I'm not sure how much help I'll be here, but I do know a few things.
The distance from the pivot point is the radius, not the diameter.
For example, if a fifty-pound boy on a see-saw sits five feet from the fulcrum, he's exerting 250 foot-pounds of torque.
If I tighten a nut with a two-foot torque wrench, I have to use twenty pounds of force to generate 40 ft-lb of torque.
What does the "3/4" signify? Is that the radius of the bolt head?
I'm probably off here, but the total FORCE acting on the bolt by the head would be the torque divided by the pivot radius.
40 ft-lb = 480 in-lb
480 in-lb/(3/4 in) = 640 lb (Twice this, if the bolt radius is 3/8")
The pressure will be the force divided by the area it's acting on. I don't know if this involves the number of threads or is simply the area of the bolt head.
I don't remember much about physics. (not sure why, it was only 40 years ago )
It seems to me, though, that the thread size is not relevant as it is the bolt head that presses the cylinder head tightly against the block. Seems to me that, no matter where on the under-surface of the bolt head you measured the pressure, the pressure would be the same, regardless of the diameter of the bolt head.
Hey, now! go easy on me! When I studied this stuff, the wheel was still a modern marvel!
How about some of our younger "science guys" like P-Mosh and Bolshevik?
Here is the question: An engine head bolt has a specified torque at 40 foot pounds on a 3/4 10 bolt. How much pressure is being exerted by the bolt on the head.
If that were a banjo head instead of an engine head,
All right piggie you don't fool me!!!!! We know you are building some sort of new pie hurling machine and just want to know how tight to torque the bolts. Math Question ....... PFFFFFF
My son and I are doing his math work. We came across a question not fully explained in his lesson. (his lessons are on dvd)
Here is the question: An engine head bolt has a specified torque at 40 foot pounds on a 3/4 10 bolt. How much pressure is being exerted by the bolt on the head.
Foot pounds was covered in a previous lesson, but not as "torque". So I googled it.
"1 foot-pound is the amount of torque created by a force of 1 pound acting at a perpendicular distance of 1 foot from a pivot point. "
Now we are trying to figure out about the "one foot perpendicular from a pivot point" Is the one foot the radius or the diameter?
I think then we set up the problem like this:
pi X diameter X 40 (lbs)
then that answer X 10 (the number of threads on the bolt)
I have the answer on the data cd sent with the program, but I want my son to see how to do the problem--but I need help myself! LOL! (We CAN call the teacher that made the dvds, too, but we couldn't get through just now)
The specified torque is 40 ft/lbs on a 3/4 bolt @ 10 threads per inch. What type of thread is it ? It is negligible but would show he knows threads. If 40 ft/lbs is exerted it should be 40 ft/lbs exerted on the bolt.
Being a twenty four year old Jet Engine mechanic and still working with torque wrenches I am curious for the answer.
It would appear the flying pig had it right from the start, according to Jim, but George makes some sense maybe ...
it is asking for pressure, which would include the area the force is acting on? like psi ... pounds/square inch
the 2 pi r just gives the distance over which the 40 foot pounds "acts" in one turn ... which is transferred (friction free) to one tenth of an inch down?
''''''''''''''' 2pi x R
Q = F x ---------
'''''''''''''''''''' P
Where Q = pressure exerted by bolt
' F = force exerted on end of wrench
' R = length of wrench
' p = thread lead
Assuming a 1 foot wrench, 40 pounds pressure and 10 threads/inch...
''''''''''''6.283 x 1
40 x ------------ = 2513.2 lbs
'''''''''''''''''' .1
so ... 40 pounds x 6.283 x 1 foot / .0083 foot = 30,163 pounds but that is not pressure?
So the 3/4" is the area the pounds works on? Whatever the area is, it seems you need that to get pressure instead of force.
Many of those physics brain cells have died for me, but their ghosts want to know the answers.
Exie--yeah, as soon as I saw the word "torque" I was lost! LOL! He's 14, but this particular math program isn't listed as grade-specific, so i don't know if he is doing 8th, 9th or 10th grade math! LOL!
Thank you all for your replies. Some of them went over my head! LOL! I think sometimes these math problems are worded weird and don't apply to real life as they are worded. But I s'pose if the kid gets the "how" of the math equation, then they have learned it right--er sompin!
The whole unit was on simple machines, and the advantages of using inclined planes, pulleys, crankwheels, etc. The last lesson was on bolts and the amount of force created by tightening a bolt and nut--say the amount of force holding two boards together.
