius $20$ cm.

 

[TonyStewart]

sorry my err pls delete.