Fundamentals Machine Elements

484 Chapter 16 Fasteners, Connections, and Power Screws
(a) Press head speed. Ans. v = 0.267 m/min.
(b) Power rating needed for the motor. Ans. hp = 2 N/µm, the stiffness of the gasket is 600 N/µm, and the
W. stiffness of the cylinder head that compresses the gas-
ket is 10,000 N/µm. By comparing the life-stress rela-
Motor tionships with those for rolling-element bearings, the car
manufacturer estimates that the stress amplitude in the
Bearings Worm screws needs to be halved to get sufficient life. How can
Spur gears that be done?

Bronze 16.62 A pressure vessel is used as an accumulator to make it
bushings possible to use a small air compressor that works contin-
uously. The stiffness parameter for the lid around each
CB Collar of the 10-mm diameter bolts is kj = 900 MN/m. The
bearing shank length is 20 mm. Because the air consumption
is uneven, the air pressure in the container varies be-
Foot A tween 0.2 and 0.8 MPa many times during a week. Af-
ter 5 years of use one of the bolts holding down the top
Sketch g, for Problem 16.58 lid of the pressure vessel cracks off. A redesign is then
made, decreasing the stress variation amplitude by 25%,
16.59 A valve for high-pressure air is shown in Sketch h. The to increase the life of the bolts to at least 50 years. The
spindle has thread M12 with a pitch diameter of 10.9 stress variation amplitude is decreased by lengthening
mm, lead l = 1.75 mm, and a thread angle of 60◦. Derive the bolts and using circular tubes with the same cross-
the relationship between torque and axial thrust force, sectional area as the solid circular cross section of the
and calculate the axial force against the seating when bolt to transfer the compressive force from the bolt head
the applied torque is 15 N-m during tightening. The co- to the lid. Calculate how long the tubes should be. Ans.
efficient of friction is 0.15. Ans. Tr = (0.00123 m)P . l = 7.089 mm.

Tr 16.63 A loading hook of a crane is fastened to a block hang-
ing in six steel wires. The hook and block are bolted to-
Airflow Thread gether with four 10-mm-diameter screws prestressed to
Seating 20,000 N each. The shank length is 80 mm and the thread
length is 5 mm. The stiffness of the material around each
Sketch h, for Problem 16.59 screw is 1 GN/m. One of the screws of the crane cracks
due to fatigue after a couple of years of use. Will it help
16.60 Derive the expression for the power efficiency of a lead to change the screws to 12-mm diameter while other di-
screw with a flat thread (thread angle β = 0◦) and find mensions are unchanged if the stress variation needs to
the lead angle α that gives maximum efficiency in terms be decreased by at least 20%? Ans. No.
of the coefficient of friction. Also, give results if µ =
0.15. Ans. eµ=0.15 = 74.2%. 16.64 Depending on the roughness of the contacting surfaces
of a bolted joint, some plastic deformation takes place on
16.61 A car manufacturer has problems with the cylinder head the tops of the roughness peaks when the joint is loaded.
studs in a new high-power motor. After a relatively The rougher the surfaces are, the more pressure in the
short time the studs crack just under the nuts, the soft bolted joint is lost by plastic deformation. For a rough-
cylinder head gasket blows out, and the motor stops. ness profile depth of 20 µm on each of the surfaces, a
To be able to analyze the problem, the car manufacturer plastic deformation of 6.5 µm can be expected for the
experimentally measures the stiffnesses of the various two surfaces in contact. For a bolt-and-nut assembly as
components. The stiffness for all bolts together is 400 shown in Fig. 16.13, three sets of two surfaces are in con-
tact. The stiffness of the two steel plates together is 700
MN/m when each is 40 mm thick. The bolt diameter is
16 mm with metric thread. The shank length is 70 mm.
The bolt is prestressed to 20 kN before plastic deforma-
tion sets in. Calculate how much of the prestress is left
after the asperities have deformed. Ans. Fp = 14.8 kN.

16.65 An ISO M12 × 1.75 class = 12.9 bolt is used to fasten
three members as shown in Sketch i. The first member
is made of cast iron, the second is low-carbon steel, and
the third is aluminum. The static loading safety factor is
3.0. Dimensions are in millimeters. Determine

(a) Total bolt length, threaded length, and threaded
length in the joint. Ans. L = 80 mm.

(b) Bolt-and-joint stiffness using a 30◦ cone. Ans. kb =
286.8 MN/m, kj = 928 MN/m.

(c) Preload for permanent connections. Ans. Pi =
73.6 kN.

(d) Maximum static load that the bolt can support.
Ans. Pmax,b = 11.54 kN.

Quantitative Problems 25 485
10
1 30 diameter is 1 m and there are 180 rivets around the cir-
2 cumference, each with a diameter of 25 mm. The riv-
3 ets are made of AISI 1020 steel (quenched and tempered
at 870◦C) and placed in three rows. Calculate the max-
imum allowable propeller torque transmitted through
the rivets for a safety factor of 5. Ans. T = 1.043 × 106
N-m.

16.69 A steel plate (sketch m) is riveted to a vertical pillar.
The three rivets have a 15 mm diameter and carry the
load and moment resulting from the external load of 8
kN. All length dimensions are in millimeters. The yield
strengths of the materials are Sy,rivet = 600 MPa and
Sy,plate = 350 MPa. Calculate the safety factors for

Sketch i, for Problem 16.65 (a) Shear of rivet when Ssy = 0.5Sy. Ans. ns = 1.79.

16.66 A pressurized cast iron cylinder shown in Sketch j is (b) Bearing of rivet. Ans. ns = 4.56.
used to hold pressurized gas at a static pressure of 10
MPa. The cylinder is joined to a low-carbon-steel cylin- (c) Bearing of plate. Ans. ns = 2.66.
der head by bolted joints. Dimensions are in millime-
ters. Determine the required number of bolts. Use (d) Bending of plate. Ans. ns = 4.26.
Grade 12.9 bolts, with M36 × 100 mm coarse threads, y 8 kN
and a safety factor of 2. Ans. Lt = 78 mm, dr = 31.67
mm, n = 39 bolts. 60 15-mm-diam bolts 15

850 30 B
0
25 60 Dx 125
30
30 C
Sketch j, for Problem 16.66
120 400
16.67 In the bolted joint shown in Sketch k the first member
is made of low-carbon steel, the second member is alu- Sketch l, for Problem 16.69
minum, and the third member is cast iron. Assuming
that the members can be rearranged and the frustum 16.70 The cylinder shown in Sketch m is pressurized up to 2
cone angle is 45◦, find the arrangement that can support MPa and is connected to the cylinder head by 32 M24
the maximum load. Dimensions are in millimeters. × 3 metric Grade 8.8 bolts. The bolts are evenly spaced
around the perimeters of the two circles with diameters
of 1.2 and 1.5 m, respectively. The cylinder is made of
cast iron and its head is made of high-carbon steel. As-
sume that the force in each bolt is inversely related to
its radial distance from the center of the cylinder head.
Calculate the safety factor guarding against failure due
to static loading. Ans. ns = 4.02.

1 35 1.5 m 25 mm
1.2 m 30 mm
2 60
1m
3 40
Sketch m, for Problem 16.70
Sketch k, for Problem 16.67
16.68 The flange of a ship’s propeller shaft is riveted in the

radial direction against the hollow shaft. The outside

486 Chapter 16 Fasteners, Connections, and Power Screws
Calculate the safety factors for
++
(a) Shear of rivet when Ssy = 0.57Sy
50
(b) Bearing of rivet
+ +
(c) Bearing of plate 25
75
(d) Bending of plate 25

16.71 An extruded beam as shown in Sketch n is welded to a 30°
thick column using shielded metal arc welding and an 100 kN
E60XX series electrode. Select the weld size for a safety
factor of 3.0, assuming the weld is placed around the en- Sketch o, for Problem 16.73
tire periphery, and indicate your recommendation using
standard weld symbols. Ans. he = 3 mm.

P

50 mm
5000 N

75 mm 120
120
250 mm 320
60 120 120 60
Sketch n, for Problem 16.71 20
15
16.72 For the beam of Problem 16.71, assume that instead of
a weld around the entire periphery, two parallel lines Sketch p, for Problem 16.74
will be used. Should the welds be placed on the hori-
zontal or vertical surfaces? To compare the geometries, 250 P
determine which would require a larger volume of elec-
trode to achieve the same safety factor. Make a sketch 20
of your recommendation using standard weld symbols. 60
Ans. he = 5 mm.
60
16.73 The bracket shown in Sketch o is attached to a beam 20 60
using 15-mm-diameter rivets, whose allowable shear
stress is 500 MPa. The bracket will support a mainly 60
vertical load, but the direction of the load could be as
much as ±30◦ from the vertical as shown. Determine the Sketch q, for Problem 16.75
safety factor based on shear of the rivets. Ans. ns = 1.47.

16.74 A rectangular steel plate is connected with rivets to a
steel beam as shown in Sketch p. Assume the steel to be
low-carbon steel. The rivets have a yield strength of 600
MPa. A load of 24 kN is applied. For a safety factor of 3,
calculate the diameter of the rivets. Ans. d = 15 mm.

Quantitative Problems 20 7 z 487
7 50 P
16.75 Repeat Problem 16.74 but with the plate and beam y
shown in Sketch q. Ans. d = 13 mm. x

16.76 The steel plate shown in sketch r is welded against a 50
wall. Length dimensions are in millimeters. The verti-
cal load W = 15 kN acts 170 mm from the left weld as
shown. Both welds are made by AWS electrode number
E8000. Calculate the weld size for a safety factor of ns =
5.0. Ans. he = 10 mm.

250 20

170
100

A B

y W Sketch t, for Problem 16.80

xG ek 16.81 The bar shown in Sketch u is welded to the wall by AWS
electrodes. A 10-kN load is applied at the top of the bar.
150 Dimensions are in millimeters. For a safety factor of 3.0
against yielding, determine the electrode number that
10 mm plate must be used and the weld throat length, which should
C Column not exceed 10 mm.

Sketch r, for Problem 16.76 350 P
60 35

16.77 Determine the weld size required if only the top (AB) 60
portion is welded in Problem 16.76. Ans. he = 40 mm.

16.78 Two medium-carbon steel (AISI 1040) plates are at- 35
tached by parallel-loaded fillet welds as shown in sketch
s. E60 series welding rods are used. Each of the welds 35
is 75 mm long. What minimum leg length must be used Sketch u, for Problem 16.81
if a load of 15 kN is to be applied with a safety factor of
3.5? 16.82 The bracket shown in Sketch v is spot welded to the
annealed AISI 1040 steel column using three 7-mm-
P diameter electrodes. If 90% of the column material shear
strength is developed by the spot welds, calculate the
A D safety factor against weld nugget shear. Dimensions are
BC in millimeters. Ans ns = 1.09.

P 8000 N
10
Sketch s, for Problem 16.78
25
16.79 The universal joint on a car axle is welded to the 60-
mm-outside-diameter tube and should be able to trans- 25
fer 1500 N-m of torque from the gearbox to the rear axle.
Calculate how large the weld leg should be to give a 25
safety factor of 10 if the weld metal is of class E70. Ans.
he = 3 mm. Sketch v, for Problem 16.82

16.80 The steel bar shown in Sketch t is welded by an E60XX 16.83 AWS electrode number E100XX is used to weld a bar,
electrode to the wall. A 1000-N load is applied in the y- shown in Sketch w, to a wall. For a safety factor of
direction at the end of the bar. Calculate the safety factor 3 against yielding, find the maximum load that can be
against yielding. Also, would the safety factor change supported. Dimensions are in millimeters. Ans. Pmax =
if the direction of load P is changed to the z-direction? 359 N.
Dimensions are in millimeters. Ans. ns = 7.14.

488 Chapter 16 Fasteners, Connections, and Power Screws

16.86 The ropes holding a children’s swing are glued into two

50 10 P plastic tubes with an inner diameter of 10 mm and a
length of 100 mm. The difference in elasticity between

the rope and the rope plus the plastic tube gives a max-

imum shear stress 2.5 times as high as the mean shear

stress in the glue. The glue is an epoxy type with an
50 ultimate shear strength of 12 MPa. How heavy can the

person on the swing be without overstressing the glue

if the speed of the swing at its lowest point is 6.5 m/s

40 6 40 and the distance from the center of gravity of the person
to the fastening points of the ropes is 2 m? The safety
factor is 10. Ans. Maximum mass is 97 kg.

200
16.87 A fishing rod is made of carbon-fiber-reinforced plastic

Sketch w, for Problem 16.83 tube. To get optimum elastic properties along the length
of the rod, and to therefore be able to make long and ac-

16.84 A cam is to be attached to an extruded channel using arc curate casts, the concentrations of the fibers in the vari-
welds. Three designs are proposed as shown in sketch ous parts of the rod have to be different. It is necessary
x. If E60 electrodes are used, find the factor of safety of to scarf joint the rod parts. The tensile strength of the
the welded joint if P =1000 N for each case. Use a = 400 epoxy glue joint is 10 MPa and its shear strength is 12.5
mm. You can assume that the extrusion fits tightly with MPa. These strengths are independent of each other.
the mating hole in the cam for the first two designs so Find the optimum scarf angle to make the rod as strong
that direct shear can be neglected. The rightmost de- as possible in bending.

sign has no mating hole, but the channel is welded to 16.88 Refer to the simple butt and lap joints shown in

the cam surface. Design based on yielding of the weld. Fig. 16.29. (a) Assuming the area of the butt joint is 5

Assume the round channel has a diameter of 50 mm, and mm x 20 mm and referring to the adhesive properties

the square channels are 50 × 50 mm. given in Table 16.14, estimate the minimum and maxi-

mum tensile force that this joint can withstand. (b) Es-

P timate these forces for the lap joint assuming that the
a members overlap by 20 mm.

16.89 A strap joint as depicted in Fig. 16.29g is used to con-

nect two strips of 2014-O aluminum with a thickness of

1 mm and a width of 25 mm. What overlap is needed

to develop the full strength of the strip using an epoxy

8 mm adhesive? Ans. l = 2.2 mm.

P Synthesis and Design
a
16.90 A threaded shaft, when rotated inside a smooth cylin-
8 mm der, can be used as a pump; the geometry is as shown
in Sketch y. Obtain an expression for the flow rate as a
P function of geometry and shaft speed. Hint: Consider
a the flow as a ribbon of fluid that is being translated by
the screw motion.

