The year is 20,018. This beautiful machine has graced many different shops and studios, The wear parts have been replaced several times. It is still in perfect condition. You are a genius and a national treasure.
It figures to 2368 pounds per bolt. Or 7104 pounds of force holding the key down. Wow I wouldn't have guessed that. Thanks for showing us your skills. Some of the old time machinist weren't always so willing to let you look over their shoulder I'm guessing. The education in the machining trade through the UA-cam community is really something. And greatly appreciated. That etching press is going to be a fine machine.
Hahaha, gotta give my wife credit for actually paying attention while I watch these videos. She said, when discussion on the cutting tools... "did he say he can cut a lot faster with the same shitload ??".... almost spewed coffee on the laptop !!
Here in Europe we recon torkforce times threadpitch. Amerika is different, Recalculating to inches screws it all up. Good luck to the winner and may Tom's swag be with him...:):)
Hey Tom, Assumptions: 13/64 drilled hole. Coefficient of friction of .16 for lubricated steel on steel for the threads and .16 for lubricated steel on bronze for the head of the SHCS. Using lubricated since you used a cutting oil and I don't think you degreased the parts. Formula from Spotts that I have in my HP48GX for the last 23 years. It has proven to be very accurate in the several times I did a physical testing check. 2196.568 pounds force per screw. ATB, Robin
Thanks Danny, I just hope the formula that Tom used is as thorough as the one in "Spotts Design of machine elements" It is not a trivial formula and most would not bother with an equation that complicated. It considers: number of leads, thread pitch diameter, thread pitch, thread pressure face angle, thread coefficient of friction, thrust collar coefficient of friction, thrust collar pitch diameter, torque. ATB, Robin
+ROBRENZ Ok, that's interesting! I guess that calculating something with so many variables there is going to be a few viable answers. Im looking forward to seeing Tom's method.
2196.568 lbs of force from 90 in.-lbs. of torque, huh? Interesting. I'm sure that's what the engineers say but the true answer is that "clamp load" is a theoretical and dimensionless "quantity". Steel on bronze pretty much self-lubricates too, by the way. You use lubricant on fasteners to prevent galling when the materials are the same or similar. Steel on steel, etc. Like metals gall and that gives you an inaccurate torque reading since a torque wrench is measuring the force required to turn the fastener. You reduce torque settings with lubricants used in conjunction with DISSIMILAR materials or ones that don't gall as easily or when a coated or plated screw is used in a natural finish steel thread to prevent over-torquing and stretching the fastener past its yield point. Its really easy to think that you want to lubricate all threads but if the manufacturer's assembly instructions don't call for lubricant, you don't use it. And when lubricant is called for you use the correct lubricant and make damn sure any screws or bolts and nuts replaced are replaced with OEM hardware or the correct aftermarket hardware. Grade, material, thread length, plating or coating, etc. Put the wrong hardware in any situation with lubrication to boot and you're asking for trouble. Especially if you're not an experienced mechanic and you don't know what X amount of torque feels like and don't recognize when a fastener should be tight and is just stretching until it breaks. As for supposed "clamping load" calculations, when they're based on factors that are rarely present in the real world when someone is fabricating things. They're based on assumptions like the number of threads engaged being consistent with the size of the fastener its length and the threads being cut and a certain percentage of "engagement" and all kinds of "constants" that may exist in the engineering world when engineers are getting out the same book and they're using it to spec fasteners and the depth of blind holes and full thread length in them, etc. The "book" may say X clamp load based on a bunch of math and theories and formulas and constants, but in reality the most honest and legitimate answer is that the "clamp load" is the sum of the torque applied to the fasteners to create it. Its still "force" and "load" and its the friction that results from them and the assembled fastener and place and torque to spec and its the only way to measure how much tension is on the fastener.
Once again, I have to say that I love the aesthetic qualities of your designs. The care and attention that you pay to not only the functionality of the machine but also adding artistic touches to achieve a consistent quality and appearance is wonderful. I think it defines you as a master craftsman. Could it be the influence from a gifted artist you married?
Geez Tom, I barely have time to keep up watching the videos, never mind doing the calculations, anyhow my guess is: Enough to perform the task confirmed by experience... lol
Very good stuff as usual.. If you you were shooting for a real precise thickness on the bars, I would done the fixture bar second after drilling holes on the bars and left it clamped in the vice after the surfacing and taping. That way you know the fixture surface is flat in the mill..
Watching you power tap those holes brought up a thought in my head. How do you know when to stop the spindle so you don't crash into the bottom of the hole and break a tap or not tap it deep enough (though that part could be remedied with a hand tap afterwards)? I'm guessing it just comes from years of doing it.
