# RE: Stopping That Truck !!

• To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
• Subject: RE: Stopping That Truck !!
• From: MKnight <MKnight(--nospam--at)Lindapter.com>
• Date: Fri, 25 Jan 2002 09:20:51 -0000
Here in the UK we have a company called Bridon who manufacture cable
restraint systems which are used as crash barriers on our main motorways,
they may be able to help. Their USA portal can be found at
http://www.bridonamerican.com/

> -----Original Message-----
> From:	John MacLean [SMTP:john_maclean(--nospam--at)pomeroy.ca]
> Sent:	Thursday, January 24, 2002 9:23 PM
> To:	seaint(--nospam--at)seaint.org
> Subject:	Re: Stopping That Truck !!
>
> Stan Caldwell wrote:
>
> <<Question:  What is the design force on the bollard?  on the wire rope?>>
>
> Interesting problem. It seems very similar to the "fall arrest force"
> problem where you have to design a fall restraint system for a guy working
> with a safety line.
>
> Here's my basic take on it which may be similar to the thought process you
>
> Your 7500 lb truck travelling at 20 mph has a kinetic energy (1/2 mv^2) of
>
> So you basically need a system that will absorb 100 kip-ft of energy.
>
> Some of the energy will be absorbed in totalling the truck.
>
> Then again it could be one of the Ford trucks built up here at the Oshawa
> truck plant in which case it will remain elastic and all the energy will
> go
> into your cables and bollards.
>
> I guess you could figure out some kind of spring constant "k" for the
> system. In that case the deflection of the system, d, would be Fmax/k.
> Fmax
> is the maximum force applied to the truck. If the "spring" is elastic, the
> work done in stopping the truck will be Fmaxd/2 because the force will
> ramp
> up linearly from 0 to Fmax. Since d = Fmax/k then the work done =
> Fmax^2/2k.
>
> So Fmax = sqrt(100kip-ft/2k) = 7 x sqrt(1/k) kip or
>
> Fmax = 100kip-ft/2d = 50 kip-ft/d
>
> So if you can stop the truck in 1 ft, Fmax = 50 kip; 2ft, Fmax = 25 kip;
> 6",
> Fmax = 100 kip; etc.
>
> If the deflection is 12" midway between the bollards (spaced at 12')  then
> Fmax = 50kip. The tension in the cable would be 50 kip x
> sqrt(12^2+72^2)/12
> = 304 kip. If the deflection includes movement of the bollards then the
> force would be even greater since the cable angle would be smaller.
>
> The hard thing is to figure out what k or d is. There's going to be some
> non-linearity in the system. Primarily, I would think, in your clay
> foundations. I've also got a feeling that the cable system will be
> non-linear due just to geometry.
>
> I would be inclined to do a rough design of a system based on a 50 kip
> then work out a spring constant for the system and cycle back through the
> design til I was happy. I would assume very little if any energy was
> absorbed by crunching sheet metal. I would tend to assume the system was
> stiff rather than flexible. I would tend not to design the bollards to
> yield
> if possible. There is a certain deterrence in the simple appearance of
> strength (as Stalin used to say "quantity has a certain quality all it's
> own").
>
> I'm not totally clear on how the bollards work. Is the cable continuous
> across several bollards and anchored intermittently or is it anchored to
> each bollard. If the cable is anchored to the bollards then the bollards
> would have to be designed for the cable force. If the cable slides through
> an eye at the bollard then it would have to be designed for the impact of
> the truck hitting the bollard itself which is a whole nother problem (but
> similar).
>
> There's my take for what it's worth.
>
> Cheers,
> John MacLean
> Vancouver, BC
>
>
>
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