Calculating L-Bend Anchor Force (Thermal Expansion) 3

[robinso5]

Hello,

I'm having a hard time calculating some anchor forces and I wondered if anyone could help me out.

An 8" schedule 40 pipe contains 140°F Heating Hot Water. The system is filled with 50°F water during the summer. Guides are located 20 ft from an elbow. The pipe has a 200 ft long straight run into one side ofthe "L Bend", which is anchored at the opposite end.

I've been using equations from the Handbook of MechE Calculations as well as the ASHRAE handbook and arrived at different results, so I'm looking for more ways to go about this.

Thank you.

 

[ColonelSanders83]

A sketch of your layout showing guides and achors would be extremly usefull in answering your question.

A question properly stated is a problem half solved.

Always remember, free advice is worth exactly what you pay for it!

http://www.ap-dynamics.ab.ca/

 

[SNORGY]

I quickly ran this (or what I think is this) in CAESAR II with the following assumptions:

* dT = 90 F
* anchor @ one end of 200'-0" leg
* pipe supports every 20'-0" up to 5'-0" short of elbow
* pipe guides every third support starting from anchor
* anchor @ 20'-0" away from elbow at "short" leg
* coefficient of friction = 0.3
* uninsulated line full of water
* 8" SCH 40 A-106-B

The computed force I got at the guide was about 1500 lbf parallel to the direction of the pipe "long leg", i.e., transverse to the axis of the "short leg".

My output, of course, is only as good as the input.

Regards,

SNORGY.

 

[BigInch]

The length of both "sides" of the "L" is important, as is the anchor type at the toe. Is the toe fully fixed? One must also know the material to find the change in lengths using the material's temperature expansion coefficient. And one must also know the material's modulus of elasticity to find the bending deflection, bending stress and temperature stress.
if E = 30,000,000 psi
[&alfa;]=0.000006 in/in-F
Length of L's foot = 20 ft.

To estimate such a problem's results, take the force in the long side of the L,
F = [&alfa;]in/in-degree *[Δ]T *E_psi
F = 0.000006 in/in-F * 90F * E
F = 16200 lbs

M = 16200 lbs * 20 ft = 324 ft-kips at the anchor
As the moment moves across the foot, it reduces due to the shear load of 16200 lbs in the L's foot

How much it reduces when it gets to the elbow is a function of the ratios of one length to the other but would everywhere be numerically less than the 323 ft-kips at the anchor, since the 200' side of the L is very flexible in relation to the short side, thereby providing little if any resistance to any moment being transferred to that elbow from the anchor, essentially converting the elbow to a near-pinned joint.

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[BigInch]

I'd agree with that low force at the guide appearing to indicate little moment at the elbow, but the moment at the anchor would result in stresses there much greater than at either the first guide, or at the elbow. There is also a 16.2 k shear load there, and a 16.2 k axial load in the 200 ft long pipe.

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying." Tony Hayward CEO BP
"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.live.co

 

[robinso5]

Thanks for the quick responses, here's what I've been working on.

I've kept the E at 29.5x10^6 psi and used thermal expansion for carbon steel and arrived at about .008in/ft of expansion at 90dT.

I used this equation for an L-bend in the 2008 ASHRAE handbook, section 45.10.

F=12*E*I*delta/(1728in^3/ft^3*L^3)

Where E is the modulus of elasticity at 29.5E6
I is the moment of inertia at 72.5in^4 for 8" schedule 40
delta is the thermal expansion which comes out to about 1.73in for the long run
and L is the length of the leg, which is 20ft in this case.

This is assuming that there is an anchor on each end of the L-bend, and the Force isn't specified in any specific direction to confuse matters even further.

I arrived at about 3200lb with this method.

Next I used the Grinnell-Spielvogel equation from the Handbook of Mechanical Engineering Calculations as well as third equation from the M.W. Kellogg Design of Piping Systems. I arrived at a force of about 320 lbs parallel to the short leg, which is probably too small.

