Page:NTSB RAR-77-10.pdf/35

 APPENDIX E _ 32 _

sc - 39.300{R2[r/2-sin'1(1-G/R)}—(R-6) (“EEK—‘5- 2)

For the leading car anticlimbers 6 - 3/4 in. and R a 49 in. I Be * 336.000 in.-1b For the remaining anticlimbers 6 - 1/4 in. and R - 49 in H

EC = 65,000 in.-lb

The total energy required to deform all the anticlimbers is given by EC = 2 EC + 24 EC or Be = 2,232,000 in.—1b The energy required to shear the shear pins is given by £5 = an Substituting data from Table l. we obtain Es - 270,000 in.-lb

Let us now determine the energy required to move the impact- ing consist. The braking rate was given at

B - 3.2 mph/sec

for a light train (42,200 lb car). Expressing this rate in terms of the acceleration due to gravity

3 - 0.1k6 g. The force resisting motion is merely

PD = VB - (42,200)(0.146) - 6,160 lb per car.

The total energy required to move the impacting consist is given by

Er - GFDD Substituting data in this equation

E? ' 11,088,000 in.-1b