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Force required to lift one end of infinitely long beam

Force required to lift one end of infinitely long beam

Force required to lift one end of infinitely long beam

(OP)
I'm trying to figure out the force required to lift up one end of a long pipe by 2m. The pipe is long enough that the entire length of it will not lift off the ground. It will bend under it's own weight and become tangent with the ground at some unknown point away from the end.

I tried to model this using various beam formulas but I end up with two unknowns (the force on one end and the length that lifts off) and can't solve them.


RE: Force required to lift one end of infinitely long beam

It's going to depend on the stiffness of what you are lifting. The stiffer it is: the more you are going to pick up.

RE: Force required to lift one end of infinitely long beam

I think you need to go back to old fashioned iteration where you guess at the beam length, solve for F knowing the tip deflection and accounting for self weight, and then check force balance as F and the total self weight of the lifted beam will cancel each other out just as the pipe touches the ground. Re-iterate until the forces balance.

RE: Force required to lift one end of infinitely long beam

Design as a cantilever. you know h.

Then F = L/2 * weight per linear foot.

RE: Force required to lift one end of infinitely long beam

If you can lift it an infinite height, the force would be infinite... or at least infinite/2.

Dik

RE: Force required to lift one end of infinitely long beam

I'm not sure if this'll get you anywhere, but do an energy calc ...
the external work done by F = the internal strain energy in the pipe (consider as a cantilever, length L).

This will give you a relationship in terms of the pipe's stiffness ... the stiffer the pipe, the longer L.

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

Quote:

I'm not sure if this'll get you anywhere, but do an energy calc ...

Why do that? I could knock this out in 5 minutes on STAAD. Just a beam on compression only springs.

RE: Force required to lift one end of infinitely long beam

if you're FEMing it, then you know the stiffness.

maybe the OP doesn't have access to a FEA ?

many ways to skin cats ...
(cats like none of them)

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

I seriously think that doing an iteration in excel would be the quickest and easiest. Using solver or goalseek would make short work of the problem.

RE: Force required to lift one end of infinitely long beam

F has to equal the weight of the pipe of of length L.

RE: Force required to lift one end of infinitely long beam

tomecki,

Try goal-seek in Excel.

--
JHG

RE: Force required to lift one end of infinitely long beam

How accurate do you need to be? Do you consider shear deformation? Do you include the distributed linear weight of the pipe to resist the force, or ignore it by assuming F >> weight? You need to know the modulus of elasticity of the pipe, the moment of inertia, and maybe some other things if you want to get super accurate. I would use singularity functions to solve this myself with a quick guess and check to hone in on the solution. The pipe is a cantilever with the fixed end location at the point where the deformation equals zero.

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

RE: Force required to lift one end of infinitely long beam

In addition, the shear/moment at the "fixed" end cannot exceed the allowable shear/moment of the pipe or you will buckle it before you get the height you want. You deflection curve would look much different if that happened!

Juston Fluckey, SE, PE, AWS CWI
Engineering Consultant

RE: Force required to lift one end of infinitely long beam

Quote:

F has to equal the weight of the pipe of of length L.

Actually no because you are creating a couple that will result in a increase in bearing pressure at the point where zero contact ends that will be in excess of its self weight. So by equilibrium, more will be picked up than the applied force. (How much will vary based on stiffness.)

RE: Force required to lift one end of infinitely long beam

(OP)
Thanks for the suggestions.

I assumed that F = 1/2 weight of pipe of length l. Then solved for l using beam formulas. I did one calc using the cantilever beam model and one using the simply supported beam model. The moment in the pipe at the location where it lifts off the ground must be small because both calcs gave very similar results.

RE: Force required to lift one end of infinitely long beam

I assume your simply supported model meant that the weight is support be F and by the reaction at the end of the pipe ?

by assuming F = wt/2, you are driving the cantilever solution to be close to the SS model (F = wt/2 acting at distance L reacts the wt acting at L/2 so the fixed end moment is small. F = wt/2 is possibly the limit of a rigid pipe.

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

Note that there are two different approaches above. If you treat the ground as compression-only springs, that's good, but it also means deflection is not zero beyond the first contact point, and that the slope is not zero at the first contact point. IE, it's a different solution than treating it as a cantilever, which assumes that at some point, it must be "fixed".

What I'd suggest for a theoretical solution- treat it as two beams, the "lifted" part being a conventional beam, the "contacted" part being a beam on elastic foundation. The "right" length is the one that shows zero deflection at the assumed contact point. If you get any "suction" beyond that point, the method fails, although it would show that beyond the first contact point is a point that lifts off the ground, which seems counter-intuitive.

RE: Force required to lift one end of infinitely long beam

I don't think anyone has mentioned the strength of the pipe. There is no guarantee that the end of the pipe can be lifted as much as 2m without yielding the pipe.

BA

RE: Force required to lift one end of infinitely long beam

That's a cute analytical problem (if we ignore shear and buckling effects), did you solve it? However I suspect the stiffness of the ground is crucial. But yeah, FEA, then plug the answer back into a hand calc as a check.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Force required to lift one end of infinitely long beam

If the ground can be assumed to be infinitely stiff then the elevated portion of the beam can be viewed as being simply supported.  If the lift-off at the end is small relative to the length of this simply supported beam then the simply supported beam can be treated as being very close to horizontal, and various small-angle simplifications will apply.  The key deflection equation (using simple beam theory) is the one for the end rotation under a UDL:
     wL³/(24EI)
Toss in F=wL/2, manipulate trivially, and you get:
     h = 2F^4/(3w^3.EI)

RE: Force required to lift one end of infinitely long beam

I thought Denial had beaten me to it, but I don't think his analysis is quite right.

Again assuming infinitely stiff support, the lifted part of the pipe can be considered as a cantilever with a point upward load at one end, an equal downward distributed force, and a point moment at the other end. The end deflection is therefore WL^4/3EI - WL^4/8EI, which gives the L required for an upward deflection dy as:

L = (dy.EI/(5W/24)))^(1/4)

Using guessed values for a concrete pipe of 0.5 m diameter of w= = 2.5 kN/m, EI = 75,000 kN.m2 and dy = 2m I get L = 23.166 m, which if I feed it into my cantilever spreadsheet gives me exactly 2 m deflection:



The end moment is about 670 kNm, which would be way over the bending capacity of an RC pipe of 500 mm diameter.

Also note that this assumes uniform stiffness. A segmented pipe with joints that allowed partial hinging would obviously allow a much shorter length, and reduced moment.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Force required to lift one end of infinitely long beam

IDS ...
"the lifted part of the pipe can be considered as a cantilever" ... ok,
"with a point upward load at one end," ... again ok, the force F, but
"... an equal downward distributed force" ... no (unless equal means uniform, which I don't think you meant from your equations); the distributed load is the weight of the pipe, not the force at the tip.

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

No, Denial, it's a cantilever, not ss. Look at the boundary conditions at the point of lift off. Deflection=0, gradient = 0. It is a cantilever encastre at the point of lift off. The analytical condition is a sum of two cases, F upwards, and a distributed load of mg/l downwards.

That ignores ground stiffness, shear effects etc.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Force required to lift one end of infinitely long beam

Quote (rb1957)

IDS ...
"the lifted part of the pipe can be considered as a cantilever" ... ok,
"with a point upward load at one end," ... again ok, the force F, but
"... an equal downward distributed force" ... no (unless equal means uniform, which I don't think you meant from your equations); the distributed load is the weight of the pipe, not the force at the tip.

Equal means equal in value to the end force. At the point of contact with the ground the shear force must be zero; if it wasn't the pipe would lift some more length of the ground, so the end upward force must be equal to the weight of pipe not in contact with the ground. The equations are the end deflection for a cantilever with an upward force W, and a cantilever with the same force W uniformly distributed over the length, downwards.

Quote (GregLocock)

That ignores ground stiffness, shear effects etc.

It assumes infinite ground stiffness, so it's an upper bound on the length. Shear effects with a pipe over that length are negligible. Variation in the pipe flexural stiffness would have a far bigger effect, but with a pipe of about 0.5 m diameter the only wayit is going to work without excessive bending moment is if the pipe joints allow sufficient rotation to get the required curvature without excessive bending.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Force required to lift one end of infinitely long beam

I think the lower bound for the lift force is 1/2 the weight, in this case the lifted portion is like a SS beam with zero end moment and 1/2 the weight supported at the ground end.

I think the end moment opposes the lift, so the lift force will be higher than 1/2 the weight.

The upper bound of the lift force ... I don't think there is one !? if the lift equals the weight, then there's no shear reaction at the ground contact point, only a fixed end moment (= WL/2). if the lift exceeds the weight then there's a down reaction against the ground and a higher fixed end moment (= WL/2 + (F-W)L).

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

I'm not convinced... if you add more force and lift the end higher, the point of tangency just moves to the right...

Dik

RE: Force required to lift one end of infinitely long beam

IDS has it. That's exactly how I was envisioning the calculations and the reasoning behind it. You can't ignore the stiffness of the pipe as suggested by others, this problem is mostly governed by the stiffness of the pipe.

RE: Force required to lift one end of infinitely long beam

jayrod:

Maybe mostly governed by the height you lift the end...

Dik

RE: Force required to lift one end of infinitely long beam

"At the point of contact with the ground the shear force must be zero; if it wasn't the pipe would lift some more"

no, I think it's not unreasonable that the shear at the ground contact is down (into the ground).

by balancing the weight of the lifted portion, then in the ground supported portion the weight is the ground contact force (UDL).
how does the ground supported portion of the pipe react the fixed end moment ?

If F > W, I see the reaction to the fixed end moment locally reducing the ground reaction DL.

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

I would tend to agree that the shear force must be zero right at the point the pipe starts contact with the ground. Obviously even an inch or two back from that point there would be load supported by the ground.

The amount you lift the end has a direct correlation to the stiffness and weight of the pipe.

RE: Force required to lift one end of infinitely long beam

to me it seems reasonable that the ground contact force is reduced near the lift-off point, rather than seeing a step change in ground contact pressure. something like ...

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

I see this structural system as following:
- Simply supported beam with span L.
- The Dy=2.0m vertical displacement in one end causes a rigid body rotation of the beam of value arcsin(Dy/L)=Dy/L;
- The pipe is loaded with a load w that causes a rotation at the end supports of wL^3/(24EI).
In order that the rotation of this point in contact with the ground is null, wL^3/(24EI)=Dy/L --> L=(24EIDy/w)^(0.25)
The internal forces are caused only by the beam self weight, so Vmax=wL/2 and Mmax=wl^2/8

For the data that IDS provided:
- EI=75000kNm2
- w=2.5kN/m
- Dy=2.0m
we should have L=34.64m; Vmax=43.3kN; Mmax=375kNm

For validation, I plugged this data into a SAP2000 model with compression only springs and the results are spot on for an end displacement of 2.0m.
I enclose a sketch of my calculations (sorry for the poor quality...).

RE: Force required to lift one end of infinitely long beam

this approach drives the lift force to be 1/2 the weight of the pipe, and zero moment at the ground contact point. I don't see that this is the only solution to the problem.

I think there's some point to the right of the contact that defines the affected part of the pipe; there'd be no shear force or moment on this section, similar to the pipe to the right of this point, the weight is continuously supported by the ground.

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

You're right. There's a small transition zone that depends on the relative stiffness of the ground and pipe. My SAP200 model was done with almost infinite spring stiffness so the extent of this area is quite limited. However, for practical purposes, I don't this as an issue.
I've run a couple of tests in SAP2000 with different compression spring stiffnesses (1E12, 1E6 and 1E3 kN/m/m) to see its influence on bending moments and shear. There wasn't a difference between maximum bending moments and shear forces greater than 0.5%. The difference lies entirely on the extent of the transitiion zone only. I've attached a couple of images showing bending moment and shear diagrams.

Bending moments:


Shear forces:

RE: Force required to lift one end of infinitely long beam

Apologies to Denial. He had it right.

For an infinitely stiff support, the curvature at the point of lift off will be zero, so the moment is zero and the pipe weight is distributed evenly to either end. With the corrected loads the pipe length required to be lifted is L = (dy.EI/(W/24)))^(1/4), and the lifting force is 34.64 kN (as found by avscorreia).

For a finite support stiffness, the lifting force and maximum moment are reduced, but not by very much. I get the following numbers from a Strand7 analysis:

Support Stiffness(kN/m/m) Lifting Force(kN) Max Moment(kNm)
100,000,000 42.8 371
10,000 42.5 371
100 37.3 290

Revised spreadsheet analysis, with the lifting force set to half the lifted pipe weight:






Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Force required to lift one end of infinitely long beam

Thanks, IDS.  I was half way through a convoluted attempt to convince you, rb1957 and GregLocock that the suspended portion of the beam is simply supported rather than cantilevered, and you have saved me from it.  I agree with your revised formula for L, provided you correct it for your unbalanced brackets:  it can be derived very easily from my result above.

RE: Force required to lift one end of infinitely long beam

Quote (GregLocock)

I'm going to have to wrap my head around that one, no BM at the point of contact with the ground?

Bending Moment -> 0 as support stiffness -> infinity

But for realistic support stiffness values the lifting force changes very little.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Force required to lift one end of infinitely long beam

Greg.  How about this argument?

Let x be measured along the beam and y be transverse to the beam.  Locate the origin at the point of contact/liftoff, with the resting part of the beam at x>=0.  Simple beam bending theory gives us that M/(EI)=y'' and V/(EI)=y''', where each successive ' denotes differentiation with respect to x.

Let x approach zero from the positive side:  the value of y is zero for all x>=0, so |y''|x=0+ is zero.
Now let x approach zero from the negative side.  If there is a bending moment there, then |y''|x=0- must be non-zero.
Therefore at x=0 y'' must undergo a step change, Heaviside function style.  In simple beam theory this cannot be achieved except by a concentrated applied moment.  There is no possible mechanism by which a concentrated applied moment could be generated in the simplified structure under consideration.  Therefore the bending moment at both x=0- and x=0+ must be zero.

Quod erat demonstrandum.

RE: Force required to lift one end of infinitely long beam

OK, I think the SS at both ends model is simplified to the point that it is no longer interesting, you've reduced it to lifting one end of a plank off the ground. In order to eliminate the impossible concentrated moment, you've replaced it with a concentrated vertical load of W/2 at x=0, and there is no downward deflection on the beam to create such a force.

However a s;ight;y more complex model is at least intuitively explaining what is happening.

Imagine the SS model, but slowly increase the torsional stiffness of the pivot at x=0. This will force downward deflection into the beam lying on the ground at x>0, which will create the upward W/2 force, and will also raise the beam off the ground at x=0.

So, I think the crucial interesting element is the interplay between the bending stiffness of the beam, and the vertical stiffness of the ground.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Force required to lift one end of infinitely long beam

Greg - the "infinitely stiff" support is just a limit to make the calculation easier, but with realistic soil stiffness the results are very close to the rigid results (as shown by both my results and those of avscorreia).

Suppose you had a pipe/plank resting on "point" supports, and were asked to calculate the span length such that if the left hand end was lifted by 2 metres, the right hand end would be rotated to horizontal. Would you still insist that we should consider the finite width of the supports at each end, and the resulting bending moments at the supports?

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Force required to lift one end of infinitely long beam

Depends what i was interested in! If I were designing the supports I might be very interested in knowing the stress distribution in the supports. Horses for courses.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

RE: Force required to lift one end of infinitely long beam

The question was what force was required to lift the pipe by 2 metres. The simply supported analysis gives a good approximation for reasonably stiff ground, and an upper bound force even if the support is very soft.

Here's the simply supported analysis, showing deflected shape for the horizontal pipe and pipe lifted by two metres, which rotates the right hand end to horizontal. I have also attached the spreadsheet.

Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/

RE: Force required to lift one end of infinitely long beam

how are you accounting for the gradual application of the ground support ?

The point to my second graph is there's a point on the infinitely long pipe where there is no effect from the lift, ground support force = weight. At this point you can have a SS beam. This is someway to the right of the lift-off point. The lift-off point is still (for all intents and purposes) a cantilever support.

This gradual variation in ground support could be a simple linear dist'n, but my money is on (as someone posted above) a more complicated non-linear dist'n (decaying cosine wave ? possibly a decaying exponential dist'n). But then goes to do you want the Right textbook solution or something near enough ?

As for displacement, how do you keep it linear for a portion, near the RH support ? by adapting the distributed load. and it needs to be linear with some relationship to the lift distance (horizontal from the SS point to the lift-off point, compared to the vertical lift of the LH end).

another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

As mentioned by others, 2 meters is a lot of deflection for a real pipe... but will put that aside.

Here's a 2-step, simple model for an accurate solution using hand calculations:

Step 1:

Turn the OP's problem upside-down (The pipe does not care, it's structural properties are symmetric and applied load is gravity. Assume the pipe "sticks" to the upside-down flat surface beginning at the tangent point).

