What to use for SMYS 1

[sopha]

thread469-194331

Hi

The formula used in this thread utilizes SMYS as allowable stress.
Our small utility designs every pipeline to operate below 30% SMYS should I use the 30% value or the allowable stress of the pipe, x52

This thread has been very helpful, I am trying to predermine a trench profile, prior to a rock breaking tender.

Thanks

 

[zdas04]

Use SMYS. If you derate it to 30% then elsewhere in your calculations you'll use that value and get a number that is 9% of SMYS.

The derations are always applied after the calculation, not during (i.e., select a standard wall thickness that is greater than the ASME calculated minimum, then calculate pipe stress, then divide predicted stress by SMYS).

David

 

[sopha]

Thanks for this.

This has been very valuable information.

Bend radius has been a problem that no-one in my circle of contacts seemed to be able to solve

 

[sopha]

If you wouldn't mind taking a look at this for me, Just to check on myself

I used the following:

R = (E*D) / ((Sa)-(P*D)/(4t)), where,
R = radius in inches
E = modulous of elasticity 29842202 psi
D = pipe diamter in inches 8.625
Sa = allowable stress (SMYS) 359 mPa= 52070 psi
P = design pressure in psi 1029
t = wall thickness in inches 0.32

and come up with:

Allowable Bend Radius (inches) R= 5702.482
Allowable Bend Radius (feet) R= 475.2068

This tells me that supported on both ends, I will get 16" natural sag in the middle of a 40' pipe.

Does that seem reasonable to you?

Thanks, Dale

 

[SNORGY]

I compute about 1.4" deflection at mid span for 8" SCH 40 full of water across a span of 40 ft.

Maybe I am doing it wrong...

Regardless, 16" seems high to me.

Regards,

SNORGY.

 

[sopha]

That's like a 1700ft radius.
What did you use as a formula?

Dale

 

[BigInch]

52 ksi is not an allowable stress for bending unless you've got MagicPipe and no internal pressure.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[SNORGY]

In response to your statement:

"This tells me that supported on both ends, I will get 16" natural sag in the middle of a 40' pipe.",

I treated the pipe full of water as a simply supported beam and calculated mid-span deflection (natural sag under its own weight) as y = [5WL^4]/[384EI].

Which, of course, I may have not done correctly...checking anyone's calculations is always prudent.



Regards,

SNORGY.

 

[BigInch]

Snorgy,
Well, the formula the OP has used is not based on a natural sag bend. That's a permanently bent pipe, as that Sa value kind'a says its a yield stress with no safety factor, and the stress formula is based on maximum remaining allowed axial stress after subtracting the axial component of pressure stress.

sopha,
Since you mention "natural sag" I think you're trying to calculate natural sag, which for a beam with pinned ends the maximum would be 5WL^3/(384EI), where W is a distributed load on the pipe, not an internal pressure (which would cause no sag). Problem is that you've used an end moment bending equation M/EI to get radius of curvature which for end moments is not 5WL^4/384EI.

So is this,
1) a natural sag bend problem
2) a permanent bend radius problem
3) an elastic bend radius problem

The formula you are using appears to be derived from a total axial stress equation with radial stress subtracted, to get the allowable bending stress. The problem with the equation is that it only shows bending stress minus pressue stress and pressure stress isn't = PD/4/t. That pressure (radial) stress is PD/2t. But since we want axial stress and axial stress due to radial stress is Poisson's ratio * radial stress, it can be reduced. Since Poisson's ratio (u) is 0.3 for steel, and 1/2 is 0.5, it seems that 1/2 the radial stress is being subtracted, which is conservative, so it could be used, if you want a conservative result.

Now we have Sa - u * PD/2/t in there and if u = 0.5 then,
Sa - 1/2 * PD/2/t

If we go back to
Bend Radius, Rb_in = E_psi * R_in / Sb_psi
and substitute the remaining allowable stress for Sb
E_psi * R_in / (Sa - 1/2 * PD/2/t)
E_psi * R_in / (Sa - PD/4/t)
So, that should be conservative, if we chose Sa properly.

Sa here is Allowable Axial Stress, which for some codes should be limited to 0.9 * SMYS, but see your design code to be sure if axial stress or combined stress, or both need to be checked and what all the allowable stresses are for whichever stress you are considering must be.

We are saying the bending stress can be maximum when equal to Allowable axial stress - pressure stress in the axial direction.

If we are checking this stress to find bend radius with 0.9 SMYS as the allowable, we are assuming that the pipe is under pressure and elastically bent, not bent permanently. Radial stress and axial stress must be checked according to the code, either individually or both combined. This also assumes no temperature change.

If you calculate the bend radius using the allowable equal to SMYS (I assume 52 ksi), its a permanent bend, but when the bending force is removed, the bend stress goes to zero, so if checking stress for the pipe allowable operating pressure, only radial stress PD/2t remains; there is no axial stress from bending, only axial stress from u * PD/2t, again assuming thermal stress is zero.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[SNORGY]

I apologize...I, too, misunderstood what the design intent is...

Regards,

SNORGY.

 

[BigInch]

Not your fault. sopha didn't say what he was trying to do, so neither of us know what value for Sa he should be using. He has a formula based on M/EI and is talking about sag bends, so its a bit confusing.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[sopha]

I apologize for my own misunderstanding of what I was asking.
I think I have enough information to arrive at my conclusion (or be more confused by it).
I misinterpretted another thread.
What I am trying to predermine is what trench bed profile an empty pipe will naturally conform to without mechanical bending or over use of welded bends.

Thanks for your input

 

[BigInch]

A straight line should be made of trench bottom with 12" of sand bedding. The pipe will conform to that, as it is a continuous support. The sand should provide a bed that has as uniform load-deflection properties at all points.

I suggest you use cold over or underbends whenever possible (see tables in B31.4 and B31.8 for minimum recommended pipe bend radii) and form the sand bedding to match. Hot bends or standard fittings can be used when a cold bend won't fit in the given right-of-way, within the available vertical profile, or into an allocated pipe route in a plant or offshore platform, building, etc.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/