Selection of Principal Stress 3

[vtmike]

Hi,

I am currently working on a cylinder with external pressure applied on it. It is a thick cylinder and I am using Lame's equations to solve for the maximum pressure at yield point.

Now the three stresses are the tangential/hoop, radial and the axial. When I combine these stresses using Von Mises criterion to solve for maximum pressure, which one should be selected as sigma_1, sigma_2 and sigma_3?

I chose the tangential stress as sigma_1 in attached solution, but am not completely sure if it is right.

Any help would be appreciated.

Thanks,
Mike

 

[prost]

Why does it matter? The expression for von Mises (or equivalent) stress has all 3 principal stresses:

Let von Mises stress==sM,
sigma_1==s1, etc.

then

2*(sM^2)=(s1-s3)^2+(s2-s3)^2+(s3-s1)^2

Then compare sM to the yield stress, sY. If sM<sY, then the "Maximum Energy of Distortion Theory of Failure" predicts no failure. Whether this is the correct Theory of Failure to use, that depends on your application. Some design codes call for one Theory of Failure to be used (for instance, max. shear); some design codes use the typical industry practice to determine Theory of Failure.

If you must select one stress over the other to be s1, s2 and s3, then use the largest stress, (hoop or tangential as you call it) as s1, s2 is the longitudinal stress (along the axis of the cylinder), and s3 is radial stress. This of course assumes the shear stresses are zero; otherwise the calculation of "sM" is more complicated.

 

[vtmike]

I would have taken the largest stress as sigma_1 but I am solving the problem to find the maximum pressure it can withstand by taking sM=sY, and which is why I do not have the numerical values of hoop/tangential, longitudinal and axial stress when I use Von Mises theory of failure.
So, in this case would you still select the tangential stress as sigma_1?

 

[vtmike]

I have assumed the tangential stress to be sigma_1 since it is usually the maximum, but am still not sure. Any comments?

 

[GBor]

Von Mises stress designations can be directly related to Mohr's circle designations. Visulaize your stress element orientation on your cylinder to help determine directions, but you can also search for Von Mises in cylindrical coordinates. So long as you are consistent, mathematically it should work out, in my VERY brief mental experiment.

 

[btrueblood]

It does not matter what order you use. Try it - you should get the same result.

 

[cbrn]

Hi,
btrueblood & others should be correct, however it is normally assumed that S1 = hoop, S2 = axial and S3 = radial. Cases in which radial becomes larger than hoop can only happen for bending-dominated problems, but not in your case of a cylindrical pressure vessel.

Regards

 

[prost]

I looked over your 'collapse3.jpg', assuming there are no small algebra errors, everything looks in order. So I still don't understand your question, looks like you can compute everything.

Here would be my back of the envelope, so I could judge orders of magnitude. We know the 'thin shell' approximations are very easy to compute--for an internal pressure on a thin cylinder, the hoop and axial stresses are approx. pr/t and 0.5*pr/t, respectively. What is the radial stress? In this case, it goes from "p" at the inside of the cylinder, to zero at the outside--the most it could be is "p", while those other two stresses, hoop and axial, are much larger. So from the thin cylinder approximation, you can guess that hoop stress>axial stress>radial stress--standard practice is to set the principal stresses sigma_1>sigma_2>sigma_3. I realize you are doing the thick cylinder analytical solution, nevertheless, I think the thin cyl. approximations are very good for estimating relative magnitudes of the stresses, which should give you pretty good idea what the relative magnitudes of the stresses computed the thick cylinder equations. Would you agree that this is a good way to ballpark the estimate of the stresses?

BTW--in the thin cylinder approximations, they write "p" but what they really mean is "delta-p"--that is, the difference between outside and inside pressures.

 

[vtmike]

Thanks for the input everyone.

btrueblood is right. I get the same answer even if I select sigma_1 = axial and sigma_2 = hoop, in both cases i.e using Von Mises failure theory and Max Shear Stress theory as I did using sigma_1 = hoop and sigma_2 = axial.
What is not clear to me is the theoretical reason behind this.

prost:
From what I got from the books is like you said, that generally the highest stress is selected as sigma_1 which is hoop here, but then why select the highest as sigma_1 if I get the same answer by switching them?

 

[CoryPad]

There are two forms of the von Mises equation - one with three principal stresses, one with three orthogonal stresses and three shears on those planes. For the former, you are taking the root of the sum of the squares of the differences, so order is immaterial.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[btrueblood]

And, at any given point, the VonMises stress is an invariant (scalar value), so computation by principal stresses or by 3 orthogonal + shear stresses will give the same value.

 

[prost]

Making the largest stress sigma_1 is just by convention normally. Think of Mohr's circle; sigma_1 is the largest stress. In your case, there are no shear stresses, so all your stresses are plotted on the x-axis.

 

[rb1957]

thx for spelling principal correctly ! (ie, not principle)