# Design of an automotive bumper

1. Sep 20, 2010

### James22

1. The problem statement, all variables and given/known data
I have attached the assignment below. The weight of the car will be 3702lbs

2. Relevant equations

3. The attempt at a solution

I am not sure if I am on the right track, but so far I have calculated the kinetic energy as;

E=1/2 mv^2
0.5 x (3702/32.2) x 7.33ft/s^2 = 3091.38 ft-lb

I have chosen to use a 6" crumple zone and 4 springs to absorb the impact.

This is where I run into trouble trying to get my head around the concept.

I know for the spring
U=0.5*k*s^2

where;
k = spring co-efficient
s = spring displacement

Since I am trying to stop the car in 6" do I have to treat the kinetic energy as double? Then divide it over the 4 springs that I will be using to stop the car?

Thanks,
Kyle
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 20, 2010

After 6" your 4 springs combined needs to absorb all the kinetic energy of the moving car.

In other words, each spring needs to absorb 1/4 of the kinetic energy.

Energy before = energy after
E = U_1 + U_2 + U_3 + U_4 , and solve for k

3. Sep 22, 2010

### James22

Thank you SirAskAlot, I completed the calculations for the spring bumper using your assistance. Now I have to design a bumper using some type of foam. My professor provided me with a few graphs of the characteristics of foam (PSI vs deflection). I have decided to use a 65.5" wide by 6" tall bumper to represent the average midsized car. That works out to have a surface area of 393in^2

What I am struggling with is the fact that I will need to calculate deceleration, so I will need to know the force (I was planning on using a=F/m).

My initial thought was to take the kinetic energy (3091.3 ft-lb) and convert it to in-lb (37092 in-lb)

Then I went;
37092 in-lb / 393 in^2 = 94.38 lb/in

Is it possible to spread the kinetic energy over the bumpers surface area like this? I still need the stopping force to calculate deceleration though.