9 mm Pitch Barrel
H Flight
Sketch x, for Problem 16.84
D θ
16.85 When manufacturing the fuselage of a commuter air- w W
plane, aluminum plates are glued together with lap
joints. Because the elastic deformation for a single plate Barrel
differs from the deformation for two plates glued to-
gether in a lap joint, the maximum shear stress in the Sketch y, for Problem 16.90
glue is twice as high as the average shear stress. The
shear strength of the glue is 20 MPa, the tensile strength 16.91 Consider the situation where large diameter bolts are in
of the aluminum plates is 95 MPa, and their thickness service for extended periods of time and are replaced to
is 4.0 mm. Calculate the overlapping length needed to ensure a high safety factor. It is proposed to refurbish
make the glue joint twice as strong as the aluminum those bolts by machining off (removing) a surface layer,
plate. Ans. L = 76 mm. and cutting new threads. Explain:

Synthesis and Design 489
because of their stable oxide film. Hence, Alclad is used
(a) whether this approach has technical merit, and commonly in aerospace structural applications. Inves-
tigate other common roll-bonded metals and their uses,
(b) any technical and ethical concerns you would have and prepare a summary table.
with this approach.
16.97 Sketch z shows a schematic of a compound joint that
16.92 Design a joint to connect two 25-mm-wide, 5-mm-thick combines rivets with adhesive bonding. List the advan-
steel members. The overlap may be as much as 25 mm, tages and disadvantages of such a design, and develop
and any one approach described in this chapter can be mathematical expressions to predict the axial force that
used. causes failure of the joint. Consider all relevant failure
modes discussed in Section 16.5.
16.93 For the same members in Problem 16.92, design a joint
using threaded fasteners arranged in one row. Do you Adhesive
advise the use of one large fastener or many small fas-
teners? Explain. Rivet

16.94 For the same members in Problem 16.92, design a joint Sketch z, for Problem 16.97
using a combination of joining techniques.
16.98 For a fast-moving power screw, modify Eq. (16.10) to in-
16.95 Design three twist-off fasteners, other than the one de- corporate inertia of the load and screw.
picted in Fig. 16.16, that can be disassembled.

16.96 Alclad stock is made from 5182 aluminum alloy and has
both sides coated with a thin layer of pure aluminum.
The 5182 provides high strength, while the outside lay-
ers of pure aluminum provide good corrosion resistance

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Chapter 17

Springs

A collection of helical compression springs. Source: Courtesy of Danly IEM. Contents
17.1 Introduction 492
It must be confessed that the inventors of the mechanical arts have 17.2 Spring Materials 492
been much more useful to men than the inventors of syllogisms. 17.3 Helical Compression Springs 495
17.4 Helical Extension Springs 502
Voltaire 17.5 Helical Torsion Springs 504
17.6 Leaf Springs 506
17.7 Gas Springs 508
17.8 Belleville Springs 509
17.9 Wave Springs 509
17.10 Summary 512

Examples
17.1 Comparison of Spring Designs 493
17.2 Analysis of a Helical Spring 498
17.3 Buckling of a Helical Spring 499
17.4 Stresses in Helical Springs 500
17.5 Analysis of a Helical Tension Spring 503
17.6 Helical Tension Spring Stiffness 504
17.7 Analysis of a Torsion Spring 505
17.8 Stress Analysis of a Torsion Spring 506
17.9 Leaf Spring 507
17.10 Design of a Leaf Spring 507

Design Procedures
17.1 Design Synthesis of Helical Springs 501

Case Study
17.1 Springs for Dickerman Feed Unit 510

Springs are among the most common machine element, and have a wide variety of forms and functions. This chapter introduces
the most common types of springs and describes their design. The chapter begins by describing the materials and properties
that are needed to produce an effective spring, and discusses the unique strength characteristics that arise from spring metals
that have been highly cold worked. Helical compression springs are analyzed, with geometric concerns and associated forces
and stresses, and the approach is then applied to extension springs. Torsion springs are then examined, and the normal stresses
that result from an applied moment are derived. Torsion springs are also unique in that the number of active coils changes as
the spring deflects. Leaf springs are then summarized; these are cantilevers that are commonly applied to vehicle suspensions
because of their compact designs. Gas springs are commonly used as counterbalances, but also introduce favorable damping
characteristics. Belleville springs, also known as Belleville lock washers, are a special conical disk spring that has a naturally
small profile, and that can be stacked in series or parallel to accentuate their performance. Finally, wave springs are discussed,
which use a helical pattern with a superimposed wave; the resulting springs result in very compact designs.

Machine elements in this chapter: Helical compression and extension springs; torsion springs, leaf springs, gas springs,
Belleville springs, wave springs.
Typical applications: General energy storage, vehicle suspensions, counterweights, preloading components.
Competing machine elements: Flywheels (Ch. 10), tension or compression members for supports (Ch. 5 and 10), retaining rings
(Ch. 11), lock washers (Ch. 16).

491

492 Chapter 17 Springs

Symbols Subscripts
A cross-sectional area, m2
Ap intercept, Pa a alternating
b width of leaf spring, m b body
bs slope c conical
C spring index, D/d d transverse (or direct)
C¯ intercept e end
c distance from neutral axis to outer fiber, m f free (without load)
D mean coil diameter, m h hook
i installed; inside; preload
Di coil inside diameter after loading, m l loop
d wire diameter, m o operating; outside
E modulus of elasticity, Pa s solid
fn lowest natural frequency, Hz t torsional; total
G shear modulus of elasticity, Pa u ultimate
g gravitational acceleration, 9.807 m/s2 w wire
ga gap in open end of extension spring, m
h height, m 17.1 Introduction
I area moment of inertia, m4
J polar area moment of inertia, m4 A spring is a flexible machine element used to exert a force or
K multiple wave factor a torque and, at the same time, store energy. Energy is stored
Kb Bergstra¨sser factor in the solid that is bent, twisted, stretched, or compressed to
Kd transverse shear factor form the spring. The energy is recoverable by the elastic re-
Ki spring stress concentration factor, turn of the distorted material. Springs must have the abil-
Kw Wahl curvature correction factor, ity to elastically withstand desired deflections. Springs fre-
K1 defined in Eq. (17.55) quently operate with high working stresses and with loads
k spring rate, N/m that are continuously varying.
kt spring rate considering torsional loading, N/m
kθ angular spring rate, Nm/rev Some applications of springs are
l length, m
M moment, N-m 1. To store and return energy, as in a gun recoil mechanism
m slope
N number of coils 2. To apply and maintain a definite force, as in relief valves
Na number of active coils and governors

Na number of active coils after loading 3. To isolate vibrations, as in automobile suspensions
Nw number of waves per turn
n number of leaves 4. To indicate and/or control load, as in a scale
ns safety factor
P force, N 5. To return or displace a component, as in a brake pedal
p pitch, m; gas pressure, Pa or engine valve
R radius used in applying torque; linear
17.2 Spring Materials
arm length for bracket, m
R¯ specific gas constant, J/kg-K Strength is one of the most important characteristics to con-
Rd diameter ratio, Do/Di sider when selecting a spring material. Figure 3.21 shows a
r radius, m plot of elastic modulus, E, against strength, S. In this fig-
Sse modified endurance limit, Pa ure, strength refers to yield strength for metals and polymers,
compressive crushing strength for ceramics, tear strength for
Sse endurance limit, Pa elastomers, and tensile strength for composites and woods.
Ssf modified shear fatigue strength, Pa As discussed by Ashby [2010], the normalized strength, S/E,
Ssu shear ultimate strength, Pa is the best parameter for evaluating lightweight strength-
Ssy shear yield strength, Pa based designs. From Fig. 3.21, note that engineering poly-
Sy yield stress, Pa mers have values of S/E in the range 0.01 to 0.1. The values
Sut tensile ultimate stress, Pa for metals are 10 times lower. Even ceramics in compression
T torque, N-m are not as strong as engineering polymers, and in tension they
t thickness, m are far weaker. Composites and woods lie on the 0.01 contour,
tm temperature, K as good as the best metals. Because of their exceptionally
U stored elastic energy, N-m low elastic modulus, elastomers have higher S/E, between
∆U change in energy, N-m 0.1 and 1.0, than does any other class of material.
v volume, m3
x Cartesian coordinate, m The loss coefficient, ∆v, is the second parameter that is
γθz shear strain due to torsional loading important in selecting a spring material. The loss coefficient
∆v loss coefficient measures the fractional energy dissipated in a stress-strain
δ deflection, m cycle. Figure 17.1 shows the stress-strain variation for a com-
ζ cone angle plete cycle. The loss coefficient is
θ angular deflection, rad
ν Poisson’s ratio ∆U (17.1)
ρ density, kg/m3
σ stress, Pa
τ shear stress, Pa

∆v = ,
2U

Spring Materials 493

Stress ∆U U Note that Eq. (17.3) approximates the shear yield stress from
Strain the material’s ultimate strength in tension. There normally is
not a generally applicable rule that allows determination of
yield strength from ultimate strength, so it should be recog-
nized that Eq. (17.3) is applicable only to spring steels and is
somewhat conservative in nature. For example, if the maxi-
mum shear stress theory (Section 6.7.1) is used, then the shear
yield stress is one-half the uniaxial yield strength.

Figure 17.1: Stress-strain curve for one complete cycle. Example 17.1: Comparison of Spring
Designs
where ∆U is the change of energy over one cycle, and U
is the stored elastic energy. The loss coefficient is a dimen- Given: A spring arrangement is being considered for open-
sionless parameter. A material used for springs should have ing the lid of a music box. There can be either one spring
a low loss coefficient. Elastomers have the highest loss co- behind the lid or one weaker spring on each side of the lid
efficients, and ceramics have the lowest, with a four-order- pushing it out after a snap mechanism is released. Except for
of-magnitude range between them. Obviously, ceramics are the spring wire diameter, the spring force and the spring ge-
not a suitable spring material, due to their brittleness. High- ometry are the same for the two options. The material chosen
carbon steels have just slightly higher loss coefficients than is music wire. For the one-spring option, the wire diameter
ceramics and are a more suitable spring material. In bulk needed is 1 mm.
form, elastomers are a good spring material. In a different
shape, such as a cantilever or helix, the high stiffness of steel Find:
is not a detriment, and its low cost and fatigue performance
make it an attractive spring material. (a) Which option is more economical if the cost of the
springs is proportional to the weight of the spring ma-
For these reasons, most commercial springs are pro- terial?
duced from the group of high-strength, low-loss-coefficient
materials that includes high-carbon steel; cold-rolled, (b) Is there any risk that the two-spring option will suffer
precipitation-hardened stainless steel; nonferrous alloys; and from fatigue if the one-spring option does not fatigue?
a few specialized nonmetallics such as laminated fiberglass.
Table 17.1 gives typical properties for common spring materi- Solution:
als (modulus of elasticity, shear modulus of elasticity, density,
and maximum service temperature) and gives additional de- (a) From Table 17.2 for music wire, m = 0.146 and Ap =
sign information. 2170 MPa. Substituting these values into Eq. (17.2)
gives
Springs are manufactured by either hot- or cold-working 2170
processes, depending on the size of the material, the spring Sut = 10.146 = 2170 MPa.
index, and the properties desired. In general, prehardened
wire should not be used if D/d < 4 or if d > 6 mm. Wind- This value is the allowable stress for the single spring
ing the spring induces residual stresses through bending, but design. From Appendix D for a concentrated load at
these are normal to the direction of torsional working stresses the end of the beam (a = x = l), the deflection at the
in a coiled spring. When a spring is manufactured, it is quite free end of the beam is
frequently stress-relieved after winding by a mild thermal
treatment. P l3 (a)
y= .
The ultimate strength of a spring material varies signifi-
cantly with wire size, so that the ultimate strength cannot be 3EI
specified unless the wire size is known. The material and its
processing also have an effect on tensile strength. Results of The spring has a circular cross section, so that the area
extensive testing show that wire strength versus wire diame- moment of inertia is
ter can be approximated by:
πr4 (b)
I= .

4

Sut = Ap , (17.2) Substituting Eq. (b) into Eq. (a) gives
dm

where Ap and m are constants. Table 17.2 gives values of P l34 4l3 P
Ap and m for selected spring materials. Equation (17.2) is y = 3Eπr4 = 3πE r4 . (c)
valid only for the limited size range given in Table 17.2. Also,
note that Ap in thousand pounds per square inch requires d The material and the spring length are the same for op-
to be in inches and that Ap in megapascals requires d to be in tions 1 and 2. Thus, if subscript 1 refers to option 1 and
millimeters. subscript 2 refers to option 2, from Eq. (c)

In the design of springs, the allowable stress is the tor- P1 = P1/2 .
sional yield strength rather than the ultimate strength. Once r14 r24
the ultimate strength is known from Eq. (17.2), the shear yield
stress can be expressed as Solving for r2,

Ssy = τall ≈ 0.40Sut. (17.3)

494 Chapter 17 Springs
Table 17.1: Typical properties of common spring materials. Source: Adapted from Relvas [1996].

Material/Specification Elastic Shear Density, Maximum Principal characteristics
High-carbon steels modulus, modulus, ρ, service
Music wire (ASTM A228) High strength; excellent fatigue life
Hard drawn (ASTM A227) E, G, kg/m3 temperature, General purpose use; poor fatigue life
Stainless steels GPa GPa 7840 ◦C Unsatisfactory for subzero applications
Martensitic (AISI 410, 420) 207 79.3 7840 120 Good strength at moderate temperatures;
Austenitic (AISI 301, 302) 207 79.3 7750 120
Copper-based alloys 200 75.8 7840 250 low stress relaxation
193 68.9 315 Low cost; high conductivity; poor
Spring brass (ASTM B134) 8520
Phosphor bronze (ASTM B159) 110 41.4 8860 90 mechanical properties
Beryllium copper (ASTM B197) 103 43.4 8220 90 Ability to withstand repeated flexures;
Nickel-based alloys 131 44.8 200
Inconel 600 8500 popular alloy
Inconel X-750 214 75.8 8250 315 High yield and fatigue strength;
Ni-Span C 214 75.8 8140 600
186 66.2 90 hardenable
Good strength; high corrosion resistance
Precipitation hardening; for high

temperatures
Constant modulus over a wide

temperature range

Table 17.2: Coefficients used in Eq. (17.2) for selected spring materials.

Material Size range, Exponent, Constant, A p
mm m MPa

Music wirea 0.10–6.5 0.146 2170
Oil-tempered wireb
Hard-drawn wirec 0.50–12 0.186 1880
Chromium vanadiumd 0.70–12 0.192 1750
Chromium silicone
0.80–12 0.167 2000
1.6–10 0.112 2000
302 stainless steel 0.33–2.5 0.146 1867
2.5–5 0.263 2065
5–10 0.478 2911
Phosphor-bronzef
0.1–0.6 0 1000
0.6–2 0.028 913
2–7.5 0.064 932
a Surface is smooth and free from defects and has a bright, lustrous finish.
b Surface has a slight heat-treating scale that must be removed before plating.
c Surface is smooth and bright with no visible marks.
d Aircraft-quality tempered wire; can also be obtained annealed.
e Tempered to Rockwell C49 but may also be obtained untempered.
f SAE CA510, tempered to Rockwell B92-B98.

r2 = r1 = 0.8409r1. (d) 4(P /2)l 2P l σ1
(2)0.25 πr23 π(0.8409)3r13 2(0.8409)3
σ2 = = = = 0.8409σ1.

The cost will change proportionally to the weight, or

πr12lρ = r12 = r12 = 0.7071. (e) Using Eq. (17.2) gives the allowable stress for option 2 as
2πr22lρ 2r22 2(0.8409r1)2

Thus, the one-spring option costs only 0.7071 times the 2170
two-spring option and is therefore 29.3% cheaper. Sut = (0.8409)0.146 = 2226 MPa.