The clamping load is the sum of the torque applied to the three screws. 270 in.-lbs. You can't possibly end up with more clamping load than you're applying in torque to the screws. That's not what the engineering textbooks are going to say with the complex formulas and calculations using all those "specs" but they're also not calculated for a clamp load delivered when a chunk of bronze is screwed to a chunk of steel or with a steel fastener head torqued against a bronze part. Steel on bronze is going to deliver a lot less friction than steel on steel would and if you torqued those screws to spec you've probably over-torqued them because a torque wrench only measures the friction required to turn the fastener and pretty much any time you see steel hardware in an aluminum or other soft metal part, you'll see a steel washer between the fastener and part so the friction is returned to a steel on steel state. Soft metals also tend to compress and its very common to see torque specs for aluminum parts to require an initial torque and then a follow-up after 10 minutes or so or maybe a heat-cool cycle because they aluminum can compress under load and lose a little torque. Especially right under the washer. Bronze is much more dense and that would be reduced but its still a good idea to follow up whenever putting steel fasteners through non-ferrous metals. No matter what the "clamp load" is the fasteners, if properly torqued, are nowhere near the "yield strength" and should never FEEL the "clamp load" anyway. If a pair of screws on say a connection rod cap are torqued to X lb.-ft. and have say 50.000 lbs of "clamping force" at that torque, as long as "pull" they're subjected to doesn't exceed 50,000 lbs., they're never actually loaded at all. Those clamp load specs are what new engineers use to decide what fasteners to use and where based upon how much load another more experienced and skilled and knowledgeable engineer tells them will be applied to the "joint" in operation. They're meaningless in the "real world" of wrenching.
As others have shown, with the basic formula: F = T/(c x d) T = 90 in-lbs (torque) c = 0.2 (coeff-dry) d = 0.190 in (diameter for 10-32) Then the answer of 2368 lbs (each bolt) falls out. Further, c=0.2 is a nominal number for unlubricated hardware, and could be lowered to perhaps 0.15 lubricated. Also, for 10-32 UNF the effective Tensile Stress Area is 0.02 in^2. So that implies an effective diameter of 0.160" Taken together these give 3750 lbs per bolt. (Which others have noted is close to the 3800 lbs rating.) I think the 2368 lbs is a comfortable guaranteed lower limit.
Another naive question: once you'd built the fixture to hold the keys while you milled the drilled faces, why did you clamp it with additional spacers around that tall plate (starting at 11:20 or so)? It isn't obvious (to me, anyway)....
Total clamping force is 7894.7 lb (2631.6 lb per bolt). I am assuming a coefficient of 0.18 like I always do, though, so this is a big unknown compared to what you might be assuming. This coefficient may be a little high but lower coefficients would result in higher loads and most people have answered lower than this so there is no use giving an answer similar to them since they answered before me.
Just did a Solidworks simulation on a 10/32 at 10.2N and it came up with 2204lbs. Is it working on a final answer basis or can we just keep guessing? Great fun though!
Mr L, I make it 2368.42 pounds per position. The efficacy of this force depends on the friction co-efficient between the screw threads and the steel block, and the helix angle of the screw, etc etc. Important question is "is it tight *enough"! ;)
Coefficient of friction for Brass-Steel is 0.5 Dry and 0.19 Lubricated. Gives a clamping force of 4219.92N Dry and 11 105.04N Lubricated per cap screw.
I hate to do this, since you tried to show the markings off in the video, but for the life of me I can't read them.. That little face mill you had out for the steel looks sweet! What was it?
Figuring that an advance of one turn of the screw at a distance of one inch requires π x 2 inches of motion and advances the screw into the work by 1/32. This gives a mechanical advantage of a bit over 200:1. Multiply times 90 gives about 18, 000 pounds of pressure (in a frictionless universe) and maybe 20% of that in the real world, so maybe 3,600 pounds. Now this is also about the ultimate tensile strength of a 10-32 SHMS (3400 pounds). I will guess somewhere in the middle at *3,500* pounds per screw. (Late guess, but I haven't checked the other answers yet.)(Checked: some answers were close but less or quite a bit less; some were more or quite a bit more. I'm comfortable with my guess.)
One million megatons, as we all failed to consider that Tom rerouted the output of the impulse engines directly to the torque wrench and extended the structural integrity field to the screws. Some 2300 pounds force sounds implausible until I compare the thought of having my finger under the front of my truck, to having it under that screw and key. Maybe so. Time for the Tom-o-Meter. A quick look at a couple of online charts suggests clamping forces in the range of 600-720 pounds for 10-32 at around 90 in. lb., depending on lubrication, bolt grade, and "things." 1800-2160 pounds total seems experientially more believable to me.
Hi Tom" I don't know how the calculator works for this one, thread, type thread pitch, % of thread depth after cutting, probably in the machinist hand book! I'm ill with flu but maybe a walk out to the workshop & get it. Do show us how to find this kind of info. Ace show.
