Effect of Tensioning a Boring Bar

[DrDaveParticipant @drdave]

Posted by Graham Meek on 22/02/2020 12:38:35:

In years gone by when I was learning my trade, mechanical engineering college courses included lab experiments.

That rang a little bell in my memory. The lab experiments that we did on my Engineering course were intended, amongst other things, to be a practical demonstration of a theory. If the results did not support the theory, then either the theory was wrong, the measurements were in error or the assumptions were wrong.

A solid cantilever bar of the same geometry as those tested should require 71.6 N to deflect 0.05 mm. A pretensioned bar cannot be stiffer than a solid bar, for the same diameter. Graham’s tests showed 37.8 N (3.85 kg) to deflect this amount and the pretensioned bar 44.4N (i.e. stiffer than the solid bar).

In the bending boring bar tests that Graham did, I have no reason to believe that the measurements were wrong. Equally, we know from decades of experience that the theory of beam bending is correct. Therefore my assumptions must have been wrong. The assumption that is probably wrong is that the cantilever is (dare I say) rigidly built in at one end. As Michael said, this is a convenient assumption because it is very difficult to achieve in practice. Any real-world flexibility at the support will decrease the load required to reach the 0.05 mm deflection.

I think that we have been concentrating on the bar in isolation, rather than considering the whole system. Small changes in the support would have a noticeable effect on deflection and vibration and could be an answer to the “why do two boring bars have such a difference in chatter” question. Can I ask, Graham, was there any difference in the way that the two bars (solid and pretensioned) that you tested in bending were supported that could have affected the results?

Dave

[DrDaveParticipant @drdave]

As an addition, to avoid cluttering my post, above, any more:

From a reference such as Roark, the maximum deflection of a cantilever loaded at the tip is delta = force x length^3/3EI.

Here, I = pi d^4/64 = 491 mm^4. E = 210 x 1000 MPa and length = 60 mm. So force to deflect 0.05 mm is 3EI delta/length^3 = 3 x 210 x 1000 x 491 x 0.05/60^3 = 71.6 N

[Martin KyteParticipant @martinkyte99762]

Posted by Kiwi Bloke on 23/02/2020 09:51:48:This makes initial deflection (lateral and torsional) minimal, until the preload is overcome. My understanding is that preloading a sprung system moves the stress/strain curve sideways, but doesn't alter its slope. So, with appropriate preloading, stress, up to the preload, can be applied without producing strain. The stress/strain curve of the preloaded system is thus initially vertical i.e. infinite stiffness (where 'stiffness' means Young's modulus). Or perhaps not. Have I gone wrong somewhere?

ummm, I think so. If you have a linear strees strain curve and preload puts you somewhere on the curve additional loads will still cause the same magnetude of extension or compression as without preload. Concrete beams are a bit of a red herring as the object is to ensure that all loads are compressive as far as possible.

I don't think the answer will be found in static deflection analysis but only by considering the dynamic system.

Boring bar chatter as far as I can make out, (and I do not set myself up as an expert in this field) appears to be be primarily modal rather than regenerative in nature. By this I mean that the bar bends into and out of the cut and also twists one way and t'other, with the tool tip oscillating in an elipse. I do not have the ability to do the maths but intuitively I can buy into the tensioning of the bar having an appreaciable effect on the energy transfer between the two modes of oscillation by causing the natural frequency of the two modes or oscillators to be non harmonic and thus making the dynamic stiffness or maybe resillience would be a better word? increase.

I do however note Graham's results on static tests which have not really been explained. Maybe the tesioned bar is twisting as well as bending as the load is applied. I don't know.

regards Martin

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Martin Kyte on 24/02/2020 09:47:41:

[…]

Concrete beams are a bit of a red herring as the object is to ensure that all loads are compressive as far as possible.

[…]

.

Concrete beams are pre-stressed by a tensioned rod, not a compressed one, but:

As I commented to Duncan, earlier : the conceptual similarity remains … it only requires a ‘change of sign’

Remember: in a bending beam … the top is in extension; the bottom in compression, and there is a neutral axis.

MichaelG.

[Martin KyteParticipant @martinkyte99762]

Hi Michael

My point was that the prestressing of concrete was not an attempt to increase stiffness or stop vibration but to ensure that the beam remained in compressive load. As you correctly say the force vectors are the same albeit with the sign change.

regards Martin

[Graham MeekParticipant @grahammeek88282]

Posted by DrDave on 23/02/2020 19:53:09:

Can I ask, Graham, was there any difference in the way that the two bars (solid and pretensioned) that you tested in bending were supported that could have affected the results?

Dave

Dave,

Both types of the boring bars were held in exactly the same holder, using exactly the same sleeve, (the sleeve is dowel located to orient the slits in the sleeve and the clamp), with the same length of bar clamped in the sleeve.