The threads of the bolt are likened to an inclined plane. The fraction, (3/8, 3/4, or whatever), is the diameter of the bolt, and the thread count is how many threads per inch (a steeper or less steep inclined plane). Most of the questions were like, "How much force will be created by tightening a 3/8-16 bolt, using a wrench that is 6 inches long and applying a force of 11 pounds to the end of the wrench"
The engine head bolt and the torque of 40 foot pounds was unexpected! LOL!
In the question about the engine head, the number of threads does make a difference, but I don't know if I can explain any further than the above eplanation of it resembling an inclined plane. The size of the bolt or the bolt head is insignificant in this problem. He also had us disregard friction in this equation.
Here was how to set up the equation.
The one foot was the radius, so 2 ft diameter, or 24 inches, because the thread is 10 turns per inch
24 X 3.14(pi) = 75.36 circumference
75.36 X 40 (lbs torque) =3014.4
3014.4 X 10 (# of threads) = 30,144 (pressure being exerted by the bolt on the engine head)
That was the correct answer to this problem. I don't know if it answers all of the questions you brainacs asked me!! LOL!
This was kinda fun-- I sat with him doing most of the work right alongside him. I sorta understood how to set up the problems, but explaining them to my dyslexic son was--er--um--interesting! LOL! (because I "get" how to set up the problem, and get the answers right, but can't explain what I just did! EEK!)
I'm sorry. I just don't understand the answer you gave.
The problem doesn't say anything about the size wrench that is used, so why assume it to be one foot? I guess it doesn't matter as long as you don't confuse torque (ft-lbs) with force (lbs). Force X circumference will be the same, because if you increase the wrench length by a certain fraction, you will reduce the amount of force by the same fraction (i.e., the torque remains the same).
The answer, though, doesn't seem to be in units of pressure, but rather of force.
Could you give the solution with all units included?
the length of the "wrench" is just from the definition of torque I think.
But they say pressure ... and Jim's definition says pressure ... but they both really just give force, as I see it.
Obviously they didn't go to U of Illinois :)
But I got it "right" on the 2nd try, except for some rounding ...
not sure why they say the size or type of bolt ... all we really did was multiply the 40 pounds force by transferring one turn of that distance into 0.1" downward.
George, I used one foot for the radius because of the definition in Wikipedia for for a unit of torque:
"It is also a unit of torque: 1 foot-pound is the amount of torque created by a force of 1 pound acting at a perpendicular distance of 1 foot from a pivot point."
I figured the pivot point was the bolt, so the radius was 1 foot (The torque being the "wrench") It was a lucky guess for me--I didn't go beyond highschool algebra, and am only having some knowledge refreshed in my kid's homeschool classes. I can't answer your question about giving the solution with all units included. The questions and answers are on a cd that comes with the curriculum, and the answers are a simple list, i.e. 1) 124,000 pounds (The weight being lifted) 2) 24 (pounds of force), 3) 30,144
I added the words in parintheses--he just lists the labeled numbers.
I don't think I would have even attempted this problem if I hadn't had the answer to check against. So what I posted above was the way I did the problem. Most of the questions on this lesson were about regular bolts and nuts and in the dvd lesson he only spoke about the pressure the bolt would be exerting on the item it was bolted to. I don't mean this disrespectfully to the teacher, I think he's wonderful, but might it be possible he is applying a math problem to something he doesn't fully understand himself? (a unit of torque being the same as using a wrench of a certain length?) He used the word "pressure" in the question and maybe he should have been asking "How much force"--maybe? I don't really know, because I barely understood what I just did through this whole lesson--and my son "got it" even less than I did.
Rhino, in every question of this lesson, he gave the size of the bolt and that never figured in-- only the number of threads on the bolt.
It was fun! Now the next unit he's doing is in negative numbers! woohoo! My son actually likes math, too and is pretty good at it, but with his dyslexic thinking, he turns the order of the equations all around--I keep thinking one of these times it WILL matter, but so far, he still comes up with the right answer.
He wants to take chemistry too. . . . (Maybe I'll learn something? HAH!)
It sounds like it was some basic math problems that they tried to make interesting ... but maybe confused some things .. but no big deal. A guess parents learn a lot that they forgot the first time through school ... I wouldn't have known about torque but for what you said.
George might be of some help in chemistry too ;) he was studying real science while I was being a WOW and a college WOW ... sigh ...
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GeorgeStGeorge
I'm not sure how much help I'll be here, but I do know a few things.
The distance from the pivot point is the radius, not the diameter.
For example, if a fifty-pound boy on a see-saw sits five feet from the fulcrum, he's exerting 250 foot-pounds of torque.
If I tighten a nut with a two-foot torque wrench, I have to use twenty pounds of force to generate 40 ft-lb of torque.
