modul 09 clutches, brakes, couplings and flywheel part i

1 MS3111 Elemen MesinMAK © 2021

Teknik Mesin - FTMD ITB

09.01. Introduction

MS3111 - Elemen Mesin

Modul 09Clutches, Brakes, Couplings

and Flywheel Part I


Modul 09 Clutches, Brakes, Couplings and Flywheel Part I

IntroductionSegment

1

Static Analysis of Clutches

and Brakes

Segment

2

Internal Expanding Rim

Clutches and Brakes

Segment

3

External Contracting Rim

Clutches and Brakes

Segment

4

Band-Type Clutches and

Brakes

Segment

5

Frictional-Contact Axial

Clutches

Segment

6

Disk BrakesSegment

7

Cone Clutches and Brakes

Segment

8



09.01. Introduction



and Flywheel Part I


Representation

𝜔2𝜔1

𝐼1𝐼2

Brakes or Clutch 𝜔 = angular velocity

𝐼 = Inertia

𝑇i, 𝜃i

Flywheel

𝐼, 𝜃

𝑇o, 𝜃o

𝑇 = Torque𝜃 = angular displacement


Performance Analysis

1. The actuating force (gaya aktivasi)

2. The torque transmitted (torsi diteruskan)

3. The energy loss (rugi-rugi)

4. The temperature rise (kenaikan temperatur)


Various Types of Devices

1. Rim types with internal expanding shoes

2. Rim type with external contracting shoes

3. Band types

4. Disk or axial types

5. Cone types

6. Miscellaneous types



09.02. Static Analysis of Clutches and Brakes



and Flywheel Part I


General procedures of analysis steps.

1. Estimate or determine the distribution of pressure on the frictional surfaces.

2. Find a relation between the maximum pressure and the pressure at any point.

3. Apply the conditions of static equilibrium to find the actuating force, the torque, and the support reactions.

9.1. Static Analysis of Clutches and Brakes

Figure 16–2 A common doorstop.


• A normal pressure distribution 𝒑(𝒖) is shown under the friction pad as a function of position 𝑢, taken from the right edge of the pad.

• Distribution of shearing frictional traction is on the surface, of intensity 𝑓𝑝(𝑢), in the direction of the motion of the floor relative to the pad, where 𝑓is the coefficient of friction.


The net force in the y-direction:

The moment about C from the pressure:

𝑁 = 𝑤2න0

𝑤1

𝑝 𝑢 𝑑𝑢 = 𝑝𝑎𝑣𝑤1𝑤2

𝑤2න0

𝑤1

𝑝 𝑢 𝑢𝑑𝑢 = ത𝑢𝑤2න0

𝑤1

𝑝 𝑢 𝑑𝑢 = 𝑝𝑎𝑣𝑤1𝑤2 ത𝑢


We sum the forces in the x-direction to obtain

𝐹𝑥 = 0 𝑅𝑥 ∓𝑤2න0

𝑤1

𝑓𝑝 𝑢 𝑑𝑢 = 0

where − or + is for rightward or leftward relative motion of the floor, respectively.

• Assuming f constant, solving for Rx gives

𝑅𝑥 = ±𝑤2න0

𝑤1

𝑓𝑝 𝑢 𝑑𝑢 = ±𝑓𝑤1𝑤2𝑝𝑎𝑣


• Summing the forces in the y-direction gives

from which

for either direction.

𝐹𝑦 = 0 −𝐹 + 𝑅𝑦 +𝑤2න0

𝑤1

𝑝 𝑢 𝑑𝑢 = 0

𝑅𝑦 = 𝐹 − 𝑤2න0

𝑤1

𝑝 𝑢 𝑑𝑢 = 𝐹 − 𝑝𝑎𝑣𝑤1𝑤2


Summing moments about the pin located at A we have

𝑀𝐴 = 0

𝐹𝑏 − 𝑤2න0

𝑤1

𝑝 𝑢 𝑐 + 𝑢 𝑑𝑢 ∓ 𝑎𝑓𝑤2න0

𝑤1

𝑝 𝑢 𝑑𝑢 = 0

A brake shoe is self-energizing if its moment sense helps set the brake, self-deenergizing if the moment resists setting the brake. Continuing:

𝐹 =𝑤2

𝑏න0

𝑤1

𝑝 𝑢 𝑐 + 𝑢 𝑑𝑢 ± 𝑎𝑓න0

𝑤1

𝑝 𝑢 𝑑𝑢


𝐹 =𝑤2

𝑏න0

𝑤1


𝑤1

𝑝 𝑢 𝑑𝑢

Can F be equal to or less than zero? Only during rightward motion of the floor when the expression in brackets in Eq. (e) is equal to or less than zero. We set the brackets to zero or less:

න0

𝑤1

𝑝 𝑢 𝑐 + 𝑢 𝑑𝑢 − 𝑎𝑓න0

𝑤1

𝑝 𝑢 𝑑𝑢 ≤ 0

from which:

𝑓𝑐𝑟 ≥1

𝑎

0𝑤1 𝑝 𝑢 𝑐 + 𝑢 𝑑𝑢

0𝑤1 𝑝 𝑢 𝑑𝑢

=1

𝑎

𝑐 0𝑤1 𝑝 𝑢 𝑑𝑢 + 0

𝑤1 𝑝 𝑢 𝑢𝑑𝑢

0𝑤1 𝑝 𝑢 𝑑𝑢

𝑓𝑐𝑟 ≥𝑐 + ത𝑢

𝑎 where ത𝑢 is the distance of the center of pressure from the right edge of the pad.


Some remarks:

➢The conclusion that a self-acting or self-locking phenomenon is present and is independent of our knowledge of the normal pressure distribution 𝑝(𝑢).

➢Our ability to find the critical value of the coefficient of friction 𝑓𝑐𝑟 is dependent on our knowledge of 𝑝(𝑢), from which we derive ത𝑢.


EXAMPLE 16–1

The doorstop depicted in Fig. 16–2a has the following dimensions: a = 4 in, b = 2 in, c = 1.6 in, w1 = 1 in, w2 = 0.75 in, where w2 is the depth of the pad into the plane of the paper.a) For a leftward relative movement of the floor, an actuating force F of 10

lbf, a coefficient of friction of 0.4, use a uniform pressure distribution pav, find Rx , Ry , pav, and the largest pressure pa.

b) Repeat part a for rightward relative movement of the floor.c) Model the normal pressure to be the “crush” of the pad, much as if it

were composed of many small helical coil springs. Find Rx , Ry , pav, and pa for leftward relative movement of the floor and other conditions as in part a.

d) For rightward relative movement of the floor, is the doorstop a self-acting brake?


𝑅𝑥 = 𝑓𝑤1𝑤2𝑝𝑎𝑣 = 0.4 1 0.75 𝑝𝑎𝑣 = 0.3𝑝𝑎𝑣

Solution

Leftward:

𝑅𝑦 = 𝐹 − 𝑝𝑎𝑣𝑤1𝑤2 = 10 − 1 0.75 𝑝𝑎𝑣 = 10 − 0.75𝑝𝑎𝑣

𝐹 =𝑤2

𝑏න0

𝑤1


𝑤1

𝑝 𝑢 𝑑𝑢

𝐹 =𝑤2

𝑏න0

1

𝑝𝑎𝑣 𝑐 + 𝑢 𝑑𝑢 + 𝑎𝑓න0

1

𝑝𝑎𝑣𝑑𝑢

𝐹 =𝑤2

𝑏𝑝𝑎𝑣𝑐 න

0

1

𝑑𝑢 + 𝑝𝑎𝑣න0

1

𝑢𝑑𝑢 + 𝑎𝑓𝑝𝑎𝑣න0

1

𝑑𝑢

𝐹 =𝑤2𝑝𝑎𝑣𝑏

𝑐 + 0.5 + 𝑎𝑓

10 =(0.75)𝑝𝑎𝑣

21.6 + 0.5 + 4(0.4)

𝑝𝑎𝑣 = 7.207 𝑝𝑠𝑖

= 0.3 7.207 = 2.162 𝑙𝑏𝑓

= 10 − 0.75 7.207= 4.595 𝑙𝑏𝑓

Answer

Answer

Answer


b) Repeat part a for rightward relative movement of the floor.

