4x6 blade life mystery

[homebrewed]

A 4x6 doesn’t have the tension spring like a vertical. That’s the tricky thing about them. That and tracking them correctly.

The "spring" is the casting where the wheels install <joke>. In truth, these small saws are not all that heavily built, but they pretty much own the low end of the bandsaw market.

I agree regarding your tracking comment. I was chasing my tail for quite awhile on that one, until I came across an older thread in this very forum: here.

 

[homebrewed]

Oooh this is a good one.

The behavior your friend observed isn't too surprising. When we look at fatigue due to a cyclic stress, it's a function of two things:

1. The peak stress. More stress means less life.
2. The stress ratio. In general, the higher this number, the longer the life.

Some explanation on stress ratio: This is the minimum stress divided by the maximum stress. Tensile stress is positive, compressive stress is negative.

• A fully reversing load (say 30 ksi tension to 30 ksi compression) would have a stress ratio of -1.
• A partially reversing load (+45 ksi to -15 ksi) would have a stress ratio of -.33
• A unidirectional load (+60 ksi to 0 ksi) would have a stress ratio of 0

As a blade goes around the wheel, the outside sees additional tension, while the inside sees reduced tension or even compression. If you know the wheel diameter, we can compute the state of stress on both sides of the blade. This is the origin of the cyclic stress which causes fatigue.

As your friend added tension to the blade, he increased the peak stress, but he also improved the stress ratio which improved the fatigue life of the blade.

Here's the S-N curve for 4130 as an example:
View attachment 398893

Interesting! The wheels on these little saws are fairly small in diameter, 186mm. So compression/tension cycling due to the small diameter probably is the main limiting factor for blade life.

Is the maximum stress on the Y axis equal to the external tension plus the tension due to bending around the wheel?

 

[KyleG]

Is the maximum stress on the Y axis equal to the external tension plus the tension due to bending around the wheel?

That's correct. Keep in mind that that chart was for 4130, and will vary based on alloy and heat treatment.

The wheels on these little saws are fairly small in diameter, 186mm. So compression/tension cycling due to the small diameter probably is the main limiting factor for blade life.

I just ran the numbers for a wheel that size, and the bending stress is shockingly high. Unless I made a mistake, 98.7 ksi! The formula for bending stress (Sparing you some algebra and assuming a thin blade) is Stress = E*bladeThickness/wheelDiameter.

Taking a preload of 15 ksi, this gives us a total stress of 118.7 ksi tension on the outside and 78.7 ksi of compression on the inside. The inside will be critical, because it sees reversing stress. This is a stress ratio of -5.6. With a preload of 25 ksi, the stress ratio is -3.0. Honestly, I don't even have S-N curves for numbers in this realm. I'll have to ask one of our fatigue pro's what happens when the ratio is significantly below -1.

 

[homebrewed]

That's correct. Keep in mind that that chart was for 4130, and will vary based on alloy and heat treatment.


I just ran the numbers for a wheel that size, and the bending stress is shockingly high. Unless I made a mistake, 98.7 ksi! The formula for bending stress (Sparing you some algebra and assuming a thin blade) is Stress = E*bladeThickness/wheelDiameter.

Taking a preload of 15 ksi, this gives us a total stress of 118.7 ksi tension on the outside and 78.7 ksi of compression on the inside. The inside will be critical, because it sees reversing stress. This is a stress ratio of -5.6. With a preload of 25 ksi, the stress ratio is -3.0. Honestly, I don't even have S-N curves for numbers in this realm. I'll have to ask one of our fatigue pro's what happens when the ratio is significantly below -1.

Tell me more about your calculations. If the bending stress is +/- 98.7 KSI, then (for example) 15 + 98.7 = 114 (about) and 15 - 98.7 = - -83.7. The ratio is -1.35, not 5.6.

 

[KyleG]

@homebrewed I did the exact same thing at first. Had it all typed up and then noticed my issue.

-1.35 is the ratio across the blade, but no atom iron is ever on the inside of the bend and then on the outside of the bend. The inner face varies from 15 ksi to -83.7 ksi, giving a ratio of -5.6. The outside of the blade varies from 25 ksi to 113.7 ksi, giving a ratio of .22.

Does that make sense?

 

[homebrewed]

@homebrewed I did the exact same thing at first. Had it all typed up and then noticed my issue.

-1.35 is the ratio across the blade, but no atom iron is ever on the inside of the bend and then on the outside of the bend. The inner face varies from 15 ksi to -83.7 ksi, giving a ratio of -5.6. The outside of the blade varies from 25 ksi to 113.7 ksi, giving a ratio of .22.

Does that make sense?

Yes, thanks for the clarification. If the outside of the blade is an apple and the inside of the blade is an orange, I was comparing apples to oranges But shouldn't the tension for both the outside and outside faces be the same when they are not wrapped around a wheel? So for the outer surface we'd see (for instance) either 15KSI or 15+97.8, and for the inner wheel either 15KSI or 15-97.8 KSI

However, I still think there's something wrong in the calculations somewhere because the result seems to fly in the face of user's experience with these saws -- the typical reason for replacing a blade isn't because it broke, but because it's become dull. A design flaw like excessively-small wheels would affect all users to a greater or lesser degree, subject to their choice of blade type and blade tension: yet blade breakage is relatively uncommon.

It would be nice if I could find a family of S-N curves for spring steel that include the stress ratio. So far I've just found single curves (or tables) so have to assume they're for a stress ratio of -1, i.e., alternating between equal amounts of compression and tension.

 

[KyleG]

But shouldn't the tension for both the outside and outside faces be the same when they are not wrapped around a wheel? So for the outer surface we'd see (for instance) either 15KSI or 15+97.8, and for the inner wheel either 15KSI or 15-97.8 KSI

Yep, my bad! That "25 ksi" should be "15 ksi".

When I ran the numbers, I was pretty incredulous too. I used a couple different methods to check my math, and they all landed at the same place. I can speculate at a few plausible explanations:

• For the inner face of the blade at a max stress of 25 ksi, you're below the "runout" stress for unnotched samples with a stress ratio of -1. The fact that we see any failure at all is because the teeth act as stress concentrations, and the stress ratio is below -1.
• For the outer face, assuming stress ratio slightly greater than 0, the runout stress is probably in the ballpark of 100 ksi, which is about what the blade on your friend's saw is seeing.
• I think this case has a particularly high blade thickness to wheel diameter ratio.
• I suspect most metal cutting blades don't see nearly 50 hours of run time before they're dull enough to be replaced.

Here's the chart for unnotched 4340 heat treated to 200 ksi for reference.

As a side note, these charts came from MMPDS-05, which is a great reference. If you PM me your email address, I'll send it to you (it's too large to attach). You can also find MIL-HDBK-5J, which is essentially the same document before the government passed the buck on standards management over to SAE. http://everyspec.com/MIL-HDBK/MIL-HDBK-0001-0099/MIL_HDBK_5J_139/

 

[homebrewed]

Yep, my bad! That "25 ksi" should be "15 ksi".

When I ran the numbers, I was pretty incredulous too. I used a couple different methods to check my math, and they all landed at the same place. I can speculate at a few plausible explanations:

• For the inner face of the blade at a max stress of 25 ksi, you're below the "runout" stress for unnotched samples with a stress ratio of -1. The fact that we see any failure at all is because the teeth act as stress concentrations, and the stress ratio is below -1.
• For the outer face, assuming stress ratio slightly greater than 0, the runout stress is probably in the ballpark of 100 ksi, which is about what the blade on your friend's saw is seeing.
• I think this case has a particularly high blade thickness to wheel diameter ratio.
• I suspect most metal cutting blades don't see nearly 50 hours of run time before they're dull enough to be replaced.

Here's the chart for unnotched 4340 heat treated to 200 ksi for reference.

View attachment 399114

As a side note, these charts came from MMPDS-05, which is a great reference. If you PM me your email address, I'll send it to you (it's too large to attach). You can also find MIL-HDBK-5J, which is essentially the same document before the government passed the buck on standards management over to SAE. http://everyspec.com/MIL-HDBK/MIL-HDBK-0001-0099/MIL_HDBK_5J_139/

I understand why you can't attach it -- I found the MIL document and it's 70MB long.

Thank you for the reference!

 

[homebrewed]

I got a little more information regarding my friend's blade life issue. I'm not sure if it is an issue or not so I'm mentioning it here for opinions.

I had suggested to him that he should carefully examine his bandsaw, looking for anything unusual and he mentioned that he had milled a shallow (~1/16") groove in both wheels for some purpose I won't go into here. I could see a number of possible issues with this.

1. The grooves slightly weakened the wheels, so they are flexing in an abrupt manner as the blade passes over them. This could translate to a temporary spike in the stress the blade sees. But how large a spike??? Determining the amount of flex might take some modelling S/W to figure out.

2. The blade is not supported over the groove so it tries to straighten out & become a chord of the wheel. This would reduce the compression/tension for a brief time, but it also means that the blade flexes slightly on either side of the groove -- increasing the effective number of flexes each time the blade completes a circuit around the wheels. I think the blade probably does "notice" the groove in this manner but the delta in stress is low....or is it??? While the change in the radius of curvature is relatively small, it takes place over a short distance so the stress delta is concentrated in a relatively small length of blade. Not good.

Since the wheels are pretty small in diameter (for bandsaws in general) that would exacerbate some of these effects, but are these more theoretical than practical issues?

 

[epanzella]

Call me cynical but I'm having a hard time giving a rat's arse what the blade life is when not following the guides that make it square to the work and not cutting anything. A previous post mentioned no tension spring but my HF 4x6 has one. I have no real way to measure the tension other than seat of my pants but I keep it set on the heavy side for narrow materials and just add weight to have the proper tooth load when cutting a larger footprint.