What does the "3/4" signify? Is that the radius of the bolt head?
I'm probably off here, but the total FORCE acting on the bolt by the head would be the torque divided by the pivot radius.
40 ft-lb = 480 in-lb
480 in-lb/(3/4 in) = 640 lb (Twice this, if the bolt radius is 3/8")
The pressure will be the force divided by the area it's acting on. I don't know if this involves the number of threads or is simply the area of the bolt head.
I hope this helps.
George
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waysider
I don't remember much about physics. (not sure why, it was only 40 years ago
)
It seems to me, though, that the thread size is not relevant as it is the bolt head that presses the cylinder head tightly against the block. Seems to me that, no matter where on the under-surface of the bolt head you measured the pressure, the pressure would be the same, regardless of the diameter of the bolt head.
Hey, now! go easy on me! When I studied this stuff, the wheel was still a modern marvel!
How about some of our younger "science guys" like P-Mosh and Bolshevik?
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rhino
I'll play along for fun :)
wouldn't twice as many threads per inch make for twice the force down, with same torque applied?
so 3/4" is the diameter of the bolt at the threads ... giving an area (or a distance)?
so the 40 foot pounds / area times threads per inch ?
needs some conversions
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excathedra
okay that's where i got lost.
--
how old is your son ? you're scaring me
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dmiller
If that were a banjo head instead of an engine head,
I might be able to help out more. :)
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WhiteDove
All right piggie you don't fool me!!!!! We know you are building some sort of new pie hurling machine and just want to know how tight to torque the bolts. Math Question ....... PFFFFFF
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Jim
From "Machinery's Handbook"
(excuse the stupid single quotes, it's the only way I could format the equations)
''''''''''''''' 2pi x R
Q = F x ---------
'''''''''''''''''''' P
Where Q = pressure exerted by bolt
' F = force exerted on end of wrench
' R = length of wrench
' p = thread lead
Assuming a 1 foot wrench, 40 pounds pressure and 10 threads/inch...
''''''''''''6.283 x 1
40 x ------------ = 2513.2 lbs
'''''''''''''''''' .1
It appears the thread diameter doesn't matter until you start worrying
about the coefficient of friction.
Disclaimer - I am not a mechanical engineer and I could be wrong....
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My son and I are doing his math work. We came across a question not fully explained in his lesson. (his lessons are on dvd)
Here is the question: An engine head bolt has a specified torque at 40 foot pounds on a 3/4 10 bolt. How much pressure is being exerted by the bolt on the head.
Foot pounds was covered in a previous lesson, but not as "torque". So I googled it.
"1 foot-pound is the amount of torque created by a force of 1 pound acting at a perpendicular distance of 1 foot from a pivot point. "
Now we are trying to figure out about the "one foot perpendicular from a pivot point" Is the one foot the radius or the diameter?
I think then we set up the problem like this:
pi X diameter X 40 (lbs)
then that answer X 10 (the number of threads on the bolt)
I have the answer on the data cd sent with the program, but I want my son to see how to do the problem--but I need help myself! LOL! (We CAN call the teacher that made the dvds, too, but we couldn't get through just now)
The specified torque is 40 ft/lbs on a 3/4 bolt @ 10 threads per inch. What type of thread is it ? It is negligible but would show he knows threads. If 40 ft/lbs is exerted it should be 40 ft/lbs exerted on the bolt.
Being a twenty four year old Jet Engine mechanic and still working with torque wrenches I am curious for the answer.
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rhino
It would appear the flying pig had it right from the start, according to Jim, but George makes some sense maybe ...
it is asking for pressure, which would include the area the force is acting on? like psi ... pounds/square inch
the 2 pi r just gives the distance over which the 40 foot pounds "acts" in one turn ... which is transferred (friction free) to one tenth of an inch down?
so ... 40 pounds x 6.283 x 1 foot / .0083 foot = 30,163 pounds but that is not pressure?
So the 3/4" is the area the pounds works on? Whatever the area is, it seems you need that to get pressure instead of force.
Many of those physics brain cells have died for me, but their ghosts want to know the answers.
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Psalm 71 one
FEATHERS! SHHHHHHHHHH!
Exie--yeah, as soon as I saw the word "torque" I was lost! LOL! He's 14, but this particular math program isn't listed as grade-specific, so i don't know if he is doing 8th, 9th or 10th grade math! LOL!
Thank you all for your replies. Some of them went over my head! LOL! I think sometimes these math problems are worded weird and don't apply to real life as they are worded. But I s'pose if the kid gets the "how" of the math equation, then they have learned it right--er sompin!
