Overturning force of stopping a moving machine

[1athoti]

Well, it's my first post though I have been reading eng-tips for a while and I appreciate the site and contributors so much!

I have a machine which rides on 2 tracks, 2 track rollers per side, and is driven by cables which pull it back and forth. The sketch would be a side view of the machine.
I would like to calculate the overturning forces involved or the reaction forces R1 and R2 for 2 scenarios, to make sure the track can withstand the R1, R2 reaction forces.

[ol 1]
[li]The machine being voluntarily stopped. The machine weighs 6500lbs and has a maximum speed of 8.3ft/s. The driving cables can stop it to a stand still within 3ft (or 0.7s).[/li]
[li]The machine being forcefully stopped (an accident, rarely but possible). The machine pushes lumber and if a beam gets caught wrong and is stiff enough that the momentum of the machine will not buckle or break it, it causes the machine to halt and bounce back. The worst case scenario for where a beam would push against the machine is at the dimensions where I have the point placed on the sketch. I'm figuring for half speed in this case so 4.15 ft/s because it is restricted when sawing or pushing lumber. Also, there are dampers in the machine to absorb the impact.[/li]
[/ol]


1. For the first scenario I'm figuring for the direction of movement as shown, since R2 would be higher than R1 in this case.
I would have a gravity resisting moment of Mr = 6.5k x 66in = 429 k*in
The kinetic energy of the machine is: Ke = 1/2 x 6.5k x (8.3ft/s)^2 = 223.9 k*(ft/s)^2
Stopping force: F = Ke/stopping distance = 223.9 / 3 = 74.63 k
Moment of Inertia: Mi = 74.63k x 66in = 4925 k*in
Total Moment: Mt = Mi-Mr = 4496 k*in
R2: Mt / 20in = 224.8 k
R2 per cam wheel: 224.8 k / 2 (sides) = 112.4 k (per cam wheel)
This seems high and I would appreciate any input on my simple calcs.

2. For the second scenario I confess I'm not sure which direction to take. Some of the impact force is taken by the dampers, but the impact basically sends the machine in the opposite direction.

Any help is greatly appreciated!

 

[BrianE22]

Scenario 2 is the tough one. "Abruptly" needs to be quantified. Do you have any accelerometers?

 

[Compositepro]

Your confusion is apparent in your questions, or rather lack of of any clear question. Your title refers to overturning but there is nothing about that in your post. Once you clarify your question you will have the answer to your question.

As for over turning, that will occur when the vector sum of all your forces acting on the center of mass points outside of your base boundary. In this case, that is the rectangle defined by where your four wheels contact the tracks. If what you hit is at the height of the center of mass, then there will be no overturning, no matter the speed.

 

[1athoti]

No, I don't have any accelerometers on the machine. The speed is given by the inputs to the drive motor and the number I've given is most likely on the high side, because of various factors, drivetrain efficiency loses, friction etc. I should take a video of the machine and get an actual speed reading. Some time ago I thought it was more like 5ft/s but here I'm going with the maximum possible speed. In any case, an accident is not wanted so I can't do tests on it, but only know the machine bounces back if the piece of wood is stiff enough. In a previous version of the machine, the cam follower actually bent the track, but it was in a weird situation where a piece of lumber got jammed between the track roller and the track and the operator saw that too late. I don't think the cam follower bent the track just from an abrupt stopping. I'm trying to improve on the first design, hence trying to figure out these forces. Thanks!

 

[LittleInch]

Apologies I'm coming a bit late to this one.

I think our calc is ok, but the unit is trying to move around the front wheel axis.

So the moment from the weight isn't x 66, it's times 32.

Also your stopping force is adding to the overturning moment as it is not in line with the front axle so you need to add on 74.36K x 3inches

I think its about right. You have 3 tonnes of something moving at 5.65 mph - a jogging speed and you're trying to stop it in less than a second with a COG 5.5 ft above the ground.

50 tonne upward force per wheel - I can see that.

Second option you just need to try some stopping times - say 0.1sec and go up in implements. This will complicated by the shock value, but force is reduced as you hitting it above the trolley height.

Probably about 5 to 10 times the static load.

Those "dampers" are you critical item to resolve and try to get the collision point closer to your CoG height.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[rb1957]

scenario 2 is an impact problem ... and the problem with impacts is always "what time duration of impact (or deceleration) ?"

whilst the body will have a deceleration at the CG (66") there will also be the force applied by the obstruction (at 36")

I think you have a mistake in your KE ... m is mass (not weight) ... mass = Wt/g = 6500/32.17

another day in paradise, or is paradise one day closer ?

 

[rb1957]

why Mr in "total moment" ? What you should be saying is that the weigh 6500 lbs is distributed over the 4 rollers, 20/52 on the front pair, and 32/52 on the rear.

your vehicle is stopping from 8.3 ft/s in 0.7s ... or 8.3/.7 = 12 ft/sec2 or about 1/3g
load on front rollers is 6500/3*66/52 = 2750 lbs
load on each roller is about 2750/2 +6500*20/52/2 = 1375 +1250 = 2625 lbs (1.2 tons)

another day in paradise, or is paradise one day closer ?

 

[LittleInch]

Maybe - when i do this in SI units I get about 2,700lb force as the force required, which isn't far away from dividing your 74k lbf by 32.

Still need to figure out the moment arm though. I think it still revolves around the front axle.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1athoti]

Greatly appreciate the input! I'm a young engineer with an appetite for learning but it's tough when you're more or less all by yourself. What wouldn't I give to have a gray haired engineer with me, full of wisdom and willing to help! So I suppose you all are doing some of this (color of hair is unknown ) Back to the basics:

Thank you for the correction of mass = Wt/g.

For whatever it's worth here is another diagram, a side view snatched from the drawing of the machine itself.
So the point of interest is the force Fb (basically to verify integrity of track and CAM wheel structure) and I keep working at it.
For case 1, I think I need to reference the sum of moments around point C, since that's where the braking is happening.
As a note on case 2, I was considering contact with piece of lumber at 36in but I really should go at the maximum height of aprox. 49in.