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Thread: Calculations for Formula SAE (which J?)

  1. #1
    Associate Engineer
    Join Date
    Apr 2020
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    2

    Calculations for Formula SAE (which J?)

    Was hoping to get some help and calcification. I am a Mechanical engineering student and I am designing a rear differential spool for the school's FSAE car. Im trying to calculate the max torque the spool (and bolts) can handle based of the design and materials used (mainly which formula for J to use). Also the amount of shear the bolts are under given the cars last dyno was 55 lb-ft of torque. Before they shut down school I had talked with a couple professors and gotten mixed answers and formulas. I have since tried email and zoom and still getting mixed results. I will try and include as much info as possible. Any help is greatly appreciated

    Car torque 55 lb ft max
    Bolts 5/16-18 UNC ASTM A574 standard 170,000 psi tensile 153,000 psi yield 140,000 psi proof
    Fit pin 5/16 harden steel approx 16,000 pound break strength
    Spool Al 7075-T6511 86,000 psi tensile 76,300 psi yield
    Attached Images Attached Images

  2. #2
    Principle Engineer
    Join Date
    May 2015
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    187
    55 ft-lbs is 660 in-lbs and acts on 5/16” pins at a 1.21” radius. The shear force on the pins is 660/1.21 = 546 lbs.


    The shear area of the 5/16” diameter pin is .0767 in^2

    If carried by a single pin the shear stress is 546/.0767 = 7120 psi

    What is the strength of the steel in shear? Don’t get fancy, just call it half of the tensile and your still at 38ksi yield for shear. So, one pin can easily handle the 55 ft-lbs of engine torque and you’ve got three. (That aluminum in the spool might be a bit closer to yield.)


    Your real problem is a bit harder to define. Say the wheel becomes airborne and spins wildly until it hits the ground and is abruptly slowed. That is the kind of shock load that you must anticipate in drivetrain design and that is why testing and development takes much more time than design for commercial products.

    You don’t have that kind of time so you build in extra strength, pay for premium materials and hope for the best.

  3. #3
    Associate Engineer
    Join Date
    Apr 2020
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    2
    Thank you for responding. Let me try to clarify my question. I am trying to calculate the amount of shear in the system using the 55 lb-ft, to do this I am using the shear stress formula, shear = (Torque * R)/J. My understanding is the shear will be applied to the bolts/pins, Torque would be divided among the 3 bolts and 3 pins, R is the 1.21in center of spool to center of bolt/pin hole location, the question is what J formula is used. Also I have been given conflicting information on if this is how you would calculate the forces in the system. Any further clarification would be greatly appreciated.

  4. #4
    Principle Engineer
    Join Date
    May 2015
    Posts
    187
    I don't know what your formula is intended to calculate.

    I calculated the total shear force on the bolts and pins based on your 55 ft-lbs.

    That force is in lbs force. I then calculated the shear stress if this force was applied to just one pin because it was obvious that it would be a low number. Exactly how the shear will be shared is kind of a gray area.

    J is sometimes used to represent a rotational inertia but it doesn't get used to answer your question about shear.

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