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Stress and deformation compressed cube

  1. Nov 4, 2015 #1
    Hi, sorry for my bad english.
    PART A
    I have a cube of 10mm x 10mm x 10mm which is compressed with a stress of 0.5 Mpa on his superior face (σz=-0.6)
    young modulus=20Mpa, poisson coef=0.5
    I have the stress tensor:
    [ 0 0 0]
    [ 0 0 0]
    [ 0 0 -0.6]

    And then, the deformation tensor is :
    [0.015 0 0]
    [ 0 0.015 0]
    [0 0 -0.3]

    Why εx+εy+εz =0? is it a question of conservation of the volume ?

    PART B
    Well, now i put this cube into an indeformable box of 10.002mm (x) x 10.004 mm (y) x 10 mm (z) and this cube is again compressed with a stress of 0.5MPa;

    1) do The face x and y of the cube will touch the faces of the box ?
    I do ε=ΔL/L and i find εx=0.0002 and εy=0.0004. So I supposed that they will touch if we compared the deformations with the deformation in the first part.
    2) Doing an hypothesis that the faces will be in contact with the face of the box, determined the stress and deformation tensor of the cube.... But here i don't know how to do that... If someone can help me.. Thanks a lots

    PART C
    IF the box has the same dimension of the cube and if will do a stress of 0.6 MPa again, determined without calculus the deformation tensor.. I think all deformation will be equal 0 is it right ?
     
    Last edited by a moderator: Nov 4, 2015
  2. jcsd
  3. Nov 4, 2015 #2
    I think you meant 0.6 MPa in the problem statement, not 0.5.

    Regarding your question, if Poisson's ratio is 0.5, that means that the material is incompressible, and the sum of the three strains is zero.
    This is a tricky thing to analyze in the case of an incompressible material. For an incompressible material, you need to go to the incompressible limiting form of Hooke's law which is:

    $$σ_x=-p+2Gε_x$$
    $$σ_y=-p+2Gε_y$$
    $$σ_z=-p+2Gε_z$$
    where p is an arbitrary compressive stress that can be determined from the problem conditions. Since you know that the material is compressible and you know the strains in the x and y directions, what is the strain in the z direction? Once you know the strain in the z direction (and you know the stress in the z direction from the problem statement), you can use the z equation to determine the compressive stress p. What value do you get? Once you know p, you can get the stresses in the x and y directions.
    You do this the same way you did part B, but here, all three strains is zero. So all you need to do is determine the compressive stress p.
     
  4. Nov 4, 2015 #3
    yes it was what I have written for the Part A and the part C. In the part B I would like to determine εz with what i have written but λ→∞...

    It is that: http://image.noelshack.com/fichiers/2015/45/1446679886-20151105-002818.jpg


    SO i can't use this method.

    I have to do that ?
    http://image.noelshack.com/fichiers/2015/45/1446680189-20151105-003354.jpg
    But as we can see σx and σy are positive then i have to take them equal to 0 if i want them to touche the box right?
     
  5. Nov 4, 2015 #4
    So λ→∞, but the sum of the strains ---> 0, and their product is finite. That finite product is what I call p.

    Now, you asked me for my help, and I've given you a method for solving this problem which I know is 100% correct. If you don't want to follow my advice, that's fine. Have a nice day.:smile:
     
  6. Nov 6, 2015 #5
    Irrespective of the value of λ, are the following 2 algebraic equations (a) correct of (b) incorrect:
    $$σ_x-σ_z=2μ(ε_x-ε_z)$$
    $$σ_y-σ_z=2μ(ε_y-ε_z)$$
     
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