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How do I determine LOAD FACTORS for calculating crush force in lbs.

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  • How do I determine LOAD FACTORS for calculating crush force in lbs.

    I forgot the rediculous OT please forgive.

    I hope I’m asking this the right way or the
    right thing, but I’m no math whiz so I’m
    not sure.


    For instance:
    A 1000 lb load
    lifted by a 2 leg bridle sling
    at 60 degrees

    So 500 x a load factor of 1.154 = 577 divided by 2 =288.5 or 289 lbs.


    A crush force in lbs for 20 degrees was given at 1374 lbs
    and I was able to determine the load factor by multiplying
    x 2 then dividing by 500 for a load factor of 5.496
    500x5.496=2748 divided by 2 =1374 lbs

  • #2
    Put that in context and someone might be able to help you. What are you trying to accomplish?
    I certainly have no idea...
    Keith
    __________________________
    Just one project too many--that's what finally got him...

    Comment


    • #3
      You want to know how much force is on the side of a box, for instance, when lifted by a sling at 60 degrees included angle (between the two parts of the sling). That is 30 deg off vertical for each half of the sling.

      So for your 1000 lb, half is on that sling part. The vertical force vector is 500, the horizontal is proportional to the tangent of the angle, which is 0.577 for the 30 deg angle. So the side force is 0.577 of the vertical, or 0.577 of 500, which is the 289 you came up with. That half-sling exerts a 289 lb force on the box. The other half will exert an equal force in the opposite direction.

      If you mean the sling is 60 degrees in some other direction, that would be different.
      1601

      Keep eye on ball.
      Hashim Khan

      Comment


      • #4
        Put someone you don't like under the load and when he/she quits screaming you have your crush load.
        The shortest distance between two points is a circle of infinite diameter.

        Bluewater Model Engineering Society at https://sites.google.com/site/bluewatermes/

        Southwestern Ontario. Canada

        Comment


        • #5
          Originally posted by J Tiers View Post
          You want to know how much force is on the side of a box, for instance, when lifted by a sling at 60 degrees included angle (between the two parts of the sling). That is 30 deg off vertical for each half of the sling.

          So for your 1000 lb, half is on that sling part. The vertical force vector is 500, the horizontal is proportional to the tangent of the angle, which is 0.577 for the 30 deg angle. So the side force is 0.577 of the vertical, or 0.577 of 500, which is the 289 you came up with. That half-sling exerts a 289 lb force on the box. The other half will exert an equal force in the opposite direction.

          If you mean the sling is 60 degrees in some other direction, that would be different.
          I think you have it right if I can undertsand
          exactly what I need to do.
          The example they give us is with an I beam
          and we have to figure the inward crush force
          in lbs at different angles from 20 to 60 degrees
          in 5 degree increments.
          The were saying to use cosine but I’m lost
          trying to figure out the crush force without
          the Load factors.
          I have all but the 25. I was able to figure out
          the load factor for the 20 degrees since the
          crush force in lbs was given.

          I’ll have to try your tangent method.

          Comment


          • #6
            Cosine is Adjacent/hypotenuse, Tangent is opposite/adjacent. Tangent made sense because you know the angle, and you also know the down force vector (weight of object), you need the horizontal component.

            Cosine could be good to find tension in the sling (hypotenuse), which you want to know for the wider, flatter angles, so you do not get past the sling rating.
            1601

            Keep eye on ball.
            Hashim Khan

            Comment

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