Bandsaw blade and its failure mode 1

[John_Vreede]

Hi all. My question relates to a bandsaw blade and its failure mode. Its a 1/2" wide x 0.025" thick blade, tensioned to a given stress level (30,000psi (about 250-300lbwt on the cross section left once teeth are ground in) for a bimetal blade or 20,000psi for a carbon steel FlexBack blade). The back roller guides are about 8" apart and the work presses into roughly the middle of that span. So its a simply supported very slender beam, loaded in the middle, that carries significant tension. In use, the back of the blade deflects upward by about 0.040" while still cutting straight.
I'm assuming the blade, as a beam, will be stable up to a certain load, allowing the blade to cut straight and square, then go into lateral torsional buckling. Then the teeth won't be pointing straight down any more and the cut will veer off from the intended line of cut.
My questions are "Does the tension on the compression side of the blade fall to zero before it deflects sideways? If not what is the limiting factor when it does deflect? Is there any way to calculate the tension on the tension side when it goes into buckling?
Regards - jv

 

[SwinnyGG]

Given the dimensions you know, you could determine what force is required to induce torsional buckling behavior and then determine if compression stress due to that load is greater in magnitude than the applied tension. At least some of the of the real torsional load on a band saw blade comes from the fact that no piece of metal is perfectly homogeneous- inclusions and areas of different hardness, even though the differences are very very small, will still cause the blade to wander. I suspect that this behavior would be difficult to separate from the torsional buckling induced by a simple model of forces, if you were to compare calculated results to real-world testing.

I think that if you're building a model of this behavior, you'd have a closer approximation by assuming the free section of the blade to be a cable under some level of preload. There's two reasons I think a cable model is closer to reality, although someone more knowledgeable may prove me wrong:

1) the cross section of the blade will support very little compression before it buckles axially, just like a cable would, because it so slender.

2) If you want to do a really deep dive, the sum of forces on the working part of the blade also includes an internal force couple between two forces- one at or near the centroid of the cross section in the direction of travel, due to the power applied to the blade to make it move; and another at or near the centroid of the tooth geometry, in the opposite direction, due to the cutting forces applied to the teeth. This is atypical of normal beam forces.

 

[Compositepro]

I believe that the parameters you are talking about are way outside the normal operating range of a band saw. Tooth loading is one of the more important operating parameters of a saw blade. This is the pounds of force pushing each tooth that is actually cutting into the cut. If more force is required this is adjusted by changing the teeth per inch (tpi) to a lower number, rather than pressing harder with the blade. As long as the toothed side of the blade in tension and the back of the blade does not touch the sides of the cut, the blade will cut straight.
You should be thinking about the blade like a string under tension. The teeth cut a kerf that is wider than the blade so that the blade will not be deflected by touching anything. only the cutting edges of the teeth contact anything in the cut.

 

[John_Vreede]

Thanks for the replies guys. I've googled LTB but all the calculations relate to I-beams or other sections not slender beams, and none feature any pre-tension. I was hoping one of you mech-guys would maybe have a formula that I could use.
I'm unclear on the mechanics of a pre-tensioned beam, other than an un-tensioned beam this slender has almost zero load carrying capacity, but quite a bit when it is tensioned, certainly more than the ~20lb that is applied.
Be that as it may, the parameters are those published by all the blade manufacturers. 30000psi seems like a lot, but 600lb pulling the bandwheels apart (1 run of the blade on either side of the wheel)is well within normal for a metal cutting bandsaw.
Yes the blade as a beam is anchored to the bottom of the cut by the set on the teeth but the sides of the blade are clear of the walls and can move and only a portion of the blade is in the cut anyway, with the rest of it in the air, and free to move in any direction, especially if there is little or no tension.
The cable analogy doesn't work for me as the blade has a direction and its the direction I'm interested in. I don't know for certain but I'm pretty sure that a bandsaw blade cuts in the direction its pointing. With proper alignment, tension and weight on the teeth it cuts directly downward every time. But when you overload it, it veers off the line it should take, so I'm surmising that the blade has buckled and is no longer pointing directly down.
I just re-read Compositepro's reply and need to add that the force pushing the teeth into the cut on these horizontal bandsaws is about 20lb (the weight of the headcasting minus counterbalance, acting at about 12" from the pivot). I didn't mean to imply the 250-300lb was the weight in the cut, that is the tension in one run of the band.
Any help with calcs or mechanics and comments on the logic greatly appreciated - jv

 

[SwinnyGG]

Veering of the blade can result from a lot of things- inconsistencies in the metal being cut being one.

The other one is the tooth loading, which Compositepro hinted at- if the teeth are overloaded, the root of each tooth can fill with chips before the tooth exits the work. This 'jams' the tooth, and creates all sorts of weird forces which can drive the blade off line. The fix for this is lower swing speed of the head.

Blades are very very hard- they are flexible because they are very slender, not because they are highly ductile material- similar to how glass fibers flexible because of extreme slenderness.

other than an un-tensioned beam this slender has almost zero load carrying capacity, but quite a bit when it is tensioned

This describes the behavior of a cable exactly.

30000psi seems like a lot

It isn't, when the blade material probably has an Fy in the 80-90,000 (or even higher) range.

 

[acspain]

Your question is bouncing around in my head so I thought I would share.
I have used three different saws, a wood shop vertical saw with the band loaded via a die spring, a wood mill saw (creating boards from logs) whose band is loaded by a bolt and the horizontal weighted metal saws (unclear how these bands are loaded). All of which I have seen the band twist and create a poor cut.
The main drivers of a twisted band, assuming all the guides are set correctly and the band is sharp, are lack of tension which provides the band "beam" stiffness, in feed direction, and feed of material or band being too fast and therefore loading the blade too much.
Based on your comments about band loading, it would seem you have the correct tension.
The feed of material or blade, is also affected by the number of teeth and how dull the teeth are. Additionally if the set of the teeth has been affected this will cause poor cut tracking too, one side cutting faster/more material than the other.
As mentioned before, the teeth should be doing the work and the feed load should be small.

Your not finding hand calculations for this problem because it is not simple, reference the following;

https://deepblue.lib.umich.edu/bits...d=00504B1A26A79CCE9C7AE3E43AB65BA4?sequence=5

Answering your remaining initial question directly; no, you will always have a tensile stress field in the cross section of the blade. Because this is maintained by a spring or bolt holding the wheels apart. In fact the tensile stress field will increase with feed loading and be non-uniform. Buckling will occur once the difference between the stress in the leading edge and the trailing edge becomes too large. The trailing edge will seek a lower energy state by twisting, moving towards the center-line of travel, thus reducing trailing edge stress.

If you want to pursue this further a Finite Element Analysis is in order. I did a quick static one with SolidWorks Simulation to verify my thoughts but vibration due to the teeth will cause buckling to occur before the static analysis prediction. Additionally the stress field will fluctuate with each contact between tooth and material.

I hope this helps.

 

[Tmoose]

A bunch of bandsaw blades have bees successfully welded like this over the years.

https://www.youtube.com/watch?v=fsOF4Zbxrc4

I have a hard time believing the material properties are still OEM be after being treated like that.
But obviously "good enough."

 

[Compositepro]

"I don't know for certain but I'm pretty sure that a bandsaw blade cuts in the direction its pointing."

This is not true except in a very superficial sense. Believing this will make diagnosing problems very difficult. The blade cuts in the direction that it is pushed, not the direction the blade is pointed. It operates just like a wire saw. The flat backing simply stiffens the blade and allows for more pressure to be applied to the teeth for faster cutting. The backing also limits the direction the blade can move, so the wider the blade the less curvature is possible in the cutting path.

The position of the blade is controlled by the blade guides and the tension in the blade between these guides. It will be disturbed by other forces like feed pressure and uneven wear or set on the teeth. Feed pressure moves the blade in the desired direction and not to the sides. Uneven blade wear will deflect a blade to the side until blade tension causes the worn side to cut at the same rate as the sharp side. The faster you cut the more deflection you get from the desired path. When the blade deflects sideways enough, it will twist and the back of the blade will contact the sides of the cut, which then pushes the blade further from the desired path. So the wider the kerf compared to the thickness of the blade the easier it is to maintain a straight cut with a high cutting speed.

I have several bandsaws and have learned this the hard way.

 

[John_Vreede]

Excellent, thankyou AC! That's exactly what I was after. Your answer squares with what I know about the way blades act, but turned my thinking on its head. Its not the leading, low tension edge that initiates the failure but the trailing high tension edge seeking to relieve excessive stress. And that's the edge that is free to move in the kerf! Also the failure looks exactly like what happens when the blade gets pushed over by swarf overfilling a gullet. When dry cutting aluminium with out lube, aluminium swarf sticks in the teeth; next time around for that tooth the new swarf pushes the stuff that's stuck in the gullet out between the wall of the cut and the back of the blade. Because the teeth are anchored in the kerf, the top of the blade tips over just like LTB and an identical curved cut results. You can see evidence of the squishing out from small pieces of swarf friction welded to the wall of the cut and sometimes the blade jamming in the cut.
The blade does vibrate, you can see the pattern it creates in the wall of the cut being exactly the pitch of the teeth on one side of the blade in from the edge of the cut. Our blades are Raker set, every third tooth is set to left or to righ,t and with an 8tpi blade the pattern is 3/8" in from the edge in the direction of cut even on round bar. As each new tooth hits the cut it send a vibration up the blade to the next tooth, which creates the pattern.
I'm setting up an experiment with a laser pointer reflecting off the blade to see at what load the buckling initiates. The static situation will be easy to determine but it remains to be seem if I can get a clear and stable enough image from the moving blade. Whatever the static experiment will give a load where I know buckling will occur and I can reduce it until the buckling disappears, as evidenced by the squareness of cut returning, to find the limit.
A big Thankyou! to everyone who replied, each one added something to my understanding - jv

 

[John_Vreede]

Thanks for those insights . Yes you're right, any damage or wear will cause the blade to not cut straight, however I am looking for the cause of an undamaged blade to not cut straight when it should. The 4x6 metal cutting bandsaw mostly cuts work held in a vice and the cut can be markedly of square with an undamaged blade being dragged or pushed off line for some reason.
My 'cutting in the direction it's pointing' thought experiment, for this sort of non square cutting, is that the neutral axis through the middle of the blade back and teeth is not pointing in the same direction as the movement of the saw frame which pivots on the end of the base that holds the vice. With the 4x6 being so cheaply made, a number of saws (mine included) have been made with the holes for the pivot bored not parallel to the vice base leading to non square cutting which cannot be adjusted out.
In this investigation I'm looking for the mechanism for non square cutting in a properly adjusted saw with a good blade. When the feed pressure is to high, I suspect the blade is buckling and that neutral axis tipping over and no longer being parallel to the arc of travel of the saw frame.
Thanks for your thoughts - jv

 

[Maui]

John,

You might find this useful:

http://www.lenoxtools.com/Pages/content-offer.aspx?Id=sawcalc

Maui

 

[John_Vreede]

Thanks mate, I have seen this - its probably the best one of any of the manufacturers, it even talks about beam strength! - jv