Clutches and brakes

Notation: $r_o$ = outer radius, $r_i$ = inner radius of the friction face, $\mu$ = coefficient of friction, $F$ = total axial force, $p$ = intensity of pressure.

Consider a thin ring of radius $r$ and width $dr$. Area $= 2\pi r\,dr$; normal force $dF = p\cdot2\pi r\,dr$; friction force $= \mu\,dF$; friction torque $dT = \mu\,dF\cdot r = 2\pi\mu p\,r^2dr$.

   ro ____
     / ri \   ring of radius r, width dr
    |  O  |   dF = p 2 pi r dr
     \____/

(a) Uniform pressure ($p$ constant)

Axial force:

$$F = \int_{r_i}^{r_o} p\,2\pi r\,dr = \pi p\,(r_o^2 - r_i^2)$$

Torque:

$$T = 2\pi\mu p\int_{r_i}^{r_o} r^2dr = \frac{2\pi\mu p}{3}(r_o^3 - r_i^3)$$

Eliminating $p$:

$$T = \frac{2}{3}\,\mu F\,\frac{r_o^3 - r_i^3}{r_o^2 - r_i^2}$$

The mean radius is $R_m = \dfrac{2}{3}\dfrac{r_o^3-r_i^3}{r_o^2-r_i^2}$, so $T = \mu F R_m$.

(b) Uniform wear

Wear is proportional to pressure times rubbing velocity, i.e. $p\,r\omega$. For uniform wear $p\,r = C$ (constant), so pressure is greatest at the inner radius: $p_{max} = C/r_i$.

Axial force:

$$F = \int_{r_i}^{r_o} \frac{C}{r}\,2\pi r\,dr = 2\pi C\,(r_o - r_i)$$

Torque:

$$T = 2\pi\mu C\int_{r_i}^{r_o} r\,dr = \pi\mu C\,(r_o^2 - r_i^2)$$

Eliminating $C$:

$$T = \mu F\,\frac{r_o + r_i}{2} = \mu F R_m$$

Design assumption

Uniform wear is used for design of clutches already run in. A new clutch follows uniform pressure, but after initial wear the pressure redistributes; the uniform wear torque is smaller (e.g. for $r_i = 0.6r_o$ it is about 2% lower), hence safer. Uniform wear also gives the maximum pressure at the inner radius, $p_{max} = F/[2\pi r_i(r_o-r_i)]$, which must not exceed the allowable value of the lining. For $n$ friction surfaces multiply the torque by $n$.

Data

$r_o = 100$ mm, $r_i = 60$ mm, $\mu = 0.12$, $p_{max} = 0.3$ N/mm², $P = 30$ kW, $N = 1200$ rpm. Uniform wear.

Torque to be transmitted

$$T = \frac{P}{\omega} = \frac{30\,000}{2\pi\times1200/60} = \frac{30000}{125.66} = 238.7\ \text{N m}$$

Axial force from the pressure limit

For uniform wear, maximum pressure is at the inner radius and $p_{max}\,r_i = C$:

$$F = 2\pi C(r_o - r_i) = 2\pi p_{max} r_i (r_o - r_i)$$ $$F = 2\pi\times0.3\times60\times(100-60) = 4524\ \text{N}$$

Torque per friction surface

Mean radius $R_m = \dfrac{r_o+r_i}{2} = 80$ mm.

$$T_1 = \mu F R_m = 0.12\times4524\times0.080 = 43.4\ \text{N m}$$

Number of friction surfaces

$$n = \frac{T}{T_1} = \frac{238.7}{43.4} = 5.5 \;\Rightarrow\; n = 6$$

Number of discs

Number of friction surfaces $= n = n_1 + n_2 - 1$, where $n_1$ and $n_2$ are the numbers of discs on the driving and driven shafts.

$$n_1 + n_2 = 7 \;\Rightarrow\; n_1 = 4,\ n_2 = 3$$

(or 3 and 4).

