Rubber - Impact - Force 3

[amorphous]

Hello All,
I am sure this question was asked in the forum before but I could not find it.

Please look at the attachement and help me figure out the force F on the wall behind rubebr, if a weight W travelling with a velocity V hits the wall reinforced with rubber (spring constant K).

FYI: I am not a student. This is not a class assignment. I am trying to design a collision post.

Any help is greatly appreciated.
Thanx in advance

 

[desertfox]

hi amorphous

0.5*M*V^2 = 0.5*k*x^2

The kinetic energy in the moving weight aassuming no losses is equal to the stored energy in the rubber after impact, so workout kinetic energy of moving block (LH side of equation) then knowing the (k) of the rubber, transpose to find x the rubber deflection.
The force on the wall will be equal and opposite to that stored in the rubber, the force on the rubber can be found by simply multiplying the rubber deflection (x) by the rubber stiffness (k)

Where M= Mass Of moving
Weight
V= Velocity of
moving weight

k= rubber
stiffness

x = defl of rubber


desertfox

 

[electricpete]

Yup. Good solution by desertfox.

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[amorphous]

Thank you desertfox and electricpete.
I tried to solve the above equations provided by desertfox with my actual values...

Weight =100 kips ; therefore mass (m) = 100,000/32.2 = 3105.6 lb/ft/sec^2
velocity= 14.6 ft/sec
rubber stiffness k = 1200 kips/ft (very stiff rubber)
(rubber thickness is only 2" or 0.166ft)

0.5*3105.6*14.6^2 = 0.5*1200,000*x^2

x = .742 ft

Now the force on the wall = k*x = 1200kips/ft * .742ft = 891.2 kips

Is it correct? if so

Question 1) Rubber is only .166 ft thick so what happens after it is fully compressed? how much % of this load is transmitted to the wall directly?

Please correct if I am not thinking right...

Thanx in advance.....

 

[electricpete]

Yes, I get the same value for X:
M:=100*1e3*lbm;
V:=14.6 *ft/sec;
KE:=0.5*M*V^2 * (lbf/(32.2*lbm*ft/sec^2)) = 330993.7888 ft lbf
K:=1200*1E3*lbf/ft;
X := sqrt(2*KE/K) = 0.7427356963*ft

So what happens if rubber is only 0.166 ft thick?
We might question whether remains linear, but if it does the energy absorbed would be:
PE:=0.5*K*(0.166*ft)^2 = 16533.6000*ft*lbf

The KE at that point is reduced to
KE_Remaining = KE-PE = 314460.1888 ft lbf

You are going to need a model of the stiff wall behind the rubber to convert that remaining KE to a force. It could be modeled as a stiff spring (which theoretically affects the first part of the calc, but not much)... in which case proceed as above. Or you can make some assumptions about length of time to bring that to a stop...



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[CostasV]

If this question is for designing a real colission post, then you need more data.
The shape and size of the contact area between the moving body and the rubber, as well the size of the rubber and the wall (or the post), are important and should be taken into consideration.

I would set up a small scale experiment, measuring the maximum force and the time needed to stop (or to zero velocity before going backwards). Then I would use these measurements for the normal scale.

mass X velocity = 1/2 X max(F) X Dt (assuming triangular force distribution relative to time)

 

[GregLocock]

Realistically do you think you can stop a truck moving at 10 mph with a 2" slab of rubber?



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[amorphous]

Thank you guys for all the replies.

Electricpete:

"The KE at that point is reduced to KE_Remaining = KE-PE = 314460.1888 ft lbf"

What is the formula to convert this energy into force on the wall behind the rubber??? any help? I dont know how much time to assume for it to come to stop????

Greglocock:

No, I am not trying to stop a 100kips car with a 2” rubber slab. I did not want to make the problem complex so I kept it simple in the post.

I have a flat car which weighs 20 kips and coupled to 3 other cars weighing almost 300 kips. This rubber is attached to the free end of flat car. What I was trying to find is when in a collision with a locomotive of 100 kips travelling at 10mph how much force is transferred to the frame past this stiff rubber. I know again how far the consist travel? rolling or sliding friction? and many more other factors to be considered... for now all I need to calculate the force on the wall behind the rubber assuming its fixed.

As the forces look awfully huge, I am intending to use a thicker rubber and I do not have all the info. right now. I am working on it and when I finish, I’ll share with you guys.

Thanks to you all once again....

 

[desertfox]

Hi amorphous

Why not use a spring and damper to absorb the impact energy.
The damper would prevent any bounce, now the force on the wall, well if you consider the rubber to be a spring and you calculate the force in the rubber absorbing the impact, then the force in the rubber is K*x where K is rubber stiffness and x is the deflection of the rubber, this force will be the force on the wall assuming the wall is considered totally rigid.

desertfox

 

[DLite30]

Keeping this problem elemetary:

using your basic equations:

xf=x0 + v*0 + 1/2 * a * t^2

and

vf=v0 + a * t

You know final velocity=0, initial velocity = 14.6 ft/s, distance traveled (which you already solved above=0.742 ft), you can solve for acceleration and time.

I get a=143.64 ft/sec^2 and t=0.102 sec

now you can use F=ma to figure the force.

Like I said this is extremely simplistic.

 

[DLite30]

Well, I mistyped the velocity equation above, but you can figure it out.

 

[MintJulep]

As the forces look awfully huge

Yeah, they typically are when you start smashing locomotives and trains together.

Smashing three train cars with a locomotive is a hugely more complex problem than is solvable with desertfox's equation.

100,000 pounds is very light for a locomotive by the way.

The loco probably has some energy absorbing structure. The flat car too.

The flat cars will not remain stationary. Many interesting things will happen at the couplings between the cars.

The locomotive will not rebound.

The flat car is unlikely to remain on the rails.

 

[berkshire]

These things have been around since about 1790

http://www.istockphoto.com/stock-photo-245312-steam-train-buffer-close-up.php

B.E.