Question

A band brake having bandwidth of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 Nm. The maximum tension (in kN) developed in the band is

• A. $6.12$
• B. $3.56$
• C. $1.88$
• D. $11.56$

Answer 1

The correct option is D $11.56$

$M_t=(T_1-T_2).R$

Given, friction torque,

$M_t=1000Nm$

$1000=(T_1-T_2)\times 0.125$

$\Rightarrow T_1-T_2=8000N\left(i\right)$

$\frac{T_1}{T_2}=e^{\mu \theta }=e^{0.25\times \frac{270\times \pi }{180}}$

$=3.248$

$T_1=3.248T_2\left(ii\right)$

Putting $T_1$ from equation (ii) into equation (i)

$T_2=\frac{8000}{2.248}=3558.7N$

and $T_1=3.248\times 3558.7=11.56kN$