Multi-Ferroics Intro Help

  • Thread starter Nickpga
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  • #1
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Homework Statement:
1D mechanical bar. Two different materials that clamped at x = 0. both are isotropic and perfectly bonded together. Young's modulus E1 = 10GPa and E2 = 50GPa. displacement of 2um at x = 2mm

Find the stresses, strains and displacements in each of the bars.
Plot the energy per unit volume in material 1 as a function of the Young's modulus in material 1, for five cases. E = 1, ,5, 10, 25 and 50
Relevant Equations:
Epsilon (stress) = sigma (strain) / E (young's modulus)
du (displacement)/ dx = * E = strain
I have a TA that is ignoring me, so I have to resort to this online forum. Also, a very unhelpful example from lecture (just 1).
I have some other equations, but I dont really know how they all tie together.

I find epsilon = 10um/1mm (both materials are 1 mm in length) = 1E-2mm/mm
sigma 2 = E1*epsilon = 50E7Pa
sigma 1 = E2*epsilon = 10E7Pa

As far as the second part, I thought I did something reasonable, but after rereading it, I realized I missed a crucial part of the problem.
I can provide anything else that you may want to help me understand this homework (in context) since I don't have anyone to explain it to me. Other than the lectures which are very unhelpful.
 

Answers and Replies

  • #2
caz
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What does the fact that your calculated stresses are different imply?
 
  • #3
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What does the fact that your calculated stresses are different imply?
That my assumption that both materials receive the same displacement, may be wrong?
 
  • #4
caz
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Yes. Think about a hard spring and a soft spring in series. If you pull it, the soft spring will extend more.

If the stresses are different at the boundary between the materials, what happens to the boundary?
 
Last edited:
  • #5
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Stress times cross sectional area is force. So at the boundary between the two materials, the forces are different. What does this mean physically?
That the stress is different between each material?
 
  • #6
caz
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I edited my previous comment.

What happens to the boundary if the stresses are different?
 
  • #7
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I edited my previous comment.

What happens to the boundary if the stresses are different?
oh. ok. now i am starting to catch on. (i am way better at programming than i ever will be at mechanics).
so the total displacement is 10um, but that is distributed between both materials?
 
  • #8
caz
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Yes, the displacements are not equally distributed.
 
  • #9
caz
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The system is at equilibrium. What does that say about the forces of the boundary?
 
  • #10
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The system is at equilibrium. What does that say about the forces of the boundary?
Wouldnt that imply that the forces are the same? Or would that be the strain that is the same?
 
  • #11
caz
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At equilibrium, the boundary does not move. The is means the net force is 0. This means that the stresses are identical in both materials.

Using this condition, do you see what you need to do?
 
  • #12
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At equilibrium, the boundary does not move. The is means the net force is 0. This means that the stresses are identical in both materials.
didnt we say that the stresses are different? oh nvm. it was more of a question.
 
  • #13
caz
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If you assume that the stresses are identical, do you see how to do the calculation?
Do you understand why they are identical?
 
  • #14
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If you assume that the stresses are identical, do you see how to do the calculation?
Do you understand why they are identical?
I am confused on how you they are in equilibrium
 
  • #15
caz
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Nothing is moving. Therefore there are no accelerations. Therefore there are no net forces.

At the boundary between the materials, each material is pulling the boundary towards itself with a force equal to the stress times the cross-sectional area. (Check your text on how to get the directions correct). This means that the stresses are identical.

Do you understand?
 
  • #16
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Alright. That makes sense. I suppose I had an imaginary force that caused the displacement. Like someone pulling it
 
  • #17
caz
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Yes.

So dtotal=dmat1+dmat2 with the condition that the stresses are identical for mat1 and mat2. Do you see where to go from here?
 
  • #18
caz
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Btw, your relevant equation for stress is wrong and you use two different lengths for the total displacement in your first post.
 
  • #19
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Btw, your relevant equation for stress is wrong and you use two different lengths for the total displacement in your first post.
really? stress = strain/ Y_modulus? Thats the equation the professor boxed and starred
 
  • #20
caz
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Stress(sigma)=strain(epsilon)*Y
(Y and stress have the same units)
 
  • #21
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Stress(sigma)=strain(epsilon)*Y
(Y and stress have the same units)
You know. Thats what happens when the professor cant be be bothered to write legibly. I thought stress was epsilon. and strain sigma.
 
  • #22
caz
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The other equation you need is strain=dl/l
 
  • #23
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The other equation you need is strain=dl/l
I have that one too. I just had it wrong. Thanks a lot for your time and effort in asking me questions to make my mind work.
I wouldn't say I am dumb, I just don't have a mind that thinks 'mechanically'.
 
  • #24
caz
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Your welcome. New concepts take time to absorb. Be patient.

You seem to know what to do now so I’ll be signing off.
 
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