Just got done watching you latest MNM103. No one is right yet?Depending on the coefficient of friction, the more lube, the higher the clamping force. Full dry(.2 coefficient of friction) is 2368.42lbs, a bit of lube (.19cof) is 2493.04lbs, a semi lubed (.18cof) is 2631.58lbs, and fully lubed could give you 2786.38lbs-2960.53lbs of clamping force. Can you tell I really want some swag Tom?
Hi Tom. While we are somewhat on the subject, could you give your opinion on which is correct, foot- pounds or pound- feet, or if it even matters. Thanks.
The foot-pound force (symbol: ft·lbf or ft·lbf), or simply foot-pound (symbol: ft·lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure A pound-foot (lb·ft or lbf·ft) is a unit of torque (a pseudovector). One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point. The name "pound-foot", intended to minimize confusion with the foot-pound as a unit of work, Wikipedia said it better than I could.
Since torque is force multiplied by the distance from the perpendicular vector to the revolving axis, it shouldn't matter. I think it's a matter of pronunciation preference. But heck, what do I know, I work in metric
𝖘 𝖍 𝖎 𝖗 𝖔 The two terms are commonly interchanged in everyday speech. You understand whether you are discussing torque or energy/work from the context of the discussion.
Hi Stephen, I have on for the surface grinder. Seems to work pretty well. They seem to get a lot of money for the darn things. Thanks for the comment. Cheers, Tom
lets give this a shot...the cutting load applied to the bolts is in shear so given a 10-32 has a shear area around 0.02in^2 (roughly the area of the tap drill size) the average SHCS has a tensile strength of 180000psi (working from memory here) so the shear strength is 0.505 of UTS...so 90900psi for shear stress and F=SxA...so 1818 lb per screw...3 screws for a grand total final answer of 5454lbf to shear them off...well above the cutting forces...
Hi Tom, Wondering why you 'rub' your files over the work when you're doing file work (especially in the baby bullet build). I'm new to machine shop and also saw Mr. Pete do this. I've learnt that you should lift the file clear off the work when pulling back. I know you ease the pressure on the back stroke but still wondering why you don't lift it, do you just not care that much about files to do it? Don't mean to smart, just curious! Aryan
On a softer material which is most if its a good quality file, light back-dragging on the file can clear some of the crag off the file teeth so they're cleaner for the next cut forward.
10-32 according to Holo-Krome's (who should know, being preeminent manufacturers of SHCS) fgures, average torque to tighten average screw to yield is 120lbs.inch which gives average tension induced in screw tightened to yield, 2800 pounds. (Average ultimate tensile strength 3600 pounds). Values are for standard steel(black finish) screws lubricated with oil clamping hardened steel plates to hardened "nuts". So under the above conditions, ignoring your under- tightening, would be somewhere in the order of 3 times 2800 lbs, 8400lbs.
I may be wrong but the screw specs give a 10/32 screw as having a tensile strength of 1200lbs so I'm guesing regardless of your torque thats the best you'll get.
My mistake, I thought I thought Tom said machine screw www.google.co.uk/url?sa=t&source=web&rct=j&url=www.fasnetdirect.com/refguide/Machinescrewthreads.pdf&ved=0ahUKEwi5se34qa3OAhVBD8AKHW6RAtAQFggbMAA&usg=AFQjCNFlnV1zGSBOyCEQ5XS_cstDSae4ag&sig2=NbKt15TIeQYTZHe33DdU0g
Well I got 2370.9167 lbs dry in steel, but not sure how the bronze affects the clamping force. Lubricated with a coefficient friction factor of .17 is 2789.3138 lbs in steel as well. Again not sure with the bronze on top of the steel on how the changes the clamping force or if that is even a factor. Metric would be 12.4115 N using the same coefficient friction factor of .17
Hi Cole, You got close to the book value. The friction coefficient has a big effect on the results and is hard to determine accurately.Thanks for the comment. Cheers, Tom
Not able to tell what type of finish was on the screws used... So - Assuming Worst case friction on one end and - Black Oxide on the other end: 2368#/ Fastener (7105# Total Clamping Force) to 4306 #/Fastener (12918# Total Clamping force) I don't think it is going to move under milling loads.
Ps based on a rule of thumb of torque = axial force x clamping diameter x 0.2. I suppose this is only going to apply *at the screw diameter and isn't considering the force at different dias. of screw head.
As a stress engineer, I find your challenge is frustratingly difficult to adequately deal with, but only because I know what I know about certain mechanical fail-safe designs where pre-load is vitally important to proper operation . Prove by test!! (edited for a bit more info...)
