mechanical design an integrated approach Part 10 pptx

The materials used for clutches and brakes are of two types, those used for the disk or drums and those used for friction materials or linings.. Tables 13.11 and 13.12 list Table 13.11

Since V belts are usually made from reinforced rubber, the required belt strength is governed mainly by the tension; that is, the effect of the additional force in the belt due to bending around the pulley may be neglected However, the tight-side tension should be multiplied by a service factor Ky The maximum tension is then

(13.22)

Service factors are listed by the manufacturers in great detail, usually based on the number

of hours per day of overload, variations in loading the driving and driven shafts, starting overload, and variations in environmental conditions Examples of the driven equipment in V-belt drives are blowers, pumps, compressors, fans, light-duty conveyors, dough mixers, generators, laundry machinery, machine tools, punches, presses, shears, printing machin- ery, bucket elevators, textile machinery, mills, and hoists Typical service factors, relying

on the characteristics of the driving and driven machinery, are given in Table 13.5 The horsepower should be multiplied by a service factor when selecting belt sizes The design

of a V-belt drive should use the largest possible pulleys As the sheave sizes become smaller, the belt tension increases for a given horsepower output Recommended pulley diameters for use with three electric motor sizes are given in Table 13.6

Table 13.5 Service factors K; for V-belt drives

CHAPTER 13 ® BELTS, CHAINS, CLUTCHES, AND BRAKES

Finally, we note that the shaft load at the pulley consists of torque T and force F, The

latter is the vector sum of tensions F; and F2, Referring to Figure 13.5a, the shaft force may

therefore be expressed as

The angle @ is defined by Eq (13.6) Equation (13.23) yields results approximately equal

to the scalar sum F -+ F2 in most cases The designer will find it useful to have the shaft load for determining the reactions at the shaft bearings

521

Design, Analysis of a V-Belt Drive

The capacity ofa V-belt drive is to bé10 kW, based on a coefficient of friction of 0.3 Determine the

required belt' tensions and:the maximum tension

Given: A driver sheave has # radius of r; = 100 mm, a speed of xj = 1800 rpm, and a contact

angle of @ = 153° The belt weighs 2.25 N/m and the included angle is 36°

Assumptions: 'The driver is a normal torqué motor and driven machine involves light shock load

SoluHon::We have @' + 153? = 2.76 rad and B = 18° The tight-side tension is estimated from

13.36 53.05

Em : = 81:5 *(sse) or 7 OPN lan n=

Then; by Eq (13,19), the slack-side tension is

53:05

Foo 655: 5 5 ôi e124 1245N Based upon a'service factor of 1.2 (Table 13.5) to F,, Eq (13.22) gives a maximum tensile force Fina 2.2655) = 786 N

applied to the belt:

EXAMPLE 13.2

The design of timing-belt drives is the same as that of flat- or V-belt drives The

A high-speed knife cutter assembly for flexible materials

(including PVC and other plastics) is shown in Figure 13.9

The unit is compact and designed for bench-top installa- tion The drive wheels are part of the automatic feeding

mechanism (not shown in the figure) These wheels drive

the material through the feed tube to the cutting wheel The

feed mechanism includes a compression spring for smooth operation The variable cut lengths and rates (as high as

1000 per minute) are accomplished by changing the num- ber of blades in the rotary cutting wheel and changing the reduction drive ratio between the motor and shaft on which

cutter wheel is keyed Determine

(a) The beit length

(b) The maximum center distance

(c) The maximum belt tension

Cutter blade Spring

Drive wheel Material — Drive wheel

Bett DESIGN OF A HIGH-SPEED CUTTING MACHINE

Design Requirements: The center distance between

the motor (driver) pulley and driven pulley should not

exceed ¢ = 17 in A belt coefficient of friction of f = 1.0

is used

Given: A 2-hp, 2, = 1800 rpm, AC motor is used The

belt weighs w = 0.007 Ib/in The driver pulley radius,

ry = 14 in Driven pulley radius, r2 = 2] in

Assumptions: The driver is a normal torque motor

The cutter, and hence the driven shaft, resists heavy shock loads The machine cuts uniform lengths of flexi-

ble materials of cross sections up to 2 in in diameter

Operation is fully automatic, requiring minimal operator

belt

Solution: See Figure 13.9 and Table 13.5

(a) The appropriate beit pitch length is determined using

Eq (13.9),

i Lederman trd+ or -ny

Substitution of the given data yields

sz 7 ~ 2sin7! x-/ạn ( 16.97 ) =z 3.024 rad

The tight-side tension in the belt is calculated, through the use of Eq (3.20) in which, for a toothed {or flat) belt, sing = i and

only maintenance required after careful alignment of the elements is proper lubrication Usu-

ally, a chain should have a sheet of metal casing for protection from atmospheric dust and to facilitate lubrication Chains should be washed regularly in kerosene and then soaked in oil

The speed ratio of a chain drive is expressed by the equation

In the preceding, we have

ng = output speed

øị = input speed

N2 = number of teeth in the output sprocket

N, = number of teeth in the input sprocket Anodd number of teeth on the driving sprocket (17, 19, 21, .) is recommended, typically

17 and 25 Usually, an odd number of sprocket teeth causes each small sprocket tooth to contact many or all chain links, minimizing wear The larger sprocket is ordinarily limited

to about 120 teeth

Center distance ¢ should be between a value that just allows the sprockets to clear It is

c = 2(r; +72); for smaller speed ratios, 71 /nạ < 3 Here, rị and r; refer to pitch radii of the

input and output sprockets, respectively When longer chains are used, idlers may be required

on the slack side of the chain For the cases in which speed ratios a; /n2 = 3, the center dis- -tance should be c = 2(m — 1) Having a tentative center distance ¢ between shafts selected, chain length L may be estimated applying Eq (13.7) Finally, the center distance is recalcu- lated through the use of Eq (13.8) The angle of contact for the chain drive is given by

Eg (13.6) Note that, for a small sprocket, the angle of contact should not be less than 120°

Chain pitch p represents the length of an individual link from pin center to pin center

‘The pitch radius of a sprocket with N teeth may be defined as

in Eqs (13.24) and (13.25), the chain pitch p is measured in in and the sprocket speed n is

in revolutions per minute, rpm The total force or tension in the chain includes transmitted force F,, acentrifugal force F,, a small catenary tension, and a force from link action There are also impact forces when link plates engage sprocket teeth, discussed in the next section

525

There are various types of power transmission chains, however, roller chains are the most widely employed Types of driven equipment with roller chain drives include bakery machinery, blowers and fans, boat propellers, compressors, conveyors, clay-working machinery, crushers, elevators, feeders, food processors, dryers, machine tools, mills, pumps, pulp grinders, printing presses, carding machinery, and wood working machinery

In some of these applications as well as in cranes, hoists, generators, ice machines, and a variety of laundry machinery, inverted chains are also used Both common chain types are used on sprockets, well suited for heavy loads, and have high efficiency

ROLLER CHAINS

Of its diverse applications, the most familiar is the roller chain drive on a bicycle A roller chain is generally made of hardened steel and sprockets of steel or cast iron Nevertheless, stainless steel and bronze chains are obtainable where corrosion resistance is needed The geometry of a roller chain is shown in Figure 13.10 The rollers rotate in bushings that are press fitted to the inner link plates The pins are prevented from turning by the outer links’

press-fit assembly Roller chains have been standardized according to size by the American National Standards Institute [14] The characteristics of representative standard sizes are

Pin diameter, dy ai Roller a

Roller Pin diameter, Link plate Minimum ultimate

Chain no Pitch, p,in Diameter d,in Width, b, in đụ in thickness, ¢ strength Ib

| SOURCE: ANSIASME Standard 829.1M-1993

listed in Table 13.7 These chains are manufactured in single (Figure 13.10a), double (Fig- ure 13.10b), triple, and quadruple strands Clearly, the use of multistrands increases the load capacity of a chain and sprocket system

Chordal Action

The instantaneous chain velocity varies from the average velocity given by Eq (13.24)

Consider a sprocket running at constant speed n and driving a roller chain in a counter- clockwise direction, as illustrated in Figure 13.11 A chord representing the link between

