Calculating Diaphragm Plates Needed to Stop 30,000 lb Object

[brambilah]

I Have an object with a mass of 30,000 lb falling at a height of 90ft. The object is to be stopped by a series of diaphragms spaced out 6" apart. I am using 1" welded plates as diaphrams. I need to find out how many plates it takes to stop the object. I have calculated my potential energy=kinetic energy to be 3.24x10^4 kip*in and the velocity at impact to be 76.101 ft*s^-1. Also each plate to have a weld capacity of 334 kip. I want to find out how many plates will the object shear before it stops.

 

[berkeman]

I Have an object with a mass of 30,000 lb falling at a height of 90ft. The object is to be stopped by a series of diaphragms spaced out 6" apart. I am using 1" welded plates as diaphrams. I need to find out how many plates it takes to stop the object. I have calculated my potential energy=kinetic energy to be 3.24x10^4 kip*in and the velocity at impact to be 76.101 ft*s^-1. Also each plate to have a weld capacity of 334 kip. I want to find out how many plates will the object shear before it stops.

Holy smokes! Is this for an episode of MythBusters or something?

 

[AlephZero]

Holy smokes!

It's' "only" a 13 ton truck doing 50 mph. That's not a ridiculous thing to contain with a few inches of mild steel.

But without knowing some more about the geometry of the situation, it's impossible to answer the question.

 

[brambilah]

It's' "only" a 13 ton truck doing 50 mph. That's not a ridiculous thing to contain with a few inches of mild steel.

But without knowing some more about the geometry of the situation, it's impossible to answer the question.

What type of geometry do you need to know??

 

[brambilah]

I used the conservation of momentum. I have initial momentum of ρi=M*a=70.959 kip*sec
*Vi=51.88mph
*Vf=0 mph
*m1=30,000 lb ======>mass of object
*m2=163 lb ======> mass of 2'x2' 1" plate
*t=1" =====>Thickness of diaphragm plate
*Δt= t/((Vi-Vf)/2) =====>Duration of Average impulse
*Δt= 2.19 x 10^-3 sec
*F=250 kip =====>Magnitude of Impulse(Chosen as a practical max.)
*ρ1=ρi-F*Δt =====>Momentum after first impact
*ρ1=70.411 kip*sec
*Δρ1= ρi-ρ1 ======>Change in momentum after first impact
*ρ1=0.548 kip*sec
*V1=ρ1/(m1+m2) ======>Velocity after first impact
*V1=75.105 ft*s^-1
*ΔV1=Vi-V1 ======>Reduction in velocity
*ΔV1=0.996 ft*s^-1

N=ρi/Δp1 =======>Plates required
N=129.6

Can someone verify that these calculations are feasible?

 

[AlephZero]

Go and look at some roadside crash barriers, then decide if "130" is a believable answer.

I would calculate how much of work you can do deforming one plate, before you either break the welds, or break the plate due to plastic deformation. Your projectile velocity is low so I don't think you need to worry about punching a "bullet hole" through the plate, or any other dynamic effects.

I would guess the answer is more than one plate (simply from the way the question is asked) but more llkely to be 2 or 3 plates, not 130.