As for further clarifying the positions of the guides and anchors and supports, I am afraid that is the entirety of the inquiry I was given. I think it is safe to assume along the lines of how Snorgy set up the problem, but to have the last guide on the long leg be a full 20' from the elbow.

Thanks again for all the help, I'm relatively new to this piping analysis and may be making some hasty assumptions/mistakes.

 

[BigInch]

It could be closer as the 20's growth length is minimal. I'd say that 320 lbs is probably quite reasonable; as I would think the 1500 lbs SNORGY got would be around the maximum I would expect. All the problems in this problem will be found only in the short leg. The long leg will snake if the axial load is too high, thereby reducing the shear and moment in the short leg too, if it did that. Why not let it do that as much as possible, not that it will. What's the reason for the guides on the long line? Its not going anywhere in the lateral direction that I can see.

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying." Tony Hayward CEO BP
"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.liv

 

[SNORGY]

BigInch's comments are valid, particularly with the assumed boundary conditions for the restraint on the short leg.

I put the guides on the long line in my model in order to compute a conservative result for the reaction force at the guide in the 20-foot leg. In essence, I was artificially forcing the axial growth along the long leg in order to compute a worst-case load at the guide by taking away some of the relaxation (Euler) that would otherwise occur due to snaking of the long leg.

When I input the same pipe data (Metric units) and treat it thus:

dX = wl^3/(3EI) = a*L*dT
w = a*L*dT*(3EI)/(l^3)
w = (1.15E-5)*(60.96)*(50)*(3*207E9*3.025E-5)/(6.096^3) = 2907 N = 654 lbf

However, since this would be the result for the free-end of a cantilever, it would not account for the stiffness and resistance to rotation of the connected pipe and elbow, so I would expect the actual applied force at the elbow to be higher than that. For the simple cantilever beam analogy that I describe above, you would have to modify the free end boundary condition somewhat to reflect reality. This leads me to suspect that the CAESAR II result might be accurate, but my assumptions in any case err on the conservative side - especially the anchor on the short leg.

Regards,

SNORGY.

 

[BigInch]

Relative flexibility factors, using the moment distribution method will give you a quick idea of the problem. Flexibility of each "beam" is E*I/L The short one is K/20, the long one is K/200, roughly saying that the short leg will carry a(1/20)/(1/20+1/200) of any moment = 90% of any bending at the elbow, the long leg about 10%, essentially making the short leg a cantilever with a point load from the long pipe equal to the expansion force. Assuming it is, and placing all the moment at the anchor is a little on the conservative side, but not much.

"We have a leadership style that is too directive and doesn't listen sufficiently well. The top of the organisation doesn't listen sufficiently to what the bottom is saying." Tony Hayward CEO BP
"Being GREEN isn't easy." Kermit

http://www.youtube.com/watch?v=hpiIWMWWVco

http://virtualpipeline.spaces.liv

 

[robinso5]

Alright guys thanks for all of the input, maybe after looking at the anchor forces I'll take a look at the bending moments and shear forces between these methods also.

From a practical perspective, how conservative/accurate is the cantilever equation? I feel like it gives a fairly good representation of the deflection problem, but I don't know from a lot of professional experience when to not apply it.

The problem is that the ASHRAE HVAC handbook takes that cantilever equation that SNORGY just used, and multiplies it by 4 arbitrarily. Is that a reasonable safety factor or are they being paranoid?

 

[SNORGY]

I guess my own thoughts would be that, in my last post, I have unwittingly pointed out the difference between the "Simple Cantilever Method" and the "Guided Cantilever Method". The modifications of the free end boundary condition that I alluded to are, in fact, the resistances to rotation accounted for in the Guided Cantilever Method. per Page 55 of L.C. Peng's book - Pipe Stress Engineering, and - I imagine - as per similar sections in other books, there is an excellent discussion of this. In the case of a Guided Cantilever, the equations would then be:

dX = wl^3/(12EI) = a*L*dT
w = a*L*dT*(12EI)/(l^3)
w = (1.15E-5)*(60.96)*(50)*(12*207E9*3.025E-5)/(6.096^3) = 11627 N = 2616 lbf

This would be "worst case" with zero rotation and no other relaxations available.

