[200q20v] Re: Strut tower maximum lateral force

Bernie Benz b.m.benz at prodigy.net
Fri Mar 9 12:44:37 EST 2001

Dave, I stand corrected!  Thanks.  Why didn't you guys catch the error when
I first posted this back in July?  So, X my figures by a factor of 3 or 4.

Compression on the EMT is insugnificant, but the EMT must be straight to
prevent buckling failure.


> From: "HAUPT,DAVID (A-Sonoma,ex1)" <david_haupt at agilent.com>
> Date: Fri, 9 Mar 2001 12:27:38 -0500
> To: "'200q20v at audifans.com'" <200q20v at audifans.com>
> Subject: [200q20v] Re: Strut tower maximum lateral force
>> From: Bernie Benz <b.m.benz at prodigy.net>
> Math check request:
> If the tire-to-road coefficient of friction is 0.3, then you're probably
> driving on snow.  Seriously, for a car to pull a 0.8G turn, the coefficient
> of friction needs to be 0.8.  Unless I've hopelessly forgotten my freshman
> courses in dynamics.
> So I think the maximum lateral force on the tire at the road is more like
> 2000 lbs.
> Assuming the remainder of the suspesion geometry evaluation is correct, this
> translates to around 1500 per strut, or 750 for the brace.  Methinks we'll
> get some compresson of the EMT at this force, but it's still orders of
> magnitude stiffer than the rubber bits.
> That value can rise if you happen to hit a curve while making that 0.8g
> turn, but if you do that, the strut brace is the least expensive item you'll
> have to replace.
> On the other hand, I am prepared to get a lesson in dynamics, as I haven't
> actually solved such a problem in 20 years.
> Dave
>> Come on guys, Lets get realistic!
>> There is just no need for anything heaver in a strut brace, as there is
> not
>> much latteral force at that point.  I went thru this with another lister
> many
>> months ago, but neglected to keep a copy.  He finely saw the light and, in
> the
>> interest of keeping the economy strong, decided to just chrome plate the
> whole
>> thing!  But, it went something like this:
>> Front axle gross vehicle weight (fully loaded) being Audi defined as 2469
> lbs,
>> Now estimate the max. weight on the front outside wheel in a max.G turn,
> the
>> ultimate being with the inside wheels off the ground, i.e. the full 2469
> lbs
>> on the outside front wheel.
>> If the tire to road coefficient of friction is 0.3, the maximum latteral
>> cornering force transmitted to the suspension at the axle would be 740
> lbs.
>> The suspension geometery divides this force between the lower link and the
>> upper strut anchor point by the ratio of these two distances from the
> axle.
>> On the type 44 chassis, Axle C/L to ball joint C/L = 4", axle C/L to strut
>> piston rod anchor w/ a compressed suspension = 20".  Therefore 4/20 x 740
> =
>> 150 lbs force.  This is  the theoretical, far out, maximum latteral force
> that
>> the strut tower will ever see!  (My strut brace further equalizes this
> force
>> between the two strut towers to 75 lbs max.)
>> Bernie
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