5. Stress and strain in 2D

Questions and discussion for this lecture live here. Fire away by hitting Reply below :fire:

Regarding the shear strain Epsilon xy = (beta1 + beta2), is beta1 = beta2 ?

Hi Tien - yes
considering the magnitudes of all of the shear stresses are the same, these angles would be equal.
SeĂĄn

When you explain biaxial-normal-stress on minute 6:30,
is this SQURE is a section of a bar or a cut along the axis of a bar?

It would be option B
imagine a square on the surface of the bar.

What’s important is that the 2D infinitesimal element is taken from a location in a structure where a state of biaxial stress exists. An infinitesimal element on the surface of the bar would be one such example.

I recommend watching this lecture where I go into this in a little bit more detail - part of my (free) course on Building a Mohr’s Circle Calculator for Stress Analysis in Python.

By “either location”, do you mean, do you mean “either orientation”?

Since the workflow will be only in 2D, can you explain this problem without the 3rd dimension?

Thank you for your answer, but I am still very confused.

You’re right, my previous answer was a little confusing
I’ve edited it to try and simplify it. Do also take a look at the other lecture I reference.

Let me know if you still have questions after looking at that lecture.

S

Thank you, I guess for 2d bar the little square is scaled in xy directions.

And the shear happens in a section of a bar or also in logitudinal direction?

The reason, I am insisting to have a clear answer, because python calculation will be explicit and there will neither this or that.

Will it involve matrices for 2 dimensions or 3 dimensions?

No - the 2D element is not scaled - it’s an infinitesimal square element - it represents a single point in space - the different sides of the element simply allow us to represents stresses in different directions at that point.

Have you watched the lecture I recommended - it’s called ‘Understanding the 2D stress element’ - it should help you.

S

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The video helps. There is a lot to learn, I will keep my fingers crossed and move on to the next videos.

Feel free to keep asking questions!
S