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} .



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.