Fundamentals Machine Elements

534 Chapter 18 Brakes and Clutches

the vertical friction force components of the lower half of the Integrating gives
shoe. For a fixed y, the vertical normal components of both
halves of the shoe are equal and opposite in direction, so that d7 cos θo sin2 π − sin2 − π + d7 sin θo π−1
the vertical reaction force is 24 4 42

Ry = 2 θ2 µ dP cos θ = µpmaxbr 2θ2 π + sin 2θ2 .
02 180◦ (18.61)
π1
= µd7 cos θo +
Making use of Eq. (18.59) gives 42

2µbr2pmax sin θ2  sin2 π sin2 − π 
Ry = d7 = µRx. 4− 4
+µd7 sin θo  2 2

Example 18.6: Pivot-Shoe Brake √ √1 − √1 .
−µr cos θo 2 + µr sin θo
22
Given: A symmetrically loaded, pivot-shoe brake has the
distance d7 shown in Fig 18.11 optimized for a 180◦ wrap an- This reduces to
gle. When the brake lining is worn out, it is replaced with a
90◦-wrap-angle lining symmetrically positioned in the shoe. d7 sin θo π−1 = µ cos θo d7 π1 √
The actuating force is 11,000 N, the coefficient of friction is 42 + −r 2 ,
0.31, the brake drum radius is 100 mm, and the brake width
is 45 mm. 42

Find: Calculate the pressure distribution in the brake shoe or
and the braking torque.
µ d7 π+1 √
Solution: The distance d7 can be expressed from Eq. (18.57) 42 −r 2
for the 180◦ wrap angle as
tan θo = π−1
42
d7

4r sin θ2 0.31 0.1273 π+1 √
(d7)180◦ = π 42 − (0.100) 2

2θ2 180◦ + sin 2θ2 = 0.1273 π − 1

4(0.1) sin 90◦ 42

= π = 0.1895,
180◦
2(90) + sin 180◦

= 0.1273 m. or θo = 10.73◦. The actuating force is

For the 90◦-wrap-angle, symmetrically loaded, pivot-shoe π/4 π/4
brake, the pressure distribution will be unsymmetrical. The
maximum pressure, which will occur at θ2, needs to be de- W= pr dθb cos θ + µpr dθb sin θ. (c)
termined from shoe equilibrium, or
−π/4 −π/4

Substituting Eq. (b) into Eq. (c) gives

π/4 π/4 π/4

pr dθ(b)(d7 sin θ) − µpr dθ(b)(d7 cos θ − r) = 0. W = rbpmax cos(θ − θo) cos θ dθ

−π/4 −π/4 −π/4

(a) π/4

If the wear rate is proportional to the pressure and the maxi- +rbpmax µ cos(θ − θo) sin θ dθ. (d)

mum pressure is at θ = θo, the pressure distribution is −π/4

p = pmax cos(θ − θo). (b) But
Substituting Eq. (b) into Eq. (a) gives
π/4 π1 cos θo, (e)
+
π/4 cos(θ − θo) cos θ dθ =
42
d7 cos (θ − θo) sin θ dθ −π/4

0= −π/4 and

π/4 π/4 π−1 sin θo. (f )
42
− µ cos(θ − θo)(d7 cos θ − r) dθ. cos(θ − θo) sin θ dθ =

−π/4 −π/4

But Substituting Eqs. (e) and (f ) into Eq. (d) while solving for the
cos(θ − θo) = cos θ cos θo + sin θ sin θo. maximum pressure gives

Therefore, W

π/4 pmax = π1 cos θo + µ π−1 sin θo
rb + 42
d7 cos θo cos θ sin θ + sin θo sin2 θ dθ
42
−π/4

π/4 11, 000
= (0.0045) [(1.285) cos 10.7◦ + (0.08835) sin 10.7◦]
= µ(cos θo cos θ + sin θo sin θ)(d7 cos θ − r)dθ.
= 1.911 × 106 Pa = 1.911 MPa.
−π/4

Band Brakes 535
From Eq. (b) the pressure distribution can be expressed as
dθ
p = 1.911 × 106 cos(θ − 10.73◦). φ

0 θ

The braking torque is Drum rotation

π/4

T = µpmaxbr2 cos(θ − θo) dθ

−π/4

= µpmaxbr2 sin π − θo + sin π
4 4 + θo

= (0.31) 1.911 × 106 (0.045)(0.1)2

× [sin 34.27◦ + sin 55.73◦] F1 F2

= 370.4 N-m.

18.9 Band Brakes (F +dF) cosd—2θ (a)
r dθ F cosd—2θ
F sind—2θ
Figure 18.16 shows the components of a band brake, which d—2θ dP d—2θ F
consists of a band wrapped partly around a drum. The brake F +dF
is actuated by pulling the band tighter against the drum, as (F +dF) sidn—2θ µdP r
shown in Fig. 18.17a. The band is assumed to be in contact dθ
with the drum over the entire wrap angle, φ in Fig. 18.17a.
The pin reaction force is given as F1 and the actuating force 0
as F2. In Fig. 18.17a, the heel of the brake is near F1 and
the toe is near F2. Since some friction will exist between the (b)
band and the drum, the actuating force will be less than the
pin reaction force, or F2 < F1. Figure 18.17: Band brake. (a) Forces acting on band; (b) forces
acting on element.
Figure 18.17b shows the forces acting on an element of
the band. The forces are the normal force, P , and the friction
force, F . Summing the forces in the radial direction while
using Fig. 18.17b gives

(F + dF ) sin dθ + F sin dθ − dP = 0;
22

dP = 2F sin dθ + dF sin dθ .
22

Since dF F , Summing the forces in the horizontal (tangential) direction
while using Fig. 18.17b gives
dθ
dP = 2F sin .
2
(F + dF ) cos dθ − F cos dθ − µdP = 0;
Since dθ/2 is small, then sin dθ/2 ≈ dθ/2. Therefore, 22

dP = F dθ. (18.62) dF cos dθ − µdP = 0.
2
Liner
Since dθ/2 is small, then cos(dθ/2) ≈ 1. Therefore,

dF − µdP = 0. (18.63)

Substituting Eq. (18.62) into Eq. (18.63) gives

F1 dF φ

dF − µF dθ = 0 or F2 F = µ dθ.
Integrating gives
0

Drum ln F1 µφπ
F2 = 180◦ ,
Band

Figure 18.16: A typical band brake. Source: Courtesy of or F1 = eµφπ/180◦ ,
Northern Tool & Equipment Company, Inc. F2
(18.64)

536 Chapter 18 Brakes and Clutches
If W = 0, the brake will self-lock, therefore:
Rotation
φ d8F2 − d10F1 = 0,

d10 = d8F2 = (0.050)(487) = 0.0195 m.
F1 1250
Cutting plane for
free-body diagram The brake will self-lock if d10 ≥ 19.5 mm.

F1 d10 W
F2

d8 18.10 Slip Clutches
d9
A clutch will often be used as a torque-limiting device, usu-
Figure 18.18: Band brake used in Example 18.7. ally to prevent machinery damage in a malfunction or unde-
sired event, although they can also be used to control peak
where φ is the wrap angle. The torque applied to the drum is torques during startup. A slip clutch, in a simple manifes-
tation, consists of two surfaces held together by a constant
T = r(F1 − F2). (18.65) force so that they slip when a preset level of torque is applied
to them. Slip clutches come in a wide variety of sizes but are
The differential normal force dP acting on the element in very compact. They are designed to be actuated only rarely
and thus the friction elements do not need to be sized for
Fig. 18.17b, with width b (coming out of the paper) and length wear. Also, slip clutches are almost always contacting disks,
mainly because it is imperative to prevent the possibility of
r dθ, is a self-energizing shoe (which would compromise the torque-
limiting control).
dP = pbr dθ, (18.66)
Slip clutches are mainly used to protect machinery el-
where p is the contact pressure. Substituting Eq. (18.66) into ements and are not relied upon to prevent personal injury.
After slip clutches spend long time periods in contact, the
Eq. (18.62) gives friction surfaces can stick or weld together, requiring a larger
torque to initiate slippage. This torque is usually not large
p= F. (18.67) enough to break a gear, for example, but can be enough of an
br increase to result in a serious injury. Regardless, a slip clutch
is an effective torque limiter and can be used instead of shear
The pressure is proportional to the tension in the band. The pins or keyways with the advantage that no maintenance is
maximum pressure occurs at the heel, or near the pin reaction required after the excessive torque condition has been cor-
force, and has the value rected.

pmax = F1 . (18.68)
br

Example 18.7: Band Brake

Given: The band brake shown in Fig. 18.18 has r = 100 mm, Case Study 18.1: Roller Coaster
b = 25 mm, d9 = 225 mm, d8 = 50 mm, d10 = 12 mm, Braking System
φ = 270◦, µ = 0.2, and pmax = 500 kPa.
Find: Determine the braking torque, actuating force, and This case study discusses some of the design decisions that
value of d10 when the brake force locks. are made for the braking system of a roller coaster such as
Solution: From Eq. (18.68), the pin reaction force is shown in Fig. 18.19. Figure 18.20 is a schematic of a typi-
cal roller coaster brake. The brake systems for automobiles
F1 = pmaxbr = 500 × 103 (0.025)(0.100) = 1250 N. or other vehicles, that is, drum and disk brakes associated
with wheels, will not work for a roller coaster. The friction
From Eq. (18.64), the actuating force is between the steel wheel and the steel rail is extremely low,
so that the cars can slide on the rails without losing much
F2 = F1e−µφπ/180◦ = 1250e−0.2(270)π/180 = 487 N. energy. This is important for roller coaster operation, as the
cars are pulled up a main hill (usually by a chain drive — see
From Eq. (18.65), the braking torque is Section 19.6), and then coast for the remainder of the ride.

T = r(F1 − F2) = (0.100)(1250 − 487) = 76.3 Nm. Roller coasters have braking stations, where the cars must
be brought to a stop so that passengers can get on or off.
Summing the moments about the hinge pin and setting the A typical brake system uses pads that bear against a flange,
sum equal to zero give or fin, on the roller coaster cars. The roller coaster operator
controls a pneumatic actuator that applies force through the
−d9W + d8F2 − d10F1 = 0. linkage to aggressively decelerate the cars. The maximum
braking force should not expose the passengers to greater
Solving for the actuating force W , than 1/2 g of deceleration, and preferably not more than 2.5
m/s2 or so.
W = d8F2 − d10F1 = (0.050)(487) − (0.012)(1250) = 41.56 N.
d9 0.225 Typically, several brakes are used at the point of access to a
roller coaster. This case study will show a four-brake system,
which is not unusual. Note that the thermal effects

Slip Clutches 537

Lining Linkage Direction of
travel

Brake shoe

Pneumatic
cylinder

(a)

Direction of
travel

(b)

Figure 18.19: A typical roller coaster. (Shutterstock) (c)

Figure 18.20: Schematic illustration of a roller coaster brake
system. (a) Components of the roller coaster and shown
when the brake is not engaged, as seen by the gap between
the liner pads; (b) top view of an engaged brake; (c) side view
of engaged brake.

that are so prominent in other brake designs are not con- the worst-case design load in the cars. No increase is neces-
sidered here because roller coaster brakes are actuated only sary from a safety standpoint for the following reasons:
intermittently, so that they avoid the thermal checking and
temperature problems typical of automotive disk or drum 1. For most cases this friction force is more than adequate.
brakes.
2. Typical installations will have a series of at least three
Brake Pad–Fin Interaction brakes spaced in alternate bays. If deceleration can
be maintained throughout the entire contact length be-
The first part of the design problem is to correlate brake tween the cars and the brakes, extra braking capacity is
system variables (actuating force, pad area, etc.) with per- inherently placed in the system.
formance (braking force developed). The coefficient of fric-
tion can be reasonably taken as at least 0.25 for sintered brass 3. Even in the event of insufficient braking (perhaps due
on steel (it is usually higher for these materials). Friction- to excessive wear of the braking elements), emergency
reducing contaminants on the roller coaster are not a serious brakes in the operator’s station can stop the cars. The
concern. This is mainly because the braking system is pneu- alternative is to generate larger braking forces with the
matic, not hydraulic, so no leaking hydraulic fluids can con- subject braking system, leading to excessive decelera-
taminate the brake pads. tion and possible injuries to riders from jarring and jerk-
ing. The target deceleration has been chosen to pro-
Figure 18.21 illustrates a three-car chain of roller coaster vide good braking potential while minimizing the risk
cars, but the manufacturer allows as many as seven cars pro- of such injuries. A stronger brake would actually lead
vided that the front and rear cars are configured as shown. to a less safe system.
Many details, especially case-specific decorations, have been
omitted for clarity. Each car can weigh 1900 lb and carry as Brake Actuation System
many as four adults. (It is more common that two adults and
two children will ride in a roller coaster car.) The 95th per- Figure 18.22 shows a detail of the brake-actuating cylinder
centile adult male weighs 216 lb, and weight of the car and and associated linkage. Recall from Fig. 18.20 that two cylin-
contents is calculated accordingly. The maximum incoming ders are used on each brake and two unpowered stabiliz-
velocity is approximately 30 mph (44 ft/s) on the basis of the ing elements are used to distribute the contact forces over
energy attainable from the first roller coaster hill. the whole shoe. The brake is based on a pneumatic cylin-
der/spring system so that the normal cylinder position is
A braking force is then determined to achieve the de- with the piston totally extended, corresponding to the brake
celeration of 0.5 g, which is then used to obtain the actuat- generating maximum brake force. The reason for such a sys-
ing force. However, the brake pad arrangement shown in tem is that many difficulties can arise with pneumatic sys-
Fig. 18.20 develops frictional force on both sides of the fin. tems: air hoses can leak, pneumatic couplings can decouple,
The normal force obtained from such a calculation represents power can fail, etc. The system is fail safe if the brakes are
engaged in such a reasonably foreseeable event.

538 Chapter 18 Brakes and Clutches
Front bumper
plate Rear drawhead Rear shock-
Coupling absorbing
Front drawhead
bumper

1.8 m 1.85 m 1.98 m
Front car Intermediate car Rear car

0.35 m 0.35 m

Figure 18.21: Schematic illustration of typical roller coaster Figure 18.23: Cross-section of the actuating pneumatic cylin-
cars. der, highlighting the helical springs incorporated into the de-
sign.
Pn /2
18.11 Summary
Pn /2 Pt
This chapter focused on two machine elements, clutches and
Figure 18.22: Detail of brake-actuating cylinder with forces brakes, that are associated with motion and have the common
shown. function of dissipating and/or transferring rotating energy.
In analyzing the performance of clutches and brakes, the ac-
Pneumatic brake cylinders are available in a range of ca- tuating force, the torque transmitted, and the reaction force
pacities from a brake manufacturer, each of which has a char- at the hinge pin were the major focuses of this chapter. The
acteristic spring force curve. The nonlinear nature of this torque transmitted is related to the actuating force, the coef-
curve is typical of pneumatic springs (as discussed in Section ficient of friction, and the geometry of the clutch or brake.
17.7), but this cylinder uses a combination of air pressure and This is a problem in statics, where different geometries were
helical springs for actuation as shown in the breakaway view studied separately.
in Fig. 18.23.
For long-shoe clutches and brakes, two theories were
A subtlety in sizing such brake cylinders must be ex- studied: the uniform pressure model and the uniform wear
plained. During assembly the piston is totally extended and model. It was found that for the same dimensionless torque
may be difficult to affix to the hinge mechanism. Therefore, a the uniform wear model requires a larger radius ratio than
piston extension is threaded for a distance on its length. The does the uniform pressure model for the same maximum
cylinder can be assembled and then the collar on the piston pressure. This larger radius ratio implies that a larger area
extension adjusted for brake actuation. is needed for the uniform wear model. Thus, the uniform
wear model was viewed as a safer approach.
Summary
Key Words
Short-shoe brakes have a long history of safe operation on
roller coasters. As discussed in this case study, the brake sys- band brake brake that uses contact pressure of flexible band
tem has a number of redundancies to make sure that failure against outer surface of drum
of any one braking station does not result in catastrophic in-
jury of the passengers. brake device used to bring moving system to rest through
dissipation of energy to heat by friction
Also, it should be noted that roller coasters have a strin-
gent maintenance schedule, and the ride is inspected before clutches power transfer devices that allow coupling and de-
the ride is placed in service every day. If a lining is worn, coupling of shafts
there will be an audible indication from the exposed rivets
on the brake shoes “grinding” against the fin, and this is a cone disk brake or clutch that uses shoes pressed against
clear indication that the pads need replacement. Any rea- convergent surface of cone
sonably attentive maintenance staff can diagnose and repair
such conditions as needed. deenergizing brake or clutch shoe where frictional moment
hinders engagement
Source: Courtesy of Brian King, Recreation Engineering LLC.
rim type brake or clutch that uses internal shoes which ex-
pand onto inner surface of drum

self-energizing brake or clutch shoe where frictional mo-
ment assists engagement

slip clutch clutch where maximum transferred torque is lim-
ited

thrust disk brake or clutch that uses flat shoes pushed
against rotating disk

Reference 539
chine Component Design, 5th ed., Wiley.
Summary of Equations Krutz, G.W., Schuelle, J.K, and Claar, P.W. (1994) Machine De-
sign for Mobile and Industrial Applications, Society of Auto-
Heat Transfer motive Engineers.
Monroe, T. (1977) Clutch and Flywheel Handbook, H.P. Books.
First Law of Thermodynamics: Qf = Qc + Qh + Qs Mott, R. L. (2014) Machine Elements in Mechanical Design, 5th
ed., Pearson.
Lining temperature rise: ∆tm = Qf Norton, R.L. (2011) Machine Design, 4th ed., Prentice Hall.
Cpma Orthwein, W.C. (1986) Clutches and Brakes: Design and Selec-
tion. 2nd ed., CRC Press.
Thrust (Disk) Brakes:
Reference
Uniform pressure model:
Juvinall, R.C., and Marshek, K.M. (2006) Fundamentals of Ma-
Actuating force: Pp = πpo ro2 − ri2 chine Component Design, 4th ed., Wiley.

Torque: Tp = 2πµpo ro3 − ri3 = 2µPp (ro3 −ri3 ) Questions
3 3(ro2 −ri2 )
18.1 What is a clutch? How is it different from a brake?
Uniform wear model:
18.2 What is the uniform pressure model? The uniform wear
Actuating force: Pw = 2πpmaxri (ro − ri) model?

Torque: Tw = πµripmax ro2 − ri2 = µPw (ro + ri) 18.3 When is a rectangular pad disk brake used?
2
Cone Clutches and Brakes: 18.4 What is a cone clutch?

Uniform pressure model: 18.5 What is a self-energizing shoe?

Actuating force: W = πpo D2 − d2 18.6 Can a short-shoe brake be self-energizing?
4
µW D3 − d3 18.7 What is a long-shoe brake? What is the main difference
Torque: T = 3 sin α (D2 − d2) between a long-shoe and a short-shoe brake?

Uniform wear model: 18.8 What is the difference between an external long-shoe
brake and a pivot-shoe brake?
Actuating force: W = πpr (D − d)
18.9 Describe the operating mechanisms of a band brake.
Torque: T = µW (D + d)
4 sin α 18.10 What is a heat check? How are they avoided?
Block (Short-Shoe) Brakes:
W d4 Qualitative Problems
Normal force: P = d3 + µd2
18.11 List the material properties that are (a) necessary, and (b)
Torque: T = µd4rW (energizing) useful for a brake or clutch lining.
d3 − µd1
18.12 List the advantages and disadvantages of cone clutches
T = µd4rW (deenergizing) compared to thrust disc clutches.
d3 + µd2
18.13 Explain the conditions when the uniform pressure
Long-Shoe, Internal, Expanding Rim (Drum) Brakes: model is more appropriate than the uniform wear
model. When is the uniform wear model more appro-
sin θ priate?
Pressure distribution: p = pmax sin θa
18.14 If a composite material is used as a brake or clutch
Normal force moment: π lining, what preferred orientation of the reinforcement
brd7pmax 180◦ would you recommend, if any? Explain your answer.
MP = 4 sin θ1 2 (θ2 − θ1) − sin 2θ2 + sin 2θ1
18.15 Explain the difference between self-energizing and self-
Friction force moment: locking.
µpmaxbr
MF = sin θa 18.16 What are the similarities and differences between band
and thrust disc brakes?
× −r (cos θ2 − cos θ1) − d7 sin2 θ2 − sin2 θ1
2 18.17 What are the main indications of an out-of-round brake
drum?
Self-energizing shoe: −W d6 − MF + MP = 0
Deenergizing shoe: −W d6 + MF + MP = 0 18.18 What are the similarities and differences between heat
checks and hard spots?
Pivot-Shoe Brakes:

Pressure distribution: p = pmax cos θ
Torque: T = 2µr2bpmax sin θ2

Band Brakes:
Forces: F1 = eµφπ/180◦
F2
F1
Maximum pressure: pmax = br

Torque: T = r(F1 − F2)

Recommended Readings

Baker, A.K. (1986) Vehicle Braking, Pentech Press.
Breuer, B. (2008) Brake Technology Handbook, Society of Auto-

motive Engineers.
Budynas, R.G., and Nisbett, J.K. (2011), Shigley’s Mechanical

Engineering Design, 9th ed., McGraw-Hill.
Crouse, W.H., and Anglin, D.L. (1983) Automotive Brakes, Sus-

pension and Steering, 6th ed., McGraw-Hill.
Juvinall, R.C., and Marshek, K.M. (2012) Fundamentals of Ma-

540 Chapter 18 Brakes and Clutches

18.19 Without the use of equations, explain why disk brakes torque when the force P = 4000 N. The friction pad is
have more heat loss through convection than drum a circular sector with the inner radius equal to half the
brakes, at least in the front axle of vehicles. outer radius. Also, a = 150 mm, b = 50 mm, D =
300 mm, and µ = 0.25. The wear of the brake lining
18.20 What are the reasons for using a slip clutch? is proportional to the pressure and the sliding distance.
Ans. T = 157 Nm.

Quantitative Problems ω Friction surface
α = 60°
18.21 The disk brake shown in Sketch a has brake pads in 0
the form of circular sections with inner radius r, outer P
radius 2r, and section angle π/4. Calculate the brake
torque when the pads are applied with normal force P .
The brake is worn in so that pu is constant, where p is
the contact pressure and u is the sliding velocity. The
coefficient of friction is µ. Ans. T = 3µrP .

ab D

r Sketch b, for Problem 18.25
2r
18.26 Three thrust disk clutches, each with a pair of frictional
P surfaces, are mounted on a shaft. The hardened-steel
clutches are identical, with an inside diameter of 100
π/4 mm and an outside diameter of 250 mm. What is the
torque capacity of these clutches based on (a) uniform
Sketch a, for Problem 18.21 wear and (b) uniform pressure? Ans. Tw = 2970 N-m,
Tp = 5513 N-m.
18.22 A plate clutch has a single pair of mating friction sur-
faces 350 mm outer diameter by 250 mm inner diameter. 18.27 A pair of disk clutches has an inside diameter of 250 mm
The mean value of the coefficient of friction is 0.25, and and an outside diameter of 420 mm. A normal force
the actuating force is 5kN. Find the maximum pressure of 18.5 kN is applied and the coefficient of friction of
and torque capacity when the clutch is new, as well as the contacting surfaces is 0.25. Using the uniform wear
after a sufficiently long run-in period. Ans. Tp = 189 and uniform pressure assumptions determine the maxi-
N-m, pmax,p = 0.127 MPa, Tw = 187.5 N-m. mum pressure acting on the clutches. Which of these as-
sumptions would produce results closer to reality? Ans.
18.23 An automotive clutch with a single friction surface is to pmax,p = 206.8 kPa, pmax,w = 277 kPa.
be designed with a maximum torque of 140 N-m. The
materials are chosen such that µ = 0.35 and pmax = 0.35 18.28 A disk clutch is made of cast iron and has a maximum
MPa. Use safety factor ns = 1.3 with respect to slippage torque of 210 N-m. Because of space limitations the out-
at full engine torque and as small an outside diameter as side diameter must be minimized. Using the uniform
possible. Determine appropriate values of ro, ri, and P wear assumption and a safety factor of 1.3, determine
by using both the uniform pressure and uniform wear
models. Ans. Pp = 8743 N, Pw = 6160 N. (a) The inner and outer radii of the clutch. Ans. ri =
0.068 m, ro = 0.118 m.
18.24 The brakes used to stop and turn a tank are built like
a multiple-disk clutch with three loose disks connected (b) The maximum actuating force needed. Ans. Pw =
through splines to the drive shaft and four flat rings con- 11.3 kN.
nected to the frame of the tank. The brake has an outer
contact diameter of 600 mm, an inner contact diameter 18.29 A disk brake for a flywheel is designed as shown in
of 300 mm, and six surface contacts. The wear of the Sketch c. The hydraulic pistons actuating the brake need
disks is proportional to the contact pressure multiplied to be placed at a radius rp so that the brake pads wear
by the sliding distance. The coefficient of friction of the evenly over the entire contact surface. Calculate the ac-
brake is 0.15, and the friction between the caterpillar tuating force P and the radius so that the flywheel can
and the ground is 0.25, which gives a braking torque of be stopped within 4 s when it rotates at 1000 rpm and
12,800 N-m needed to block one caterpillar track so that has a kinetic energy of 4 × 105 N-m. The input param-
it slides along the ground. Calculate the force needed eters are µ = 0.3, α = 30◦, ro = 120 mm, ri = 60 mm.
to press the brake disks together to block one caterpillar Ans. P = 35.4 kN, rp = 0.089 m.
track. Also, calculate the force when the brake is new.
Ans. Pw = 63.2 kN, Pp = 60.9 kN.

18.25 A disk brake used in a printing machine is designed as
shown in Sketch b. The brake pad is mounted on an arm
that can swivel around point 0. Calculate the braking

Quantitative Problems 541

PP wear, determine the actuating force and the contact pres-
sure. Ans. Pp = 854 N, pmax,p = 101 kPa.

ro α rp 18.36 The coefficient of friction of a cone clutch is 0.25. It can
ri ω support a maximum pressure of 410 kPa while transfer-
ring a maximum torque of 280 N-m. The width of the
clutch is 65 mm. Minimize the major diameter of the
clutch. Determine the clutch dimensions and the actuat-
ing force.

18.37 A block brake is used to stop and hold a rope used to

transport skiers from a valley to the top of a mountain.

Sketch c, for Problem 18.29 The distance between cars used to transport the skiers

18.30 A disk clutch produces a torque of 125 N-m and a max- is 100 m, the length of the rope from the valley to the
imum pressure of 315 kPa. The coefficient of friction of
the contacting surfaces is 0.28. Assume a safety factor of top of the mountain is 4 km, and the altitude difference
1.8 for maximum pressure and design the smallest disk
clutch for the above constraints. What should the nor- is 1.4 km. The rope is driven by a V-groove wheel with
mal force be? Ans. Pw = 4.39 kN.
a diameter of 2 m. The rope is stopped and held with a
18.31 A cone clutch is to transmit 150 Nm of torque. The half-
cone angle α = 10◦, the mean diameter of the friction block brake mounted on the shaft of the V-groove wheel,
surface is 300 mm, and face width b = 50 mm. For coef-
ficient of friction µ = 0.20 find the normal force P and shown in Sketch d. Neglect all friction in the different
the maximum contact pressure p by using both the uni-
form pressure and uniform wear models. Ans. Pp = 868 parts of the ropeway except the friction in the driving
N, pmax,p = 104.6 kPa, pmax,w = 107.8 kPa.
sheave, and assume that the slope of the mountain is
18.32 The synchronization clutch for the second gear of a car
has a major cone diameter of 50 mm and a minor diam- constant. Dimension the brake for 20 passengers with
eter of 40 mm. When the stick shift is moved to second
gear, the synchronized clutch is engaged with an axial each passenger’s equipment weighing 100 kg, and as-
force of 100 N, and the moment of inertia of 0.005 kg-
m2 is accelerated 200 rad/s2 in 1 s to make it possible sume that all ropeway cars descending from the top are
to engage the gear. The coefficient of friction of the cone
clutch is 0.10. Determine the smallest cone clutch width empty of people. The direction of rotation for the drive
that still gives large enough torque. Assume the clutches
are worn in. Ans. b = 21.6 mm. motor is shown in the figure. Calculate the braking force

18.33 A safety brake for an elevator is a self-locking cone W needed to hold the ropeway still if all passengers are
clutch. The minor diameter is 120 mm, the width is 60
mm, and the major diameter is 130 mm. The force ap- on their way up. Do the same calculation if all pas-
plying the brake comes from a prestressed spring. Cal-
culate the spring force needed if the 2-ton (2000 kg) el- sengers continue on down to the valley with the rope-
evator must stop from a speed of 3 m/s in a maximum
distance of 3 m while the cone clutch rotates five revo- way. The coefficient of friction in the brake is 0.25. Ans.
lutions per meter of elevator motion. The coefficient of
friction in the cone clutch is 0.20. Ans. P = 4781 N. Wup = 2.197 MN, Wdown = 671 kN.

18.34 A cone clutch is used in a car automatic gearbox to fix 0.45 m 0.90 m W
the planet wheel carriers to the gearbox housing when
the gear is reversing. The car weighs 1300 kg with 53% 0.15 m
loading on the front wheels. The gear ratio from the
driven front wheels to the reversing clutch is 16.3:1 (i.e., ω
the torque on the clutch is 16.3 times lower than the
torque on the wheels if all friction losses are neglected). R = 0.50 m
The car wheel diameter is 550 mm, the cone clutch ma-
jor diameter is 85 mm and the minor diameter is 80 mm, Sketch d, for Problem 18.37
and the coefficients of friction are 0.3 in the clutch and
1.0 between the wheel and the ground. Dimension the 18.38 The motion of an elevator is controlled by an electric mo-
width of the cone clutch so that it is not self-locking. Cal- tor and a block brake. On one side the rotating shaft
culate the axial force needed when the clutch is worn in. of the electric motor is connected to the gearbox driv-
Ans. b < 0.0083 m, P = 2647 N. ing the elevator, and on the other side it is connected
to the block brake. The motor has two magnetic poles
18.35 A cone clutch has a major diameter of 328 mm and a and can be run on either 60- or 50-Hz electricity (3600 or
minor diameter of 310 mm, is 50 mm wide, and trans- 3000 rpm). When the elevator motor is driven by 50-Hz
fers 250 N-m of torque. The coefficient of friction is 0.33. electricity, the braking distance needed to stop is 52 cm
Using the assumptions of uniform pressure and uniform when going down and 31 cm when going up with max-
imum load in the elevator. To use it with 60-Hz elec-
tricity and still be able to stop exactly at the different
floor levels without changing the electric switch posi-
tions, the brake force at the motor should be changed.
How should it be changed for going up and for go-
ing down? The brake geometry is like that shown in
Fig. 18.5 with d1 = 0.030 m, d3 = 0.100 m, d4 = 0.400
m, r = 0.120 m, and µ = 0.20. Only the inertia of the el-
evator needs to be considered, not the rotating parts as-

542 Chapter 18 Brakes and Clutches

sociated with the drive or other components. Make the 18.42 The brake on the rear wheel of a car is the long-shoe in-
brake self-energizing when the elevator is going down. ternal type. The brake dimensions according to Fig. 18.7
are θ1 = 10◦, θ2 = 120◦, r = 95 mm, d7 = 73 mm,
18.39 An anchor winding is driven by an oil hydraulic mo- d6 = 120 mm, and d5 = 30 mm. The brake shoe lining is
tor with a short-shoe brake to stop the anchor machin- 38 mm wide, and the maximum allowable contact pres-
ery from rotating and letting out too much anchor chain sure is 5 MPa. Calculate the braking torque and the frac-
when the wind moves the ship. The maximum force tion of the torque produced from each brake shoe when
transmitted from the anchor through the chain is 1.1 MN the brake force is 5000 N. Also, calculate the safety factor
at a radius of 2 m. Figure 18.9 describes the type of against contact pressure that is too high. The coefficient
block brake used, which is self-energizing. Calculate the of friction is 0.25. Ans. Ttot = 534 N-m, Tse = 359 N-m.
brake force W needed when the the brake dimensions
are d1 = 0.9 m, d3 = 1.0 m, d4 = 6 m, r = 3 m, and 18.43 Sketch g shows a long-shoe, internal expanding shoe
µ = 0.25. Also, calculate the contact force between the brake. The inside rim diameter is 280 mm and the di-
brake shoe and the drum. Ans. W = 379 kN, P = 2.93 mension d7 is 90 mm. The shoes have a face width of 30
MN. mm.

18.40 The hand brake shown in Sketch e has an average pres- (a) Find the braking torque and the maximum pres-
sure of 600 kPa across the shoe and is 50 mm wide. The sure for each shoe if the actuating force is 1000 N,
wheel runs at 150 rpm and the coefficient of friction is the drum rotation is counterclockwise, and the co-
0.25. Dimensions are in millimeters. Determine the fol- efficient of friction is 0.30. Ans. T = 218.7 N-m.
lowing:
(b) Find the braking torque if the drum rotation is
(a) If x = 150 mm, what should the actuating force clockwise.
be? Ans. Pn = 2.97 kN.

(b) What value of x causes self-locking? Ans. x = 2.0
m.

(c) What torque is transferred? Ans. T = 303 N-m.

(d) If the direction of rotation is reversed, how would
the answers to parts (a) to (c) change?

500 600 WW Rotation
100 Pn 30°

20°

x 80° 150

ω R = 170 110° Drum
Sketch e, for Problem 18.40 d7 Lining

18.41 The short-shoe brake shown in Sketch f has an average 30° 10°
pressure of 1 MPa and a coefficient of friction of 0.32.
The shoe is 250 mm long and 45 mm wide. The drum A
rotates at 310 rpm and has a diameter of 550 mm. Di-
mensions are in millimeters. Sketch g, for Problem 18.43.

(a) Obtain the value of x for the self-locking condition. 18.44 Sketch h shows a 450-mm inner diameter brake drum
Ans. x = 1.81 m. with four internally expanding shoes. Each of the hinge
pins A and B supports a pair of shoes. The actuating
(b) Calculate the actuating force if x = 275 mm. Ans. mechanism is a linkage which produces the same actuat-
P = 3982 N. ing force W on each shoe as shown. The lining material
is to be sintered metal. The face width of each shoe is 70
(c) Calculate the braking torque. Ans. T = 3.44 kN-m. mm. Using the mid-range values of friction coefficient
(d) Calculate the reaction at point A. and contact pressure,

P (a) Identify the self-energizing and de-energizing
shoes and label them.
xω
(b) Determine the maximum pressure on the deener-
A gizing shoe.Ans. pmax = 1.071 MPa.

x (c) Determine the maximum force W and the braking
220 torque T at this force. Ans. T = 3430 Nm.

P

580 650 160
Sketch f , for Problem 18.41

Quantitative Problems 543

y 575
18° 18°

WW 100 60° 60°
100 100
120 mm
210 mm 500 30
60° 60°
8° 200 mm 200 mm 8° x 1000 N Rotation
8°
8° A B 475 425

r 210 mm
120 mm

WW Sketch i, for Problem 18.47
ω

18° 18°
Sketch h, for Problem 18.44

18.45 The maximum volume of the long-shoe internal brake 18.48 A long-shoe external brake as shown in Fig. 18.15 has a
on a car is given as 10−3 m3. The brake should have pivot point such that d7 = 4r, d6 = 2r, θ1 = 5◦, and θ2 =
two equal shoes, one self-energizing and one deenergiz- 45◦. Find the coefficient of friction needed to make the
ing, so that the brake can fit on both the right and left brake self-lock if the rotation is in the direction shown in
sides of the car. Calculate the brake width and radius Fig. 18.15. If the shaft rotates in the opposite direction,
for maximum braking power if the space available in- calculate the drum radius needed to get a braking torque
side the wheel is 400 mm in diameter and 100 mm wide. of 180 N-m for the actuating force of 10,000 N. Ans. r =
The brake lining material has a maximum allowed con- 43.32 mm.
tact pressure of 4 MPa and a coefficient of friction of
0.18. Also, calculate the maximum braking torque. Ans. 18.49 An external, long brake shoe is mounted on an elastic
T = 730 N-m. arm. When a load is applied, the arm and the brake lin-
ing bend and redistribute the pressure. Instead of the
18.46 A long-shoe brake in a car is designed to give as high a normal sine pressure distribution the pressure becomes
braking torque as possible for a given force on the brake constant along the length of the lining. For a given actu-
pedal. The ratio between the actuating force and the ating force calculate how the brake torque changes when
pedal force is given by the hydraulic area ratio between the pressure distribution changes from sinusoidal to a
the actuating cylinder and the cylinder under the pedal. constant pressure. Also, assume that d7 = 110 mm,
The brake shoe angles are θ1 = 10◦, θ2 = 170◦, and r = 90 mm, b = 40 mm, θ1 = 20◦, θ2 = 160◦, d6 = 220
θa = 90◦. The maximum brake shoe pressure is 5 MPa, mm, µ = 0.25, and W = 12 kN.
the brake shoe width is 40 mm, and the drum radius is
100 mm. Find the distance d7 that gives the maximum 18.50 A long-shoe external brake has two identical shoes cou-
braking power for a coefficient of friction of 0.2 at any pled in series so that the peripheral force from the first
pedal force. What braking torque would result if the co- shoe is directly transferred to the second shoe. No ra-
efficient of friction were 0.25? dial force is transmitted between the shoes. Each of the
two shoes covers 90◦ of the circumference, and the brake
18.47 An external drum brake assembly (see sketch i) has a linings cover the central 70◦ of each shoe, leaving 10◦ at
normal force P = 1000 N acting on the lever. Dimensions each end without lining as shown in Sketch j. The actu-
are in mm. Assume that coefficient of friction µ = 0.25 ating force is applied tangentially to the brake drum at
and maximum contact pressure p = 1 MPa. Determine the end of the loose shoe, 180◦ from the fixed hinge point
the following from long-shoe calculations: of the other shoe. Calculate the braking torques for both
rotational directions when d7 = 150 mm, r = 125 mm,
(a) Free-body diagram with the directionality of the b = 50 mm, W = 14, 000 N, and µ = 0.2. Also show
forces acting on each component a free-body diagram of these forces acting on the two
shoes. Ans. T = 717 N-m. If reversed, T = 1197 N-m.
(b) Which shoe is self-energizing and which is deen-
ergizing

(c) Total braking torque Ans. T = 2632 Nm.

(d) Pad width as obtained from the self-energizing
shoe (deenergizing shoe width equals self-
energizing shoe width) Ans. b = 26.3 mm.

(e) Pressure acting on the deenergizing shoe

544 Chapter 18 Brakes and Clutches
How much does the brake torque change? Ans. 15%
10° 10° decrease.
70° 70° 18.54 The band brake shown in Sketch l is activated by a
compressed-air cylinder with diameter dc. The brake
10° 10° cylinder is driven by air pressure p = 0.7 MPa. Calcu-
W = 14,000 N late the maximum possible brake torque if the coefficient
d7 = 150 mm of friction between the band and the drum is 0.20. The
r = 125 mm mass force on the brake arm is neglected, dc = 50 mm,
ω r = 200 mm, l1 = 500 mm, l2 = 200 mm, and l3 = 500
mm. Ans. T = 719 N-m.
Sketch j, for Problem 18.50
r
18.51 A special type of brake is used in a car factory to hold the
steel panels during drilling operations so that the forces dc
from the drill bits cannot move the panels. The brake is l1
shown in Sketch k. Calculate the braking force PB on the
steel panel when it moves to the right with the speed ub l3 l2
between drilling operations. The actuating force is PM . Sketch l, for Problem 18.54
The brake lining is thin relative to the other dimensions.
9 18.55 The band brake shown in Sketch m has wrap angle
Ans. Pb = 7 µPM . φ = 225◦ and cylinder radius r = 80 mm. Calculate
the brake torque when the lever is loaded by 100 N and
PM coefficient of friction µ = 0.3. How long is the braking
time from 1200 rpm if the rotor mass moment of inertia
2L is 2.5 kg-m2? Ans. T = 54 N-m, t = 5.825s.

LL

ω

PB P = 100 N
a 2a
Sketch k, for Problem 18.51
Sketch m, for Problem 18.55
18.52 Redo Problem 18.39 for long-shoe assumptions. The
average contact pressure occurs at 40◦. Determine the 18.56 The band brake shown in Sketch n is 40 mm wide and
maximum contact pressure and its location. Assume can take a maximum pressure of 1.1 MPa. All dimen-
that the distance x is 150 mm. What is the braking sions are in millimeters. The coefficient of friction is 0.3.
torque? Also, repeat this problem while reversing the Determine the following:
direction of rotation. Discuss the changes in the results. (a) The maximum allowable actuating force.
(b) The braking torque. Ans. T = 1.26 kN-m
18.53 A symmetrically loaded, pivot-shoe brake has a wrap (c) The reaction supports at 01 and 02.
angle of 180◦ and the optimum distance d7, giving a
symmetrical pressure distribution. The coefficient of
friction of the brake lining is 0.30. A redesign is con-
sidered that will increase the breaking torque without
increasing the actuating force. The wrap angle is de-
creased to 80◦ (+40◦ to –40◦) and the d7 distance is de-
creased to still give a symmetrical pressure distribution.

Design and Projects B 545
W 50 200
02
C 01 A

350 135 85 180

Sketch n, for Problem 18.56

(d) Whether it is possible to change the distance 01A φ
in order to have self-locking. Assume point A can r
be anywhere on line C01A.
120 mm
18.57 A brake (see Sketch o) consists of a drum with a brake
shoe pressing against it. The drum radius is r = 80 mm.
Calculate the brake torque when P = 7000 N, µ = 0.35,
and brake pad width b = 40 mm. The wear is propor-
tional to the contact pressure times the sliding distance.
Ans. T = 205 Nm.

