Dear All,

As we know that

Expansion stress allowable Sa= 1.5(Sc+Sh)

=> Sa = 3/2 (Sc + Sh)------------(1)

As per ASME B 31.3 Code

Sc= min( 2/3 Sy or 1/3 Sut)
Sh= min( 2/3 Sy or 1/3 Sut)

If I substitue Sc and Sh values in EQ-1

then Sa= 3/2((2/3*Sy)+(2/3*Sy))
= 3/2(2/3(Sy+Sy))
=(Sy+Sy)
=2Sy

Means for Expansion stress we are going up to two times yield value. I dont think so. Some where I did a mistake.

Please Correct me.

Is it because of stress range?? the value is 2Sy.

Durga,

Expansion stress allowable as per eq. 1a of the B31.3 Code is (1.25Sc+ 0.25 Sh), considering the stress range factor to be 1.

What you are quoting is the operating stress SoA = 1.5(Sc + Sh) as per equation P1a (i.e. equation used in appendix P, prior to 2014 edition of the code). Now along with this equation appendix P itself stands deleted in 2014 edition of ASME B31.3.

Correct, as per DineshK, if you put up the values your result will be 1*Sy.

Regards,
R.K.

In addition to my previous post, let me share you with the derivation/fixing of the allowable for this expansion stress range.

As proposed by A.R.C. Markl, in hot condition (i.e.conditions conducive to creep), the stresses will limit itself to the YEILD strength before producing any flow in the hot condition.
Simmilarly, at cold temperature the stresses will limit itself to the YEILD strength, before producing any flow in the cold condition.

So, the FLOW RANGE for the material gets confined within the YEILD STRENGTH of the material, both in hot and cold direction. Meaning to say, the range in which the material will undergo UN-YEILDED is twice of yeild strength (taking account of both the directions – hot & Cold).

Thus mathematically, Limit = (Sy + Sy) = 2Sy. …… (1)
But again, THIS INCLUDES ALL THE STRESSES CREATED BY THERMAL EXPANSION, PRESSURE & WEIGHT.

Also, as you pointed out correctly that,
Sc= min( 2/3 Sy or 1/3 Sut) and Sh= min( 2/3 Sy or 1/3 Sut), Substituting them in above equation (1), the limit becomes
3/2 Sc + 3/2 Sh = 3/2 (Sc + Sh)

i.e. 1.5 (Sc + Sh) …….. (2)

Now, by taking some suitable factor in account, the range established by code is
1.25 (Sc + Sh) ….. (3)

Now, recalling one of point mentioned above i.e. this includes ALL the stresses created by thermal expansion, pressure & weight, hence with the above statement, as the maximum allowed sustain stress by B31.3 is limited to Sh; this entire sustained value is subtracted from what was calculated in previous statement, making equation to be

1.25Sc + 0.25Sh ……. (4)

which is simmilar to conservative equation (1a) of the ASME B31.3 Code and stated in my previous post.

Again recalling that what I did in the previous statement was to subtract the entire sustain stress i.e. the maximum sustain stress a system can have i.e. equal to its allowable (Sh).
Actually, if there is a computed sustain stress for any system, we can subtract only that computed sustain stress (called SL) by the code instead of subtracting the maximum sustain stress value that a system is permitted to have.

This liberty is what is provided in equation 1b of the B31.3 code, and is what is temed as “Liberal Stress allowable” in CAESAR II terminology.

(The above explanation was based on considering the cyclic stress range factor (f) be equal to 1, which is another factor included in the calculation of expansion stress allowable computation)

Hope this clears your doubt!

Thanks for your reply,

But... What I understand is:

Code Applicable SIFs is only one-half of the theoretical SIFs.

Which means=>

Code stress is only one-half of the theoretical stress when an SIF is involved

When an SIF is involved, the stress calculated using the ASME B31 code formula is only one-half of the theoretical stress. This does not cause problems if everything is done within the range specified by the code, because the allowable stress has also been adjusted accordingly.

I think that is the reason the benchmark stress range SEB is 2 Sy
SEB = 1.5(Sc+Sh)

To adjust the allowable,
Markl suggested a total allowable stress range SA+SPW = 1.25(Sc +Sh)

Now,The expansion stress range allowable SA is half of that i.e 1 Sy

Any comments, Please share

Hi Durga,

Would you please share or state the basis which refers that "Code Applicable SIFs is only one-half of the theoretical SIFs which leads to the conclusion that Code Applicable SIFs is only one-half of the theoretical SIFs".

I'd also like to look into it.

hi Mr Dinesh
hope this will help you
The failure mode of self-limiting expansion stress is fatigue due to repeated operations. Therefore, to validate these SIFs, the most direct and logical approach is the fatigue test. After many tests and researches, Mark! found that theoretical SIFs are consistent with the test data. However, tests performed on commercial pipe also revealed an SIF of almost 2.0 against a polished homogeneous tube with regard to fatigue failure. This factor is mainly due to the unpolished weld effect, or clamping effect at fixing points, combined with the less than Perfectly homogeneous commercial pipe. To simplify the analysis procedure, the applicable SIFs are taken based on the commercial girth welded pipe as unity. This, in effect, reduces the applicable SIFs to just one-half of the theoretical SIFs(Reference Peng)

if you need more detials mail me jaypipingstress@gmail.com

EPJ

Dinesh,

Ref: L.C peng book; chapter-3; 3.4 STRESS INTENSIFICATION AND FLEXIBILITY FACTORS.

Well, addressing the concern directly i.e,

why computed code stress is half of the theoretical computed stress and similarly why the theoretical allowable is not halved accordingly as it remains the same as is derived based on yield limits, lets have this comparison:

(Commercial pipe) (Polished homogeneous tube) (Theoretically calculation)

SIF 2 1 1

Or

SIF 1 ½ ½

As the commercial pipes are used practically and CODES being meant to simplify the processes as simple as possible, have judiciously enforced the half of the theoretically calculated SIF values.

Actually, when the CODE is dealing with the commercial, practical conditions, it needs to have THOSE DATAS (SIF’s) directly put in instead of entering into the loop of theoretical or perfect/ idealized conditions and back calculating the results. This same is referenced by Sir L.C. Peng in his coveted work when he says “By using the commercial pipe with an unpolished girth weld as the basis, the code SIFs as given by eq. (3.11) & (3.12) are only one half of the theoretical SIFs. THE ADOPTION OF THIS BASIS IS MAINLY ATTRIBUTED TO PRACTICALITY.”

However, as the above things are mentioned in section 3.4 of the book, Sir L.C. Peng CONTINUES to write in section 3.5 of the same saying:
“the the stress allowable derivation of non-yeilding benchmark stress range REQUIRE FURTHER JUSTIFICATION before being used”.

Here it’s been clearly stated that why the STRESS ALLOWABLES derived from the yeild limits are not being factored.
“The first thing….” adresses the discrepancy arised due to SIF but the “Other thing….” clearly emphasise the practical aspect of fatigue cycle.

He further adds that “with f=1 at 7000 cycles, the allowable stress has already reached the benchmark stress limit. Any stress beyond that might produce gross yielding in the system, thus invalidating the elastic analysis."

Hence as pointed out by Durga, "I think that is the reason the benchmark stress range SEB is 2 Sy i.e. SEB = 1.5(Sc+Sh)",
half of code used SIF for COMMERCIAL PIPES is ONLY A factor which further JUSTIFIES the stress allowable derivation of non-yeilding benchmark stress range.

Hope this clears the doubt!