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Thread: Bandsaw tension = new math????

  1. #1
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    Bandsaw tension = new math????

    Just got a chuckle out of reading the tension gauge thread and seeing it was good for 15,000 psi. I've never paid much attention to bandsaw blade tension numbers as they just seem completely irrational. Someone want to explain the math required to come up with these numbers? Apparently it's an arbitrary scale bearing little relevance to actual measurable spring tensions. That 3/8" chain I use to tie down loads on my trailer is rated at 6600 pounds, but a tiny little bandsaw blade is good for 15,000 pounds per square inch??? That little tiny tension spring on a Delta 14" saw will hold up three big old cars??? That's where the laughing starts in my mind.

    Someone explain?

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    The important part is "per square inch" Take a 1/4" bandsaw blade that is .025" thick, for example. If my mental arithmetic is correct the cross section of the blade is 1/160th of an inch. This comes to a bit under 100 pounds at 15,000 PSI. I have never tried to measure blade tension, other than by pulling the blade sideways (when the saw is not running, of course). I get the blade so it is tight enough to not jump off the wheels, and then if it seems to want to wander I tighten it up some more. If the blade breaks I back off the tension slightly till it repairs itself.

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    Not sure if it is still true but it used to be that spider web silk was the strongest material known. But a easy hand push would break the strands. Similar to a daddy long legs spider has the most poisonous venom of any spider but is harmless to man.
    Bill D

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    Steve, the math is simple. We start with the stiffness of steel, known as Young's modulus, which is 29,000,000 psi. Yes, 29 million. Wood, for comparison has a value of around 1.5 million psi, for comparison. We can discuss how Young's Modulus is measured if you find those numbers hard to believe but they are real values and measured every day by steel and other producers to verify product properties.

    OK, Young's modulus (Y) = Stress / strain. Stress is the tension we want to apply to our bandsaw blade, the 15,000 psi you referenced, for example. Strain is the amount the blade has to be stretched to put 15,000 psi on it. Strain is measured as stretch per unit length, so inches/inch, for example.

    Since you picked 15,000 psi for the stress, we can use the above equation to calculate how much we need to stretch the blade to get that stress.

    Strain = Stress / Y = 15,000 psi / 29,000,000 psi = 0.0005 inches/inch. Most all the tension meters you see measure strain over some defined distance, known as gage length. My tension meter has a gage length of 12 inches, so to put 15,000 psi on a blade I need to see a stretch of 0.0005 inches/inch x 12 inches = 0.006" on the dial gage. I typically run 24,000 psi on my carbide blade, which means I'm looking for the dial gage to move 0.010".

    You'll note the tension on the blade is irrelevant to the size of the blade. It doesn't matter if the blade is 1/4" wide x 0.020" gage or 1-1/4" x 0.042", if my meter shows a stretch of 0.006" over 12" it will have 15,000 psi on it. Of course, the spring force required to apply 15,000 psi to those two blades will be very different. But how much? OK, now we need to know the cross sectional area of the blade. If we forget about teeth and just say a 1/4" x 0.020" blade is a rectangle of those dimensions then the cross sectional area of it is 0.005 square inches. The 1-1/4" x 0.042" blade, again, disregarding the teeth, has a cross sectional area of 0.0525", more than 10X higher.

    OK, we're nearly there. Tension = Force/Area, or Spring Force = Blade Tension x 2 x Cross Sectional Area. For your 15,000 psi on the 1/4" x 0.020" blade, 15,000 psi x 0.005 sq. inches x 2 = 150 lbs. Where'd the 2 come from you ask? Because the blade is a loop the spring has to apply 15,000 psi of tension to both sides of the loop so it takes twice the force to do that. So you need a spring capable of applying 150 lbs of force to the blade loop to get a tension of 15,000 psi. To do the same on the 1-1/4" x 0.042" blade the spring would have to be capable of applying a force of 1575 lbs. No car lifting involved, but still a lot of force.

    Nothing irrational about it once you understand the math.

    John

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    Well at least the math is simple and obvious! I'm going to have to ponder this a little...

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    Quote Originally Posted by John TenEyck View Post
    Steve, the math is simple. We start with the stiffness of steel, known as Young's modulus, which is 29,000,000 psi. Yes, 29 million. Wood, for comparison has a value of around 1.5 million psi, for comparison. We can discuss how Young's Modulus is measured if you find those numbers hard to believe but they are real values and measured every day by steel and other producers to verify product properties.

    OK, Young's modulus (Y) = Stress / strain. Stress is the tension we want to apply to our bandsaw blade, the 15,000 psi you referenced, for example. Strain is the amount the blade has to be stretched to put 15,000 psi on it. Strain is measured as stretch per unit length, so inches/inch, for example.

    Since you picked 15,000 psi for the stress, we can use the above equation to calculate how much we need to stretch the blade to get that stress.

