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 




“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)



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.




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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