Actual axial force and pressure with $n = 6$

$$F = \frac{T}{n\mu R_m} = \frac{238.7}{6\times0.12\times0.080} = 4145\ \text{N}$$ $$p_{max} = \frac{F}{2\pi r_i(r_o - r_i)} = \frac{4145}{2\pi\times60\times40} = 0.275\ \text{MPa} < 0.3\ \text{MPa}$$

Answer: 6 friction surfaces (4 discs on one shaft, 3 on the other); axial force 4145 N; $p_{max} = 0.275$ MPa.

Notation: $\alpha$ = semi-cone angle, $R$, $r$ = outer and inner radii of the cone face, $\mu$ = friction coefficient, $p_n$ = normal pressure, $F_n$ = total normal force on the cone surface, $F_a$ = axial force.

         axial force Fa
              |
        ______v_____
       /  \  alpha  \
      R    \ slant   r
       \____\_______/

Torque (uniform wear)

Take an elementary ring of radius $x$ and slant width $dx/\sin\alpha$. Area $= 2\pi x\,dx/\sin\alpha$. Normal force $dF_n = p_n\,2\pi x\,dx/\sin\alpha$. Friction torque:

$$dT = \mu\,dF_n\,x = \frac{2\pi\mu p_n}{\sin\alpha}x^2dx$$

For uniform wear $p_n x = C$:

$$F_n = \frac{2\pi C}{\sin\alpha}\int_r^R dx = \frac{2\pi C (R-r)}{\sin\alpha}$$ $$T = \frac{2\pi\mu C}{\sin\alpha}\int_r^R x\,dx = \frac{\pi\mu C (R^2-r^2)}{\sin\alpha}$$

Eliminating $C$:

$$T = \mu F_n R_m, \qquad R_m = \frac{R+r}{2}$$

The axial force transmitted by the cone face in steady running is $F_a = F_n\sin\alpha$, so $T = \dfrac{\mu F_a R_m}{\sin\alpha}$.

(For uniform pressure, $R_m = \dfrac{2}{3}\dfrac{R^3-r^3}{R^2-r^2}$ and $F_n = \pi p_n(R^2-r^2)/\sin\alpha$.)

Axial force for engagement

While engaging, friction acts along the cone surface opposing the motion of the cone into the seat. Resolving along the axis:

$$F_e = F_n(\sin\alpha + \mu\cos\alpha)$$

Hence the clutch spring force needed to engage is $F_e = F_n(\sin\alpha+\mu\cos\alpha)$; once engaged and running, $F_a = F_n\sin\alpha$ is enough to hold it.

Advantages over plate clutch

• The wedge action gives a normal force $F_n = F_a/\sin\alpha$ much larger than the axial force, so for the same $F_a$ a cone clutch transmits more torque (about $1/\sin\alpha$ times).
• Smaller diameter for the same torque.
• No need for many plates.

Disadvantages: more expensive to make, needs close alignment, tends to stick.

Condition to avoid sticking

The cone must release easily: $\alpha > \tan^{-1}\mu$ (the semi-cone angle must be greater than the friction angle). Practical values are $\alpha = 12.5^\circ$ (for leather and cork linings with $\mu\approx0.2$ to $0.35$), up to 20 degrees.

Data

$D = 500$ mm ($r = 0.25$ m), $N = 300$ rpm, $P = 15$ kW, $\mu = 0.25$, $\theta = 270^\circ = 4.712$ rad, $a = 60$ mm, $l = 700$ mm.

(a) Torque and tensions

$$\omega = \frac{2\pi\times300}{60} = 31.42\ \text{rad/s}, \qquad T_B = \frac{P}{\omega} = \frac{15000}{31.42} = 477.5\ \text{N m}$$

Net braking force on the band: $T_1 - T_2 = \dfrac{T_B}{r} = \dfrac{477.5}{0.25} = 1910\ \text{N}$.