    Strain = Stress / Y = 15,000 psi / 29,000,000 psi = 0.0005 inches/inch. Most all the tension meters you see measure strain over some defined distance, known as gage length. My tension meter has a gage length of 12 inches, so to put 15,000 psi on a blade I need to see a stretch of 0.0005 inches/inch x 12 inches = 0.006" on the dial gage. I typically run 24,000 psi on my carbide blade, which means I'm looking for the dial gage to move 0.010".

    You'll note the tension on the blade is irrelevant to the size of the blade. It doesn't matter if the blade is 1/4" wide x 0.020" gage or 1-1/4" x 0.042", if my meter shows a stretch of 0.006" over 12" it will have 15,000 psi on it. Of course, the spring force required to apply 15,000 psi to those two blades will be very different. But how much? OK, now we need to know the cross sectional area of the blade. If we forget about teeth and just say a 1/4" x 0.020" blade is a rectangle of those dimensions then the cross sectional area of it is 0.005 square inches. The 1-1/4" x 0.042" blade, again, disregarding the teeth, has a cross sectional area of 0.0525", more than 10X higher.

    OK, we're nearly there. Tension = Force/Area, or Spring Force = Blade Tension x 2 x Cross Sectional Area. For your 15,000 psi on the 1/4" x 0.020" blade, 15,000 psi x 0.005 sq. inches x 2 = 150 lbs. Where'd the 2 come from you ask? Because the blade is a loop the spring has to apply 15,000 psi of tension to both sides of the loop so it takes twice the force to do that. So you need a spring capable of applying 150 lbs of force to the blade loop to get a tension of 15,000 psi. To do the same on the 1-1/4" x 0.042" blade the spring would have to be capable of applying a force of 1575 lbs. No car lifting involved, but still a lot of force.

    Nothing irrational about it once you understand the math.

    John

    Very good piece John. Thanks.

    J.

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    I tension my blade until it is stable and add one turn.

  8. #8
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    Steve

    Believe it or not, the math is actually pretty simple. It's just math
    Many years ago I did some destructive testing on bandsaw blades. I was using 1/2" and 5/8" wide blades. Nothing special, just some T-wolves.
    I promise you that the blade will take much, much, more than 15,000 psi.
    The upper yoke on a Delta, or Jet, 14" bandsaw will yield at about 50,000-60,000 psi. At 80,000 psi the blade was cutting through the fixture bolts I was using to affix the blade to the test stand. I never snapped a blade on my test stand from force applied, but I stopped when the bolts I was using failed.

    That tiny little spring on a 14" Delta, or one of the clones, will probably only take about 9,000 to 10,000 psi, force to fully compress it, as read with any available tension gauge.

    As for the "Magic 15,000 psi" value? The closest I got was tracing the original Starrett tension gauges from the 30's and 40's that were being used when the 14" Delta were made for metal fabrication. No one at Delta, or Starrett, could validate the absolute origin of the 15,000 psi value, even though Starrett still makes the tension gauge.
    Bottom line, If you tighten the bandsaw blade on a 14" Delta, or clone, and there is no space in between the coils of the spring, you need to back off, regardless of what the tension gauge reads, or you're getting ready to break something, and it won't be the blade. I promise.
    Last edited by Mike Cutler; 04-11-2021 at 9:14 AM.
    "The first thing you need to know, will likely be the last thing you learn." (Unknown)

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    Isn’t it just less stressful to accept things that are known and proven , even if you don’t fully understand them ?

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    Quote Originally Posted by Dave Sabo View Post
    Isn’t it just less stressful to accept things that are known and proven , even if you don’t fully understand them ?
    Dave

    Yes it is.
    I spent months doing the testing I referred to in my post.
    I didn't use a tension meter before I did all that work and I still don't to this day. I crank it up, look at the spring and look at the cut.
    I actually started out doing all that work to prove a well known author wrong. In the end, I validated his position. Rather ,the math validated his position.
    To many people get hung up on small things. This is suppossed to be fun. At the end of the day it's wood. Cut it today, it's a different dimension tomorrow.
    "The first thing you need to know, will likely be the last thing you learn." (Unknown)

  11. #11
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    Quote Originally Posted by John TenEyck View Post
    Steve, the math is simple. We start with the stiffness of steel, known as Young's modulus, which is 29,000,000 psi. Yes, 29 million. Wood, for comparison has a value of around 1.5 million psi, for comparison. We can discuss how Young's Modulus is measured if you find those numbers hard to believe but they are real values and measured every day by steel and other producers to verify product properties.

    OK, Young's modulus (Y) = Stress / strain. Stress is the tension we want to apply to our bandsaw blade, the 15,000 psi you referenced, for example. Strain is the amount the blade has to be stretched to put 15,000 psi on it. Strain is measured as stretch per unit length, so inches/inch, for example.

    Since you picked 15,000 psi for the stress, we can use the above equation to calculate how much we need to stretch the blade to get that stress.

    Strain = Stress / Y = 15,000 psi / 29,000,000 psi = 0.0005 inches/inch. Most all the tension meters you see measure strain over some defined distance, known as gage length. My tension meter has a gage length of 12 inches, so to put 15,000 psi on a blade I need to see a stretch of 0.0005 inches/inch x 12 inches = 0.006" on the dial gage. I typically run 24,000 psi on my carbide blade, which means I'm looking for the dial gage to move 0.010".

