How do I calculate shock loading

begginning time should be 0 (Zero) t2-t1 is 1/8 seconds absolute. inreality -1/8 second. the calculation is deceleration not velocity.

think a little differently. what additional G loading is needed to stop in 1/8 sec from that speed? 2A*s= (V2*V2)- (V1*V1). V1 = 1 ft/sec, V2 =0 s= 1/(8*2)ft assumes uniform deceleration over the 1/8 sec. solve for A (deceleration g load.)

F=M*a = the deceleration load

or use simple comparison: 1 G will slow load 32 ft/sec/sec, or 4 ft/sec in 1/8 of a sec.

you need to slow only 1 ft/sec in the same time, therefore you need only 1/4 G deceleration. Again, simply 1 G exerts 5000#, therefore 1/4 G exerts 1250 # additional load to the lifting system.

remember impulse law?

use it to solve your problem.

The 5000 pounds is a force weight, (mass * acceleration due to gravity). The mass which you must decelerate is 5000lbf/32.2 ft/s^2 = 155 lbm. Your formula should then be

F = 155 lbm * (1ft/s / 1/8 s)

F= 155 * 8

F = 1240 lbf

This is in addition to the weight due to gravity, for a total force of 6240 lbf.

I believe this is correct.

This issue comes up periodically in my work. My problem is always with the stopping time. The questioner speculates on 1/8th of a second, but, how do we know without some time lapse photography? And it's critical to know as this is the ultmate determining factor.

Hello,

I calculate like this: (in metric)

Mass = 5000 lbs / 2.205 = 2267.6 kg. vel = 12" /s = .305 m/s.

Impulse and Momentum. Impulse = resultant force * t in sec; and Momentum = M*v,

Momentum: 2267.6 kg * .305 m/s = 691.6 kg.m/s, this equals R*t.

Resultant force R = 691.6 kg.m/s / 0.125 sec. = 5532.9 kg * 2.205 = 12200 lbs.

It was pointed out before that the time is crucial. If the 1/8 second changes the result will be different. However, I am not sure how some people calculated some 1250 lbf?

Rolf

Sorry, I guess I made a mistake.

The answer above is in Newton not kg. Dividing by 9.81 results in 1243 lbf.

My apologies. I was answering too fast. Rolf

First of all, Thank you all for replying so fast.

I found the my first error in that the extremely high forces I calculated were due to having changed to inch/sec and therefor had inch-pounds not foot pounds. A 25% increase in load due to emergency stop would not exceed my brakemotor selection given a healthy safety factor that was applied.

The time of reaction, 1/8 sec was dirived by two means. 1st, observation of the load's travel distance after appling the Estop. It appeared to travel approximately an inch and a half . 2ndly, the gearmotors manufacturing specs. Both of which are still only approximations and gives a good starting place. I know the machine is not drifting much further so the time is not greatly increased and I know it is drifting some so it is not instantainous.

Again thank you all for your inputs. Ken