A simple band brake is operated by lever 40 cm long. The brake drum

diameter is 40 cm and brake band embrance 5/8 of its circumference. One end

of band is attached to a fulcrum of lever while other end attached to pin 8 cm

from fulcrum. The co-efficient of friction is 0.25. The effort applied at the end

of lever is 500 N. Find braking torque applied if drum rotates anti-clockwise

and acts downwards.

Simple band brake:

Given: Length of lever l = 40 cm = 0.4m, diameter d = 40 cm = 0.4, µ = 0.25, b =0 .08 m ϴ = Angle of wrap = 5/8 x 360 = 225 x π /180 = 3.93 rad Braking torque = (T1 –T2) x r T1/T2 = e µϴ = e 0.25 x 3.93 = 2.67 Taking moments about fulcrum P x l = b x T1 500 x 0.40 = 0.08 x T1 T1 = 2500 N T2 = 2500 / 2.67 = 936.3 N Braking Torque = (2500 – 936.3) x 0.2 = 312.74 N-m