(b) From Eq. (4.45), the bending (design) stress for one Thus, the allowable stress is higher for option 2 than for op-
spring (option 1) is
tion 1 and the design stress is lower for option 2 than for op-
σ1 = Mc = P lr1 = 4P l (f ) tion 1; therefore, the safety factor for option 2 is much larger
I πr14 πr13 . than that for option 1. Recall from Eq. (1.1) that the safety
factor is the allowable stress divided by the design stress.
4 Therefore, if there is no risk of fatigue failure for option 1,
there is no risk of fatigue failure for option 2.
For two springs (option 2), the design stress is

Helical Compression Springs 495
D

RP

dd
P

T = PR (a) (b)
Spring Sparxiinsg
PR axis
P
dd
(a) (b)
D/2 D/2
Figure 17.2: Helical coil. (a) Coiled wire showing applied (c) (d)
force; (b) coiled wire with section showing torsional and di-
rect (vertical) shear acting on the wire. Figure 17.3: Shear stresses acting on wire and coil. (a) Pure
torsional loading; (b) transverse loading; (c) torsional and
17.3 Helical Compression Springs transverse loading with no curvature effects; (d) torsional
and transverse loading with curvature effects.
The helix is the spiral form of spring wound with constant
coil diameter and uniform pitch. Pitch is the distance, mea- 17.3.3 Combined Torsional and Transverse
sured parallel to the coil axis, from the center of one coil to the Shear Stress
center of the adjacent coil. In the most common form of the
helical compression spring, a round wire is wrapped into a The maximum shear stress resulting from summing the tor-
cylindrical form with a constant pitch between adjacent coils. sional and transverse shear stresses, using Eqs. (17.4) and
(17.5), is
Figure 17.2a shows a wire that has been formed into a
helix of N coils with mean coil radius R. The coiled wire is in 8P D 4P 8DP d
equilibrium under the action of two equal and opposite forces τmax = τt,max + τd,max = πd3 + πd2 = πd3 1+ .
P . In Fig. 17.2b, a section has been taken through a wire, and 2D
the statically equivalent force and torque are shown. Note
from Fig. 17.2b that the wire experiences both transverse and (17.6)
torsional shear. Torsional shear in a helical coil spring is the
primary stress, but transverse (or direct) shear is significant, Figure 17.3c shows that the maximum shear stress occurs
so that it cannot be neglected.
at the midheight of the wire and at the coil inside diameter.

Curvature effects are not considered in Eq. (17.6) and are not

shown in Fig. 17.3c. The spring index, which is a measure of

coil curvature, is

D (17.7)
C= .

d

17.3.1 Torsional Shear Stress For most springs, the spring index is between 4 and 12. Equa-
tion (17.6) can be rewritten as

From Eq. (4.33), the maximum torsional shear stress is, using τmax = 8DKdP , (17.8)
Table 4.1: πd3

T c T d(32) 16P R 8P D (17.4) where Kd is the transverse shear factor, given by
τt,max = J = (2)πd4 = πd3 = πd3 ,

where D is the mean coil diameter and d is the wire diameter. Kd = C + 0.5 (17.9)
Figure 17.3 shows the stress distribution across the wire cross .
section. From Fig. 17.3a, for pure torsional stress, the shear C
stress is a maximum at the outer fiber of the wire and zero at
the center of the wire. Note from Eq. (17.6) that if the transverse shear were small
relative to the torsional shear, then as it relates to Eq. (17.8),
17.3.2 Transverse Shear Stress Kd would be equal to 1. Any contribution from the transverse
shear term would make the transverse shear factor greater
The maximum transverse (also called direct) shear stress than 1.

shown in Fig. 17.3b can be expressed for a solid circular cross Also, recall that the spring index C is usually between
4 and 12. If that range is used, the range of transverse shear
section as factor is 1.0417 to 1.1667. Thus, the contribution due to trans-
verse shear is indeed small relative to that due to torsional
P 4P (17.5) shear.
τd,max = A = πd2 .
Recall from Section 4.5.3 that, for a curved member, the
In Fig. 17.3b, the maximum stress occurs at the midheight of stresses can be considerably higher at the inside surface than
the wire. at the outside surface. Thus, incorporating curvature can
play a significant role in the spring design. Curvature effects

factor

496 Kw Chapter 17 Springs
1.5 Kb
where C is the spring index, D/d, Na is the number of active
1.4 coils, and G is the elastic shear modulus. The circumferential
deflection due to torsional and transverse shear loading can
1.3 be derived by using Castigliano’s theorem (Section 5.6). The
total strain energy from Table 5.2 is
1.2
Spring T2l P2l (17.16)
U= + .

2GJ 2AG

1.1 Kd The first term on the right of the equal sign corresponds to
1.0 3 torsional loading and the second term corresponds to trans-
6 9 verse shear for a circular cross section.
Spring C 12 Applying Eq. (17.16) to a helical coil spring of circular
Index,cross section gives the total strain energy as

Figure 17.4: Comparison of the Wahl and Bergstra¨sser curva- U = (P D/2)2(πDNa) + P 2(πDNa)
ture correction factors used for helical springs. The transverse 2G(πd4/32) 2G(πd2/4)
shear factor is also shown.

= 4P 2D3Na + 2P 2DNa .
Gd4 Gd2
can be included in Eq. (17.8) by simply replacing the trans-
verse shear factor with another factor that accounts for stress Using Castigliano’s theorem, Eq. (5.30), gives
concentrations. A curvature correction factor attributed to
A.M. Wahl results in the following: δ= ∂U = 8P D3Na + 4P DNa = 8P C3Na 0.5 .
∂P Gd4 Gd2 Gd 1 + C2
8DKw P
τmax = πd3 , (17.10) (17.17)

Comparing Eq. (17.15) with (17.17) shows that the second

where 4C − 1 0.615 term in Eq. (17.17) is the transverse shear term and that, for a
Kw = 4C − 4 + C .
(17.11) spring index in the normal range 3 ≤ C ≤ 12, the deflection

due to transverse shear is extremely small.

Another curvature correction factor in common use is the From Eq. (17.17), the spring rate is
Bergstra¨sser factor, given by
P Gd
k= = . (17.18)
4C + 2 δ 0.5
Kb = . (17.12) 1+
4C − 3 8C 3 Na
C2

Selecting a curvature correction factor requires consider- The difference between spring index, C, and spring rate,
ation of a number of factors, including the application, pre- k, is important. Spring index is a dimensionless geometric
vious design history, and anticipated loading. In general, a variable whereas spring rate has units of newtons per me-
spring that is to be designed for static failure without consid- ter and incorporates material shear modulus. Also, a stiff
eration of multiple cycles can often use the transverse shear spring has a small spring index and a large spring rate. (Re-
factor, Kd. For more demanding applications, or when fa- call that the spring index for most conventional springs varies
tigue is a possibility, either Kw or Kb provides good results; between 4 and 12.) A spring with excessive deflection has a
they are very close but Kw is slightly more conservative (see large spring index and a small spring rate.
Fig. 17.4).
17.3.5 End Conditions and Spring Length
Figure 17.3d shows the stress distribution when curva-
ture effects and both torsional and transverse shear stresses Figure 17.5 shows four types of ends commonly used in com-
are considered. The maximum stress occurs at the midheight pression springs. Some end coils produce an eccentric ap-
of the wire and at the coil inside diameter. This location is plication of the load, increasing the stress on one side of the
where failure should first occur in the spring. spring, while others result in more uniform loading. The im-
portance of end conditions is self-evident; plain ends are less
17.3.4 Deflection expensive than squared and ground ends, for example, but
are not as uniformly loaded and therefore more susceptible
From Eq. (4.27), the shear strain due to torsional loading is to fatigue failure. Note that the end coils can remain flush
with the element they are bearing against. Thus, a spring can
γθz = rθ = δt = D θ. (17.13) have an active number of coils that is different from the to-
l l 2 l tal. It is difficult to identify just how many coils should be
considered end coils, as this can vary with spring index, solid
Making use of Eq. (4.31) gives length, and specific manufacturing parameters. However, an
average number based on experimental results is used in Ta-
D Tl (17.14) ble 17.3 and is useful for designers.
δt = 2 .
Figure 17.5a shows plain ends that have a noninter-
JG rupted helicoid; the spring rates for the ends are the same
as if they were not cut from a longer coil. Figure 17.5b shows
Applying this equation to a helical coil spring, using Fig. 17.2, a plain end that has been ground. In Fig. 17.5c, a spring with
and assuming that the wire has a circular cross section results plain ends that are squared (or closed) is obtained by deform-
in ing the ends to a 0◦ helix angle. Figure 17.5d shows squared

D (D/2)P (2π)(D/2)Na = 8P D3Na = 8P C3Na
δt = 2 G(πd4/32) Gd4 Gd

(17.15)

Helical Compression Springs 497

(a) (b) (c) (d)

Figure 17.5: Four end types commonly used in compression springs. (a) Plain; (b) plain and ground; (c) squared; (d) squared
and ground.

(P = 0)

Pr

Po
Ps

lf

li

lo ls
ga

(a) (b) (c) (d)

Figure 17.6: Various lengths and forces applicable to helical compression springs. (a) Unloaded; (b) under initial load; (c) under
operating load; (d) under solid load.

and ground ends. A better load transfer into the spring is 17.3.6 Buckling and Surge
obtained by using ground ends, but at a higher cost.
Relatively long compression springs should be checked for
Table 17.3 also shows useful formulas for the pitch, buckling. Figure 17.8 shows critical buckling conditions for
length, and number of coils for compression springs. Recall parallel and nonparallel ends. The critical deflection where
that the pitch is the distance measured parallel to the spring buckling will first start to occur can be determined from this
axis from the center of one coil to the center of an adjacent figure. If buckling is a problem, it can be prevented by plac-
coil. The solid length is the length of the spring when all ad- ing the spring in a cage or over a rod. However, the coils
jacent coils are in metal-to-metal contact. The free length is rubbing on these guides will take away some of the spring
the length of the spring when no external forces are applied force and thus reduce the load delivered at the spring ends.
to it. Figure 17.6 shows the various lengths and forces appli- There is also the potential for wear at the cage or rod contact
cable to helical compression springs. points with the spring.

Figure 17.7 shows the interdependent relationships be- A longitudinal vibration that should be avoided in
tween force, deflection, and spring length for four distinc- spring design is a surge, or pulse of compression, passing
tively different positions: free, initial, operating, and solid. through the coils to the ends, where it is reflected and re-
The deflection is zero at the top right corner and moves to the turned. An initial surge is sustained if the natural frequency
left for positive lengths. Also, the free length, lf , is equal to of the spring is close to the frequency of the repeated loading.
the solid deflection plus the solid length, or The equation for the lowest natural frequency in cycles per

lf = ls + δs. (17.19)

498 Chapter 17 Springs

Table 17.3: Useful formulas for compression springs with four end conditions.

Term Plain Type of spring end Squared and ground
Number of end coils, Ne 0 Plain and ground Squared or closed 2
Total number of active coils, Na Nt
Free length, lf 1 2 Nt − 2
Solid length, ls pN a + d Nt − 1 Nt − 2 pN a + 2 d
Pitch at free length, p d(Nt + 1) p(Na + 1) pN a + 3 d
(lf − d)/N a d(Nt + 1) dN t
dN t ( lf − 3d)/N a (lf − 2d)/N a
lf / (Na + 1)

Spring force, P lf 0.80 Stable Unstable
Ratio of deflection to free length, δ/lf 0.60
li

lo

ls

Ps Length, l 0 Unstable
Stable
0.40

0.20 Parallel ends
Nonparallel ends

Po 0
3 4 5 6 7 8 9 10
Pi Ratio of free length to mean coil diameter, lf /D

Figure 17.8: Critical buckling conditions for parallel and non-
parallel ends of compression springs. Source: Adapted from
Engineering Guide to Spring Design, Barnes Group, Inc. [1987].

0 δo δs
0 δi Deflection, δ

Figure 17.7: Graphical representation of deflection, force, and
length for four spring positions.

second is or Na = 7.843 coils. From Eq. (17.9), the transverse shear
factor is
2d G (17.20)
fn = πNa D2 , C + 0.5 5 + 0.5
32ρ Kd = C = 5 = 1.10.

where G is the shear modulus of elasticity and ρ is the mass For τmax = τall = 480 N/mm2, Eq. (17.8) gives the maximum
density. Vibrations may also occur at whole multiples, such force as
as two, three, and four times the lowest frequency. The spring
design should avoid these frequencies. Pmax = πd3τmax = π(0.010)3 480 × 106 = 3427 N.
8DKd (8)(0.050)(1.10)

Example 17.2: Analysis of a Helical The maximum deflection just to bring the spring to solid
Spring length is

Given: A helical compression spring with plain ends is to be δs = δmax = Pmax = 3427 = 34.27 mm.
designed with a spring rate of approximately 100,000 N/m, k 105
a wire diameter of 10 mm, and a spring index of 5. The max-
imum allowable shear stress is 480 N/mm2 and the shear From Table 17.3, the solid length for plain ends is
modulus of elasticity is 80 GPa.
ls = d(Na + 1) = (0.010)(8.843) = 0.08843 m = 88.43 mm.
Find: The number of active coils, the maximum allowable
static load, and the manufactured pitch so that the maximum From Eq. (17.19), the free length is
load just compresses the spring to its solid length.
lf = ls + δs = 88.43 + 34.27 = 122.7 mm.
Solution: From Eq. (17.18), the number of active coils is

Gd 80 × 109 (0.010) From Table 17.3, the pitch is

Na = 0.5 =,
8C 3 k 1 + C2 0.5 p = (lf − d) = (122.7 − 10) = 14.37 mm.
8(5)3(105) 1+ Na 7.843
25

Helical Compression Springs 499

Example 17.3: Buckling of a Helical ls = dNt = (0.808)(8.41) = 5.17 mm.
Spring Equation (17.19) gives the free length as

Given: A compression coil spring is made of music wire with lf = ls + δs = 5.17 + 48 = 53.17 mm.
squared and ground ends. The spring is to have a spring rate
of 1250 N/m. The force corresponding to the solid length is The pitch is
60 N. The spring index is fixed at 10. Static loading condi-
tions are assumed with few loading cycles, so that the trans- p = (lf − 2d) = (53.17 − 1.736) = 7.75 mm.
verse shear factor can be used. Na 6.64

Find: Find the wire diameter, mean coil diameter, free and Making use of Fig. 17.8 when
solid lengths, and indicate whether buckling is a problem.
Analyze the nominal case (ns = 1) and give a design recom- lf = 53.17 = 0.634,
mendation. D 83.8
and
Solution: The tensile ultimate strength can be determined δs = 48.00 = 0.9027,
from Eq. (17.2) and Table 17.2 as lf 53.17
and, assuming parallel ends, shows that stability should not
Sut = Ap = 2.170 × 109 be a problem.
dm d0.146 ,

where Sut is in pascals if d is in millimeters. From Eq. (17.3)
the shear yield stress is

868.0 × 106 (a) 17.3.7 Cyclic Loading
Ssy = 0.40Sut = .
d0.146
Because spring loading is most often continuously fluctuat-
For static loading and given that the spring index is 10, the ing, allowance must be made in designing for fatigue and
transverse shear factor is, from Eq. (17.9), stress concentration. The material developed in Ch. 7 is ap-
plied in this section and should be understood before pro-
0.5 ceeding. Helical springs are never used as both compression
Kd = 1 + C = 1.05. and extension springs, as is readily apparent from their end
coils. Furthermore, springs are assembled with a preload in
From Eq. (17.8) the maximum design shear stress is addition to the working stress. Thus, fatigue is based on non-
zero mean stress approaches considered in Section 7.3. The
τmax = 8C Kd P = 8(10)(1.05)(60) = 1604 (b) worst condition would occur if there were no preload (i.e.,
πd2 πd2 d2 . when τmin = 0).

Equating Eqs. (a) and (b) gives For cyclic loading, the Wahl or Bergstra¨sser curvature
correction factors given in Eqs. (17.11) and (17.12) should
868.0 × 106 1604 (c) be used instead of the transverse shear factor defined in
= d2 , Eq. (17.8), which is used for static conditions. The Wahl and
d0.146 Bergstra¨sser curvature correction factors can be viewed as a
fatigue stress concentration factor. However, a major differ-
or d = 0.808 mm. Normally, the diameter would be rounded ence for springs in contrast to other machine elements is that
to a convenient value for design purposes, but this analysis for springs the curvature correction factor is applied to both
can proceed with d = 0.808 mm for illustration purposes. the mean stress and the stress amplitude. The reason is that in
The mean coil diameter is, from Eq. (17.7), its true sense, the curvature correction factors are not fatigue
stress concentration factors (as considered in Section 7.7), but
D = Cd = (10)(0.838) = 83.8 mm. a way of calculating the shear stress inside the coil. Plastic de-
formation is to be avoided in springs, as this does not reduce
Since the spring rate is 1250 N/m and the force correspond- stress concentration factors as is the case with other geome-
ing to a solid length is 60 N, tries, but instead tends to increase stress.

Ps 60 The alternating and mean forces can be expressed as
k 1250
δs = = = 0.0480 m = 48 mm. Pmax − Pmin ,
2
Pa = (17.21)

From Table 17.1, the shear modulus of elasticity for music Pm = Pmax + Pmin . (17.22)
wire is G = 79.29 × 109 Pa. Therefore, from Eq. (17.18), the 2
number of active coils is found to be
The alternating and mean stresses, from Eq. (17.10), are then

Gd 79.29 × 109 (0.000808) = 6.41 coils. 8DKwPa ,
Na = 8ktC3 = πd3
8 (1250 × 103) τa = (17.23)

From Table 17.3, for squared and ground ends, the total num- τm = 8DKw Pm . (17.24)
ber of coils is πd3

For springs, the safety factor guarding against torsional en-

Nt = Na + 2 = 8.41 coils. durance limit fatigue is

The solid length is ns = Sse . (17.25)
τa

500 Chapter 17 Springs
Against torsional yielding it is

ns = Ssy , (17.26) Either the Wahl or Bergstra¨sser factors could be used,
τa + τm and will give very similar results. Arbitrarily selecting
the Wahl factor, Eq. (17.11) yields

and against torsional fatigue strength (when infinite life can- 4C − 1 0.615 4(10) − 1 0.615
not be attained — see Section 7.6) it is Kw = 4C − 4 + C = 4(10) − 4 + 10 = 1.145.

ns = Ssf . (17.27) From Table 17.3 for squared and ground ends
τa

where Ssf is the modified shear fatigue strength. Most man- Nt = Na + 2 = 14 + 2 = 16 coils.
ufacturers provide information regarding the fatigue proper-
ties of their springs. The best general purpose data for the tor- The solid length is
sional endurance limits of spring steels are those of Zimmerli
[1957]. The surprising fact about these data is that size, mate- ls = dNt = (1)(16) = 16 mm.
rial, and tensile strength have no effect on the endurance lim-
its of spring steels with wire diameters under 10 mm. Zim- The pitch is
merli’s results are

Sse = 310 MPa for unpeened springs (17.28) p = lf − 2d = 32 − 2 = 2.143 mm.
Sse = 465 MPa for peened springs Na 14

These results are valid for all materials in Table 17.2. Shot Also, from Eq. (17.19), the deflection to solid length is

peening is working the surface material to cause compres- δs = lf − ls = 32 − 16 = 16 mm.

sive residual stresses that improve fatigue performance (see

Section 7.8.5). The endurance limit given in Eq. (17.28) is cor- From Eq. (17.2) and Table 17.2, the ultimate strength in
tension is
rected for all factors given in Eq. (7.18) except the reliability

factor.

The shear ultimate strength, Ssu, for spring steels can be Ap 2170
dm (1)0.146
expressed as Sut = = = 2170 MPa.

Ssu = 0.60Sut. (17.29)

From Eq. (17.3), the torsional yield stress, or the allow-
able shear stress for static loading, is

Example 17.4: Stresses in Helical τall = Ssy = 0.40Sut = (0.40) (2170) = 868 MPa
Springs
Table 17.1 gives G = 79.3 GPa. Therefore, from
Given: A helical compression spring has 14 active coils, a Eq. (17.15), the force required to compress the coils to
free length of 32 mm, and an outside diameter of 11 mm. a solid length is
The ends of the spring are squared and ground and the end
plates are constrained to be parallel. The material is music Ps = Gdδs
wire with a diameter of 1 mm. 8C 3 Na

Find: 79.3 × 109 (0.001)(0.016)
=
(a) For static conditions compute the spring rate, the solid
length, and the stress when the spring is compressed 8(10)3(14)
to the solid length. Will static yielding occur before the
spring is compressed to its solid length? = 11.33 N.

(b) For dynamic conditions with Pmin = 4 N and Pmax = 12 From Eq. (17.8), the maximum design stress is
N, will the spring experience torsional endurance limit
fatigue, torsional yielding, or torsional fatigue failure? τmax = 8DKdPs = 8C Kd Ps
Assume a survival probability of 90%, unpeened coils, πd3 πd2
and fatigue failure based on 50,000 cycles.
8(10)(1.05)(11.33)
= π(0.001)2

= 303 MPa.

Solution: The safety factor guarding against static yielding is

(a) From Eq. (17.7), the spring index is ns = τall = 2170 = 7.16.
τmax 303
C = D = Do − d = 11 − 1 = 10.
dd 1 Thus, failure should not occur due to static yielding.
Checking for buckling gives

From Eq. (17.9), the transverse shear factor is

0.5 0.5 δs = 16 = 0.5.
Kd = 1 + =1+ = 1.05. lf 32
10
C

Helical Compression Springs 501

lf = 32 = 3.2. Ssf = 10C¯ (Nt)bs = 109.498 (50, 000)−0.1754
D 11 − 1
or Ssf = 472 MPa. From Eq. (17.27), the safety factor guard-
From Fig. 17.8, buckling is not a problem. ing against torsional fatigue strength failure is

(b) The alternating and mean forces are ns = Ssf = 472 = 4.048
τa 116.6

Pa = Pmax − Pmin = 12 − 4 = 4.0 N, Therefore, failure should not occur under either static or dy-
2 2 namic conditions.

Pm = Pmax + Pmin = 12 + 4 = 8.0 N.
2 2

From Eq. (17.10), the alternating and mean stresses are Design Procedure 17.1: Design
Synthesis of Helical Springs
τa = 8C Kw Pa
πd2 The approach described in Section 17.3 is very useful for de-
sign analysis purposes. However, it is often the case that a
8(10)(1.145)(4) designer needs to specify parameters of a spring and obtain
= a reasonable design that is then suitable for analysis and op-
timization. The following are important considerations for
π(0.001)2 synthesis of springs. The considerations are strictly applica-
ble to helical compression springs, but will have utility else-
= 116.6 MPa, where as well.

τm = 8C Kw Pm = τa Pm = 233.3 MPa. 1. The application should provide some information re-
πd2 Pa garding the required force and spring rate or total de-
flection for the spring. It is possible that the solid and
The safety factor guarding against torsional yielding is free lengths are also prescribed. Usually, there is sig-
nificant freedom for the designer, and not all of these
ns = Ssy = 868 = 2.48. quantities are known beforehand.
τa + τm 116.6 + 233.3
2. Select a spring index in the range of 4 to 12. A spring in-
Therefore, failure should not occur from torsional yield- dex lower than 4 will be difficult to manufacture, while
ing. From Table 7.4 and Eq. (17.28), the modified en- a spring index higher than 12 will result in springs that
durance limit is are flimsy and tangle easily. Higher forces will require
a smaller spring index. A value between 8 and 10 is
Sse = krSse = (0.9) (310) = 279 MPa. suitable for most design applications.

The safety factor against torsional endurance limit fa- 3. The number of active coils should be greater than 2 in
order to avoid manufacturing difficulties. The number
tigue is of active coils can be estimated from a spring stiffness
design constraint.
ns = Sse = 279 = 2.39.
τa 116.6 4. For initial design purposes, the solid height should be
specified as a maximum dimension. Usually, applica-
Therefore, failure should not occur from torsional en- tions will allow a spring to have a smaller solid height
durance limit fatigue. From Eq. (17.29), the modulus of than the geometry allows, so the solid height should not
rupture for spring steel is be considered a strict constraint.

Ssu = 0.60Sut = 0.60 (2170) = 1302 MPa. 5. When a spring will operate in a cage or with a central
rod, a clearance of roughly 10% of the spring diameter
From Eq. (7.11), the slope used to calculate the fatigue must be specified. This is also useful in compensating
strength is for a coating thickness from an electroplating process,
for example.
bs = − 1 log 0.72Ssu
3 Sse 6. At the free height, the spring has no restraining force,
and therefore a spring should have at least some
= − 1 log (0.72) 1302 × 106 , preload.
3 279 × 106
7. To avoid compressing a spring to its solid length, and
or bs = −0.1754. From Eq. (7.12), the intercept used to the impact and plastic deformation that often result, a
calculate the fatigue strength is clash allowance of at least 10% of the maximum work-
ing deflection should be required before the spring is
C¯ = log (0.72Ssu)2 = log (0.72)2 1302 × 106 2 , compressed solid.
Sse 279 × 106

or C¯ = 9.498. From Eq. (7.13), the fatigue strength is

502 P Chapter 17 Springs
P
8. Consider the application when designing the spring d r3 A
and the amount of force variation that is required. r1 d
Sometimes, such as in a garage door counterbalance
spring, it is useful to have the force vary significantly, r4r2
because the load changes with position. For such ap- B
plications, a high spring rate is useful. However, it is
often the case that only small variations in force over
the spring’s range of motion are desired, which sug-
gests that low spring rates are preferable. In such cir-
cumstances, a preloaded spring with a low stiffness will
represent a better design.

17.4 Helical Extension Springs (a) (b)
P P
There are often circumstances when a spring is needed to
support tensile loads, and the ends of these springs will have d r3 A d
coils or hooks, as appropriate. In applying the load in he- r1
lical extension springs, these hooks, or end turns, must be r4r2
designed so that the stress concentration caused by the pres- B
ence of sharp bends are decreased as much as possible. In
Fig. 17.9a and b the end of the extension spring is formed (c) (d)
by merely bending up a half-loop. If the radius of the bend
is small, the stress concentration at the cross section, at B in Figure 17.9: Ends for extension springs. (a) Conventional de-
Fig. 17.9b, will be large. sign; (b) side view of Fig. 17.9a; (c) improved design over
Fig. 17.9a; (d) side view of Fig. 17.9c.
The most obvious method for avoiding these severe
stress concentrations is to make the mean radius of the hook,
r2, larger. Figure 17.9c shows another method of achieving
this goal. Here, the hook radius is smaller, which increases
the stress concentration. The stress is not as severe, however,
because the coil diameter is reduced. A lower stress results
because of the shorter moment arm. The largest stress occurs
at cross section B shown in Fig. 17.9d, a side view of Fig. 17.9c.

Figure 17.10 shows some of the important dimensions
of a helical extension spring. All the coils in the body are
assumed to be active. The total number of coils depends on
the design of the hooks, but generally it can be stated that

Nt = Na + 1. (17.30) The preferred range of 4 to 12 for the spring index is
The length of the body is
equally valid for extension and compression springs. The

initial tension wound into the spring, giving a preload, is de-

lb = dNt. (17.31) scribed by

The free length is measured from the inside of the end loops, Pi = πτid3 = πτid2 . (17.35)
8D 8C

or hooks, and is Recommended values of τi depend on the spring index and
are given in Fig. 17.11. Springs should be designed for mid-
lf = lb + lh + ll. (17.32) way in the preferred spring index range in Fig. 17.11.

When extension springs are wound, they can be twisted dur- The critical stresses in the hook occur at sections A and
ing the manufacturing process so that a preload exists even B, as shown in Fig. 17.9. At section A the stress is due to
when the spring is at its solid length. The result is that for bending and transverse shear, and at section B the stress is
close-wound extension springs, this initial force needs to be due to torsion. The bending moment and transverse shear
exceeded before any deflection occurs, and after the initial stresses acting at section A can be expressed as
force the force-deflection curve is linear. Thus, the force is

δGd4 (17.33) Mc r1 + PA = 32PAr1 r1 + 4PA .
P = Pi + 8NaD3 , σA = I r3 A πd3 r3 πd2

(17.36)

where Pi is the preload. The spring rate is The shear stress acting at section B is

k = P − Pi = d4G = dG (17.34) τB = 8PB C r2 . (17.37)
δ 8NaD3 8NaC3 . πd2 r4

The shear stresses for static and dynamic conditions are given Radii r1, r2, r3, and r4 are given in Fig. 17.9. Recommended
by Eqs. (17.8) and (17.10), respectively. The preload stress τi design practice is that r4 > 2d. Values of σA and τB given in
is obtained by using Eq. (17.4). Eqs. (17.36) and (17.37) are the design stresses. These stresses

Helical Extension Springs 503
do
200 28
di Preload stress, MPa175 24
Preload stress, ksi15020
ll 125 Preferred 16
12
range 8
100
lf lb
ga lh 75
50

25 4 6 8 4
10 12 14 16

Spring index

Figure 17.11: Preferred range of preload stress for various
spring indexes.

Figure 17.10: Important dimensions of a helical extension Ap 1750 × 106
spring. dm (2)0.192

Sut = = = 1.532 GPa.

are compared with the allowable stresses, taking into account From Eq. (17.3), the torsional yield stress, which in this
the type of loading to determine if failure will occur. The case is the allowable shear stress, is
allowable shear stress to be used with Eq. (17.37) is the shear
yield stress given in Eq. (17.3). The allowable stress to be used Ssy = 0.40Sut = 0.40 1.532 × 109 = 0.6128 GPa.
with Eq. (17.36) is the yield strength in tension, given as
From Eq. (17.38), the tensile yield strength is

Sy = 0.60Sut. (17.38) Sy = 0.60Sut = 0.60 1.532 × 109 = 0.9192 GPa.

