×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Lateral Force From Stacked Pipes

Lateral Force From Stacked Pipes

Lateral Force From Stacked Pipes

(OP)
I've got an analysis task to evaluate a pipe storage system at a tube mill. The basic rack geometry is two vertical posts extending up from a horizontal wideflange beam at the base. Think of "U" shaped frames spaced regularly with steel pipes stacked directly on top of one another within these frames between the vertical rack members. My question is what is the lateral force produced by a stack of pipes? Pipes stored within a given rack will all be the same diameter and wall thickness but there are multiple pipe sizes that will be stored in different racks. I'm wondering if anybody knows of a resource that solves this general case. Default method may be to calculate the "density" of the stacked pipes and treat them as a fluid which would create a triangular force distribution on the wall but that may be too conservative. Any input is appreciated.

RE: Lateral Force From Stacked Pipes

I assume the stack will be stacked triangularly? which would just put the force acting on the verticals at the bottom.

If the stacks are stacked in such a way that the side vertical walls are resisting them then i would look at it as a vector analysis. Draw a free body diagram of three pipes stacked in a pyramid. and look at how the weight of the top pipe flows through the bottom two to be resisted laterally. This is your thrust. Now increase the weight of the top pipe to match a height of total stack (ie 20 pipes) and see what that thrust is. When you expand the model to 3-levels of pipes it should make complete sense and you can tell how it works with N-levels without any more models. These models can just be mental exercises. There is no friction.

I would not assume that the pipes sit directly on top of each other, not in perfect vertical/horiz rows, but rater each row will be offset 1/2-Diameter. They will stack in a dynamically stable position.

And you may also want to check if blocking is added between rows, I have seen it with 4x4s between every row to increase ease of lifting with forks.

RE: Lateral Force From Stacked Pipes

Without crunching numbers, you need to be triple whatever static load you come up with for impact loading of dropped pipes and such. Look at forks on a log truck. Obviously sized for dynamic loads.

I'm totally guessing here, but I'd guess the force is opposite of fluid pressure. The lateral loading the top pipe places against the side wall is much higher that the load placed at the bottom. The force at the bottom is very small. I would start by looking at what pipes would fall if you release the side but held the bottom pipe in place. Some sort of triangular shape. I'd directly apply the vertical weight as a horizontal load to the side wall. Than I'd multiply by 3. Or 6. Maybe 10.

There is no extra credit for saving material costs. Assume the operator is lazy and throws pipe against the sidewall. Just thinking out loud-I have no experience with this. I would want to play around with this till I was comfortable.

_________________________
Tony Krempin, PE
TopKnot Engineering

RE: Lateral Force From Stacked Pipes

Based on a free-body diagram for one pipe sitting on two others, the lateral thrust is (tan 30)(W/2) where W is the weight of the pipe per foot. You can extrapolate this for a bin of any width and determine the pressure to the side wall. Seems to me it would be triangular at the top for a depth down half of the width of the U, then rectangular from there to the base.

Mike McCann
MMC Engineering

RE: Lateral Force From Stacked Pipes

Also think about loading the rack. Are they dropping these things on and then pushing them into place? If they load like that, or otherwise push things around once they're on the rack, you'll have a force equal to the friction factor between pipe and rack times the weight of the pipe

RE: Lateral Force From Stacked Pipes

Can you sketch this as I'm not clear if we're talking about a standdard pyramid shape of pipes or something different. What is the size range of these and how high do they stack?

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

RE: Lateral Force From Stacked Pipes

(OP)
I'm not having luck attaching a pdf sketch. The rack structure is a series of "U' shaped frames with the horizontal part of the "U" sitting on the floor and the vertical elements of the "U" spaced at about 5 feet apart. The verticals are 15 feet tall. Thus, the pipes being stored are stacked between the uprights creating a rectangular block of pipes 5 feet wide x 15 feet tall. There is no dunnage or spacers between rows of pipes, they are just piled in to fill the space. Straps are placed across the stack every so often to allow an overhead crane to lift out bundles of pipes.

The size of pipes being stored ranges from 3" diameter to 12". Each rack will contain pipes of the same size at a given time.

RE: Lateral Force From Stacked Pipes

(OP)
Yes Crackerjack47, that sketch describes the situation.

RE: Lateral Force From Stacked Pipes

I'd drop the question in a forum for geotechincal problems. If you know the density of the pipes per cubic foot, the friction angle of the stack, a soil mechanics formula will give the result as if these were round rocks placed against a wall. A rough estimate of friction angle would be what angle does a stack of pipes sit at with the bottom row restrained from sideways movement. The impact part would come from assuming the stack is maybe 5 times the height you have to design for. Then with a bunch of small diameter pipes the density would be much higher than for large diameter pipes. So you need to know the sizes.

RE: Lateral Force From Stacked Pipes

OK, maybe impact is not times 6. However, the active equivalent fluid density horizontally against a wall is:

P=density x 1 - sine of friction angle / 1 + sine of friction angle. For equal spheres that friction angle is near 26 degrees.

Therefore, an equivalence fluid unit weight (P) would be 0.39 times what ever unit density you decide you have. This is a triangular pressure distribution. For impact, the lateral load imposed would be 0.39 times the effective vertical load from the drop. I'd just add that as a horizontal load at the top of the vertical brace.

RE: Lateral Force From Stacked Pipes

Crackerjack47's diagram presupposes that the inside width of the "U" frame is an exact multiple of the outside diameter of the stored piping.  This will seldom, if ever, be the case.  As a consequence, any form of neat geometric stacking will be impossible (even if the person doing the stacking was prepared to spend the time to try to stack neatly).

RE: Lateral Force From Stacked Pipes

I agree with Denial. I think the solution will be highly variable depending on the way the pipes stack. In the second diagram shown by Crackerjack47, the pipes align perfectly, one above the other and there is no lateral pressure on the vertical arms. If they don't align, the lateral pressure will depend on the angle between centers of abutting pipes and as the deflection of the vertical arms changes, those angles will change.

Offhand, I would say it is not a good idea to stack pipe in this way.

BA

RE: Lateral Force From Stacked Pipes

BA

What is a better way to stack pipes? Not only pipes but steel rods, logs, Newspaper rolls, Steel sheet rolls, etc, get this treatment. Very common. One way to answer his question for sure would be to do a trial model with light load measuring scales. The width of the "U" also affects the result. A test easily done at his shop. However, my method has been tested plenty. It needs a few "( )"'s

P= pipe density x (1 - sine of friction angle) / (1 + sine of friction angle). Where P is the density of an equivalent fluid. The higher the stack, the higher the loading.

This formula is for infinite width of stack. Worst case. It assumes a horizontal top of pile. Going higher, without spilling, adds to this "equivalent" density. It can add 12 percent to total loading, for a 26 degree friction angle.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources