Force to Straighten Curved Beam
Force to Straighten Curved Beam
(OP)
I have a 25' round steel rod with a diameter of 1.125". This rod is not straight, but rather has a radius of curvature of "p". Does anyone know how to calculate the axial force necessary to straighten the rod? I don't know a formula to determine lateral deflection in a curved beam due to axial tension.
When I say "straighten" I mean elastically deform the rod into a straight beam, not permanently yield the rod to a straight condition.
When I say "straighten" I mean elastically deform the rod into a straight beam, not permanently yield the rod to a straight condition.





RE: Force to Straighten Curved Beam
Somethig simmilar I can see on "Spotts Design of machine elements" 4-th edition on page 197 as "Belleville Spring"
may be you can use that theory, check here with other proffesionals.
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
Assume your beam of length L may be considered as bent to a constant radius of curvature R (as you suggest): if that is not exactly true the following will give anyway a good approximation.
If the deflection at center with respect to a straight line joining the two ends is f then R=L2/8f and the energy that must be spent to go elastically from the bent shape to the straight one is (the same that would be required to bend it from straight to the deformed shape) W=EJL/2R2=32EJf2/L3.
In going from the deformed shape to the straight one the distance between the end points raises by ΔL=L3/24R2=8f2/3L.
Now if you call F the average value of the force by which you pull axially the beam to straighten it one has
F=W/ΔL=12EJ/L2
Hoping that all the above contains no mistake, if you take the double of that force I guess you'll be quite close to a straight shape.
prex
http://www.xcalcs.com
Online tools for structural design
RE: Force to Straighten Curved Beam
I would use a strain energy method or Castigliano's theorem
which can be found in most mechanics books.
If you assume one end is fixed and you apply your load at the other end parallel to that end face you should be able to work out the force required.
It would help if you could describe what angle the radius 'p' goes through and also the actual value of 'p'
regards desertfox
RE: Force to Straighten Curved Beam
i did like prex's observation that this is a two force member, so the bar looks like a shallow arch being pulled in tension. it's also symetrical so you can consider just 1/2, cutting at the crest of the arch. this end has a moment (due to the off-set). two questions come to mind ...
the off-set is reduced as the arch flattens, reducing the moment which is the only thing trying to straighten the arch. i suspect that the arch will want to form a plastic hinge, and will adopt a double arch shape as it is stretched.
also, this is a large displacement problem because as the displacements (that flatten the arch) have a significant impact on the internal loads.
possibly you could analyze the starting geometry, a cantilevered semi-arch. radial sections will be sufficient to determine the internal loads along the arch. apply the load in increments, determine the deformed shape, at some stage you'll need to reform the internal loads (depending on how you've formed the expressions (maybe you can iterate and converge on a solution for eah load step).
alternatively use some non-linear FE code (MARC).
whatever happens, i doubt that you can elastically straighten an arch (depending on your definition of straight!)
good luck
RE: Force to Straighten Curved Beam
F=(dW/df)/(dΔL/df)
Now dW/df=64EJfdf/L3
and dΔL/df=16fdf/3L
Now making the ratio as above the term fdf cancels out and one finds that the pulling force F=12EJ/L2 is constant throughout the straightening process. Hence, contrary to what almost everyone above was expecting, the required load does not tend to infinity, it is on the contrary constant!
So I must come back on my conclusion in the preceding post: you don't need to double the value given by the equation above, just take it as F=12EJ/L2.
prex
http://www.xcalcs.com
Online tools for structural design
RE: Force to Straighten Curved Beam
but also the problem is non-linear (each increment in load will not have the same transverse displacement) and the displacement is large (particularly in its effect on the internal loads). i believe that your analysis includes these assumptions, which are violated.
i did a quick linear run (a cantilevered semi-arch). i had 1" offset between the ends, radius of beam = 50", a steel bar 1/2" radius, applied 1000 lbs and got 0.03" transverse displacement with a max stress of 12ksi; so the linear (wrong) answer is 30,000 lbs with a stress of 360 ksi (a bit above yield).
looking at your equation F = 12EI/L^2, I = A^2/(4pi),
so axial stress = F/A = 12EA/(4piL^2), about EA/L^2
then stress/E = A/L^2, or E/stress*A = L^2,
E=30Mpsi, stress=100ksi, A=1in2, then L=sqrt(30000)=173"
but you previous comments still apply (i don't think this is right)
RE: Force to Straighten Curved Beam
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Force to Straighten Curved Beam
At any point in the rod, the moment is equal to tension at the ends times the deflection away from that line.
The curvature of the rod due to applied forces is proportional to the moment at each point.
In order to get the rod straight, the curvature has to be zero, which requires some finite moment to offset the inital curvature. But if rod is straight, there IS no moment. Or alternatively, as the centerline rod approaches a straight line, the required end force increases without limit. And note that the moments at the ends of the rod are always zero, so the ends of the rod are never straight.
One problem I see with the energy method is that it assumes the rod will take a certain shape, only the assumed shape is not one that can be formed by applying end tension. If the problem was restated as "What shape will be formed by applying end tension?", I think this would be more evident. You could assume that rod deflected into a pretzel shape and calculate the tension required to do that using energy methods, but I don't think you'd have a meaningful result.
RE: Force to Straighten Curved Beam
Wes C.
RE: Force to Straighten Curved Beam
For those that question this result (that surprises me the first) without providing a different route, I'll try to detail here the assumptions on which it is based.
The first assumption is that no account is made for the axial strain of the beam: this one is quite correct till the beam approaches the straight shape, after that any increase in the axial load will of course go into an axial strain.
The second assumption is that we may use the common formula for the energy of deformation in bending: W=(∫EJy'2dx)/2 the integral being extended over beam length and y being the deflection. I think no one can question this formula, that is known to work well for quite high deflections (though we are not dealing with a semi circle of course) and is valid for any deformed shape.
The third assumption is that the beam bent shape may be approximated with an arc of a circle: this one will be fairly good for most beam deflections that resemble an arc of a circle (not an S shape for instance).
The fourth assumption is that the arc is quite shallow. In this case we have the following approximations (the symbols are as in my first post above):
y'=1/R
R=L2/8f : this one comes from simple geometry and includes the approximations f<<L<<R that are of course acceptable for a shallow arc
ΔL=L3/24R2 : this one is obtained by making the difference α-sinα, 2α being the angle subtended by beam length from the center of curvature, and using the first two terms of the Taylor expansion for sinα (which is again good for a shallow arc).
From all the above, based on the relationship dW=FdΔL (in this differential form it is of course one of the least questionable principles of the classical physics), one obtains the formula F=12EJ/L2.
In all this:
- to rb1957: you are correct that this is a problem in the so called large deformation field (that is where the stress depends on the strain), but nevertheless all the above stays correct; the proof? The large deformation theory is not linear and indeed we have (surprisingly) a constant load for a changing deformation!
- to JStephen: I too was expecting a force tending to infinite when the shape approaches the straightness, but, as you certainly know, one has to be careful when dividing two quantities that both approach zero. It is true that the lever arm of the axial force tends to zero, but the same does the resistance of the beam to bending (or in other words the energy required to deform it), and the surprising result is that the force stays constant over the whole deformation (assuming as stated above a shallow arc)
prex
http://www.xcalcs.com
Online tools for structural design
RE: Force to Straighten Curved Beam
"As ryldbl provided all the required data, I can calculate the required load with my formula: F=1390 Newtons (sorry for not using american units)."
um, i didn't see a radius posted (only a length (25') at some radius, p).
um, this is a solid steel (which flavour; never mind, doesn't matter much) bar 1.125" diameter ... and you're going to straighten it with a force of 1400N (= 300 lbs) ...
i don't think so ... 300psi ??? you'd get this load just by hanging the bar from one end ! (25' = 300"*1in2 = 300in3*0.3 = 90 lbf ... ok, you get only 1/3 the force !!
again i don't think you're equations account for large displacement in a non-linear problem, and i think a quick sanity check on your result bears this out.
RE: Force to Straighten Curved Beam
Also what my formula says is that the load doesn't depend on the radius of curvature (provided this one is large with respect to length).
As of course I can't personally be 100% sure of a result obtained on the fly (no literature references that I know of), why don't you try to discuss the merit of my arguments or to obtain an independent result?
prex
http://www.xcalcs.com
Online tools for structural design
RE: Force to Straighten Curved Beam
i've run a couple of quick models. curved beams, about 30" long, 1" diameter, steel, loaded with 1,000lbs (about 3* your load).
one model had about 1" rise, and the load produced about 0.05" lateral displacement.
the 2nd model had 0.05" rise, and produced 0.0015" lateral displacement.
first, the lateral displacement is non-linear, the same force produced different effects.
2nd, the result is proportionally constant (which i thought was quite intereting). what i mean is that in both cases the load produced a lateral displacement = 3% of the rise.
maybe i should try my models with 300" length; but basically, i think that trying to straighten something by teniosn isn't particularly effective.
i did think your fR = L^2/8 is a neat observation, and a way to avoid f and R, so long as you're always using (fR) and not f or R. but think about the effect of 300psi (=2MPa, =1400N/625mm2) on a steel bar.
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
so the external work done is applied force P, displaced by L-2Rsinx; note we can't say sinx =x (even tho' x is small) because L=2Rx.
if x = 0.01c (L=300", R=15000", f=0.75")
then delta = 0.005" (well, 0.004999975")
so the external work is 0.005P
if x = 0.1c (L=300", R=1500", f=7.49375")
delta = 0.49975006"
then the external work is about 50*x^2*P
and the internal work is ...
RE: Force to Straighten Curved Beam
And EnglishMuffin I see no confusion: y'=1/R is of course the second derivative with respect to x of the deflection y.
rb1957 I agree of course that the relationship between loads and deflection is not linear in a beam beam bent by transverse loads and with an axial one, if that is what you mean: this is a well known result (a cubic equation is found for the deflection if I recall correctly) and you can see calculation sheets for this condition in the site below.
Coming to your models, I can easily confirm your results with my formulae (though .0015" is perhaps .0025"?): for a 30" long beam your load of 1000 lbs is about 1/20th of the straightening load I calculate, and in fact your calculated lateral displacement is about 1/20th (or close to that) of the initial rise. You should apply a load 20 times higher (or use a length some 5 times higher) and you should see in both models that the residual deflection goes close to zero.
I can check your results also in a more detailed way, using only the first model for brevity.
In the initial configuration the radius of curvature is (excuse me if I will use the SI) R1=2857.5 mm and after applying the load R2=3007.9 mm.
This requires a work of deformation given by W=EJL(R2-R1)2/R14/2=544 Joules
The displacement of the end under the load (you could certainly confirm that in your calculation) is ΔL=L3(1/R12-1/R22)/24=0.22 mm
and the force to do that is F=W/ΔL=2473 N = 556 lbs
This is not exactly what you find, but it must be recognized that the radius of curvature in your first model is not really big, so there is quite a big error in differentiating 1/R2. Suppose that the second model would give a better result, can you check and confirm the data?
And thanks to you rb1957 I think I discovered an error in my second post above (not in the first one however), but I'll leave to find it as an exercise...
prex
http://www.xcalcs.com
Online tools for structural design
RE: Force to Straighten Curved Beam
i notice you've linearised your equations ...
f=L^2/8R is an asymtope for f=R(1-cosx)
delta = L^3/(24R^2) uses only the first two terms of the expansion of sinx (=x-x^3/6+x^5/5! ... )
maybe this distorts the math ?
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
Perhaps in your energy equations, you are assuming that the applied energy is equally distributed along the bar? It is not, of course, the moment generated by the tension is maximum at the center and goes to zero at the ends.
The reverse problem is similar: given an initially straight bar, apply compressive end forces to bend it into a curved shape. Of course, it doesn't bend into exactly a cylindrical shape, because of the moment-variation problem mentioned above. And in fitting up tank shell plates with this method, considerable work with bull pins and key plates is required to induce that end moment. I believe your energy method would say this was not required; that X amount of force would bend it into a perfectly cylindrical shape.
I am very skeptical of a model that requires "the resistance of the beam to bending" to go to zero.
This problem does seem to me very similar to a suspended-chain problem. If you have a suspended chain, and want to know how much tension is required to make it hang perfectly straight, you can easily calculate it using energy methods. But in fact, as you apply more tension, you just get a flatter and flatter catenary; you don't reach some specific load where it is now "flat" and wasn't one pound before then. The problem, once again, is that you're applying energy methods between two shapes, but one of those shapes is not achievable.
RE: Force to Straighten Curved Beam
Of course, if this is a real-world application, you'll have to consider how the bar was curved in the first place. Are there residual stresses? Work hardening? Inhomogeneous properties? Also, how the heck are you gonna hold on to this thing? End conditions are pretty important.
Hopefully I'll play with this later if I don't lose interest.
-Josh
RE: Force to Straighten Curved Beam
Note that the actual length dropped out near the beginning. This makes sense, because with the circular arc of the beam, everything is proportional. Make t=0 and force required goes to 0. Makes sense. The larger R1 is, the smaller the force. As R1 approaches infinity, force approaches 0. Again, these all make sense.
Now to address the round bar issue. To find max stress, just replace t with 2r in the above derivation, where r is the bar radius. The biggest headache is in the integral of stress over area. With a rectangular section, the infinitesimal area dA is w(dy). For a circle, the infinitesimal dA=2(sqrt(r^2-(y-r)^2))dy. I don't really care to do that integration, though!
Of course, this is neglecting any gravity effects. If this thing is actually suspended between to ends, the catenary thing comes into play, although the actual shape will not be a catenary since a bar can support moment while a chain/rope/cable cannot.
This was fun...
-Josh
RE: Force to Straighten Curved Beam
however, i can't believe ('cause i haven't done the sums yet) that it's impossible to straight a minutely bowed beam with a tension load; damned inefficient but impossible ?
i liked your approach about making lengths the same ... we've had energy methods, simple statics ... so many ways to skin the cat !!
thinking about your rectangular bar, side W, so the outer fiber length is 2theta(R+W) (theta is the semi-angle of the bowed beam). as you say, this will change to be the same as the mid-fiber length (2thetat(R+W/2), a change in length of thetaW, and a strain (for the outer fiber) of thetaW/(2theta(R+W)) = W/2(R+W) = 1/2(R/W+1). let's use a strain of 0.004 as the proportional limit, so R/W = (1/.008-1) = 124. so that a rectangular bar bowed with a radius more than 124W can be straightened elastically, no ?
a circular bar should floow a similar result ...
RE: Force to Straighten Curved Beam
There really needs to be some sort of sketchpad or something on this type of forum!
RE: Force to Straighten Curved Beam
we'll obviously be straining the faces in opposite directions (compressing the outer face, tensing the inner), but the applied load is axial (when the beam is straight).
well, the mid-fiber is also straining, so that the final length is 2(R+W/2)theta(1+strain); and the inner fiber will extend to this length, an extension of thetaW(1+strain)+2Rtheta*strain. then inner fiber strain is ...
W/2R*(1+strain) +strain = 0.004 (proportional limit).
so if the axial strain (=(P/Wt)/E) is 0.002, then a beam with a bow of more than W/2*1.002/0.002 = 250.5 (a bit more than double the previous result) would be straightened by a load of 0.002*E*(Wt) !
RE: Force to Straighten Curved Beam
Decreasing that distance makes the proportion of fibers in compression decrease, until at some point all fibers are in tension. At some force the attachment point will actually be on the end of the beam. As force increases to infinity the distance from the beam centerline to the attachment point goes to 0, but of course you will yield the bar long before that.
So in the real world, we are sort of back into the realm of the impossible. You might get close enough with a carefully designed fixture with an attachment point at some offset from the centerline.
Of course, if you pull any harder than the specified calculated force, you will bend the beam in the opposite direction.
Basically, this is what JStephen said above:
"If the bar has a uniform curvature, then the only way to bring it to exactly a straight shape is to apply end moments such that the curvature induced is exactly equal to (and opposite from) the initial curvature. You can't do this by applying end forces."
I would append that final statement with "...at the beam centerline."
Still fun... Does it make me a complete geek to think that?
-Josh
RE: Force to Straighten Curved Beam
If the bar is bent has something yielded?
In a real world situation one approach to straightening the bar would be to apply a very slight twisting motion by some means as a force is applied to end of the bar. A collet is normally used to grip the bar a close to end as possible. This will take care of the rigid plate.
Depending on the method of manufacture some bars will rotate around the centerline axis while being bent.
RE: Force to Straighten Curved Beam
no, well at least that's not the way i see it ...
i have a symmetric arch, pinned at one end, pulled at the other (normal to the axis of symmetry). so effectively we cantilever sagging. i picture the applied force (and its reaction) as being on the CL.
and if you pull harder, won't you just be adding axial strain (the internal bending stresses cancel the displaced shape).
RE: Force to Straighten Curved Beam
maybe it was stress relieved afterwards !?
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
"That's how I pictured it at first too. But in order to have tensile stresses in one side and compressive in the other, there _must_ be an external moment supplied somehow."
... no, the load and the reaction are co-linear; for convenience we're looking at 1/2 the arch which has a single applied force, a parallel reaction (offset at the crest of the arch) and an internal moment reacting the couple.
"Otherwise we are ignoring the requirement that the sum of the moments = 0. A theoretical pinned connection is a point force incapable of supplying that moment. Surely you can picture the FBD of a slice of the bar and see the internal moment. The end of the bar must support that somehow."
... it is free body, in equilibrium.
"If the bar were straight to begin with, a tensile force pinned at the center would act along the centerline of the bar. However, the pinned arch's force acts along a straight line between the two pull points, not somehow along the centerline."
... yes, the external forces (load and reaction are co-linear and not directed along the the arch. the CL i was referring to is the neutral axis of the cross-section (the poster i was replying to was discussing offseting the load from the center line to create an end moment, which i don't think is required).
"The bar will strain and straighten to an extent, but the force can never act along the centerline of the bar."
... i think it can (in fairly special geometries) if the arch flattens completely.
RE: Force to Straighten Curved Beam
Looking at your half-arch, the initial moment arm is the height of the arch at center, and your force and reaction are parallel and horizontal. I'm sure you'll agree that as the bar bends toward straight it requires more and more internal moment. But at the same time, the moment arm (arch height) is decreasing! With moment required increasing and moment arm decreasing, force is shooting off towards infinity pretty quickly.
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
i can see your point (now) about the free body of the flattened 1/2 arch. i think the moment to flatten the arch is fixed, think about making a three-point bend load, the transverse load is easily calculated, no?
that said, the offset is reducing as load is applied so that eventually you'll have a very large force acting over a very small moment arm to create this moment.
i have a little problem saying we can't straighten any beam, what if the load was applied very quickly ? wouldn't the beam straighten ? how about the argument about reversing the question and trying to bow a straight beam (essentially buckle the beam) ?
maybe that's the point ... the strain argument i'll been posting only considers the effects of the loading (the strains and moments) without considering how the loads are going to produce those effects. i mean it's easy to straighten the beam with transverse loads, and this will produce the same effects we've talked about (strains, internal moments).
say we applied a very large force very quickly; if i'm right about the fixed moment required (which could be calculated by consideration of transverse loads), apply a force that'll produce a larger couple about the crest. won't the beam straighten ? i see it working that way. but you still have an internal stress distribution that's inconsistent with the external loading ??
RE: Force to Straighten Curved Beam
i wish i knew !
RE: Force to Straighten Curved Beam
As force increases, the arch does move toward straight. The harder you pull, the straighter it gets until you go plastic.
It is as you say, the internal stress distribution that we calculate for a straightened beam is inconsistent with the external loading conditions.
I sure wish ryldbl would chime in with some more information about the reason he asked the question in the first place. If the end conditions are something other than pinned, we're barking up the wrong tree anyway. Guess we just like to hear ourselves bark....
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
Presumably, if the bar was made perfectly straight and is not curved, then it has yielded.
You get the faces of the bar or plate to the yield stress when you roll it. But it will spring back somewhat, and the residual stresses are not necessarily close to the yield point afterwards. In other words, you could apply some amount of tension without anything yielding.
In a real world situation, this wouldn't be a problem- get 90% of the deflection out and call it done.
RE: Force to Straighten Curved Beam
Well, I'm glad nobody called me on the superposition thing, as I've forgotten how I did it!
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Force to Straighten Curved Beam
The application is a continous sucker rod string in an oil well. For those who are unfamiliar, a contious sucker rod string is a length of steel rod ~3,000 to 7,000 ft long which is used to connect a prime mover at surface to a pump underground. When the rod is transported to the well site, it is wrapped onto a large spool. When this is done, a slight amount of yielding occurs resulting in a slightly bent rod string.
It is imperative that the rod string be as straight as possible when placed in service. Every rod string will have a certain amount of axial load applied by the sub-surface pump. This has lead to the question: how much load is neccessary to straighten a rod with a given radius of curvature?
The 25' length was arbitrary and I only proposed it because this is the length that I am trying to get some tests done on. My thinking was that if I could get a methodology from you all I could apply that to any length.
In short, the load conditions are: The beam is pin supported at the top end and hangs under its own weight. I know that the weight of the rod alone is not enough to straigten it. I do not know the exact radius of curvature.
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
RE: Force to Straighten Curved Beam
I'll keep looking
RE: Force to Straighten Curved Beam
also, surely this problem has been uncountered before, as it sounds (to the uninitiated) to be a stadnard operation.
also, is the rod really axially loaded ? sure there is the weight to be considered, and sure the rod is being "forced" into its hole, but is "driven" a better description ... isn't the force applied to the rod being reacted dynamically (with movement of the rod) ?
i think unclesyd will come up with the answer (whihc i think is a couple of idler wheels.
RE: Force to Straighten Curved Beam
f=ML2/8EJ
This represents the center deflection of a simply supported beam under the action of two equal end couples M, so I suppose that in your reference the eccentricity e is at both ends of the beam, from what M=Pe.
Don't want to propose again my conclusions, that were clearly completely wrong
However ryldbl I understand now that with such a load and no initial eccentricity (except beam deflection) one would reduce the deflection to some 20% of the initial value, so that the beam would hardly be defined as straight. The load to straighten it to say 1% is really much higher and of course it is impossible to have it fully straight.
prex
http://www.xcalcs.com
Online tools for structural design
RE: Force to Straighten Curved Beam
Yes, there is axial load applied to the rod. The application considered here is not the reciprocating up and down beam pumps that you typically see in oil fields, but rather a PC pump where the rod is rotated to operate the pump. Thus the rods is exposed to a constant torque - something which I do not want to consider in this discussion - and a significant axial pulling force caused by both rod weight and pump operation. This is key, when the pump is operational, is directly applies a large axial load to the rod string.
RE: Force to Straighten Curved Beam
also, we're getting into the words a little more ... how much is "slightly", describing the plastic deformation ? i suspect that the 7000' of rod is coiled for transport, say a diameter of 25', and uncoiled at the site; now we're into a non-linear model of how the rod coils and relaxes. and how straight is "somewhat"; we've spent quite some time discussing gettting a mathematically (perfectly) straight rod from a deformed rod ... i suspect we're into another problem ... how much force is required to push the slightly deformed rod down its hole ? how much bigger is the hole (than the rod) ?
prex, you sounded a little defensive; i don't think there's any need to be, we're all trying to solve a question with little information (that's not a shot, ryldbl) and i think we're all taking our best shots and learning from them. hopefully we're not learning not to shoot !