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what the torisonal strenth of a 5" 1
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EnglishMuffin
Mechanical
The equation for the shear stress at the outside surface of the tube is :
16*T*Do/(pi*(Do^4-Di^4))
Where T is the torque
Do is the outside diameter
Di is the inside diameter
This gives for the large tube :
16*14000*5/(3.14*(5^4-4.25^4)) = 1193 lbf/in^2
This doesn't look too bad for steel - or even a decent aluminum alloy - not counting stress concentrations if present.
Would need more data for the spline on the small tube.
I'm not sure where boo1 gets his equation from - maybe I'm missing something.
Incidentally, as a point of form, torque should be specified with the force first, and distance second, with a dot between them. Work or energy should be specified with the distance first, force second.
16*T*Do/(pi*(Do^4-Di^4))
Where T is the torque
Do is the outside diameter
Di is the inside diameter
This gives for the large tube :
16*14000*5/(3.14*(5^4-4.25^4)) = 1193 lbf/in^2
This doesn't look too bad for steel - or even a decent aluminum alloy - not counting stress concentrations if present.
Would need more data for the spline on the small tube.
I'm not sure where boo1 gets his equation from - maybe I'm missing something.
Incidentally, as a point of form, torque should be specified with the force first, and distance second, with a dot between them. Work or energy should be specified with the distance first, force second.
EnglishMuffin
Mechanical
Who - me ?
Torsional shear stress = Radius*Torque/(Polar second moment of area)
Polar second moment of area = pi/32*(Do^4-Di^4)
Hence my equation
Perhaps a bit simplistic, but good enough for this problem I would have thought.
I don't inderstand what boo1 has typed.
Torsional shear stress = Radius*Torque/(Polar second moment of area)
Polar second moment of area = pi/32*(Do^4-Di^4)
Hence my equation
Perhaps a bit simplistic, but good enough for this problem I would have thought.
I don't inderstand what boo1 has typed.
EnglishMuffin
Mechanical
Sorry - I goofed - its lbf.ft not lbf.in - too much of a hurry.
Should have said :
16*14000*12*5/(3.14*(5^4-4.25^4)) = 14316 lbf/in^2
That looks a bit high to me
Still need more data about the spline and the material
Sorry
But I still don't understand what boo1 has typed.
Should have said :
16*14000*12*5/(3.14*(5^4-4.25^4)) = 14316 lbf/in^2
That looks a bit high to me
Still need more data about the spline and the material
Sorry
But I still don't understand what boo1 has typed.
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what i am trying figure out is if i have an outside housing with a 5" o.d. and 4 1/4" i.d. it also has 1/4" splines, with a 2 1/2" o.d. with 2" i.d. and a inside sprocket that fits between these to splines. what kinda of torque it can handle. 4340 material
EnglishMuffin
Mechanical
pso311
I can't see what you are complaining about - I think your equation is exactly the same as mine - based on your last reply. The shear stress at any radius r is given by :
Shear stress = r* T/J
This equation is only valid for rotational symmetry - ie round tubes.
which gives Max Stress = 16*T*Do/(pi*(Do^4-Di^4))
(as I said in my first reply)
This gives a stres of 14316 lbf/in^2
What stres do you get ?
I can't see what you are complaining about - I think your equation is exactly the same as mine - based on your last reply. The shear stress at any radius r is given by :
Shear stress = r* T/J
This equation is only valid for rotational symmetry - ie round tubes.
which gives Max Stress = 16*T*Do/(pi*(Do^4-Di^4))
(as I said in my first reply)
This gives a stres of 14316 lbf/in^2
What stres do you get ?
EnglishMuffin
Mechanical
runutts
We need to be very specific here - we don't want to get anyone killed.
1. What type of spline is it ? (for example - involute stub ?)
2. What Standard if known ?
3. Spline dimensions (pitch, OD, length etc)
4. Are there relief grooves for manufacturing ?
The weakest link in this set-up is probably going to be the inner sprocket (shaft ?), not the large tube. Your spline size has changed from your original post. Can you send a drawing ?