Of course torque is a very inaccurate indicator of fastener preload. Especially with a dry fastener. Further increasing the inaccuracy is using a multiple step torquing procedure. The fasteners should all be snugged down, then brought to the final spec in one fluid motion. Another way to increase accuracy is to cycle the fastener a few times to burnish the contact surfaces before the final torquing procedure. All that aside, those small cap screws can produce a very impressive preload
You could just about lift my car with that bolt! I would only have guessed a couple hundred kilogram for that size bolt, but then you are giving it some with 10Nm on what is essentially an out of tolerance M5 bolt! I would have expected half that torque, and then my guess isn't so far off. You must have some pretty high grade bolts there.
10-32 screw = .19 in. friction coefficient for steel = .2 90 in-lbs torque 90 in-lbs / (.2 x .19) = 2368.42 lbs. per screw 2368.42 x 3 = 7105.26 lbs total clamping force
Continued, 8880 is the combined axial clamping force of the screws. The part appears to be 12" long and is 1/2" wide for a total surface area of 6 square inches. This translates into 1480 lbs per square inch.
2368lb per screw.. I'm not too old for "the following is left to the reader..." type problems. For clarity my answer is based on a rapid fire Google of how to figure it out.
Around 4,327 lbs. F=(T/cD)*N where: F=Force T=Torque (90 inlb) c=Coefficient of friction (Steel v Steel is around .2) D=Bolt head diameter (.312in) N=# of bolts (3) Have no idea if this is right
The load is probably higher for a socket cap screw as the head is smaller (smaller torque radius for the head friction) and the bronze will slightly reduce the head friction, so the actual tension is probably more like 35-37 kN for all 3 (7875-8325 lb). There are a half dozen equations in the machinery handbook and all are BS insofar as the precision of the equation vastly exceeds the precision of the data we have (the precise friction coefficients under each part). The simplest equation is good enough which is torque = 0.2 x diameter x tension for UNC threaded bolts. If you want a finer answer you either need to measure fastener stretch and calculate the elastic force resulting from that or use a Wilhelm Skidmore type machine to calibrate the torque tension relationship for the specific fasteners you are using.
Thank you for pointing this out. Mechanics is not my field, but in electronics I see the same thing, people making calculations to the n-th decimal place, using wildly approximated parameters. If you don't have a sense for the margin of error you don't have any business even writing down a number.
The year is 20,018. This beautiful machine has graced many different shops and studios, The wear parts have been replaced several times. It is still in perfect condition. You are a genius and a national treasure.
It figures to 2368 pounds per bolt. Or 7104 pounds of force holding the key down. Wow I wouldn't have guessed that. Thanks for showing us your skills. Some of the old time machinist weren't always so willing to let you look over their shoulder I'm guessing. The education in the machining trade through the UA-cam community is really something. And greatly appreciated. That etching press is going to be a fine machine.
Throwing in a technical calculation adds an interesting dimension to the video, Tom. Great idea.
At 19:24, the Randy Richard fastener fumble. lol. I love this series Tom.
Hahaha, gotta give my wife credit for actually paying attention while I watch these videos. She said, when discussion on the cutting tools... "did he say he can cut a lot faster with the same shitload ??".... almost spewed coffee on the laptop !!
Approx. 10.5kN clamping force per screw, assuming steel fasteners into steel fixture plate.
Love your videos.
Cheers from Australia.
Chris
Nice video, interesting setup. Keep on keeping on.
your kicking ass Tom, love it!
T=KDP
90=.2 (.190)(X)
90/.2 (.190)=X
2368.42 lb.
Great work Tom. I always look forward to it.
Times three fasteners of course but I'm not sure if that is strictly linear.
Hi Tom 1
Really nice machining and setup ... I love it !
Very nice work Tom as always, my school of thought on what cutter to use; you use what you have available in your shop.
That right there is what you call an Abom size clamping load! ;) Great video Tom
Here in Europe we recon torkforce times threadpitch. Amerika is different, Recalculating to inches screws it all up.
Good luck to the winner and may Tom's swag be with him...:):)
Your right there tom , SWEET ! Thumbs up man ..
Hey Tom,
Assumptions: 13/64 drilled hole. Coefficient of friction of .16 for lubricated steel on steel for the threads and .16 for lubricated steel on bronze for the head of the SHCS. Using lubricated since you used a cutting oil and I don't think you degreased the parts. Formula from Spotts that I have in my HP48GX for the last 23 years. It has proven to be very accurate in the several times I did a physical testing check. 2196.568 pounds force per screw.
ATB, Robin
I'm putting my money on you Robin!!!
Thanks Danny, I just hope the formula that Tom used is as thorough as the one in "Spotts Design of machine elements" It is not a trivial formula and most would not bother with an equation that complicated. It considers: number of leads, thread pitch diameter, thread pitch, thread pressure face angle, thread coefficient of friction, thrust collar coefficient of friction, thrust collar pitch diameter, torque.
ATB, Robin
+ROBRENZ Ok, that's interesting! I guess that calculating something with so many variables there is going to be a few viable answers. Im looking forward to seeing Tom's method.
