Question 13.10: Design of a Long-Shoe Drum Brake The long-shoe drum brake is......

The angle of contact is \phi=90^{\circ} \text { or } \pi / 2  rad . From the geometry, \alpha=\tan ^{-1}(200 / 150)=53.13^{\circ} . Hence,

\theta_1=8.13^{\circ}, \quad \theta_2=98.13^{\circ}

c=\sqrt{200^2+150^2}=250  mm

Inasmuch as \theta_2>90^{\circ},(\sin \theta)_m=1 .

a. Through the use of Equation (13.53),

M_n=\frac{w r c p_{\max }}{4(\sin \theta)_m}\left[2 \phi-\sin 2 \theta_2+\sin 2 \theta_1\right]         (13.53)

\begin{aligned} M_n & =\frac{(0.075)(0.15)(0.25) p_{\max }}{4(1)}\left[2\left(\frac{\pi}{2}\right)-\sin 196.26^{\circ}+\sin 16.26^{\circ}\right] \\ & =2.6\left(10^{-3}\right)  p_{\max } \end{aligned}

From Equation (13.54),

M_f=\frac{f w r p_{\max }}{4(\sin \theta)_m}\left[c\left(\cos 2 \theta_2-\cos 2 \theta_1\right)-4 r\left(\cos \theta_2-\cos \theta_1\right)\right]         (13.54)

\begin{aligned} M_f & =\frac{0.35(0.075)(0.15) p_{\max }}{4} \\ & {\left[(0.25)\left(\cos 196.26^{\circ}-\cos 16.26^{\circ}\right)-4(0.15)\left(\cos 98.13^{\circ}-\cos 8.13^{\circ}\right)\right] } \\ & =0.196\left(10^{-3}\right)  p_{\max } \end{aligned}

Applying Equation (13.55), we then have

F_a=\frac{1}{a}\left(M_n \mp M_f\right)       (13.55)

4000(0.45)=(2.6+0.196)\left(10^{-3}\right)  p_{\max }

or

p_{\max }=644  kPa

b. Using Equation (13.57),

T=\frac{f w r^2 p_{\max }}{(\sin \theta)_m}\left(\cos \theta_1-\cos \theta_2\right)       (13.57)

\begin{aligned} T & =\frac{(0.35)(0.075)\left(0.15^2\right)\left(0.644 \times 10^2\right)}{1}\left(\cos 8.13^{\circ}-\cos 98.13^{\circ}\right) \\ & =430.3  N \cdot m \end{aligned}

By Equation (1.15), the corresponding power is

kW =\frac{F V}{1000}=\frac{T n}{9549}         (1.15)

kW =\frac{T n}{9549}=\frac{430.3(250)}{9549}=11.28