𝑅𝑥 = −𝑓𝑤1𝑤2𝑝𝑎𝑣 = −0.4 1 0.75 𝑝𝑎𝑣 = −0.3𝑝𝑎𝑣

𝑅𝑦 = 𝐹 − 𝑝𝑎𝑣𝑤1𝑤2 = 10 − 1 0.75 𝑝𝑎𝑣 = 10 − 0.75𝑝𝑎𝑣

𝐹 =𝑤2

𝑏න0

𝑤1


𝑤1

𝑝 𝑢 𝑑𝑢

𝐹 =𝑤2

𝑏න0

1

𝑝𝑎𝑣 𝑐 + 𝑢 𝑑𝑢 − 𝑎𝑓න0

1

𝑝𝑎𝑣𝑑𝑢

𝐹 =𝑤2

𝑏𝑝𝑎𝑣𝑐 න

0

1

𝑑𝑢 + 𝑝𝑎𝑣න0

1

𝑢𝑑𝑢 − 𝑎𝑓𝑝𝑎𝑣න0

1

𝑑𝑢

𝐹 =𝑤2𝑝𝑎𝑣𝑏

𝑐 + 0.5 − 𝑎𝑓

10 =(0.75)𝑝𝑎𝑣

21.6 + 0.5 − 4(0.4)

𝑝𝑎𝑣 = 53.33 𝑝𝑠𝑖

= 0.3 53.33 = −16 𝑙𝑏𝑓

= 10 − 0.75 5.33 = −30 𝑙𝑏𝑓

Answer

Answer

Answer


c) Model the normal pressure to be the “crush” of the pad, much as if it were composed of many small helical coil springs. Find Rx , Ry , pav, and pa for leftward relative movement of the floor and other conditions as in part a.

From similar triangles 𝑦1𝑟1∆𝜙

=𝑐

𝑟1

𝑦2𝑟2∆𝜙

=𝑐 + 𝑤1𝑟2

𝑦1 = 𝑐∆𝜙 𝑦2 = (𝑐 + 𝑤1)∆𝜙

This means that y is directly proportional to the horizontal distance from the pivot point A; that is, 𝑦 = 𝐶1𝜈, where 𝐶1 is a constant

Assuming the pressure is directly proportional to deformation, then 𝑝 𝜈 = 𝐶2𝜈, where 𝐶2 is a constant. In terms of 𝑢, the pressure is 𝑝 𝑢 = 𝐶2 𝑐 + 𝑢 = 𝐶2 1.6 + 𝑢


𝐹 =𝑤2

𝑏න0

𝑤1

𝑝 𝑢 𝑐 + 𝑢 𝑑𝑢 + 𝑎𝑓න0

𝑤1

𝑝 𝑢 𝑑𝑢

𝐹 =𝑤2

𝑏න0

𝑤1

𝑝 𝑢 𝑐𝑑𝑢 + න0

𝑤1

𝑝 𝑢 𝑢𝑑𝑢 + 𝑎𝑓න0

𝑤1

𝑝(𝑢)𝑑𝑢

𝑝 𝑢 = 𝐶2 𝑐 + 𝑢 = 𝐶2 1.6 + 𝑢

𝐹 =0.75

2න0

1

𝐶2 1.6 + 𝑢 (1.6)𝑑𝑢 + න0

1

𝐶2 1.6 + 𝑢 𝑢𝑑𝑢 + 𝑎𝑓න0

1

𝐶2 1.6 + 𝑢 𝑑𝑢

10 = 0.375𝐶2 1.6 + 0.5 1.6 + (0.8 + 0.3333) + 4(0.4)(1.6 + 0.5)

𝐶2 = 3.396 𝑝𝑠𝑖/𝑖𝑛 𝑝 𝑢 = 3.396 1.6 + 𝑢

The average pressure is given by

𝑝𝑎𝑣 =1

𝑤1න0

𝑤1

𝑝 𝑢 𝑑𝑢 =1

1න0

1

3.396(1.6 + 𝑢)𝑑𝑢 = 3.396 1.6 + 0.5 = 7.132 𝑝𝑠𝑖

The maximum pressure occurs at 𝑢 = 1 𝑖𝑛, and is 𝑝𝑎 = 3.396 1.6 + 1 = 8.83 𝑝𝑠𝑖

𝑅𝑥 = 0.3𝑝𝑎𝑣 = 0.3 7.132 = 2.139 𝑙𝑏𝑓 𝑅𝑦 = 10 − 0.75𝑝𝑎𝑣 = 10 − 0.75 7.132 = 4.652 𝑙𝑏𝑓

The reaction at support:


d) For rightward relative movement of the floor, is the doorstop a self-acting brake?

To evaluate ത𝑢 we need to evaluate two integrations

න0

𝑐

𝑝 𝑢 𝑢𝑑𝑢 = න0

1

3.396(1.6 + 𝑢)𝑢𝑑𝑢 = 3.396 0.8 + 0.3333 = 3.849 𝑙𝑏𝑓

න0

𝑐

𝑝 𝑢 𝑑𝑢 = න0

1

3.396(1.6 + 𝑢)𝑑𝑢 = 3.396 1.6 + 0.5 = 7.132 𝑙𝑏𝑓/𝑖𝑛

Thus, ത𝑢 =3.849

7.132= 0.5397 𝑖𝑛 𝑓𝑐𝑟 ≥

𝑐 + ത𝑢

𝑎𝑓𝑐𝑟 ≥

1.6 + 0.5397

4= 0.535

The doorstop friction pad does not have a high enough coefficient of friction to make the doorstop a self-acting brake. The configuration must change and/or the pad material specification must be changed to sustain the function of a doorstop.



09.03. Internal Expanding Rim Clutches and Brakes



and Flywheel Part I


9.2. Internal Expanding Rim Clutches & Brakes

(a) Clutch (b) BrakeFigure 16–3 (a) internal expanding centrifugal-acting rim clutch (b) internal expanding brake


❑ Depending upon the operating mechanism, such clutches are further classified as: expanding-ring, centrifugal, magnetic, hydraulic, and pneumatic.

❑ Internal-shoe rim expanding type consist of 3 elements:

• the mating frictional surface,

• the means of transmitting the torque to and from the surfaces,

• the actuating mechanism.

❑ The expanding-ring clutch is often used: in textile machinery, excavators, and machine tools.

❑ In braking systems, the internal-shoe or drum brake is used mostly for automotive applications.


Force Analysis▪ Let us consider the unit pressure p acting upon

an element of area of the frictional material

located at an angle from the hinge A.

▪ We designate the maximum pressure by pa

located at the angle a from the hinge pin A.

▪ The mechanical arrangement permits no pressure to be applied at the heel (point A) ➔the pressure at this point is assumed to be zero.

▪ In some designs the hinge pin is made movable to provide additional heel pressure.

In this case, as long shoe, the uniform distribution of pressure is not valid anymore !

p

Figure 16–5 The geometry associated with an arbitrary point on the shoe.


The following assumption are implied by the following analysis:

1. The pressure at any point on the shoe is assumed to be proportional to distance from the hinge pin, being zero at the heel.

2. The effect of centrifugal force has been neglected. In the case of brakes, the shoes are not rotating, and no centrifugal force exists. In clutch design, the effect of centrifugal force must be considered in writing the equations of static equilibrium.

