Is Stress dependent on the material properties?

[Astronaut]

I have this fundamental question about stress and strain.

If we apply same Force on two different objects of same geometry in the same configurations, will they experience the same stress?
If yes, then does it imply that stress is independent of the internal material properties?

 

[jack action]

By definition stress is a force acting on an area, ##\sigma= \frac{F}{A}##. So material properties does not come into play.

Though, strength is a material property in itself which corresponds to how much stress a material can support.

 

[Dr.D]

As Jack Action hinted at with his comment about strength, stress will be the same, provided the material properties function alike in both cases. Specifically, if one of the bodies is of a low yield point material while the other has a high yield point, for very small loads the stresses will be the same. If the stress exceeds the yield point in one of the bodies but not in the other, then the stresses will be different. By definition of yielding, in the softer material, the stress cannot rise (actually it usually does in the plastic range, but that just complicates the picture).

Provided both bodies remain in the linear elastic range, then stress should be expected to be the same in both bodies.

 

[Astronaut]

As Jack Action hinted at with his comment about strength, stress will be the same, provided the material properties function alike in both cases. Specifically, if one of the bodies is of a low yield point material while the other has a high yield point, for very small loads the stresses will be the same. If the stress exceeds the yield point in one of the bodies but not in the other, then the stresses will be different. By definition of yielding, in the softer material, the stress cannot rise (actually it usually does in the plastic range, but that just complicates the picture).

Provided both bodies remain in the linear elastic range, then stress should be expected to be the same in both bodies.

What if I had a perfectly rigid block and applied a force on it, will stress be defined in this case?

 

[Dr.D]

If you had a perfectly rigid block, you should file a claim against the guarantee. There is no perfectly rigid material.

 

[Astronaut]

If you had a perfectly rigid block, you should file a claim against the guarantee. There is no perfectly rigid material.

I am trying to relate some things by this hypothesis. So, ASSUMING you somehow have a perfectly rigid material.. there will be no strain on the block no matter what the load applied.

So, will stress be defined or not?

 

[Astronaut]

By definition stress is a force acting on an area, ##\sigma= \frac{F}{A}##. So material properties does not come into play.

Though, strength is a material property in itself which corresponds to how much stress a material can support.

I have read a different definition which involves RESTORING FORCE, not the force applied. That's why confusions are arising.

 

[Dr.D]

What is the point in discussing hypotheticals that cannot possibly exist in reality? There are no perfectly rigid materials, so that assuming one is pointless.

 

[jack action]

I have read a different definition which involves RESTORING FORCE, not the force applied. That's why confusions are arising.

I don't know this definition, can you provide a source?

If you examine closely the definition of stress, you will notice it is the same one as pressure. In both cases, you are pushing against something that is pushing back, thus maybe the concept of restoring force.

For a perfectly rigid material the modulus of elasticity ##E## is infinite thus the strain ##\epsilon## is defined by ##\epsilon= \frac{\sigma}{E}= \frac{\sigma}{\infty}##. Hence, for any value of stress ##\sigma##, the strain will always be zero. So you can see that the stress is still present, it is just that any value will result with the same strain. Even if such material doesn't exist, one can always assume it would still have a maximum strength, i.e. a value of stress for which it would break.

Compare to the following stress-strain curves, a perfectly rigid material would have a vertical line aligned on the stress-axis, stopping at some maximum value.

 

[Chestermiller]

As Jack Action hinted at with his comment about strength, stress will be the same, provided the material properties function alike in both cases. Specifically, if one of the bodies is of a low yield point material while the other has a high yield point, for very small loads the stresses will be the same. If the stress exceeds the yield point in one of the bodies but not in the other, then the stresses will be different. By definition of yielding, in the softer material, the stress cannot rise (actually it usually does in the plastic range, but that just complicates the picture).

Provided both bodies remain in the linear elastic range, then stress should be expected to be the same in both bodies.

I disagree with this. In my judgment, even if one of the materials yields, at the same applied force, the engineering stress (based on the initial cross sectional area) will be the same for both materials. However, the true stress (calculated as the force divided by the current cross sectional area) will be higher in the material that strains more (assuming the same Poisson ratio).

 

[Astronaut]

I disagree with this. In my judgment, even if one of the materials yields, at the same applied force, the engineering stress (based on the initial cross sectional area) will be the same for both materials. However, the true stress (calculated as the force divided by the current cross sectional area) will be higher in the material that strains more (assuming the same Poisson ratio).

Please Can you elaborate this ??

 

[Chestermiller]

Please Can you elaborate this ??

If the deformation is large, the cross sectional area of the sample is not the same after the deformation as it was initially. This is because of the Poisson effect. The force divided by the initial cross sectional area of the sample is called the engineering stress. The force divided by the current cross sectional area of the sample (i.e., after the deformation) is called the true stress. At small strains, there is not significant difference between the engineering stress and the true stress. However, for large tensile deformations the true stress is larger than the engineering stress.

 

[Astronaut]

If the deformation is large, the cross sectional area of the sample is not the same after the deformation as it was initially. This is because of the Poisson effect. The force divided by the initial cross sectional area of the sample is called the engineering stress. The force divided by the current cross sectional area of the sample (i.e., after the deformation) is called the true stress. At small strains, there is not significant difference between the engineering stress and the true stress. However, for large tensile deformations the true stress is larger than the engineering stress.

Got it! Thanks