An automotive type internal-expanding double-shoe brake is shown in Fig. 12.18. The face width of the friction lining is 40 mm and the maximum intensity of normal pressure is limited to 1 N/mm². The coefficient of friction is 0.32. The angle $\theta_{l}$ can be assumed to be zero. Calculate:
(i) the actuating force P; and
(ii) the torque-absorbing capacity of the brake.

Question

An automotive type internal-expanding double-shoe brake is shown in Fig. 12.18. The face width of the friction lining is 40 mm and the maximum intensity of normal pressure is limited to 1 N/mm². The coefficient of friction is 0.32. The angle [latex] heta_{l} [/latex] can be assumed to be zero. Calculate:
(i) the actuating force P; and
(ii) the torque-absorbing capacity of the brake.

Accepted Answer

[latex] ext { Given } w=40 mm \quad \mu=0.32 \quad p_{\max }=1 N / mm ^{2} [/latex].

R = 125 mm
Step I Actuating force
It is assumed that the maximum normal pressure will occur between the lining on the right-hand shoe and the brake drum. For the right-hand shoe,

[latex] heta_{1}=0 \quad heta_{2}=120^{\circ} \quad \phi_{\max .}=90^{\circ} \quad \sin \phi_{\max .}=1 [/latex].

The distance h of the pivot from the axis of the brake drum is given by

[latex] h=\sqrt{86.6^{2}+50^{2}}=100 mm [/latex].

From Eq. (12.19),

[latex] M_{f}=\frac{\mu p_{\max } \cdot R w\left[4 R\left(\cos heta_{1}-\cos heta_{2} ight)-h\left(\cos 2 heta_{1}-\cos 2 heta_{2} ight) ight]}{4 \sin \phi_{\max .}} [/latex] (12.19).

[latex] M_{f}=\frac{\mu p_{\max } \cdot R w\left[4 R\left(\cos heta_{1}-\cos heta_{2} ight)-h\left(\cos 2 heta_{1}-\cos 2 heta_{2} ight) ight]}{4 \sin \phi_{\max } .} [/latex].

[latex] =\frac{0.32(1)(125)(40)\left[4(125)\left(1-\cos 120^{\circ} ight)-100\left(1-\cos 240^{\circ} ight) ight]}{4(1)} [/latex].

= 240 000 N-mm
From Eq. (12.20),

[latex] M_{n}=\frac{p_{\max } R w h\left[2\left( heta_{2}- heta_{1} ight)-\left(\sin 2 heta_{2}-\sin 2 heta_{1} ight) ight]}{4 \sin \phi_{\max }} [/latex] (12.20).

[latex] M_{n}=\frac{p_{\max } R w h\left[2\left( heta_{2}- heta_{1} ight)-\left(\sin 2 heta_{2}-\sin 2 heta_{1} ight) ight]}{4 \sin \phi_{\max .}} [/latex].

[latex] =\frac{1(125)(40)(100)\left[2\left(\frac{120 \pi}{180} ight)-\sin \left(240^{\circ} ight) ight]}{4(1)} [/latex].

= 631 851.95 N-mm
From Eq. (12.22),

[latex] P=\frac{M_{n}-M_{f}}{C} [/latex] (12.22).

[latex] P=\frac{M_{n}-M_{f}}{C} [/latex].

[latex] =\frac{631851.95-240000}{100.9+86.6}=2089.88 N [/latex] (i).

Step II Torque-absorbing capacity
From Eq. (12.21), the torque [latex] \left(M_{t} ight)_{R} [/latex] for the righthand
shoe is given by

[latex] M_{t}=\frac{\mu R^{2} p_{\max .} w\left(\cos heta_{1}-\cos heta_{2} ight)}{\sin \phi_{\max }} [/latex] (12.21).

[latex] \left(M_{t} ight)_{R}=\frac{\mu R^{2} p_{\max } \cdot w\left(\cos heta_{1}-\cos heta_{2} ight)}{\sin \phi_{\max }} [/latex].

[latex] =\frac{0.32(125)^{2}(1)(40)\left(1-\cos 120^{\circ} ight)}{1} [/latex].

= 300 000 N-mm.

The maximum intensity of pressure for the left-hand shoe is unknown. For identical shoes, it can be seen from the expressions of [latex] M_{n} ext { and } M_{f} [/latex], that both are proportional to [latex] \left(p_{\max } ight) [/latex]. For left-hand shoe, the maximum intensity of pressure is taken as [latex] \left(p_{\max }^{\prime} ight) [/latex]. Therefore, for the left-hand shoe

[latex] M_{f}^{\prime}=\frac{240000 p_{\max }^{\prime}}{p_{\max }}=\frac{(240000) p_{\max }^{\prime}}{(1)} [/latex].

[latex] =240000 p_{\max }^{\prime} [/latex].

Similarly,

[latex] M_{n}^{\prime}=\frac{631851.95 p_{\max }^{\prime}}{p_{\max }}=\frac{(631851.95) p_{\max .}^{\prime}}{(1)} [/latex].

[latex] =631851.95 p_{\max }^{\prime} [/latex].

For the left-hand shoe,

[latex] P=\frac{M_{n}^{\prime}+M_{f}^{\prime}}{C} [/latex].

[latex] ext { or } \quad 2089.88=\frac{(240000+631851.95) p_{\max }^{\prime}}{(100.9+86.6)} [/latex].

∴ [latex] p_{\max .}=0.45 N / mm ^{2} [/latex].

Since the shoes are identical,

[latex] \left(M_{t} ight)_{L}=300000\left(\frac{p_{\max }^{\prime}}{p_{\max }} ight)=300000\left(\frac{0.45}{1} ight) [/latex].

= 135000 N-mm.

The total torque-absorbing capacity of the brake is given by,

[latex] M_{t}=300000+135000=435000 N - mm [/latex].

or 435 N-m (ii).

Question 12.8: An automotive type internal-expanding double-shoe brake is s...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions