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Posted:
1 decade ago
26 mai 2014, 06:52 UTC−4
Hi Andreas,
With strain in th xz direction you mean shear deformation? So there can be shear deformation, but the average of this value throughout your cube should be zero?
Hi Andreas,
With strain in th xz direction you mean shear deformation? So there can be shear deformation, but the average of this value throughout your cube should be zero?
Andreas Emmanouil Tzatzanis
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Posted:
1 decade ago
26 mai 2014, 06:59 UTC−4
Hi Andreas,
With strain in th xz direction you mean shear deformation? So there can be shear deformation, but the average of this value throughout your cube should be zero?
Hi again and thanks for your answer
Well i am not sure i need a shear deformation.
I really can't imagine how is this possible.
But yes i need the average volume strain in both xz and xy zero.
Thanks in advance
[QUOTE]
Hi Andreas,
With strain in th xz direction you mean shear deformation? So there can be shear deformation, but the average of this value throughout your cube should be zero?
[/QUOTE]
Hi again and thanks for your answer
Well i am not sure i need a shear deformation.
I really can't imagine how is this possible.
But yes i need the average volume strain in both xz and xy zero.
Thanks in advance
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Posted:
1 decade ago
26 mai 2014, 07:02 UTC−4
Forgive me that I do not understand but what do you mean with xy and xz? This is an expression that is used in shear deformation right?
Forgive me that I do not understand but what do you mean with xy and xz? This is an expression that is used in shear deformation right?
Andreas Emmanouil Tzatzanis
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Posted:
1 decade ago
26 mai 2014, 07:05 UTC−4
Well i mean the resultant strain between x and z.
Well i mean the resultant strain between x and z.
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Posted:
1 decade ago
28 mai 2014, 07:26 UTC−4
So you mean the strain in z and y resulting from a stress in x?
If that is what you want it can be achieved in many ways I think. You can choose a random point in your cube and impose on it a change in displacement proportional to the average volume strain in your cube. So I think you have to think about what the physical reason is for imposing this condition.
I think you would also achieve this condition if you would restrict all boundaries except the two facing towards the positive and negative x direction to movement parallel to their plane. It is also not so difficult to think of a physical reason for such a restriction.
So you mean the strain in z and y resulting from a stress in x?
If that is what you want it can be achieved in many ways I think. You can choose a random point in your cube and impose on it a change in displacement proportional to the average volume strain in your cube. So I think you have to think about what the physical reason is for imposing this condition.
I think you would also achieve this condition if you would restrict all boundaries except the two facing towards the positive and negative x direction to movement parallel to their plane. It is also not so difficult to think of a physical reason for such a restriction.