Stress and deformation compressed cube

In summary: These equations are correct. Note that they are just special cases of the general equation I gave you. I don't know what you mean by λ→∞. I think that you are trying to use the finite strain equations. You can do that, but you have to be very careful in how you do it. I would recommend that you use the equations I gave you. The reason is that I know that they are correct, and I know how to use them.(b) These equations are incorrect. These equations do not account for the fact that the material is incompressible. If you use these equations, you will not get the correct answer.
  • #1
zDrajCa
8
0
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 ?
 
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  • #2
zDrajCa said:
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 ?
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.
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
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.
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 ?
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.
 
  • #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.jpgSO 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?
 
  • #4
zDrajCa said:
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 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:
 
  • #5
Irrespective of the value of λ, are the following 2 algebraic equations (a) correct of (b) incorrect:
$$σ_x-σ_z=2μ(ε_x-ε_z)$$
$$σ_y-σ_z=2μ(ε_y-ε_z)$$
 

1. What is the definition of stress and deformation?

Stress is the force applied per unit area of an object, while deformation is the change in shape or size of an object in response to applied stress.

2. How is stress and deformation measured in a compressed cube?

Stress is typically measured in units of pressure, such as pounds per square inch (psi) or pascals (Pa). Deformation is measured as a percentage of the original size of the cube.

3. What factors can affect stress and deformation in a compressed cube?

The material properties of the cube, such as its elasticity and strength, as well as the magnitude and direction of the applied force, can affect stress and deformation. Temperature and environmental conditions can also play a role.

4. How does stress and deformation impact the structural integrity of a compressed cube?

If the stress applied to a cube exceeds its strength, it can lead to permanent deformation or failure of the cube. However, a certain level of stress and deformation can also be beneficial in strengthening the cube's structure and making it more resistant to future stress.

5. How can we use stress and deformation data to improve the design of a compressed cube?

By analyzing stress and deformation data, we can identify weak points in the cube's design and make adjustments to improve its strength and durability. This information can also help us determine the optimal amount of stress to apply for maximum structural integrity.

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