The discrepancy in results in theory over practical can be best explained by a sentence on the Sandvik website. The sentence relates to the actual flexibility of the machine tool itself during boring operations. While my Emco Maximat Super 11 is fairly rigid it would not produce the same results as in the test lab. Under lab conditions the block holding the bar would be bolted to an anchorage which would be known not to move.

Unfortunately my workshop does not run to this level of lab equipment and so the lathe is pressed into service. The Maximat has clearances which are necessary in order for the slides to operate, (I did however lock those I could). As both tests were carried out under the same initial conditions I feel some continuity exists in my method.

The practical versus theory results were as I expected.

As I said initially the structure needs to be looked at as an overall package.

There is no doubt in my mind that the push rod is a pre-tensioned Torsion bar. The pre-tensioned boring bar is being presented to the work with a preset loading. Which is opposite to the load which is about to be applied by the cutting action. The stress in a tube in torsion is said to be concentrated in the surface layer, this is also pre-loaded by the reaction to the load on the push rod.

Neither of these additional loaded elements are present in the plain bar.

How this all relates to Martin Cleeve's version I have yet to fathom.

Regards

Gray,

[Graham MeekParticipant @grahammeek88282]

Following a little light reading a reference was made to Machinery's Handbook, which led me to the following.

In my 15th edition, at the bottom of page 356, is a paragraph entitled, "Shear Stresses Combined with Tension or Compression Stresses"

The paragraph opens saying the complicated calculations associated with the above can be avoided by using the accompanying table on page 357. The paragraph goes on to show an example where the Shear Stress (S) is divided by the Tension or Compression Stress (T). The example gives a product of 0.75 for S/T.

From the table on page 357, 0.75 gives factors of x = 1.401 and y= 1.2. Quoting from the book, "This means that in this case the maximum combined tension will be 1.401 times what it would have been if there had been no shear". The passage goes on to say the that the maximum Shear stress is 1.2 times what it would have been if there had been no tension or compression stresses.

If the product of the S/T ratio had been 0.05 then x = 1.002, but y would be a factor of 10.05. Similarly the product of S/T at 1.5 gives x = 2.08 and y = 1.05. (0.05 and 1.5 are the extent of the table)

For my part the above gives a clear indication that the addition of the tension loading in the boring bar is a win, win situation. Not only is the boring bar experiencing this combined effect but also the push-rod, albeit the push-rod is in compression.

Regards

Gray,

 

Edited By Graham Meek on 27/02/2020 11:36:02

[Michael GilliganParticipant @michaelgilligan61133]

Posted by duncan webster on 14/02/2020 22:48:31:

OK I've changed it and sent it off to the FE man. as I said, don't hold your breath

.

That was wise advice, Duncan

Just curious, though … Has there been any progress ?

MichaelG.

[duncan webster 1Participant @duncanwebster1]

haven't seen him for obvious reasons

[duncan webster 1Participant @duncanwebster1]

Reluctant as I am to set this going again I decided to have a little experiment. I took the little boring bar shown in the first picture, clamped it in a vice and set a clock against it. With no load clock adjusted to zero. Then add a 5.6kg weight. Clock now reads 3.5 thou. The bar is 8.74mm diameter and the overhang was 54 mm. If the mounting was absolutely rigid and the bar was solid the calculated deflection is just short of 2 thou (0.0019&quot, so the combination of mounting flexibilty and the hole down the middle is having an effect. I then slackened off the centre rod, no difference, tightened it again, no difference. This is what I expected to happen.

This is not quite as Gray's tests as there is no torsion involved.

[Michael GilliganParticipant @michaelgilligan61133]

Thanks for doing that, Duncan

I have one question, after which I am happy to let this rest until someone does the FEA

How much end-load is on that central rod when it’s tight ?

MichaelG.

[duncan webster 1Participant @duncanwebster1]

Haven't a clue, it was 'tight'. As it made no difference I didn't see any point quantifying it.

Edited By duncan webster on 28/04/2020 11:35:15

[Michael GilliganParticipant @michaelgilligan61133]

Posted by duncan webster on 28/04/2020 11:33:24:

Haven't a clue, it was 'tight'. As it made no difference I didn't see any point quantifying it.

Edited By duncan webster on 28/04/2020 11:35:15

.

I think that nicely demonstrates the difference in our approach to this

… In the absence of any evidence: I would imagine that your ‘tight’ is in the context of what would be appropriate to hold the cutter in place.

To see any difference on a simple static load test; I would guess that you need at least ten [and possibly a hundred] times more tension than you are effecting. … Which is one of the reasons why FEA would be so informative.

Further discussion seems pointless; but thanks anyway.

MichaelG.

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