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

Design of a Long-Shoe Drum Brake

The long-shoe drum brake is actuated by a mechanism that exerts a force of F_{a} = 4 kN (Figure 13.23). Determine

a. The maximum pressure

b. The torque and power capacities

Design Decision: The lining is a molded material having a coefficient of friction f = 0.35 and a width w = 75 mm.

F13.23
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The angle of contact is \phi = 90° or \pi / 2 rad. From the geometry, \alpha=\tan ^{-1}(200 / 150)=53.13^{\circ} .
Hence,

\begin{array}{l} \theta_1=8.13^{\circ}, \quad \theta_2=98.13^{\circ} \\ c=\sqrt{200^2+150^2}=250  mm \end{array}

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\lgroup \frac{\pi}{2} \right\rgroup -\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 10^6\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

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