Hexcel Calculations for NHTSA Side Impact Test

From bottom of  page 13 of Hexcel Application Note

 

Total Energy of motion must be totally absorbed by the honeycomb material.

 

½ mV2   =  fcr Acr s 

 

and

 

“G” load = V2 / 2gs    which can be derived from the

mks version for constant acceleration

 

v2 = v2 init + 2ax   where G=a/g in English units, and v2  is the final velocity squared ( = 0 ),  where a = acceleration, and where x = distance

 

 

From page 15 of  Hexcel Application Note

 

(Note that the weight is now 3000 lbs, and the max G-force is now 20)

 

Situation:

A 3000 pound object is traveling at

30 mile per hour and must be stopped.

 

W=3000, g=32.2

 

Problem: The object must not be subjected

to more than 20 G’s.

 

Calculations:

 

V= 30 mph x 1.467 = 44.0 fps

 

G= V2 / 2gs  where  s= feet  :   where G=20 is the maximum G force

 

  s(reqd) =(44x44) / 2(32.2)(20)  = 1.5 ft = 18 inches absorbing depth required

 

 

 

 

 

 

 

 

 

s=70% Tc    -  the honeycomb cannot be compressed to zero inches.

                       

            I add extra honeycomb to final design, so here I assume  s = 18 inches

 

________________________________________________________

 

 

    

Using ACG- ¼ - 4.8  which has crush strength of 245 psi

 

 

(I will not increase its strength for dynamic force situation.

Increasing the strength would help me.)

 

fcr static = 245 psi    

    

 

Total Energy of motion must be totally absorbed by the honeycomb material.

 

WV2 /2g = fcr Acr s        where s = 1.5 ft.

 

 

0.5(3000)(44x44)/32.2 = 245 (Acr)1.5

 

 

Acr  =   245.4 square inches of honeycomb that is 1.5 ft. deep will completely stop the  3000 lb object.

 

Assume that the actual honeycomb depth is 2.0 feet to accommodate the hardware geometry.

 

A block of honeycomb that is 4 feet wide and 4.26 inches high

will yield the required 245.4 square inch cross section.

 

If it is 2.0 feet deep, it will have a volume of 3.4 ft sq. and will weigh

 

(3.4 ft sq)  x (4.8 lbs/ft sq) = 16.35 lbs

 

This quantity of material costs less than $400 retail.

 

The steel boxes to enclose the material, and the shafts and bumpers will weigh approx. ? lbs and cost ?

 

If the bumpers are hit at an oblique angle, they are unlikely to be pushed into the box.

 

It will be necessary to design a bumper shaft that will deform with enough force through a distance that the G forces will be kept low.


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