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Biaxial forces

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the cube with sides a=b=c=10cm . This block is tested under biaxial forces that are applied in the X and y directions. Assume that the forces applied have equal magnitudes of Fx=Fy=2x10^6 and that the modulus and poissons ratio of the block material is E=2x10^11 Pa and poissons ratio is = 0.3
    I) find the new dimensions of sides if both are tensile and I) if Fx is tensile and fy is compressive

    2. Relevant equations
    σ=F/A
    εX=1/E(σx-νσy)

    3. The attempt at a solution
    σX=2x10^6/0.1=2x10^7
    Same for σy as same variables

    εX= 1/2x10^11(2x10^7-0.3(2x10^7))

    εX = 7x10^-5

    εY= same as same variables

    => a=0.100007m
    b=0.100007m
    C=0.09993m

    I know that this is wrong but I can't seem to figure out why and I can't seem to do when one is tensile and one is compressive could really use help
     
  2. jcsd
  3. Aug 8, 2017 #2

    haruspex

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    Check the area term in there.

    Your work would be much easier to follow if you were to use parentheses as appropriate.
     
  4. Aug 8, 2017 #3
    Sorry yes your right it should be 0.01
    I mistyped as I only have my phone it's quite hard to reread probably
     
  5. Aug 8, 2017 #4

    haruspex

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    It's not just that you mistyped that line, though. You carried the error through the rest of the calculation, no?
     
  6. Aug 8, 2017 #5
    Also, C is wrong. What is the equation for epsilonC?
     
  7. Aug 9, 2017 #6
    $$\epsilon_C=-\frac{\nu(\sigma_x+\sigma_y)}{E}$$
     
  8. Aug 9, 2017 #7
    And what is the difference then between tensile and compressive ? Also the equations εx=1/E(σx-μ(σy+σz))
    εy=1/E(σy-μ(σx+σz))
    εz=1/E(σz-μ(σy+σx))

    Sorry I got thrown all these equations but no explanation on which or why use them
     
  9. Aug 9, 2017 #8
    No. That is just the result of substituting ##\sigma_z=0## into the equation below.
    Yes. These equations are correct. You can choose whichever directions you please for the three stresses. In your problem, two of them are non-zero (and equal), and the third is zero.
     
  10. Aug 9, 2017 #9

    haruspex

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    Then let me break this down: εz=1/E(σz-μ(σy+σx))
    With only a force in the x direction, the consequence for the z direction would be εz=(1/E)(-μσx). Similarly for a force in the y direction only.
    For a force in the z direction only, the consequence for that direction is (1/E)(σz)
    With forces in all three directions, to a first approximation, you can just add them together.
     
  11. Aug 9, 2017 #10
    i did it out again , does this look more correct??
     

    Attached Files:

  12. Aug 9, 2017 #11
    Both calculations of C are incorrect. In part (a), you made an algebra error and in part (b) you calculated ##\epsilon_z## incorrectly.
     
  13. Aug 9, 2017 #12
    On the final page is it ? A and B dimensions are correct thou?
     
  14. Aug 9, 2017 #13
    Yes, A and B are correct. But, there is no calculation on either page for ##\epsilon_z## in case (b)
     
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