Trouble understanding stress and strain problems

In summary, stress is the force applied to an object while strain is the resulting deformation or change in shape of the object. They are directly proportional to each other and are commonly measured in units of force per unit area and as a dimensionless quantity respectively. The common types of stress include tensile, compressive, and shear stress, while the common types of strain include tensile, compressive, and shear strain. The primary factors that affect stress and strain in a material are the type of material, the amount of force applied, and the geometry of the object, but other factors such as temperature and external conditions can also play a role.
  • #1
VortexLattice
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Hi everyone,

so I'm trying to understand how to use the (Cauchy I think?) stress tensor. The definition I have is that the element [itex]\sigma_{\alpha\beta}[/itex] of the stress tensor σ is the force per unit area in the α direction on a surface that's normal to the β direction.

I also have a definition of the strain tensor ε, where [itex]\epsilon_{\alpha\beta} = (M_{\alpha\beta} + M_{\beta\alpha})/2[/itex], and [itex]M_{\alpha\beta} = \frac{\partial u_\alpha}{\partial r_\beta}[/itex], where u is the displacement of the solid at position r.

We also have the relation between stress and strain: [itex]\sigma_{\alpha\beta} = \sum\limits_{\gamma \delta} C_{\alpha \beta \gamma\delta} \epsilon_{\gamma\delta}[/itex]

This all makes a little sense to me, but it's kind of convoluted and I'm having trouble using it. For example, wikipedia seems to say that the stress tensor can have different values at different points in the solid, generally. But let's say we squeeze two opposite sides of a cube. Then, the stress tensor is the same at both of those sides, right? If the cube is otherwise uniform, is the stress tensor the same everywhere in the cube?

I found an example online but some of the stuff in it is unexplained to me. It's a cube with side A of a cubic crystal, and the faces of the cube are the faces of the crystal planes. A force per unit area σ is applied to one side of the cube away from the cube (call this direction +z) and another σ is applied to the opposite side, in the opposite direction (so the cube is basically being stretched).

They use the Voigt notation, and say that only [itex]\sigma_3[/itex] is nonzero, which makes sense. Then they say that only [itex]\epsilon_1, \epsilon_2, \epsilon_3[/itex] will be nonzero, and thus we get the equations (from the stress/strain relation):

[itex]\sigma_1 = C_{11} \epsilon_1 + C_{12} \epsilon_2 + C_{13} \epsilon_3 = 0[/itex]
[itex]\sigma_2 = C_{21} \epsilon_1 + C_{22} \epsilon_2 + C_{23} \epsilon_3 = 0[/itex]
[itex]\sigma_3 = C_{31} \epsilon_1 + C_{32} \epsilon_2 + C_{33} \epsilon_3[/itex]

Then, because it's a cubic crystal, we have the relations [itex]C_{11} = C_{22} = C_{33}[/itex] and [itex]C_{12} = C_{23} = C_{31}[/itex], so that let's us solve for [itex]\epsilon_1[/itex], [itex]\epsilon_2[/itex], & [itex]\epsilon_3[/itex] in terms of the C's and σ.

Then they say that the change in the length of the crystal in the z direction is [itex]A\epsilon_3[/itex]. I know that [itex]\epsilon_{3} = M_{33} = \frac{\partial u_3}{\partial r_3}[/itex], but what exactly are we doing to find this change in length? Are we really just doing:

[itex]\epsilon_{3} = \frac{\partial u_3}{\partial r_3} \rightarrow \partial u_3 = \epsilon_{3} \partial r_3 \rightarrow u(A) = \int \partial u_3 = \int_0 ^A \epsilon_{3} \partial r_3 = A\epsilon_3[/itex]

? It seems like I'm doing something wrong... I saw another problem similar to this, but one side is held stationary (while the opposite side is the same as before, still stretching). The answer comes out to be the same, that the change is length is the unstretched length times the strain element in that direction. Why? It seems like the one that's being pulled from both directions should stretch twice as much. (Unless the side being held stationary is held stationary by the same force as before, so it's actually identical...?)

Does anyone have any good resources for reading about this? I'm having trouble learning this stuff from the few sources I have.

Thank you!
 
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  • #2
</code>Yes, you are correct. The change in the length of the crystal in the z direction is A\epsilon_3.The equation \epsilon_{3} = M_{33} = \frac{\partial u_3}{\partial r_3} is correct, and the equation you derived from it (u(A) = \int \partial u_3 = \int_0 ^A \epsilon_{3} \partial r_3 = A\epsilon_3) is also correct.In the problem you mentioned with one side held stationary, the amount of strain is still the same as before, since the forces on each side are equal and opposite. That is, the force on the stationary side is equal in magnitude to the force on the stretching side, so the amount of strain will be the same in both cases.As for resources, a good place to start is the Wikipedia page on Stress–strain relationships. It contains a thorough explanation of the concept, as well as some examples. Additionally, there are a number of books available on the topic, such as Mechanics of Materials by Ferdinand Beer and John DeWolf.
 

What is the difference between stress and strain?

Stress is the force applied to an object, while strain is the resulting deformation or change in shape of the object. In other words, stress is the cause and strain is the effect.

How are stress and strain related?

Stress and strain are directly proportional to each other, meaning that as stress increases, strain increases as well. This relationship is known as Hooke's Law and is often represented by the equation: stress = Young's modulus x strain.

What are the common types of stress and strain?

The common types of stress include tensile stress (pulling force), compressive stress (pushing force), and shear stress (opposite forces in parallel directions). The common types of strain include tensile strain (lengthening), compressive strain (shortening), and shear strain (distortion).

How is stress and strain measured?

Stress is typically measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa). Strain is a dimensionless quantity and is often expressed as a percentage or in units of length per length, such as inches per inch or millimeters per millimeter.

What factors affect stress and strain in a material?

The primary factors that affect stress and strain in a material include the type of material, the amount of force applied, and the geometry of the object. Other factors such as temperature, time, and external conditions can also impact stress and strain in a material.

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