Brake deceleration rate

[swedish]

Hey guys,
just had a bit of a brain snap. Have been asked to go through and verify some calculations from 1992 and have got to a part where he calculated the brake deceleration rate. Can't seem to figure out how he went about it, if someone could help me that would be great.
Here is the information given
Brake Torque = 295.1002 kNm
Drum Speed at brake tip = 3.68 m/s
Drum diameter at rope = 3400.8 mm
total system inertia at drum = 465109 Kg m2
Rope load at drum = 114.9 kN
brake path width = 203mm
brake path thickness = 38.1 mm
brake path length = 1438 mm

deceleration rate at brake = ?
deceleration rate at drum = ?

Thanks

 

[GregLocock]

Need a picture of the system.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Oldhydroman]

Yes, need a picture/drawing diagram to be sure.

If you have a rope tension of 115 kN working at a radius of 1.7 m then this is an applied torque of 195.5 kNm. Your brake can apply a deceleration torque of 295.1 kNm of which ~100 kNm is left over after braking the load. This remaining torque (T) will decelerate the rotating mass of the drum.

Now you can work out how the drum will decellerate based on its moment of inertia (I). Use the rotational equivalent of Newtons 2nd law: T=Ia.

Your acceleration (a) equates to -0.214 rad/sec².

[Here's where some guess work is involved regarding the layout of your system.] The rope speed is 3.68 m/s and the drum diameter is 3.4 m, so the drum circumference is 10.68 metres and your drum speed must be 3.68/10.68 = 0.344 revs per second (20.7 rpm - does this sound resonable?). A drum speed of 0.344 revs per second is 2.16 rad/sec and you have a deceleration rate of 0.214 rad/sec² … so your deceleration time will be 2.16/0.214 = 10.1 seconds.

Those figures about the width of the brake track etc. are to do with evaluating the heat dissipated in the brakes. There’s no point being super precise in the calculations because the brake performance depends heavily on too many external factors: the state of the discs (rusty, wet, greasy), the state of the pads (cold/hot, new/worn) and the speed of the disc (which varies from max to zero [you hope] during the braking process). It is also difficult to assume that the load will remain constant during the deceleration process.

DOL

 

[rb1957]

i'd also draw free body diagrams. at t=0 the drum is stationary, the rope tension is applied and reacted by the brake.
then the brake is released, the drum accelerates, rope is paid out changing the torque applied to the drum and the inertia of the drum.
then the brake is applied, and it's all change.

btw i assume "acceleration rate" = "acceleration", yes?

 

[Twoballcane]

“deceleration rate at brake = ?”
Isn’t the brake stationary? If so, acceleration is zero…no?


Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”
"People get promoted when they provide value and when they build great relationships"

 

[swedish]

here is the system (see link), and the value recieved for KL=105.326.
his deceleration rate at brake = 0.3921137 m/s/s
deceleration rate at drum = 0.3645834 m/s/s; both of these values i don't know how he obtained.

 

[chicopee]

Torque= (Rotational inertia of rotating drum+ linear inertia of load being retarded)X deceleration rate. Since you have the equation to calculate torque, you need to figure out the inertias then calculate the deceleration which will have units in radians per second^2.

 

[IRstuff]

Why does the "system" look like it comes from a textbook?

TTFN
faq731-376

 

[chicopee]

The system looks more like from a design manual. I have a similar example in my Kent ME handbook but the analysis is done with the brakes internally positioned to the drum.

 

[zekeman]

Swedfish,

Showing us a picture of a generic brake doesn't help us help you.

 

[MotionGirl]

Nice to see SI units for a change.
And yes, rb1957, I do get annoyed when term 'acceleration rate' is used instead of straight 'acceleration'. Acceleration rate would mean 'jerk', or the third derivative of position or angle etc.

TwoBallCane - it is very common for objects to have acceleration without velocity - drop anything and it accelerates with a starting velocity of 0m/s.

Can't offer much for the calcs though - sorry - looks too much like a brake calliper and text book.

 

[chicopee]

Yes Motion girl you are correct about the term "rate" I should have kept it to decelaration in my first reply.

 

[swedish]

Sorry guys not in the office for a couple of days. I put the previous picture up as the drawing I am working with are from the 1960s. I'll try my best to scan in a decent drawing when im back in the office

 

[swedish]

Here is what I am working with guys

 

[chicopee]

From your second drawing, you can calculate the drum moment of inertia, the inertia from the load being decelerated as mR^2, the rope inertia which you'll have to base on a length, the torque from friction of the bearings, the torque from the equation given in your first drawing; from all these variable you can calculate the angular deceleration from which you can determine time if an initial velocity is given.

 

[zekeman]

Here is what I am with guys

http://files.engineering.com/getfile.aspx?folder=8a3592b6-fe23-4e59-b44a-da"

If that is all that you have to work with, I think coming up with a valid answer would make you or any responder here a clairvoyant genius.

 

[PersonalProfile]

Much the same comments as every one else, though if you provide a diameter (I for one, could not read it on the scan) of the brake drum I think we can work something out.

From what I have seen and have assumed, I understand the brake and drum would be a rigid body and propose the original designer has calculated the "stand alone" deceleration rates of the individual (drum and brake) components, if they were left to their "own effects".

I myself use a similar concept for conveyor drives and winders if I want to know ("accurately" cf factoring the system torque) the actual torque between the drive and the driven element (during de / acceleration the torque is not constant from the drive to driven equipment - due to inertia).

It would be recommended to check with someone suitability qualified (this may be statutory), particularly as it is a winder and safety is critical, who can do all of the appropriate checks whatever they may be for the current question, whatever it may be.

Regards,
Lyle