Determing torque so that I may determine the diameter of my axle

  • Thread starter blinddog22
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  • #1
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I am a little rusty at the moment but let's say I have an axle with two loads being applied to it. Both 500 lb. I am trying to figure out the various stresses on it so i can determine a suitable diameter for the axle connecting the tires. The two tires and axle are all one piece. The bearing locations are indicated by teh boxes and 500 lb forces.
I made a FBD and calculated reaction forces, moment, shear, and bending. However, I am stuck with the torsion or torque. I don't know which one it is. I believe it is torque since the whole axle is one piece. How do i determine the torque given
diameter of wheel = 20in
axle is driven at a constant speed of 3 ft/s
friction coeff = 0.3

I will be using von-mises afterwards to determine the proper diameter of the axle.

Please see attached picture.

Thank you.
Can't seem to get picture to work, but here is an imjur link. http://imgur.com/a/HHmgn
 

Answers and Replies

  • #2
haruspex
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axle is driven at a constant speed of 3 ft/s
Ft/s is not a rotation rate. Do you mean the vehicle moves at that speed? On the level or is this going uphill?
friction coeff = 0.3
Do you mean rolling resistance? 0.3 is very high for that. Friction between tyre and road is not relevant.
 
  • #3
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Ft/s is not a rotation rate. Do you mean the vehicle moves at that speed? On the level or is this going uphill?

Do you mean rolling resistance? 0.3 is very high for that. Friction between tyre and road is not relevant.
I was given that the velocity was 3 ft/s, this is of the vehicle.

I believe it is rolling resistance. I was told to assume around 0.2, as long as I state my assumption. So i just choose 0.3 arbitrarily.
 
  • #4
haruspex
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I was given that the velocity was 3 ft/s, this is of the vehicle.

I believe it is rolling resistance. I was told to assume around 0.2, as long as I state my assumption. So i just choose 0.3 arbitrarily.
Ok. If the coefficient is 0.3, what is the magnitude of it as a force?
 
  • #5
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Ok. If the coefficient is 0.3, what is the magnitude of it as a force?
To be honest, I don't understand what you are asking.

This is for a trailer that is holding a 1000 pound load. The load is bearing on the axle at the indicated positions. I am not even sure where the 0.3 friction coeff comes into play calculation wise, unless it is needed to find the torque.
 
  • #6
haruspex
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To be honest, I don't understand what you are asking.

This is for a trailer that is holding a 1000 pound load. The load is bearing on the axle at the indicated positions. I am not even sure where the 0.3 friction coeff comes into play calculation wise, unless it is needed to find the torque.
Rolling tesistance acts , like friction, to oppose relative motion. If the axle load is 500 lb and the coefficient of rolling resistance is 0.3, how hard would you have to push horizontally to get that wheel turning?
 
  • #7
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Rolling tesistance acts , like friction, to oppose relative motion. If the axle load is 500 lb and the coefficient of rolling resistance is 0.3, how hard would you have to push horizontally to get that wheel turning?
You would have to push with a force greater than the friction force multiplied by the normal force.
 
  • #8
haruspex
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You would have to push with a force greater than the friction force multiplied by the normal force.
Rolling resistance coefficient multiplied by normal force.
So what is the strength of force you have to overcome, and where does it act?
 
  • #9
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I suppose in this case, it would be 1000lb since there is two 500 lb forces acting on the axle, and it would be acting on the axis of the axle. So you would need to overcome 1000(0.3) = 300 lb?
 
  • #10
haruspex
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I suppose in this case, it would be 1000lb since there is two 500 lb forces acting on the axle, and it would be acting on the axis of the axle. So you would need to overcome 1000(0.3) = 300 lb?
Each wheel gets its own share.
And where does the resisting force act?
 
  • #11
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Oh, ok. So then the force that needs to be overcome is 500(0.3) = 150 lb. Since each wheel gets its own share.
The resisting force acts between the wheel and the ground.
 
  • #12
haruspex
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Oh, ok. So then the force that needs to be overcome is 500(0.3) = 150 lb. Since each wheel gets its own share.
The resisting force acts between the wheel and the ground.
Right, so what torque does it exert on the axle?
 
  • #13
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150 lb x 9in (radius of tire) = 1350 lb-in?
 
  • #15
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Alright, then that is my torque applied to the axle so that I may calculate my maximum shear stress? As shown in my initial picture.
 
  • #16
haruspex
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Alright, then that is my torque applied to the axle so that I may calculate my maximum shear stress? As shown in my initial picture.
Yes.
 
  • #17
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Excellent. Thank you very much.

If you look at my diagram, the maximum moment i calculated was 5000 lb-in. Does that seem correct? When i calculate my sigma and tau and use the von mises criterion, I feel like the value seem awfully big. In the end I ended up with a diameter a little shorter than 1.42 inches. Which seems kind of skinny.
 
  • #18
haruspex
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Excellent. Thank you very much.

If you look at my diagram, the maximum moment i calculated was 5000 lb-in. Does that seem correct? When i calculate my sigma and tau and use the von mises criterion, I feel like the value seem awfully big. In the end I ended up with a diameter a little shorter than 1.42 inches. Which seems kind of skinny.
I'm not an engineer, so I don't know about von Mises' work.
As far as I am aware, you have calculated the minimum diameter under ideal conditions. Real axles have to cope with bumps in the road, wear and tear, and a safety margin.
 
  • #19
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Alright, no worries.

Ok, I suppose that makes sense then. A 1.5 diameter thick beam of 1018 steel sound like it should handle a 1000lb load.

Also, I just thought of something during my calculation. When i plug my torque into my shear stress equation to determine the torsion, I only need to plug in 1350 lb-in, correct? Not 1350 lb-in x 2 for the other wheel as well? Do i worry about the torque from the other wheel?
 
  • #20
haruspex
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Alright, no worries.

Ok, I suppose that makes sense then. A 1.5 diameter thick beam of 1018 steel sound like it should handle a 1000lb load.

Also, I just thought of something during my calculation. When i plug my torque into my shear stress equation to determine the torsion, I only need to plug in 1350 lb-in, correct? Not 1350 lb-in x 2 for the other wheel as well? Do i worry about the torque from the other wheel?
Somewhere along the axle must be the drive (differential?). The torque is between there and the wheels. The two sides are independent.
 
  • #21
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Its just a straight axle with 2 wheels. Meant for a trailer to carry a load. Treating it as one solid piece essentially. Same concept?
 
  • #22
haruspex
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Its just a straight axle with 2 wheels. Meant for a trailer to carry a load. Treating it as one solid piece essentially. Same concept?
In that case there is no torque.
 
  • #23
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Very well. No shear stress then.
 
  • #25
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Alright, I will review that.

I have another question if you are willing and able. Do bolts affect the shear and moment force diagrams for beam loading/deflection at all? I have dug through 4 of my text books and scoured the internet but haven't found anything to explain it.

See attached pic if you don't mind. http://imgur.com/a/MuJTc

There are 4 beams, the two smaller ones on top of the two longer ones and bolted together using two bolts per connection. In the center of each beam there is a load acting on it. How are the shear/moment diagrams affected, or are they?
 

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