Example 17.5: Analysis of a Helical (b) For static loading, the transverse shear factor is given
Tension Spring by Eq. (17.9) as

Given: A helical extension spring similar to that shown in 0.5 0.5
Figs. 17.9a and b is made of hard-drawn wire with a mean Kd = 1 + C = 1 + 5 = 1.1.
coil diameter of 10 mm, a wire diameter of 2 mm, and 120 Using Eq. (17.35) gives
active coils. The hook radius is 6 mm (r1 = 6 mm) and the
bend radius is 3 mm (r2 = 3 mm). The preload is 30 N and Pi = πτid2 = πτi(0.002)2 = 30 N,
the free length is 264 mm. 8C 8(5)

Find: Determine the following:

(a) Tensile and torsional yield strength of the wire or τi = 95.49 MPa.

(b) Initial torsional shear stress of the wire (c) From Table 17.1 for hard-drawn wire, G = 79.29 × 109
Pa. From Eq. (17.34), the spring rate is
(c) Spring rate
dG (0.002) 79.29 × 109 = 1322 N/m.
(d) Force required to cause the normal stress in the hook to k= =
reach the tensile yield strength 8Na C 3 8(120)(5)3

(e) Force required to cause the torsional stress in the hook (d) The critical bending stress in the hook, from Eq. (17.36)
to reach the yield stress and the yield stress in tension given above, is

(f) Distance between hook ends if the smaller of the two σA = 32PAr1 r1 + 4PA = Sy .
forces found in parts d and e is applied πd3 r3 πd2

Solution: From Fig. 17.9a, note that r3 = r1 − 1 = d/2 = 5 mm.
Therefore,
(a) From Table 17.2 for hard-drawn wire, m = 0.192 and Ap
= 1750 MPa. From Eq. (17.2), the ultimate strength in 32PA(0.006)2 + 4PA = 0.9192 × 109.
tension is π(0.002)3(0.005) π(0.002)2

504 Chapter 17 Springs
P P
Solving for PA gives the critical load where failure will
first start to occur by normal stress in the hook as PA = a d
96.90 N.
D
(e) The critical torsional shear stress in the hook can be de-
termined from Eq. (17.37), or

τB = 8PB C r2 = Ssy. Figure 17.12: Helical torsion spring.
πd2 r4

Therefore,

8PB (5) 3 = 0.6128 × 109, From Eq. (17.14), the deflection in the main part of the spring
π(0.002)2 2 is

δ = Do Tl Do P DoπDo(20) = 5πP Do3 .
=
so that PB = 128.3 N. The smaller load is PA = 96.91 N, 2 GJ 2 2GJ GJ
which indicates that failure will first occur by normal
stress in the hook. Therefore,

(f) Using PA in Eq. (17.34) gives the deflection as 5π + 75π P Do3
δt = δ + δc = 81 GJ
δ = PA − Pi = 96.91 − 30 = 0.05061 m = 50.61 mm. δδ = 1.185.
k 1322 5π P Do3
GJ
The distance between hook ends is
Thus, the spring becomes 18.5% more compliant by includ-
lf + δ = 264 + 50.61 = 314.6 mm. ing the conical ends.

Example 17.6: Helical Tension 17.5 Helical Torsion Springs
Spring Stiffness

Given: The ends of a helical extension spring are manufac- The previous section demonstrated that the helical coil spring
tured according to the improved design shown in Fig. 17.9c. can be loaded either in compression or in tension. That is, he-
The conical parts have five coils at each end with a winding lical springs can support axial loads. This section considers
diameter changing according to a spring that supports a torsional load. Figure 17.12 shows
a typical example of a helical torsion spring. The coil ends
D = Do 1− ζ , can have a great variety of shapes to suit various applications.
15π The coils are usually close wound like an extension spring, as
shown in Fig. 17.12, but differ from extension springs in that
where ζ is a coordinate. The cylindrical part of the spring torsion springs do not have any initial tension. Also, note
has an outside diameter Do and 20 coils. from Fig. 17.12 that the ends are shaped to transmit torque
rather than force as is the case for compression and tension
Find: How much softer does the spring become with the con- springs. The torque is applied about the axis of the helix. The
ical parts at the ends? torque actually acts as a bending moment on each section of
the wire, so that the primary stress in a torsional spring is a
Solution: Referring to Fig. 17.2, the torque in the spring is normal stress due to bending. (In contrast, in a compression
T = P R = P D/2. From Eq. (4.31), the change of angular or extension spring, the load produces a torsional stress in the
deflection for a small length is wire.) During manufacture, the residual stresses built in dur-
ing winding are in the same direction, but of opposite sign,
T as the working stresses that occur during use. These residual
dθ = dl. stresses are useful in making the spring stronger by opposing
the working stress, provided that the load is always applied
GJ to cause the spring to wind up. Because the residual stress
The axial compression of the conical part of the spring is opposes the working stress, torsional springs are designed to
operate at stress levels that equal or even exceed the yield
= = RT P R2 P R3 dζ . strength of the wire.
dδc Rdθ dl = R dζ =
GJ GJ GJ The maximum bending stress occurs at the inner fiber of
the coil, and, for wire of circular cross-section, it is given as
The total deflection of the two conical ends is

δc = 2 10π P R3
dζ
0 GJ σ = KiM c = 32KiM ,
I πd3
= 2P 10π Do 3 1 − ζ 3 4C2 − C − 1 (17.39)
GJ 0 2 15π (17.40)
dζ Ki = 4C(C − 1) .

= 75π P Do3 . where
81 GJ

Helical Torsion Springs 505
The angular deflection in radians is
Solution:

θrad = M lw , (17.41) (a) The mean coil diameter is
EI

where lw = πDNa is the length of wire in the spring. The D = Do − d = 16.5 − 1.2 = 15.3 mm.
angular deflection in revolutions is

θrev = M πDNa 1 rev = 32M DNa = 10.19M DNa . The coil inside diameter without a load is
πd4
E 2π rad πEd4 Ed4 Di = D − d = 15.3 − 1.2 = 14.1 mm.

64

(17.42) The spring index is

The angular spring rate, in Nm/rev, is

M Ed4 D 15.3
C= = = 12.75.
kθ = = . (17.43) d 1.2
θrev 10.19DNa

The number of active coils is From Eq. (17.40),

Na = Nb + Ne, (17.44) 4C2 − C − 1 4(12.75)2 − 12.75 − 1
Ki = 4C(C − 1) = 4(12.75)(12.75 − 1) = 1.062.
where

Nb = number of coils in body From Eq. (17.2) and Table 17.2 for music wire,

Ne = number of end coils = (l1 + l2)
3πD
l1, l2 = length of ends, m Ap 2170
Sut = dm = (1.2)0.146 = 2113 MPa.

The inside diameter of a loaded torsion spring is

NaDi , From Eq. (17.38),
Na
Di = (17.45)

Sy = 0.6Sut = 1268 MPa.

where After equating the bending stress to the yield strength,
Na = number of active coils at no load Eq. (17.39) gives
Di = coil inside diameter at no load, m
Na = number of active coils when loaded M = πd3Sy = π(0.0012)3 1268 × 106 = 0.2026 Nm.
Di = coil inside diameter when loaded, m

Torsion springs are frequently used over a round bar. 32Ki 32(1.062)
When a load is applied to a torsion spring, the spring winds
up, causing a decrease in the inside diameter. For design pur- (b) The number of active coils is, from Eq. (17.44),
poses, the inside diameter of the spring must never become 100
equal to the diameter of the bar; otherwise, a spring failure
will occur. On the other hand, a locating device or spring- Na = Nb + Ne = 6 + 3π(15.3) = 6.693.
loaded clip can use this feature to secure the spring in a de-
sired location; when load is applied to the spring, its inner di- From Table 17.1, the modulus of elasticity for music
ameter will increase, allowing the spring to slide freely over wire is 207 GPa. From Eq. (17.43), the angular spring
a bar or tube. When load is removed, it will develop a restric- rate is
tive force due to the reduction in spring diameter.

Example 17.7: Analysis of a Torsion Ed4 207 × 109 (0.0012)4
Spring kθ = 10.18DNa = 10.18(0.0153)(6.693)

Given: A torsion spring similar to that shown in Fig. 17.12 or kθ = 0.4118 Nm/rev. The angular deflection is
is made of 1.2-mm-diameter music wire and has six coils in
the body of the spring and straight ends. The distance a in θrev = M = 0.2026 = 0.4920 rev.
Fig. 17.12 is 50 mm. The outside diameter of the coil is 16.5 kθ 0.4118
mm.
(c) The number of coils when loaded is
Find:
Na = Na + θrev = 6.693 + 0.4920 = 7.185.
(a) If the maximum stress is set equal to the yield strength
of the wire, what is the corresponding moment? From Eq. (17.45), the inside diameter after loading is

(b) With load applied, what is the angular deflection in rev- Di = Na Di = 6.693 (14.1) = 13.1 mm
olutions? Na 7.185

(c) When the spring is loaded, what is the resulting inside (d) Thus, if the bar inside the coils is 13 mm in diameter or
diameter? smaller, the spring should not fail.

(d) What size bar should be placed inside the coils?

506 Chapter 17 Springs

Example 17.8: Stress Analysis of a σ = (5.914) 1.142 = 1.126 GPa. (e)
Torsion Spring 6

Given: A hand-squeeze grip training device with two 100- From Eq. (17.2), the allowable stress is
mm-long handles is connected to a helical torsion spring
with 4.375 coils. The handles and coils are made of music σall = Sut = Ap = 2170 × 106 = 1.852 GPa.
wire. When actuated, the helical torsion spring has 4.5 coils dm (2.956)0.146
and has a spring rate of 2000 N/m.
The safety factor is

Find: Determine the spring dimensions for a safety factor of ns = σall = 1.852 = 1.645.
2 against breakage. σ 1.126

Solution: From Eq. (17.39), the maximum bending stress is

Since the safety factor is too low, try C = 7.5. From Eq. (17.40),

σ = KiM c = KiM r . (a) 4C2 − C − 1 4(7.5)2 − 7.5 − 1
II Ki = 4C(C − 1) = 4(7.5)(6.5) = 1.110.

Using Eq. (17.41) gives the moment needed to deform the From Eq. (d)
spring to 4.5 coils as

M = EIθrad = EI(4.5 − 4.375)2π = 0.25πEI . (b) d3 = 4.303 × 10−9(7.5) = 32.27 × 10−9,
lw lw lw

Substituting Eq. (b) into Eq. (a) gives d = 3.184 × 10−3 m = 3.184 mm.
Therefore,
σ = π KidE .
8 lw D = 7.5 3.184 × 10−3 = 23.88 × 10−3 m = 23.88 mm.

But lw = πDNa, so that From Eq. (c), the maximum design stress is

σ = πKidE = KiE . σ = 1.110 = 0.8753 GPa.
8πDNa 8CNa (5.914)

For music wire, E = 207 GPa, and it was given that Na = 7.5
4.375. Therefore, the stress is
From Eq. (17.2), the allowable stress is

σ = Ki 207 × 109 = 5.914 Ki . σall = Sut = Ap = 2170 × 106 = 1.832 GPa.
dm (3.184)0.146
(c)
C 8(4.375) C

Solving for the safety factor,

To get the spring rate of 2000 N/m at the ends of the handles ns = σall = 1.832 = 2.09.
(see Fig. 17.12), the angular spring rate is σ 0.8753

kθ = (2000)(0.1)(0.1) = 20 N-m/rad. Since ns > 2, the design is adequate according to the con-
straint given in the problem statement.
From Eq. (17.43),

d3 = (10.18)C Na kθ (10.18)(4.375)(20) 17.6 Leaf Springs
E = C 207 × 109 ,

or (d) Multiple-leaf springs are in wide use, especially in the rail-
d3 = 4.303 × 10−9C. way and automotive industries, such as the truck suspension
arrangement shown in Fig. 17.13. A multiple-leaf spring can
As a first try, assume C = 6. From Eq. (17.40), be considered as a simple cantilever type, as shown schemat-
ically in Fig. 17.14b. It can also be considered as a triangu-
4C2 − C − 1 4(6)2 − 6 − 1 lar plate, as shown in Fig. 17.14a. The triangular plate is cut
Ki = 4C(C − 1) = 4(6)(5) = 1.142. into n strips of width b and stacked in the graduated manner
shown in Fig. 17.14b. The layers can slide over each other, as
From Eq. (d), long as they are not welded, clamped, or fastened together.
Some interaction between layers is desirable, however, as is
d3 = 25.82 × 10−9, discussed below.

d = 2.956 × 10−3 m = 2.956 mm. Before analyzing a multiple-leaf spring, first consider
a single-leaf, cantilevered spring with constant rectangular
Therefore, cross section. For bending of a straight beam, from Eq. (4.45),
the magnitude of the bending stress is given by
D = 6 2.956 × 10−3 = 17.73 × 10−3 = 17.73 mm.

From Eq. (c) the maximum design stress is σ = Mc.
I
(17.46)

Leaf Springs 507
Rear shock
absorber lP
x
Spring shackle t

Spring eye Brake drum (a)

Leaf spring

Figure 17.13: Illustration of a leaf spring used in an automo-
tive application.

For a rectangular cross section of base, b, height, t, and mo-
ment, M = P x, Eq. (17.46) becomes

6M 6P x (17.47) P
σ= = . x

bt2 bt2 t
l
The maximum moment occurs at x = l and at the outer fiber
of the cross section, or b

6P l (17.48) (b)
σmax = bt2
Figure 17.14: Leaf spring. (a) Triangular plate, cantilever
From Eq. (17.47) the stress is a function of position along the spring; (b) equivalent multiple-leaf spring.
beam. One method of design optimization requires that the
stress be constant all along the beam. To accomplish this, ei-
ther t is held constant and b is allowed to vary, or vice-versa,
so that the stress is constant for any x. It is convenient from
a manufacturing perspective to use a constant thickness, so
that cold-rolled shapes can be used. Thus,

b(x) = 6P = Constant. (17.49)
x t2σ

Equation (17.49) is linear, giving the triangular shape shown Solution: From Eq. (17.50),
in Fig. 17.14a and a constant stress for any x.
6P l3
The triangular-plate spring and the equivalent multiple- δ= ,
leaf spring have identical stresses and deflection characteris-
tics with two exceptions: Enbt3

1. Interleaf friction provides damping in the multiple-leaf or 6P l3
spring. so that ,

2. The multiple-leaf spring can carry a full load in only one t3 = δEnb
direction.

The deflection and spring rate for the ideal leaf spring t3 = 6(2000)(1)3
3 (207 × 109) (8)(0.05)
are
6P l3
δ= , (17.50) or t = 2.00 mm. From Eq. (17.48), the maximum bending
stress is
Enbt3

P Enbt3 (17.51) 6P l 6(2000)(1)
k= = 6l3 . nbt2 (8)(0.050)(0.002)2
δ σmax = = = 1250 MPa.

Example 17.9: Leaf Spring Example 17.10: Design of a Leaf
Spring
Given: A 1-m-long cantilever spring is composed of eight
graduated leaves. The leaves are 50 mm wide. A load of Given: A leaf spring for a locomotive wheel axle is made
2000 N at the end of the spring causes a deflection of 75 mm. of spring steel with a thickness of 20 mm and an allowable
The spring is made of steel with modulus of elasticity of 207 bending stress of 1050 MPa. The spring has a modulus of
GPa. elasticity of 207 GPa, is 1.6 m long from tip to tip, and carries
a weight of 12,500 N at the middle of each leaf spring.
Find: Determine the thickness of the leaves and the maxi-
mum bending stress.

