Calculating L-Bend Anchor Force (Thermal Expansion)
Calculating L-Bend Anchor Force (Thermal Expansion)
(OP)
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.
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.





RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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/
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
* 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.
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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 *ΔT *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.
"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." [b]Tony Hayward CEO BP
http://www.youtube.com/watch?v=hpiIWMWWVco
"Being GREEN isn't easy." Kermit
http://virtualpipeline.spaces.live.co
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
"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." [b]Tony Hayward CEO BP
http://www.youtube.com/watch?v=hpiIWMWWVco
"Being GREEN isn't easy." Kermit
http://virtualpipeline.spaces.live.co
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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.
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
"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
http://www.youtube.com/watch?v=hpiIWMWWVco
"Being GREEN isn't easy." Kermit
http://virtualpipeline.spaces.liv
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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.
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
"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
http://www.youtube.com/watch?v=hpiIWMWWVco
"Being GREEN isn't easy." Kermit
http://virtualpipeline.spaces.liv
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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?
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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.
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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?
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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.
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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.
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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
RE: Calculating L-Bend Anchor Force (Thermal Expansion)
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.