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

In summary: GPa) a compressive force of ##F = 2940N##.In summary, the column is displaced by 10μm due to the weight of the rock.
  • #1
Hamiltonian
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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}##
1682198429592.png


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##?
 
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  • #2
Seems like they over constrained the problem to me.
 
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  • #3
erobz said:
Seems like they over constrained the problem to me.
are you saying there's a mistake in the question?
 
  • #4
Hamiltonian said:
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...
 
  • #5
a) the compressive strain = the displacement under force.
This was given and does not need to be computed !
 
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  • #6
TonyStewart said:
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:

Hamiltonian said:
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?
 
  • #7
TonyStewart said:
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.
 
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  • #8
Hamiltonian said:
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.
 
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  • #9
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.
 
  • #10
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.
 
  • #11
Hamiltonian said:
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?
 
  • #12
haruspex said:
Are you sure it gives the displacement as 10mm, not 10μm?
Are you saying it is multiple choice? If so, what are the options
1682234182990.png

its 10mm.
It's not multiple choice.
 
  • #13
Maybe we are not on Earth anymore.
 
  • #14
Hamiltonian said:
View attachment 325359
its 10mm.
It's not multiple choice.
Then I give up. Even the grammar is faulty. What book is this?
 
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  • #15
TonyStewartEE75 said:
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}##.

TonyStewartEE75 said:
Compressive force applied to the column, F = 400 kg = 400 * 9.8 N = 3920 N
 
Last edited:
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  • #16
haruspex said:
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:/
 
  • #17
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?
 
  • #18
:wink:
 
  • #19
TonyStewart said:
Diameter of the column, d = 40 mm
Why are you repeating that claim?
Hamiltonian said:
Homework Statement: …. with radius ##20## cm.
 
  • #20
sorry my err pls delete.
 

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

What is stress in the context of a vertical column with a rock on top?

Stress is defined as the internal force per unit area within a material. In the context of a vertical column with a rock on top, stress is the force exerted by the weight of the rock divided by the cross-sectional area of the column.

How do you calculate the stress on the vertical column due to the rock?

To calculate the stress on the vertical column, you use the formula: Stress (σ) = Force (F) / Area (A). The force is the weight of the rock, which can be calculated by multiplying the mass of the rock by the acceleration due to gravity (F = mg). The area is the cross-sectional area of the column.

What is strain in this context?

Strain is the measure of deformation representing the displacement between particles in the material body. In this context, strain is the ratio of the change in length of the vertical column to its original length when the rock is placed on top.

How do you calculate the strain in the vertical column?

Strain (ε) is calculated using the formula: Strain (ε) = Change in Length (ΔL) / Original Length (L). The change in length is the amount by which the column deforms under the weight of the rock.

What factors can affect the stress and strain in the vertical column?

Several factors can affect the stress and strain in the vertical column, including the material properties of the column (such as Young's modulus), the cross-sectional area of the column, the weight of the rock, and environmental conditions such as temperature and humidity.

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