Good thread. All good comments. I wonder if the OP thought there would be a simple answer.
It’s clear there is a difference of opinion among published industry experts, so not surprising that there are different opinions on the board. I am by no means claiming to know “the answer”. But that doesn’t stop me from trying to talk through it.
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Thanks John for explaining why gears are doweled as near the high-speed pinion as possible.
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Sam – I think most would agree Piatrowski went over the deep edge in specifying periodic checks. But the important thing is that it seems he believes machines can change alignment state even when no change in conditions (temperature etc), which I believe is emerging as the central question in this thread.
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Dan - I always respect your comments. I have zero doubt you have a much better knowledge of mechanical stuff than me. I’d like to play Devil’s advocate and explore the opposite position (that movement can occur).
Attached is my analysis regarding the potential for a motor to move for one particular motor at our plant. I’d say it’s not conclusive, but gives you something to think about.
The motor is 800hp, 1800rpm vertical motor. It is held in place by four 1.25” 7 threads-per-inch bolts (no dowels, no boss fit).
Tab “photo” shows indeed the oil had penetrated between the motor and the stool during operation.
We torqued these hold-down bolts to 470 ft-lbf based on info provided by the OEM. There is potential to go much higher based on bolt strength, but we were a little bit leery of thread-stripping in the tapped female threads. Analysis of bolt torque is included in tab “bolttorqueworksheet (revisiting this 2 years later, I’m wondering if we could drill out those tapped holes and use longer bolts with nuts down below... not sure if there is enough room to put nuts below or not... anyway we are working with OEM configuration so it’s not unreasonable to presume others can have similar OEM configuration).
As shown in tab “comparison”, each of the bolts generates around 22,000 pounds of clamping force. When adding in motor and pump rotor weight, we get a total of approx 100,000 psi downward force.
If we select a friction factor 0.05 for lubricated steel, that gives approx 5,000 pounds resisting force.
Motor full-load torque is 2350 ft-lbf. Motor breakdown torque (applied accross this joint during start) is approx 250% of full-load, or 5882 ft-lbf. This corresponds to a force of approx 4,300 lbf acting at the bolt-circle radius of approx 1.3 foot. 4,300 lbf applied vs 5,000 psi resisting static friction. This is not a lot of margin, especially considering all the uncertainties:
#1 – friction factor is notoriously unpredictable
#2 – estimating bolt preload from torque in field conditions can be somewhat unpredictable
#3 – Influence of vibration over time (?)
#4 ** Torque from unusual electrical transients can approach 20 times full load torque (much more than the 2.5 times full load torque used in this analysis).
Let me talk about #4 some more. Transients in the electrical system can cause momentary torques from 10 – 20 times full load torque (4 to 8 times the 2.5*FLT analysed above). These transients include:
1 – momentary interruption and re-supply of power. This can occur as a result of break-before-make transferring among sources or as a result of trip/reclose of utility lines.
2 – fault on the bus which feeds the motor (turns the motor into a generator).
There have been at least 2 plants that experienced step-increase of RCP vibration after grid transient. The cause was found to be shifting of the flywheel due to transient torques.
You may say
“Wait a minute: if the motor generates 10 – 20 times full load torque, that would break the shaft.” . But it is important to recognize that the torque is not transferred down the shaft, instead it goes primarily into accelerating the motor inertia. A quasi-static analysis gives:
Tshaft = Tpump + (Tmotor-Tpump) * Jpump/(Jmotor+Jpump)
Tpump is torque used by pump to pump fluid and will not change during transient (depends on speed, which does not change much). Jpump includes pump inertia plus some amount of water that is assumed to move with the pump. Typically Jpump<Jmotor and often Jpump << Jmotor, which means the majority of the torque is not transmitted through the coupling to the pump.
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And, each bolt in a clamped joint can be counted on to share the load, unlike multiple pins or dowels even with gorgeously precise machining.
I’m not sure why you say that. If the bolts in a clamped joint were all right next to each other, I would say they share the load because they share the compression. But for holddown bolts spread out, tightening one does not compress the joint below the other very much. And why would dowel pins not share load? They do have the benefit of precise machining because the practice is to drill the hole in place once the machine is aligned. And there will be either a tight fit or a tapered pin so no slop. How does it not carry/share the shear load?
quote:
Dowels by themselves, like shaft keys are simply incapable of preventing motion and resisting alternating forces long term (or even short term).
No-one suggested to use them by themselves. They are a diverse/redundant means of accomplishing the same task. Diversity among redundant components tends to improve reliability because the components tend not to be susceptible to the same degradation/failure modes. For example cylindrical dowels should not be affected by vibration in the same way that bolted joint may be affected.
And to see the relative effectiveness of holddown bolts vs dowels, let’s look at holding power of 1.25” bolt vs 1.25” dowel. In the spreadsheet calculation, using the (low) torque that OEM specified, we came up with 22,000 pounds clamping force for this bolt. And let’s not use the lubricated minimum friction factor 0.05, let’s be more generous and give it credit for 0.2 friction factor. So this bolt with clamping force 22,000 can resist a shear-direction force of 22,000 * 0.2 = 4,400 pounds before the joint moves. Let’s compare that to the force required to shear a 1.25” dowel pin. Bolt tensile yield stress = 81,000. Assume bolt shear yield stress is 50% or 40,500 psi. Area is pi*1.25”^2 / 4 = 1.22 inch^2. Shear Force before yield is 1.22*40,500 = 49,400 pounds. In this particular case, one dowel pin can resist about 10 times the force of one hold-down bolt. (49,400 ~ 10*4,400). Now I’ll admit dowel pins are often made as a smaller diameter than holddown bolts, but if we are comparing effectiveness of one vs the other, comparing similar diameter seems like a fair comparison to me.
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There are not a lot of times when we go in and check the
as-found alignment after a period of time. Usually we do in and do maintenance and aligne afterwards. So there is not a lot of data available to judge this. I am curious:
For people that check as-found alignments, what fraction of the time do you find movement occurs? (If the answer is as high as 10%, that might argue dowel is worthwhile on critical equipment imo).
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]SPREADSHEET ATTACHMENT MOVED TO NEW POST 10 October 2011 11:50 AM 