50 mm P = 70 N

60° P Sketch q, for Problem 18.59

Design and Projects

Sketch o, for Problem 18.57 18.60 Many brake systems are designed so that they will have
one component wear preferentially to the others. For
18.58 For the band brake shown in Sketch p, the following example, a pad will wear faster than a disk in a thrust
conditions are given: d = 350 mm, pmax = 1.2 MPa, disk clutch. Explain why this feature is incorporated
µ = 0.25, and b = 50 mm. All dimensions are in mil- into brakes, instead of designing the entire system with
limeters. Determine the following: the same intended life.

(a) The braking torque. Ans. T = 1.26 kN-m. 18.61 Excessive heating of brake materials leads to brake
checking, warping, and out-of-roundness for brake
(b) The actuating force. Ans. P = 2.57 kN. drums. Conduct a literature search and investigate the
highest temperature that exists between contacting as-
(c) The forces acting at hinge 0. Ans. Ox = 0, Oy = perities of brake systems.
2.57 kN.
18.62 Write a computer program to determine the optimum
Rotation 210 150 130 angular coverage of lining material for the pivot-shoe
brake of Example 18.5.
AB
0 18.63 Electric vehicles often use regenerative braking to recover
some of the kinetic energy when vehicles are brought to
P rest. Write a one-page summary of the design features
and disadvantages of regenerative braking.

18.64 What are anti-lock brakes? Conduct an Internet search
and explain their operation.

Sketch p, for Problem 18.58

18.59 The band brake shown in Sketch q has wrap angle φ =
215◦ and cylinder radius r = 60 mm. Calculate the
brake torque when coefficient of friction µ = 0.25. How
long is the braking time from 1500 rpm if the rotor mo-
ment of inertia is J = 2 kg-m2? Ans. t = 20.0 s.

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Chapter 19

Flexible Machine Elements

Contents
19.1 Introduction 548
19.2 Flat Belts 548
19.3 Synchronous Belts 551
19.4 V-Belts 551
19.5 Wire Ropes 555
19.6 Rolling Chains 559
19.7 Summary 566

Examples 561
19.1 Forces on a Flat Belt 549
19.2 Optimum Pitch Diameter for a

Timing Belt 551
19.3 V-Belt Design 554
19.4 Analysis of Wire Rope 558
19.5 Power and Force in Rolling Chains
19.6 Chain Drive Design 564

Design Procedures
19.1 V-Belt Drives 553
19.2 Design of Chain Drives 564

A rolling chain on a sprocket. Source: Shutterstock. Case Study
19.1 Design of a Gantry for a Dragline 565
Scientists study the world as it is, engineers create the world that
never has been.

Theodore von Karman

Rubber belts, roller chains, and wire ropes are three examples of power transmission devices that are widely used in many
industries. These machine elements have the common characteristic of flexibility. Rubber belts are used around pulleys, and
transmit a torque based on friction associated with the belt-pullet interface. Flat belts can be constructed from soft or hard
rubber, depending on the power to be transmitted, but synchronous and V-belts are advanced composite materials. Reinforced
rubber belts are economically attractive and reliable machine elements that have further advantages of reducing impact stresses,
but they will slip if torque becomes too large. Synchronous (timing) belts have teeth that prevent slip, and have the further
advantages of low preload and associated reduced bearing loads. Roller chains are longer-lived and more robust devices than
belts, but are more expensive. Roller chains and their sprockets require effective lubrication to achieve long lives. There are
a number of design variables that can be modified with roller chains, and conventional and silent chains are presented in this
chapter. Wire rope consists of many strands of wire carefully wound around a core; such rope is widely used to hoist loads or
provide a force through a winch or equivalent device. Smaller diameter ropes are often used as cables to transmit a force in order
to open a clutch or engage a brake, for example. Wire ropes are especially useful when power needs to be transferred efficiently
over long distances.

Machine elements in this chapter: Flat belts, V-belts, synchronous belts, roller and silent chains, and wire ropes.
Typical applications: Power transmission from electric motors or internal combustion engines to power other devices; bicycle
power transmission; cranes, hoists, and derricks; elevators.
Competing machine elements: Shaft couplings (Ch. 11), gears (Ch. 14 and 15).

547

548 Chapter 19 Flexible Machine Elements

Symbols A α B
α
Am cross-sectional area of metal strand in rope, m2 φ1 α φ2
a link plate thickness, m D1 D
a1 service factor 01 D2
a2 multiple-strand factor in rolling chain α
c distance from neutral axis to outer fiber, m 02
cd center distance, m
D sheave or pulley diameter, m α
d diameter, m
dw wire diameter, m cd
E modulus of elasticity, Pa Figure 19.1: Dimensions, angles of contact, and center dis-
F force, N tance of open flat belt.
Fa force due to acceleration, N
Ff fatigue allowable force, N 19.1 Introduction
Fh static force, N
Fi initial tensile force, N The machine elements considered in this chapter, like the ma-
Fr rope weight, N chine elements considered in Chapter 18, use friction as a use-
Ft total friction force, N ful agent to produce a high and uniform force and to transmit
Fw deadweight, N power. A belt, rope, or chain provides a convenient means for
f1 overload service factor transferring power from one shaft to another, and are espe-
f2 power correction factor cially useful for transfer of power over long distances. Table
g gravitational acceleration, 9.807 m/s2 19.1 compares belt, chain, and gear power transmission ap-
gr velocity ratio proaches.
HB Brinell hardness, kg/mm2
hp power transmitted, W 19.2 Flat Belts
hpb rated power per belt, W
hpr rated power, W Flat belts find considerable use in applications requiring
ht V-belt height, m small pulley diameters, high belt surface speeds, low noise
I area moment of inertia, m4 levels, low weight, and/or low inertia. They cannot be
L belt length, m used where absolute synchronization between pulleys must
M bending moment, N-m be maintained because they rely on friction for their proper
m mass, kg functioning, so some slip is inevitable. Flat belts must be kept
m mass per unit length, kg/m under tension to function, and therefore they require tension-
N number of teeth in sprocket or number of belts ing devices.
Na angular speed, rpm
ns safety factor There are many applications where belts are used to
p bearing pressure, Pa transfer very low power levels. This book emphasizes ma-
pall allowable bearing pressure, Pa chinery applications, where the power that is transferred is
pt pitch or datum, m generally larger than 100 Nm/s (75 ft-lb/s). Smaller belts are
r sheave or pulley radius, m readily available, and usually can be selected based solely on
∆r chordal rise, m kinematic constraints.
rc chordal radius, m
Su ultimate strength, Pa 19.2.1 Belt Length
t thickness, m
u belt velocity, m/s Figure 19.1 shows the dimensions, angles of contact, and cen-
wt belt width, m ter distance of an open flat belt. The term “open” is used
wz weight, N to distinguish it from the geometry of a crossed belt, where
wz weight per unit length, N/m the belt forms a figure eight when viewed on end. Note that
α angle used to describe loss in arc of contact, deg distance 02D is equal to r1 = D1/2, distance BD is equal to
θr angle of rotation to give chordal rise, deg r2 − r1 = (D2 − D1)/2, and distance AD is the center distance
µ coefficient of friction cd. Also, triangle ABD is a right triangle so that
σ normal stress, Pa
σb bending stress, Pa AB2 + BD2 = AD2,
σt tensile normal stress, Pa
φ wrap angle; gantry line angle, deg
ω angular velocity, rad/s

Subscripts

1 driver pulley or sheave
2 driven pulley or sheave

so that

AB2 + (r2 − r1)2 = c2d;

AB = cd2 − (r2 − r1)2 . (19.1)

Flat Belts 549

Table 19.1: Comparison of selected power transmission devices.

Constraint Flat belt V-belt Power transmission device Silent chain Spur gear
1 Synchronous belt Roller chain 4 4
Synchronization 1 1 4 4
Efficiency 1 4 14 3 1
Anti-shock 4 4 24 3 1
Noise/vibration 4 32
Compactness 2 3 32 3
1 1 2
High speed/low load None None 3 1 4 Required
Low speed/high load 2 1 2 4 3 4
Lubrication 1 2 None Required Required 4
Bearing loads 2 4 3
Longevity 2 3 3

1, Poor; 2, Fair; 3, Good; 4, Excellent.

The length of the open flat belt can be expressed as In obtaining the preceding equations, it was assumed that the
coefficient of friction on the belt is uniform over the entire
L = 2AB + r1φ1 ππ (19.2) angle of wrap and that centrifugal forces on the belt can be
+ r2φ2 , neglected.
180 180
In transmitting power from one shaft to another by
where φ is the wrap angle. The wrap angles can be expressed means of a flat belt and pulleys, the belt must have an initial
as tensile force, Fi. The required initial belt tension (or tensile
force) depends on the elastic characteristics of the belt and
φ1 = 180◦ − 2α and φ2 = 180◦ + 2α. (19.3) friction, but can be approximated by

Also, from right triangle ABD in Fig. 19.1, the angle used to Fi = F1 + F2 . (19.6)
describe the loss in arc of contact is 2

sin α = D2 − D1 or α = sin−1 D2 − D1 . (19.4) Also, from Eq. (18.65), note that when power is being trans-
2cd 2cd mitted, the tensile force, F1, in the tight side exceeds the ten-
sion in the slack side, F2.
The angle α is in degrees in order to be consistent with
Eq. (19.3), and is equal to zero only if the pulleys have a 1-to-1 The power transmitted is
ratio. By substituting Eqs. (19.3) and (19.4) into Eq. (19.2), the
belt length can be expressed as hp = (F1 − F2)u, (19.7)

where u is the belt velocity. The centrifugal force acting on
the belt can be expressed as

L = (2cd)2 − (D2 − D1)2 Fc = m u2 = wz u2, (19.8)
+ D1π 180◦ − 2 sin−1 D2 − D1 g
360 2cd
+ D2π 180◦ + 2 sin−1 D2 − D1 , where
360 2cd m = mass per unit length, kg/m
u = belt velocity, m/s
or wz = weight per unit length, N/m

π When the centrifugal force is considered, Eq. (18.64) becomes
2 (D1
L= (2cd)2 − (D2 − D1)2 + + D2) F1 − Fc = eµφπ/180◦
F2 − Fc
+ π(D2 − D1) sin−1 D2 − D1 . (19.9)
180 2cd
(19.5) Of course, Eq. (18.65) is unchanged when centrifugal forces
are considered.

19.2.2 Belt Forces Example 19.1: Forces on a Flat Belt

The basic equations developed for band brakes are also ap- Given: A flat belt is 150 mm wide and 8 mm thick and trans-
plicable here. From Section 18.9, the following torque and mits 12 kW. The center distance is 2.5 m. The driving pulley
friction equations can be written: is 150 mm in diameter and rotates at 2000 rpm such that the
loose side of the belt is on top. The driven pulley has a diam-
F1 = eµφπ/180◦ ; (18.64) eter of 450 mm. The belt material weighs 970 kg/m3.
F2
Find: Determine the following:
T = (F1 − F2)D1 , (18.65)
2 (a) If µ = 0.30, determine F1 and F2.

where (b) If µ is reduced to 0.20 because of oil getting on part of
φ = wrap angle, deg the pulley, what are F1 and F2? Would the belt slip?
µ = coefficient of friction
F1 = tight-side or driver force, N (c) What is the belt length?
F2 = slack-side or driven force, N

550 Chapter 19 Flexible Machine Elements
Solution:
(b) If µ = 0.20 instead of 0.30,
(a) The belt angular velocity is ω = 2000 rpm = 209.4
rad/s. Therefore, F1 − 286.4 = e(0.2)(173.1)π/180 = 1.830.
F2 − 286.4
0.150 so that (d)
2
u = rω = (209.4) = 15.7 m/s.

From Eq. (19.7), F1 = 1.830F2 −(1.83)(286.4)+286.4 = 1.830F2 −237.7,
(e)
F1 − F2 = hp = 12, 000 = 764 N. (a)
u 15.7 Since the initial belt tension is still Fi = 1186 N, substi-
tuting Eq. (e) into Eq. (19.6) gives

The mass per volume was given, so that the weight per 1.830F2 − 237.7 + F2 = 1186,
unit volume is (970)(9.81) = 9515 N/m3, therefore, 2

wz = 9515wtt = (9515)(0.150) (0.008) or F2 = 922 N. Substituting this result into Eq. (e) pro-
L duces

= 11.4 N/m.

The centrifugal force acting on the belt is, from F1 = 1.830(922) − 237.7 = 1450 N.
Eq. (19.8),
From Eq. (19.7) the power transmitted is

Fc = wz u2 = (11.4)(15.7)2 = 286.4 N. hp = (F1 − F2) u
L g (9.81) = (1450 − 922)(15.7)
= 8.29 kW.
From Eq. (19.4),

α = sin−1 D2 − D1 Since this is less than the required power, the belt will
2cd slip.

= sin−1 450 − 150 (c) From Eq. (19.5) while making use of Eq. (c),
(2)(2500)
L = 2 (2)(2500)2 − (450 − 150)2 + π (450 + 150)
= 3.440◦. 2

The wrap angle is given by Eq. (19.3) as + π(450 − 150) sin−1 450 − 150
φ = 180◦ − 2(α) = 180◦ − 2(3.440◦) = 173.1◦. (b) 180 2(2500)

= 4483 mm = 4.483 m.

Making use of Eq. (19.9) gives

F1 − 286.4 = e(0.3)(173.1)π/180 = 2.475. 19.2.3 Slip
F2 − 286.4
Slip is detrimental for a number of reasons. It reduces belt ef-
Therefore, ficiency and can cause thermal damage to the belt, leading to
elongation (and resulting in lower belt tension and more slip)
F1 − 286.4 = 2.475F2 − 708.9, or chemical degradation. To eliminate slip, the initial belt ten-
sion needs to be retained. But as the belt stretches over time,
so that some of the initial tension will be lost. One solution might be
to have a very high initial tension, but this would put large
F1 = 2.475F2 − 422.5. (c) loads on the pulley, shaft, and bearings, and also shorten the
belt life. Some better approaches are the following:
Substituting Eq. (c) into Eq. (a) gives
1. Develop means of adjusting tension during operation.
2.475F2 − F2 = 422.5 + 764;
2. Increase the wrap angle.
F2 = 1186 = 804 N.
1.475 3. Change the belt material to increase the coefficient of
friction.
Therefore, from Eq. (a),
4. Use a larger belt section.
F1 = 764 + 804 = 1568 N.
5. Use an alternative design, such as a synchronous belt
From Eq. (19.6), the initial belt tension is (see Section 19.3) or a chain as discussed in Section 19.6.

Fi = F1 + F2 = 1568 + 804 = 1186 N. There are many tensioning devices involving spring-loaded
2 2 pulleys or weights to apply loads to belts. Figure 19.2 illus-
trates one way of maintaining belt tension. The slack side of
the belt is on the top, so that the sag of the belt acts to increase
its wrap angle.

V-Belts 551

Idler hp = u 2000 − 0.180 u2 = u 2000 − 0.24u2 .
0.750
Pivot

The optimum power transmitted occurs when

Weight ∂hp = 2000 − (0.24)(3)u2 = 0.
Tight ∂u
side
Solving for the velocity yields u = 52.70 m/s. The pulley
Figure 19.2: Weighted idler used to maintain desired belt ten- diameter that produces the maximum power transfer is
sion.

ωD
=u

2

Tooth included Circular Backing A Neoprene-encased 2u 2(52.70) = 0.2013 m = 201.3 mm.
angle pitch tension member D= =

ω 2π
(5000)
60

Facing Section A–A

A 19.4 V-Belts

Neoprene
tooth cord

Pulley face radius V-belts are an extremely common power transmission de-
Pulley pitch radius vice, used on applications as diverse as blowers, compres-
sors, mixers, machine tools, etc. One or more V-belts are used
Figure 19.3: Synchronous, or timing, belt. to drive accessories on an automobile and transfer power
from the internal-combustion engine. V-belts are made to
19.3 Synchronous Belts standard lengths and with standard cross-sectional sizes, as
shown in Fig. 19.4a. V-belts A through E are standard shapes
Synchronous belts, or timing belts, are basically flat belts that were standardized as early as the 1940s, but the more
with a series of evenly spaced teeth on the inside circumfer- modern 3V, 5V, and 8V belts shown have higher power rat-
ence, thereby combining the advantages of flat belts with the ings for the same cross-sectional area. V-belts have a fiber-
excellent traction of gears and chains (see Section 19.6). A reinforced construction, as shown in Fig. 19.4b and c, so they
synchronous belt is shown in Fig. 19.3. should be recognized as advanced composite materials. The
flexible reinforcement is often a cord, not an individual fiber,
Unlike flat belts, synchronous belts do not slip or creep, and can be made from nylon, kevlar (aramid), polyester, etc.
and the required belt tension is low, resulting in very small The impregnated woven jacket shown protects the interior
bearing loads. Synchronous belts will not stretch and require and provides a wear- and oil-resistant surface for the belt.
no lubrication. Speed is transmitted uniformly because there The compressive side of the belt is produced from a high-
is no chordal rise and fall of the pitch line as in rolling chains strength, fatigue-resistant rubber. Because their interior ten-
(Section 19.6). The equations developed for flat-belt length sion cords are stretch and creep resistant, V-belts (unlike flat
and torque in Section 19.2 are equally valid for synchronous belts) do not require frequent adjustment of initial tension.
belts. The grooved pulleys that V-belts run in are called sheaves.
They are usually made of cast iron, pressed steel, or die-cast
Example 19.2: Optimum Pitch metal. Figure 19.5 shows a V-belt in a sheave groove.
Diameter for a Timing Belt
V-belts can also be used in multiples as shown in
Given: A timing belt is used to transfer power from a high- Fig. 19.4c. The obvious advantage of such a belt is that it
speed motor to a grinding wheel. The timing belt is 750 mm can transmit higher power; a two-belt design can transmit
long and weighs 180 g. The maximum allowable force in the twice the power of a single belt, etc. The tie band construc-
belt is 2000 N, and the speed of the turbine is the same as the tion shown ensures that the belt maintains alignment when
speed of the grinding wheel, 5000 rpm. it is outside of the sheaves, which helps the belt run over the
sheave properly.
Find: Calculate the optimum pulley pitch diameter for max-
imum power transfer. V-belts find frequent application where synchronization
between shafts is not important; that is, V-belts can slip (see
Solution: If the total maximum force in the belt is F1 and the Section 19.2.3). V-belts are easily installed and removed, quiet
centrifugal force is Fc, the maximum power transmitted is in operation (but not quite as quiet as flat belts), low in main-
tenance, and provide shock absorption between the driver
hp = u(F1 − Fc). and driven shafts. V-belt drives normally operate best at belt
velocities between 7.5 and 30 m/s (1500 and 6500 ft/min).
Letting F1 be equal to the maximum allowable force and
making use of Eq. (19.8) for the centrifugal force gives The velocity ratio is similar to the gear ratio given in
Eq. (14.19), or

gr = Na1 = r2 . (19.10)
Na2 r1

V-belts can operate satisfactorily at velocity ratios up to ap-
proximately 7 to 1. V-belts typically operate at 90 to 98% effi-
ciency, lower than that found for flat belts.

552 22.2 Chapter 19 Flexible Machine Elements
16.7
12.7 Rubber impregnated
A8 B 10.3 C 13.5 woven cover

38.1 Flexible reinforcing cords

31.75

D 19 E 23 Durable rubber
(b)

9.5 25 Tie band construction
3V 9 Flexible reinforcing cords
15.9
5V 14.7 8V 23

Durable rubber

(a) (c)

Figure 19.4: Design and construction of V-belts. (a) Standard V-belts cross sections with dimensions in millimeters; (b) typical
single-belt, showing reinforcing cords and wear-resistant exterior; (c) double V-belt, used for higher power transmission than
single belts. Up to five belts can be combined in this fashion.

wt Speed of Faster Shaft, rpm 10,000
5000
–ds–iNn–/2β–
ht 2000
1000
3V
500
dN
2 200
100
5V 5VP 8V 8VP
2
2β
=36°

5 10 20 50 100 200 500
Design power, kW

Figure 19.5: V-belt in sheave groove.

Figure 19.6: Guide to selection of belt cross section as a func-
tion of power transmitted and shaft speed.

