Impact loads - Navigation Piles 6

[Myoho]

Hi I’m designing some navigation piles (free ended, single laterally loaded piles) to protect a water inlet structure for a water treatment works. The inlet is constructed on the River Thames (UK). The design impact load from a vessel on the river is 100 tonnes at 2.5 meters/second. However I’m not quite sure to convert that force into kN. Any ideas please!

 

[nicam]

100 tonnes = 100000kg = 1000kN
momentum = P=mv mass = 1000kN * 2.5 = 2500kN

hope this helps!

 

[gwynn]

I would look at this in terms of energy, not momentum. Equate the kinetic energy of the vessel to the energy absorbed by deformation of the piles. 0.5mv^2 = kd.

Coming up with the spring stiffnes of your piles will require geotech input. If the soil strata are uniform it becomes a fairly straight forward problem. It may be worth looking at ductility of the system as well, depending on frewuency of occurence and the replacement cost of the piles versus the cost of piles large and deep enough to survive the impact.

Also, 100 tonnes only equals ~1000kN when accelerated due to gravity. The acceleration due to a vessel hitting a pile will be at a much different rate.

 

[nicam]

The vessel is not accelerating, it's travelling at a constant velocity of 2.5m/sec, that's why I used the momentum equation.
The guy was asking to convert the force into kN.

 

[nicam]

Also the acceleration due to hitting the pile would be 2.5m/s to 0, if we assume the pile does not move.

 

[GregLocock]

Your equation des not seem to make any sense to me.

Cheers

Greg Locock

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

 

[nicam]

not sure which equation your refering to, however both momentum and kinetic energy equations are correct.
To get the KE equation you simply intergate the momentum equ.

 

[GregLocock]

Sorry, what I meant was that your arithmetic is correct, you are missapplying the equations.

Cheers

Greg Locock

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

 

[nicam]

oh you mean when I wrote mass = 1000kN * 2.5 = 2500kN ?
yeah it didn't come out right! I meaning to write mass = 1000kN and V = 2.5m/sec, then apply the momentum equ.

sorry for poor description.

 

[GregLocock]

That is still a most unlikely solution to this problem. Perhaps a search for impact loads might reveal the MANY threads on closely related subjects.

Trust me, the complexities are immense.

Cheers

Greg Locock

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

 

[nicam]

Not sure why momentum is unlikely, sure you can go into the heavies of physics but as Engineers we want the easiest solution.

You can calculate KE, deflection, stresses etc but I think the guy wanted to know how to convert one unit into another.

 

[hokie66]

The momentum is not 2500 kN, it is 2500 kNm/sec, but like Greg, I'm not sure that helps.

 

[GregLocock]

Good grief. There is NO easy solution.

Cheers

Greg Locock

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

 

[nicam]

ok so I got lazy with units,.....

 

[slickdeals]

I think the way to go about this is to set up your maximum tip displacement. You can never design it to remain rigid. And then use this displacement as the distance at which the velocity will change from 2.5 m/s to ZERO. And once you have this information, you can compute the equivalent force. This is what we do when we are designing barrier cables for vehicle impact.

Use Force * distance = 1/2 * mass * velocity ^ 2
where distance = tip displacement of pile

You know all variables except Force.

HTH

 

[Myoho]

Thanks lads. I'll try both the Momentum and Kinetic Energy equations. Your responses are very much appreciated.

Nam Myoho Renge Kyo.

 

[JStephen]

On occasion, I ride some ferries. At the landings, they have big masses of timber pilings. When they just barely bump those with the ferry, that whole mass will shift 6" or more- things tend to be very flexible when hit by ferries. I would think you'd want to assume substantial movement, rather than a rigid condition, unless you know otherwise.

 

[GregLocock]

Well, let's do it properly.

The correct form of the impulse equation is F*dt=m*dv

for constant mass

If you can assume constant F, then this becomes F=ma, but constant F is hard to achieve with a structure. So that still is not much help.

Rather more usefully if we can assume that the system can be represented by a 1 dof spring damper system, then the instantaneous work for a small period of time is

d(1/2m*v^2)/dt+d(1/2k*x^2+c*v*x)/dt =0

where dx/dt=v

But in order to get a solution you need to make some assumptions about k and c, the latter in particular is quite problematical, for your system. If you set it to zero then in theory the boat will just bounce off, at -2.5 m/s, but this will at least give you a force that is easy to calculate for a known k




Cheers

Greg Locock

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

 

[swearingen]

There are many things to consider for fendering systems. The industry uses the energy calculation. A decent example of how to do the complete calculation is presented on pages 74 to 93 of this catalog:

http://www.jhmenge.com/fentek/fentek_catalog.pdf

However, they only provide a factor for compliance of the structure. In designing monopile fenders, you can certainly benefit from using the flexibility of the pile itself - it reduces the resultant maximum load. Take note of the load/deflection curves for the fenders in the catalog. If you don't put a fender on your pile, your "curve" will be a straight line. I do recommend the fenders on page 36 for this application, though.



If you "heard" it on the internet, it's guilty until proven innocent. - DCS

 

[graybeach]

I have designed fenders using the slickdeal's method. It worked out ok with a bid of hand waving as to the pile's point of fixity.