Results 1 to 3 of 3

Thread: How to measure force generated by a tyre when it fails?

  1. #1
    Associate Engineer
    Join Date
    Jan 2018

    How to measure force generated by a tyre when it fails?

    Hello everyone,

    I am trying to work out how much force is generated when a tyre fails at a certain pressure. The tyre will be inside an OSHA section 1910.177 compliant tyre cage.

    What calculation would be used to solve the following?:

    The tyre size is 35 x 11.5-17 with a rated inflation of 255 psi.

    Maximum supply available to the tyre is 300 psi.

    I need to know what force/shockwave/energy is released if the tyre were to fail inside the cage.

    Worst case scenario is the tyre sidewall splits with all stored pressure/energy being released in one direction.

    Thanks for your time.

  2. #2
    Administrator Kelly_Bramble's Avatar
    Join Date
    Feb 2011
    Bold Springs, GA
    I'm not sure what the specific requirements are but I would start with the potential energy of the compressed gas and then define the failure mode and calculate from there.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  3. #3
    Associate Engineer
    Join Date
    Jan 2018


    Since pressure is force per area, then to know the total force contained (not the impulse to be delivered by failure), you have to multiply by the inside area of the tire, not the outside area. Get another tire that size, and start measuring, but also you must include the force exerted where the tire ends, and the wheel (steel, or aluminum, etc) begins (at the bead) If you have or had access to the cross-sectional area traced out by the inner surface of the tire and the wheel, you could presumably idealize this contained semi-toroid of pressure to a congruently sized pure toroid, divide the cross-sectional area by pi, take the square root, and have a radius to use. Radius doubled, multiplied by pi gives the small circumference of the donut (toroid), then you could apply calculus to that (take the integral of rotation), and have a pretty good approximation of the inner surface area (container).

    To calculate the impulse, you would have to assess the duration of the force released, and its directionality.

    Knowing the energy stored, is also a route to answers you are seeking, I think, since the rate of energy release can be equated with power during the event, and this can be related back to the impulse calculation.
    Last edited by jastewart; 01-30-2018 at 02:02 PM.

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts