Normal stresses 2

[Everydaylearning]

Should there by sigma y and sigma z stresses in this plate due to poisons ratio effects? How would I go about finding exactly what they are?

 

[Everydaylearning]

Should this go in the mechanical engineering section?

 

[MIStructE_IRE]

In all my years of first principle structural calculations, I would never be bothered to work this out... stress = Load/area is good enough for me and in that case subtract the diameter of the hole when getting the area.

Many will disagree with me I’m sure!

 

[Everydaylearning]

Very well. This is more of a theoretical question.

 

[ThomasH]

Hi
If you remove the hole and constrain the plate in the y and/or z direction there will be stresses. It's a simple calculation. With the hole it gets more complicated and no constraints are required to get stresses.

Any FEM-software should be able to show you this. And I would not consider is "only" a theoretical issue.

Thomas

 

[Everydaylearning]

so technically there are constraints in the y and z direction. The face on the far left is fixed. FEA simulation is showing the following.

I’m not sure how to explain why there are sigmay stresses, or show what they should be in hand calcs

 

[Everydaylearning]

Thanks for your input ThomasH it is much appreciated. See below for the sigmaz and sigma x results. As you can see sigmaZ is fairly large actually

 

[ThomasH]

Hi
I see that you posted something new while I updated my post.

Remove the hole, add load on the short side (like you hve it now) and constrain in the perpendicular direction. You will obviously have stresses in x-direction sigma x = P / (100 mm * t).

You will also have stresses in the y-direction, sigma y = poissons ratio * sigma x. The same should apply for sigma z but that will not work if you use plate elements (plane stress means sigma z = 0).

The hole makes it more complicated to verify by a simple hand calculation.

Thomas

 

[Everydaylearning]

So it should be sigmay=Kt*v*sigmax if Kt=stress concentration, v=poissons ratio?

 

[ThomasH]

Hi
A simple example: Plate 500 mm x 200 mm x 20 mm.
Load in x-dirextion: 100 kN gives sigma x = 100000/(0.2*0.02) = 25 MPa
Poissons ratio (steel) 0.3: sigma y = 0.3 * 25 = 7.5 MPa

Thomas

 

[Everydaylearning]

Thanks Thomas. My only challenge now is to decide how the hole effects this!

 

[ThomasH]

Actually, it does not have to be that difficult.

You have two effects to consider. The hole and Poissons effect. Set Poissons ratio = 0 and that effect dissapears.

Thomas

 

[Everydaylearning]

This is the question that is posed

There’s clearly something a little different going on here than poissons*sigmax which can be identified by looking at the stress contours I posted

 

[ThomasH]

That sounds a bit like a school assignment, but never mind .

Below you see another figure. It is a platee with a 2d tensor plot showing the Major and Minor stresses with orientation.

At the left and right edge of the plate tha stresses are fairly uniform. In the middle they are definetly not uniform. But notice thar the stress near the hole (left and right edge) is small. That "stress" has moved sideways resulting in a sigma y component.

If you look at sigma y (not included here) the difference in results with Poisson = 0.3 and 0.0 is fairly small. That is as expected since the plate is not constrained in that direction.

/Thomas

 

[JAE]

DonMises is this some kind of class exercise? What’s the story behind your question?

Check out Eng-Tips Forum's Policies here:
faq731-376

 

[skeletron]

When would the force be applied at the edge of the plate in a practical scenario? I see this as a lifting plate situation, where the force would be applied at the hole. This seems too much like an academic problem...

 

[ThomasH]

Hi
Over the years I have more than once met situations where a hole in a plate in tension has been required. That is the reason why I did the first simple analysis, to show a "pure" example of Poisson effect. It's not that difficult to understand if you keep the example clean.

My second example with vectors was intended to show why sigma y differs from zero. The stresses/forces has to "move" sideways which results in a sigma y component, this has little to do with Poisson effects. Hopefully not that hard to see either. It also shows that stress=force/area on the sides doesn't work. There will be effects due to excentricity.

The question in DonMises latest post also made me think of a school assigment.

Thomas