3. The shoe is assumed to be rigid. Since this cannot be true, some deflection will occur, depending upon the load, pressure, and stiffness of the shoe. The resulting pressure distribution may be different from that which has been assumed.

4. The entire analysis has been based upon a constant coefficient of friction (does not vary with pressure). Actually, the coefficient may vary with a number of conditions including temperature, wear, and environment.


Step 1Make the assumption that the pressure at any point is proportional to the vertical distance from the hinge pin.

This vertical distance is proportional to sin➔ 𝑝 ≈ sin 𝜃 ⇒𝑝

sin 𝜃= 𝑐𝑜𝑛𝑠𝑡

Step 2

To find the pressure distribution on the periphery of the internal shoe, consider point B on the shoe.

As in Ex. 16–1, if the shoe deforms by an infinitesimal rotation Δ𝜙 about the pivot point A, deformation perpendicular to B is ℎΔ𝜙.

▪ From triangle AOB, ℎ = 2𝑟 sin( Τ𝜃 2), so


▪ The deformation perpendicular to the rim is ℎΔ𝜙 cos Τ𝜃 2 , which is

Thus, the deformation, and consequently the pressure, is proportional to sin 𝜃.

ℎΔ𝜙 cos Τ𝜃 2 = 2𝑟Δ𝜙 sin Τ𝜃 2 cos Τ𝜃 2 = 𝑟∆𝜙 sin 𝜃

𝑝

sin 𝜃=

𝑝𝑎sin 𝜃𝑎

𝑝 =𝑝𝑎

sin 𝜃𝑎sin 𝜃or

▪ In terms of the pressure at B and where the pressure is a maximum, this means

(a) (1)


▪ The useful characteristics of the previous pressure distribution are:

➢The pressure distribution is sinusoidal.

➢ If the shoe is short, the largest pressure on the shoe is 𝑝𝑎 occurring at the end of the shoe, 𝜃𝑎 .

➢ If the shoe is long, the largest pressure on the shoe is 𝑝𝑎 occurring at 𝜃𝑎 = 90° .

▪ In choosing friction material, the designer should think in terms of 𝑝𝑎 and not about the amplitude of the sinusoidal distribution that addresses locations off the shoe.

Figure 16–6

𝑝 =𝑝𝑎

sin 𝜃𝑎sin 𝜃


Figure 16–7Forces on the shoe

Drum’s rotation

Step 3

▪ At the angle from the hinge pin, there acts a differential normal force dN whose magnitude is.

where b is the face width of the friction material.

𝑑𝑁 = 𝑝𝑏𝑟𝑑𝜃 (b)

𝑝 =𝑝𝑎

sin 𝜃𝑎sin 𝜃

𝑑𝑁 =𝑝𝑎𝑏𝑟 sin 𝜃 𝑑𝜃

sin 𝜃𝑎(c)

▪ Substituting the value of the pressure, eq. (1), we find


𝒅𝑵

𝜃

𝑑𝑁 cos 𝜃

𝑑𝑁 sin 𝜃𝑓𝑑𝑁 cos 𝜃

𝑓𝑑𝑁 sin 𝜃

𝑭

𝑭𝒚

𝑭𝒙

Step 3 (continued)

▪ The normal force, dN and other forces act on the friction material can be decomposed into horizontal and vertical component, as shown on Fig. 16–7.

▪ The actuating force F can be found by using the condition that the summation of the moments about the hinge pin is zero.




(3)𝑀𝑁 = න𝑑𝑁 𝑎 sin 𝜃 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃

The moment arm of the normal force 𝑑𝑁 about the pin is 𝑎 sin 𝜃. Designating the moment of the normal forces by 𝑀𝑁 and summing these about the hinge pin give

𝐹 =𝑀𝑁 −𝑀𝑓

𝑐(4)

The actuating force F must balance these two moments:

The frictional forces have a moment arm about the

pin of 𝑟 − 𝑎 cos 𝜃 . The moment 𝑀𝑓 of these frictional forces is:

𝑀𝑓 = න𝑓𝑑𝑁 𝑟 − 𝑎 cos 𝜃 =𝑓𝑝𝑎𝑏𝑟

sin 𝜃𝑎න

𝜃1

𝜃2

sin 𝜃 𝑟 − 𝑎 cos 𝜃 𝑑𝜃

(2)


Step 3 (continued)

▪ If we make MN = Mf , self-locking is obtained, and no actuating force is required.

▪ To obtain self-energizing condition, the dimension a in figure must be such that

(5)𝑀𝑁 > 𝑀𝑓

▪ The torque 𝑇 applied to the drum by the brake shoe is the sum of the frictional forces 𝑓𝑑𝑁 times the radius of the drum:

𝑇 = න𝑓𝑟 𝑑𝑁 =𝑓𝑝𝑎𝑏𝑟

2

sin 𝜃𝑎න

𝜃1

𝜃2

sin 𝜃 𝑑𝜃 =𝑓𝑝𝑎𝑏𝑟

2 cos 𝜃1 − cos 𝜃2sin 𝜃𝑎

(6)

Braking capacity


Step 3 (continued)

▪ The hinge pin reactions are found by taking a summation of the horizontal and vertical forces, thus we have

(d)𝑅𝑥 = න𝑑𝑁 cos 𝜃 − න𝑓𝑑𝑁 sin 𝜃 − 𝐹𝑥 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝐴 − 𝑓𝐵 − 𝐹𝑥

(e)𝑅𝑦 = න𝑑𝑁 sin 𝜃 + න𝑓𝑑𝑁 cos 𝜃 − 𝐹𝑦 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝐵 + 𝑓𝐴 − 𝐹𝑦

𝐴 = න

𝜃1

𝜃2

sin 𝜃 cos 𝜃 𝑑𝜃 = อ1

2sin2 𝜃

𝜃1

𝜃2

𝐵 = න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃 = อ𝜃

2−1

4sin 2 𝜃

𝜃1

𝜃2

where:


Step 3 (continued)

▪ The direction of the frictional forces is reversed if the rotation is reversed. Thus, for counterclockwise rotation of drum, the actuating force become

𝐹 =𝑀𝑁 +𝑀𝑓

𝑐(7)

▪ Since both moments have the same sense, the self-energizing effect is lost and also self-locking.


Step 3 (continued)

(g)

𝑅𝑦 = න𝑑𝑁 sin 𝜃 − න𝑓𝑑𝑁 cos 𝜃 − 𝐹𝑦

=𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝐵 − 𝑓𝐴 − 𝐹𝑦

(f)𝑅𝑥 = න𝑑𝑁 cos 𝜃 + න𝑓𝑑𝑁 sin 𝜃 − 𝐹𝑥

=𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝐴 + 𝑓𝐵 − 𝐹𝑥

For counterclockwise

rotation

▪ Also, for counterclockwise rotation the signs of frictional terms in the equilibrium for the pin reactions change, and equation (d) and (e) become:


Figure 16–8Brake with internal expanding shoes; dimensions in millimeters.

EXAMPLE 16–2

The brake shown in Fig. 8 has 300 mm in diameter and is actuated by a mechanism that exerts the same force F on each shoe. The shoes are identical and have a face width of 32 mm. The lining is a molded asbestos having a coefficient of friction of 0.32 and a pressure limitation of 1000 kPa. Estimate the maximum

(a) Actuating force F.(b) Braking capacity.(c) Hinge-pin reactions.


Known

𝑏 = 32 𝑚𝑚𝑓 = 0.32𝑝𝑎 = 1000 𝑘𝑃𝑎

𝜃1 = 0°𝜃2 = 126°𝜃𝑎 = 90°

𝑎 = 1122 + 502 = 122.7 𝑚𝑚


𝜃1 = 0°

𝜃2 = 126°

𝜃𝑎 = 90°

𝑑𝑁

𝑓𝑑𝑁

Solution

𝑀𝑓 =𝑓𝑝𝑎𝑏𝑟

sin 𝜃𝑎න

0

𝜃2


The moment of the frictional force :

=𝑓𝑝𝑎𝑏𝑟

sin 𝜃𝑎−𝑟 cos 𝜃 0

𝜃2 − 𝑎1

2sin2𝜃

0

𝜃2

=𝑓𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝑟 −𝑟 cos 𝜃2 −

𝑎

2sin2𝜃2

=(0.32)(1000 × 103) (0.032)(0.150)

sin 90°

× 0.150−0.150 cos 126° −0.1227

2sin2126°

= 304 𝑁𝑚

(a) Actuating force F.


The moment of the normal forces

𝑀𝑁 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎න

0

𝜃2

sin2 𝜃 𝑑𝜃 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎

𝜃

2−1

4sin 2𝜃

0

𝜃2

=𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎

𝜃22−1

4sin 2𝜃2

=(1000 × 103)(0.032)(0.150)(0.1227)

sin 90°

126° ×𝜋

180°2

−1

4sin 2(126°)

= 788 𝑁𝑚

The actuating force is


𝑐=788 − 304

100 + 112= 2.28 𝑘𝑁 Answer (a)


(b) Braking capacity.

𝑇𝑅 =𝑓𝑝𝑎𝑏𝑟


=(0.32)(1000 × 103)(0.032) (0.150)2 cos 0 − cos 126°

sin 90= 366 𝑁𝑚

The torque contributed by the left-hand shoe cannot be obtained until we learn its maximum operating pressure. Equations (16–2) and (16–3) indicate that the frictional and normal moments are proportional to this pressure. Thus, for the left-hand shoe,

𝑀𝑁 =788𝑝𝑎1000

𝑀𝑓 =304𝑝𝑎1000


2.28 =

788𝑝𝑎1000 +

304𝑝𝑎1000

100 + 112

𝐹 = 2.28 𝑘𝑁

𝑝𝑎 = 443 𝑘𝑃𝑎

𝑇𝐿 =𝑓𝑝𝑎𝑏𝑟


=(0.32)(443 × 103)(0.032) (0.150)2 cos 0 − cos 126°

sin 90= 162 𝑁𝑚

The braking capacity is the total torque:

Answer (b)

𝑇 = 𝑇𝑅 + 𝑇𝐿 = 366 + 162 = 528 𝑁𝑚


𝑐Left shoe:


(c) Hinge-pin reactions.Right shoe:

𝐴 = อ1

2sin2 𝜃

0°

126°

= 0.3273

𝐵 = อ𝜃

2−1

4sin 2 𝜃

0°

126°

= 1.3373

𝐷 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎=(1000 × 103)(0.032)(0.15)

sin 90°= 4.8 𝑘𝑁

𝐹𝑥 = 2.28 sin 24° = 0.93 𝑘𝑁

𝐹𝑦 = 2.28 cos 24° = 2.08 𝑘𝑁

𝑅𝑥 = 𝐷 𝐴 − 𝑓𝐵 − 𝐹𝑥 = −1.410 𝑘𝑁

𝑅𝑦 = 𝐷 𝐵 + 𝑓𝐴 − 𝐹𝑦 = 4.839 𝑘𝑁

𝑅 = 𝑅𝑥2 + 𝑅𝑦

2 = 5.04 𝑘𝑁


(c) Hinge-pin reactions.Left shoe:

𝐴 = อ1

2sin2 𝜃

0°

126°

= 0.3273

𝐵 = อ𝜃

2−1

4sin 2 𝜃

0°

126°

= 1.3373

𝐷 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎=(443 × 103)(0.032)(0.15)

sin 90°= 2.13 𝑘𝑁

𝐹𝑥 = 2.28 sin 24° = 0.93 𝑘𝑁

𝐹𝑦 = 2.28 cos 24° = 2.08 𝑘𝑁

𝑅𝑥 = 𝐷 𝐴 + 𝑓𝐵 − 𝐹𝑥 = 0.678 𝑘𝑁

𝑅𝑦 = 𝐷 𝐵 − 𝑓𝐴 − 𝐹𝑦 = 0.538 𝑘𝑁

𝑅 = 𝑅𝑥2 + 𝑅𝑦

2 = 0.866 𝑘𝑁



09.04. External Contracting Rim Clutches

and Brakes



and Flywheel Part I


9.3. External Contracting Rim Clutches & Brakes

The patented clutch-brake of figure has external contracting friction elements, but the actuating mechanism is pneumatic.

The mechanism can be classified as:

1. Solenoids

2. Levers, linkages, or toggle devices

3. Linkages with spring loading

4. Hydraulics and pneumatic devices

Figure 16–10An external contracting clutch-brake


The notation for external contracting shoes is shown in Figure 11.

The moments of friction and normal forces about the hinge pin are the same as for the internal expanding shoes.

Equations (2) and (3) apply and repeated here for convenience:

Force Analysis

Figure 16–11Notation of externalcontracting shoes.

(3)

𝑀𝑁 = න𝑑𝑁 𝑎 sin 𝜃 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃

The moment of the normal forces by 𝑀𝑁 and summing these about the hinge pin give

The moment 𝑀𝑓 of frictional forces is:

𝑀𝑓 = න𝑓𝑑𝑁 𝑟 − 𝑎 cos 𝜃 =𝑓𝑝𝑎𝑏𝑟

sin 𝜃𝑎න

𝜃1

𝜃2


(2)


(3)𝑀𝑁 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃

The moment of the normal forces by 𝑀𝑁 and summing these about the hinge pin give

The moment 𝑀𝑓 of frictional forces is:


sin 𝜃𝑎න

𝜃1

𝜃2

sin 𝜃 𝑟 − 𝑎 cos 𝜃 𝑑𝜃 (2)

Both these equations give positive values for clockwise moments (Fig. 16–11) when used for external contracting shoes. The actuating force must be large enough to balance both moments:


𝑐(11)


The horizontal and vertical reactions at the hinge pin are:

(d)𝑅𝑥 = න𝑑𝑁 cos 𝜃 + න𝑓𝑑𝑁 sin 𝜃 − 𝐹𝑥 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝐴 + 𝑓𝐵 − 𝐹𝑥

(e)𝑅𝑦 = න𝑓𝑑𝑁 cos 𝜃 − න𝑑𝑁 sin 𝜃 − 𝐹𝑦 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝑓𝐴 − 𝐵 − 𝐹𝑦

𝐴 = න

𝜃1

𝜃2

sin 𝜃 cos 𝜃 𝑑𝜃 = อ1

2sin2 𝜃

𝜃1

𝜃2

𝐵 = න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃 = อ𝜃

2−1

4sin 2 𝜃

𝜃1

𝜃2

where:


If the rotation is counterclockwise, the sign of the frictional term in each equation is reversed. Thus equation for actuating force becomes


𝑐(13)

and self-energization exist for counterclockwise rotation.

(14b)𝑅𝑦 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎−𝑓𝐴 − 𝐵 + 𝐹𝑦

(14a)𝑅𝑥 =𝑝𝑎𝑏𝑟

sin 𝜃𝑎𝐴 − 𝑓𝐵 − 𝐹𝑥

The horizontal and vertical reaction are found to be:

•When external contracting designs are used as clutches, the effect of centrifugal force is to decrease the normal force. Thus, as the speed increases, a larger value of the actuating force 𝐹 is required.


dNf dN

A special case arise when the pivot is symmetrically located and also placed so that the moment of the friction forces about the pivot is zero. The geometry of such a brake will be similar to that of Fig. 12 (a).

Figure 16–12(a) Brake with symmetricalpivoted shoe; (b) wear ofbrake lining.


• To get a pressure-distribution relation, we assume that the lining wear is such to retain its cylindrical shape, much as a milling machine cutter feeding in the x direction would do to the shoe held in a vise, see Fig. 12 (b).

• This means the abscissa component of wear is 𝑤0 for all positions 𝜃. If wear in the radial direction is expressed as 𝑤(𝜃), then.

𝑤(𝜃) = 𝑤0 cos 𝜃

• The radial wear 𝑤(𝜃) can be expressed as

where K is a material constant, P is pressure, V is rim velocity, and t is time

𝑤(𝜃) = 𝐾𝑃𝑉𝑡


• Denoting 𝑃 as 𝑝(𝜃) above and solving for 𝑝(𝜃) gives

𝑝 𝜃 =𝑤(𝜃)

𝐾𝑉𝑡=𝑤0 cos 𝜃

𝐾𝑉𝑡

• Proceeding to the force analysis, we observe from Fig. 12 (a) that

• Since all elemental surface areas of the friction material see the same rubbing speed for the same duration, Τ𝑤0 (𝐾𝑉𝑡) is a constant and

where 𝑝𝑎 is the maximum value of 𝑝(𝜃).

𝑝 𝜃 = 𝑐𝑜𝑠𝑡𝑎𝑛𝑡 cos 𝜃 = 𝑝𝑎 cos 𝜃 (c)

𝑑𝑁 = 𝑝𝑏𝑟𝑑𝜃

𝑑𝑁 = 𝑝𝑎𝑏𝑟 cos 𝜃 𝑑𝜃

or (d)

(e)


• The distance a to the pivot is chosen by finding where the moment of the frictional forces Mf is zero.

• First, this ensures that reaction Ry is at the correct location to establish symmetrical wear.

• Second, a cosinusoidal pressure distribution is sustained, preserving our predictive ability. Symmetry means 𝜃1 = 𝜃2, so

𝑀𝑓 = 2න

0

𝜃2

𝑓𝑑𝑁 𝑎 cos 𝜃 − 𝑟 = 0

Substituting Eq. (e) gives

2𝑓𝑝𝑎𝑏𝑟න

0

𝜃2

𝑎 cos2𝜃−𝑟 cos 𝜃 𝑑𝜃 = 0

𝑎 =4𝑟 sin 𝜃2

2𝜃2 + sin 2𝜃2from which : (15)


• The distance a depends on the pressure distribution. Mislocatingthe pivot makes Mf zero about a different location, so the brake lining adjusts its local contact pressure, through wear, to compensate.

• With the pivot located according Eq.(15), the moment about the pin is zero.


• Note, too, that 𝑅𝑥 = −𝑁 and 𝑅𝑦 = −𝑓𝑁, as might be expected for the

particular choice of the dimension a.

• Therefore the torque capacity is

𝑇 = 𝑎𝑓𝑁 (18)

න𝑓𝑑𝑁 sin 𝜃 = 0

where, because of symmetry :

න𝑑𝑁 sin 𝜃 = 0

𝑅𝑦 = 2න0

𝜃2

𝑓𝑑𝑁 cos 𝜃 =𝑝𝑎𝑏𝑟𝑓

22𝜃2 + sin 2𝜃2 (17)

The horizontal and vertical reactions are

(16)𝑅𝑥 = 2න0

𝜃2

𝑑𝑁 cos 𝜃 =𝑝𝑎𝑏𝑟

22𝜃2 + sin 2𝜃2


The block-type hand brake shown in the figure has a face width of 1.25 in and a mean coefficient of friction of 0.25. For an estimated actuating force of 90 lbf, find the maximum pressure on the shoe and find the braking torque.

Q 16–5

Known:

𝑏 = 1.25 𝑖𝑛𝐹 = 90 𝑙𝑏𝑓𝑓 = 0.25


𝑎 = 62 + 82 = 10 𝑖𝑛

𝜃1

𝛼

𝜃1 = 45° − 𝛼

= 45° − tan−16

8= 8.13°

𝜃2 = 8.13° + 90° = 98.13°

𝜃𝑎 = 90°

Solution


dN

fdN

𝜃

r

𝑎 cos 𝜃

𝑟 − 𝑎 cos𝜃

𝑎 sin 𝜃

The moment of frictional force:


sin 𝜃𝑎න

𝜃1

𝜃2


=(0.25)𝑝𝑎(1.25)(6)

sin 90°න

8.13°

98.13°

sin 𝜃 6 − 10 cos 𝜃 𝑑𝜃

= 3.728𝑝𝑎 𝑙𝑏𝑓. 𝑖𝑛

𝑀𝑁 =𝑝𝑎𝑏𝑟𝑎

sin 𝜃𝑎න

𝜃1

𝜃2

sin2 𝜃 𝑑𝜃

The moment of normal force:

=𝑝𝑎(1.25)(6)(10)

sin 90°න

8.13°

98.13°

sin2 𝜃 𝑑𝜃 = 64.405𝑝𝑎 𝑙𝑏𝑓. 𝑖𝑛


𝐹𝑐 = 𝑀𝑁 −𝑀𝑓

(90)(2) = (69.405 − 3.728)𝑝𝑎

𝑝𝑎 = 27.4 𝑝𝑠𝑖 Answer

(b) Braking capacity.

𝑇𝑅 =𝑓𝑝𝑎𝑏𝑟


=(0.25)(27.4)(1.25) (6)2 cos 8.13° − cos 98.13°

sin 90°= 348.7 𝑙𝑏𝑓. 𝑖𝑛 Answer



09.05. Band-Type Clutches and Brakes



and Flywheel Part I


Application, mostly:

• Power excavators

• Hoisting machinery

• Others

9.4. Band-Type Clutches and Brakes

Broderson IC-80-1D Carry Deck

• Engine: Continental TM27, 4 cyl., 165 C.I.D., 64 hp at governed speed.

• Transmission: Borg Warner, 72T/T18, Reversing

Gearbox and Manual Shift Three Speed Gearbox.

• Brakes: Service - 4-wheel drum type brakes, Parking - Band Type Mounted on Transmission


(𝑃 + 𝑑𝑃) sin𝑑𝜃

2+ 𝑃 sin

𝑑𝜃

2− 𝑑𝑁 = 0

or 𝑑𝑁 = 𝑃𝑑𝜃

(a)

(b)

for small angles,

∗) sin𝑑𝜃

2=𝑑𝜃

2∗)𝑑𝑃 × 𝑑𝜃 ≈ 0

Note:Forces in vertical direction gives:

𝑃1 > 𝑃2

Why?

Figure 16–13Forces on a brake band.


Forces in horizontal direction gives

(𝑃 + 𝑑𝑃) cos𝑑𝜃

2− 𝑃 cos

𝑑𝜃

2− 𝑓𝑑𝑁 = 0

or𝑑𝑃 − 𝑓𝑑𝑁 = 0

(c)

(d)

Substituting the value of dN from Eq. (b) in (d) and integrating gives

න

𝑃2

𝑃1𝑑𝑃

𝑃= 𝑓න

0

𝜙

𝑑𝜃 ⇒ ln𝑃1𝑃2

= 𝑓𝜙

and𝑃1𝑃2

= 𝑒𝑓𝜙

(19)


𝑃𝑑𝜃 = 𝑝𝑏𝑟𝑑𝜃

The torque may be obtained from the equation

𝑇 = 𝑃1 − 𝑃2𝐷

2(20)

The normal force dN acting on element is

𝑑𝑁 = 𝑝𝑏𝑟𝑑𝜃 (e)

Substitution of dN from Eq. (b) gives

Therefore:

𝑝 =𝑃

𝑏𝑟=2𝑃

𝑏𝐷(21)

The maximum pressure will occur at the toe with the value 𝑝𝑎 =

2𝑃1𝑏𝐷

(22)

The pressure is therefore proportional to the tension in the band.


Q 16–11

The maximum band interface pressure on the brake shown in the figure is 620 kPa. Use a 350 mm diameter drum, a band width of 25 mm, a coefficient of friction of 0.30, and an angle-of-wrap of 270◦. Find the band tensions and the torque capacity.

Known:𝐷 = 350 𝑚𝑚𝑏 = 100 𝑚𝑚𝑝𝑎 = 620 𝑘𝑃𝑎𝑓 = 0.30𝜙 = 270°


Solution

The band tensions

𝑝𝑎 =2𝑃1𝑏𝐷

Eq. (22)

𝑃1𝑃2

= 𝑒𝑓𝜙Eq. (19)

𝑃1 =𝑝𝑎𝑏𝐷

2=(620)(0.1)(0.35)

2= 10.85 𝑘𝑁 Answer

𝑃2 =𝑃1𝑒𝑓𝜙

=10.85

𝑒(0.3)(270°×𝜋

180°)= 2.64 𝑘𝑁 Answer

The torque capacity.

𝑇 = 𝑃1 − 𝑃2𝐷

2Eq. (20) = 10.85 − 2.64

0.350

2= 1.437 𝑘𝑁𝑚 Answer



09.06. Frictional-Contact Axial Clutches



and Flywheel Part I


• Mating frictional members are moved in a direction parallel to shaft.

• Most application: automotive

9.5. Frictional-Contact Axial Clutches


Figure 16–15An oil-actuated multiple-disk clutch-brake for operation in an oil bath or spray

Figure 16–14Cross-sectional view of a single-plate clutch; A, driver; B, driven plate (keyed to driven shaft); C, actuator


Advantages of the disk clutch:

Free from centrifugal effects

Large frictional area which can be installed in small space

More effective heat-dissipation surfaces

Favorable pressure distribution

Two methods for the analysis:

➢ Uniform wear

➢ Uniform pressure


𝑝𝑟 = 𝑝𝑎𝑑

2or 𝑝 = 𝑝𝑎

𝑑

2𝑟

Uniform wear

After initial wear has taken place and the disks have worn down to the point where uniform wear becomes possible, the greatest pressure must occur at 𝑟 = Τ𝐷 2 in order for wear to be uniform. Denoting the maximum pressure by 𝑝𝑎 , we can then write.(see the explanation in Shigley, page 847)

Figure 16–16Disk friction member.(a)


Figure 16–16Disk friction member.

Referring to Fig. 16, we have an element of area of radius 𝑟 and thickness 𝑑𝑟 . The area of this element is 2𝜋𝑟𝑑𝑟 , so that the normal force acting upon this element is 𝑑𝐹 = 2𝜋𝑝𝑟𝑑𝑟. Thus, the total normal force become:

𝐹 = න

Τ𝑑 2

𝐷/2

2𝜋𝑝𝑟𝑑𝑟 = 𝜋𝑝𝑎𝑑 න

Τ𝑑 2

𝐷/2

𝑑𝑟 =𝜋𝑝𝑎𝑑

2𝐷 − 𝑑 (23)

The torque is found by integrating the product of the frictional force and the radius:

𝑇 = න

Τ𝑑 2

𝐷/2

2𝜋𝑓𝑝𝑟2𝑑𝑟 = 𝜋𝑓𝑝𝑎𝑑 න

Τ𝑑 2

𝐷/2

𝑟𝑑𝑟 =𝜋𝑓𝑝𝑎𝑑

8𝐷2 − 𝑑2

𝑇 =𝐹𝑓

4𝐷 + 𝑑

(24)

(25)


Uniform Pressure

When uniform pressure can be assumed over the area of the disk, the actuating force 𝐹 is simply the product of the pressure and the area

𝐹 =𝜋𝑝𝑎4

𝐷2 − 𝑑2 (26)

Note:Equations (26) and (28) are for single pair of mating surface.This value must be multiplied by the number of pairs of surfaces in contact.

As before, the torque is found by integrating the product of the frictional force and the radius:

𝑇 = 2𝜋𝑓𝑝𝑎 න

Τ𝑑 2

𝐷/2

𝑟2𝑑𝑟 =𝜋𝑓𝑝𝑎24

𝐷3 − 𝑑3

𝑇 =𝐹𝑓

3

𝐷3 − 𝑑3

𝐷2 − 𝑑2(28)

(27)


Figure 16–17Dimensionless plot of Eqs. (b) and (c).

Difference characteristic of uniform pressure and uniform wear

𝑈𝑛𝑖𝑓𝑜𝑟𝑚𝑊𝑒𝑎𝑟 (𝑜𝑙𝑑 𝑐𝑙𝑢𝑡𝑐ℎ):𝑇

𝑓𝐹𝐷=1

41 +

𝑑

𝐷

𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 (𝑛𝑒𝑤 𝑐𝑙𝑢𝑡𝑐ℎ):

𝑇

𝑓𝐹𝐷=1

3

1 −𝑑𝐷

3

1 −𝑑𝐷

2

(b)

(c)


Q 16–16

A plate clutch has a single pair of mating friction surfaces 250-mm OD by 175-mm ID. The mean value of the coefficient of friction is 0.30, and the actuating force is 4 kN.a) Find the maximum pressure and the torque capacity using the

uniform-wear model.b) Find the maximum pressure and the torque capacity using the

uniform-pressure model.

Known:

𝑂𝐷 = 250 𝑚𝑚𝐼𝐷 = 175 𝑚𝑚𝑓 = 0.3𝐹 = 4 𝑘𝑁


Solution

a) Find the maximum pressure and the torque capacity using the uniform-wear model.

𝐹 =𝜋𝑝𝑎𝑑

2𝐷 − 𝑑Eq. (23)

𝑇 =𝐹𝑓

4𝐷 + 𝑑

Eq. (25)

𝑝𝑎 =2𝐹

𝜋𝑑 𝐷 − 𝑑=

2(4000)

𝜋(0.175) 0.250 − 0.175= 194 𝑘𝑃𝑎 Answer

=(4000)(0.3)

40.250 + 0.175 = 127.5 𝑁𝑚 Answer


b) Find the maximum pressure and the torque capacity using the uniform-pressure model.

𝐹 =𝜋𝑝𝑎4

𝐷2 − 𝑑2

Eq. (26)

𝑇 =𝐹𝑓

3

𝐷3 − 𝑑3

𝐷2 − 𝑑2

Eq. (28)

𝑝𝑎 =4𝐹

𝜋 𝐷2 − 𝑑2=

4(4000)

𝜋 0.2502 − 0.1752= 159 𝑘𝑃𝑎 Answer

=(4000)(0.3)

3

0.2503 − 0.1753

0.2502 − 0.1752= 128 𝑁𝑚 Answer



09.07. Disk Brakes



and Flywheel Part I


• There is no fundamental difference

between a disk clutch and a disk

brake.

• The analysis of preceding section

applies to disk brakes too.

• Disk brake has no self-energization,

and hence is not so susceptible to

changes in the coefficient of friction.

9.6. Disk Brakes


• Fig. 18 shows a floating caliper disk brake.

• The caliper support a single floating piston

actuated by hydraulic pressure.

• The action is like a screw clamp, with the

piston replacing the function of the screw.

• The floating action compensates for wear

and ensures a fairly constant pressure over

the area of the friction pads.

Figure 16–18An automotive disk brake.


Figure 16–19Geometry of contact area of an annular-pad segment of a caliper brake.

Fig. 19 is the geometry of an annular-pad brake contact area. The governing axial wear equation is (see Eq. 12-27, p. 663 -Shigley)

𝑤 = 𝑓1𝑓2𝐾𝑃𝑉𝑡

Of interest also is the effective radius 𝑟𝑒 , which is the radius of an equivalent

shoe of infinitesimal radial thickness.

Annular-pad brake contact area.

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


If 𝑝 is the local contact pressure, the actuating force 𝐹 is

𝐹 = න𝜃1

𝜃2

න𝑟𝑖

𝑟𝑜

𝑝𝑟𝑑𝑟𝑑𝜃 = 𝜃2 − 𝜃1 න𝑟𝑖

𝑟𝑜

𝑝𝑟𝑑𝑟

(29)

Figure 16–19Geometry of contact area of an annular-pad segment of a caliper brake.

and the friction torque 𝑇 is

𝑇 = න𝜃1

𝜃2

න𝑟𝑖

𝑟𝑜

𝑓𝑝𝑟2𝑑𝑟𝑑𝜃 = 𝜃2 − 𝜃1 𝑓න𝑟𝑖

𝑟𝑜

𝑝𝑟2𝑑𝑟

(30)

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

𝑟𝑒 =𝑇

𝑓𝐹=𝑟𝑖𝑟𝑜 𝑝𝑟2𝑑𝑟

𝑟𝑖𝑟𝑜 𝑝𝑟𝑑𝑟

(31)


The locating coordinate ҧ𝑟 of the activating force is found by taking moments about the x axis:

𝑀𝑥 = 𝐹 ҧ𝑟 = න𝜃1

𝜃2

න𝑟𝑖

𝑟𝑜

𝑝𝑟 𝑟 sin 𝜃 𝑑𝑟𝑑𝜃 = cos 𝜃1 − cos 𝜃2 න𝑟𝑖

𝑟𝑜

𝑝𝑟2𝑑𝑟

ҧ𝑟 =𝑀𝑥

𝐹=

cos 𝜃1 − cos 𝜃2𝜃2 − 𝜃1

𝑟𝑒 (32)


Uniform Wear

Using that pressure distribution 𝑝 = Τ𝑝𝑎𝑟𝑖 𝑟. Eqs. (29) to (32) become

𝐹 = 𝜃2 − 𝜃1 න𝑟𝑖

𝑟𝑜 𝑝𝑎𝑟𝑖𝑟

𝑟𝑑𝑟 = 𝜃2 − 𝜃1 𝑝𝑎𝑟𝑖 𝑟𝑜 − 𝑟𝑖 (33)

𝑇 = 𝜃2 − 𝜃1 𝑓න𝑟𝑖

𝑟𝑜 𝑝𝑎𝑟𝑖𝑟

𝑟2𝑑𝑟 =1

2𝜃2 − 𝜃1 𝑓𝑝𝑎𝑟𝑖 𝑟𝑜

2 − 𝑟𝑖2 (34)

𝑟𝑒 =𝑇

𝑓𝐹=𝑟𝑖𝑟𝑜 𝑝𝑎𝑟𝑖

𝑟𝑟2𝑑𝑟

𝑟𝑖𝑟𝑜 𝑝𝑎𝑟𝑖

𝑟𝑟𝑑𝑟

=𝑟𝑜 + 𝑟𝑖2

(35)

ҧ𝑟 =𝑀𝑥

𝐹=


𝑟𝑒 =cos 𝜃1 − cos 𝜃2

𝜃2 − 𝜃1

𝑟𝑜 + 𝑟𝑖2

(36)


Uniform Pressure

In this situation, approximated by a new brake, 𝑝 = 𝑝𝑎 , Eqs. (29) to (32) become

𝐹 = 𝜃2 − 𝜃1 න𝑟𝑖

𝑟𝑜

𝑝𝑎𝑟𝑑𝑟 =1

2𝜃2 − 𝜃1 𝑝𝑎 𝑟𝑜

2 − 𝑟𝑖2 (37)

𝑇 = 𝜃2 − 𝜃1 𝑓න𝑟𝑖

𝑟𝑜

𝑝𝑎𝑟2𝑑𝑟 =

1

3𝜃2 − 𝜃1 𝑓𝑝𝑎 𝑟𝑜

3 − 𝑟𝑖3 (38)

𝑟𝑒 =𝑇

𝑓𝐹=𝑟𝑖𝑟𝑜 𝑝𝑎𝑟

2𝑑𝑟

𝑟𝑖𝑟𝑜 𝑝𝑎𝑟𝑑𝑟

=2

3

𝑟𝑜3 − 𝑟𝑖

3


2(39)

ҧ𝑟 =𝑀𝑥

𝐹=


𝑟𝑒 =cos 𝜃1 − cos 𝜃2

𝜃2 − 𝜃1

2

3


3


2(40)


EXAMPLE 16–3

Two annular pads, 𝑟𝑖 = 3.875 in, 𝑟𝑜 = 5.50 in, subtend an angle of 108°, have a coefficient of friction of 0.37, and are actuated by a pair of hydraulic cylinders 1.5 in in diameter. The torque requirement is 13000 lbf · in. For uniform weara) Find the largest normal pressure 𝑝𝑎 .b) Estimate the actuating force 𝐹.c) Find the equivalent radius 𝑟𝑒 and force location ҧ𝑟.d) Estimate the required hydraulic pressure.

Known 𝑟𝑖 = 3.875 𝑖𝑛𝑟𝑜 = 5.500 𝑖𝑛𝑓 = 0.37𝜃1 = 36°𝜃2 = 144°


Solution Two annular pads, total torque = 13000 lbf.in.For each pad, 𝑇 = Τ13000 2 = 6500 𝑙𝑏𝑓. 𝑖𝑛

𝑇 =1

2𝜃2 − 𝜃1 𝑓𝑝𝑎𝑟𝑖 𝑟𝑜

2 − 𝑟𝑖2Eq. (34)

a) Find the largest normal pressure 𝑝𝑎 for uniform wear.

𝑝𝑎 =2𝑇

𝜃2 − 𝜃1 𝑓𝑟𝑖 𝑟𝑜2 − 𝑟𝑖

2

=2(6500)

(144° − 36°) ×𝜋180

(0.37)(3.875)(5.52 − 3.8752)= 315.8 𝑝𝑠𝑖

Answer

b) Estimate the actuating force 𝐹 for uniform wear.

𝐹 = 𝜃2 − 𝜃1 𝑝𝑎𝑟𝑖 𝑟𝑜 − 𝑟𝑖Eq. (33)

= 144° − 36°𝜋

180315.8 3.875 5.500 − 3.875 = 3748 𝑙𝑏𝑓

Answer


c) Find the equivalent radius 𝑟𝑒 and force location ҧ𝑟 for uniform wear.

𝑟𝑒 =𝑟𝑜 + 𝑟𝑖2

=5.500 + 3.875

2= 4.688 𝑖𝑛Eq. (35)

ҧ𝑟 =cos 𝜃1 − cos 𝜃2

𝜃2 − 𝜃1𝑟𝑒 =

cos 36° − cos 144°

144° − 36° ×𝜋180

4.688 = 4.024 𝑖𝑛Eq. (36)

Answer

d) Estimate the required hydraulic pressure.

Each cylinder supplies the actuating force, 3748 lbf

𝑝ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 =𝐹

𝐴𝑝=

3748𝜋4(1.5)2

= 2121 𝑝𝑠𝑖 Answer


Circular (Button) Pad Caliper Brake

Figure 20 displays the circular pad geometry. Numerical integration is necessary to analyze this brake since the boundaries are difficult to handle in closed form. Table 1 gives the parameters for this brake as determined by Fazekas.

Table 1 Parameters for a Circular-Pad Caliper Brake

Figure 16–20Geometry of circular pad of acaliper brake.

Effective radius, 𝑟𝑒 = 𝛿𝑒


Figure 16–20Geometry of circular pad of acaliper brake.

Effective radius, 𝑟𝑒 = 𝛿𝑒

The effective radius is given by

𝑟𝑒 = 𝛿𝑒 (41)

The actuating force is given by

𝐹 = 𝜋𝑅2𝑝𝑎𝑣 (42)

and the torque is given by𝑇 = 𝑓𝐹𝑟𝑒 (43)


EXAMPLE 16–4

A button-pad disk brake uses dry sintered metal pads. The pad radius is ½ in, and its center is 2 in from the axis of rotation of the 312 -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.

Known

𝑅 = 0.5 𝑖𝑛𝑒 = 2 𝑖𝑛𝑝𝑚𝑎𝑥 = 350 𝑝𝑠𝑖𝑓 = 0.31


𝑅

𝑒=0.5

2= 0.25

By interpolation:

𝛿 = 0.963𝑝𝑚𝑎𝑥

𝑝𝑎𝑣= 1.290

𝑟𝑒 = 𝛿𝑒 = 0.963 2 = 1.926 𝑖𝑛Eq. (41)

𝑝𝑎𝑣 =𝑝𝑚𝑎𝑥

1.290=

350

1.290= 135.7 𝑝𝑠𝑖

𝐹 = 𝜋𝑅2𝑝𝑎𝑣 = 𝜋 0.5 2 135.7= 106.6 𝑙𝑏𝑓

Eq. (42)

Answer

𝑇 = 𝑓𝐹𝑟𝑒 = 0.31 106.6 1.926 = 63.65 𝑙𝑏𝑓. 𝑖𝑛Eq. (43) Answer

Solution



09.08. Cone Clutches and Brakes



and Flywheel Part I

../Videos/Cone clutch works in racing cars.MP4


• Consist of a cup and a cone.

• Cone angle, the diameter and face width of the cone are the important geometric design parameters.

• If the cone angle is too small, say, less than about 8o, then the force required to disengage the clutch may be quite large.

• Wedging effect lessens rapidly when larger cone angles are used.

• A good compromise can usually be found between 10o – 15o.

8.7. Cone Clutch and Brakes

../Videos/Cone clutch works in racing cars.MP4


Actuating force and torque transmittedcan be found using Fig. 22.

Uniform Wear

The pressure relation is the same as for the axial clutch

𝑝 = 𝑝𝑎𝑑

2𝑟(a)

The element area 𝑑𝐴 of radius 𝑟 and width Τ𝑑𝑟 sin𝛼 is

𝑑𝐴 = Τ(2𝜋𝑟𝑑𝑟) sin 𝛼 (b)

Figure 16–22Contact area of a cone clutch.


Figure 16–22Contact area of a cone clutch.

As shown in Fig. 22, the operating force will be the integral of the axial component of the differential force 𝑝𝑑𝐴. Thus

𝐹 = න𝑝𝑑𝐴 sin 𝛼 = න

Τ𝑑 2

Τ𝐷 2

𝑝𝑎𝑑

2𝑟

2𝜋𝑟 𝑑𝑟

sin 𝛼(sin 𝛼)

𝐹 = 𝜋𝑝𝑎𝑑 න

Τ𝑑 2

Τ𝐷 2

𝑑𝑟 =𝜋𝑝𝑎𝑑

2𝐷 − 𝑑 (44)

which is the same result as in Eq. (23)


The differential friction force is 𝑓𝑝𝑑𝐴, and the torque is the integral of the product of this force with the radius. Thus:

𝑇 = න𝑟𝑓𝑝𝑑𝐴 = න

Τ𝑑 2

Τ𝐷 2

(𝑟𝑓) 𝑝𝑎𝑑

2𝑟

2𝜋𝑟𝑑𝑟

sin 𝛼=𝜋𝑓𝑝𝑎sin 𝛼

න

Τ𝑑 2

Τ𝐷 2

𝑟𝑑𝑟 =𝜋𝑓𝑝𝑎𝑑

8 sin 𝛼𝐷2 − 𝑑2 (45)

Note that Eq. (24) is a special case of Eq. (45), with = 90o.

Using Eq. (44), the torque can also be written

𝑇 =𝐹𝑓

4 sin 𝛼𝐷 + 𝑑 (46) Note that Eq. (25) is also a special case of Eq.

(46), with = 90o.


Uniform Pressure

Using 𝑝 = 𝑝𝑎, the actuating force and torque are found to be

𝐹 = න𝑝𝑎𝑑𝐴 sin𝛼 = න

Τ𝑑 2

Τ𝐷 2

𝑝𝑎2𝜋𝑟𝑑𝑟

sin𝛼sin𝛼 =

𝜋𝑝𝑎4

𝐷2 − 𝑑2 (47)

𝑇 = න𝑟𝑓𝑝𝑎𝑑𝐴 = න

Τ𝑑 2

Τ𝐷 2

𝑟𝑓𝑝𝑎2𝜋𝑟𝑑𝑟

sin𝛼=

𝜋𝑓𝑝𝑎12 sin𝛼

𝐷3 − 𝑑3 (48)

(49)𝑇 =𝐹𝑓

3 sin𝛼

𝐷3 − 𝑑3

𝐷2 − 𝑑2


As in the case of the axial clutch, we can write Eq. (43) dimensionlessly as

and write Eq. (49) as

This time there are six (T, α, f, F, D, and d) parameters and four pi () terms:

As in Fig. 17, we plot T sin α/( f FD) as ordinate and d/D as abscissa. The plots and conclusions are the same ➔ Students are recommended to do the plot.

𝑇 sin 𝛼

𝑓𝐹𝐷=1 +

𝑑𝐷

4

𝑇 sin 𝛼

𝑓𝐹𝐷=1

3

1 −𝑑𝐷

3

1 −𝑑𝐷

2

𝜋1 =𝑇

𝐹𝐷𝜋2 = 𝑓 𝜋3 = sin 𝛼 𝜋4 =

𝑑

𝐷


A cone clutch has D = 12 in, d = 11 in, a cone length of 2.25 in, and a coefficient of friction of 0.28. A torque of 1800 lbf.in is to be transmitted. For this requirement, estimate the actuating force and pressure by both models.

Q 16–19

Known

𝐷 = 12 𝑖𝑛𝑑 = 11 𝑖𝑛𝐿 = 2.25 𝑖𝑛𝑓 = 0.28𝑇 = 1800 𝑙𝑏𝑓. 𝑖𝑛

CLNot to scale

5.5

6.02.25

𝛼𝛼 = tan−1

0.5

2.25= 12.53°


Solution

𝑇 =𝜋𝑓𝑝𝑎𝑑

8 sin 𝛼𝐷2 − 𝑑2Eq. (45)

Uniform Wear

(1800) =𝜋(0.28)𝑝𝑎(11)

8 sin 12.53°122 − 112

𝐹 =𝜋𝑝𝑎𝑑

2𝐷 − 𝑑 =

𝜋(14.04)(11)

212 − 1 = 243 𝑙𝑏𝑓Eq. (44) Answer

𝑝𝑎 = 14.04 𝑝𝑠𝑖 Answer


Uniform Pressure

𝑇 =𝜋𝑓𝑝𝑎12 sin 𝛼

𝐷3 − 𝑑3Eq. (48)

1800 =𝜋(0.28)𝑝𝑎12 sin 12.53°

123 − 113 𝑝𝑎 = 13.42 𝑝𝑠𝑖 Answer

𝐹 =𝜋𝑝𝑎4

𝐷2 − 𝑑2 =𝜋(13.42)

4122 − 112 = 242 𝑙𝑏𝑓Eq. (47) Answer


1 2 3 4 5 6 7 8 9 10 11 12 13 14

Thank YouLecturers

Faculty of Mechanical and Aerospace Engineering

Institut Teknologi Bandung

Modul 09 Clutches, Brakes, Couplings and Flywheel Part I

modul 09 clutches, brakes, couplings and flywheel part i

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