2196.568 lbs of force from 90 in.-lbs. of torque, huh? Interesting. I'm sure that's what the engineers say but the true answer is that "clamp load" is a theoretical and dimensionless "quantity". Steel on bronze pretty much self-lubricates too, by the way. You use lubricant on fasteners to prevent galling when the materials are the same or similar. Steel on steel, etc. Like metals gall and that gives you an inaccurate torque reading since a torque wrench is measuring the force required to turn the fastener. You reduce torque settings with lubricants used in conjunction with DISSIMILAR materials or ones that don't gall as easily or when a coated or plated screw is used in a natural finish steel thread to prevent over-torquing and stretching the fastener past its yield point. Its really easy to think that you want to lubricate all threads but if the manufacturer's assembly instructions don't call for lubricant, you don't use it. And when lubricant is called for you use the correct lubricant and make damn sure any screws or bolts and nuts replaced are replaced with OEM hardware or the correct aftermarket hardware. Grade, material, thread length, plating or coating, etc.
Put the wrong hardware in any situation with lubrication to boot and you're asking for trouble. Especially if you're not an experienced mechanic and you don't know what X amount of torque feels like and don't recognize when a fastener should be tight and is just stretching until it breaks. As for supposed "clamping load" calculations, when they're based on factors that are rarely present in the real world when someone is fabricating things. They're based on assumptions like the number of threads engaged being consistent with the size of the fastener its length and the threads being cut and a certain percentage of "engagement" and all kinds of "constants" that may exist in the engineering world when engineers are getting out the same book and they're using it to spec fasteners and the depth of blind holes and full thread length in them, etc.
The "book" may say X clamp load based on a bunch of math and theories and formulas and constants, but in reality the most honest and legitimate answer is that the "clamp load" is the sum of the torque applied to the fasteners to create it. Its still "force" and "load" and its the friction that results from them and the assembled fastener and place and torque to spec and its the only way to measure how much tension is on the fastener.
Have you considered adding a playlist of videos that have workholding tips and tricks?
Once again, I have to say that I love the aesthetic qualities of your designs. The care and attention that you pay to not only the functionality of the machine but also adding artistic touches to achieve a consistent quality and appearance is wonderful. I think it defines you as a master craftsman. Could it be the influence from a gifted artist you married?
Geez Tom, I barely have time to keep up watching the videos, never mind doing the calculations, anyhow my guess is: Enough to perform the task confirmed by experience... lol
Very good stuff as usual.. If you you were shooting for a real precise thickness on the bars, I would done the fixture bar second after drilling holes on the bars and left it clamped in the vice after the surfacing and taping. That way you know the fixture surface is flat in the mill..
Watching you power tap those holes brought up a thought in my head. How do you know when to stop the spindle so you don't crash into the bottom of the hole and break a tap or not tap it deep enough (though that part could be remedied with a hand tap afterwards)? I'm guessing it just comes from years of doing it.
have you ever talked about the ( allen wrench ) you were using , I like the design
Good call to make the fixture, take it out, machine holes in the part, then put the fixture back in.
did you lube the holes before inserting the screws? Because that makes a world of difference in what 90lbs*in really means
The clamping load is the sum of the torque applied to the three screws. 270 in.-lbs. You can't possibly end up with more clamping load than you're applying in torque to the screws. That's not what the engineering textbooks are going to say with the complex formulas and calculations using all those "specs" but they're also not calculated for a clamp load delivered when a chunk of bronze is screwed to a chunk of steel or with a steel fastener head torqued against a bronze part. Steel on bronze is going to deliver a lot less friction than steel on steel would and if you torqued those screws to spec you've probably over-torqued them because a torque wrench only measures the friction required to turn the fastener and pretty much any time you see steel hardware in an aluminum or other soft metal part, you'll see a steel washer between the fastener and part so the friction is returned to a steel on steel state. Soft metals also tend to compress and its very common to see torque specs for aluminum parts to require an initial torque and then a follow-up after 10 minutes or so or maybe a heat-cool cycle because they aluminum can compress under load and lose a little torque. Especially right under the washer. Bronze is much more dense and that would be reduced but its still a good idea to follow up whenever putting steel fasteners through non-ferrous metals.
No matter what the "clamp load" is the fasteners, if properly torqued, are nowhere near the "yield strength" and should never FEEL the "clamp load" anyway. If a pair of screws on say a connection rod cap are torqued to X lb.-ft. and have say 50.000 lbs of "clamping force" at that torque, as long as "pull" they're subjected to doesn't exceed 50,000 lbs., they're never actually loaded at all.
Those clamp load specs are what new engineers use to decide what fasteners to use and where based upon how much load another more experienced and skilled and knowledgeable engineer tells them will be applied to the "joint" in operation. They're meaningless in the "real world" of wrenching.