508 Chapter 17 Springs
Find:
(a)
(a) The width of the spring for a safety factor of 3
High pressure
(b) How high the locomotive has to be lifted during over- nitrogen gas
hauls to unload the springs
chamber Metering orifice
Solution:

(a) From Eq. (1.1), the design stress is

σmax = σall = 1050 × 106 = 0.350 GPa. (a)
3 3

From Eq. (17.48), the maximum bending stress is Integral grease
chamber

6P l
σmax = bt2 .

Because P is the force applied at the tip, the load is Seals Polished steel rod
6250 N and the length is 0.8 m in the locomotive spring.
Therefore, Oil zone for end position
damping and lubrication
6P l 6(6250)(9.807)(0.8) 0.7355 × 109
σmax = bt2 = = .
b(0.020)2 b (b)

Making use of Eq. (a) results in b = 2.101 m. Splitting Figure 17.15: Gas springs. (a) A collection of gas springs.
into 10 leaves gives the width of the leaf spring as Note that the springs are available with a wide variety of end
attachments and strut lengths. Source: Courtesy of Newport
b = 0.2101 m = 210.1 mm. Engineering Associates, Inc. (b) Schematic illustration of a
10 typical gas spring.

(b) The locomotive has to be lifted at least as high as the motion produces a damping effect, which is a fundamental
spring’s deflection. Thus, from Eq. (17.51),
characteristic of such springs. The amount of damping can

P 6P l3 6(6250)(9.807)(0.8)3 be tailored to a particular application by changing the diam-

δ = k = Ebt3 = (207 × 109) (2.101)(0.020)3 eter and length of the orifice in the piston. A small amount of

oil is added to one side of the piston to provide end position

= 0.05412 m = 54.1 mm. damping. This can be obtained at either the end of the exten-

sion stroke or compression stroke by changing the mounting

orientation of the gas spring.

Notice in Fig. 17.15b that there is also an integral grease

17.7 Gas Springs chamber behind the rod seals. The rod is polished, and the

grease helps to prevent wear of the rod and also assists in

Gas springs, as shown in Fig. 17.15, are very commonly ap- sealing the pressurized gas. Clearly, the rod cannot be sub-
plied as support struts or counterweights. These are com-
monly seen in applications such as automobile liftgate or jected to bending stresses or damage from impact or abra-
hood counterbalances or support structures that allow height
adjustment for office chairs and hospital beds. sion from dirt or other debris, or else the seal will be compro-

A cross-section of a typical gas spring is shown in mised, pressure will be lost, and uneven rod wear will result.
Fig. 17.15b. There are many potential mounting options,
ranging from threaded shafts to incorporation of eyelets or There is considerable flexibility in the design of gas
integral brackets. The gas spring has a high-pressure (typi-
cally 10 to 20 MPa) nitrogen or air chamber that acts against springs, and a very long stroke can be achieved with min-
the piston as shown. The piston has small diameter orifices
that allow the pressurized gas to flow in order to equalize imal change in force. By changing the length of the cylin-
pressure on both sides of the piston. Referring to Fig. 17.15,
the pressurized gas on the left side of the piston acts on the der and/or rod, long strokes can be achieved over rela-
entire internal (bore) area. On the right side, the pressurized
gas acts on the annular region around the rod. For slow op- tively long distances. However, gas springs have a minimum
eration, a force on the moving rod is opposed by a force
length which must be accommodated by the application. Gas

springs are rated at room temperature, but changes in am-

bient temperatures will affect the spring capacity. The Ideal

Gas Law can be used to estimate the change in pressure and

therefore the change in piston force as a function of temper-

ature. The Ideal Gas Law using the specific gas constant is

given by p = R¯tm ,
v
(17.53)

F = pA, (17.52) where
p = pressure, N/m2
where p is the initial pressurization and A is the area of the R¯ = specific gas constant, 296.8 J/kg-K for nitrogen,
rod. Note that for rapid operation, a much higher force
is needed because the pressure in the two chambers is not 286.9 J/kg-K for air
equal, and one of the chambers will be compressed by the
sudden movement of the piston. This increased resistance to tm = absolute temperature, K
v = specific volume, m3/kg

Wave Springs 509

200 Height-to-

Di thickness
h ratio, 2.828
160 2.275 1.000

t Percent force to flat 120
Do 1.414

(a) (b) 80

Figure 17.16: Typical Belleville spring. (a) Isometric view of 40 0.400
Belleville spring; (b) cross section, with key dimensions iden-
tified. 0
0 40 80 120 160 200
Equation (17.53) can be used to obtain the resisting force of
a gas spring for a wide temperature range. However, most Percent deflection to flat
gas spring manufacturers will provide estimates, such as a
3.5% increase in force with a 10◦C increase in temperature, Figure 17.17: Force-deflection response of Belleville spring
etc., and these approximations may be sufficient for design given by Eq. (17.54).
purposes.

17.8 Belleville Springs

Belleville springs are named after their inventor, (a) (b)
J.F. Belleville, who patented their design in 1867. Shaped
like a coned disk, these springs are especially useful where Figure 17.18: Stacking of Belleville springs. (a) In parallel; (b)
large forces are desired for small spring deflections. In in series.
fact, because many lock washers used with bolts follow
this principle to obtain a bolt preload (see Section 16.4.5), The force required to totally flatten a Belleville spring is given
these springs are often referred to as Belleville washers.
Typical applications include clutch plate supports, gun recoil by
mechanisms, and a wide variety of bolted connections.
Pflat = 4Eht3 (17.57)
Figure 17.16 shows the cross section of a Belleville .
spring. Two of the critical parameters affecting a Belleville K1Do2(1 − ν2)
spring are the diameter ratio, Rd = Do/Di, and the height-
to-thickness ratio, h/t. From Fig. 17.17, the behavior of a The forces associated with a Belleville spring can be multi-
Belleville spring is very complex and varies considerably plied by stacking them in parallel, as shown in Fig. 17.18a.
with a change in h/t. For low h/t values, the spring acts al- The deflection associated with a given force can be increased
most linearly, whereas large h/t values lead to highly non- by stacking the springs in series, as shown in Fig. 17.18b. Be-
linear behavior. A phenomenon occurring at large h/t is cause such a configuration may be susceptible to buckling, a
“snapthrough buckling.” In snapthrough, the spring de- central support is necessary.
flection is unstable once the maximum force is applied; the
spring quickly deforms, or snaps, to the next stable position.

The force and deflection for Belleville springs are related
by

4Eδ (h − δ) h − δ t + t3 (17.54)
P=
K1Do2 (1 − ν2) 2

where E is the elastic modulus, δ is the deflection from the 17.9 Wave Springs
unloaded state, Do is the coil outside diameter, ν is Poisson’s
ratio for the material, h is the spring height, and t is the spring Figure 17.19 shows a wave spring, which consists of a heli-
thickness. The factor K1 is given by cal pattern with an incorporated wave. Wave springs have
been used for a variety of industrial applications, such as in
6 (Rd − 1)2 , (17.55) valves, preloading bearings, etc. Wave springs are used in-
K1 = π ln Rd Rd2 stead of helical compression springs when compact designs
are required, as wave springs generate the same spring rates
where Rd is the diameter ratio, given by as helical springs but require up to one-half the axial space.
The stress in a wave spring is given by

Rd = Do . (17.56) 3πP D (17.58)
Di σ= ,

4bt2Nw2

510 Chapter 17 Springs

(a) (b) (c) Ram Piercing
Blanking punch
Figure 17.19: Examples of common wave spring configura- Stripper
tions. (a) Common crest-to-crest orientation; (b) crest-to-crest punch
orientation with shim ends; (c) nested wave springs. Source: Pilot Strip
Courtesy of Smalley Co.
Scrap Slug
Die Part

Stop Strip

Table 17.4: Multiple wave factor, Kw, used to calculate wave Finished Scrap First
spring stiffness. Source: from Lee Spring, Inc. washer operation

Waves per turn, Multiple wave factor, (a)
Nw Kw
3.88
2.0–4.0 2.9
4.5–6.5 2.3
7.0–9.5 2.13
> 9.5

where (b)
P = applied force, N
D = mean spring diameter, m Figure 17.20: Illustration of a simple part that is produced
b = width of wave, m by a progressive die. (a) Schematic illustration of the two-
t = thickness of material, m station die set needed to produce a washer; (b) sequence of
Nw= number of waves per turn. operations to produce an aerosol can lid. Source: Kalpakjian
and Schmid [2010]

Note that the number of waves per turn must be in 1/2 turn Case Study: Springs for a Dicker-
increments, or the wave spring will not stack properly. Wave man Feed Unit
springs are linear over a design range, but should not be com-
pressed to their solid length. Thus, it is preferred that the Figure 17.20 illustrates a simple part that is manufactured
spring be compressed so that the height per turn is not less through progressive dies. As can be seen, this small part
than twice the material thickness. Most wave springs have has a number of operations performed on it, each needing
between 2 and 20 turns, with a width between 3 and 10 times specific features in the associated die. Such parts are pro-
the material thickness. Care must be taken in prescribing the duced economically in large numbers by using progressive
cages for wave springs, as these springs expand slightly (usu- or multi-staged dies, and by using a feed mechanism that
ally less than 10% of the spring diameter) when compressed. advances the material a controlled and repeatable distance
with every press stroke.
The stiffness of a wave spring is given by
Figure 17.21 shows a Dickerman feed unit, which is one
k= P = Ebt3Nw4 Do , (17.59) of the most widely used approaches for feeding material by
d Kw D3 Na Di a controlled distance. During a stroke of a mechanical press,
a cam attached to the press ram or punch lowers, pushing
where the follower shown to the right and causing one or more
E = elastic modulus, N/m2 springs to compress. When the ram rises, the compressed
spring forces the gripping unit to the left, and since the grip-
Do = outer diameter, m ping unit mechanically interferes with notched strip or sheet
stock, it forces new material into the press.
Di = inner diameter, m
Dickerman feeds are among the oldest and least expensive
Kw= multiple wave factor, as given in Table 17.4 feeding devices in wide use. They can be mounted to feed
material in any direction, on both mechanical and hydraulic
Na = number of active turns presses, and the feed length can be adjusted by changing the
low-cost, straight cams. Grippers can be blades (as shown)
Wave springs have an advantage over stacked Belleville or cylinder assemblies.
springs discussed in Section 17.8 and shown in Fig. 17.18 in
that a wave spring is a single component and therefore as-
sembly is simplified.

Wave Springs 511

3000

Cam Maximum force,Pmax, N

Gripping unit Gripping unit 2000
(sliding) (fixed)
1000
Spring Fixed rear
guide

Figure 17.21: Dickerman feed unit. Safety factor, ns0
0 1.0 2.0 3.0 4.0 5.0
The case study will investigate the design alternatives in Wire diameter,d, mm
selecting a helical compression spring for a Dickerman feed
unit. It is desired to use a spring that will provide at least 1.4
100 N and have a 150 mm stroke. Buckling of the spring is a 1.2
concern with this combination of load and stroke (see Section 1.0
17.3.6), so a 15-mm-diameter shaft will be used to guide the 0.8
spring. 0.6
0.4
Not much more information is usually given before select- 0.2
ing a spring. Noting that the Dickerman drive can be used
for different applications, it is necessary that 100 N be pro- 0 1.0 2.0 3.0 4.0 5.0
vided whether the deflection is 150 mm, 100 mm, or a very Wire diameter, d, mm
small distance. Thus, the spring will be chosen so that at a
length of 200 mm, it is loaded with 100 N of compressive Figure 17.22: Performance of the spring in Case Study 17.1.
force, and a relatively small stiffness will be selected while
satisfying other design constraints, as discussed in Design 1. The more active coils in the spring, the lower the spring
Procedure 17.1. Another requirement on the spring is that it stiffness and hence the lower the maximum force. In ad-
must have a solid length not greater than 100 mm, or it will dition to reducing the stress levels encountered by the
bottom out with every stroke of the press. spring, a large number of coils makes a more smoothly
operating drive for the reasons discussed above.
Economics is not a critical issue here. As long as the spring
is selected with standard dimensions and made from spring 2. The spring diameter is not specified. However, the
steel, the cost difference between candidate springs will be guide rod in the center of the spring is 15 mm in diame-
low, especially when spread over the total number of parts ter, so that the minimum spring diameter is 15 mm + d.
that will be produced with the Dickerman feed. It is much Normally, the spring diameter would be treated as an-
more essential that a highly reliable spring be chosen because other parameter to be varied, but in this problem it has
an hour of downtime (required to replace a spring and set been set at 20 mm.
up the Dickerman drive) costs far more than a spring. Thus,
the ends will be squared and ground, as discussed in Section The reasons for this are twofold: First, Design Proce-
17.3.5, for better load transfer into the spring. The material dure 17.1 recommended a 10% clearance based on the
used will be music wire, an outstanding spring material. spring diameter, which would suggest a minimum di-
ameter around 16.5 mm. However, a standard size will
No maximum force requirement is given; indeed, it ap- be more readily available, and the use of a 20-mm-
pears that the maximum force is inconsequential. However, diameter spring will lead to lower stresses and better
if the compressed spring contains too large a force, it could reliability.
buckle the press stock, or else dynamic effects could cause
too much material to feed with each stroke. An inconsequen- A design approach that combines the Goodman failure cri-
tial effect is the reduction of press capacity; loads on the or- terion (see Section 7.10.1), the use of Wahl correction factor
der of tens of pounds are not high enough to be significant. given by Eq. (17.11), wire properties from Eq. (17.2), and
the number of active coils for squared and ground ends (see
Since mechanical presses used for progressive die work Table 17.3) results in the performance prediction shown in
are operated unattended for long periods and typically will Fig. 17.22.
operate at 100 strokes per minute or faster, the Dickerman
feed unit must be designed for fatigue. The important equa- From the charts, the safety factor is highest for a spring us-
tions are (17.21) to (17.27). The solution method involves a ing a wire diameter of approximately 2.03 mm. Thus, a wire
variation of the unknown independent parameters (i.e., d, diameter of 2 mm is chosen. The resulting spring will have
D, and Na). All other quantities can be calculated from these 47.4 active coils and result in a maximum force (when com-
parameters. However, the problem is actually much simpler, pressed to a length of 100 mm) of 108 N, with a free length of
for the following reasons: approximately 300 mm. The safety factor for this case is 1.23.