Pitch circle

Figure 13.11 Aroller chain and

sprocket engagement

CHAPTER 13 @ Bers, CHAINS, CLUTCHES, AND BRAKES centers has the pitch length of p The chord subtends the pitch angle of the sprocket, 360°/N From the geometry, sin(180°/2) = (p/2)/r We therefore have

ry = rcos(i80°/N), after the sprocket is turned the angle of articulation The change of velocity AV is called the chordal action:

AV == 27rn{l — cos(180°/N)]

Rotation of the link through the angle of articulation causes impact among the rollers and the sprocket teeth as well as wear in the chain joint The movement of the link up and down with rotation through the articulation angle develops an uneven chain exit velocity Conse- quently, the driven shaft of a roller chain drive may be given a pulsating motion, particularly

at high-speed operations The angle of articulation (hence the impact and chordal action) should be reduced as much as practicable, by increasing the number of sprocket teeth [1, 15] THe Power CapaciTy OF ROLLER CHAINS

Equations and tables for roller-chain power capacity and selection were developed through the American Chain Association (ACA), as the result of many years of laboratory testing and field observation Rated horsepower capacities are usually given in tabular form for each type of single-strand chain corresponding to a life expectancy of 15 khr for a variety

of sprocket speeds Table [3.8 is an example for ANSI No 60 roller chains Listed in Table 13.9 are the service factors that account for the abruptness associated with load application Table 13.10 shows multiple-strand factors

At lower speeds, the power capacity of roller chains is based upon the fatigue strength

of the link plate On the other hand, at higher speeds, the power relies on roller and bushing impact life At extremely high speeds, the power capacity is on the basis of the galling or welding between pin and bushings The design power capacity may be expressed as follows:

where

H, = horsepower rating (from Table 13.8)

Kị = service factor (from Table 13.9)

K2 = multiple-strand factor (from Table 13.10) Usually a medium or light mineral oil is used as the lubricant We observe from

Table 13.8 that the proper lubrication of roller chains is essential to their performance As

04 S585

ace fee

Ayoedeo semodasioy paiey

Assumptions: ’ The input power type is an internal combustion (L.C.} engine, mechanical drive

js moderate shock With the exception of the tensile force, all forces are taken

to be negligible

Solution: See Tables 13.7 through 13.10

(a): For driver sprocket H, = 20.6 hp, type B lubrication is required (Table 13.8) Service fac- tor Ky = L4 (Table 13.9), From Table 13.10 for three strands, K2 = 2.5 Applying

“The inverted-tooth chain, also referred to as the silent chain, is composed of a series a

toothed link plates that are pin connected to allow articulation The chain pitch is define

in Figure 13.12 An inverted-tooth chain ordinarily has guide links on the sides oe he center to keep it on the sprocket To increase the chain life, different details of joint

Sprocket Figure 13.12 Portion of an inverted tooth or “silent” chain

CHAPTER 13 ® = Betts, CHAINS, CLUTCHES, AND BRAKES

construction are used Enclosures for the chain are customarily needed Therefore, silent chains are more expensive than roller chains Usually, when properly lubricated, at full load, drive efficiency is as high as 99% :

As the name suggests, these chains are quieter than roller chains They may be run at higher speeds, because there is minor impact force when the chain link engages the sprocket

The inverted-tooth chain has a smooth flat surface, which can be conveniently used for con- veying Power capacities of silent chains are listed in tables analogous to those for roller chains However, these chains reach maximum power at maximum speed, while roller chains reach highest power far below their maximum speed Most of the remarks in the fore- going paragraphs relate as well to inverted-tooth chains and sprockets, which are also stan- dardized by ANSI [16] Regular pitches vary between 3 and 2 in Sprockets may have

21 to 150 teeth Center distance adjustment is periodically needed to compensate for wear

tion torque Performance analysis of clutches and brakes involves the determination of the

actuating force, torque transmitted, energy absorption, and temperature rise The transmit- ied torque is associated with the actuating force, the coefficient of friction, and the geome- try of the device The temperature rise is related to energy absorbed in the form of friction heat during braking or clutching

The materials used for clutches and brakes are of two types, those used for the disk or drums and those used for friction materials or linings In the design of these devices, the selection of the friction materials is critical Most linings are attached to the disks or drums by either riv- eting or bonding The former has the advantage of low cost and relative ease in installation

