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Metal bandsaw tension math ?

Rafael Rak

Plastic
Joined
Feb 14, 2022
Time to get understand how much tension do I need. I have homemade horizontal metal bandsaw with 19mm saw(0.9mm wide).I'v found tension parameters from european manufacturers.For my saw it's 23500 PSI(1600 bar !).
Everything nice and shinny ,but something seems to be unreal with calculations.
Calculations with Young's Module. (for steel 29000000) and looking for stretch dimension in 120mm.

S-stretching the blade in mm
D-Lenght measured in mm
(S/D) x 29000000
(0.01/120) x 29000000 =2410
23500/2410 = 9.54
9.54x0.01 = 0.095 mm (stretch at distance 120mm ! )

I have tested that calculations and tightened the saw(digital caliper adjusted to 120mm and clamped to saw) , almost breaking my hand with 15' wrench.
Every video with tensioning operation we can see that operator is stretching saw with 3' handle without using crazy force.
Wrong calculations? Or manufacturer parameters.
 
If you have gotten to the point of attempting proper tension I give you respect on your saw. 11 out of eight saws do not and can not have enough tension. Yes, it is a lot...

If you are struggling with big wrench your free wheel guide might be binding or you need more flinestones.
The blade will stretch as used, finding what is binding now save calories later.
 
Time to get understand how much tension do I need. I have homemade horizontal metal bandsaw with 19mm saw(0.9mm wide).I'v found tension parameters from european manufacturers.For my saw it's 23500 PSI(1600 bar !).
Everything nice and shinny ,but something seems to be unreal with calculations.
Calculations with Young's Module. (for steel 29000000) and looking for stretch dimension in 120mm.

S-stretching the blade in mm
D-Lenght measured in mm
(S/D) x 29000000
(0.01/120) x 29000000 =2410
23500/2410 = 9.54
9.54x0.01 = 0.095 mm (stretch at distance 120mm ! )

I have tested that calculations and tightened the saw(digital caliper adjusted to 120mm and clamped to saw) , almost breaking my hand with 15' wrench.
Every video with tensioning operation we can see that operator is stretching saw with 3' handle without using crazy force.
Wrong calculations? Or manufacturer parameters
Sticking with either english or metric would probably make life easier for one thing, but your calcs look correct to me.

Gonna take 622lbs to get that tension on 19mmx0.9mm blade. Should be able to calculate if you are getting in that ballpark depending on how big tightening screw is.

I imagine though that part of your issue is trying to measure ~4 thou with a pair of calipers.
 
My old Wilton saw used the same blade, 3/4" x .035" thick x 108" lg. It had a handle on the tensioner about 5" in diameter. Tighten the blade by hand. Never had a problem with improper tension on the blade in the twenty something years using that bandsaw. Did have to replace a few rollers and a broken guide over the years. And that was not from improper tension.

Oh, should be using 30,000,000 not 29,000,000 in your calculations. I think you slipped the decimal place a couple of notches to the right too far! Don't forget the frictional bite on the screw thread, too!

Ken
 
I think you’ve got your calculations a bit off whack , Europeans use metric and pascals and all sorts mut it seems a saw was mating with a tensile test machine
Try pedrazolis site or kaltenbach, not sure which you’ve got but if your winding it with a 15” wrench, somethings wrong, I’m surprised the band didn’t twang
Mark
 
Rafael, I think your math is right. Using inches, I get about .004", about .1mm. But in my experience, and I have a bandsaw blade tension gauge, many saws are unable to get the correct tension, require really stout construction.

My Powermatic 20" vertical saw can barely get a 1/2" x .025" blade to tension, have to bottom out the spring and then keep cranking. On the other hand my old 36" cast iron Tannewitz did it with ease, and was capable of much more.
 
Rafael, I think your math is right. Using inches, I get about .004", about .1mm. But in my experience, and I have a bandsaw blade tension gauge, many saws are unable to get the correct tension, require really stout construction.

My Powermatic 20" vertical saw can barely get a 1/2" x .025" blade to tension, have to bottom out the spring and then keep cranking. On the other hand my old 36" cast iron Tannewitz did it with ease, and was capable of much more.
I think the hi-tension is for metal cutting blades with power feed.
In the case of a vertical contour (hand feed) saw, many times flex back blade is used, I can see where hi-tension is not really needed.
 
If you wish to get anal about the tension, you need to measure the stretch of the blade, not force applied to the adjustment mechanism. For 30,000,000psi young's modulus, if you want a tension of 20,000 psi, you need to stretch the blade by .00067"/inch measured length. There are store bought gages that mount on the blade to measure stretch, or its an easy project to build one using a dial indicator.
 
Basic rule of thumb I have learned is that virtually nobody tightens their blades enough, unless they have a 50k machine that has built in gages and power tightening.
And second rule- It doesnt really matter.
I am sure that if you are running 1 1/2" blades on a full auto saw 24/7, you will notice a better cut and longer blade life if you use a cheater bar and a torque wrench- but for basically any smaller bandsaw, just tighten it as much as you can.
 
I got my tension gauge because as a woodworker I was sawing .050" veneers from lumber, and the correct tension made a big difference in accuracy of the cut. The .050 veneers cleaned up at .035 in the widebelt. With expensive and rare stuff, a few extra veneers made a big $ difference
 
Good topic. I've been afraid of over tightening....guess I was wrong.

This is why I come here....to learn stuff I didn't know I needed to learn
 
So your stress on the blade should be 23500 PSI(1600 bar !). This btw is at the high end of a recommended range (15000-25000psi). It probably is half, or less, than the yield strength.
Blade is 19mm x 0.9mm, or 0.75 x 0.035 inches, for an area of 0.026 in^2.
This means blade tension is 622 lb.
This is consistent with the data provided by Forrest. It implies a force of 1244 pounds on the wheels, which seems high to me but may not be.
Now 23,500/29,000,000 equals a strain of 0.00081
For the stretch in 120mm (about five inches - why is five inches relevant?) you should have a stretch of of 0.00081 x 120mm, or 0.097mm
Why is 0.097mm stretch so surprising?
 








 
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