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# Bump Stop

## Bump Stop

(OP)
Am I right in thinking that the load on a bump stop from say a large pothole or brick at speed would be the vertical G value multiplied by the wheel assy mass, rather than the wheel load. Would 3G not seem unreasonable?
Replies continue below

### RE: Bump Stop

No, you are wrong, and no, that value is not even close to the load seen. You can use a 2 degree of freedom model to estimate the wheel and bump stop loads to within a factor of two.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

### RE: Bump Stop

(OP)
Greg, when you say that value is not even close, can you give me a clue - more or less??

### RE: Bump Stop

Consider that going into potholes will sometimes cause thing to break.

TTFN

### RE: Bump Stop

Scarecrow, perhaps if you wrote out the reasoning which brought you to your original supposition, it would be easier to give you a clue.  Then again, somebody might just point and laugh.  I'm a bit baffled as to how you came up with the figure - it sounds more like the average tire load than a bump-stop load to me - my gut feel is that your guess is WAY too low.

### RE: Bump Stop

Conversely, just consider the placement of the bump stop relative to where the normal spring compression is.  Compression force is kx^2

TTFN

### RE: Bump Stop

A good cross check might be to work out the sort of force it takes to bend a wheel rim.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

### RE: Bump Stop

(OP)
Thanks Guys, the bump stop is on a trailing arm suspension on a little trailer for measuring distances.  I'd seemed to remember years ago that truck axles were designed for fatigue loads using normal wheel loads, then a "double check" that normal wheel loads X 3G didn't bring the stresses over the yield. So I assumed 3G was the max that axles may see in service.

### RE: Bump Stop

A senior project that I had to do while at the university in order to learn how to use the analog computer was to design an automotive spring/shock system for a car when the wheel encountered a step function (curb).  I can only picture the wave form if the suspension spring constant had been changed to the (much stiffer) spring constant of the (normally hard rubber) bump stop.

Based on later life experience with heavy duty tractor/trailer axles/suspensions, I can tell you that once the bump stop is encountered, irrespective of whatever spring effect of the bump stop rubber might have, there is no suspension, and the axle in question begins to lift the entire trailer load by itself, while unloading the other axles, and if carried to infinity, would then support the entire trailer weight and transmit to that trailer any acceleration loads that the wheel encountered (pothole is a step function) that it could survive with or without bending or breaking said wheels, tires, or axles.

I've seen some pretty stiff axles badly bent when this problem was encountered.

Problem was fixed when the suspension was redesigned so that the parts that made contact with the frame were modified. The 'nuts up' U-bolts that held the suspension spring packs were the offending parts, not the bump stops-there were none.  They became bump stops with no spring constant other than what it took to do the mushrooming that was evident, however.  Changing to a 'nuts down' U-bolt type of suspension and taller spring shackles did away with the bent axles and mushroomed U-bolts.

rmw

### RE: Bump Stop

So now you are saying three times the axle load? that is much more reasonable than 3g times the unsprung mass, but still might be a little on the low side.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

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