Ratio of tensions (belt friction): $\dfrac{T_1}{T_2} = e^{\mu\theta} = e^{0.25\times4.712} = e^{1.178} = 3.248$.

$$T_2 = \frac{1910}{3.248-1} = 849.5\ \text{N}, \qquad T_1 = 3.248\times849.5 = 2759.4\ \text{N}$$

  fulcrum O    pin A (a=60)         force F
     |-----------|------------------->| l=700
      \band ends/
        drum with tight T1, slack T2

(b) Lever force

Taking moments about the fulcrum, $F\,l = T_{pin}\,a$.

• Slack end on the pin ($T_{pin} = T_2$):

$$F = \frac{T_2\,a}{l} = \frac{849.5\times60}{700} = 72.8\ \text{N}$$

• Tight end on the pin ($T_{pin} = T_1$):

$$F = \frac{T_1\,a}{l} = \frac{2759.4\times60}{700} = 236.5\ \text{N}$$

The slack-end arrangement needs about one third of the force, so it is used in the direction of main rotation.

(c) Band width

Maximum tension is $T_1$. With joint efficiency $\eta$:

$$T_1 = \sigma\,b\,t\,\eta \;\Rightarrow\; b = \frac{2759.4}{60\times3\times0.8} = 19.2\ \text{mm} \;\Rightarrow\; 20\ \text{mm}$$

Answer: $T_B = 477.5$ N m, $T_1 = 2759$ N, $T_2 = 850$ N; lever force 72.8 N (slack end on pin) or 236.5 N (tight end on pin); band width 20 mm.

Method

For a pivoted shoe, the normal pressure is proportional to $\sin\theta$ ($\theta$ from the pivot line): $p = p_a\sin\theta/\sin\theta_a$, where $\theta_a$ is the angle of maximum pressure. As $\theta_2 = 120^\circ > 90^\circ$, $\theta_a = 90^\circ$ and $\sin\theta_a = 1$.

Data: $r = 0.15$ m, $a = 0.11$ m, $b = 0.04$ m, $\theta_1 = 10^\circ$, $\theta_2 = 120^\circ$, $p_a = 0.8\times10^6$ Pa, $\mu = 0.28$, $c = 0.2$ m.

         drum centre O
            |  a = 110
   pivot A -+--- shoe from 10 deg to 120 deg
            |
        F at c = 200 from A

Braking torque

$$T = \mu p_a b\,r^2(\cos\theta_1 - \cos\theta_2)$$ $$T = 0.28\times0.8\times10^6\times0.04\times0.15^2\times(0.9848 + 0.5) = 299.3\ \text{N m}$$

Moment of the normal forces about the pivot

$$M_N = p_a b\,r\,a\int_{\theta_1}^{\theta_2}\sin^2\theta\,d\theta = p_a b\,r\,a\left[\frac{\theta_2-\theta_1}{2} - \frac{\sin2\theta_2-\sin2\theta_1}{4}\right]$$

The bracket $= \dfrac{1.9199}{2} - \dfrac{(-0.8660) - 0.3420}{4} = 0.9599 + 0.3020 = 1.2619$ (angles in radians: $\theta_2-\theta_1 = 1.9199$ rad).

$$M_N = 0.8\times10^6\times0.04\times0.15\times0.11\times1.2619 = 666.3\ \text{N m}$$

Moment of friction forces about the pivot

$$M_f = \mu p_a b\,r\int_{\theta_1}^{\theta_2}\sin\theta\,(r - a\cos\theta)\,d\theta = \mu p_a b\,r\left[r(\cos\theta_1-\cos\theta_2) - \frac{a}{2}(\sin^2\theta_2 - \sin^2\theta_1)\right]$$ $$M_f = 0.28\times0.8\times10^6\times0.04\times0.15\,[\,0.15\times1.4848 - 0.055\times(0.75-0.0302)\,] = 246.1\ \text{N m}$$

Actuating force

• Self-energising (friction moment assists the actuating force): $F\,c = M_N - M_f$

$$F = \frac{666.3 - 246.1}{0.2} = 2101\ \text{N}$$

• Non-self-energising (friction moment opposes): $F\,c = M_N + M_f$

$$F = \frac{666.3 + 246.1}{0.2} = 4562\ \text{N}$$

The self-energising (leading) shoe needs less than half the force for the same maximum pressure.

