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

will a pic help show what I am banging on about http://i976.photobucket.com/albums/ae248/cooper12000/carramp.jpg

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.

thanks for the reply. My issue seems to be the huge load on the unsupported triangle once the ramp is level. no matter what I do due to the length of my car I cant seem to get a better safety factor then 0.4. admittedly I am using the lower grade s275 steel but only because that's all I seem to be able to find. (parkers etc). I have decided to bite the bullet and buy them ready made for 1200 that way I know they will work and all the safety stuff has been factored in and I don't have to rely on my welding skills. many thanks anyways coop