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Thread: Inturra High Tension Band Saw Springs question

  1. #1
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    Inturra High Tension Band Saw Springs question

    Does anyone know the spring rate or maximum force that a 2-1/2" Inturra High Tension spring can exert?

    Here's what I know. The cross-sectional area of my blade is 0.01469 square inches.

    Please check my 50 years from college memory of stress calculations. If you need 15,000 psi to tension the blade properly, then the spring would need to exert somewhere near 220 pounds of force to create that much tension. Since two sides of the blade are in tension that would be 440 pounds total force. Can any 2-1/2" x .730 dia. spring do that? And, can the average 14" band saw handle that much tension?
    Lee Schierer
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    Your math is correct, Lee. If the spring you are talking about is for a 14" cast iron Delta, or clone, you would only be able to apply 15 ksi to a 3/8" or narrower blade. While the spring may be capable of supporting the higher force needed to put 15 ksi on a 1/2" blade the machine itself is not. You'll find the upper casting starts bending, enough to throw the upper guides significantly out of alignment. More importantly, you can easy bend/break the top wheel yoke. I almost lost mine seeing how much tension I could apply with the Iturra spring. I can report that the spring is capable of 18 ksi on a 1/2" blade. I also can report that you do not want to do it. I now run 12 ksi on a 1/2" blade and accept that as all the machine is capable of. For that reason, I normally run a 3/8" blade because it will give me 15 ksi at the same spring force and no additional stress on the frame/components.

    FWIW, 15 ksi is the bottom end of where I would like to run the tension. 20 ksi would be better, and 25 ksi would be ideal and is where I run the carbide resaw blade on my larger BS.

    John

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    Thanks John, I appreciate the verification that I still remember some basics.

    I understand what you are saying, but there is more than just the blade size at play. Not all band saw blades are the same thickness. Thickness makes a significant difference on achieving the proper tension. Materials used range from .022-.035 thick. A half inch blade from .025 material only needs about 288 pounds to tension where as a .035 thick blade needs 403 pounds. A 3/4" blade made from .035 material needs about 670 pounds of force to tension while a blade made from 022 material only needs about 440 pounds.
    tension.JPG

    I have a 3/4" blade that is .022 thick, it cuts a wonderfully small kerf and does a really good job resawing. I do not have a tension gauge. I use the markings that came on the upper casting as a guide. My saw does not have a riser block. I tension this blade to the same level as my 1/2" timberwolf blade and get excellent results. When I measure the deflection of the upper arm right under the guide support rod, it moved just .010 inches, the same deflection I get on the 1/2" timberwolf.
    Last edited by Lee Schierer; 05-14-2021 at 7:42 PM.
    Lee Schierer
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  4. #4
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    Just FYI - Iturra and Carter are actually selling a "Gold" 400 lbs./in. rated Raymond-style die spring https://www.mcmaster.com/die-springs...-diameter~3-4/

    Pretty inexpensive from McMaster, could always try the lower or higher load rated ones if you're getting too much deflection or inadequate tension.

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    Your math skills seem fine to me Lee and, yes, the thickness of the blade is important as it's half of the parameters required to calculate the cross sectional area of the blade and, therefore, the spring force required to tension it to whatever value you desire. No matter what spring you put on a 14" Delta, however, bad things start to happen if you put too much stress on it. As I said earlier, the practical limit of a 1/2" x 0.025" blade is 12 ksi. If you get good results with a 3/4" blade that's great, but you can't put much tension on it without serious risk of damaging the upper wheel yoke.

    Rather than guess at how much tension you have, why not measure it? You can do it with nothing more than a set of Vernier calipers and two little clamps.



    Set it up for a 5" gap. The actual value doesn't matter, just write it down after you have the blade snug but with no tension and reset it to zero. Now add tension. Every 0.001" of movement =

    29 x 10^6 x 0.001/5 = 5800 psi. This works for any blade of any thickness and any width.

    Or do it the other way around. Tension a blade how you normally do it, install the Verniers while it's still under tension. Again, write down the actual opening of the jaws, reset it to zero, and then relax the tension. When the Verniers stop moving record that number. Plug both values into the equation above and that's the tension you had on the blade. Where'd the equation come from? Young's modulus = Stress / Strain.

    John

  6. #6
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    Quote Originally Posted by John TenEyck View Post
    Your math skills seem fine to me Lee and, yes, the thickness of the blade is important as it's half of the parameters required to calculate the cross sectional area of the blade and, therefore, the spring force required to tension it to whatever value you desire. No matter what spring you put on a 14" Delta, however, bad things start to happen if you put too much stress on it. As I said earlier, the practical limit of a 1/2" x 0.025" blade is 12 ksi. If you get good results with a 3/4" blade that's great, but you can't put much tension on it without serious risk of damaging the upper wheel yoke.

    Rather than guess at how much tension you have, why not measure it? You can do it with nothing more than a set of Vernier calipers and two little clamps.



    Set it up for a 5" gap. The actual value doesn't matter, just write it down after you have the blade snug but with no tension and reset it to zero. Now add tension. Every 0.001" of movement =

    29 x 10^6 x 0.001/5 = 5800 psi. This works for any blade of any thickness and any width.

    Or do it the other way around. Tension a blade how you normally do it, install the Verniers while it's still under tension. Again, write down the actual opening of the jaws, reset it to zero, and then relax the tension. When the Verniers stop moving record that number. Plug both values into the equation above and that's the tension you had on the blade. Where'd the equation come from? Young's modulus = Stress / Strain.

    John
    John, I rigged up my digital caliper per your instructions and what I discovered is that both the 1/2" blade and the 3/4" blade give a reading of 1.5 thousandths of an inch deflection before the threads on the tensioning screw bottom out just as the tension indicator reaches the 1/2" mark on the casting. This would indicate that both blades are too long for my saw, even though they were ordered at 93-1/2" per the manual. I am actually only compressing the stock spring or Inturra spring 1/2", neither is bottomed out. Since all the blades have seen little use, I won't be ordering new blades anytime soon. So I guess I will continue doing what I have been doing since I get good cut quality with the blades I have and I'm not deflecting the upper arm more than 0.010".
    Lee Schierer
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  7. #7
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    Lee you just need a longer thread on the tension screw. Put a metal shim under it, where the screw sits against the upper casting. That will give you more effective screw length. 93-1/2" is the correct blade length for that saw.

    FWIW, 0.010" of deflection of the upper blade guides is a LOT, especially at such low spring force. I would retighten the bolt holding the two casting together, with no tension on the blade.

    John

  8. #8
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    Quote Originally Posted by John TenEyck View Post
    Lee you just need a longer thread on the tension screw. Put a metal shim under it, where the screw sits against the upper casting. That will give you more effective screw length. 93-1/2" is the correct blade length for that saw.

    FWIW, 0.010" of deflection of the upper blade guides is a LOT, especially at such low spring force. I would retighten the bolt holding the two casting together, with no tension on the blade.

    John
    This saw was purchased new in Canada. I thought of just cutting additional threads on the rod, but everything seems to be working just fine. The end of the adjustment rod is pointed and the hole it rides in is cone shaped, so a simple shim isn't going to work. The threads are metric, which I don't have a die for.

    The very large bolt holding the upper arm to the lower part of the frame is a metric size. It is extremely tight.
    Lee Schierer
    Captain USNR(Ret)

    My advice, comments and suggestions are free, but it costs money to run the site. If you found something of value here please give a little something back by becoming a contributor! Please Contribute

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