512 Chapter 17 Springs

The fact that the safety factor is small for extremely thin Key Words
wires should not be surprising, since the stresses are in-
versely proportional to the cube of wire diameter. However, Belleville spring coned disk spring
it is not immediately obvious that the safety factor would
eventually increase for increasing wire diameters. The ulti- free length length of spring when no forces are applied to it
mate strength of the wire decreases rapidly with increasing
wire diameter, so that there is in fact an optimum wire di- gas spring a spring that utilizes a compressed gas and mov-
ameter. Indeed, the finding that there is an optimum wire ing piston to achieve fairly uniform force over long
diameter is typical. lengths

17.10 Summary helical compression spring most common spring, wherein
round wire is wrapped into cylindrical form with con-
When a machine element needs flexibility or deflection, or stant pitch between adjacent coils and is loaded in com-
needs to store energy, some form of spring is often selected. pression
Springs are used to exert forces or torques in a mechanism or
to absorb the energy of suddenly applied forces. Springs fre- helical extension spring spring wherein round wire is
quently operate with high stresses and continuously varying wrapped into cylindrical form with constant pitch be-
forces. tween adjacent coils and is loaded in tension

This chapter provided information about spring design helical torsion spring helical coil spring loaded in torsion
while considering popular types of spring designs avail-
able to engineers. Different spring materials were presented. leaf spring spring based on cantilever action
Spring strength is a material parameter that is obviously im-
portant in spring design. The loss coefficient, which measures pitch distance, measured parallel to coil axis, from center of
the fractional energy dissipated in a stress-strain cycle, is an one coil to center of adjacent coil
equally important parameter.
solid length length of spring when all adjacent coils are in
This chapter emphasized helical compression springs. metal-to-metal contact
Both torsional shear stress and transverse shear stress were
considered. Transverse shear is small relative to torsional spring flexible machine element used to exert force or torque
shear. The maximum stress occurs at the midheight of the and, at the ame time, to store energy
wire at the coil inside diameter.
spring index ratio of coil to wire diameter, measure of coil
The spring index is dimensionless and is the mean coil curvature
diameter divided by the wire diameter. The spring rate is
the force divided by the deflection, thus having the unit of spring rate ratio of applied force to spring deflection
newtons per meter. The difference between these terms is
important. Different conditions cause different amounts of surge stress pulse that propagates along spring
eccentric loading that must be compensated for in the design
of a compression spring. Relatively long compression springs wave spring a helical coil spring with a superimposed wavi-
must be checked for buckling. Also, a surge, or longitudinal ness
vibration, should be avoided in spring design. Avoiding the
natural frequency is thus recommended. Spring loading is Summary of Equations
most often continuously fluctuating, so this chapter consid-
ered the design allowance that must be made for fatigue and Spring Materials:
stress concentrations.
∆U
In helical extension springs, it was found that the hooks Loss coefficient: ∆v = 2U
need to be shaped so that the stress concentration effects are Ap
decreased as much as possible. Two critical locations within a Wire strength: Sut = dm
hook were analyzed. In one, the designer must consider nor-
mal stress caused by the bending moment and the transverse Wire shear strength: Ssy = 0.40Sut
shear stresses; and in the other, shear stress must be consid-
ered. Both should be checked. Fatigue strength:

A spring can be designed for axial loading, either com- Sse = 310 MPa for unpeened springs,
pressive or tensile, or it can be designed for torsional loading, 465 MPa for peened springs
as discovered when helical torsion springs were considered.
The ends of torsion springs are shaped to transmit a torque Helical Compression Springs:
rather than a force as for compression and tensile springs.
The torque is applied about the axis of the helix and acts as a Spring index: C = D
bending moment on each section of the wire. d

The leaf spring, used extensively in the automobile and 8DP d
railway industries, was also considered. An approximate Maximum shear stress: τmax = πd3 1+
analysis considered a triangular-plate, cantilever spring, and 2D
an equivalent multiple-leaf spring. Finally, Belleville springs
were considered, and their advantages explained. Transverse shear factor: Kd = C + 0.5
C
4C −1 0.615
Wahl factor: Kw = 4C −4 + C

Bergstra¨ sser factor: Kb = 4C + 2
4C − 3
Deflection: δ = 8P C3Na
Gd 0.5
1 + C2

P Gd
Spring rate: k = =
δ 0.5
8C 3 Na 1+

C2

References 513

Lowest natural frequency: fn = 2 d G References
πNa D2 32ρ

Extension Springs: Ashby, M.J. (2010) Materials Selection in Mechanical Design, 4th
ed., Butterworth-Heinemann.
Spring index: k = P − Pi = d4G
δ 8NaD3 Design Handbook (1987) Engineering Guide to Spring Design.
Associated Spring Corp., Barnes Group Inc.
Helical Torsion Springs:
Kalpakjian, S., and Schmid, S.R. (2010), Manufacturing Engi-
Maximum bending stress: σ = 32KiM neering and Technology, 6th ed., Pearson.
πd3
Relvas, A.A. (1996) Springs, Chapter 28 in Mechanical Design
where Ki = 4C2 − C − 1 Handbook, H.A. Rothbart, Ed., McGraw-Hill.
4C(C − 1)
Zimmerli, F.P. (1957) “Human Failures in Spring Applica-
tions,” The Mainspring, no. 17, Associated Spring Corp.,
Barnes Group, Inc.

Angular spring rate: kθ = M = Ed4 Questions
θrev 10.186DNa
17.1 What is a spring? What are springs used for?
Spring diameter: Di = NaDi 17.2 What is music wire?
Na 17.3 What kind of stress is developed in a compression

Leaf Springs: spring? What about a torsion spring?
17.4 What is the Wahl factor?
Maximum stress: σmax = 6P l 17.5 How is buckling avoided with springs?
bt2 17.6 What is spring surge?
17.7 What is a leaf spring?
6P l3 17.8 Explain why gas springs provide almost uniform forces
Deflection: δ = Enbt3
over their entire stroke.
Spring rate: k = P = Enbt3 17.9 Why do gas springs provide damping?
δ 6l3 17.10 What is a Belleville spring?

Gas Springs:
Ideal gas law: p = R¯tm
v

Belleville Springs:

Force-deflection relation: h− δ
4Eδ 2
P = K1Do2 (1 − ν2) (h − δ) t + t3

6 (Rd − 1)2
where K1 = π ln Rd Rd2

and Rd = Do
Di
4Eht3 Qualitative Problems
Flattening force: Pflat = K1Do2(1 − ν2)
17.11 What is the loss coefficient? Why is steel used for springs
Wave Springs: even though other materials have a superior loss coeffi-
cient?
Maximum stress: σ = 3πP D
4bt2Nw2 17.12 Explain why the strength of a wire increases as the di-
Ebt3Nw4 Do ameter decreases.
Stiffness: k = P = Kw D3 Na Di
d 17.13 Using appropriate sketches, explain why the spring in-
dex for helical compression springs is usually between 4
Recommended Readings and 12.

Budynas, R.G., and Nisbett, J.K. (2011), Shigley’s Mechanical 17.14 What are the advantages and disadvantages of using
Engineering Design, 9th ed., McGraw-Hill. plain ends on helical compression springs? What about
for plain and ground ends?
Carlson, H. (1978) Spring Designers Handbook, Marcel Dekker.
Juvinall, R.C., and Marshek, K.M. (2012) Fundamentals of Ma- 17.15 Review Design Procedure 17.1 and list applications
where a low spring index is preferable. Create a second
chine Component Design, 5th ed., Wiley. list for applications where a high spring index is prefer-
Krutz, G.W., Schuelle, J.K., and Claar, P.W. (1994) Machine De- able.

sign for Mobile and Industrial Applications, Society of Auto- 17.16 What is detrimental about compressing a spring to its
motive Engineers. solid length?
Mott, R. L. (2014) Machine Elements in Mechanical Design, 5th
ed., Pearson. 17.17 Explain why helical extension springs are not used to
Norton, R.L. (2011) Machine Design, 4th ed., Prentice Hall. support compressive loads, and why helical compres-
Rothbart, H.A., Ed. (2006) Mechanical Design Handbook, 2nd sion springs are not used to support tensile loads.
ed., McGraw-Hill.
SAE (1982) Manual on Design and Application of Leaf Springs, 17.18 Without the use of equations, explain why the diame-
Society of Automotive Engineers. ter of torsion springs changes as the spring is loaded or
SAE (1996) Spring Design Manual, 2nd ed., Society of Auto- unloaded.
motive Engineers.

514 Chapter 17 Springs
15
17.19 Referring to Fig. 17.14, explain what machining opera-
tions you would apply to the triangular plate to create
the equivalent multi-leaf spring. What is the base of the
triangular plate?

17.20 Why do wave springs need to have multiples of 0.5
waves per turn?

Quantitative Problems 5

17.21 Sketch a shows a guide wire, used to deliver catheters Sketch b, for Problem 17.23
in the human body during angeoplasty operations. The
compression spring at the end is used as a flexible sup- 17.24 A vehicle has individual wheel suspension in the form
port for the nose as it opens the artery and prevents its of helical springs. The free length of the spring lf = 360
puncture by the guide wire. The spring has an outside mm and the solid length ls = 160 mm at a compressive
diameter of 0.625 mm, a wire diameter of 0.075 mm, force of 5000 N. The shear modulus G = 80 GPa. Use
a free length of 2.75 mm and 30 total coils, and has D/d = 10 and calculate the shear stress for pure tor-
squared and ground ends. The wire is a stainless steel sion of the spring wire. The spring ends are squared and
(E = 193 GPa, G = 75.8 GPa), but has a strength so that ground. Find Na, p, d, and D, and τmax. Ans. τmax = 370
Ap = 700 MPa, m = 0.0474). Determine MPa, Na = 7.06, p = 38.5 mm, d = 17.65 mm.

(a) The solid length and spring rate. Ans. ls = 2.25 17.25 Two equally long cylindrical helical compression
mm, k = 63.85 N/m. springs are placed one inside the other (see Sketch c)
and loaded in compression. How should the springs
(b) The force needed to compress the spring to its solid be dimensioned to get the same shear stresses in both
length. Ans. F = 0.0319 N. springs?

(c) The safety factor against static overload if the P
spring is compressed to its solid length. Use the
Wahl factor. Ans. ns = 3.45.

Guide wire Compression spring Nose 0.625 mm

Polymer liner 2.75 mm

Sketch a, for Problem 17.21

17.22 A helical compression spring with square and ground P
ends has an outside diameter of 14 mm, 20 total coils, Sketch c, for Problem 17.25
a wire diameter of 2 mm, and a free length of 105 mm.
Calculate the spring index, the pitch, the solid length, 17.26 A mechanism is used to press as hard as possible against
and the shear stress in the wire when the spring is com-
pressed to its solid length. Ans. C = 6, p = 5.722 mm, a moving horizontal surface shown in Sketch d. The
ls = 40 mm, τmax = 1560 MPa.
mechanism consists of a stiff central beam and two flex-
17.23 An overflow valve, shown in Sketch b, has a piston di-
ameter of 15 mm and a slit length of 5 mm. The spring ible bending springs made of circular rods with length
has mean coil diameter D = 10 mm and wire diameter
d = 2 mm. The valve should open at 1 bar (0.1 MPa) l, diameter d, modulus of elasticity E, and allowable
pressure and be totally open at 3 bar (0.3 MPa) pres-
sure when the spring is fully compressed. Calculate the stress σall. Wheels are mounted on these rods and can
number of active coils, the free length, and the pitch of
the spring. The shear modulus for the spring material roll over a bump with height f . Calculate the diameter
G = 80 GPa. The spring ends are squared and ground.
Determine the maximum shear stress for this geometry. of the springs so that the prestress of the wheels against
Ans. Na = 22.6, lf = 56.7 mm.
the moving surface is as high as possible without plas-

tically deforming the springs when the bump is rolled

over. The deflection of a spring is shown in Sketch e.

Ans. =. σl2 , P = .πσ4 l5
Ef
128E3f 3

Quantitative Problems l ll 515
l dB is desired to have a total deflection of 175 mm, and each
spring should exert 1000 N on the die when the spring
Rigid Bending is compressed and 500 N when extended. Squared and
beam spring ground ends will be used to ensure good performance.
If 30 active coils are used in each spring, what is the
Sketch d, for Problem 17.26 required wire diameter and spring solid length? Use
G = 79.3 MPa. Ans. d = 3.409 mm, ls = 92 mm.
llll
δ1 17.32 Design a helical compression spring made of music wire
with squared and parallel ends. Assume a spring index
δ2 C = 12, a spring rate k = 300 N/m, and a force to solid
length of 60 N. Find the wire diameter and the mean coil
P1 + ∆P1 P2 + ∆P2 diameter while assuming a safety factor ns = 1.5. Also,
check whether buckling is a problem. Assume steady
Sketch e, for Problem 17.26 loading. Ans. d = 1.90 mm, D = 22.8 mm.

17.27 A helical compression spring is used as a catapult. Cal- ∆
culate the maximum speed of a body weighing 5 kg be-
ing thrown by the catapult, given that τmax = 500 MPa, k1 k2
D = 50 mm, d = 8 mm, Na = 20, and G = 80 GPa. Ans.
Using a Wahl factor, v = 6.51 m/s.

17.28 A spring is preloaded with force Pi and is then exposed

to a force increase whereby the shear stress increases to

a certain value τ . Choose the mean coil diameter D to

maximize the energy absorption caused by the force in-

crease. Ans. D = √π τ d3
.
8 3 Pi

17.29 A compression spring made of music wire is used for Sketch f , for Problem 17.32
static loading. Wire diameter d = 1.5 mm, coil out-
side diameter Do = 12.1 mm, and there are eight active
coils. Also, assume that the spring ends are squared and
ground. Find the following:

(a) Spring rate and solid length. Ans. k = 5214 N/m, 17.33 Two helical springs with spring rates of k1 and k2 are
ls = 15 mm.
mounted one inside the other as shown in Sketch f .
(b) Greatest load that can be applied without causing
a permanent set in excess of 2%. Ans. Pmax = 95.5 The difference in unstressed length is ∆ and the second
N.
spring is longer. A loose clamping device is mounted
(c) The spring free length if the load determined in
part (b) causes the spring to compress to a solid at the top of the springs so that they are deformed and
state. Ans. 32.3 mm.
become equal in length. Calculate the forces in the
(d) Whether buckling is a problem. If it is, recommend
what you would change in the redesign. two springs. An external force P is the load applied

thereafter to the spring. In a diagram show how the

spring forces vary with P . Ans. P1 = Po + k1P ,
k1 + k2
k2P .
17.30 A helical compression spring will be used in a pressure P2 = Po − k1 + k2
relief valve. When the valve is closed, the spring length
is 50 mm, and the spring force is to be 7 N. The spring 17.34 Using Eqs. (17.4) and (17.18) describe how the spring di-
uses a wire diameter of d = 1.5 mm and a spring diam- ameter can be chosen to give as low a maximum shear
eter of D = 10 mm, and has a solid length of ls = 30 stress as possible.
mm. Find the force needed to compress the spring to
its solid length, the number of active coils, the spring’s 17.35 Consider a helical compression spring with plain ends
free length, the spring rate and the safety factor against made of hard-drawn wire with a spring index of 12 and
torsional yielding. Use hard drawn ASTM A227 wire, a stiffness of 0.3 kN/m. The applied load on the spring is
squared and ground ends, and ignore buckling since 60 N. For a safety factor of 1.5 guarding against yielding
the spring is in a cage. Ans. Fs = 22.54 N, Na = 18, find the spring diameter, the wire diameter, the num-
lf = 52.54 mm, k = 2752 N/m, ns = 3.12. ber of coils, and the free and solid lengths of the spring.
Does this spring have buckling and/or dynamic insta-
17.31 Two compression springs will be used in a stamping bility problems? Ans. d = 2.18 mm, D = 26.16 mm,
press to open a die after it has been pushed closed by Na = 41.5.
the action of the press ram. The springs will stay in a
retainer, which means buckling is not an issue and also 17.36 An 18-mm mean diameter helical compression spring
means that the springs will be allowed to be preloaded. has 20 coils, has 2-mm wire diameter, and is made of
Also, this retainer geometry requires that D = 25 mm. It chromium vanadium. Determine the following:

516 Chapter 17 Springs

(a) The maximum load-carrying capacity for a safety (a) The number of coils, the free length, the maximum
factor of 1.5 guarding against yielding. Ans. P = and minimum lengths during cyclic loading, and
78.5 N the spring rate. Ans. Na = 38.0, lf = 73.5 mm.