Self-locking

Self-locking occurs when $M_f \ge M_N$ ($F \le 0$). $M_f$ is proportional to $\mu$:

$$\mu_{lock} = 0.28\times\frac{666.3}{246.1} = 0.76$$

Answer: $T = 299$ N m per shoe; $F = 2.10$ kN (self-energising), $4.56$ kN (non-self-energising); self-locking at $\mu \ge 0.76$, so the brake is not self-locking at $\mu = 0.28$.

Data

$m = 1400$ kg, $v = 90$ km/h $= 25$ m/s, $t = 6$ s.

(a) Energy and power

Kinetic energy to be absorbed:

$$E = \tfrac12 m v^2 = \tfrac12\times1400\times25^2 = 437\,500\ \text{J}$$

Per brake: $E_1 = 437\,500/4 = 109\,375$ J $= 109.4$ kJ.

Average retardation: $a = v/t = 25/6 = 4.17\ \text{m/s}^2$.

Stopping distance: $s = \dfrac{v t}{2} = \dfrac{25\times6}{2} = 75$ m.

Average power per brake: $P_1 = \dfrac{E_1}{t} = \dfrac{109\,375}{6} = 18.2$ kW (the peak is twice as large at the start for uniform retardation).

(b) Temperature rise (no heat loss)

$$E_1 = m_d\,c\,\Delta T \;\Rightarrow\; \Delta T = \frac{109\,375}{7\times500} = 31.3\ ^\circ\text{C}$$

A drum of larger mass, finned surface or repeated braking needs the cooling to be checked.

(c) Descending a slope

Slope angle: $\sin\theta = 1/20 \approx 0.05$ and $v = 36$ km/h $= 10$ m/s. At steady speed the brakes absorb the loss of potential energy:

$$P = m g\,v\sin\theta = 1400\times9.81\times10\times0.04994 = 6859\ \text{W}$$

Per brake: $6859/4 = 1715$ W.

At steady state, heat generated equals heat dissipated: $Q = hA\,\Delta T$:

$$\Delta T = \frac{1715}{18} = 95.3\ ^\circ\text{C}$$

This is acceptable for cast-iron drums, but on long descents the temperature may rise further and cause brake fade (the coefficient of friction falls with temperature), so engine braking is used.

Answer: (a) 109.4 kJ per brake, retardation 4.17 m/s², distance 75 m, 18.2 kW; (b) 31.3 °C; (c) 1.72 kW per brake, steady rise 95 °C.

Clutch vs brake

Self-energising

A brake is self-energising when the friction force on the shoe produces a moment about the pivot that helps the actuating force press the shoe against the drum. This increases the normal force and therefore the friction and the braking torque for the same actuating force. Self-energising occurs when the friction moment $M_f$ acts in the same sense as the actuating moment.

Self-locking

If the friction moment exceeds the moment of the normal force, $M_f \ge M_N$, the shoe pulls itself into the drum even with zero (or negative) actuating force and the brake locks the drum. It is dangerous: it gives jerky braking or cannot be released. It occurs for a large coefficient of friction or when the pivot is badly placed.

Leading and trailing shoes

In an internal expanding brake:

• The leading (primary) shoe has its pivot at the trailing end, so the drum drags the shoe into contact. It is self-energising: more braking torque for the same force.
• The trailing (secondary) shoe has the pivot at the leading end, so the drum rotation tends to drag the shoe away from the drum. It is non-self-energising: friction reduces the pressure.

Hence the leading shoe does more work and wears faster; a brake with a leading and a trailing shoe is stable, which avoids the self-locking risk of two leading shoes.