We need to be very specific here - we don't want to get anyone killed.
1. What type of spline is it ? (for example - involute stub ?)
2. What Standard if known ?
3. Spline dimensions (pitch, OD, length etc)
4. Are there relief grooves for manufacturing ?
The weakest link in this set-up is probably going to be the inner sprocket (shaft ?), not the large tube. Your spline size has changed from your original post. Can you send a drawing ?
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- #14
I am a newbie at this just looking to see if it is possible. Let me tell you what i am trying to figure out. I have a 5" o.d. max on the outerhousing and i want to keep a 2" i.d. on a mandrill that rotates. i want a gear in between the outerhousing and the inner mandril that can engage and carry 14000 ft/lbs. looking for the wall thicknesses for the outerhousing and wall thickness of the mandrill. looking for like 20 teeth on the outside and the inside doesen't matter.
EnglishMuffin
Mechanical
So, let me see if I understand this correctly:
Starting from the inside, we have a hollow shaft (mandrel) with a 2 inch ID. The outside diameter of this shaft has a male spline on it - of some (as yet) unspecified pitch and OD. Then, fitting over that, we have an intermediate hollow spline coupling (gear as you call it) with an unspecified number of internal teeth and 20 external spline teeth (suggested). And over that we have a 5 inch OD hollow tube, or housing, with an (as yet) unspecified ID, with internal spline teeth meshing with the external teeth of the intermediate coupling. All the parts are made from 4130, heat treatment unspecified as yet, and the complete assembly should be capable of carrying a maximum continuous torque of 14000 lbf.ft., no cyclic loading specified as yet. The method of axially locating the spline coupling is unspecified.
Have I understood you correctly ?
Is there any limitation on the OD of the hollow shaft (Mandrel) ? This may turn out to be the weakest link, as I said before.
This is actually a complete design problem, not a simple calculation. Quite a bit different from your original post, wouldn't you say ? If you can confirm my description, or even expand upon it, we will see what can be done.
Starting from the inside, we have a hollow shaft (mandrel) with a 2 inch ID. The outside diameter of this shaft has a male spline on it - of some (as yet) unspecified pitch and OD. Then, fitting over that, we have an intermediate hollow spline coupling (gear as you call it) with an unspecified number of internal teeth and 20 external spline teeth (suggested). And over that we have a 5 inch OD hollow tube, or housing, with an (as yet) unspecified ID, with internal spline teeth meshing with the external teeth of the intermediate coupling. All the parts are made from 4130, heat treatment unspecified as yet, and the complete assembly should be capable of carrying a maximum continuous torque of 14000 lbf.ft., no cyclic loading specified as yet. The method of axially locating the spline coupling is unspecified.
Have I understood you correctly ?
Is there any limitation on the OD of the hollow shaft (Mandrel) ? This may turn out to be the weakest link, as I said before.
This is actually a complete design problem, not a simple calculation. Quite a bit different from your original post, wouldn't you say ? If you can confirm my description, or even expand upon it, we will see what can be done.
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- #16
thats it exactly would like to have as much area from the o.d of the inner mandrill including splines to the outside id of the housing including splines as possible. looking at other downhole tools in the industry the wall thickness on mandrills are like 1/4" and outerhousing like 3/8". thats why that was my starting point.
EnglishMuffin
Mechanical
Well - I'll tell you right now - that 1/4" wall thickness for the mandrel is not going to cut it. If you look at my formula, just considering a 2.5" OD tube with a 2"ID, you will get, for the max shear stress :
16*14000*12*2.5/(3.14*(2.5^4-2^4)) = 92750 lbf/in^2
According to my "ASM Engineering properties of steel", the very best you can do for 4340 (oil quenched and tempered) would be a tensile yield strength of about 139000 lbf/in^2, which would imply a shear strength of about 70000 lbf/in^2, and this doesn't include stress concentrations or factors of safety etc. 4130 is a case hardening steel, so if we went that route, might be able to do a little better - but it depends on whether that is feasible (accuracy considerations etc) or desirable.