    You'll note the tension on the blade is irrelevant to the size of the blade. It doesn't matter if the blade is 1/4" wide x 0.020" gage or 1-1/4" x 0.042", if my meter shows a stretch of 0.006" over 12" it will have 15,000 psi on it. Of course, the spring force required to apply 15,000 psi to those two blades will be very different. But how much? OK, now we need to know the cross sectional area of the blade. If we forget about teeth and just say a 1/4" x 0.020" blade is a rectangle of those dimensions then the cross sectional area of it is 0.005 square inches. The 1-1/4" x 0.042" blade, again, disregarding the teeth, has a cross sectional area of 0.0525", more than 10X higher.

    OK, we're nearly there. Tension = Force/Area, or Spring Force = Blade Tension x 2 x Cross Sectional Area. For your 15,000 psi on the 1/4" x 0.020" blade, 15,000 psi x 0.005 sq. inches x 2 = 150 lbs. Where'd the 2 come from you ask? Because the blade is a loop the spring has to apply 15,000 psi of tension to both sides of the loop so it takes twice the force to do that. So you need a spring capable of applying 150 lbs of force to the blade loop to get a tension of 15,000 psi. To do the same on the 1-1/4" x 0.042" blade the spring would have to be capable of applying a force of 1575 lbs. No car lifting involved, but still a lot of force.

    Nothing irrational about it once you understand the math.

    John
    Very nice explanation, John. Hadn't thought about this since college. Couldn't have explained it now if you tortured me.

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    Ok, I get the math, with the exception of Young's Modulus. The rational part of my mind says there are no presses that can exert 29,000,000 psi. I've seen large hammer forging presses rated at 50 tons, well short of the 13,500 of Young's modulus numbers. I also use steel and wood in my construction business every day, and I know that a 1"x1" piece of wood can easily be obliterated at far less than 1,500,000 pounds of force.

    I see that Young's modulus is used as a constant in a common equation to compare strengths of different materials, but how is Young's Modulus calculated?

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    Quote Originally Posted by Steve Rozmiarek View Post
    Ok, I get the math, with the exception of Young's Modulus. The rational part of my mind says there are no presses that can exert 29,000,000 psi. I've seen large hammer forging presses rated at 50 tons, well short of the 13,500 of Young's modulus numbers. I also use steel and wood in my construction business every day, and I know that a 1"x1" piece of wood can easily be obliterated at far less than 1,500,000 pounds of force.

    I see that Young's modulus is used as a constant in a common equation to compare strengths of different materials, but how is Young's Modulus calculated?
    Just a little more math, Steve. You saw the equation for Young's Modulus is Stress/Strain. There are different ways to measure Y, but with respect to steel and our discussion on bandsaw blades, it's measured by pulling a steel bar in tension. When you do that you get a force vs. elongation curve like this:



    Young's Modulus is calculated from the slope of the initial linear portion of the curve. In that linear portion, the elastic region, you can increase and decrease tension and the steel will remain unchanged. That's why you can tension, untension, and then retension your bandsaw blade to the same spring setting and get the same tension again and again. But if you increase the force beyond that linear portion of the curve the steel will go beyond the elastic limit and into the yield portion of the curve. At that point, it hasn't broken yet but it is permanently stretched. If you release the tension now it won't follow the curve back down.

    You are confusing lbs/square inch and lbs force. As I discussed in the last post, your bandsaw blade only requires 150 lbs of spring force to apply 15,000 lbs/square inch of tension to it because the cross sectional area is so small. It's not hocus pocus, just math, but you have to keep the units straight.

    You can read up on Young's Modulus and how it's used in all manner of ways by doing a Google search on "Measuring Young's Modulus". Ever wonder how recommended bolt loads are derived?

    John

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    John, thank you for the detailed explanation, that makes more sense now. I suspect bolts use the same approach?

    I see building codes used that require nails in situations where screws are obviously stronger, I suppose there is some interesting math going on there too.

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    Quote Originally Posted by Steve Rozmiarek View Post
    John, thank you for the detailed explanation, that makes more sense now. I suspect bolts use the same approach?

    I see building codes used that require nails in situations where screws are obviously stronger, I suppose there is some interesting math going on there too.
    Yes, bolt loads are related to Young's modulus, too, in so far as you have to stay within the elastic region of the curve. Shear loads play a big role in how bolts are sized, too, but that's a topic for another day.

    Nails aren't hardened so they can bend more w/o breaking compared to screws, so it makes sense they can better withstand racking forces in framing. But there are screws rated for structural applications now, too, and I used them when I built my shed a few years ago. They can withstand multiple bendings just like nails, so maybe the codes just haven't caught up to these advancements.

    Beyond the math, there is a lot of testing that is done to prove what the tensile, breaking, shear, etc. strength of both materials and components made from those materials is. Together, most everything we build comes from those underpinnings.

    John

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