The whole unit was on simple machines, and the advantages of using inclined planes, pulleys, crankwheels, etc. The last lesson was on bolts and the amount of force created by tightening a bolt and nut--say the amount of force holding two boards together.
The threads of the bolt are likened to an inclined plane. The fraction, (3/8, 3/4, or whatever), is the diameter of the bolt, and the thread count is how many threads per inch (a steeper or less steep inclined plane). Most of the questions were like, "How much force will be created by tightening a 3/8-16 bolt, using a wrench that is 6 inches long and applying a force of 11 pounds to the end of the wrench"
The engine head bolt and the torque of 40 foot pounds was unexpected! LOL!
In the question about the engine head, the number of threads does make a difference, but I don't know if I can explain any further than the above eplanation of it resembling an inclined plane. The size of the bolt or the bolt head is insignificant in this problem. He also had us disregard friction in this equation.
Here was how to set up the equation.
The one foot was the radius, so 2 ft diameter, or 24 inches, because the thread is 10 turns per inch
24 X 3.14(pi) = 75.36 circumference
75.36 X 40 (lbs torque) =3014.4
3014.4 X 10 (# of threads) = 30,144 (pressure being exerted by the bolt on the engine head)
That was the correct answer to this problem. I don't know if it answers all of the questions you brainacs asked me!! LOL!
This was kinda fun-- I sat with him doing most of the work right alongside him. I sorta understood how to set up the problems, but explaining them to my dyslexic son was--er--um--interesting! LOL! (because I "get" how to set up the problem, and get the answers right, but can't explain what I just did! EEK!)
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GeorgeStGeorge
I'm sorry. I just don't understand the answer you gave.
The problem doesn't say anything about the size wrench that is used, so why assume it to be one foot? I guess it doesn't matter as long as you don't confuse torque (ft-lbs) with force (lbs). Force X circumference will be the same, because if you increase the wrench length by a certain fraction, you will reduce the amount of force by the same fraction (i.e., the torque remains the same).
The answer, though, doesn't seem to be in units of pressure, but rather of force.
Could you give the solution with all units included?
George
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rhino
the length of the "wrench" is just from the definition of torque I think.
But they say pressure ... and Jim's definition says pressure ... but they both really just give force, as I see it.
Obviously they didn't go to U of Illinois :)
But I got it "right" on the 2nd try, except for some rounding ...
not sure why they say the size or type of bolt ... all we really did was multiply the 40 pounds force by transferring one turn of that distance into 0.1" downward.
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Psalm 71 one
George, I used one foot for the radius because of the definition in Wikipedia for for a unit of torque:
"It is also a unit of torque: 1 foot-pound is the amount of torque created by a force of 1 pound acting at a perpendicular distance of 1 foot from a pivot point."
I figured the pivot point was the bolt, so the radius was 1 foot (The torque being the "wrench") It was a lucky guess for me--I didn't go beyond highschool algebra, and am only having some knowledge refreshed in my kid's homeschool classes. I can't answer your question about giving the solution with all units included. The questions and answers are on a cd that comes with the curriculum, and the answers are a simple list, i.e. 1) 124,000 pounds (The weight being lifted) 2) 24 (pounds of force), 3) 30,144
I added the words in parintheses--he just lists the labeled numbers.
I don't think I would have even attempted this problem if I hadn't had the answer to check against. So what I posted above was the way I did the problem. Most of the questions on this lesson were about regular bolts and nuts and in the dvd lesson he only spoke about the pressure the bolt would be exerting on the item it was bolted to. I don't mean this disrespectfully to the teacher, I think he's wonderful, but might it be possible he is applying a math problem to something he doesn't fully understand himself? (a unit of torque being the same as using a wrench of a certain length?) He used the word "pressure" in the question and maybe he should have been asking "How much force"--maybe? I don't really know, because I barely understood what I just did through this whole lesson--and my son "got it" even less than I did.
Rhino, in every question of this lesson, he gave the size of the bolt and that never figured in-- only the number of threads on the bolt.
It was fun! Now the next unit he's doing is in negative numbers! woohoo! My son actually likes math, too and is pretty good at it, but with his dyslexic thinking, he turns the order of the equations all around--I keep thinking one of these times it WILL matter, but so far, he still comes up with the right answer.
He wants to take chemistry too. . . .
(Maybe I'll learn something? HAH!)
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rhino
Thanks for the entertainment psalmie :)
It sounds like it was some basic math problems that they tried to make interesting ... but maybe confused some things .. but no big deal. A guess parents learn a lot that they forgot the first time through school ... I wouldn't have known about torque but for what you said.
George might be of some help in chemistry too ;) he was studying real science while I was being a WOW and a college WOW ... sigh ...
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