Inertia of rotating drum vs. Braking requirements 1

[TheReg]

First, I havent done these calcuations since school.

I have a drum, which weighs 400lbs (24"Dia. x 12"W) supported by two stub shafts (1.688"Dia. x 12&quot rotating at 1750 rpm.

I am trying to determine the intertia of the drum, so I can determine the torque required to stop the load, immediately, if the drive belt breaks.

Any help would be greatly appreciated.

Thanks,

Doug

 

[ivymike]

you haven't stated whether the mass is evenly distributed throughout the drum.

You might take a look at

http://www.efunda.com/

for the basic equations.

 

[TheReg]

Yes, the weight is evenly distributed, throughout the drum.

 

[harrisj]

You can't stop it immediately!

It will need a finite length of time - the difference between 0.5 seconds and 10 seconds dictates what sort of solution you need.

Most conventional braking systems (eg a disc brake) are not constant torque, but extract energy at a constant rate.

Remeber you will have a lot of energy (heat) to dissipate.

Regards - John

 

[pso311]

I was attempting to calculate this out, but I was unsure had accurate you had to be. A really good reference is Mott's "machine Elements in Mechanical Design" 3rd edition.

pg 766 gives the equivalent load by combining all the inertia, gyration mumbo-jumbo into a couple simple equations.

The formula is:

Wk^2= [(R^4)- (r^4)(L)] / 323.9

Wk^2 is the inertia given in lb*ft
323.9 is a special case for steel (density)
R is outside radius in inches
r is inside radius in inches (if solid, this is zero)
L is the length of the component in inches

do this for the shafts and drum. the shafts inertia will be magnitudes smaller than the drum.
don't forget the T= f x D for the drum.
(1/2 diameter = distance)

After you find your Wk^2, plug into this eq. to find stopping torque:

T= [Wk^2 * RPM change]/ 308t

Since you want a total stop, your rpm change will be 1750. t is the time (in seconds) you want to stop the drum.

 

[desertfox]

Hi dougwdicar

try this thread its got the basic equation for change in angular momentum which can be used to find braking torque or starting torque.
The thread is for a starting torque situation.

thread404-54137

regards desertfox

 

[SERVOCAM]

If your drum is solid, the formula for inertia is:

Wr^2
Jdrum = ------
2g

Where:
Jdrum = Inertia of drum (lb-in-sec^2)
W = weight of drum (lbs)
r = radius of drum (Inches)
g = gravity (386 in/sec^2)

Given:
W = 400 lbs
r = 12
g = 386

Solve:

400 x 12^2
Jdrum = ----------
2 x 386

Jdrum = 74.61 lb-in-sec^2


For the inertia of the stub shafts, since we don't know the weight, we use a differnet formula (or use same and just find the weight of the stubs).

(pi)(L)(p)(r^4)
Jstub = --------------
2g

Where:
Jstub = Inertia of stub shaft (lb-in-sec^2)
pi = pi, 3.1415927
L = Length of stub shaft (inches)
p = material density (lb-in^3)
r = radius of stub shaft (inches)
g = gravity (386 in/sec^2)

Given: (1.688"Dia. x 12&quot
L = 12 inches
p = 0.28 lb-in^3 (Steel)
r = 0.844 inches

Solve:

(3.14..)(12)(0.28)(0.844^4)
Jstub = ---------------------------
(2)(386)

Jstub = 0.006838 lb-in-sec^2


Now add them all up:

Jtotal = Jdrum + (N)Jstub
Jtotal = 74.61 + (2)0.006838
Jtotal = 74.623676 lb-in-sec^2


Now as pointed out by others, in order to find the torque, we need to know the Decel rate or the time you want to take to stop.

If we make it simple and ignor friction, Decel Torque is defined as Inertia x rotational deceleration, where the deceleration rate is in radians per second squared

Td = J x Drad

Where:
Td = Decel torque (lb-in)
J = system inertia (lb-in-sec^2)
Drad = decel rate (rad/sec^2)

I will run through it for an example, and once you define your decel rate, you can re do it.

1750 rpm = 29.1666....but I will use 1800 rpm or 30 rev/sec. Decel rate can be found several ways, simpelest is Initial Velocity minus Final velocity all divided by decel time.

Vi - Vf
Decel rate = ----------
t

If we want to decel in 1 second from 30 rev/sec to 0 rev/sec, or Decel rate will be

30 - 0
Decel rate = ----------
1

Decel rate = 30 rev/sec^2

Now I need to convert this in to radians per second squared. There are 2pi radians per rev, so to convert, just multiply by 2 pi

Drad = 30 x 2pi
Drad = 188.5 rad/sec^2

Now we can plug this into our torque formula:


Td = J x Drad

Where:
Td = Decel torque (lb-in)
J = system inertia (lb-in-sec^2)
Drad = decel rate (rad/sec^2)

Td = 74.6 x 188.5
Td = 14062.1 lb-in

If you want lb-ft, just divide by 12

14062 / 12 = 1172 lb-ft of torque to stop the load.


Hope that helped.

Cameron Anderson - Sales & Applications Engineer
Aerotech, Inc. -

http://www.aerotech.com

"Dedicated to the Science of Motion"

 

[hacksaw]

good grief, I hope that helps a lot!