As others have shown, with the basic formula:
F = T/(c x d)
T = 90 in-lbs (torque)
c = 0.2 (coeff-dry)
d = 0.190 in (diameter for 10-32)
Then the answer of 2368 lbs (each bolt) falls out.
Further, c=0.2 is a nominal number for unlubricated hardware, and could be lowered to perhaps 0.15 lubricated.
Also, for 10-32 UNF the effective Tensile Stress Area is 0.02 in^2. So that implies an effective diameter of 0.160"
Taken together these give 3750 lbs per bolt. (Which others have noted is close to the 3800 lbs rating.)
I think the 2368 lbs is a comfortable guaranteed lower limit.
Another naive question: once you'd built the fixture to hold the keys while you milled the drilled faces, why did you clamp it with additional spacers around that tall plate (starting at 11:20 or so)? It isn't obvious (to me, anyway)....
Total clamping force is 7894.7 lb (2631.6 lb per bolt). I am assuming a coefficient of 0.18 like I always do, though, so this is a big unknown compared to what you might be assuming. This coefficient may be a little high but lower coefficients would result in higher loads and most people have answered lower than this so there is no use giving an answer similar to them since they answered before me.
Just did a Solidworks simulation on a 10/32 at 10.2N and it came up with 2204lbs. Is it working on a final answer basis or can we just keep guessing? Great fun though!
Nice work
Tom,
What kind/make of counterbore was that? The flutes looked more open (for lack of a better term) than I'm used to!
Were you running that counterbore at 1/3 the drilling speed? Great work as always.
Marc
Mr L, I make it 2368.42 pounds per position. The efficacy of this force depends on the friction co-efficient between the screw threads and the steel block, and the helix angle of the screw, etc etc. Important question is "is it tight *enough"! ;)
Coefficient of friction for Brass-Steel is 0.5 Dry and 0.19 Lubricated. Gives a clamping force of 4219.92N Dry and 11 105.04N Lubricated per cap screw.
I hate to do this, since you tried to show the markings off in the video, but for the life of me I can't read them..
That little face mill you had out for the steel looks sweet! What was it?
Figuring that an advance of one turn of the screw at a distance of one inch requires π x 2 inches of motion and advances the screw into the work by 1/32. This gives a mechanical advantage of a bit over 200:1. Multiply times 90 gives about 18, 000 pounds of pressure (in a frictionless universe) and maybe 20% of that in the real world, so maybe 3,600 pounds. Now this is also about the ultimate tensile strength of a 10-32 SHMS (3400 pounds). I will guess somewhere in the middle at *3,500* pounds per screw. (Late guess, but I haven't checked the other answers yet.)(Checked: some answers were close but less or quite a bit less; some were more or quite a bit more. I'm comfortable with my guess.)
One million megatons, as we all failed to consider that Tom rerouted the output of the impulse engines directly to the torque wrench and extended the structural integrity field to the screws.
Some 2300 pounds force sounds implausible until I compare the thought of having my finger under the front of my truck, to having it under that screw and key. Maybe so. Time for the Tom-o-Meter.
A quick look at a couple of online charts suggests clamping forces in the range of 600-720 pounds for 10-32 at around 90 in. lb., depending on lubrication, bolt grade, and "things." 1800-2160 pounds total seems experientially more believable to me.
Hi Tom" I don't know how the calculator works for this one, thread, type thread pitch, % of thread depth after cutting, probably in the machinist hand book! I'm ill with flu but maybe a walk out to the workshop & get it. Do show us how to find this kind of info. Ace show.
I could be ignorant, but i do feel the need to know the grade of the bolts, or is it just me not getting all this inch stuff...?
Just got done watching you latest MNM103. No one is right yet?Depending on the coefficient of friction, the more lube, the higher the clamping force. Full dry(.2 coefficient of friction) is 2368.42lbs, a bit of lube (.19cof) is 2493.04lbs, a semi lubed (.18cof) is 2631.58lbs, and fully lubed could give you 2786.38lbs-2960.53lbs of clamping force. Can you tell I really want some swag Tom?
Hi Tom. While we are somewhat on the subject, could you give your opinion on which is correct, foot- pounds or pound- feet, or if it even matters. Thanks.
The foot-pound force (symbol: ft·lbf or ft·lbf), or simply foot-pound (symbol: ft·lb) is a unit of work or energy in the Engineering and Gravitational Systems in United States customary and imperial units of measure
A pound-foot (lb·ft or lbf·ft) is a unit of torque (a pseudovector). One pound-foot is the torque created by one pound force acting at a perpendicular distance of one foot from a pivot point.
The name "pound-foot", intended to minimize confusion with the foot-pound as a unit of work,
Wikipedia said it better than I could.
Since torque is force multiplied by the distance from the perpendicular vector to the revolving axis, it shouldn't matter. I think it's a matter of pronunciation preference. But heck, what do I know, I work in metric
𝖘 𝖍 𝖎 𝖗 𝖔 The two terms are commonly interchanged in everyday speech. You understand whether you are discussing torque or energy/work from the context of the discussion.
Did you ever use a cold air gun, use one on my Bridgeport with great success with all the tough materials?
Hi Stephen,
I have on for the surface grinder. Seems to work pretty well. They seem to get a lot of money for the darn things. Thanks for the comment.
Cheers,
Tom
I get 2368.4 lbs per fastener or 7105 lbs total. I did have to assume a coefficient of friction for steel-on-steel, screw to fixture.
lets give this a shot...the cutting load applied to the bolts is in shear so given a 10-32 has a shear area around 0.02in^2 (roughly the area of the tap drill size) the average SHCS has a tensile strength of 180000psi (working from memory here) so the shear strength is 0.505 of UTS...so 90900psi for shear stress and F=SxA...so 1818 lb per screw...3 screws for a grand total final answer of 5454lbf to shear them off...well above the cutting forces...
thanks for the vid very nice
in the 8k lb range depending on grade of fastener and coefficient of friction n such
Hi Tom,
Wondering why you 'rub' your files over the work when you're doing file work (especially in the baby bullet build). I'm new to machine shop and also saw Mr. Pete do this. I've learnt that you should lift the file clear off the work when pulling back. I know you ease the pressure on the back stroke but still wondering why you don't lift it, do you just not care that much about files to do it? Don't mean to smart, just curious!
Aryan
On a softer material which is most if its a good quality file, light back-dragging on the file can clear some of the crag off the file teeth so they're cleaner for the next cut forward.
10-32 according to Holo-Krome's (who should know, being preeminent manufacturers of SHCS) fgures, average torque to tighten average screw to yield is 120lbs.inch which gives average tension induced in screw tightened to yield, 2800 pounds. (Average ultimate tensile strength 3600 pounds). Values are for standard steel(black finish) screws lubricated with oil clamping hardened steel plates to hardened "nuts".
So under the above conditions, ignoring your under- tightening, would be somewhere in the order of 3 times 2800 lbs, 8400lbs.
Been meaning to ask ya Tom... who makes those Allen Wrenches or what are they called?
I believe it is shop made
Yes. I am impressed by them the number of times they are used. They look REALLY handy. Surely they are Lipton manufactured !!
Just tight enough to hold securely, but not break the screws?
@ 18:59 love the TOM LIPTON stop clamp (it's all about class)
2368.4 pounds of clamping force per screw, considering you installed it dry. It's amazing what a little #10 screw will hold by itself:)
Dry = 2368 and change per fastener, fully lubed is 3157 and change... So your clamping force total ranges from "plenty power" to "waaay plenty"
GREAT VIDEO !! CLAMPING FORCE = 1.73 ABOMS :-)
I came up with 2368 pounds of clamp force, although 90 in-lbs torque is a bit high for a 10-32 screw.
I may be wrong but the screw specs give a 10/32 screw as having a tensile strength of 1200lbs so I'm guesing regardless of your torque thats the best you'll get.
3330 lb for UNC and 3800 lb for UNF socket cap screws according to Unbrako fastener guide.
My mistake, I thought I thought Tom said machine screw www.google.co.uk/url?sa=t&source=web&rct=j&url=www.fasnetdirect.com/refguide/Machinescrewthreads.pdf&ved=0ahUKEwi5se34qa3OAhVBD8AKHW6RAtAQFggbMAA&usg=AFQjCNFlnV1zGSBOyCEQ5XS_cstDSae4ag&sig2=NbKt15TIeQYTZHe33DdU0g
In Oxtoolworld, the answer is 3 x1032 x 90 = 278640 = 278,640KL (Kilo-Liptons) or in English, enough!
about 2300 lbs pr. screw (no lube), nice video :-)
Well I got 2370.9167 lbs dry in steel, but not sure how the bronze affects the clamping force.
Lubricated with a coefficient friction factor of .17 is 2789.3138 lbs in steel as well. Again not sure with the bronze on top of the steel on how the changes the clamping force or if that is even a factor.
Metric would be 12.4115 N using the same coefficient friction factor of .17
Hi Cole,
You got close to the book value. The friction coefficient has a big effect on the results and is hard to determine accurately.Thanks for the comment.
Cheers,
Tom
The clamping load is in the order of sufficient and a bit
2125 lbs per screw, using F=T/(KD), where K was estimated at 0.22 using Shigley's equations.
Per screw at 90 in\lbs
2370 lbs dry,
2960 lbs with lube
^ This is what I got also
10549.8N/bolt x 3 bolts gives 31649.4N total clamping force.
Cheers.
Not able to tell what type of finish was on the screws used... So - Assuming Worst case friction on one end and - Black Oxide on the other end:
2368#/ Fastener (7105# Total Clamping Force) to 4306 #/Fastener (12918# Total Clamping force)
I don't think it is going to move under milling loads.