So, the answer to the question posed by robinso5 appears to be that the constant "4" is not arbitrary; it's the correct number.

Regards,

SNORGY.

 

[robinso5]

Alright I believe I understand the problem now, and I will check out that book to compare and contrast the difference in the cantilever methods.

Judging by BigInch's post we will most likely be able to assume that there is a good deal of rotation and relaxation, and that our model will more closely align with a simple cantilever rather than a guided one.

However it does appear that while 600lbf is probably more accurate, 2400lbf is a very conservative and safe estimate.

Finally this raises the question, why doesn't the Grinnell-Spielvogel method in the Handbook of Mechanical Engineering Calculations, or Kellogg's Design of Piping, account for a lack of rotation?

 

[CRG]

An important consideration in determining the real load on anchors and guides is the flexibility of the anchors and guides. Often, these supports are considered rigid; however, they rarely are. If you include the flexibility of anchors and guides in your analysis, you may see a significant change in loading.

 

[pipengg306]

HI,
I MODELED in CAESER II, 16" PIPE, 20FT LONG WITH WALL THICKNESS OF 0.375in, T1=550F, P1= 550lb/sqr in, Material A106 GrB. It is Anchored at one end. When i check the Sustain load conditions and stresses that are developed in it. The following results has been genertated.


if anyone could help me out, how bending stress at node 10 = 2135.3

OPE STRESS: 7592.8 @NODE 10
BENDING STRESS: 2135.3 @NODE 10
TORSIONAL STRESS: 0.0 @NODE 15
AXIAL STRESS: 5457.4 @NODE 15
HOOP STRESS: 11183.3 @NODE 15
3D MAX INTENSITY: 14383.0 @NODE 15



10 2135. 0. 1.000 / 1.000 7593. 0. 0.
15 0. 0. 1.000 / 1.000 5457. 0. 0.


if anyone could help me out

 

[pipengg306]

HI,
I MODELED in CAESER II, 16" PIPE, 20FT LONG WITH WALL THICKNESS OF 0.375in, T1=550F, P1= 550lb/sqr in, Material A106 GrB. It is Anchored at one end. When i check the Sustain load conditions and stresses that are developed in it. The following results has been genertated.


if anyone could help me out, how bending stress at node 10 = 2135.3

OPE STRESS: 7592.8 @NODE 10
BENDING STRESS: 2135.3 @NODE 10
TORSIONAL STRESS: 0.0 @NODE 15
AXIAL STRESS: 5457.4 @NODE 15
HOOP STRESS: 11183.3 @NODE 15
3D MAX INTENSITY: 14383.0 @NODE 15



10 2135. 0. 1.000 / 1.000 7593. 0. 0.
15 0. 0. 1.000 / 1.000 5457. 0. 0.

 

[richay]

D = 16 in
t = 0.375 in
M = 12502.6 ft-lb = 150031.2 in-lb

I = PI/64*(Do^4-Di^4)
Do = 16
Di = 15.25
I = 562.0841165

Z = I / c
c = Do/2 = 8
Z = 70.26051456


Sb = M/Z 2135.355839


Richard Ay
COADE, Inc.

 

[pipengg306]

Hi richay,

how M(moment)= 12502.6 ft-lb = 150031.2 in-lb

because wt of pipe is 62.25 per ft. the total length of pipe is 20ft so total wt of pipe =1250.3 lb

Can u clarify how the bending moment = 12502.6 ft-lb

 

[richay]

Actually the pipe is 62.513 lb/ft.

From simple statics:

M = w * L^2 / 2
M = 62.513 * 20^2 / 2
M = 12,502.6 ft-lb

and
12,502.6 ft-lb * 12 in/ft = 150031.2 in-lb

Richard Ay
COADE, Inc.