Consider the upside-down pipe to be a cantilever beam. While upside-down, for any value of deflection "h", force "F" = 0 (when upside-down, the uniformly distributed self-weight of the pipe, over length "L", causes deflection "h").

For a pipe with known properties, "w", "E" and "I", calculate length "L" for the proposed 2 meter deflection:



Step 2:

Flip the model back right-side-up. Now the the pipe is represented by a "deflected propped cantilever".

When the problem is right-side-up, force "F" is no longer zero. Instead, "F" is the force needed to deflect the pipe (of length "L", calculated above) into the correct shape (calculated above).

Using the pipe's known properties and length "L", calculate force "F" (shown as R1 on the diagram below)



I'm not skilled with metric, so an example with modest deflection in US customary units is attached. Here is an image of the attachment:




www.SlideRuleEra.net idea
www.VacuumTubeEra.net r2d2

RE: Force required to lift one end of infinitely long beam

Sorry, but I disagree that it's a cantilever. The fact that the rotation is zero at the support doesn't mean that you have any stiffness to generate a moment. The ground does not generate a bending moment at the point of contact as it only allows compression, as Denial so elegantly demonstrated and as the numerical results show.

Regarding the force distribution at the contact with the ground, here are some additional results for the analysis cases I presented earlier:



RE: Force required to lift one end of infinitely long beam

is the force to lift fairly consistent ? From the graphs, I'd expect 1E12 and 1E6 to be very similar, and 1E3 to be a little lower (since slightly less pipe is being lifted).

what happens if you don't constrain the model at 60m ? (or if you did at 120m ?) If think the results show not much difference (since things look well behaved at 50m).

Is it interesting that peak soil pressure seems to happen at the same point, regardless of soil stiffness ? could this be an equivalent cantilever length ?



another day in paradise, or is paradise one day closer ?

RE: Force required to lift one end of infinitely long beam

avscorreia - For calculations, I'm assuming the tangent point to the surface is a "point of fixity"... somewhat analogous to the (hypothetical) "point of fixity" used for lateral load calcs on piling.

www.SlideRuleEra.net idea
www.VacuumTubeEra.net r2d2

RE: Force required to lift one end of infinitely long beam

The force to lift is very consistent. I got 42.3 kN for all cases (as I think it should, since the portion of the pipe that is not in contact with the ground is a statically determinate system).

There are some variations in soil pressure along the entire length for 1E3, so a larger length could probably stabilize it a bit, but I don't think it would be too significant.

The resultants of the peak stresses at ground contact do seem to be located around the same point. I don't know about an equivalent cantilever, as there is only support on one side, but maybe we should think of an equivalent support length.

RE: Force required to lift one end of infinitely long beam

SlideRuleEra - I know, but you can't have a moment there as there is nothing there to properly constrain the pipe, so the concept is different from lateral load calcs on piling.

RE: Force required to lift one end of infinitely long beam

I modeled Avscorreia's problem in STAAD.Pro and got the same results as he did. Tomecki, if you will give us an actual pipe size and material I can plug those in and see what I get. Unless, of course, you're just wondering how to solve this problem but you have no real world application.

RE: Force required to lift one end of infinitely long beam

(OP)
WOW! I just came back to this thread after moving on to something else for a while. Thanks everyone for contributing. I thought it was an interesting problem too, but I had no idea it would generate so much discussion. Might be a good one for a physics contest - simple question, takes a lot of thought to analyze.

Dozer, this is actually a real problem. The pipe is 36” DR9 HDPE:

OD: 36”
ID: 28”
HDPE SG: 0.958
Contained Fluid SG: 1.1
E: 130,000 psi

The manufacturer says that the pipe can be bent to a 60 ft radius so I don’t think this lift will overstress the pipe, although I haven’t actually checked the stresses in the pipe yet.

My results using the simply supported model were: L = 770 in and F = 14,770 lb.
Using the cantilever model with F = W/2: I got L = 763 so F is similar.

Bonus points: This pipe is actually floating in water, but I thought I’d start working on a solid support model for a conservative result.

Extra bonus points: The combined SG of the contained fluid and pipe make it sink, so it’s supported by floats.

Thanks again for your help everyone!

RE: Force required to lift one end of infinitely long beam

I thought you might be floating a pipe. I used to do towed arrays which are neutrally buoyant,floppy and compressible, so if a loop of the streamer starts to sink, it sinks some more, and suddenly you have an entire array on the seabed. That's very expensive.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

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