So I say again - what is the limitation on the OD of the Mandrel ? I'm sure you must have some limit - that should be the starting point, as its the weakest part of the design.
16*14000*12*2.5/(3.14*(2.5^4-2^4)) = 92750 lbf/in^2
According to my "ASM Engineering properties of steel", the very best you can do for 4340 (oil quenched and tempered) would be a tensile yield strength of about 139000 lbf/in^2, which would imply a shear strength of about 70000 lbf/in^2, and this doesn't include stress concentrations or factors of safety etc. 4130 is a case hardening steel, so if we went that route, might be able to do a little better - but it depends on whether that is feasible (accuracy considerations etc) or desirable.
So I say again - what is the limitation on the OD of the Mandrel ? I'm sure you must have some limit - that should be the starting point, as its the weakest part of the design.
EnglishMuffin
Mechanical
Sorry - I goofed again - I meant to say 4340 is a case hardening steel.
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- #19
EnglishMuffin
Mechanical
The bearing ID is actually probably 70mm - am I right?
If so, that means it might be possible to design the shaft so that it is nowhere smaller than 2 3/4".
But I doubt it - how is the bearing retained ? If you use a locknut, the threads will be smaller, and you'll end up with about 2 5/8" probably - depending on the locknut,or snap ring groove, relief groove etc.
Since I don't have enough detail to give you a definitive answer to this problem, and the devil is in the details, I am going to give you a number of different answers, which involve various worst case scenarios.
1. Assuming a plain tube, 2.75" OD (Taking you at your word), 2" ID, 4340 with a yield shear stress of 70000 lbf/in^2 - with a static torque application - no shock.
For static torque failure of ductile materials, you can normally ignore stress concentrations.
For local yielding at the extreme fiber :
Rearranging my formula a bit,
T = stress*pi*(Do^4-Di^4)/(16*Do)
T = 70000*pi*(2.75^4-2^4)/(16*2.75) = 205874 lbf.in
= 17156 lbf.ft
For complete failure across the whole section, it will take more torque than this. The multiplier for this is the "Limit factor" (L). (see Peterson, "stress concentartion factors" for example)
L = 4/3*(1- (Di/Do)^3)/(1-(Di/Do)^4)
= 4/3*(1 - (2/2.75)^3)/(1-(2/2.75)^4) = 1.14
Torque for complete failure = 19558 lbf.ft
2. For cyclic loading conditions, fatigue must be taken into account. There is likely to be a stress concentration factor of at least 3, but it depends on your exact geometry, which I do not know. For infinite life, we must stay below the endurance limit. According to my "atlas of fatigue curves", for 4340 (150000 lb/in^2 tensile strength), assuming we alternate between 0 and 29000 lb/in^2 shear stress, with a stress concentration notch of 3.3, we would have a life of about 10^7 cycles.
At that stress range, this corresponds to an alternating torque of
T = 29000*pi*(2.75^4-2^4)/(16*2.75) = 85290 lbf.in
= 7108 lbf.ft
(ie the torque alternates between 0 and 7108 lbf ft)
I don't know what this component is, but it sounds as though it has something to do with oil drilling, and probably rotates. At 60 rpm in continuous operation, 10^7 cycles means it would last about 115 days. In practice, I expect your torque will either be constant or will alternate through a much smaller range than this, and then it would last a lot longer.
We therefore can say that your component is likely to be able to withstand anywhere from about 7000 to 19000 lbf.ft of torque, depending on the conditions, and on what is meant by "failure".
So you see, it is impossible to give a definitive answer without a lot more detail, such as :
1. Exact heat treatment
2. Exact part geometry, groove details if any etc
3. Exact details of how the torque will vary over time - if at all.
Regarding the splines, if the parts are perfectly concentric, it should be possible to design the spline such that it has as much shear strength as the shaft, and such that it does not introduce a worse stress concentration than those already present - assuming that the root diameter can be made greater than 2.75". It is normal in such cases as this to use the design rule of thumb that half the spline teeth carry the shear load.
If the spline is there to cater for significant angular misalignment, then this will affect the spline geometry and all bets are off - it depends on how much angular misalignment you are trying to accomodate.
The rest of the design can be made a lot stronger than this shaft. If the shaft can be made strong enough for what you need - then the design is feasible.
I hope this has given you some idea of how strong your part will be.
If so, that means it might be possible to design the shaft so that it is nowhere smaller than 2 3/4".
But I doubt it - how is the bearing retained ? If you use a locknut, the threads will be smaller, and you'll end up with about 2 5/8" probably - depending on the locknut,or snap ring groove, relief groove etc.
Since I don't have enough detail to give you a definitive answer to this problem, and the devil is in the details, I am going to give you a number of different answers, which involve various worst case scenarios.
1. Assuming a plain tube, 2.75" OD (Taking you at your word), 2" ID, 4340 with a yield shear stress of 70000 lbf/in^2 - with a static torque application - no shock.
For static torque failure of ductile materials, you can normally ignore stress concentrations.
For local yielding at the extreme fiber :
Rearranging my formula a bit,
T = stress*pi*(Do^4-Di^4)/(16*Do)
T = 70000*pi*(2.75^4-2^4)/(16*2.75) = 205874 lbf.in
= 17156 lbf.ft
For complete failure across the whole section, it will take more torque than this. The multiplier for this is the "Limit factor" (L). (see Peterson, "stress concentartion factors" for example)
L = 4/3*(1- (Di/Do)^3)/(1-(Di/Do)^4)
= 4/3*(1 - (2/2.75)^3)/(1-(2/2.75)^4) = 1.14
Torque for complete failure = 19558 lbf.ft
2. For cyclic loading conditions, fatigue must be taken into account. There is likely to be a stress concentration factor of at least 3, but it depends on your exact geometry, which I do not know. For infinite life, we must stay below the endurance limit. According to my "atlas of fatigue curves", for 4340 (150000 lb/in^2 tensile strength), assuming we alternate between 0 and 29000 lb/in^2 shear stress, with a stress concentration notch of 3.3, we would have a life of about 10^7 cycles.
At that stress range, this corresponds to an alternating torque of
T = 29000*pi*(2.75^4-2^4)/(16*2.75) = 85290 lbf.in
= 7108 lbf.ft
(ie the torque alternates between 0 and 7108 lbf ft)
I don't know what this component is, but it sounds as though it has something to do with oil drilling, and probably rotates. At 60 rpm in continuous operation, 10^7 cycles means it would last about 115 days. In practice, I expect your torque will either be constant or will alternate through a much smaller range than this, and then it would last a lot longer.
We therefore can say that your component is likely to be able to withstand anywhere from about 7000 to 19000 lbf.ft of torque, depending on the conditions, and on what is meant by "failure".
So you see, it is impossible to give a definitive answer without a lot more detail, such as :
1. Exact heat treatment
2. Exact part geometry, groove details if any etc
3. Exact details of how the torque will vary over time - if at all.
Regarding the splines, if the parts are perfectly concentric, it should be possible to design the spline such that it has as much shear strength as the shaft, and such that it does not introduce a worse stress concentration than those already present - assuming that the root diameter can be made greater than 2.75". It is normal in such cases as this to use the design rule of thumb that half the spline teeth carry the shear load.
If the spline is there to cater for significant angular misalignment, then this will affect the spline geometry and all bets are off - it depends on how much angular misalignment you are trying to accomodate.
The rest of the design can be made a lot stronger than this shaft. If the shaft can be made strong enough for what you need - then the design is feasible.
I hope this has given you some idea of how strong your part will be.
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