Calculating bandsaw blade tension from deflection 3

[John_Vreede]

Hi guys
I’m trying to work out the mechanics of ‘tension measurement by deflection’ in a bandsaw blade.

The bandsaw has wheels spaced 20.57” apart and a 'properly tensioned’ blade takes 27.3lb ±0.5lb to deflect ½” in the middle of its span measured with a cheap digital airport luggage weigh scale jury-rigged to the saw (see attachment). I’m pretty sure that info should be enough to calculate the initial tension in the blade, but how?

My first thought was that both the deflection and the force are vectors and should be comparable, ie. The ratio (deflection / half the span) should be the same as the ratio (deflection force / initial tension). Solve for initial tension and it works out to 828lbf. Unfortunately this is wrong.

It is out by a factor of more than 2 as I ‘properly tensioned’ the blade to 300lbf as measured by an hydraulic load cell. (For the blade I‘m using the manufacturer specified a stress of 25-30000psi and 300lbf is within that range. The gauge reads 320~325psi when the blade is deflected the ½”.

My question is 'How do I calculate tension from the force needed to deflect the blade a given amount?'. Thanks in advance - jv

 

[3DDave]

What does a free-body diagram look like? You have tension on both sides trying to pull at an angle to the applied side load. Your initial description of deflection/half-span seems to ignore that factor of two. I don't know which way your error is because of the lack of seeing how the tension was measured by the load cell. However, I still have confidence in basic statics and the application of free-body diagrams.

 

[SwinnyGG]

You're off by a factor of ~2 because you only took one side of the blade into account when you calculated tension.

If your 27.3 lb load was applied in the exact center of the blade span, there are THREE force vectors to be accounted for, not two. One force vector is your applied load, the other two vectors are applied by the two halves of the blade, on each side of the load.

You can calculate a correct resultant either way- you either use 27.3 lb and calculate the magnitude of both unknown vectors, which you can do because they have equal magnitudes in both x and y; or you halve the load and solve for the single unknown vector.

There's also some additional error incorporated, because the blade is not static. When you deflect it from the free state, you are applying additional tension and stretching the blade.

To arrive at actual tension, you need to calculate the tension in the deflected state, and then subtract the tension you added by stretching the blade into the deflected state.

By my math, you stretch the blade about .015" to get to the deflected position. If you know the dimensions of the blade you can calculate its axial spring rate. Do the math and I bet you'll be very close to what your load cell says.

 

[BrianE22]

On something like a saw blade, wouldn't there also be bending loads in addition to the tension loads? Maybe they're insignificant.

 

[John_Vreede]

Hi Guys
Sorry haven't got back to you before now.
Used your suggestion straight away. Of course that's the right way to calculate it , but the answer didn't turn out as I thought it would and I'm still trying to figure out where the errors are.
If 27.3lbf pulls the blade over 1/2" at the centre of a span of 20.57", that give the tension in the blade at 281lbf. Yet my hydraulic load cell gives it as closer to 325lbf in the pulled over state (started at 300lbf not pulled over).
So either the luggage weigh scale is wrong, the pressure gauge on hydraulic load cell is wrong, or both are. I'm in the process of getting them both checked now. Both are cheap instruments so is very possible that they're out.
The reason I'm doing this is to get away from a strain measurement. The bandsaw blade manufacturers all sell strain gauges that measure the elongation of the blade, which is like you say of the order of a few thou". However bandsaw users, like the guys at 'Fine Woodworking' magazine, say they have very poor repeatability using these gauges, which also cost more than a 4x6 bandsaw is worth. I believe the error in measuring such small distances is likely to be of the same order as the measurements themselves, so the %error would be high. I figure that the gross measurements in a deflection setup like this ought to have a lower error. Not looking that way so far.
Thanks for straightening me out, I'll come back once I've checked out the gauges.
Rgds - jv

 

[PackardV8]

From one jv to another, I like to understand the first principles as well as the next member, but on a 4x6 import bandsaw, aren't you overthinking this? Does it make good straight cuts? If not, buy a higher quality blade and/or add more tension. If the blade breaks or the spring bottoms out, use less tension. Worked for me.

jack vines