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• # Car ramp deflection - help pls compleatly confused

Discussion in 'Calculations' started by cooper12000, Mar 13, 2014.

1. ### cooper12000New Member

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I wish to make a car ramp out of angle. the whole car will go on the ramp and the ramp is constructed of a rectangle and a triangle. the rectangle is 1500mm long and 500 mm high with a vertical brace at 750 mm. the triangle is 3000mm long and 500mm high again with vertical braces every 750mm. Therefor the total length of the ramp will be 4500 mm. I will pivot at the point the triangle and rectangle join.

my question is how do I work out the deflection in the material assuming the wheel is half way between the vertical supports and therefor work out what size angle is needed.

any thanks coop

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4. ### LochnagarWell-Known Member

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I would be working out the stress first - rather than the deflection.

In the worst case as the vehicle goes up the ramp - your wheel will be resting on just one angle - 350mm wide. (It is possible if the angles are more closely spaced - depending on the size of the tyre and tyre pressure etc - that the angles on either side - may share a small percentage of the load - as the tyre sits directly above one angle). Cars vary in weight - I think the Range Rover is about 2500Kg - so that is about 625Kg per wheel (ignoring the inclination effect - and traction effect). Usually, the angles on which the tyre "rolls over" are "inverted" so that "90 degree corner" of the angle is facing the tyre.

If you assume a 45 x 45 x 5 thick EN 10025 S355 angle (not S275 which is a lower grade). The Ixx with the angle as I have described it above (i.e. 90 degree corner facing tyre) - is approximately 33091mm^4 - and the distance to the extreme fibre is approximately 18.5mm in this orientation.

If you assume the tyre width is say 200mm - then analysing it - as a simply supported beam with a uniformly distributed load from the tyre - means you have (625 x 10)/2 = 3125N - are the two reactions at either side of the simply supported beam (taking g = 10m/s^2).

So the distance from the edge of the tyre to the reaction is (350-200)/2 = 87.5mm. Thus the bending moment at mid span is 3125 x 350/2 - (3125/200 x 100) x 200/2/2 = 468750Nmm.

The equation for bending stress is My/I. So we have 468750 x 18.5/33091 = 262N/mm^2. The steel S355 - has a yield strength of 355N/mm^2. So you have got a safety factor of roughly 1.35. I am unsure if there is any particular standard governing the design of vehicle ramps (there might be) - so you should check that out too - to see if it provides guidance on safety factors. I personally would not go lower than the above safety factor - in the absence - of any statutory requirements.

However, most ramps have a width - which is much closer to the width of the tyre - which will obviously minimize the bending moment.

Hopefully this gives you the general idea - from which you can "customize" the design to suit the particular vehicle, etc.

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