Band brakes

A simple band brake operates on a drum of 600 mm in diameter *Also
rotating counter clockwise and is running at 200 rpm. The coefficient F1  3.2482 F2  3.2482  2477.723 N 
of friction is 0.25. The brake band has a contact of 270°, one end is
fastened to a fixed pin and the other end to the brake arm 125 mm F1  8048.14 N
from the fixed pin. The straight brake arm is 750 mm long and placed
perpendicular to the diameter that bisects the angle of contact. *Taking the sum of moments at pivot O:
Determine (a) the pull (P) necessary on the end of the brake arm to  MO  0
stop the wheel if 35 kW is being absorbed and (b) the width of steel
band of 2.5 mm thick if the maximum tensile stress is not to exceed P  750 mm   F2 OD  0 
50 MPa?
* Solving for distance OD:
  mm
sin 45 
OD
OD  88.388 mm

* Therefore, solving for the required pull (P):

P
    2477.723 N 88.388 mm 
F2 OD
750 mm 750 mm
P  292 N
Given:
Dd = 600 mm (Rd = 300 mm) CCW θ = 270o * The width of the steel band can be solved from its maximum stress:
n = 200 rpm Pbrake = 35 kW
F F
μ = 0.25 in. Smax = 50 MPa S max  1  1
A wt
F1 8048.14 N
Solution: w 
S max t  50 N/mm   2.5 mm 
2



* From the power absorbed by the brake, we can solve for the w  64.385 mm
braking torque by:
Pbrake  2 Tf n

Pbrake 35 kJ/s
Tf  
2 n 
2  200
rev  1min 
 
 min  60 sec 
Tf  1.6711 kJ  1671.125 J

* The belt tension ratio is given by:
F1
e

e
0.25  
270 
180

F2

F1
 3.2482
F2
F1  3.2482 F2

* Also, the braking torque can be expressed as:
Tf   F1  F2  r

Tf 1671.125 J
F  F  
1 2

r 0.3 m
 F1  F2   5570.4167 N
*Substituting:
 3.2482F2  F2   5570.4167 N
F2  2477.723 N