Question 13.9: Design of a Short-Shoe Drum Brake The brake shown in Figure ......
Design of a Short-Shoe Drum Brake
The brake shown in Figure 13.21 uses a sintered metal lining having design values of f =0.4 and p_{max} =150 psi. Determine:
a. The torque capacity and actuating force.
b. The reaction at pivot A .
Given: a=12 \text { in., } w=3 \text { in., } b=5 \text { in., } c=2 \text { in., } r=4 \text { in., } \phi=30^{\circ} \text {. }
a. From Equation (13.47), we have
F_n=p_{\max }\left[2\left(r \sin \frac{\phi}{2}\right)\right] w (13.47)
F_n=150\left[2\left(4 \sin \frac{30^{\circ}}{2}\right)\right] 3=931.7 lb
Equation (13.48) yields
T=f F_n r (13.48)
T=(0.4)(931.7)(4)=1.491 kip \cdot in
Applying Equation (13.49),
F_a=\frac{F_n}{a}(b-f c) (13.49)
F_a=\frac{931.7(5-0.4 \times 2)}{12}=326.1 lb
b. The conditions of equilibrium of the horizontal (x) and vertical (y) forces give
R_{A x}=931.7(0.4)=372.7 lb \quad R_{A y}=605.6 lb
The resultant radial reactional force is
R_A=\sqrt{372.7^2+605.6^2}=711.1 lb