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 mV^{2 }=
f_{cr} A_{cr }s

and

“G” load = V^{2} /
2gs which can be derived from the

mks version for constant
acceleration

v^{2}
= v^{2 }_{init }+ 2ax
where G=a/g in English units, and v^{2 }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= V^{2 }/ 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.)

f_{cr static }= 245
psi

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

WV^{2 }/2g = f_{cr}
A_{cr }s where s = 1.5 ft.

0.5(3000)(44x44)/32.2 = 245 (A_{cr})1.5

_{ }

A_{cr } = 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.