1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 16, 2016 #1
    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
     
  2. jcsd
  3. Dec 16, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  4. Dec 16, 2016 #3
    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.
     
  5. Dec 16, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Ok. If the coefficient is 0.3, what is the magnitude of it as a force?
     
  6. Dec 16, 2016 #5
    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.
     
  7. Dec 17, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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?
     
  8. Dec 17, 2016 #7
    You would have to push with a force greater than the friction force multiplied by the normal force.
     
  9. Dec 17, 2016 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Rolling resistance coefficient multiplied by normal force.
    So what is the strength of force you have to overcome, and where does it act?
     
  10. Dec 17, 2016 #9
    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?
     
  11. Dec 17, 2016 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Each wheel gets its own share.
    And where does the resisting force act?
     
  12. Dec 17, 2016 #11
    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.
     
  13. Dec 17, 2016 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right, so what torque does it exert on the axle?
     
  14. Dec 17, 2016 #13
    150 lb x 9in (radius of tire) = 1350 lb-in?
     
  15. Dec 17, 2016 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
     
  16. Dec 17, 2016 #15
    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.
     
  17. Dec 17, 2016 #16

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes.
     
  18. Dec 17, 2016 #17
    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.
     
  19. Dec 17, 2016 #18

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  20. Dec 17, 2016 #19
    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?
     
  21. Dec 17, 2016 #20

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Somewhere along the axle must be the drive (differential?). The torque is between there and the wheels. The two sides are independent.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Determing torque so that I may determine the diameter of my axle
Loading...