Dynamics problem
Dynamics problem
(OP)
I'm ashamed to say that it's been a long time since I have needed to solve any dynamics problems but I now have one that I am stumped with. Can anyone help?
I have a joint of A-36 pipe that is sitting on 2 A-500 GrB tubing racks. The outside dimeter of the pipe will be pushed with an arm (at the midpoint of the length of pipe) that has a roller mounted on the end of it. The pipe weighs 43.7 kips and is 30" OD by 1 1/2" wall. What I am trying to calculate is the force that it will take to start the pipe rolling and the force that it will take to keep it rolling.
Thanks in advance for any suggestions for this problem.
I have a joint of A-36 pipe that is sitting on 2 A-500 GrB tubing racks. The outside dimeter of the pipe will be pushed with an arm (at the midpoint of the length of pipe) that has a roller mounted on the end of it. The pipe weighs 43.7 kips and is 30" OD by 1 1/2" wall. What I am trying to calculate is the force that it will take to start the pipe rolling and the force that it will take to keep it rolling.
Thanks in advance for any suggestions for this problem.
Tony Billeaud
Mechanical Engineer
Franks Casing Crew





RE: Dynamics problem
The actual rolling friction would, I think, be similar to that encountered by railroad wheels. Any railroad engineerers on here?
RE: Dynamics problem
The previous post is correct rolling friction it is and its not so simple. I have a formula here which might give an approximate minimum force to make the pipe roll ie:-
resistance to rolling, in pounds = (W*f)/r
where W = total weight of pipe
r = radius of pipe
f = coeff of rolling friction approx (0.02 inch)
for iron on iron
The answer given by the formula should give the approximate mimimum force to make the tube just roll.
There are lots of arguments regarding the correctness of this formula so it should be used as a gude only.
best regards
desertfox
RE: Dynamics problem
RE: Dynamics problem
I also did a quick calculation of the torque required to start the pipe rolling based on the polar mass moment of inertia and some assumed velocity. I assumed 10 ft/min. This converts to .021 rev/sec or .13 rad/sec. Assuming 2 seconds to accelerate to final speed gives an angular acceleration of .067 rad/s^2. Next the pipe's polar mass moment of inertia was calculated at 2417 in-lb-s^2. Torque equals polar mass moment of inertia times angular acceleration to get 162 lb-in. If the force is applied horizontally toward the center of the pipe then I believe the force (due to acceleration only) would be (162 lb-in / 15 in radius = 10.8 lb).
Overcoming the deformation of the pipe (both flattening discussed above and also deflection due to the support) and the rolling resistance on the pipe rack would seem intuitively to be a much larger force however. So I think the experimental approach seems best.
RE: Dynamics problem
RE: Dynamics problem
As to rolling resistance from the deflection in the pipe- I'm not sure there would be any if the deflection is elastic.
RE: Dynamics problem
Nominal Yield strength is 36 KSI.
Nominal Modulus of Elasticity is 29 x 10^6 PSI. (Like most steels).
Nominal Weight is 490 Lb/Ft^3 (Like most steels).
A36 has been in common use for over 40 years, but is now being replaced with higher strength structural steel, A992.
RE: Dynamics problem
In the absense of experimental data, you would need the Herzian deformations of the two bodies in contact.
See, for example, Timoshenko and Goodier, "The Theory of Elasticity", 2nd ed., McGraw Hill, 1951.
If you get the dimension of the " deformation ellipse" in the direction of the rolling, then, the weight vector would be displaced 1/2 of that dimension from the front, which means the rolling force at the center would be e/R *weight, where e is 1/2 the deformation dimension.
RE: Dynamics problem
The coefficient of rolling resistance in my earlier thread is unlike the normal coefficient of friction in that it has
units ie 0.02 inch. This is because the deformation pattern just ahead of a rolling cylinder is like a bow wave of a ship through water and the angle of the resultant contact force from this bow wave contact acts at some angle to the vertical but passes through the centre of the rolling cylinder, because the angle at which this resultant force acts along with its contact area is very small the angle is replaced by a horizontal distance at which this resultant force taken from the cylinder centre. The value quoted 0.02 inch is taken from experimental data and thus I feel could be used to find an approximate minimum force to start the pipe rolling. Using my formula I calculate 58.26lb of force
to start rolling, the force being applied at the cylinder centre. it is interesting to note that according to my formula that I need a higher force than that needed according to DVD's calculations in which is included an acceleration force to reach a particular velocity.
I will study this problem further and post later.
regards
Desertfox
RE: Dynamics problem
zekeman,
The racks are oriented such that the axis of the racks is parallel with the direction of pipe movement. i.e. The axis of the racks are perpendicular to the axis of the pipe. The racks are of 4x6x1/2 rectangular tubing with the 4" side in contact with the bottom of the pipe. The racks are not on an incline.
Thanks again for the help.
Tony Billeaud
Mechanical Engineer
Franks Casing Crew
RE: Dynamics problem
I believe that the way the force is being applied makes this a simple f=m*a formula. The only reason that the pipe will roll is it's resistance to sliding. No torque is being applied. I believe that the point of application of the pusher arm to the pipe and the point of contact of the pipe to the rack forms a couple; not a moment. There is the problem of the pipe rack deflecting. for example, if the pipe was resting midway between supports on 36" centers the rack would deflect .02" meaning the pipe would need to be pushed uphill.
Using the above assumption as well as the assumptions made by DVD for the speeds, I come up with the following values for the forces: Force to accelerate from 0 to 10 fpm in 2 secs = 113 lb; force to overcome rolling friction = 49 lb; force to roll uphill = 50 lb. Therefore, the total force required to bring it up to speed = 113 + 49 + 50 = 212 lb. The force required to keep it moving = 49 + 50 = 99 lb.
Of course if the racks were perfectly level and did not deflect, such as if they were sitting on a concrete floor, then 50 lb could be deducted from both.
RE: Dynamics problem
I would suggest you need to build something strong enough and not worry about theoretical matters. The initial push will require the greatest force so start with something with enough force to slide the pipe without rolling. That is the worst case. Now look at the equipment required and start stepping down in size of ram and see if you are saving anything by minimizing the ram size. If you are fabricating something for long term use, you will want to build in enough extra capacity to make up for wear and deformation. If it is too strong, who will know? If it is too weak, everyone will know. The material cost between an optimized solution and a too strong solution may not be significant.