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Does torque translate into linear force? 4
- Thread starter hanksboy
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1
- #2
Hi hanksboy
The torque on a bolt is transmitted through the bolt and hence the parts its mounted through as a tensile force.
Have a look at this link here's another site
regards desertfox
The torque on a bolt is transmitted through the bolt and hence the parts its mounted through as a tensile force.
Have a look at this link here's another site
regards desertfox
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2
- #3
I think you are saying that the bolt end (or "point") presses against a third object. In that case, the torque doesn't translate into a force. Rather, the rotation of the bolt translates into a deflection of the third object, and its geometry and elastic modulus determines the force generated.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
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1
- #4
hanksboy: The compressive force will be approximately
P = T/[0.159*p + 0.375(mu1)(1.54*D - p) + 0.40*mu2*D],
where T = bolt installation torque, D = bolt nominal diameter, p = thread pitch, mu1 = thread coefficient of friction, and mu2 = coefficient of friction at bolt tip. For lubricated steel on steel, mu = 0.11 isn't uncommon. For "dry" steel on steel, mu = 0.15 isn't uncommon.
P = T/[0.159*p + 0.375(mu1)(1.54*D - p) + 0.40*mu2*D],
where T = bolt installation torque, D = bolt nominal diameter, p = thread pitch, mu1 = thread coefficient of friction, and mu2 = coefficient of friction at bolt tip. For lubricated steel on steel, mu = 0.11 isn't uncommon. For "dry" steel on steel, mu = 0.15 isn't uncommon.
Hi hanksboy
Sorry I misread your thread however instead of getting a tensile force you get a compressive force in the bolt along the lines of what vonluke as posted ie- the bolt torque applied is transmitted as a axial force but instead of the bolt going into tension its in compression.
regards
desertfox
Sorry I misread your thread however instead of getting a tensile force you get a compressive force in the bolt along the lines of what vonluke as posted ie- the bolt torque applied is transmitted as a axial force but instead of the bolt going into tension its in compression.
regards
desertfox
Quote from Cory:
"I think you are saying that the bolt end (or "point") presses against a third object. In that case, the torque doesn't translate into a force. Rather, the rotation of the bolt translates into a deflection of the third object, and its geometry and elastic modulus determines the force generated.
Regards,
---------------------------------------------------
I respectfully disagree. The force on the third object has nothing to do with deformation. Deformation is a manifestation of the force. The force is obtained from the energy equation as follows;
1)T*@=F*x+F*f*R*@
where
T=torque applied to the screw
@= angle of input torque
x=transverse motion of screw corresponding to @
f=friction coefficient
R =an average radius over which the friction takes place
fR are usually determined empirically.
1st term is input energy;second term is output energy; third term is frictional loss of energy
Now dividing the energy equation by @=2pi,(which corresponds to x=p and solving for F we get
F=2pi*T/(p+f*R*2pi)
"I think you are saying that the bolt end (or "point") presses against a third object. In that case, the torque doesn't translate into a force. Rather, the rotation of the bolt translates into a deflection of the third object, and its geometry and elastic modulus determines the force generated.
Regards,
---------------------------------------------------
I respectfully disagree. The force on the third object has nothing to do with deformation. Deformation is a manifestation of the force. The force is obtained from the energy equation as follows;
1)T*@=F*x+F*f*R*@
where
T=torque applied to the screw
@= angle of input torque
x=transverse motion of screw corresponding to @
f=friction coefficient
R =an average radius over which the friction takes place
fR are usually determined empirically.
1st term is input energy;second term is output energy; third term is frictional loss of energy
Now dividing the energy equation by @=2pi,(which corresponds to x=p and solving for F we get
F=2pi*T/(p+f*R*2pi)
There seems to be some wild assumptions made about the geometry of this system. What is the definition of stiff? How thick is object three? It could be a stiff panel, in which case, it will deflect significantly. Or, it could be a large, rigid block. What are the constraints on objects two and three?
This certainly is a displacement controlled problem rather than a force controlled problem. The force between screw point and the third object develops due to the contact between them.
There are at least two sources of friction: between the screw threads of the screw and object two; and between the screw's point and object three.
I think some tips have been provided, but you would need to define the problem much more precisely for a true solution.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
This certainly is a displacement controlled problem rather than a force controlled problem. The force between screw point and the third object develops due to the contact between them.
There are at least two sources of friction: between the screw threads of the screw and object two; and between the screw's point and object three.
I think some tips have been provided, but you would need to define the problem much more precisely for a true solution.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
desertfox,
I understand what you are saying. Superbolt is an example of a product that uses this principle. I just think that there is a lot of interpretation to this particular problem.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
I understand what you are saying. Superbolt is an example of a product that uses this principle. I just think that there is a lot of interpretation to this particular problem.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
- Thread starter
- #14
Thanks for all the help.
A lot more help than I expected.
I'm an obsolete 'sparkey' who spent nearly all
of my career in programming.
I was hoping for a much simpler solution. Here's the source of the problem.
I was reading one of those Haynes Techbooks from Autozone on overhauling your own transmission.
They present 5 or 6 different models to work on.
ON each one they require special, expensive factory tools to make measurements. I'm wondering
how I might make work-around homemade tools for the process.
In one case the movement of a clutch band piston is made by directly applying a known force, with a special tool,
to the piston, and measuring the movement. The duplication of this tool is not possible without the original specs.
In another case, the movement of a clutch band piston is made by applying a screw to the piston through a special cap, using
a specified torque, and aqain measuring the movement. The tool for this method is easily duplicated.
In each case a return spring of unknown specifications is under the piston.
So I'm wondering, is the method for one applicable to the other.
The bolt would screw through the cap by hand until contact with the piston, then tightened with a torque wrench to a specified value. A dial indicator would be set on the piston and measure movement as the screw is backed out.
Without getting into impossible to measure parameters such as friction coefficients, is there a simple equation to give
me the force applied to the piston, given the size of the bolt, and the torque applied to it?
Thanks again for your input.
A lot more help than I expected.
I'm an obsolete 'sparkey' who spent nearly all
of my career in programming.
I was hoping for a much simpler solution. Here's the source of the problem.
I was reading one of those Haynes Techbooks from Autozone on overhauling your own transmission.
They present 5 or 6 different models to work on.
ON each one they require special, expensive factory tools to make measurements. I'm wondering
how I might make work-around homemade tools for the process.
In one case the movement of a clutch band piston is made by directly applying a known force, with a special tool,
to the piston, and measuring the movement. The duplication of this tool is not possible without the original specs.
In another case, the movement of a clutch band piston is made by applying a screw to the piston through a special cap, using
a specified torque, and aqain measuring the movement. The tool for this method is easily duplicated.
In each case a return spring of unknown specifications is under the piston.
So I'm wondering, is the method for one applicable to the other.
The bolt would screw through the cap by hand until contact with the piston, then tightened with a torque wrench to a specified value. A dial indicator would be set on the piston and measure movement as the screw is backed out.
Without getting into impossible to measure parameters such as friction coefficients, is there a simple equation to give
me the force applied to the piston, given the size of the bolt, and the torque applied to it?
Thanks again for your input.
I saw a special bolt that had a tensioning element through the axis connected to a knob on the head. When the bolt was tightened to a certain level, the tensioning element prevented the knob from turning. One could check for adequate tightness by checking the knob.
This scheme made sense, as it measured actual bolt tension, but the bolts were expensive.
This scheme made sense, as it measured actual bolt tension, but the bolts were expensive.
Hi hanksboy
I don't know of any formula that doesn't introduce friction coeficients, the simplist formula I know is :-
T*0.2*d= tensile force
where T = torque applied
0.2 friction factor for dry threads
d = bolt dia.
having said this specifying a torque is only about +/- 20%
accurate at best therefore the second method specified by the Haynes manual might not be that accurate and you might get away with your home made tool. The class of spring will also play apart depending what class its made too.
If memory serves me correct class to springs have a tolerance on spring rate of +/- 10% and I think about +/-3% on a specified load at length so there accuracy varies too.
Best regards
desertfox
I don't know of any formula that doesn't introduce friction coeficients, the simplist formula I know is :-
T*0.2*d= tensile force
where T = torque applied
0.2 friction factor for dry threads
d = bolt dia.
having said this specifying a torque is only about +/- 20%
accurate at best therefore the second method specified by the Haynes manual might not be that accurate and you might get away with your home made tool. The class of spring will also play apart depending what class its made too.
If memory serves me correct class to springs have a tolerance on spring rate of +/- 10% and I think about +/-3% on a specified load at length so there accuracy varies too.
Best regards
desertfox
Tick,
That's a Rotabolt:
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
That's a Rotabolt:
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
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