The latter affords more friction area and greater effective thickness but is more expensive

Drums are ordinarily made of cast iron with some alloying material added Materials like stainless steei, Monel, and so on are used when good heat conduction is important

Many railroad brakes employ cast iron shoes, which are bearing on cast iron wheels or drums Friction, thermal conductivity, resistance to wear, and thermal fatigue characteris- tics of drums are very important They must have a sufficiently smooth surface finish to minimize wear of the lining

Linings are often made of molded, woven, sintered, or solid materials They are com- posed mainly of reinforcing fibers (to render strength and ability to resist high tempera- tures), metal particles (to obtain wear resistance and higher coefficient of friction), and bonding materials The binder is ordinarily a thermosetting resin or rubber, In addition, a friction material should have good heat conductivity and impenetrability to moisture

scoring of the drum and disks These materials are the most commonly used in drum brakes and the least costly Sintered metal pads are made of a mixture of copper or iron particles having friction modifiers molded then heated to blend the material They are the most costly but also the best suited for heavy-duty applications Sintered metal-ceramic friction pads are similar, except that ceramic particles are added prior to sintering

For sufficient performance of the brake or clutch, the requirements imposed on friction materials include the following: a high coefficient of friction having small variation on changes in pressure, velocity, and temperature; resistance to wear, seizing, and the tendency to grab; heat and thermal fatigue resistance Tables 13.11 and 13.12 list

Table 13.11 Properties of common brake and clutch friction materials, operating dry

Maximum pressure Maximum drum

*WWhen rubbing against smooth cast iron or steel

Table 13.12 Values of friction coefficients of common

brake/clutch friction materials, operating In oil

CHAPTER 13 ° BELTS, CHAINS, CLUTCHES, AND BRAKES

qpproxuate data related to allowable pressures and the coefficient of friction for a few

fining or longer life, the lower values of the maximum pressure given should be used

s seen from the tables, the coefficients of friction are much smaller in oi! than under dry

friction, as expected For more , ore accurate accurate informati information, Cr COnSU: t the 5 i

The centrifugal clutch is in widespread use for automatic operation, such as to coupl

an engine to the drive train When the engine speed increases, it automaticall engage: the clutch This is particularly practical for electric motor drives, where during startin ì the driven machine comes up to speed without shock Used in chain saws for the same ap centrifugal clutches serve as an overload release that slips to allow the tc continue

Magnetic, hydraulic, and pneumatic drum clutches are also useful in drives with co:

plex loading cycles and in automatic machinery or robots The expanding drum clut nis frequently used in textile machinery, excavators, and machine tools Inasmuch as the analysis of drum clutch is similar to that for dru m brakes, to be take i i

and 13.14, we do not discuss them at this time up in Sections 13.18

Figure 13,13 An internal expanding centrifugal-acting drum clutch (Courtesy of Hillarcl Corporation.)

Basic disk clutches and brakes are considered in this section The former transmits torque

from the input to the output shaft by the frictional force developed between the two disks

or plates when they are pressed together The latter is basically the same device, but one of

the shafts is fixed One of the friction surfaces of the clutch or brake is typically metal (cast iron or steel) and the other is usually a friction material or lining Magnetic, hydraulic, and

pneumatic operating mechanisms are also available in disk, cone, and multiple disk

clutches and brakes

Uniform pressure and uniform wear are two basic conditions or assumptions that may

occur at interface of the friction surfaces The designer must decide on which assumption

more closely approximates the particular clutch or brake being analyzed The uniform- wear assumption leads to lower calculated clutch or brake capacity than the assumption of uniform pressure, as observed in Example 13.4 Hence, disk clutches and brakes are ordi- narily designed on the basis of uniform wear that gives conservative results The preceding analysis can be used as a guide Design considerations also include the characteristics of the machine whose brakes or clutches are to be a part and the environment in which the ma- chine operates

Disk CLUTCHES Friction clutches reduce shock by slipping during the engagement period The single-plate

or disk clutch, shown schematically in Figure 13.14, is employed in both automotive and industrial service These devices are larger in diameter to give adequate torque capacity

Note that, in an automotive-type disk clutch, the input disk (flywheel) rotates with the crankshaft The hub of the clutch output disk is spline-connected to the transmission shaft

Clearly, the device is disengaged by depressing the clutch pedal The torque that can be transferred depends on the frictional force developed between the disks, coefficient of fric- tion, and the geometry of the clutch The axial force typically is quite large and can be ap- plied mechanically (by spring, as in the figure), hydraulically, or electromagnetically An advantage of the disk clutch over the drum clutch is the absence of centrifugal effects and efficient heat dissipation surfaces

Piston Bushing

Output

Input Oil passage shaft

Figure 13.15 Half-section view of a multiple-disk clutch, hydraulically

operated

Muttiple-disk clutches can have the friction lining on facing sides of a number (as many

as 24) of alternative driving and driven disks or plates The disks are usually thin (about

15 mm) with small diameters, Thus, additional torque capacity with only a small increase

in axial length is obtained When the clutch is disengaged, the alternate disks are free to slide axially to disjoin After the clutch is engaged, the disks are clamped tightly together to pro- vide a number of active friction surfaces N Disk clutches can be designed to operate either dry” or “wet” with oil The advantages of the latter are reduced wear, smoother action, and lower operating temperatures As a result, most multiple-disk clutches operate either im- mersed in oil or in a spray Multiple disk clutches are compact and suitable for high-speed operations in various machinery They are often operated automatically by either air or hydraulic cylinders (e.g., in automotive automatic transmissions)

/ Figure 13.15 shows a hydraulically operated clutch In this device, the axial piston mo-

tion and force are produced by oil in an annular chamber, which is connected by an oil pas-

sage to an external pressure source We see from the figure that, with the housing keyed to the input shaft, two disks and the piston are internally splined and an end plate is fastened

These are the driving disks The six driven disks are externally splined to the housing keyed

to the output shaft

; We develop the torque capacity equations for a single pair of friction surfaces, as in Figure 13.14, However, they can be modified for multiple disks by merely multiplying the

values obtained by the number of active surfaces N For example, N = 6 in the device

depicted in Figure 13.15

Uniform Wear When the clutch disks are sufficiently rigid, it can be assumed that wear over the lining is

uniform This condition applies after an initial wearing-in has occurred The uniform wear

rate, which is taken to be proportional to the product of pressure and velocity pV is con- stant Note that the velocity at any point on the clutch surface varies with the radius and the angular velocity Therefore, assuming a uniform angular velocity,

pr=C

535

where p = préssure, r = radius, and C = constant This equation indicates that the maxi- _ mum pressure Pmax takes place at the inside radius r = d/2 (Figure 13.14) Hence,

d

The total normal force that must be exerted by the actuating spring in Figure 13.14 is found by multiplying the area 2zzr dr by the pressure p = Pmaxd/2r and integrating over the friction surface Hence, the actuating force Fy required equals

Fas [Poa or = 57P nn d(D ~€) d/2 ee 2 cee (13.28)

The friction torque or torque capacity is obtained by multiplying the force on the element

by the coefficient of friction f and the radius and integrating over the area It follows that

An expression relating the torque capacity to actuating force is obtained by solving

Eq (13.28) for Pmax and inserting its value into Bg (13.29) In so doing, we have

where rave is the average disk radius Note the simple physical interpretation of this equa- tion As previously pointed out, for a multiple-disk clutch, Eqs (13.29) and (13.30) must

be multiplied by the number of active surfaces N

In the design of clutches, the ratio of inside to outside diameters is an important param~

eter It can be verified, applying Eq (13.29), that the maximum torque capacity for a prescribed outside diameter is attained when

CHAPTER 13 © BELTS, CHAINS, CLUTCHES, AND BRAKES

These can be combined to yield the torque as a function of actuating force:

The torque capacity for a multiple-disk clutch is obtained on multiplying EB is (13.33

(13.34) by the number of active surfaces N piving Bas Pane

537

Design of a Disk Clutch

A = iD with a-single: friction: surface has’an: outer diameter D' and: inner diameter d (Fig- ure SES 13.14) Determine ‘the torque: that can be: transmitted and: thẻ actuati ing force i

Design Decisions: Molded: friction: material: and: a steel’ di es cisions : § isk: are used, ha’ i = 0.3

7 _ z7(035/0500)(057 022) =:16.08 kÑ: m

"By Eq 13.32),

Fa = 71500) 0.5" — 0.2?) = 247.4 kN

Comment: The preceding: results indicate that: the uniform wear condition yielded a smaller

torque and actuating force; it is therefore the more conservative of the two assumptions in terms of

clutch capacity, :

EXAMPLE 13.4

Figure 13.16 Caliperdisk brake, hydraulically operated (Courtesy of Ausco

Products, Inc., Benton Harbor, MI.)

Disk BRAKES

A disk brake is very similar to the disk clutch shown in Figure 13.14, with the exception that one of the shafts is replaced by a fixed member Loads are balanced by locating fric- tion linings or pads on both sides of the disk Servo action can be obtained by addition of several machine parts The torque capacity and actuating force requirements of disk brakes may readily be ascertained through the use of the foregoing procedures The equations for the disk clutch can be adapted to the disk brake, if the brake pad is shaped like a sector of

a circle and calculations are made accordingly, as illustrated in the next example

The caliper disk brake includes a disk-shaped rotor attached to the machine to be con-

trolled and friction pads The latter cover only a small portion of the disk surface, allowing the remainder exposed to dissipate heat The linings are contained in a fixed-caliper assembly and forced against the disk by air pressure or hydraulically (Figure 13.16) Disk brakes have been employed in automotive applications, due to their equal braking torque for either direction of rotation as well as greater cooling capacity than drum brakes (see Section 13.14) They are also often preferred in heavy-duty industrial applications Caliper disk brakes are also widely used on the front wheel of most motorcycles The common bicycle is another example, where the wheel rim forms the disk

EXAMPLE 13.5 -> Design of a Disk Brake

A disk brake has two pads of included angle y = 60° each, D = 10 in., d = 5 in (Figure 13.14)

Determine

(a): The actuating force required to apply one shoe

CHAPTER 13 e BELts, CHains, CLUTCHES, AND BRAKES

Design Decision: Sintered-metal pads and cast iron disk are used with f = 0.2 and Pax = 200 psi

The cone clutch, Figure 13.17, can be considered as the general case of a disk clutch hav- ing a cone angle of 90° Due to the increased frictional area and the wedging action of the parts, cone clutches convey a larger torque than disk clutches with the identical outside diameters and actuating forces Practically, a cone clutch can have no more than one fric- tion interface; hence, N = I, Cone clutches are often used in low-speed applications They could also be employed as cone brakes with some slight modifications

UNIFORM WEAR

The presupposition is made that the normal wear is proportional to the product of the nor- mal pressure p and the radius Let the radius r in Figure 13.17 locate the ring element run- ning around the cone The differential area is then equal todA = 2xr dr/sina The normal force in the element equals dF, = pdA, in which p = pmaxd/2r As before, p represents the maximum pressure Hence, the total normal force is m

F.= [ PP Dax Inv dr n= —- = wd Pmax (b~a _

Figure 13.17 Cone clutch

The torque that can be transmitted by the ring element is equal to ar =ủF,ƒr = 2wpƒr? dr/sina The torque capacity of the clutch T is obtained by integrating the forgo- ing expression over the conical surface In so doing, we obtain

CHAPTER 13 ® BELTS, CHAINS, CLUTCHES, AND BRAKES 541

The band brake, the simplest of many braking devices, is employed.in power excavators and hoisting and other machinery Usually, the band is made of steel and lined with a woven friction material for flexibility The braking action is secured by tightening the band wrapped around the drum that is to be slowed or halted The difference in tensions at each end of the band ascertains the torque capacity

Figure 13.18 shows a band brake with the drum rotating clockwise For this case, fric- tion forces acting on the band increase the tight-side tension F', and decrease the slack-side tension F2 Consider the drum and band portion above the sectioning plane as a free body

Then, summation of the moments about the center of rotation of the drum gives the torque

capacity, which is the same as for a belt drive:

to the fixed support and the slack side attached to the lever, as shown in the figure