A major advantage of V-belts over flat belts is their ure 19.6 shows the belt profiles that are commonly applied as
wedging action, which increases the normal force from dN a function of design power rating.
for flat belts to (dN/2)/ sin β (as shown in Fig. 19.5) for V-
belts, where β is the sheave angle. Because the V-belt has 19.4.2 Sheave Size
a trapezoidal cross section, the belt rides on the side of the
groove and a wedging action increases the traction. Pressure The materials and construction of V-belts have improved dra-
and friction forces act on the side of the belt. The force equa- matically in the past few decades. For example, a B-Section in
tions developed in Section 19.2 for flat belts are equally appli- a 175-mm sheave was rated at 3.1 kW in 1945, while a mod-
cable for V-belts if the coefficient of friction µ is replaced with ern belt under the same conditions has a rating over 8 kW.
µ/ sin β. Also, the belt length equation for flat belts given by This is due in large part to the transition from multiple re-
Eq. (19.5) is applicable to V-belts. inforcing cords centrally located in the belt to a single ten-
sile cord line located higher in the cross section. One sub-
19.4.1 Input Normal Power Rating tle complication occurs with such belt designs: the pitch di-
ameter (from which velocities are calculated) shifts upward
It is essential that the maximum possible load conditions be when a modern belt replaces a conventional design. For this
considered in designing a V-belt. The design power rating of reason, the International Standards Organization introduced
the belt needs to consider the service factor, or the term datum line to differentiate pitch lines associated
with new and conventional belt designs; this standard was
hpr = f1hp, (19.11) adopted in the United States in 1988. Modern sheaves are

where f1 is the overload service factor for various applica-
tions as given in Table 19.2, and hp is the input power. Fig-

V-Belts 553

Table 19.2: Typical overload service factors, f1. Source: Courtesy of the Gates Corporation.

Power source
Normala Demandingb

Driven machine Service (hr/day) Service (hr/day)
3–5 8–10 16–24 3–5 8–10 16–24

Dispensing, display equipment, measuring equipment, office and pro- 1.0 1.1 1.2 1.1 1.2 1.3
jection equipment.
Liquid agitators, appliances, sewing machines, sweepers, light-duty 1.1 1.2 1.3 1.2 1.3 1.4
conveyors, fans, light duty machine tools (drill presses, lathes, saws),
woodworking equipment.
Semi-liquid agitators, centrifuges, centrifugal compressors, heavy-duty 1.1 1.2 1.4 1.2 1.3 1.5
conveyors, dough mixers, generators, laundry equipment, heavy-duty
machine tools (boring mills, grinders, mills, shapers), presses, shears,
printing machinery, centrifugal and gear pumps.
Brick machinery, piston compressors, screw conveyors, bucket ele- 1.2 1.3 1.5 1.4 1.5 1.6
vators, extractors, hammer mills, paper pulpers, pulverizers, piston
pumps, extruders, rubber calendar mills, textile machinery.
Jaw crushers, hoists, ball mills, rod and tube mills, sawmill machinery. 1.3 1.4 1.6 1.5 1.6 1.8

a Includes normal torque, squirrel cage, synchronous and split phase AC motors; shunt wound DC motors;
multiple cylinder internal combustion engines.
b Includes high torque, high slip, repulsion-induction, single phase, series wound AC motors; series wound,
compound wound DC motors; single piston internal combustion engines.

Table 19.3: Recommended minimum datum diameters, in 19.4.3 Design Power Rating
millimeters, of sheaves for general purpose electric motors.
Source: Courtesy of the Gates Corporation. Table 19.5 gives the rated power for selected belt types. The
belts considered are the 3V and 5V cross sections, but the
Motor power, Motor rpm data for the other cross sections shown in Fig. 19.4a are read-
ily available in manufacturers’ literature. It should be noted
kW 575 690 870 1160 1750 3450 that Table 19.5 represents a very small sampling of avail-
able V-belts; there are many more cross-sections, and many
0.375 64 64 56 — — — more lengths and capacities depending on specific materi-
0.50 75 64 61 56 — — als and belt construction. Table 19.5 is useful for illustration
0.75 75 75 61 61 56 — purposes, but for detailed design, a wider selection of belts
1.1 75 75 61 61 61 56 should be considered, as can be readily obtained from manu-
1.5 96 75 75 61 61 61 facturers’ web sites.
3.75 115 115 96 75 75 66
7.5 150 132 117 112 96 75 Design Procedure 19.1: V-Belt
11.2 175 150 5.4 117 112 96 Drives
15 210 175 150 137 117 112
22.5 250 230 175 175 137 — It will be assumed that a belt drive will be designed for
37.3 280 250 230 208 175 — power transmission where the shaft speeds (and hence speed
55 355 330 267 250 230 — ratio) and desired center distance are known. The power
75 457 380 320 280 250 — available can be obtained from the rating of the motor, or
150 560 560 560 — — — else it can be obtained from design requirements. Based on
these quantities, this design procedure provides a methodol-
specified based on a datum diameter; the concepts of datum ogy for selecting a cross-section of a belt, choosing sheaves
diameter and pitch diameter are the same, and the sheave and number of belts required.
datum diameter can be used for velocity calculations. For
standard belt cross-sections, the datum diameter is nearly the 1. Estimate the overload service factor from Table 19.2 and
same as the outside diameter of the sheave, since this is very use it to obtain the required belt power rating using
close to the value of the reinforcing line for most belts. Eq. (19.11).

The design of a V-belt drive should use the largest pos- 2. Select a cross section of the belt from the required belt
sible sheaves. Unfortunately, large sheaves are usually more power rating and the shaft speed using Fig. 19.6.
expensive, and there is an inherently larger center distance
associated with their use compared to smaller sheaves. Small 3. Obtain the minimum allowable sheave datum diameter
sheaves are less efficient, require larger belt preload (and as- from Table 19.3.
sociated higher load on bearings and shafts), and greatly re-
duce belt life because of slip and extreme flexing of the belt.
Table 19.3 shows minimum recommended sheave datum di-
ameters for general purpose electric motors.

Usually, much larger sheave datum diameters are used
than suggested by Table 19.3. For some applications, it may
be economically justifiable to select sheave designs based on
design constraints and manufacture them to the desired di-
mensions. However, it is often preferable to utilize standard
sheaves. Table 19.4 presents some combinations of standard
sheave sizes, and also provides a power correction factor, f2,
which will modify the belt power rating as discussed in De-
sign Procedure 19.1.

554 Chapter 19 Flexible Machine Elements
Table 19.4: Center distance and power correction factor, f2, for standard sheaves. Source: Courtesy of the Gates Corporation.

Belt Speed Datum diameter, mm Belt length, mm

type ratioa Small sheave Large sheave 635 762 875 1020 1140 1250 1520
274 338 401 465 528 655
33VV 1.00 67 67 211 262 325 389 452 516 643 Power Correction
76 76 198 246 310 373 437 500 627 Factor:
1.25 85 85 183 201 264 328 391 455 582
1.5 114 114 137 269 333 396 460 523 650 f2 = 0.8
2.0 64 80 206 218 282 345 409 472 599 f2 = 0.9
2.5 93 114 155 178 244 307 371 434 561 f2 = 1.0
3.0 114 142 — 165 229 292 355 420 546
120 152 — 241 305 371 434 498 625
71 105 178 231 295 358 422 485 612
76 114 173 213 277 340 404 467 594
85 127 150 — — 241 305 368 495
135 203 — 221 284 348 411 485 333
67 135 155 198 262 325 391 465 569
76 152 132 — 196 262 325 391 594
105 203 — — — — 244 310 582
135 269 — 193 257 323 386 450 577
67 165 — 180 246 310 373 439 566
71 175 — — 213 279 343 409 536
80 203 — — ——— — 356
142 356 — 152 221 287 353 417 546
67 203 — — — 203 272 340 470
93 269 — — ——— — 368

121 356 — 1770 Belt length, mm 3175 3810
617 2030 2290 2540 1303 1621
Belt Speed Datum diameter, mm 582 732 859 986 1267 1585
544 696 823 950 1229 1547
type ratioa Small sheave Large sheave 1500 404 658 785 912 1090 1407
523 518 645 772 1212 1529
55VV 1.00 180 180 478 434 640 767 894 1123 1440
203 203 442 — 549 676 805 963 1280
1.5 229 229 404 472 — 516 643 1163 1608
2.0 318 318 — 411 589 716 843 1105 1549
3.0 191 287 384 — 528 655 785 729 1181
235 356 — — ——— 937 1260
318 475 — — — 470 607 1006 1328
180 356 330 — 411 549 681 — 848
203 406 — ———
356 711 —
203 599 —
180 538 —
318 953 —

a Nominal speed ratio, actual value may vary slightly.

4. Locate the sheave diameter combinations in Table 19.4 Example 19.3: V-Belt Design
that are suitable for a desired speed ratio. Disregard
from consideration any candidates that are smaller than Given: A 7.5-kW squirrel cage electric motor operating at
the minimum values obtained in Step 3. From the re- a speed of 2000 rpm drives a centrifugal air compressor in
maining candidates, select a sheave size that is consis- normal service at 1000 rpm. It is desired to have a center
tent with space requirements. distance around 375 mm.

5. From Table 19.4, locate the center distance that most Find: Determine the proper belt cross section for this appli-
closely matches design constraints, and obtain the cation, select pulleys, and choose the number of belts.
power correction factor, f2. Note that the belt length
can be calculated from Eq. (19.5) or read directly from Solution: This approach will closely follow Design Proce-
Table 19.4. dure 19.1.

6. From Table 19.5, locate the proper belt cross section 1. Referring to Table 19.2, note from the footnote that the
and center distance, and obtain the basic power rating squirrel cage motor represents a normal power source.
per belt, h1. Note that for very high speeds or small Therefore, an overload service factor of f1 = 1.2 is se-
sheaves, an additional power may be required. This is lected as typical for a compressor, assuming the pump
usually a small amount and is neglected in this design is operated during a normal 8-hr shift. Therefore, the
procedure. design power is, from Eq. (19.11),

7. The rated power per belt is given by hpr = f1hp = (1.2)(7.5) = 9.0 kW.

hpb = f2h1. (19.12) 2. For 7.5 kW and a shaft speed of 2000 rpm, Fig. 19.6 sug-
gests a 3V cross-section is proper. Note, however, that
8. The number of belts required can be obtained from the the application is very close to the limit of 3V effective-
required power from Step 1: ness, so a 5V belt may also provide a reasonable design
solution. However, this solution will continue using a
N = hpr . (19.13) 3V belt as recommended by Fig. 19.6.
hpb

Wire Ropes 555

Table 19.5: Rated power in kW per belt for selected 3V and 5V cross sections. Source: Courtesy of the Gates Corporation.

Belt Speed Small sheave outside diameter, mm

(rpm) 65 71 76 80 85 93 105 114 120 127 135 142

33VV 200 0.20 0.23 0.26 0.29 0.32 0.37 0.46 0.52 0.56 0.60 0.66 0.71

400 0.37 0.41 0.48 0.53 0.60 0.69 0.85 0.97 1.05 1.13 1.22 1.32
600 0.51 0.58 0.68 0.75 0.85 0.99 1.22 1.40 1.51 1.63 1.77 1.90
800 0.64 0.74 0.87 0.96 1.08 1.27 1.56 1.80 1.95 2.10 2.28 2.46
1000 0.77 0.89 1.04 1.16 1.31 1.54 1.89 2.18 2.36 2.55 2.78 2.99
1200 0.89 1.03 1.21 1.35 1.53 1.80 2.22 2.56 2.78 2.99 3.25 3.51
1400 1.01 1.16 1.37 1.53 1.74 2.05 2.53 2.92 3.17 3.42 3.72 4.01
1600 1.11 1.29 1.53 1.71 1.94 2.29 2.83 3.27 3.54 3.83 4.16 4.48
1800 1.22 1.42 1.68 1.88 2.14 2.53 3.13 3.60 3.92 4.22 4.59 4.95
2000 1.31 1.54 1.83 2.04 2.33 2.75 3.41 3.93 4.27 4.60 5.00 5.39
2400 1.51 1.76 2.10 2.36 2.69 3.19 3.95 4.54 4.94 5.32 5.77 6.21
3000 1.75 2.07 2.48 2.78 3.18 3.77 4.68 5.38 5.83 6.27 6.80 7.30
4000 2.09 2.48 3.00 3.37 3.87 4.60 5.68 6.51 7.02 7.53 8.06 8.65
5000 2.33 2.78 3.38 3.81 4.38 5.18 6.36 7.23 7.76 — — —

Belt Speed Small sheave outside diameter, mm

(rpm) 180 190 200 215 230 235 250 260 287 317 355 405

55VV 100 1.01 1.10 1.22 1.34 1.46 1.52 1.63 1.76 1.99 2.27 2.61 3.07
200 1.88 2.06 2.28 2.51 2.73 2.84 3.07 3.31 3.75 4.27 4.92 5.78
300 2.69 2.95 3.28 3.60 3.93 4.10 4.42 4.77 5.41 6.17 7.12 8.36
400 3.45 3.80 4.23 4.66 5.08 5.29 5.71 6.17 7.00 7.98 9.18 10.82
500 4.20 4.62 5.15 5.67 6.18 6.45 6.96 7.53 8.50 9.77 11.26 13.20
600 4.91 5.41 6.03 6.65 7.26 7.53 8.13 8.80 10.00 11.41 13.13 15.44
800 6.27 6.92 7.68 8.50 9.33 9.70 10.44 11.34 12.83 14.62 16.79 19.62
1000 7.53 8.36 9.33 10.29 11.19 11.71 12.61 13.65 15.44 17.53 20.14 23.35
1500 10.37 11.49 12.83 14.10 15.44 16.04 17.31 18.65 20.96 23.65 26.71 30.29
2000 12.76 14.02 15.67 17.23 18.72 19.47 20.81 22.31 24.84 — — —

19.5 Wire Ropes

3. From Table 19.3, note that the 7.5-kW motor operat- Wire ropes are used instead of flat belts or V-belts when
ing at 2000 rpm will require sheaves that are slightly
smaller than 96 mm. power must be transmitted over long distances as in hoists,

4. In Table 19.4, there are four sheave combinations that elevators, ski lifts, etc. Figure 19.7 shows cross sections of se-
are suitable for this application. Either the 105 mm or
135 mm smaller sheave are above the minimum sheave lected wire ropes. The center portion (dark section) is the core
dimensions and will yield reasonable designs. This so-
lution will assume that the bearing reaction forces are of the rope and is often made of hemp (a tall Asiatic herb), but
not as much of a concern as low center distance, so the
105- and 203-mm sheave combination will be selected. can also be constructed of a polymer such as polypropylene

5. For the required nominal center distance of 375 mm, a or steel strands. The purposes of the core are to elastically
1250-mm belt is most suitable. From the color code in
Table 19.4, note that the application factor for this belt support the strands and to lubricate them to prevent exces-
is f2 = 0.90.
sive wire wear.
6. Referring to Table 19.5, a 105 mm outer diameter sheave
operating at 2000 rpm has a basic power rating of h1 = The strands are groupings of wires placed around the
3.41 kW.
core. In Fig. 19.7a there are six strands, and each strand con-
7. The rated power per belt is therefore, from Eq. (19.12),
sists of 19 wires. Wire ropes are typically designated as, for
hpb = f2h1 = (0.90)(3.41) = 3.07 kW.
example, “1 1 , 6 × 19 hauling rope.” The 1 1 gives the wire
8. The number of belts required is given by Eq. (19.13) as 8 8
rope diameter in inches, designated by the symbol d. The 6
N = hpr = 9.0 = 2.93.
hpb 3.07 designates the number of strands and 19 designates the num-

Therefore, the number of belts are chosen as N = 3. ber of wires in a strand. The wire diameter is designated by

In summary, three 1.25-m-long, 3V belts operating on the symbol dw. The term “hauling rope” designates the ap-
sheaves of 105 and 203 mm are needed for this application.
Since three belts are needed, a triple belt design (see Fig. 19.4) plication in which the wire rope is to be used. Generally, wire
and associated sheaves are specified.
ropes with more but smaller wires are more resistant to fa-

tigue, whereas ropes with fewer but larger wires have greater

abrasive wear resistance.

When a wire rope is bent and unbent around a sheave,

the strands and wires slide over each other at a small length

scale. The sliding velocities are low, but the pressure is high,

so that wire wear is a concern. For this reason, wire rope

lubrication is essential for prolonged life, and lubricant needs

to be periodically replaced. Lubricants need to be matched to

core materials to ensure proper wetting, and are also essential

for corrosion protection.

Figure 19.8 shows two lays of wire rope. The regular

lay (Fig. 19.8b) has the wire twisted in one direction to form

strands and the strands twisted in the opposite direction to

form the rope. Visible wires are approximately parallel to the

rope axis. The major advantage of the regular lay is that the

rope does not kink or untwist and is easy to handle. The Lang

lay (Fig. 19.8a) has the wires in the strands and the strands in

the rope twisted in the same direction. This type of lay has

556 Chapter 19 Flexible Machine Elements
Core (a)
Strand
(b)
Wire
(a)

(b)

Figure 19.8: Two lays of wire rope. (a) Lang; (b) regular.

(c) (d) where Am is the cross-sectional area of metal strand in stan-
dard hoisting and haulage ropes. The allowable stress is ob-
Figure 19.7: Cross sections of selected wire rope. (a) 6 × 19 tained from Table 19.6.
fiber core; (b) 1 × 19; (c) 6 × 36 wire core; (d) 18 × 7 fiber core.
From Eq. (1.1), the safety factor is

ns = σall . (19.16)
σt

more resistance to abrasive wear and bending fatigue than Table 19.7 gives minimum safety factors for a variety of wire
the regular lay. Lang-lay ropes are, however, more suscepti- rope applications. The safety factor obtained from Eq. (19.16)
ble to handling abuses, pinching in undersized grooves, and should be larger than the safety factor obtained from Table
crushing when improperly wound on drums. Also, the twist- 19.7.
ing moment acting in the strand tends to unwind the strand,
causing excessive rope rotation. Lang-lay ropes should there- 19.5.2 Bending Stress
fore always be secured at the ends to prevent the rope from
unlaying. From Eqs. (4.46) and (4.47), the bending moment applied to
the wires in a rope passing over a pulley is
Although steel is most popular, wire rope is made of
many kinds of metal, such as copper, bronze, stainless steel, M = EI and M = σI . (19.17)
and wrought iron. Table 19.6 lists some of the various ropes r c
that are available, together with their characteristics and
properties. The cross-sectional area of the metal strand in Equating these expressions and solving for the bending stress
standard hoisting and haulage ropes is Am ≈ 0.38d2.

gives

σb = Ec
19.5.1 Tensile Stress ,
r
The total force acting on the rope is
where r is the radius of curvature that the rope will experi-

ence and c is the distance from the neutral axis to the outer

Ft = Fw + Fr + Fa + Fh, (19.14) fiber of the wire. A rope bearing against a sheave may flatten

slightly, but the radius of curvature that the rope experiences

where is very nearly the sheave radius D/2. Similarly, c can be taken
Fw = weight being supported, N
Fr = rope weight, N as dw/2, yielding
Fa = force due to acceleration, N
Fh = static load, often due to a headache ball, N σb = Edw , (19.18)
D
The static load or dead weight, Fh is essential for most appli-
cations to maintain tension in the wire rope and to prevent it where dw is the wire diameter and D is the pulley diameter.
from slipping off of a sheave. The tensile stress is Note from Eq. (19.18) that if D/dw is very large, the bending
stress will be small. Suggested minimum sheave diameters
σt = Ft , (19.15) in Table 19.6 are based on a D/dw ratio of 400. If possible,
Am the sheaves should be designed for a larger ratio. If the ratio
D/dw is less than 200, heavy loads will often cause a perma-
nent set of the rope. Thus, for a safe design with very long
service-free life, it is recommended that D/dw ≥ 400.

Wire Ropes 557

Table 19.6: Wire rope data. Source: Shigley and Mitchell [1983].

Rope Weight Minimum Rope Material Size of Stiffness,b Strength,c
per lengtha , sheave, diameter, outer GPa MPa
diameter, wires
kg/m d, mm

6 × 7 Haulage 3465 d2 42d 6 − 12 Monitor steel d/ 9 96 690

Plow steel d/ 9 96 600

6 × 19 Standard 3700 d2 26d − 34d 6 − 70 Mild plow steel d/ 9 96 525
Monitor steel d/ 13 − d/ 16 83 730

hoisting Plow steel d/ 13 − d/ 16 83 640

6 × 37 Special 3580 d2 Mild plow steel d/ 13 − d/ 16 83 550
18d 6 − 90 Monitor steel d/ 22 75 690

flexible 3350 d2 21d − 26d 6 − 40 Plow steel d/ 22 75 600
8 × 19 Extra Monitor steel d/ 15 − d/ 19 69 630

flexible 3930 d2 Plow steel d/ 15 − d/ 19 69 550
7 × 7 Aircraft 4040 d2 850
4970 d2 — 1.5 − 10 Corrosion-resistant steel — — 850
7 × 9 Aircraft Carbon steel —— 930
985
19-Wire aircraft — 3 − 35 Corrosion-resistant steel — — 1140
Carbon steel —— 1140

— 0.75 − 8 Corrosion-resistant steel — —
Carbon steel ——

a Weight per length in kg/m for d in meters.
b The stiffness is only approximate; it is affected by the loads on the rope and, in general, increases with the life of the rope.
c The strength is based on the nominal area of the rope. The figures given are only approximate and are based on 25-mm rope sizes

and 6-mm aircraft cable sizes.

However, these design rules are hardly ever followed for Strength loss, percent 50
some applications, for a number of reasons. First, consider a 40
50-mm-diameter wire rope, such as would typically be used 30 10 20 30 40
in the crane or in the dragline discussed in Case Study 19.1. 20 D/d ratio
According to the rules just stated, the sheave on the crane 10
would have to be around 2 m in diameter. Further, since the
wire rope is wound on a drum and failure can occur in the 0
rope adjacent to the drum, such a large diameter would also 0
be needed for the drum. Because the required motor torque is
the product of the hoist rope tension and the drum radius, a Figure 19.9: Percent strength loss in wire rope for different
very large motor would be required for lifting relatively light D/d ratios.
loads. As a result, the entire crane would be much larger and
more expensive, and other design challenges would arise. Relative service life, percent 100
For example, ground collapse below the crane would be more 80
common, energy consumption would increase, and it would 60
be more difficult to move cranes to different sites. 40
20
The design rules just stated are recommendations for ap- 00
plications where the wire rope should attain infinite life. The
economic consequences of infinite-life wire rope are usually
too large to bear, and smaller pulleys are usually prescribed.
To prevent failures that result in property damage or personal
injury, the wire ropes are periodically examined for damage.
Since a broken wire will generally be easily detected and will
snag a cotton cloth dragged over the rope’s surface, rope life
requirements are often expressed in terms of the number of
broken wires allowed per length of wire rope. For example,
the American Society of Mechanical Engineers [2007] requires
inspections of crane wire ropes every 6 months, and if any
section has more than six broken wires within one lay (revo-
lution of a strand) of the rope, or three in any strand within
a lay, the entire wire rope must be replaced. The same stan-
dard calls for pulley-to-wire diameter ratios of 12:1, which
obviously results in wire rope with finite service life. Figures
19.9 and 19.10 relate the decrease in strength and service life
associated with smaller pulley diameters.

19.5.3 Bearing Pressure 20 40 60
D/d ratio
The rope stretches and rubs against the pulley, causing wear
of both the rope and pulley. The amount of wear depends on Figure 19.10: Service life for different D/d ratios.
the pressure on the rope in the pulley groove, or

p = 2Ft , (19.19)
dD

558 Chapter 19 Flexible Machine Elements

Table 19.7: Minimum safety factors for a variety of wire rope Table 19.8: Maximum allowable bearing pressures for various
applications. Note that the use of these safety factors does not sheave materials and types of rope. Source: From Shigley and
preclude a fatigue failure. Source: From Shigley and Mitchell Mitchell [1983].
[1983].
Maximum bearing pressure for
Safety sheave material, MPa
factor,
Application Rope Cast Cast Chilled Manganese
Track cables ns iron a steelb cast ironc steel d
Guys 3.2
Mine shafts, m 3.5 Regular lay
8.0 6× 7 2.07 3.80 4.48 10.1
Up to 150 7.0 6 × 19 3.31 6.20 7.58 16.5
300–600 6.0 6 × 37 4.04 7.41 9.13 20.7
600–900 5.0 8 × 19 4.69 8.69 10.7 24.1
Over 900 5.0 Lang lay
Hoisting 6.0 6× 7 2.41 4.13 4.93 11.4
Haulage 6.0 6 × 19 3.79 6.89 8.34 19.0
Cranes and derricks 7.0 6 × 37 4.55 8.14 10.0 22.8
Electric hoists 5.0 a On end grain of beech, hickory, orgum.
Hand elevators 7.5 a For minimum HB = 125.
Private elevators 4.5 b 0.30–0.40% carbon; HB (min.) = 160.
Hand dumbwaiters 7.5 c Use only with uniform surface hardness.
Grain elevators d For high speeds with balanced sheaves having ground surfaces.
Passenger elevators 7.60
Up to 0.25 m/s 9.20 Example 19.4: Analysis of Wire Rope
0.25–1.5 11.25
1.5–4.0 11.80 Given: Lighting fixtures in a large theater are to be raised
4.0–6.0 11.90 and lowered by a wire rope, and need to be moved a max-
6.0–1500 imum of 30 m. The maximum load to be hoisted is 12 kN
Freight elevators 6.65 at a velocity not exceeding 0.5 m/s and an acceleration of 1
Up to 0.25 m/s 8.20 m/s2. Use 25-mm plow steel in the form of 6 × 19 standard
0.25–1.5 10.00 hoisting ropes. Use a rope cross-sectional area of 2.41 × 10−4
1.5–4.0 10.50 m2.
4.0–6.0 10.55
6.0–7.5 Find: Determine the safety factor while considering
Powered dumbwaiters 4.8
Up to 0.25 m/s 6.6 (a) Tensile stress
0.25–1.5 8.0
1.5-4.0 (b) Bending stress

where d is the rope diameter. The pressure obtained from (c) Bearing pressure
Eq. (19.19) should be less than the maximum pressure ob-
tained from Table 19.8 for various pulley materials and types (d) Fatigue
of rope.

Solution: From Table 19.6 for 6 × 19 standard hoisting wire
rope, assuming the use of only one rope,

Fr = 3700d2h2 = 3700(0.025)2(30) = 69.4 kg/rope,

19.5.4 Fatigue Fw = Wmax = 12, 000 N.
Note that Fr is 680 N/rope. The force due to acceleration is
For the rope to have a long life, the total force, Ft, must be
less than the fatigue allowable force, Ff , where Fa = ma = W a = (12, 000 + 680)(1) = 1290 N
g 9.81

The total force on the rope is

Ff = SudD . (19.20) Ft = Fw + Fr + Fa = 12, 000 + 680 + 1290 = 14, 000 N.
2000

(a) Tensile stress:

The ultimate strength given by Eq. (19.20) for three materials σt = Ft = 14, 000 = 58.1 MPa.
is Am 2.41 × 10−4

From Table 19.6 for 6 × 19 standard hoisting wire rope
made of plow steel, σall = 640 MPa. The safety factor is

Monitor plow steel 1650 ≤ Su ≤ 1930 MPa σall 640
σt 58.1
Plow steel 1400 ≤ Su ≤ 1650 MPa (19.21) ns = = = 11.0

Mild plow steel 1250 ≤ Su ≤ 1400 MPa

Rolling Chains 559

From Table 19.7, the recommended safety factor for Roller link Link Pin Pin a
hoisting applications is 5.0. Thus, one rope is sufficient Roller plate link
as far as the tensile stress is concerned.
d
(b) Bending stress: From Table 19.6, the minimum pulley di-
ameter for 6 × 19 standard hoisting wire rope is 26d Bushing Pitch Pitch
to 34d. No design constraints have been given regard-
ing the pulley, so assign the conservative value of D = Figure 19.11: Various parts of rolling chain.
34d = 850 mm. Also, from the same table, the wire
diameter should be between d/13 and d/16. Choose
dw = d/16 = 1.56 mm, so that

D 850
= = 545.

dw 1.56

Permanent set should not be a concern, since D/dw ≥
400. From Table 19.6, the stiffness is 83 GPa. The bend-
ing stress is, from Eq. (19.18),

σb = E dw = 83 × 109 = 153 MPa.
D 544

The safety factor due to bending is

ns = σall = 640 = 4.18.
σb 153

This safety factor is less than the 5.0 recommended, but (a) (b)
the safety factor is for static tensile loading. Changing
the material from plow steel to monitor steel would pro- Figure 19.12: Typical rolling chain. (a) One-strand rolling
duce a safety factor of 4.81, closer to 5, but this is an chain; (b) three-strand chain.
indication that the wires will not fail in bending. The
tensile overload factor from Table 19.7 is not really ap- 19.6 Rolling Chains
plicable to other failure modes.
Rolling chains are used to transmit power between two
Note that increasing the number of ropes will not alter sprockets rotating in the same plane. The machine element
the results, since the rope will still be wrapped around that it most resembles is a timing belt (shown in Fig. 19.3).
the same sized sheave. The obvious solution is to use The major advantage of using a rolling chain compared to a
an even larger pulley, but this solution has design im- belt is that rolling chains do not slip. Large center distances
plications, as discussed in Section 19.5.2. can be dealt with more easily with rolling chains with fewer
elements and in less space than with gears. Rolling chains
(c) Bearing pressure: From Eq. (19.19), also have high efficiency. No initial tension is necessary and
shaft loads are therefore smaller than with belt drives. The
p= 2Ft = 2(14, 000) = 1320 kPa. only maintenance required after careful alignment of the ele-
dD (0.025)(0.85) ments is periodic lubrication, and with proper lubrication, a
long life can be attained.
From Table 19.8 for a 6 × 19 Lang lay for a cast steel
pulley, pall = 6.20 MPa. The safety factor is 19.6.1 Operation of Rolling Chains

ns = pall = 6.20 = 4.70. Figure 19.11 shows the various parts of a rolling chain with
p 1.32 pins, bushings, rollers, and link plates. The rollers turn on
bushings that are press fitted to the inner link plates. The pins
(d) Fatigue: For monitor steel the ultimate strength is at are prevented from rotating in the outer plates by the press-fit
most 1930 MPa from Eq. (19.21). The allowable fatigue assembly. Table 19.9 gives dimensions for standard sizes. The
force from Eq. (19.20) is size range given in Table 19.9 is large, and multiple strands
can be used (see Fig. 19.12), so that chains can be used for both
Ff = SudD = (1930 × 106)(0.025)(0.85) = 20 kN. large and small power transmission levels. A large reduction
2000 2000 in speed can be obtained with rolling chains if desired. The
tolerances for a chain drive are larger than for gears, and the
The safety factor is installation of a chain is relatively easy.

ns = Ff = 20, 000 = 1.67. The minimum wrap angle of the chain on the smaller
Ft 12, 000 sprocket is 120◦. A smaller wrap angle can be used on idler
sprockets, which are used to adjust the chain slack where the
Thus, fatigue failure is the most likely failure to oc- center distance is not adjustable. Horizontal drives (the line
cur, since it produced the smallest safety factor. Using connecting the axes of sprockets is parallel to the ground) are
four ropes instead of one would produce a safety factor
greater than 5.

560 Chapter 19 Flexible Machine Elements
Table 19.9: Standard sizes and strengths of rolling chains.

ANSI ISO Pitch, Roller Pin Link Average Weight
Chain Chain pt , mm Diameter, Width, diameter, plate tensile per meter,
numbera number thickness, strength,
mm mm d, mm a , mm S u , kN kg/m

b 25 — 6.350 3.302 3.175 2.30 0.762 3.89 0.12
b 35 1.27 9.34 0.31
— 9.525 5.080 4.763 3.58 1.50 13.90 0.591
40 08A-1 12.700 7.93 7.85 3.98 2.03 21.80 1.00
50 10A-1 15.875 10.15 9.55 5.09 2.42 31.13 1.36
60 12A-1 19.050 11.91 12.65 5.96 3.25 55.60 2.58
80 16A-1 25.400 15.88 15.87 7.94 4.00 87.00 3.88
100 20A-1 31.750 19.05 19.05 9.53 4.80 125.00 5.56
120 24A-1 38.100 22.22 25.40 11.10 5.60 170.00 7.44
140 28A-1 44.450 25.40 25.40 12.70 6.40 223.00 10.0
160 32A-1 50.800 28.58 31.75 14.27 8.00 347.00 16.7
200 40A-1 63.500 39.67 38.10 19.85 9.50 500.00 23.7
240 48A-1 76.200 47.60 47.60 23.80

a The pitch, in inches, can be obtained from the chain number by taking the left digits and dividing by 8.
b Without rollers.

pt/2 pt The chordal rise while using Eq. (19.23) gives
BCA
∆r = r − rc = r (1 − cos θr) = r 1 − cos 180 , (19.24)
θr Pitch N
circle
rc where N is the number of teeth in the sprocket. Note also
0 from triangle 0CA

sin θr = pt/2 or pt = 2r sin θr = D sin θr. (19.25)
r

19.6.4 Chain Length

The number of links is

Figure 19.13: Chordal rise in rolling chains. Note that the L = 2cd + N1 + N2 + (N2 − N1)2 , (19.26)
chain link travels upwards as well as horizontally when mov- pt pt 2 4π2 cd
ing from position A to position B.
pt

recommended; vertical drives (the line connecting the axes of where cd is the center distance between sprockets. It is nor-
sprockets is perpendicular to the ground) are less desirable. mally recommended that cd/pt lie between 30 and 50 pitches.
Vertical drives, if used, should be used with idlers to prevent If the center distance per pitch is not given, the designer can
the chain from sagging and to avoid disengagement from the fix cd/pt and calculate L/pt from Eq. (19.26). The next larger
lower sprocket. even integer L/pt should be chosen. With L/pt as an integer,
the center distance per pitch becomes

19.6.2 Kinematics cd = A + A2 − B2 , (19.27)
pt 2

The velocity ratio, comparable to the gear ratio given in where
Eq. (14.19), is
A = 1 L − N1 + N2 ,
d2 ω1 N2 , 4 pt 2 (19.28)
d1 ω2 N1 (19.22)
gr = = =

where N is the number of teeth in the sprocket, ω is the an- B = N2 − N1 . (19.29)
gular velocity in rad/s, and d is the diameter. For a one-step 2π
transmission it is recommended that gr < 7. Values of gr be-
tween 7 and 10 can be used at low speed (below 650 ft/min). The value of cd/pt obtained from Eq. (19.27) should be de-
creased by about 1% to provide slack in the nondriving chain
strand.

The chain velocity is

19.6.3 Chordal Rise u = NaptN, (19.30)

An important factor affecting the smoothness of rolling chain where
drive operation, especially at high speeds, is chordal rise, Na = angular speed, rev/min
shown in Fig. 19.13. From right triangle 0CA pt = chain pitch, m
N = number of teeth in the sprocket
rc = r cos θr. (19.23)

Rolling Chains 561

Table 19.10: Service factor, a1, for rolling chains. Example 19.5: Power and Force in
Rolling Chains
Type of input power
Given: A four-strand, ANSI No. 25 rolling chain transmits
Type of Internal Electric Internal power from a 25-tooth driving sprocket that turns at 900
driven load combustion motor combustion rpm. The speed ratio is 4:1.
Smooth engine with engine with
Moderate shock hydraulic or mechanical Find: Determine the following:
Heavy shock turbine
drive drive (a) Power that can be transmitted for this drive
1.0 1.0 1.2
1.2 1.3 1.4 (b) Tension in the chain
1.4 1.5 1.7
(c) Chain length if center distance is about 250 mm.
Table 19.11: Multiple-strand factor for rolling chains.
Solution:
Number Multiple-strand
of factor, (a) From Table 19.12 for a smaller sprocket of 25 teeth and
a2 a speed of 900 rpm, the power rating is 0.9 kW and
strands 1.0 bath lubrication is required. From Table 19.11 for four
1 1.7 strands, a2 = 3.3. Nothing is mentioned about the type
2 2.5 of input power or drive load, so assume that a1 = 1.
3 3.3 From Eq. (19.31), the power that can be transmitted is
4 3.9
5 4.6
6

19.6.5 Power Rating hp = hpr a2 = (0.9) 3.3 = 2.97 ≈ 3.0 kW.
a1 1
The required power rating of a chain is given by
(b) From Table 19.9 for No. 25 chain, the pitch is 6.35 mm.
a1 The velocity can be calculated from Eq. (19.30) as
a2
hpr = hp , (19.31) u = NaptN = (900)(0.00635)(25) = 142.9 m/min

where hp is the power that is transmitted or input, a1 is the or u = 2.38 m/s. Therefore, the tension in the chain is
service factor obtained from Table 19.10 and a2 is the strand
factor obtained from Table 19.11. P = hp = 3000 = 1260 N.
u 2.38
Rolling chains are available in a wide variety of sizes;
Fig. 19.14 and Design Procedure 19.2 provide some guide- (c) The number of teeth on the larger sprocket given a
lines for selection of rolling chains. Note that Fig. 19.14 has speed ratio of 4 is 4(25) = 100 teeth. Note that for a cen-
four scales, depending on the number of strands in the chain ter distance of 250 mm,
(see Fig. 19.12). For each type of roller chain, there is a char-
acteristic shape to the power curve, where the power rating cd = 250 = 39.4,
increases up until a certain sprocket size, and then decreases. pt 6.35
This is attributed to a change in failure mode; at low speeds,
link plate fatigue will be the dominant failure mode, while at which is between the 30 and 50 pitches recommended.
higher speeds, wear of the pins becomes dominant. From Eq. (19.26),

Table 19.12 provides power ratings for selected standard- L = 2 cd + N1 + N2 + (N2 − N1)2
sized roller chains. Similar data can be readily obtained pt pt 2 4π2 cd
for other chain configurations from manufacturers’ literature
and web sites. The four types of lubrication given in Table pt
19.12 are 25 + 100 (100 − 25)2
= 78.8 + 2 + 4π2(39.4) = 144.9.
• Type A – Manual or drip lubrication, oil applied peri-
odically with brush or spout can or applied between the Since it is good practice to specify the next largest even
link plate edges with a drop lubricator. number of links, L is chosen as 146 pitches, or 927 mm.

• Type B – Oil bath or oil slinger, where the oil level is 19.6.6 Silent Chain
maintained in a casing at a predetermined height, and
the chain is immersed into the bath during at least a part An interesting type of chain is the inverted tooth or silent
of its travel. chain, shown in Fig. 19.15, and made up of stacked rows
of flat, tooth shaped links that mesh with sprockets hav-
• Type C – Oil stream lubrication, where oil is supplied by ing compatible tooth spaces, much the way gear teeth mesh
a circulating pump inside the chain loop or lower span. (see Fig. 14.8). Because of the geometry advantages, these
chains are much quieter than conventional roller chains, es-
pecially at high speeds, but they are somewhat more expen-
sive. Washers or spacers may be present in some chain con-

562 Chapter 19 Flexible Machine Elements

Table 19.12: Power rating of selected standard roller chains, in kW.

ANSI Maximum speed of small sprocket (rpm) Required
Chain No. of Lubrication:
No. Teetha 50 100 300 500 900 1500 2100 3000 4000 5000 6000
0.56 Manual/drip
25 11 0.02 0.04 0.14 0.22 0.40 0.65 0.90 1.26 1.03 0.74 0.64 Bath or disc
12 0.03 0.05 0.15 0.25 0.43 0.71 0.98 1.37 1.17 0.84 0.90 Oil stream
15 0.04 0.07 0.19 0.31 0.54 0.88 1.22 1.72 1.64 1.17 1.17
18 0.04 0.08 0.22 0.37 0.65 1.06 1.46 2.06 2.16 1.54 1.37
20 0.04 0.09 0.25 0.41 0.72 1.18 1.63 2.29 2.52 1.81 1.58
22 0.05 0.10 0.28 0.45 0.79 1.29 1.79 2.51 2.91 2.08 1.92
25 0.06 0.11 0.31 0.51 0.90 1.47 2.03 2.86 3.53 2.52 2.27
28 0.06 0.12 0.35 0.57 1.01 1.65 2.28 3.21 4.18 2.99 2.52
30 0.07 0.13 0.37 0.61 1.08 1.77 2.44 3.43 4.53 3.32 3.18
35 0.08 0.16 0.44 0.72 1.26 2.06 2.84 4.01 5.28 4.18 3.66
40 0.09 0.17 0.50 0.82 1.44 2.35 3.25 4.58 6.03 5.11 1.03
45 0.10 0.19 0.57 0.92 1.62 2.65 3.66 5.15 6.79 6.09 0.78
0.88
35 11 0.08 0.16 0.46 0.76 1.34 2.19 3.02 2.19 1.42 1.02 1.23
12 0.09 0.18 0.51 0.83 1.46 2.39 3.30 2.50 1.62 1.16 1.62
15 0.11 0.22 0.63 1.04 1.83 2.98 4.12 3.49 2.27 1.62 1.90
18 0.13 0.27 0.76 1.25 2.19 3.58 4.95 4.59 2.98 2.13 2.19
20 0.15 0.30 0.85 1.39 2.43 3.98 5.49 5.37 3.49 2.50 2.66
22 0.16 0.33 0.93 1.52 2.68 4.37 6.04 6.20 4.03 2.88
25 0.19 0.37 1.06 1.73 3.04 4.97 6.86 7.51 4.88 3.49 –
28 0.22 0.41 1.18 1.94 3.41 5.57 7.69 8.90 5.78 4.14 –
30 0.23 0.44 1.27 2.08 3.66 5.97 8.24 9.87 6.41 4.59 –
35 0.27 0.51 1.48 2.42 4.26 6.96 9.62 12.44 8.08 0.25 –
40 0.31 0.59 1.73 2.77 4.87 7.95 10.99 15.20 8.24 – –
45 0.34 0.66 1.91 3.12 5.48 8.94 12.36 17.40 2.32 – 0.78
–
50 11 0.42 0.81 2.34 3.83 6.73 6.23 3.76 2.20 1.43 1.02 –
12 0.52 1.02 2.93 4.78 8.41 8.71 5.26 3.08 2.00 1.43 –
15 0.63 1.22 3.51 5.74 10.09 11.44 6.91 4.04 2.63 0.04 –
18 0.70 1.36 3.91 6.38 11.22 13.41 8.09 4.74 3.08 – –
20 0.77 1.50 4.30 7.02 12.34 15.46 9.34 5.47 3.55 – –
22 0.87 1.70 4.89 7.97 14.02 18.73 11.31 6.62 – – –
25 0.98 1.90 5.47 8.93 15.70 22.21 13.41 7.85 – – –
28 1.05 2.04 5.86 9.57 16.82 24.63 14.87 8.71 – – –
30 1.22 2.38 6.83 11.17 19.63 31.03 18.73 0.70 – – –
35 1.40 2.72 7.81 12.76 22.43 36.64 22.89 – – – –
40 1.57 3.06 8.79 14.35 25.24 41.21 27.31 – – – –
–
80 11 1.54 3.01 8.62 14.08 17.14 8.03 4.87 2.81 1.83 – –
12 1.69 3.27 9.41 15.36 19.52 9.15 5.56 3.21 2.08 – –
15 2.10 4.10 11.76 19.20 27.29 12.79 7.77 4.48 – – –
18 2.53 4.92 14.11 23.04 35.87 16.81 10.21 5.89 – – –
20 2.80 5.46 15.67 25.60 42.01 19.69 11.96 – – – –
22 3.09 6.01 17.25 28.16 48.47 22.72 13.79 – – – –
25 3.51 6.83 19.60 32.00 56.26 27.51 15.40 – – – –
28 3.93 7.65 21.95 35.84 63.01 32.62 – – – – –
30 4.13 8.18 23.51 38.40 67.51 36.17 – – – – –
35 4.91 9.56 27.44 44.80 78.76 45.58 – – – – –
–
100 11 2.95 5.75 16.52 26.97 20.49 9.60 6.18 3.36 – – –
12 3.22 6.27 18.02 29.42 23.34 10.94 7.05 3.83 – – –
15 4.04 7.84 22.52 36.78 32.62 15.29 9.85 – – – –
18 4.84 9.41 27.03 44.13 42.88 20.10 12.94 – – – –
20 5.38 10.46 30.03 49.03 50.22 23.54 15.16 – – – –
22 5.92 11.50 33.03 53.94 57.94 27.15 17.49 – – – –
25 6.72 13.07 37.53 61.29 70.19 32.89 – – – – –
28 7.53 14.64 42.04 68.65 83.18 38.99 – – – – –
30 8.06 15.69 45.04 73.55 92.28 42.20 – – – – –
35 9.41 18.30 52.55 85.79 116.30 – – – – – –
–
160 11 11.20 21.81 62.60 72.05 29.83 13.98 9.00 – – – –
12 12.23 23.78 68.29 82.06 34.00 15.93 10.26 – – – –
15 15.29 29.74 85.34 114.73 47.51 22.27 – – – – –
18 18.34 35.68 102.43 150.84 62.46 – – – – – –
20 20.38 39.64 113.84 176.65 73.15 – – – – – –
22 22.42 43.60 125.18 203.81 84.37 – – – – – –
25 25.48 49.56 142.26 232.30 102.20 – – – – – –
28 28.53 55.50 159.35 260.20 121.15 – – – – – –
30 30.57 59.46 170.76 278.78 134.35 – – – – – –
35 35.67 69.38 199.18 325.26 84.00 – – – – – –

200 11 20.54 39.97 114.76 86.13 35.67 16.71 – – – –
12 22.42 43.60 125.19 98.14 40.64 19.04 – – – –
15 28.02 54.51 156.50 137.16 56.79 – – – – –
20 37.36 72.68 208.66 211.17 87.44 – – – – –
25 46.70 90.85 260.97 295.12 122.20 – – – – –

a The use offewer than 17 teeth is possible, but should be avoided when practical.

Rolling Chains 563

Number of Strands Chain number
160
4321 140
200 100 120

200 100
80
200 100 50 60
100 50
40
100 50
50 20 35

Chain power rating, kW 50 20 10 25
20

20 10 5.0
10

10 5.0
5.0 2.0

5.0
2.0 1.0

2.0

2.0 1.0 0.5
1.0

1.0 0.5
0.5 0.2

0.5 0.1
0.2

10 20 50 100 200 500 1000 2000 5000 10,000

Speed of smaller sprocket, rpm

Figure 19.14: Design guideline for standard roller chains.

(a)

(a) (b) (b)

Figure 19.15: A silent chain drive. (a) Silent chain with (c)
sprockets; (b) detail of silent chain links. Source: Courtesy
of Ramsey Products Corp. Figure 19.16: The use of guide links in silent chains. (a) One
guide link in center of chain; (b) two center guide links; (c)
structions, especially for conveyor applications. All of these two side guide links. Source: Courtesy of Ramsey Products
components are held together by riveted pins located in each Corp.
chain joint. Although all silent chains have these basic fea-
tures, there are still many different styles, designs, and con- any of these guide types will be satisfactory but it is essential
figurations. The width of a silent chain can be varied accord- that sprockets be selected with the same guide type as the
ing to horsepower constraints. Silent chains are available for chain.
power transmission, as considered in this section, but they
can also be modified to incorporate links that provide a sta- Most silent chains have teeth that engage sprockets on
ble and wear resistant surface for conveyor applications. only one side of the chain. However, duplex silent chain
designs are available that have teeth on both sides and are
Guide links are used to maintain tracking on sprockets, designed for use in serpentine drives, where sprockets are
as shown in Fig. 19.16. The guide links may be located within driven from both sides of the chain.
the chain (center guide) or along the outer edges of the chain
(side guide). A chain with two rows of guide links within the Silent chains are able to transfer higher powers at higher
chain is referred to as two center guide. In many applications, speeds than roller chains. Essentially, the width of the chain

564 Chapter 19 Flexible Machine Elements

can be increased to accommodate any power. Because they Example 19.6: Chain Drive Design
do not have a chordal rise, there is less vibration and longer
sprocket life as well. However, the main advantage of silent Given: A chain will transmit power from a 7.5-kW motor to
chains is that they are much quieter than conventional rolling a drive roller on a belt conveyor. The motor is connected to
chains. a speed reducer so that the input speed is 100 rpm, and the
desired conveyor drive roller speed is 25 rpm. The conveyor
Silent chains can be made in large widths and used as has some starts and stops, so it should be considered to have
conveyor belts. In such cases, it is common to use a spacer moderate shock. The center distance of the shafts should be
between links to produce a more open belt. around 1.25 m.

Design Procedure 19.2: Design of Find: Select a roller chain and sprockets for this application.
Chain Drives
Solution: This solution will closely follow Design Procedure
This Design Procedure presents a method for designing 19.2.
power transmission devices consisting of chains and sprock-
ets. While the approach is intended for roller chains, the De- 1. Referring to Table 19.11, note that for an electric mo-
sign Procedure can be used for silent chains, with caveats as tor with a driven load with moderate shock, the service
noted. It will be assumed that operating conditions and de- factor is a1 = 1.3. The required power rating, assuming
sign constraints are adequately described. For the purposes a2 = 1 is obtained from Eq. (19.31) as
of this Design Procedure, the power transmitted (or chain
force and speed), power source, speed ratio, and loading en- hpr = hp a1 = (7.5)(1.3) = 9.75 kW.
vironment need to be known, or at least be somewhat con- a2
strained.
2. Noting that the smaller sprocket will have a speed of
1. Obtain the service factor from Table 19.10. Calculate the 100 rpm, Fig. 19.14 suggests that a single strand of stan-
chain’s required power rating from Eq. (19.31), taking dard No. 100 chain can be used for this application.
a2 = 1.0. Note that a No. 120 chain may also lead to a reason-
able design, and may be worth investigating, but this
2. Select a chain size from Fig. 19.14 using the required solution will continue with a No. 100 chain.
power rating and the small sprocket speed. Note that
using the fewest number of chain strands while satis- 3. From Table 19.11, the strand factor is a2 = 1.0.
fying power requirements usually results in the most
economical design. 4. A single strand chain is being analyzed, so that the
power rating of 9.75 kW can still be used. Note that the
3. Obtain the strand factor, a2, from Table 19.11. power rating would need to be recalculated if a multi-
ple strand chain was selected.
4. The required power rating, given by Eq. (19.31), needs
to be recalculated if a multiple strand chain is to be 5. Referring to Table 19.12, for a No. 100 chain with a small
used. sprocket speed of 100 rpm, 20 teeth are required in order
to exceed the modified power rating of 9.75 kW.
5. Referring to Table 19.12, identify the column of the table
that corresponds to the small sprocket’s speed. Read- 6. Table 19.12 also indicates that a bath-type of lubrication
ing down from the top, find the number of teeth in the is required, and will need to be incorporated into the
smaller sprocket that produces the required modified chain system.
power rating, hpr. This is the minimum number of teeth
that are required for the application. Larger sprockets 7. From Eq. (19.22), the number of teeth on the larger
can be used if desired. sprocket can be obtained as:

6. If a multiple strand chain is being considered, record gr = ω1 = N2 ,
the modified power rating from Table 19.12, and mul- ω2 N1
tiply by the strand factor to obtain the chain’s power
rating. 100 = N2 ,
25 20
7. Note the required lubrication method in Table 19.12.
Variation from the lubrication approach may compro- or N2 = 80.
mise chain longevity.
8. Note from Table 19.9 that the pitch of No. 100 chain is
8. The number of teeth on the larger sprocket can be
calculated from the desired velocity ratio by using 31.75 mm, so that cd = 1.25 = 39.3. This is within
Eq. (19.22). pt 0.03175

9. If the center distance has not been prescribed, it can be the standard range, so no further modification (chang-
estimated by recognizing that cd/pt should be between
30 and 50, although larger lengths can be allowed if ing chain size, using more strands, etc.) is required.
chain guides are incorporated into the design. If the
center distance exceeds space limitations, increase the
number of strands or select the next largest pitch chain
and return to Step 4.

10. The number of links in the chain can be calculated from
Eq. (19.26), rounded up to the next highest even integer.

9. From Eq. (19.26),

Rolling Chains 565

Hoist rope Gantry line system (Single-part line)

Drag rope φ Mast (6 m)
θ Multi-part line

1.5 m
5m

Figure 19.17: Typical dragline.

L = 2cd + N1 + N2 + (N2 − N1)2 fore, the range of boom angles that must be achieved is some-
pt pt 2 4π2 cd what limited.

pt Just as with cranes, however, the gantry line system serves
to eliminate bending forces on the boom. Thus, the gantry
2(1250) 20 + 80 (80 − 20)2 line must attach at the tip of the boom, known as the boom
= ++ = 131. point. Attaching the gantry line to the crane’s superstruc-
31.75 2 4π2 1250 ture near the boom pivot point will result in extremely large
forces, as can be seen from moment equilibrium. A mast is
31.75 used to offset the gantry line system and obtain reasonably
low forces in the gantry line.
or L = 4.16 m. Since it is good practice to specify an even in-
teger of links, a value of L/pt = 132 is selected. This would A number of clever design features have been incorpo-
result in an actual center distance of cd = 1.32 m as obtained rated into draglines in the past (Fig. 19.17). The lines closest
from Eq. (19.27). Note that the center distance should be re- to the crane superstructure have multiple pulleys (and hence
duced by 1% to provide some slack, so that an actual center are called a multiple-part line) to share the load and reduce the
distance of around 1.3 m should be used. stress on the cable. Above the multiple-part line is a single
line attached by proper couplings to the boom. Given the op-
Case Study 19.1: Design of a Gantry erating characteristics of a dragline, this single line is never
for a Dragline wound over a pulley and its size is determined by the rated
cable strength and does not have to be reduced for fatigue
A dragline, often used for mining or dredging operations, effects (see Fig. 19.10).
uses a large bucket to remove material and transport it else-
where, usually into a trailer or a train boxcar. The gantry line Because the mast rotates as the boom angle changes, the
system is the portion of the dragline that supports the boom load supported by the gantry lines will change with the
and fixes the boom angle, as shown in Fig. 19.17. The boom boom angle. The gantry line tension is highest at a boom an-
typically weighs over 40 kN, is around 30 m long, and can gle of 0◦. Draglines are not operated at such low boom angles
be considered to have its center of gravity at its geometric because the lifting capacity is severely limited, but analyzing
center. The dragline tips if the moment from the hoist rope a boom angle of 0◦ is important. Draglines are shipped in
and the boom exceeds the moment from the dragline about multiple parts and assembled at the construction site, so that
the tipping point. Tipovers are to be avoided, of course, but the boom needs to be lifted, at least for the first time that the
it should be recognized that the load applied to the gantry dragline is placed into service.
cable cannot be larger than that resulting from a tipover con-
dition. A number of equalizer pulleys (a multiple-part line) are
used in Fig. 19.17 so that the load in the gantry line is re-
Figure 19.17 shows the dimensions of a typical dragline. duced. The load in the gantry line depends on the number of
The gantry line lengths shown were selected so that the 6- pulleys or parts in the line, each of which supports an equal
m mast is vertical at a boom angle of 30◦. Since draglines share of the load. The earlier discussion (Section 19.5.2) re-
have booms, many people confuse them with cranes. There garding sheave sizing is relevant. If the pulley and drum are
are significant differences, however. Dragline load is limited too large, the motor capacity will be excessive. Per industry
by the size of the bucket, whereas a crane can attempt to lift standard requirements, pulley and drum diameters of 12d
extremely large loads. Further, a dragline will operate at a are to be chosen. These diameters reduce gantry line strength
set boom angle for extended periods of time and will rarely, by approximately 12% (Fig. 19.9), but a maintenance proce-
if ever, operate at very high or very low boom angles. There- dure of checking for broken wires in the wire rope must still
be followed. This leads to a very safe system, since draglines
rarely change their boom angle, so the gantry line does not
run over pulleys often.

566 Chapter 19 Flexible Machine Elements

19.7 Summary Force: F1 = eµφπ/180◦
F2
Belts and ropes are machine elements (like brakes and F1 − Fc
clutches) that use friction as a useful agent, in contrast to Including centrifugal force: F2 − Fc = eµφπ/180◦
other machine elements in which friction is to be kept as low
as possible. Belts, ropes, and chains provide a convenient Torque: T = (F1 − F2)D1
means for transferring power from one shaft to another. This 2
chapter discussed flat, synchronous, and V-belts. All flat belts Na1 r2
are subject to slip, that is, relative motion between the pulley Velocity ratio: gr = Na2 = r1
surface and the adjacent belt. For this reason, flat belts must
be kept under tension to function and require tensioning de- Design power rating: hpr = f1hp
vices. One major advantage of V-belts over flat belts is that
the wedging action of V-belts increases the normal force from Rated power per belt: hpb = f2h1
dN to dN/ sin β, where β is the sheave angle. V-belts also can
transfer much higher power than a similarly sized flat belt. Number of belts required: N = hpr
hpb
Wire ropes are used when power must be transmitted
over very long center distances. Wire rope is widely used Wire Rope:
in hoisting applications including cranes and elevators. The
major advantage of using rolling chains over flat or V-belts is Total force: Ft = Fw + Fr + Fa + Fh
that rolling chains do not slip. Another advantage of rolling Wire stress: σ = Edw
chains over belt drives is that no initial tension is necessary D
and thus the shaft loads are smaller. The required length, 2Ft
power rating, and modes of failure were considered for belts, Bearing pressure: p = dD
ropes, and rolling chains.
Fatigue: Ff = SudD
Key Words 2000
Rolling Chains:
core center of wire rope, mainly intended to support outer
strands Velocity ratio: gr = d2 = ω1 = N2
d1 ω2 N1
flat belts power transmission device that consists of loop of
rectangular cross section placed under tension between Chordal rise: ∆r = r 1 − cos 180
pulleys N

inverted tooth chain see silent chain Number of links: L = 2cd + N1 + N2 + (N2 − N1)2
pt pt 2 4π2 cd
lay type of twist in wire rope (regular or Lang); distance for pt
strand to revolve around wire rope
Center distance: cd = A + A2 − B2
rolling chains power transmission device using rollers and pt 2
links to form continuous loop, used with sprockets
A = 1 L − N1 + N2
sheaves grooved pulleys that V-belts run in 4 pt 2

silent chain a chain that consists of specially formed links B = N2 − N1
that mesh with a gear-shaped sprocket 2π
Power rating: hpr = hp a1
strands grouping of wire used to construct wire rope a2

synchronous belt flat belt with series of evenly spaced teeth Recommended Readings
on inside circumference, intended to eliminate slip and
creep American Chain Association, Chains for Power Transmission
and Material Handling, Marcel Dekker.
timing belt same as synchronous belt
Budynas, R.G., and Nisbett, J.K. (2011), Shigley’s Mechanical
V-belt power transmission device with trapezoidal cross Engineering Design, 9th ed., McGraw-Hill.
section placed under tension between grooved sheaves
Dickie, D.E. (1985) Rigging Manual, Construction Safety As-
wire rope wound collection of strands sociation of Ontario.

Summary of Equations Juvinall, R.C., and Marshek, K.M. (2012) Fundamentals of Ma-
chine Component Design, 5th ed., Wiley.
Belts:
Belt length: Krutz, G.W., Schuelle, J.K., and Claar, P.W. (1994) Machine De-
sign for Mobile and Industrial Applications, Society of Auto-
L = (2cd)2 − (D2 − D1)2 motive Engineers.
π
Mott, R. L. (2014) Machine Elements in Mechanical Design, 5th
+ 2 (D1 + D2) ed., Pearson.
+ π(D2 − D1) sin−1 D2 − D1
Rossnagel, W.E., Higgins, L.R., and MacDonald, J.A. (1988)
180 2cd Handbook of Rigging, 4th ed., McGraw-Hill.

Shapiro, H.I. et al. (1991) Cranes and Derricks, 2nd ed.,
McGraw-Hill.

Wire Rope Users Manual (1972) Armco Steel Corp.

References

ASME (2007) B30.5 Mobile and Locomotive Cranes, American
Society of Mechanical Engineers.

Shigley, J.E., and Mitchell, L.D. (1983), Mechanical Engineering
Design, 4th ed., McGraw-Hill.

Quantitative Problems 567
19.28 Explain why the curves in Fig. 19.14 have their shape.
Questions
19.29 A vendor approaches you and asks to replace your steel
19.1 What is the difference between a flat belt and a V-belt? roller chain with a wear-resistant polymer chain. What
questions would you ask before considering the change?
19.2 What is a synchronous belt?
19.30 Explain why the reinforcing cords are located near the
19.3 What do the terms 3V, A, E, 5V, and 8V have in common? exterior of a V-belt.

19.4 Define slip as it relates to machine elements of this chap- Quantitative Problems
ter.
19.31 Repeat Example 19.3 using a 5V cross-section. Ans. 1
19.5 What is the difference between a strand and a wire? belt needed, with 191 mm and 287 mm sheaves.

19.6 Describe the difference between Lang and regular lay. 19.32 An open flat belt 200 mm wide and 4 mm thick connects
a 400-mm-diameter pulley with a 900-mm-diameter pul-
19.7 Why is a headache ball (static weight) used with wire ley over a center distance of 4.5 m. The belt speed is 10
rope? m/s. The allowable preload per unit width of the belt is
17.5 kN/m, and the weight per volume is 11.4 kN/m3.
19.8 What is larger, a rope or a strand? A strand or a wire? The coefficient of friction between the belt material and
the pulley is 0.8. Find the length of the belt, the max-
19.9 What is the difference between a sheave and a sprocket? imum forces acting on the belt before failure is experi-
enced, and the maximum power that can be transmitted.
19.10 Why is a preload needed for belts? Ans. L = 11.08 m, hp,max = 60.5 kW.

19.11 Should there be a preload on roller chains? Explain your 19.33 The driving and the driven pulley in an open-flat-belt
answer. transmission each have a diameter of 160 mm. Calculate
the preload needed in the belt if a power of 7 kW should
19.12 What are the components of a roller chain? be transmitted at 1000 rpm by using only half the wrap
angle on each pulley (that is, φ = π/2). The coefficient
19.13 What is a silent chain? of friction is 0.20, the density of the belt material is 1500
kg/m3, and the allowable belt preload stress is 5 MPa.
19.14 What are the consequences of having a vertical chain Also, calculate the cross-sectional area of the belt. Ans.
drive? Fi = 2740 N.

