asce 7 - impact loading for bulkheads

huangyhg

asce 7 - impact loading for bulkheads
i have to check a bulkhead for the impact of debris based upon section 5.3.3.5 in asce 7 - 98 edition and a 1 kip object.  i specifically have a question regarding the last statement of that section.  it states, "it shall be assumed that the velocity of the object is reduced to 0 in 1 sec."  if i make this assumption with the wave speed provided (approx. 25 ft/sec), the deflection to allow for the object to come to rest over a full second is not practical.  i am attempting to use a conservation of energy approach to determine the actual force based upon some assumed deflections, but the forces are getting extremely high.  has anyone had experience with this analysis and are there any other references i could use for help?  thank you.

the commentary in asce 7-98 helps a little (see p. 241-242). it indicates that rigid structures of concrete or steel may reduce the velocity to zero time interval to 0.1 - 0.5 sec. but even at t = 0.1 second your distance to zero is 1.25 ft., which could still translate into a large force on your barrier.
the real solution of course is a balance between deceleration of the object, and spring stiffness of your barrier, complicated by plastic deformation of both your barrier and the debris itself. and as a dynamic load, even the natural frequency of your barrier system could impact the solution (much as it does in blast/seismic loads).
maximum plastic deformation of the debris is desirable but unpredictable. plastic deformation of the barrier is limited to what you can tolerate. if it is for a one time load, and repairs can be effected afterwards, then 2 to 3 times yield deformation or more might be allowable. if the loads are frequent and/or repair is not easy to effect, then 1 times yield deformation is probably your limit. ignoring natural frequency vs. load duration for now, blast design indicates that allowing 3 times yield plastic deformation could reduce your design load by a factor of about 1.75, or in other words a structure designed for a 1 kip static load could handle an impact load of about 1.75 kips.
since 5.3.3.5 stipulates that the load impact over an area of 1 sq. ft., lets quickly look at a simple model. assume that we have a 4 ft. high by 1 ft. wide section of steel plate cantilevering up from a foundation acting as your barrier (the barrier could be continuous but we will only analyse a 12" wide section and ignore any help from adjacent portion of the barrier - a conservative assumption). let's assume that the structure/foundation of this barrier is infinitely rigid. the debris will impact your barrier at 3.5 ft. above the base for a moment of 3.5 ft. kips.
load factor for this design will be 1 since we are attempting plastic design. and for dynamic we can actually use a material phi factor greater then 1 (perhaps as high as 1.25) - this is based on blast resistant design again. so assuming 50 ksi plate material the plastic section modulus required to support this load will be m*12/(phi*50) = (3.5*12)/(1.25*50) = 0.672 in^3. this gives a required plate thickness of sqrt(z*4/12) (4 because this is the plastic modulus, assuming full rectangular stress distribution, not the triangular associated with elastic design) --> t = 0.473". lets be practical and go with 0.5".
the moment of inertia i = 12*0.5^3/12 = 0.125 in^4
based on a 1 kip load at 3.5 ft. above the base, yield deformation would be pl^3/3ei = 1*1728*3.5^3/(3*29000*.125) = 6.81"
if we decide to allow 3 times yield deformation then deltamax = 20.43"
if we use this deflection as the debris stopping distance, then using the motion equations d = do + vo*t + a*t^2/2 and v = vo + at to solve for t, vo = 25 ft/sec or 300 in/sec. & v = 0. therefore at = -300, and a = -300/t. substituting into the other equation:
20.43 = 0 + 300*t - 300*t/2 --> t = 0.1362 sec.
then a = -300/.1362 = 2202 in/sec^2 or 183.55 ft/sec^2
f = m*a = (1000/32.2)*183.55 = 5700 lb. or 5.7 kips
natural period of the plate would be t = 2*pi*sqrt(m/k)
m = 1000/32.2 = 31 lb*sec^2/ft.
k = 1000/(6.81/12) = 1762 lb/ft.
t = 2*pi*sqrt(31/1762) = 0.8334 sec.
again based on blast design, the required pseudo static design load would be based on the dynamic load (5.7 kips) reduced by the following factor (i'll give you the necessary reference later): 4.48 which gives a pseudostatic design load of 5.7/4.48 = 1.27 kips, not too far off the 1.0 kip that we started with. the final solution is arrived at by iteration of the above process. and since we used 0.5" thick plate instead of 0.473" then the capacity of this plate will actually be about 1.1 kips.
note though that if you cannot tolerate 3 * yield deformation then your reduction factor would be lower (yielding a higher necessary design load and a higher applied load due to the shorter stopping distance).
the formula that i used came from asce manual no. 42, which was the bible in the past for blast resistant design. this manual is no longer in publication (although you may be able to find a copy at a nearby technical university library), but has been replaced by another asce publication (i can't recall the title right now and don't have access to a copy at this time).
now you may not be able to use this approach, but it gives you an example of how you might deal with this type of problem. it would all depend upon what local code authorities would allow, what your client is willing to accept, and then also i'm not certain that the blast methodology is directly translatable to your debris impact scenario.
having said that, the k reduction factor in seismic design is very similar, in that while you can design for static loads reduced by the k factor, deflections are not so reduced, and therefore plastic deformation occurs.
thank you for the detailed and helpful response.  i have taken into account everything you have discussed except for taking the analysis well into the plastic limit.  this is based upon restraints from the client causing the bulkhead to take the impact within the yield limit so future repairs will not be required.  this limits the deflection and greatly increases the force.  i have detailed it to account for the large impact relying on mass to lower the k factor and engaging a larger portion of the wall to resist the impact.  thanks again for all of your help.
oops, one mistake in my post when i said k factor for seismic, i meant r factor.
glad i was able to help. your response was the only positive in a bad day where i got sideswiped by the cancellation of a $2.5 billion (going onto $8.5 billion - which kind of explains the cancellation) project.
bantrelstructural,
i try to get the asce publication from libary but fail to get it. is there any other way to get this referance. thanks.