Chapter 13
Q. 13.8
Design of a Differential Band Brake
A differential band brake similar to that in Figure 13.20 uses a woven lining having design values of f =0.3 and p_{max} =375 kPa. Determine:
a. The torque capacity.
b. The actuating force.
c. The power capacity.
d. The value of dimension s that would cause the brake to be self-locking.
Given: The speed is 250 rpm, a=500 mm , c=150 mm , w=60 mm , r=200 mm , s=25 mm , \text { and } \phi=270^{\circ} .
Step-by-Step
Verified Solution
a. Through the use of Equation (13.45), we obtain
F_1=w r p_{\max } (13.45)
F_1=w r p_{\max }=(0.06)(0.2)(375)=4.5 kN
Applying Equation (13.44),
\frac{F_1}{F_2}= e ^{f \phi} (13.44)
F_2=\frac{F_1}{e^{f \phi}}=\frac{4.5}{e^{0.3(1.5 \pi)}}=1.095 kN
Then, Equation (13.42) gives T =(4.5 − 1.095)(0.2)=0.681 k N · m.
T=\left(F_1-F_2\right) r (13.42)
b. By Equation (13.46),
F_a=\frac{1}{a}\left(c F_2-s F_1\right) (13.46)
F_a=\frac{150(1.095)-25(4.5)}{0.5}=103.5 N
c. From Equation (1.15),
kW =\frac{F V}{1000}=\frac{T n}{9549} (1.15)
kW =\frac{T n}{9549}=\frac{681(250)}{9549}=17.8
d. Using Equation (13.46), we have F_a =0 for s=150(1.095)/4.5=36.5 mm.
Comment: The brake is self-locking if s ≥ 36.5 mm.