Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Pipe Flange Bolt Stress - How does corrosion + reduced diameter affect original applied bolt stress?

Status
Not open for further replies.

flare9x

Petroleum
Aug 30, 2018
24
Ok so lets say we have the following:

B7 Bolt UNC 8, 105ksi yield
Bolt Diameter = 1.5 (D)
Lets say target bolt stress = 60% of yield, 63ksi.
K = .16

Using ASME PCC-1 bolt root area - that gives Target Torque , T =(K D F / 12) ft-lbs = 1772.788205 ft-lbs

So in theory - we apply the bolt root area torque of 1772.788205 ft-lbs and we arrive at a bolt stress of 63ksi - not accounting for relaxation.

So Force = Stress / Area. In-service our Area has reduced due to corrosion. Lets say bolt D now is 1.35 due to corrosion reduction.
Initially Force: Bolt root area, F = 88639.41023 lbs

So if we start with a Force of 88639.41023 lbs now that we reduce in area - then the force must drop also correct, as the area is now dropping also?

That is true - but Bolt Stress is unchanged - does that mean if I achieve 63ksi bolt stress then reduce bolt diameter to 1.35 - the bolt stress should remain the same? Less D = Less torque to achieve the same stress in larger D...




 
Replies continue below

Recommended for you

first off, way too many decimal places !

so as corrosion attacks the bolt, do you think you're relieving the preload ? IMO, no ... the original preload is reacted over a smaller area, hence higher stress. More importantly, bolt critical load is not preload in isolation. Simply put, corrosion loses area, which increases stress.

Now, if you have a corroded bolt and reinstall it, and say you can determine (somehow) that stress due to torque, so you limit the torque to 63ksi. Well, now you've applied less preload (less torque) and the joint may underperform (gap earlier).

another day in paradise, or is paradise one day closer ?
 
Ok so the corrosion and reduction in bolt diameter are not 100% even - so lets say we have full D, 1.5 at the tops and bottoms within the nuts.

So in the middle, exposed area its reduced to 1.35 - so stress over an area at the 1.5 and 1.3 sections are different. So a static bolt stress, 63ksi acting over a reduced section over time now 1.35 - this stress on this area will increase... hrm - well S = F/ A

Stress = 63ksi
A = 1.406974766 (root area at 1.5)
A = 1.109087891(root area at 1.35)

This shows if reduce area - then we reduce stress too.... OR it says if we want 63ksi over that area we need LESS force. but if froce already in, and we lose section - then higher stress?
 
Perhaps: Corroded Diameter Stress = Original Force Applied / Corroded Reduced (Root or Tensile Area)
 
that's more like it !

another day in paradise, or is paradise one day closer ?
 
Get real. You have only two significant digits past the decimal place. The equations are no more accurate than that.
 
Personally, I would replace all the bolts with new ones when corrosion is a problem. Meanwhile during the original installation of new bolts, the bolts would undergo preloading which stretches the bolts beyond their elastic limit and on that basis, I would replace with new bolts. The thing to watch out for is the condition of the internal threads when replacement with new bolts is expected.
 
Yes thats the only way to mitigate the risk because the bolt inside the flanges you never know what the condition is.
 
I'm not convinced that the pre-load is not reduced as the section corrodes away.

Although I've thoroughly confused myself thinking about this over my lunch hour.

I think I've settled on this: The pre-load has put the bolt into a strain. If the area reduces, stress increases. If stress increases, strain increases. If strain increases, the elongation would increase except that we have fixed length. So for the length to remain the same, the pre-load must decrease. Problem here is that the length of the decreased area also impacts the reduction in force as L becomes a function in the delta F. (As L reaches zero however, this whole thing doesn't matter)

I've thought more about this more as I've typed. My confusion probably has something to do with strain energy and is more than I'm going to re-learn during lunch.

How long is this theoretical 1.5" bolt?
 
[pre]
f_1 is the bolt tension or clamped member compression force before corrosion
f_2 is the bolt tension or clamped member compression force after corrosion
k_m is the clamped member stiffness
k_b_1 is the bolt stiffness before corrosion
k_b_2 is the bolt stiffness after corrosion
x_m_1 is the clamped member compression distance before corrosion
x_m_2 is the clamped member compression distance after corrosion
x_b_1 is the bolt stretch distance before corrosion
x_b_2 is the bolt stretch distance after corrosion

f_1 = k_b_1 * x_b_1
f_1 = k_m * x_m_1
f_2 = k_b_2 * x_b_2
f_2 = k_m * x_m_2
x_b_2 + x_m_2 = x_b_1 + x_m_1

f_2 / k_b_2 + f_2 / k_m = f_1 / k_b_1 + f_1 / k_m

f_2 * k_b_1 * (k_m + k_b_2) = f_1 * k_b_2 * (k_m + k_b_1)

f_2 = f_1 * k_b_2 / k_b_1 * (k_m + k_b_1) / (k_m + k_b_2)
[/pre]

If the bolt stiffness is small compared to the clamped member stiffness, a further reduction in bolt stiffness due to corrosion would lead to a small increase in bolt stress but a large decrease in bolt tension.

For an extreme example, imagine the bolt is replaced with five rubber bands stretched in parallel. Cut one band to simulate loss of bolt material due to corrosion. The total tension drops by almost 20%, but the stress in the remaining bands hardly increases at all.


pylfrm
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor