Force to Straighten Curved Beam 4

[ryldbl]

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.

 

[rb1957]

prex,

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 ?

 

[EnglishMuffin]

Prex : There is nothing wrong with saying y" = 1/R (approximately). But that y" cannot be the same as the y" in the energy integral you have quoted - the second differential of y for a curved beam is not zero when the deflection is zero, although the value of the strain energy integral should then be zero. Whether or not you made this assumption without realizing it I do not know, but you must have made an error somewhere since your conclusion is incorrect. I repeat - it is inconceivable that you can completely straighten a bar using axial tension if you cannot completely straighten a cable !

 

[JStephen]

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.

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.

 

[handleman]

I definitely agree with JStephen. The ludicrosity of an elastic structure with step response is mind-boggling. This has to be a case of bad assumptions and good math triumphing over common sense. (BTW, has anyone noticed that ryldbl is MIA? The man starts a fire and walks away....) Instead of this energy thing, think about it in terms of classical mechanics of materials. What would have to happen to the structure of the bar in order for it to be straightened? Essentially, the bar is a portion of a torus. Assuming the mentioned radius of curvature 'p' was for the centerline of the torus, (and say 'd' is the diameter of the bar), the length of the inside "edge" of the torus portion is 2*pi*(p-(d/2)) and the length of the outside "edge" is 2*pi*(p+(d/2)). Let's call this L1 and L2. In order for the bar to be straight, L1 must be "stretched" equal to L2. Therefore the gradient of the strain across a circular cross section must be constant from the inside of the radius of curvature to the outside. That means that the stress gradient must also be constant. I think the integrals etc to go from strain gradient to stress gradient to force required for a bar of rectangular cross section wouldn't be too bad. A circular cross section is a bit more complicated. I don't have time to play with either here at work, though.

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

 

[handleman]

OK, starting with a rectangular beam of width w and thickness t (t being parallel to the radius), we can call the inside and outside lengths concentric circles. L1 is the length of the outside and L2 is the length of the inside. Of course, R2=R1-t. The chord lengths are proportional to the radii to the inside and outside, such that L2=(L1)(R2)/(R1). The strain 's' required to make the inside length equal to the outside (and therefore straighten the beam) is s=(L1-L2)/L2. Combining the above, we get s=(R1/(R1-t))-1. Let's call this S for max strain. This is the maximum strain at the inside of the curve when the beam is straigtened. Of course, at this point you must stop and check the stress required to produce this strain. If it's above the yield stress then you will move into plastic deformation. As per my previous post, strain (and therefore stress) varies linearly across a cross section, from 0 at the outside to maximum on the inside. Let y be an axis with origin at the outside edge going towards the center of the radius of curvature. Then stress as a function of y is SEy/t (E=Young's). The force P will be the integral of stress times area over the cross section of the beam (y = 0 to t). Evaluated, this is SEwT/2. Taking all of the formulae above and combining to express P in terms of knowns, we have P = (R1/(R1-t)-1)(Ewt/2)

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

 

[rb1957]

i agree josh, a fun analysis.

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 ...

 

[handleman]

Yes, I was thinking impossible at first as well by comparing to the hanging catenary principles, but when you look deeper it's definitely possible (again, assuming this thing isn't suspended in gravity). rb1957, it looks like the W in your analysis is the t in mine.

There really needs to be some sort of sketchpad or something on this type of forum!

 

[rb1957]

yeas, i was keeping t for thickness, BUT ...

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) !

 

[handleman]

Uh-oh. Wait a moment. Get it? A moment? We simply cannot bend a beam (straight to curve or curve to straight) without an end condition that can support some kind of moment. Your mention of straining faces in opposite directions smacked me in the forhead with that fact. There would have to be some kind of perfect rigid plate on the end of our rod. Tensile force would have to be applied somewhere on this plate away from the centerline of the rod. And this is where the transition from theory to reality comes in. Anybody have a perfect rigid plate? Assuming we could find one, the question then becomes how far from the bar centerline you want to pull from. As the distance from centerline to attachment increases, the proportion of fibers in compression increases, and force required goes toward 0 at some rate.

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

 

[unclesyd]

Real World:
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.

 

[rb1957]

josh,

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).

 

[rb1957]

unclesyd,

maybe it was stress relieved afterwards !?

 

[handleman]

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. 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. 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. The bar will strain and straighten to an extent, but the force can never act along the centerline of the bar.

 

[rb1957]

sorry handleman, "don't get your banter ..."

"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.

 

[handleman]

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.

 

[handleman]

Hmm, seems like we're going in circles, rb. I just went back and re-read the entire thread. My last post says almost exactly what your first post does! What convinced you that your initial assessment was incorrect?

 

[rb1957]

handleman,

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 ??

 

[rb1957]

handleman,

i wish i knew !

 

[handleman]

I guess if you pulled the ends fast and hard enough, you could cause the arch to flatten dynamically in a vibrational sort of way, but once the system reaches static equilibrium it will be back to some kind of bent shape. As you said, the arch can certainly be flattened if loaded transversely, and the internal moment can easily be calculated. However, that is the moment that must be applied to the _straightened_ arch, and you can't apply that moment without the moment arm that the curved arch supplies.

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....

 

[rb1957]

like wrestling with a pig ...