An expression relating band tensions F, and Fy is derived on following the same procedure used for flexible belts, with the exception that the centrifugal force acting on

belts does not exist Hence, referring to Section 13.3, the band tension relationship has the form

(13.44)

_~ Band of width, w

Figure 13.18 | Simple band brake

- # = the larger tensile force

Fy = the smaller tensile force

f = coefficient of friction

@ = angle of contact between band and drum or the angle of wrap Let the analysis of belt shown in Figure 13.5b be applied to the band at the point of tan- gency for F; We now have F, = 0 and dN = pmaxwr d0 The inward components of the band forces are equal to dN = F dé These two forces are set equal to each other to yield

The quantity Pinax is the maximum pressure between the drum and lining, and w represents

the width of band An expression similar to this can also be written for the slack side

The differential band brake is analogous to the sirnple band brake, with the exception that the tight-side tension helps the actuating force (Figure 13.19) A brake of this type is termed self-energized, since the friction force assists in applying the band Por a differential brake, Eq (13.43) becomes

In the case of a self-locking brake, the product s F' is greater than cF) Note that, when a brake is designed to be self-locking for one direction of rotation, it can be free to rotate in the opposite direction A self-locking brake can then be employed when rotation is in one direction only

EXAMPLE 13.6 | Design of a Differential Band Brake

” A‘differential band brake similar to that in Figure 13.19 uses a woven lining having design values of

f= 0.3 and pmax = 375 kPa, Determine (a) The torque capacity : (b) The actuating force

CHAPTER 13 @ BELTS, CHAINS, CLUTCHES, AND BRAKES (c) The power capacity

(d) The value of dimension s that would cause the brake to be self-locking

Given: The speed is 250 rpm, a = 500 mm, ¢ = 150 mm, w = 60 mm, r = 200 mm, s = 25 mm,

and @ = 270°

Solution:

(a) Through the use of Eq (13.45), we obtain

Fy = wrprax == (0.06)(0.2)(375) = 4.5 kN Applying Eq (13.44),

(d) Using Eq (13.46), we have F, = 0 for s = 150(1.095)/4.5 = 36.5 mm

Comment: The brake is self-locking if s > 36.5 mm

A short-shoe drum brake consists of a short shoe pressed on the revolving drum by a lever

The schematic representation of a brake of this type is depicted in Figure 13.20 Inasmuch

as the shoe is relatively short (i.c., the angle of contact is small, @ < 45°), a uniform pres- sure distribution may be taken between drum and shoe Accordingly, the resultant normal force and the friction force act at the center of contact

The projected area A of the shoe is the width multiplied by the chord length subtended

by a ° arc of the radius of the drum From the geometry of the figure, A = 2[r sin(¢/2)]w

Hence, the normal force on the shoe is

Figure 13.20 Short-shoe drum brake

In the foregoing, we have

F, = normal force Pmax = Maximum pressure between the drum and shoe

r = radius of the drum

= angle of contact w_s= width of the shoe The value of the friction force is fF,, The sum of moments about point O for the free-body diagram of the drum yields the torque capacity of the brake as

The quantity f represents the coefficient of friction

We now consider the lever as the free body Then taking moments about the pivot A,

in which a, b, and ¢ represent the distances shown in Figure 13.20

SELE-ENERGIZING AND SELF-LOCKING BRAKES For the brake with the direction of the rotation shown in the figure, the moment of the fric- tion force assists in applying the shoe to the drum; this makes the brake self-energizing If b= fe or b < fe, the force F, required to actuate the brake becomes 0 or negative,

respectively The brake is then said to be self-locking when

A self-locking brake requires only that the shoe be brought in contact with the drum (with

F, = 0) for the drum to be “loaded” against rotation in one direction The self-energizing feature is useful, but the self-locking effect is generally undesirable To secure proper uti- lization of the self-energizing effect while avoiding self-lock, the value of b must be at least

Note that, if the brake drum rotation is reversed from that indicated in Figure 13.20,

the sign of fc in Eq (13.49) becomes negative and the brake is then self-deenergized Also,

if the pivot is located on the other side of the line of action of fF,, as depicted by the dashed lines in the figure, the friction force tends to unseat the shoe Then, the brake would not be self-energizing Clearly, both pivot situations discussed are reversed if the direction

of rotation is reversed

545

Design of a Short-Shoe: Drum Brake

The brake shown in Figure 13.20 uses'a cork lining having design values of f = 0.4 and Dinax =

‘150 psi, Determine

(a): The torque capacity and ‘actuating force

(6): The reaction at pivot A:

Givens: @:12 ins w.= 4 in; b= Sin; e=2in, r=4in, ó=30°

Solution:

(a): From Eq: (13.47), we have

F, = 150 [2 (4si 5 | 3.2 931.7 Ib Equation (13.48) yields

T:.(0:4)(931:7);:=.1:491.kip-in:

Applving:Eg: (13.49),

16-04% 2 Fos 931.76 = 0.4% 2) 396-1 Ib

A short-shoe brake is used on the drum, which is keyed to

the center shaft of the high-speed cutter as shown in

Figure 13.9 The driven pulley is also keyed to that shaft

For details see Case Study 13-1 Determine the actuating force F,

Assumptions: The brake shoe material is molded as-

bestos The drum is made of iron The lining rubs against

the smooth drum surface, operating dry

Given: The drum radius r = 3 in,, torque T = 270 Ib-in

(CW), a= 12 in, b = 1.2 in., d = 2.5 in., the width of shoe w = 1.5 in (Figure 3.9) By Table 13.11, pmax =

200 psi and f = 0.35

Requirement: Shoe must be self-actuating

Solution: The normal force, through the use of Eq

(13.48), is

R= fr

270

= = 257.1 lb (0.35)3

Case Study 13-2 | THE BRAKE DESIGN OF A HIGH-SPEED CUTTING MACHINE

The angle of contact, applying Eq (13.47), is then

= aa ~ 0.5 x 2.5)

= —1.071b

Comments: Since ¢ < 45°, the short-shoe drum brake

approximations apply A negative value of F, means that the brake is self-energizing, as required

shoe equations can lead to appreciable errors Most shoe brakes have contact angles of 90°

or greater, so a more accurate analysis is needed The obvious problem relates the determi- nation of the pressure distribution The analysis that includes the effects of deflection is complicated and not warranted here In the following development, we make the usual sim- plifying assumption: the pressure varies directly with the distance from the shoe pivot point This is equivalent to the presupposition made earlier, that the wear is proportional to the product of pressure and velocity

EXTERNAL LONG-SHOE DRUM BRAKES

Figure 13.21 illustrates an external long-shoe drum brake The pressure p at some arbitrary angle @ is proportional to c sin @ However, since c is a constant, p varies directly with sin @

CHAPTER 13 @ Bevrs, Cains, CLUTCHES, AND BRAKES

ý: Roelaion Figure 13.21 External long-shoe drum brake

02 < 120°, and @ = 90°, where ¢ is the angle of contact

Let w represent the width of the lining Then, the area of a small element, cut by two radii an angle d@ apart, is equal to wr dé, Multiplying by the pressure p and the arm c sin 0 and integrating over the entire shoe, the moment of normal forces, M,, about pivot A results:

548 PART iÌ,- ®' ÁPPLICATIONS

In a like manner, the moment of friction forces, My, about A is written in the form

The torque capacity of the brake is found by taking moments of the friction forces about the center of the drum O In so doing, we have

®

Ts Spwr d@r

6 ƒWF max Ỉ :

(sin®)m Jo, 5

from which

fe ? Daiax ;

Finally, pin reactions at A and O can readily be obtained from horizontal and vertical force equilibrium equations Note that reversing the direction of the rotation changes the sign of the terms containing the coefficient of friction in the preceding equations

CHAPTER 13 ® Bets, CHAINS, CLUTCHES, AND BRAKES Design of a Long-Shoe Drum Brake

The: long-shoe drum brake is actuated by a mechanism that exerts a, force of F, = 4 kN

(Figure 13.22) Determine (a) The maximum pressure

(b).: The torque and power capacities

(a) Through the use of Eq (13.53),

= 2.6(1079) pax

From Eq, (13.54), 0.35(0.075)(0.15) Dmx

— 4(0.15)(cos98,13° — cos8.13°)]

= 0.196107?) Pana

EXAMPLE 13.8