disk brakes

Budynas−Nisbett: Shigley’s

Mechanical Engineering

Design, Eighth Edition

III. Design of Mechanical

Elements

16. Clutches, Brakes,

Couplings, and Flywheels

826 © The McGraw−Hill

Companies, 2008

Clutches, Brakes, Couplings, and Flywheels 829

Figure 16–18

An automotive disk brake.(Courtesy DaimlerChryslerCorporation.)

WheelCaliper

Boot Seal

Piston

Brake fluid

Adapter

Mounting bolt

Splash shield

Braking disk

Outer

bearing

Spindle

Wheel stud

Shoe and

lining

Steering

knuckle

Inner

bearing Seal

16–6 Disk BrakesAs indicated in Fig. 16–16, there is no fundamental difference between a disk clutch

and a disk brake. The analysis of the preceding section applies to disk brakes too.

We have seen that rim or drum brakes can be designed for self-energization. While

this feature is important in reducing the braking effort required, it also has a disadvantage.

When drum brakes are used as vehicle brakes, only a slight change in the coefficient of

friction will cause a large change in the pedal force required for braking. A not unusual 30

percent reduction in the coefficient of friction due to a temperature change or moisture, for

example, can result in a 50 percent change in the pedal force required to obtain the same

braking torque obtainable prior to the change. The disk brake has no self-energization, and

hence is not so susceptible to changes in the coefficient of friction.

Another type of disk brake is the floating caliper brake, shown in Fig. 16–18. The

caliper supports a single floating piston actuated by hydraulic pressure. The action is

much like that of a screw clamp, with the piston replacing the function of the screw. The

floating action also compensates for wear and ensures a fairly constant pressure over

the area of the friction pads. The seal and boot of Fig. 16–18 are designed to obtain

clearance by backing off from the piston when the piston is released.

Caliper brakes (named for the nature of the actuating linkage) and disk brakes

(named for the shape of the unlined surface) press friction material against the face(s)





Elements



827© The McGraw−Hill

Companies, 2008

830 Mechanical Engineering Design

of a rotating disk. Depicted in Fig. 16–19 is the geometry of an annular-pad brake con-

tact area. The governing axial wear equation is Eq. (12–27), p. 643,

w = f1 f2 K PV t

The coordinate r̄ locates the line of action of force F that intersects the y axis. Of

interest also is the effective radius re, which is the radius of an equivalent shoe of infin-

itesimal radial thickness. If p is the local contact pressure, the actuating force F and the

friction torque T are given by

F =∫ θ2

θ1

∫ ro

ri

pr dr dθ = (θ2 − θ1)

∫ ro

ri

pr dr (16–29)

T =∫ θ2

θ1

∫ ro

ri

f pr2 dr dθ = (θ2 − θ1) f

∫ ro

ri

pr2 dr (16–30)

The equivalent radius re can be found from f Fre = T , or

re = T

f F=

∫ ro

ri

pr2 dr

∫ ro

ri

pr dr

(16–31)

The locating coordinate r̄ of the activating force is found by taking moments about the

x axis:

Mx = Fr̄ =∫ θ2

θ1

∫ ro

ri

pr(r sin θ) dr dθ = (cos θ1 − cos θ2)

∫ ro

ri

pr2 dr

r̄ = Mx

F= (cos θ1 − cos θ2)

θ2 − θ1

re (16–32)

Uniform Wear

It is clear from Eq. (12–27) that for the axial wear to be the same everywhere, the prod-

uct PV must be a constant. From Eq. (a), Sec. 16–5, the pressure p can be expressed

in terms of the largest allowable pressure pa (which occurs at the inner radius ri ) as

Figure 16–19

Geometry of contact area ofan annular-pad segment of acaliper brake.

y

xro

1

2ri

FF

r





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p = pari/r . Equation (16–29) becomes

F = (θ2 − θ1)pari (ro − ri ) (16–33)

Equation (16–30) becomes

T = (θ2 − θ1) f pari

∫ ro

ri

r dr = 1

2(θ2 − θ1) f pari

(

r2o − r2

i

)

(16–34)


re =pari

∫ ro

ri

r dr

pari

∫ ro

ri

dr

= r2o − r2

i

2

1

ro − ri

= ro + ri

2(16–35)


r̄ = cos θ1 − cos θ2

θ2 − θ1

ro + ri

2(16–36)

Uniform Pressure

In this situation, approximated by a new brake, p = pa . Equation (16–29) becomes

F = (θ2 − θ1)pa

∫ ro

ri

r dr = 1

2(θ2 − θ1)pa

(

r2o − r2

i

)

(16–37)