19.15 Should a V-belt be lubricated? Why or why not? 19.34 A belt transmission is driven by a motor hinged at point
0 in Sketch a. The motor mass is 50 kg and its speed is
Qualitative Problems 1500 rpm. Calculate the maximum transferable power
for the transmission. Check, also, that the largest belt
19.16 Define the term datum and explain its importance. stress is lower than σall when the bending stresses are
included. Belt dimensions are 100 by 4 mm, the coeffi-
19.17 What are the main advantages of V-belts over spur cient of friction is 0.32, the modulus of elasticity is 100
gears? MPa, the belt density is 1200 kg/m3, and the allowable
stress is 20 MPa. Ans. hp = 10.88 kW.
19.18 List the advantages and disadvantages of V-belts com-
pared to chain drives. 150 mm 150 mm
Motor
19.19 List the reasons that a wire rope requires lubrication.
What kind of lubricant would you recommend? T
125 mm
19.20 What are the similarities and differences between V-
belts, wire ropes, and roller chains? 400 mm 0

19.21 Using proper sketches, determine if there is a chordal Sketch a, for Problem 19.34
rise with synchronous belts.
19.35 The tension in a flat belt is given by the motor weight, as
19.22 Explain why there is a minimum recommended sheave shown in Sketch b. The mass is 80 kg and is assumed to
diameter for V-belts. be concentrated at the motor shaft position. The motor
speed is 1405 rpm and the pulley diameter is 400 mm.
19.23 Explain the similarities and differences between rein- Calculate the belt width when the allowable belt stress is
forcing strands in V-belts and wire rope. 6.00 MPa, the coefficient of friction is 0.5, the belt thick-
ness is 5 mm, the modulus of elasticity is 150 MPa, and
19.24 Review Figures 3.19 through 3.22 and identify materials the density is 1200 kg/m3. Ans. wt = 33.84 mm.
that would be useful for wire ropes other than steel.

19.25 Plot power rating for V-belts as a function of small
sheave diameter, and describe your observations.

19.26 List the characteristics and advantages of (a) regular lay,
and (b) Lang lay.

19.27 Why is steel popular for wire rope?

568 Chapter 19 Flexible Machine Elements

500 mm R = 100 mm

R L2 = R1.5

800 mm ω
250 mm
L1 = R0.8
0

700 mm Sketch d, for Problem 19.38

Sketch b, for Problem 19.35 19.39 A flat belt drive, shown in Sketch e, has the top left pul-
ley driven by a motor and driving the other two pul-
19.36 Calculate the maximum possible power transmitted leys. Calculate the wrap angles for the two driven pul-
when a wrap angle is used as shown in Sketch c. The leys when the full wrap angle is used on the driving pul-
belt speed is 5 m/s and the coefficient of friction is 0.25. ley and the power taken from pulley 2 is twice the power
Ans. hp = 585 W. from pulley 3. The coefficient of friction is 0.3 and l = 4r.
Ans. α3 = 37.2◦.

l Power ph2
r
r ω
Drive l/2
r/2

210° Mass, 5 kg l/2 l/2
180° Power ph3

Driver Sketch e, for Problem 19.39

Sketch c, for Problem 19.26 19.40 A flat belt 6 mm thick and 60 mm wide is used in a belt
transmission with a speed ratio of 1. The radius of the
19.37 Equation (19.9) gives the belt forces, where F1 is the pulleys is 100 mm, their angular speed is 50 rad/s, the
largest belt force. What is the maximum power the belt center distance is 1000 mm, the belt modulus of elastic-
can transmit for a given value of F1? Hint: Express the ity is 200 MPa, and the density is 1000 kg/m3. The belt
power as a function of speed, and then take the deriva- is pretensioned by increasing the center distance. Cal-
tive with respect to speed and set the derivative equal to culate this increase in center distance if the transmission
zero. should use only half the wrap angle to transmit 1 kW.
Centrifugal effects should be considered. The coefficient
19.38 In a flat belt drive with parallel belts the motor and pul- of friction is 0.3. Ans. ∆a = 8 mm.
ley are pivoted at point 0 as shown in Sketch d. Calcu-
late the maximum possible power transmitted at 1200 19.41 A 120-mm-wide and 5-mm-thick flat belt transfers
rpm. The motor and pulley together have a 50-kg mass power from a 250-mm-diameter driving pulley to a 700-
at the center of the shaft. The centrifugal forces on the mm-diameter driven pulley. The belt has a mass per
belt have to be included in the analysis. The coefficient length of 2.1 kg/m and a coefficient of friction of 0.25.
of friction is 0.2, belt area is 300 mm2, and density is 1500 The input power of the belt drive is 60 kW at 1000 rpm.
kg/m3. Ans. hp = 5.69 kW. The center distance is 3.5 m. Determine

(a) The maximum tensile stress in the belt. Ans. σ =
15.04 MPa.

(b) The loads for each pulley on the axis. Ans. Fx =
12.7 kN.

19.42 A flat belt drive is used to transfer 100 kW of power. The
diameters of the driving and driven pulleys are 300 and

Quantitative Problems 569

850 mm, respectively, and the center distance is 2 m. The F2 4r
belt has a width of 200 mm, thickness of 10 mm, speed α1
of 20 m/s, and coefficient of friction of 0.40. For a safety r β
factor of 2.0 for static loading determine the following: ω 6r

(a) The contact angles and belt length for an open α2 F1
configuration. Ans. φ1 = 164.2◦, φ2 = 195.8◦,
L = 5.84 m. Sketch f , for Problem 19.47

(b) The loads for each pulley on the axis. Ans. Fx = 19.48 A small lathe (consider the power source to be ‘normal’,
9.56 kN, Fy = 687 N. with a service of 8 hrs/day) uses a belt drive to transmit
power from a 5 kW electric motor to a spindle. The max-
19.43 A synchronous belt transmission is used as a timing belt imum motor speed is 1000 rpm, and the maximum spin-
for an overhead camshaft on a car motor. The belt is elas- dle speed is 1000 rpm. Select a belt and sheaves for this
tically prestressed at standstill to make sure it does not application, as well as the center distance. Ans. 3V cross-
slip at high speeds due to centrifugal forces. The belt section with four belts, 105 mm small sheave, cd = 196
is 1100 mm long and weighs 200 g. The elastic spring mm.
constant for 1 m of belt material is 105 N/m. The belt
prestress elongates it 2 mm. It needs to elongate 4 mm 19.49 The input power to shaft A, shown in Sketch g, is trans-
more to start slipping. Calculate the maximum allow- ferred to shaft B through a pair of mating spur gears
able motor speed if the pulley on the motor shaft has a and then to shaft C through a 3V -section V-belt drive.
diameter of 100 mm. Ans. ω = 1095 rad/s. The sheaves on shafts B and C have 76- and 200-mm di-
ameters. For the maximum power the belt can transfer,
19.44 A timing belt for power transfer should be used at a determine the following:
velocity ratio of one-third, so that the outgoing speed
should be three times as high as the incoming speed. (a) The input and output torques of the system. Ans.
The material for the belt reinforcement can be chosen TB = 98.9 N-m.
from glass fiber, carbon fiber, and steel wire. These ma-
terials give different belt densities as well as different (b) The belt length for an approximate center distance
tensile strengths. The glass-fiber-reinforced belt has a of 550 mm. Ans. L = 1.32 m.
density of 1400 kg/m3 and an allowable stress of 300
MPa. The carbon-fiber-reinforced belt has a density of 50T
1300 kg/m3 and an allowable stress of 600 MPa. The
steel-wire-reinforced belt has a density of 2100 kg/m3 B
and an allowable stress of 400 MPa. Calculate the maxi-
mum power for each belt type if the belt speed is 30 m/s, V-belt 1800 rpm A
the belt cross-section size is 50 mm2, and the safety fac- 20T
tor is 12. Ans. For glass fibers, hp = 37.34 kW.
C
19.45 A 10 kW, 1750-rpm electric motor drives a machine
through a multiple V-belt drive. The driver sheave is Sketch g, for Problem 19.49
94 mm in diameter and the wrap angle of the driver is
165◦. The weight per length is 2.0 N/mm. The maxi- 19.50 An automobile fan is driven by an engine through a V-
mum belt preload is 600 N, and the coefficient of fric- belt drive. The engine’s sheave has a 200 mm diameter
tion of the belt material acting on the sheave is 0.2. How and is running at 880 rpm. The fan’s sheave has an 80
many belts should be used? Assume a sheave angle of mm diameter. The power required to move air is 1 kW.
18◦. Ans. Two are needed. Select the size of the V-belt to obtain a compact design.
Ans. Single 3V belt, L = 875 mm.
19.46 A V-belt drive has r1 = 200 mm, r2 = 100 mm, 2β =
36◦, and cd = 700 mm. The speed of the smaller sheave
is 1200 rpm. The cross-sectional area of the belt is 160
mm2 and its density ρ = 1500 kg/m3. How large is
the maximum possible power transmitted by six belts
if each belt is prestressed to 200 N? The coefficient of
friction µ = 0.30. If 15 kW is transmitted, how large a
part of the periphery of the smaller wheel is then active?
Ans. φ = 85.0◦.

19.47 A combined V-belt and flat belt drive, shown in Sketch
f , has a speed ratio of 4. Three V-belts drive the sheave
in three grooves but connect to the larger sheave like
flat belts on the cylindrical surface. The center distance
is six times the smaller sheave radius, which is 80 mm
with a speed of 1500 rpm. Determine which sheave can
transmit the largest power without slip. Determine the
ratio between the powers transmitted through the two
sheaves. The angle 2β is 36◦ and the coefficient of fric-
tion is 0.30. Ans. hp,small = 2.64hp,large.

570 Chapter 19 Flexible Machine Elements

19.51 A hoist uses a 50 mm, 6 × 19 monitor steel wire rope 19.56 End and front views of a hoisting machine are shown in
with Lang lay and dw = d/14 on cast steel sheaves with Sketch i. Calculate the lifting height per drum revolu-
D/d = 30. The rope is used to haul loads up to 35 kN a tion. Calculate the maximum force in the wire when the
distance of 150 m. drum is instantly started with angular speed ω to lift a
mass m. The free length of the wire is L and its cross-
(a) Using a maximum acceleration of 0.5 m/s2, deter- sectional area is A.
mine the tensile and bending stresses in the rope
and their corresponding safety factors. Ans. σt = Small drum Large drum
53.8 MPa, σb = 198 MPa.
Anchor Anchor
(b) Determine the bearing pressure in the rope and
the safety factor when it interacts with the sheave. r
Also, determine the stretch of the rope. Ans. p = R
1.36 MPa, δ = 97.2 mm.
Drum shaft
(c) Determine the safety factor due to fatigue and an-
ticipate the number of bends until failure. PP
Cable
19.52 A steel rope for a crane has a cross-sectional area of 31
mm2 and an ultimate strength of 1500 N/mm2. The Sheave
rope is 12 m long and is dimensioned to carry a maxi-
mum load of 1000 kg. If that load is allowed to free-fall End view Front view
1 m before the rope is tightened, how large will the stress
be in the rope? The modulus of elasticity is 68 GPa. Ans. Sketch i, for Problem 19.56
σ = 2236 MPa.
19.57 A hauling unit with 25-mm, 8 × 19 wire rope made of
19.53 Two 2-cm, 6 × 19 plow-steel wire ropes are used to haul fiber-core plow steel is used to raise a 45-kN load to a
mining material to a depth of 150 m at a speed of 8 height of 150 m at a speed of 5 m/s and an acceleration
m/s and an acceleration of 2 m/s2. The pulley diame- of 2 m/s2. Design for a minimum pulley diameter while
ter is 80 cm. Using a proper safety factor, determine the determining the safety factors for tensile strength, bend-
maximum hauling load for these ropes. Are two ropes ing stress, bearing pressure, and fatigue endurance. Ans.
enough for infinite life? Ans. Fw = 3580 lb. For D/d = 21, nst = 2.10, nsb = 2.99.

19.54 Using six 16-mm, 6 × 19 wire ropes, an elevator is to lift 19.58 A building elevator operates at a speed of 5 m/s and
a 2500-kg weight to a height of 81 m at a speed of 4 m/s 1.2-m/s2 acceleration and is designed for 10-kN dead-
and an acceleration of 1 m/s2. The 726-mm-diameter weight, five 70-kg passengers, and 150-kg overload. The
pulley has a strength of 6.41 MPa. Determine the safety building has twelve 4-m stories. Using the proper safety
factors for tension, bending, bearing pressure, and fa- factor, design the wire rope required for the elevator. A
tigue failure. Also, calculate the maximum elongation maximum of four ropes may be used.
of the ropes. Ans. ns,tension = 11.1, ns,fatigue = 1.80.
19.59 A three-strand, ANSI 50 roller chain is used to transfer
19.55 Sketch h shows a wire rope drive used on an elevator power from a 20-tooth driving sprocket that rotates at
to transport dishes between floors in a restaurant. The 500 rpm to a 60-tooth driven sprocket. The input power
maximum mass being transported is 300 kg and hangs is from an internal combustion engine, and the chain ex-
by a steel wire. The line drum is braked during the periences moderate shock. Determine the following:
downward motion by a constant moment of 800 N-m.
The rotating parts have a mass moment of inertia of 13 (a) The power rating. Ans. hp = 5.85 kW per chain.
kg-m2. Calculate the maximum force in the wire and
find when it first appears. Ans. Fmax = 3960 N. (b) The length of the chain for an approximate center
distance of 550 mm. Ans. L = 1762 mm.
Line drum

r = 0.2 m 19.60 To transfer 100 kW of power at 400 rpm, two strands
of roller chain are needed. The load characteristics are
L = 12 m heavy shock with an internal combustion engine and
A = 15 m2m mechanical drive. The driving sprocket has 12 teeth and
the driven sprocket has 42 teeth. Determine the kind,
E = 70 GPa length, and size of the chain for a center distance close
to 50 pitches. Ans. For ANSI 160 chain, L = 6.451 m.

Motor and brake m = 300 kg 19.61 A two-strand AISI 50 roller chain system is used to
transmit the power of an electric motor rotating at 500
rpm. The driver sprocket has 12 teeth and the driven
sprocket has 60 teeth. Calculate the power rating per
strand, the power that can be transmitted, and the chain
length if cd = 50pt. Ans. hp = 6.71 hp per strand,
hpr = 4.78 kW, L = 2.175 m.

Sketch h, for Problem 19.55 19.62 A roller chain is used to transfer 8 kW of power from
a 20-tooth driving sprocket running at 500 rpm to a 40

Design and Projects 571

tooth sprocket. Design the chain drive for heavy shocks, 19.66 One form of a V-belt drive uses two grooved sheaves
an electric motor and single-strand chain. Also, deter- with a flat pulley in between, so that the flat side of the
mine the required chain length for a center distance of V-belt runs against the pulley. What concerns would
approximately 70 pitches. Ans. For ANSI 100 chain, you have about such an arrangement? What design con-
hpr = 12 kW per chain, L = 5.4 m. siderations would you have regarding the belt?

Design and Projects 19.67 A mobile crane is being used to help construct a dam. In
this particular application, wooden molds and steel re-
19.63 It has been proposed to use thin metal foils moving over inforcement (rebar) is located, then concrete is poured in
smooth spools as power transmission devices instead of place. After the concrete has set, the molds are attached
rubber belts on sheaves. List the advantages and dis- to a crane hook, unbolted and pried off of the concrete.
advantages of metal foils for this application and de- Often the molds will fall 5 to 10 ft because of slack in the
termine under what conditions you would expect metal load line, and then will be caught by the wire rope. List
foils would be preferable. your concerns regarding this situation.

19.64 Assume you are investigating a failure of wire rope on 19.68 Wire rope is often used for high-voltage electrical power
a mobile crane. What would you expect to find at the transmission. These are often called high-tension lines
fracture surfaces of the wires if because the tension is very high to avoid excessive sag
in the rope. Conduct an Internet search, and summa-
(a) the rope was overloaded; rize the theory of operation for cable pullers that apply
(b) the rope was in service for too long; a tension to wire rope.
(c) a kink was put in the rope when it was installed.
19.69 Assume you are asked to investigate a ski lift to make
Make appropriate sketches and describe what you certain that it is in proper operating condition for the
would do to determine which mode was the cause of start of winter. Construct an inspection protocol outlin-
the wire rope failure. ing the necessary steps you would take regarding the
wire rope.
19.65 Explain the function of the tie layer in a multiple belt
arrangement. Describe what steps you would take to 19.70 In the footnote to Table 19.6, it states that the stiffness
determine the load on the tie layer, and how this would of wire ropes is affected by load and increases with the
allow you to design the tie layer properly. life of the rope. Explain why this would be the case.
How would you expect the stiffness of V-belts and roller
chains to change, if at all, during their service lives? Ex-
plain your answer.

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Appendix A

Physical and Mechanical Properties of
Materials

A.1 Mechanical Properties of Selected Alloy Steels

AISI Yield Ultimate Elongation Reduction Brinell
Number Conditiona Strength, Tensile in 50 mm, in Area, Hardness,
Strength, %
MPa % 56 HB
MPa 59 156
41 197
Chromium-Molybdenum 43 467
4130 Annealed 361 560 28 49 430
N 870◦ C 436 670 25 57 380
Q&T 205◦ C 1460 1630 10 64 315
Q&T 315◦ C 1380 1500 11 57 245
Q&T 425◦ C 1190 1280 13 47 200
Q&T 540◦ C 910 1030 17 38 300
Q&T 650◦ C 703 814 22 43 510
4140 Annealed 417 655 26 49 445
N 870◦ C 655 1020 18 58 370
Q&T 205◦ C 1640 1770 8 63 285
Q&T 315◦ C 1430 1550 9 50 230
Q&T 425◦ C 1140 1250 13 40 160
Q&T 540◦ C 834 951 18 50 235
Q&T 650◦ C 655 758 22 40 220
44 486
Nickel-Chromium-Molybdenum 580 50 430
4320 Annealed 425 790 29 60 360
N 870◦ C 460 745 21 60 280
4340 Annealed 470 1720 22 59 150
Q&T 315◦ C 1590 1470 10 50 183
Q&T 425◦ C 1360 1170 10 30 190
Q&T 540◦ C 1080 965 13 20 270
Q&T 650◦ C 855 536 20 350
8620 A 357 632 31
N 870◦ C 385 26

Chromium-Vanadium 407 662 23
6150 Annealed 615 940 22
N 870◦ C 1160 1200 15
Q&T 540◦ C

Note: a N, normalized; Q&T, quenched and tempered

573

574 Appendix A Physical and Mechanical Properties of Materials

A.2 Mechanical Properties of Selected Carbon Steels

AISI Yield Ultimate Elongation Reduction Brinell
Number Conditiona Strength, Tensile in 50 mm, in Area, Hardness,
Strength, %
MPa % 55 HB
MPa 45 85
40 95
1006 Hot Rolled 170 300 30 50 105
Cold Drawn 280 330 20 40 95
1010 Cold Drawn 305 365 20 50 111
Hot Rolled 180 325 28 40 100
1015 Cold Drawn 325 385 18 50 126
Hot Rolled 190 340 28 60 116
1018 Cold Drawn 370 440 15 40 100
Hot Rolled 220 400 25 50 121
1020 Q&T 870◦ C 295 395 37 60 110
Cold Drawn 350 420 15 61 130
Hot Rolled 205 380 25 47 150
1030 Annealed 317 430 30 53 495
N 925◦ C 345 520 32 60 400
Q&T 205◦ C 648 848 17 65 300
Q&T 315◦ C 621 800 19 70 250
Q&T 425◦ C 579 731 23 35 210
Q&T 540◦ C 517 669 28 40 149
Q&T 650◦ C 441 586 32 57 130
Cold Drawn 440 525 12 55 150
Hot Rolled 260 70 20 48 170
1040 Annealed 350 520 30 54 262
N 900◦ C 374 590 28 65 240
Q&T 205◦ C 593 779 19 35 192
Q&T 425◦ C 552 758 21 40 170
Q&T 650◦ C 434 634 29 40 149
Cold Drawn 490 585 12 39 190
Hot Rolled 290 525 18 27 220
1050 Annealed 365 636 24 36 514
N 900◦ C 427 748 20 65 444
Q&T 205◦ C 807 1120 9 30 235
Q&T 425◦ C 793 1090 13 35 197
Q&T 650◦ C 538 717 28 38 179
Cold Drawn 580 690 10 37 179
Hot Rolled 340 620 15 40 230
1060 Annealed 372 626 22 45 310
N 900◦ C 421 776 18 40 280
Q&T 425◦ C 765 1080 14 30 230
Q&T 540◦ C 669 965 17 30 201
Q&T 650◦ C 524 800 23 25 255
Hot Rolled 370 680 12 21 229
1080 Q&T 800◦ C 380 615 25 13 190
Hot Rolled 420 770 10 30 293
1095 Annealed 380 658 13 32 375
N 900◦ C 500 1010 9 37 360
Q&T 315◦ C 813 1260 10 30 320
Q&T 425◦ C 772 1210 12 25 269
Q&T 540◦ C 676 1090 15 20 248
Q&T 650◦ C 552 896 21 197
Hot Rolled 455 825 10
Cold Drawn 525 680 10

Note: a N, normalized; Q&T, quenched and tempered

Material Properties 575

A.3 Mechanical Properties of Selected Cast Irons

ASTM class Conditiona Tensile Ultimate Compressive Hardness
yield tensile strength HB
strength strength MPa 156
MPa MPa 572 174
669 210
Gray cast irons As cast — 152 752 212
20 As cast — 179 855 235
25 As cast — 214 965 262
30 As cast — 252 1130 302
35 As cast — 293 1293 160
40 As cast — 362 359 174
50 As cast — 431 365 228
60 385 325
924
Ductile (nodular or spheroidal) cast irons 448
60-40-18 Annealed 324 462
65-45-12 Annealed 331 565
80-55-06 Annealed 365 965
120-90-02 Q&T 827

Note: a Q&T, quenched and tempered

A.4 Mechanical Properties of Selected Stainless Steels

Yield Ultimate Elongation
AISI Manufacturing strength, tensile in 50 mm,
no. historyc MPa strength, Hardness
MPa % 85 HRB
301 Annealed 205 40 95 HRB
1/16-hard 310 515 40 20 HRC
1/8-hard 380 620 40 25 HRC
1/4-hard 515 690 25 32 HRC
1/2-hard 758 860 18 37 HRC
3/4-hard 930 1034 12 41 HRC
Full-hard 965 1205 9 150 HB
302 Annealed 207 1276 40 310 HB
1/4-hard 517 517 10 320 HB
1/2-hard 758 860 9 335 HB
Full-hard 965 1034 3 201 HB
304 Annealed 215 1275 40 201 HB
304L Annealed 170 505 40 201 HB
304H Annealed 205 485 40 150 HB
316 Annealed 240 515 55 145 HB
316L Annealed 207 585 55 95 HB
410 Annealed 275 538 20 241 HB
420 Annealed 345 485 25 444 HB
Q&T 204°C 1360 655 12 461 HB
Q&T 427°C 1420 1600 10 262 HB
Q&T 650°C 680 1620 20 285 HB
431 Annealed 655 895 20 388 HB
Q&T 204°C 1055 862 20 388 HB
Q&T 427°C 1080 1345 19 277 HB
Q&T 650°C 695 1350 20 41 HRC
631 Cold rolled 1275 960 6
1030

a Q&T, quenched and tempered

576 Appendix A Physical and Mechanical Properties of Materials

A.5 Mechanical Properties of Selected Aluminum Alloys

Alloy Temper Yield Ultimate Elongation
Wrought O Strength Tensile in 50 mm,
Strength
1100 H14 MPa %
1350 O 35 MPa 40
2011 120 90 25
2014 H19 28 125 —
2017 T3 165 83 —
2024 T8 296 186 15
O 310 379 12
2219 T4 97 407 18
3003 T6 290 186 20
T4 414 427 13
3004 O 276 483 22
T3 75 427 20
3105 T4 345 190 18
O 325 483 20
5005 T87 76 470 20
5052 O 393 179 10
H12 40 476 35
5056 H14 117 110 20
5083 H16 145 131 12
5086 O 165 150 14
H34 69 179 20
5454 H38 186 179 6
O 234 234 6
6061 H14 55 276 24
6063 H18 152 117 5
7050 H25 193 172 3
7075 H34 159 214 —
Cast O 138 179 8
319.0 H32 90 159 25
333.0 H34 186 190 62
335.0 H36 215 234 12
O 234 260 10
H18 152 269 35
O 407 290 10
H321 145 434 22
H32 228 290 16
H34 207 317 12
H112 255 290 10
O 131 324 14
H32 117 269 22
H34 207 248 10
H112 241 276 10
O 124 303 18
T4 55 248 25
T6 145 125 22
T5 275 241 15
T6 145 310 12
T7651 214 186 12
O 490 241 —
T6 105 552 16
T6 500 230 11
T5 165 570 2.0
T6 172 248 1.0
T6 207 234 1.5
T7 172 289 3.0
248 241 0.5
262

Material Properties 577

A.6 Mechanical Properties of Selected Magnesium Alloys

Alloy Yield Ultimate Elongation Hardness
Strength Tensile in 50 mm, 69 HB
Temper MPa Strength 73 HB
% 50 HB
Sand and permanent-mold castings MPa 1.0 55 HB
AM100-A T61 150 275 5.0 62 HB
AZ63A T6 130 275 2.0 70 HB
EZ233A T5 110 160 8.0 —
HK31A T6 105 220 3.5 63 HB
ZC63A T6 125 210 10.0 —
ZK61A T6 195 310 4.0 60 HB
220 3.0 —
Die castings F 150 230 15 88 HB
AS41A F 150 260 16.0 73 HB
AZ91A 310 7 68 HB
380 11.0 —
Extrusions F 200 365 15.0 —
AZ31B F 230 290 9.0 —
AZ61A T5 275 255 16
AZ80A T5 305 263 19
ZK60A 311 11
365
Sheet and Plate H24 220
AZ31B 200
HK31A H24 163
ZE10 F 308
ZEK199 F 300
ZK60A T5

A.7 Mechanical Properties of Selected Wrought Copper and Copper Alloys

Alloy UNS Temper Yield Ultimate Elongation Hardness
Pure copper No. — Strength Tensile in 50 mm, —
C10200 Strength
0FHC C17200 Annealed MPa % 60 HRB
High-copper alloys Hardened 69-365 MPa 55-4 42 HRC
C21000 Annealed 221-455 35 52 HRF
Beryllium-copper C23000 — 64 HRB
Brass C26000 Hard 1050 490 2 64 HRF
C28000 Annealed 77 1400 45 73 HRB
Gilding, 5% Zn C35300 350 245 5 72 HRF
Red brass, 15% Zn Hard 91 392 47 82 HRB
Cartridge brass, 30% Zn C51000 Annealed 406 280 5 80 HRF
Muntzmetal, 40% Zn CS52400 133 434 55 75 HRB
High lead brass, 36% Zn, 2% Pb C60800 Hard 441 357 8 68 HRF
Bronze C63000 Annealed 119 532 45 80 HRB
Phosphor bronze, 5% Sn C65500 350 378 15 40 HRB
Phosphor bronze, 10% Sn Hard 119 490 52 90 HRB
Aluminum bronze C71500 Annealed 318 350 7 62 HRB
C75700 175 420 55 96 HRB
High-silicon bronze Hard 581 350 9 49 HRB
Other alloys Annealed 250 588 63 94 HRB
658 483 16 96 HRB
Cupronickel, 30% Ni Hard 175 707 66 98 HRB
Nickel Silver Annealed 441 420 8 66 HRB
414 700 15 95 HRB
Hard 517 690 15 40 HRB
Annealed 210 814 55 86 HRB
Cold rolled 406 441 8 55 HRB
Extruded 126 658 36 89 HRB
Halfh ard 553 385 3
Annealed 196 588 35
525 427 4
Hard 595
Annealed
Cold rolled
Annealed

Hard

578 Appendix A Physical and Mechanical Properties of Materials

A.8 Mechanical Properties of Selected Nickel-base Alloys at 25◦C

Alloy Yield Ultimate Elongation Hardness
Strength Tensile in 50 mm, 109 HB
Strength 129 HB
MPa % —
MPa 30 HRC
110 HB
Commercially pure and low-alloy nickels 300 HB
Nickel 200 148 462 47 —
Nickel 201 103 403 50 —
Nickel 211 240 530 40
Duranickel 301 862 1170 25 93 HRB
92 HRB
Nickel-copper alloys 240 550 40
Alloy 400 790 1100 20 —
Alloy K-500 230 525 35 35 HRC
Monel R-405 750 1050 30 75 HRB
Monel K-500 173 HB
88 HRB
Nickel-molybdenum and Nickel-silicon alloys 352 HB
Hastelloy B 138 HB
As cast 345 586 10 209 HB
Sheet 386 834 63 79 HRB
Hastelloy C-4 400 785 54 190 HB
Hastelloy D — 793 —
—
Nickel-chromium-iron alloys
Inconel 600 310 655 40
Inconel 617 350 755 58
Inconel 690 348 725 41
Inconel 751 976 1310 22
Inconel 800 295 600 44

Other alloys 372 785 62
Hastelloy C-276 320 690 50
Hastelloy G 517 930 42.5
Inconel 625 310 690 45
Inconel 825

A.9 Properties of Selected Nickel-based Superalloys at 870◦C (1600◦F)

Alloy Condition Ultimate Yield Elongation
Astroloy Wrought tensile strength, in 50 mm
Hastelloy X Wrought strength,
IN-100 MPa MPa %
IN-102 Cast 690 25
Inconel 625 Wrought 770 180 50
Inconel 718 Wrought 255 695 6
MAR-M 200 Wrought 885 200 110
MAR-M 432 215 275 125
Rene´ 41 Cast 285 330 88
Udimet 700 Cast 340 760 4
Waspaloy Wrought 840 605 8
Wrought 730 550 19
Wrought 620 635 27
690 515 35
525

A.10 Mechanical Properties of Selected Zinc Alloys

Alloy Ultimate Elongation Hardness
Tensile in 50 mm, 100 HB
Strength 82 HB
% 91 HB
MPa 76 HB
103 HB
Die-casting alloys 100 HB
Z35541 359 7 119 HB
Z33520 283 10 43 HB
Z33531 329 7 52 HB
Z33523 283 14 61 HB
Z35635 374 8
Z35630 404 5
Z35840 426 2

Wrought alloys (hot rolled) 52
Z21220 150 50
Z44330 170 38
Z41320 221

Material Properties 579

A.11 Mechanical Properties of Selected Titanium Alloys

Alloy Condition Temperature Yield Ultimate Elongation
Unalloyed grades Annealed ◦C Strength Tensile in 50 mm,
25 Strength
99.5% Ti 300 MPa %
25 240 MPa 30
ASTM Grade 1 Annealed 25 95 330 —
ASTM Grade 4 Annealed 25 170 150 —
Alpha and near-alpha alloys Annealed 300 480 240 —
Ti-5Al-2.5Sn 25 810 550 16
25 450 860 —
Ti-8Al-1Mo-1V Annealed 25 830 565 —
Ti-2.25Al-11Sn-5Zr-1Mo Annealed 25 900 900 —
Ti-6Al-2Sn-4Zr-2Mo Annealed 300 830 1000 —
Alpha-beta alloys Annealed 25 925 900 10
Ti-6Al-4V 300 650 1000 —
25 1100 725 10
Solution + age 25 900 1175 —
25 1100 980 —
Ti-6Al-2Sn-4Zr-6Mo Solution + age 25 830 1170 —
Beta alloys Annealed 25 965 900 —
300 1100 1000 —
Ti-3Al-8V-6Cr-4Mo-4Zr Solution + age 1210 1170 8
Ti-15V-3Cr-3Al-3Sn Annealed 830 1275 —
Ti-10V02Fe-3Al 1100
Ti-13V-11Cr-3Al Solution + age

A.12 Mechanical Properties of Selected Powder Metal Alloys

Yield Ultimate Elastic Elongation Density
strength Tensile modulus in 25 mm (g/cm3 )
strength
Material MPa GPa Hardness (%) 5.8
MPa 85 35 HRB <1 7.0
Ferrous 170 200 140 70 HRB 6.3
F-0008-20 260 390 115 22 HRC 1 7.1
F-0008-35 450 150 35 HRC <1 5.8
F-0008-55HT 240 660 85 50 HRB <1 7.2
F-0008-85HT 450 240 155 84 HRB <1 7.1
FC-0008-30 170 520 150 43 HRC <1 6.6
FC-0008-60 280 720 115 44 HRB <1 7.4
FC-0008-95 340 280 170 78 HRB 7.4
FN-0205-20 480 170 78 HRB 1 7.3
FN-0205-35 1280 160 82 HRB 5 7.3
FN-0205-180HT 530 160 38 HRC <1 6.9
FX-1005-40 830 115 70 HRB 4 6.4
FX-1005-110HT 470 105 61 HRB <1 6.4
300 105 59 HRB 5 6.9
Stainless Steel 310 280 140 65 HRB <1 7.6
SS-303N1-38 260 480 80 65 HRH <1 8.1
SS-304N1-30 230 120 100 80 HRH 131 8.0
SS-316N1-25 310 160 90 88 HRH 9 7.2
SS-316N2-38 220 60 82 HRH 12 2.7
150 — 47 HRB 16 2.7
Copper and Copper Alloys 270 — 72 HRB 4 2.7
CZ-1000-9 70 400 — 55 HRB 3 2.7
CZ-1000-11 80 220 — 77 HRB <1 2.8s
CZP-3002-14 110 320 — 55 HRB 1 2.8
CT-1000-13 110 300 — 80 HRB <1 13
470 — 5 <1
Aluminum Alloys 200 917 — — 2
Ax 123-T1 390 1035 — —
Ax 123-T6 200 49 HRC
Ax 231-T6 310
Ax 231-T6 270
Ax 431-T6 440
Ax 431-T6

Titanium Alloys 827
Ti-6Al-4V (HIP)

Superalloys —
Stellite 19

580 Appendix A Physical and Mechanical Properties of Materials

A.13 Mechanical Properties of Selected Engineering Plastics at Room Tem-
perature

Material Ultimate Elastic Elongation Poisson’s
Thermoplastics: tensile modulus % ratio,
strength 75-5 ν
Acrylonitrile-butadiene-styrene (ABS) MPa GPa - -
ABS, reinforced 28-55 1.4-2.8 0.35
Acetal 75-25 -
Acetal, reinforced 100 7.5 -
Acrylic 55-70 1.4-3.5 0.35-0.40
Cellulosic 135 50-5 -
Fluorocarbon 40-75 10 100-5 -
Nylon 10-48 1.4-3.5 300-100
Nylon, reinforced 7-48 0.4-1.4 200-60 0.46-0.48
Polycarbonate 55-83 0.7-2 10-1 0.32-0.40
Polycarbonate, reinforced 70-210 1.4-2.8 125-10
Polyester 55-70 6-4 -
Polyester, reinforced 110 2-10 300-5 0.38
Polyethylene 2.5-3
Polypropylene 55 3-1 -
Polypropylene, reinforced 110-160 6 1000-15 0.38
Polystyrene 2 500-10
Polyvinyl chloride 7-40 8.3-12 -
Thermosets: 20-35 0.1-1.4 4-2 0.46
Epoxy 40-100 0.7-1.2 60-1
Epoxy, reinforced 14-83 3.5-6 450-40 -
Phenolic 7-55 1.4-4 10-1 -
Polyester, unsaturated 35-140 0.014-4 4-2 0.35
Elastomers: 70-1400 3.5-17 2-0 -
Chloroprene (neoprene) 28-70 21-52 1-0 -
Natural rubber 2.8-21 100-500 -
Silicone 30 5-9 75-650 -
Styrene-butadiene 15-25 1-2 100-1100 -
Urethane 17-25 1.3 250-700 0.5
5-8 1-5 300-450 0.5
10-25 2-10 0.5
20-30 2-10 0.5
0.5

A.14 Mechanical Properties of Selected Ceramics at Room Temperature

Material Symbol Transverse Compressive Elastic Hardness Poisson’s
rupture strength modulus HK ratio, ν
strength MPa
MPa GPa 2000–3000 0.26
4000–5000 —
Aluminum oxide Al2 O3 140–240 1000–2900 310–410 7000–8000 —
Cubic boron nitride cBN 725 7000 850 0.25
Diamond — 1400 7000 830–1000 550 0.14
Silica, fused SiO2 — 1300 70 2100–3000 0.24
Silicon carbide SiC 100–750 700–3500 240–480 2000–2500 —
Silicon nitride Si3 N4 480–600 — 300–310 1800–3200 —
Titanium carbide TiC 1400–1900 3100–3850 310–410 1800–2400 0.30
Tungsten carbide WC 1030–2600 4100–5900 520–700
Partially stabilized zirconia PSZ 620 — 200 1100

Note: These properties vary widely depending on the condition of the material.

Material Properties 581

A.15 Mechanical Properties of Selected Materials used in Rapid Prototyp-
ing.

Material Tensile Elastic Elongation Characteristics
Stereolithography strength modulus in 50 mm
Transparent; good general-purpose material for
Accura 60 MPa GPa (%) rapid prototyping
Somos 9920 68 3.10 5 Transparent amber; good chemical resistance; good
9 1.81 15 fatigue properties; used for producing patterns in
WaterClear Ultra rubber molding
WaterShed 11122 55 2.9 6–9 Optically clear resin with ABS-like properties
53 2.5 15 Optically clear with a slight green tinge; mechanical
DMX-SL 100 properties similar to those of ABS; used for rapid
Polyjet 32 2.4 12–28 tooling
Opaque beige; good general-purpose material for
FC720 60 2.9 20 rapid prototyping
FC830 50 2.49 20
FC 930 1.4 0.185 218 Transparent amber; good impact strength, good
Fused-deposition modeling paint adsorption and machinability
Polycarbonate 52 2.0 3 White, blue, or black; good humidity resistance;
Ultem 9085 72 2.2 5.9 suitable for general-purpose applications
ABS-M30i 36 2.4 4 Semiopaque, gray, or black; highly flexible material
used for prototyping of soft polymers or rubber
PC 68 2.28 4.8
Selective laser sintering White; high-strength polymer suitable for rapid
78 7.32 2.6 prototyping and general use
WindForm XT Opaque tan, high-strength FDM material, good
45 3.3 6 flame, smoke and toxicity rating.
Polyamide PA 3200GF Available in multiple colors, most commonly white;
– 0.015 110 a strong and durable material suitable for general
SOMOS 201 305 137 10 use; biocompatible
ST-100c 970 120 12–16 White; good combination of mechanical properties
Electron-beam melting and heat resistance
Ti-6Al-4V
Opaque black polymide and carbon; produces
durable heat- and chemical-resistant parts; high
wear resistance.
White; glass-filled polyamide has increased stiff-
ness and is suitable for higher temperature appli-
cations
Multiple colors available; mimics mechanical prop-
erties of rubber
Bronze-infiltrated steel powder

Can be heat - treated by hot isostatic pressing to ob-
tain up to 600 MPa fatigue strength

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Appendix B

Stress-Strain Relationships

Symbols y C pny n, ’ z
x’
C elastic material coefficients, Pa τ z ’x ’ σ z'
E modulus of elasticity, Pa 0A P
G shear modulus, Pa pnz 0’
K bulk modulus, Pa
x, y, z Cartesian coordinate system, m τ ’,y ’ zB pnx
x , y , z rotated Cartesian coordinate system, m
γ shear strain z y’ x
δ deformation, m
Figure B.1: Small tetrahedral element cut from body at point
normal strain P . The three perpendicular stress components σz , τz x , and
λ Lame’s constant, Pa τz y acting on the inclined plane are shown.
ν Poisson’s ratio
σ normal stress, Pa σz , τz x , and τz y acting on the inclined plane. The laws of
τ shear stress, Pa stress transformation give these stresses in terms of σx, σy,
σz, τxy, τyz, and τzx. The previously established rules for
Subscripts shear stress subscripts are equally applicable here. For exam-
ple, τz x is the shear stress directed in the x direction and
x, y, z Cartesian coordinates acting in a plane through P whose normal is directed in the
x , y , z rotated Cartesian coordinates z direction.
1, 2, 3 principal axes
Because the normal direction coincides with the z direc-
B.1 Introduction tion,

This Appendix presents selected equations of elasticity, in- pnz = σz cos(z , z) + τxz cos(z , x) + τyz cos(z , y) (B.1)
cluding derivations and definitions of terms that are of im- pny = τzx cos(z , z) + σx cos(z , x) + τyx cos(z , y)
portance to the design of structures and machine elements. pnx = τzy cos(z , z) + τxy cos(z , x) + σy cos(z , y)
The derivations in this Appendix are much more mathemat-
ical than typical treatment elsewhere in the book, with much where
less effort expended to explain approaches. It is hoped that cos(z , x) = cosine of angle between z and x
the statement and rapid derivation of equations will suffice to
demonstrate their existence; the interested reader is encour- The stress σz must equal the sum of the projections of pnz,
aged to find far more in-depth treatment in the classic text by pny, and pnx onto the z axis,
Timoschenko and Goodier [1970], as well as any of the other
Recommended Readings at the end of the Appendix. σz = pnz cos(z , z) + pnx cos(z , x) + pny cos(z , y) (B.2)

This appendix presents the laws of stress transformation Similarly,
that are used in Chapter 2, the Generalized Hooke’s Law in
three dimensions, defines elastic constants by their proper τz x = pnz cos(x , z) + pnx cos(x , x) + pny cos(x , y) (B.3)
names, and gives generalized equations for stress and strain.
By substituting Eqs. (B.1) into Eqs. (B.2) and (B.3) while
B.2 Laws of Stress Transformation also making use of the fact developed earlier that the shear

Let a new orthogonal coordinate system 0 x y z be placed
having origin P, having the z axis coincident with the nor-
mal n to the plane, and having the x and y axes (which must
parallel the plane with normal n ) coincident with the desired
shear stresses directions. Such a coordinate system is shown
in Fig. B.1, where the plane ABC is imagined to pass through
point P , since the tetrahedron is very small. Figure B.1 also
shows the three mutually perpendicular stress components

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