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stresses and corrosion allowance

stresses and corrosion allowance

stresses and corrosion allowance

(OP)
B31.1 uses tn (pipe nominal wall thickness) for the calculation Z and therefor stess for all loads other than hoop stress. This means corrosion is not considered for sustained and thermal loads.

But

Paragraph 102.4.1 requires the design to include corrosion allowance and increase wall thickness over the "that required by other design requirements". I would take this mean corrosion has to be considered when calculating all stresses. It certainly means you can not ignore it.

So how is corronsion allowance to be applied to sustained and thermal loads?

 

RE: stresses and corrosion allowance

Corrosion allowance is an addition to the wall thickness above what is actually required to resist hoop stress.

You can use the hoop stress equation, temporarily ignoring the corrosion allowance term, arrive at an answer for the wall thickness actually required for hoop stress, then add the corrosion allowance to the previous answer.

Likewise, you can do the same for thermal load cases if you like, but those equations are geared to checking allowable stress levels of a pipe that already has an assumed wall thickness, rather than for a direct determination of what wall thickness should be used for a known stress, so its not so convenient.

It might make more sense to totally ignore corrosion allowance until you have finished the stress analysis for all load cases using some assumed wall thickness and, if all stress checks pass, add the corrosion allowance to your assumed wall thickness.      

 

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

RE: stresses and corrosion allowance

(OP)
I figure there are two ways to consider corrosion allowance for stresses other that hoop.

1. Most conservate and used by programes like Autpoipe. Calculate the forces and moments using tn then then calculate the stesses using tn-A

2. Calcaulte forces and moments and stresses using tn-A. Note you will need to also calculate support loads using tn.

I have come across some designers not using A at all when calculating stresses and this does not seen right!

 

RE: stresses and corrosion allowance

Autopipe isn't really correct either, if they do that for applied loads.  Increasing section properties reduces those stresses.

In thermal cases (since pipe inertias and x-sect area ratios hold with wt changes) stress analysis results in equal stresses no matter what wall thickness is used, but would increase end forces, since F = S * A.  If they subtract A before they calculate Anchor Forces, no problem.  For stress checks, if they subtract A, they are checking the same stress not using A, so its the same thing.  

 

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

RE: stresses and corrosion allowance

(OP)
I agree that thermal stresses are much the same wheather tn or tn-A is used for load and stress calculation but.

Susstainted stresses are much higher when tn-A is used as content weight, pressure and external loads do not depend on pipe thickness.

The allowable thermal stress equations use the susstained load margin (Sh-SL) which come from the Susstained stress calculation above.

The stress intenfifcatin factors are higher for tn-A.

RE: stresses and corrosion allowance

Lets look at a more detailed case.

12" pipes one is 0.375" wt and the other is 0.406 wt.
25ºF temperature change.

Beam fixed at both ends, End moment = wL^2/10, End moments equal for all wall thicknesses.  Axial force = 0 all wt. Bending Stress is 8% lower for 0.406 wt pipe.  If there were any axial forces they would be equal, but axial stress would e 8% lower for 0.406 pipe too.
External loads:  bending moments equal, bending stress less, axial stress 0. Are there capped pipe ends tension forces?

Thermal.  Applied Bending moment = 0, End Loads = S * A = 0.0000065 * 25 * 14.58 = 710775 lbs for 0.375 wt., 76878 lbs for 0.406 wt pipe.  1.08 x higher in the 0.406 wt.  Stresses are equal @ 4,875 psi.

Assume a 2" thermal displacement (eccentricity) at pipe centerline. Moment = 71077 x 0.1667 ft = 11849 in-lb for 0.375" wt pipe and 12815 in-lbs (8% higher) for 0.406 wt pipe.  Bending Stress Mc/I = 3245 psi for 0.375" wt pipe and 3265 psi for 0.406 wt pipe.  
Thermal case: Forces higher, axial stresses equal, Moments and bending stresses roughly equal in this case, but deflections might not always be proportional only to M/EI, as eccentricities and resulting secondary bending moments also depend on changes in axial lengths and bending caused by axial forces converting from axial to shear loads at bends.  That gets hard to generalize.

"Sustained stresses are higher when tn-A is used."
All stresses calculate higher when tn-A is used.

SL stress has to be calculated minus corrosion allowance.
 

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

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