Stress Concentration Factor

The pipe wall can be assumed with the plate situation. The problem is which stress level you should consider. If the ratio wall thickness/pipe diameter is low then you an use exactly same conditions as for the plate. If the ratio is important the wall is not any more to be considered as under an uniform load so that you should compute with help of Lamé equations the stress at the notch root and work with this one which is higher.

As a start, I would simply calculate the pressure capacity of the pipe by assuming its OD to be the ID plus 2X the wall thickness between the bottom of the U-groove and the ID. A U-groove is not much of a stress riser (notch).

I agree with the 1st part as 1st step but I do totally DISAGREE with the2nd statement since it is purely qualitative and when one has to deal with stress levels only a quantitative approach is valid. I shall give an example:

Let assume a pipe with a ratio k=Ri/Re=0.8 and with a groove such that the root is at a radius 0.95 of Re of pipe (Re=De/2). The stress σtan= pi*k^2/(1-k^2)

If there is no groove the stress is pi*0.64/(1-0.64)=1.78 pi

If the Re value is reduced to 0.95 of initial value then k'= 0.842 the stress becomes 2.44 pi which means an increase of 37%.

Let us assume the wall is 10 mm thick which for above values corresponds to a De=100mm and a Di= 80mm. The groove has only 1.5 mm thick with a root radius of 1mm. With all computations done the notch factor will be around 2.3 !

Do you think it is a low value ?

Combining the 2 computations the result will be a stress at the root 5.61 pi which represents an increase of 215%!

So conclusion never use qualitative estimations when VALUES are involved.

I know such comments are not appreciated but I feel obliged to under stress once more that if we write some thing it must have a solid ground. People who ask have confidence in our answers and we have to maintain it we must do all to avoid to disappoint them.

The method presented in #2 is not a qualitative estimate. It is an engineering calculation that ignores the effect of stress concentration. Under certain conditions this may cause inaccuracy as pointed out in answer #3. Under other conditions it may be fine. If OP provides details of his pipe problem, we can advise. Answer #4 is beyond my capabilities. I learned that stress and strain are inextricably linked by Young's modulus. To suggest that one can exist without the other doesn't compute.

I shall try to explain the differences. As you know there is the Poisson coefficient which is the bond between strain in the 2 directions due to the volume changes. For instance rubber which is not modifying its volume under strain has ν=0.5. Steel has ν=0.285...0.3 depending on type. When you tension in one direction a thin sheet it can get slim in the normal direction, when the sheet is no more thin this freedom is limited. This leads to an other strain-stress structure. This is the difference between plane stress and plane strain. Nobody wrote that the stress is NOT related to Young' s module. It is of course but it is also related to the transverse changes related to the ν values. Think about the steel samples for the traction machine under load they become slimmer till they break this is the effect of transversal strict ion.

I was may be to rough but you understand that the differences are so important that I had to react. If there is only a groove the problem is even more complex since the wall thickness is from the stress point of view not any more constant but changes from a minimum (at the groove) to a maximum where the wall is not touched. The equations are not any more axis symmetric and to solve them is very difficult. It is possible to assume that there are 2 grooves at 180° from each other but then it is also compulsory to consider the notch effect since the stress in the groove section is related also to a local bending.

If I am not clear enough please say it and I shall try to give a better explanation.

This is not a full answer. However, I think that is may clarify the question a little bit.

The tables in Peterson's are for thin elements. These are in Plane Stress.

The situation you have with the longitudinal groove in the pipe is Plane Strain.

The thin element is unrestrained in the through thickness direction - it is free to constrict under in-plane tension. Therefore there is no stress in the through thickness direction, but there is a strain.

The prismatic solid (pipe with longitudinal groove) is restrained in the longitudinal direction - it is not free to constrict under in-plane tension. Therefore there is no strain in the longitudinal direction, but there is stress.

What the Peterson's stress concentration factors for thin elements don't take into consideration, if they are extended for use with prismatic solids, is the effect of the stress in the longitudinal direction. However, for most plane strain problems, the longitudinal stress is ignored.

I did a quick FEA of what I believe is the situation.

My model is of a 4" XS pipe (4.5" OD x 0.337" thick) with a longitudinal groove 1/8" wide by 1/8" deep with a 1/16" radius bottom. For a 100 psig internal pressure, P*R/t = 568 psi hoop stress.

I modeled it two ways:

1. plane stress - which corresponds to the "thin element" assumption of Peterson's stress concentration factor table, but not the pipe model
2. plane strain - which corresponds to the pipe model, but not Peterson's

The FEA results show that the peak stress at the bottom of the groove (Stress Tensor YY) is approximately equal for the two assumptions (3538 psi for plane stress and 3562 psi for plane strain). The difference is in the direction perpendicular to the plane shown. For plane strain, the stress is 0 and for plane stress, the stress is 1111 psi. Not that this is approximately 0.3*3528 psi; 0.3 psi being Poisson's ratio. The von Mises stresses for plane strain are less than for the plane stress assumption.

From this, I would say the the thin element stress concentration factors (plane stress) are applicable to prismatic situations (plane strain). Because the out-of-plane stresses are non-zero, the von Mises stresses are lower, which will benefit a fatiuge analysis.

Plane Strain SXX

Plane Strain SYY (Stress Concentration)

Plane Strain von Mises (used for fatiuge analysis)

Plane Stress SYY (Stress Concentration)

Plane Stress von Mises (used for fatiuge)

Using the model in #7, the hoop stress for the un-notched pipe would be 1780 psi and for a pipe with a 4.25" OD as described, 2830 psi. The FEA analysis indicates 3528 psi. This is be about 25% higher than the "back of envelope" calculation which took less than 1 minute to perform. It makes sense to me, and I always recommend this type of calculation as a sanity check. I can't count the number of times my analysts have provided answers that were a few orders of magnitude off because an assumed boundary condition was not realistic.

I do not know how you compute the hoop stress but if you use the equation σ=0.5*di/t your results do not fit.

Without the notch the stress is 568 with the stress is 902.4 and if the Lamé equations are used the values will be (on the internal wall where they are maximal) 621 and 953. The differences are not important the ratios di/de being 0.85 and 0.9 (near to 1).

The picture shows the bending I mentioned so that a simple approach could be considered a stress sum of the hoop stress and the bending. This leads for the hoop stress without notch of 568 to a stress in the groove of 2500. Even so the FEA result which the nearest to reality is about 50% higher. One sees that the notch factor cannot be avoided which ever source it has.

Although much more available due to drop of cost FEA soft are not every where so that the classical method has still a right to be used.

As a mention there is a FEA soft which is quite correct and does cost ONLY $50! It is not so easy to use as other but if one does not need to make a computation very often it is worth to have it. The name is "LISA".