T = (θ2 − θ1) f pa

∫ ro

ri

r2 dr = 1

3(θ2 − θ1) f pa

(

r3o − r3

i

)

(16–38)


re =pa

∫ ro

ri

r2 dr

pa

∫ ro

ri

r dr

= r3o − r3

i

3

2

r2o − r2

i

= 2

3

r3o − r3

i

r2o − r3

i

(16–39)


r̄ = cos θ1 − cos θ2

θ2 − θ1

2

3

r3o − r3

i

r2o − r2

i

= 2

3

r3o − r3

i

r2o − r2

i

cos θ1 − cos θ2

θ2 − θ1

(16–40)

EXAMPLE 16–3 Two annular pads, ri = 3.875 in, ro = 5.50 in, subtend an angle of 108◦, have a co-

efficient of friction of 0.37, and are actuated by a pair of hydraulic cylinders 1.5 in in

diameter. The torque requirement is 13 000 lbf · in. For uniform wear

(a) Find the largest normal pressure pa .

(b) Estimate the actuating force F.

(c) Find the equivalent radius re and force location r̄ .

(d) Estimate the required hydraulic pressure.





Elements



829© The McGraw−Hill

Companies, 2008

832 Mechanical Engineering Design

Solution (a) From Eq. (16–34), with T = 13 000/2 = 6500 lbf · in for each pad,

Answer pa = 2T

(θ2 − θ1) f ri

(

r2o − r2

i

)

= 2(6500)

(144◦ − 36◦)(π/180)0.37(3.875)(5.52 − 3.8752)= 315.8 psi

(b) From Eq. (16–33),

Answer F = (θ2 − θ1)pari (ro − ri ) = (144◦ − 36◦)(π/180)315.8(3.875)(5.5 − 3.875)

= 3748 lbf

(c) From Eq. (16–35),

Answer re = ro + ri

2= 5.50 + 3.875

2= 4.688 in

From Eq. (16–36),

Answer r̄ = cos θ1 − cos θ2

θ2 − θ1

ro + ri

2= cos 36◦ − cos 144◦

(144◦ − 36◦)(π/180)

5.50 + 3.875

2

= 4.024 in

(d ) Each cylinder supplies the actuating force, 3748 lbf.

Answer phydraulic = F

AP

= 3748

π(1.52/4)= 2121 psi

Circular (Button or Puck) Pad Caliper Brake

Figure 16–20 displays the pad geometry. Numerical integration is necessary to ana-

lyze this brake since the boundaries are difficult to handle in closed form. Table 16–1

gives the parameters for this brake as determined by Fazekas. The effective radius is

given by

re = δe (16–41)

The actuating force is given by

F = π R2 pav (16–42)

and the torque is given by

T = f Fre (16–43)

Figure 16–20

Geometry of circular pad of acaliper brake.

e

R





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EXAMPLE 16–4 A button-pad disk brake uses dry sintered metal pads. The pad radius is 12

in, and its

center is 2 in from the axis of rotation of the 3 12-in-diameter disk. Using half of the

largest allowable pressure, pmax = 350 psi, find the actuating force and the brake

torque. The coefficient of friction is 0.31.

Solution Since the pad radius R = 0.5 in and eccentricity e = 2 in,

R

e= 0.5

2= 0.25

From Table 16–1, by interpolation, δ = 0.963 and pmax/pav = 1.290. It follows that the

effective radius e is found from Eq. (16–41):

re = δe = 0.963(2) = 1.926 in

and the average pressure is

pav = pmax/2

1.290= 350/2

1.290= 135.7 psi

The actuating force F is found from Eq. (16–42) to be

Answer F = π R2 pav = π(0.5)2135.7 = 106.6 lbf (one side)

The brake torque T is

Answer T = f Fre = 0.31(106.6)1.926 = 63.65 lbf · in (one side)

16–7 Cone Clutches and BrakesThe drawing of a cone clutch in Fig. 16–21 shows that it consists of a cup keyed or splined

to one of the shafts, a cone that must slide axially on splines or keys on the mating shaft,

and a helical spring to hold the clutch in engagement. The clutch is disengaged by means

of a fork that fits into the shifting groove on the friction cone. The cone angle α and the

diameter and face width of the cone are the important geometric design parameters. If the

R

e

re

e

pmax

pav

0.0 1.000 1.000

0.1 0.983 1.093

0.2 0.969 1.212

0.3 0.957 1.367

0.4 0.947 1.578

0.5 0.938 1.875

Table 16–1

Parameters for a

Circular-Pad Caliper

Brake

Source: G. A. Fazekas, “OnCircular Spot Brakes,” Trans.ASME, J. Engineering forIndustry, vol. 94, Series B,No. 3, August 1972,pp. 859–863.

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