garage19
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| posted on 24/10/06 at 12:23 PM |
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Working out the braking forces on a wishbone? Help!
Can someone who knows theyre physics double check my workings?
I'm trying to work out the forces on a wishbone under braking.
acceleration (m/s2)= change in velocity (m/s)
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time taken (s)
therefore if i slow my car from 100mph to 0 in 6 secs, 1 mile = 1609.3m, change in velocity =2682.167m/s
acceleration = 2682.167
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6
447.0278 m/s2 = 2682.167
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6
Then if:
Force = mass x acceleration (or deceleration in this case)
force = 800 (kg) x 2682.167
357622 N (357.622 kN) = 800 (kg) x 2682.167
So if i divide 357 kN by 2 i have the force on each wheel??
178.5 kN per wheel???
Please help me if i am wrong as i has been many years since i did physics at high school. Thanks.
[Edited on 24/10/06 by garage19]
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kikiturbo
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| posted on 24/10/06 at 12:32 PM |
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100 mph is 44.704 m/s
so acceleration is 7.45 m/s2
force - cca 6000 N divided by 2, 3000 N.... but
I guess you want to calculate max forces on the wishbone... but you should better use some safety factors... for a road car you want to use max loads
of 5g vertically, 4g longitudinally, 2g laterally. So, get your corner weights, myltyply them by the 5,4,2 g and you will get your max loads...
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oily85
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| posted on 24/10/06 at 12:34 PM |
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I think that acceleration = change in velocity/time,
so
change in velocity = acceleration x time
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garage19
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| posted on 24/10/06 at 01:08 PM |
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quote:
Originally posted by kikiturbo
100 mph is 44.704 m/s
so acceleration is 7.45 m/s2
force - cca 6000 N divided by 2, 3000 N.... but
I guess you want to calculate max forces on the wishbone... but you should better use some safety factors... for a road car you want to use max loads
of 5g vertically, 4g longitudinally, 2g laterally. So, get your corner weights, myltyply them by the 5,4,2 g and you will get your max loads...
Good point on the m/s. I minor slip in my calcs. doh
Quite high safety factors there then.
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kikiturbo
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| posted on 24/10/06 at 01:19 PM |
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yes, but that is for a road going vehicle... you could read around a bit to get the safety factors for the track... production cars use even higher
safety factors...
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chriscook
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| posted on 24/10/06 at 06:05 PM |
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Also don't forget that the wishbones transmit the brake torque at the same time. Unless the brakes are inboard of course.
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omega 24 v6
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| posted on 24/10/06 at 08:24 PM |
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Also
Will the top bones be carrying some of the load as well??
The will the top bones be trying to resist being forced forward (or backwards) due to the torque from the brakes (depends if the calipers are front or
rear mounted)??
Just my thoughts/ input perhaps someone may be so kind as to tell me if my reasoning is flawed
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chriscook
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| posted on 24/10/06 at 10:20 PM |
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Both top and bottom wishbones will take the load.
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matt_claydon
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| posted on 24/10/06 at 10:54 PM |
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Top and bottom 'bones will share longitundinal load of the car decellerating as described above proportionally (depending on the position of the
axle between top and bottom 'bones - moments). They will also share the torque from the brake caliper (independent of whether it is front or
rear mounted) such that the top 'bone is forced forward and the rear 'bone is forced back. Again this is dependent on distances from the
axle and this will subtract from (top) and add to (bottom) the linear forces.
[Edited on 25/10/06 by matt_claydon]
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MikeRJ
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| posted on 24/10/06 at 11:43 PM |
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quote:
Originally posted by matt_claydonAgain dependent on distances from the axle and this will subtract/add from the linear forces.
It will always add to the bottom wishbone and subtract from the top, hence why people get away with having top wishbones made with curves in them
(i.e. pre-failed according to Chapman!).
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matt_claydon
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| posted on 25/10/06 at 07:35 AM |
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quote:
Originally posted by MikeRJ
quote:
Originally posted by matt_claydonAgain dependent on distances from the axle and this will subtract/add from the linear forces.
It will always add to the bottom wishbone and subtract from the top, hence why people get away with having top wishbones made with curves in them
(i.e. pre-failed according to Chapman!).
Sorry, that's what I meant - just came across a little confusingly 
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chriscook
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| posted on 25/10/06 at 06:52 PM |
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I thought this link might be useful to some on here. It is a pdf of a presentation giving an overview of a CAE approach for structural analysis. The
bit that might be useful is on page 5 where there is a table of standard loadcases.
http://www.ulsab.org/ulsas/General/CAEStructuralApproach2.pdf
The figures it gives are essentially industry standard for basic analyses - there's nothing clever in there. Just used it as its public domain,
we use similar figures where I work.
As a rough rule-of-thumb you'd want to keep the stresses below 1/2 UTS or 2/3 yield if its a 'normal' steel with a decent surface
finish and no stess concentrations. The fatigue properties for high strength steels are different so you'd need a lower stress limit in relation
to its UTS or yield.
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