Ps based on a rule of thumb of torque = axial force x clamping diameter x 0.2. I suppose this is only going to apply *at the screw diameter and isn't considering the force at different dias. of screw head.
So, according to my calculations, each screw would give you around 2368lb of clamping pressure. Amazing what mechanical advantage can do.
My physics teacher used to say "It's not a torque wrench it's a moment wrench but don't tell that to a mechanic as he'll brain you with it"
As a stress engineer, I find your challenge is frustratingly difficult to adequately deal with, but only because I know what I know about certain mechanical fail-safe designs where pre-load is vitally important to proper operation . Prove by test!! (edited for a bit more info...)
The answer is: More than enough!
Specs at 13:29
Of course torque is a very inaccurate indicator of fastener preload. Especially with a dry fastener. Further increasing the inaccuracy is using a multiple step torquing procedure. The fasteners should all be snugged down, then brought to the final spec in one fluid motion. Another way to increase accuracy is to cycle the fastener a few times to burnish the contact surfaces before the final torquing procedure. All that aside, those small cap screws can produce a very impressive preload
I'm not entirely sure, but it should be roughly 31.6 kN clamping force
8437.5 lbs of total clamping force.
Approximately 2368 lbs depending on lubrication. The majority of the torque is lost to friction.
Assuming there is lubrication from the tapping operation, 9,529.41 lbs of clamping force.
the total clamping load is pretty scoocum, or to be in metric 3 Chooches!
Hi Tom
My guess is 925.44 N-mm of clamping force.
installation bolt load is: 21 N
close to 1060 Newtons. 1059.375 each to be exact.
Tom.... Doesn't your test ( Meatloaf 103 ) confirm how close I am with the 2960.526 lbs I calculated ?? :) Take Care.
You could just about lift my car with that bolt! I would only have guessed a couple hundred kilogram for that size bolt, but then you are giving it some with 10Nm on what is essentially an out of tolerance M5 bolt! I would have expected half that torque, and then my guess isn't so far off. You must have some pretty high grade bolts there.
Juhuu!!!
2,960.52 lbs......what are we missing?
10-32 screw = .19 in.
friction coefficient for steel = .2
90 in-lbs torque
90 in-lbs / (.2 x .19) = 2368.42 lbs. per screw
2368.42 x 3 = 7105.26 lbs total clamping force
8,640lbs total clamping force, or 2,880lbs per screw.
Assuming a lubricated thread it would be 2960 X 3 = 8880 Lbs.
Continued, 8880 is the combined axial clamping force of the screws. The part appears to be 12" long and is 1/2" wide for a total surface area of 6 square inches. This translates into 1480 lbs per square inch.
2368lb per screw.. I'm not too old for "the following is left to the reader..." type problems.
For clarity my answer is based on a rapid fire Google of how to figure it out.
I get 2368 Lb per machine screw (7104# total for 2 screws)
Approx. 2800 lb. clamping force.
The answer is adequate clamping force. 😬
Figuring the actual answer is above my "ciphering" capabilities.
Around 4,327 lbs.
F=(T/cD)*N where:
F=Force
T=Torque (90 inlb)
c=Coefficient of friction (Steel v Steel is around .2)
D=Bolt head diameter (.312in)
N=# of bolts (3)
Have no idea if this is right
more then need
How about 2368 pounds each? 7105 total clamping force.
~2368.4lbs per so 7105.2lbs total?
93.3 (Axial Screw Clamp Force) X 3 = 279.9 Axial Clamp Force (if that lower bar is steel)
42. because 42 the answer to everything. XD
I got 12,216 kilos! That seems like a miscalculation XD
10.55 N of hold down (metric)
4180.36 N per screw for a total of 12541.07 N!
I think I'm wrong... ?
Total clamping force: 8360 lbs
10.55 kN per screw x 3 = 31.65 kN (7114 lb).
The load is probably higher for a socket cap screw as the head is smaller (smaller torque radius for the head friction) and the bronze will slightly reduce the head friction, so the actual tension is probably more like 35-37 kN for all 3 (7875-8325 lb). There are a half dozen equations in the machinery handbook and all are BS insofar as the precision of the equation vastly exceeds the precision of the data we have (the precise friction coefficients under each part). The simplest equation is good enough which is torque = 0.2 x diameter x tension for UNC threaded bolts. If you want a finer answer you either need to measure fastener stretch and calculate the elastic force resulting from that or use a Wilhelm Skidmore type machine to calibrate the torque tension relationship for the specific fasteners you are using.
Thank you for pointing this out. Mechanics is not my field, but in electronics I see the same thing, people making calculations to the n-th decimal place, using wildly approximated parameters. If you don't have a sense for the margin of error you don't have any business even writing down a number.