(b) The maximum deflection of the spring. Ans. δt = (b) For infinite life with 99% reliability, the safety fac-
0.0577m. tors guarding against static and fatigue failure.
Ans. ns,static = 1.66.
(c) The free length for squared and ground ends. Ans.
lf = 0.1075 m. 30°
ll lb lh
17.37 A 60-mm mean diameter helical compression spring
with plain ends is made of hard-drawn steel and has a 30 mm
wire diameter of 2.5 mm. The shear strength of the wire
material is 500 MPa and the spring has 20 coils. The free Sketch g, for Problem 17.42
length is 500 mm. Find the following:
17.43 Calculate the safety factor guarding against fatigue fail-
(a) The required load needed to compress the spring ure if the spring given in Problem 17.42 is designed
to its solid length. Ans. P = 40.11 N. for 50,000 strokes of motion with 50% reliability. Ans.
ns = 11.8.
(b) By applying the load found in part (a) and then
unloading it, whether the spring will return to its 17.44 The extension spring shown in Sketch h is used in a car
free length. braking system. The spring is made of 2-mm-diameter
wire of hard drawn steel, has a mean coil diameter of 10
17.38 A desk lamp has four high-carbon steel helical extension mm, and has a free length of 106 mm. The spring has
springs mounted on a mechanism that makes it possi- 12 active coils and a deflection under braking force of 11
ble to position it over different parts of the desk. Each mm. Determine the following:
spring is preloaded to 15 N, so that no deflection takes
place for forces below 15 N. Above 15-N force the spring (a) The safety factor of the spring. Ans. ns = 1.01.
rate is 100 N/m. The mean coil diameter is 12 mm and
the wire diameter is 1 mm. Calculate the torsional de- (b) The torsional and normal stresses at the hook. Ans.
flection needed on the wire during manufacturing to get τ = 577 MPa, σ = 815 MPa.
the correct preload, and calculate how many coils are
needed to obtain the correct spring rate. Ans. θ/l = 9.63 Lo D/2
rad/m, Na = 57.36.
N
17.39 A spring balance for weighing fish needs to be dimen- d
sioned. The weighing mechanism is a sharp hook hang-
ing in a helical extension spring. To make it easy to read Sketch h, for Problem 17.44
the weight of the fish in the range from 0 to 10 kg, the
length of the scale should be 100 mm. The spring mate- 17.45 A torsion rod is used as a vehicle spring. The torque on
rial is music wire. For d = 2 mm and C = 10, determine the rod is created by a force P = 1500 N acting on radius
the number of active coils and the safety factor. Ans. R = 200 mm. Maximum torsion angle is 30◦. Calculate
Na = 20.1, ns = 1.10. the diameter and length of the rod if the maximum shear
stress is 500 MPa. Ans. d = 14.50 mm, L = 0.606 m.
17.40 A muscle-training device consists of two handles with
three parallel 500-mm-long springs in between. The 17.46 A torsional spring, shown in Sketch i, consists of a steel
springs are tightly wound but without prestress. When cylinder with a rubber ring glued to it. The dimensions
the springs are fully extended to 1600 mm, the force in of the ring are D = 45 mm, d = 15 mm, and h = 20
each spring is 100 N. The springs are made of music mm. Calculate the torque as a function of the angular
wire. Dimension the springs. Ans. For C = 10 and deflection. The shear modulus of elasticity for rubber is
d = 2 mm, ns = 1.08. 150 N/cm2. Ans. T = (23.9 N-m/rad)φ.

17.41 A 45-mm-diameter extension spring (similar to that
shown in Fig. 17.9a and b) has 102 coils and is made of
4-mm-diameter music wire. The stress due to preload is
equivalent to 10% of the yield shear strength. The hook
radius is 5 mm and the bend radius is 2.5 mm. Deter-
mine the following:

(a) The solid length of the spring and the spring stiff-
ness. Ans. k = 273 N/m, ls = 0.412 m.

(b) The preload and the load that causes failure. Ans.
Pi = 39.6 N.

17.42 The extension spring shown in Sketch g is used in a
cyclic motion in turning on and off a power switch. The
spring has a 15-mm outer diameter and is made of a 1.5-
mm-diameter wire of hard-drawn steel. The spring has
no preload. In a full stroke of the spring the force varies
between 25 and 33 N. Determine the following:

Quantitative Problems 517
M
(a) Using a safety factor of 2, find the maximum force
Rubber and the corresponding angular displacement. Ans.
h P = 12.5 N, θ = 0.18 rad.

(b) What would the coil inside diameter be when the
maximum load is applied? Ans. Di = 0.0194 m.

(c) For 100,000 loading cycles calculate the maximum
moment and the corresponding angular displace-
ment for a safety factor of 2.5 guarding against fa-
tigue failure. Ans. θ = 0.180 rev.

PP

d 35 35
D
Sketch i, for Problem 17.46

17.47 To keep a sauna door shut, helical torsion springs are 22
mounted at each of the two door hinges. The friction
torque in each hinge is 0.2 N-m. It should be possible Sketch j, for Problem 17.51
to open the door 180◦ without plastically deforming the
spring. Dimension the springs so that a 10-kg door 700 17.52 The helical torsion spring shown in Sketch k has a coil
mm wide will shut itself in 2 s. The length of the wire outside diameter of 22 mm, 10.25 turns, and a wire di-
ends can be neglected. The air drag on the door is also ameter of 2 mm. The spring material is hard-drawn
neglected. The springs are manufactured from music steel. What would the applied moment be if the maxi-
wire and have a diameter of 2.5 mm. Use C = 10. Ans. mum stress equaled the yield limit? Calculate the inside
Na = 10.03. diameter of the loaded spring and the corresponding an-
gular displacement. Dimensions are in millimeters. Ans.
17.48 A helical torsion spring used in a door handle is made Di = 17.29 mm, θ = 0.421 rev.
of hard-drawn steel wire with a diameter of 3 mm. The
spring has 10 active coils and a coil diameter of 21 mm. P
After 5 years of use the spring cracks due to fatigue. To
get longer fatigue life, the stresses should be decreased 25
by 5%, but the only available space is that used by the
present spring, which has an outside diameter of 24 mm, P 25
an inside diameter of 18 mm, and a length of 30 mm. The
spring rate for the new spring should be the same as for Sketch k, for Problem 17.52
the old one. Dimension a new spring with a rectangular
cross-section to satisfy the new stress requirement. Ans. 17.53 A helical torsion spring made of music wire has a coil di-
w = 3 mm, h = 2.36 mm. ameter of 17.5 mm and a wire diameter of 1.5 mm while
supporting a 0.15-N-m moment with a 20% fluctuation.
17.49 A mouse trap uses two anti-symmetric torsion springs. The maximum number of turns is 12, and the load is 22
The wire has a diameter of 2 mm and the inside diameter mm from the center of the spring. For infinite life with
of the spring in the unset position is 12 mm. Each spring 99% reliability find the safety factors guarding against
has 10 turns when unset. About 50 Newtons are needed static and fatigue failure. Also, determine the inside di-
to set the trap (not to start setting, but rather the largest ameter of the spring when load has been applied. Ans.
load needed to deflect the spring to the set position). ns,static = 3.52, ns,fatigue = 2.51.

(a) Find the number of coils and diameter of the 17.54 A fishing rod is made like an ideal leaf spring with
spring prior to assembly into the mouse trap, that rectangular cross sections. It is made of carbon-fiber-
is, in unstressed state. Ans. NA = 9.58. reinforced plastic with a 150-GPa modulus of elasticity.
The thickness is constant at 8 mm and the length is 2 m.
(b) Find the maximum stress in the spring wire when Find how large the cross section must be at the handle to
the trap is set.Ans. σ = 3010 MPa. carry a 0.3-kg fish by the hook without bending the top
of the rod more than 200 mm. Neglect the weight of the
17.50 The oven door on a kitchen stove is kept shut by a he- lure. Also, calculate the bending stress. Ans. b = 9.19
lical torsion spring. The spring torque is 1 N-m when mm, σ = 60.0 MPa.
the door is shut. When the oven door is fully open, it
must stay open by gravitational force. The door height
is 450 mm and it weighs 4 kg. Dimension the helical
torsion spring using music wire as the spring material
and a wire diameter of 4.5 mm. Is it also possible to
use 3-mm-diameter wire? Ans. For D/d=10, Na = 5.89,
lw = 0.83 m.

17.51 A helical torsion spring, shown in Sketch j, is made from
hard-drawn steel with a wire diameter of 2.2 mm and 8.5
turns. Dimensions are in millimeters.

518 Chapter 17 Springs

17.55 A trampoline is made like a leaf spring with variable The wave spring should support 1000 N, and assume
width so that the maximum bending stress in each sec- that 4 waves per turn are to be used. If the wave spring
tion of the trampoline is constant. The material is glass- is to be constructed from steel with a width of b = 3 mm
fiber-reinforced plastic with a modulus of elasticity of and an allowable stress of 800 MPa, determine the thick-
28 GPa and a bending strength of 300 MPa. Calculate ness of the wave ribbon. How much will the spring com-
the spring rate at the tip of the trampoline and the cor- press under the 1000 N load if there are 25 turns? Com-
responding safety factor if a swimmer weighing 80 kg pare this spring with that analyzed in Problem 17.31.
jumps onto the trampoline from a height of 2 m. The Ans. t = 1.16 mm, d = 3.163 mm.
active length of the trampoline is 3 m, its width is 1.2
m, and its thickness is 38 mm. Ans. k = 11.38 kN/m, 17.60 Given a steel wave spring with an outer diameter of 150
ns = 4.24. mm and a width of 10 mm, with a thickness of 2 mm,
plot the stiffness of the spring as a function of the num-
17.56 The leaf spring of a truck should be able to accommo- ber of waves per turn if the spring has 10 active turns.
date 55-mm deflections (up and down) of the wheels
from an equilibrium position when the truck is driven Design and Projects
on a rough road. The static load in the middle of the
leaf spring is 50,000 N. Assume an allowable stress of 17.61 As shown in the opening photo for the chapter, helical
1050 MPa, a safety factor of 3, leaf half-length of 0.8 m, compression springs can have a rectangular cross sec-
a leaf thickness of 0.02 m, and that there are 10 leaf lay- tion. Derive the equation for maximum stress for such a
ers. Determine the width of the spring. The modulus of cross section. Conduct an Internet search and obtain the
elasticity for the spring is 207 GPa. Ans. b = 0.1714 m. equivalent to the Wahl or Bergstra¨sser factor for such a
geometry.
17.57 A Belleville spring is formed from cold-rolled steel (AISI
1040) with t = 1 mm, h = 2.5 mm, Do/Di = 2.0 and 17.62 Create a table that lists every type of spring discussed in
Di = 7 mm. Calculate the force needed to flatten the the chapter, the size or capacity of such springs, and two
spring to hf = 1.5 mm. What force is needed to fully or three typical applications.
flatten the spring? Explain your answer. Ans. For hf =
1.5, P = 10.1 kN, Pflat = 6.79 kN. 17.63 Review Fig. 17.15 and explain what design features you
would incorporate to allow for modification of the gas
17.58 A Belleville spring is to be used to control the pretension charge pressure.
applied to a bolt by a nut. Thus, it is desired that the
inner diameter be set equal to the crest diameter and that 17.64 It is possible to create a compression spring by laser cut-
Rd = 1.5 so that the Belleville spring will have the same ting a pattern into a tube. This can result in a wide vari-
basic size as the bolt head. If the head height is to be ety of spring shapes, but consider the case when a helix
h = 3 mm, what thickness is needed for a metric grade is machined into the tube. Make a sketch of the result-
4.6 bolt with a 12-mm crest diameter if the connection is ing spring and list the geometric variables that define
to be reused? Is this a feasible design? Ans. t = 1.08 the spring’s geometry. Explain how the stress analysis
mm. would differ, if at all, from the discussion in Section 17.3.
Would there be any advantages to having a helix with
17.59 Consider a wave spring to be used to open a forging die more than one start (see Section 15.4)?
after it has been pushed closed by the action of the press
ram. The retainer allows for an outer diameter of 25 mm.

Chapter 18

Brakes and Clutches

A truck brake drum with cooling fins around the periphery for extended life and Contents
improved performance. Source: Courtesy of Webb Wheel Products, Inc. 18.1 Introduction 520
18.2 Thermal Considerations 520
Nothing has such power to broaden the mind as the ability to 18.3 Thrust Pad Clutches and Brakes 522
investigate systematically and truly all that comes under thy ob- 18.4 Cone Clutches and Brakes 525
servation in life. 18.5 Block or Short-Shoe Brakes 526
18.6 Long-Shoe, Internal, Expanding
Marcus Aurelius, Roman Emperor
Rim Brakes 528
18.7 Long-Shoe, External, Contracting

Rim Brakes 532
18.8 Symmetrically Loaded Pivot-Shoe

Brakes 533
18.9 Band Brakes 535
18.10 Slip Clutches 536
18.11 Summary 538

Examples
18.1 Optimum Size of a Thrust Disk Clutch

Designs 524
18.2 Cone Clutch 526
18.3 Short-Shoe Brake 527
18.4 Long-Shoe Internal Brake 530
18.5 External Long-Shoe Brake Spring 532
18.6 Pivot-Shoe Brake Stiffness 534
18.7 Band Brake 536

Design Procedures
18.1 Long-shoe, Internal,

Expanding Brake Analysis 530

Case Study
18.1 Roller Coaster Braking System 536

This chapter deals with clutches and brakes, two machine elements that are very similar in function and appearance. Brakes
convert mechanical energy to heat, and are widely used to bring all types of machinery to rest. Examples are the thrust pad
and long-shoe, internal expanding (drum) brakes on automobiles, and external expanding and pivot shoe brakes in machinery.
Clutches serve to bring one shaft to the same speed as another shaft, and are available in a wide variety of sizes. Both brakes and
clutches use opposing surfaces, and rely on friction to fulfill their function. Friction causes heat generation, and it is important
that brakes and clutches be designed properly to ovoid overheating. For this reason, clutches for high power applications are
often operated while submerged in a fluid (wet clutches). However, thermal and material considerations are paramount for
effective brake and clutch designs. The chapter begins by analyzing thrust brakes and clutches, as well as the related cone
clutches. Block or short-shoe brakes introduce the concept of self-energizing brake shoes, which is further examined with drum
brakes and clutches. Band brakes, pivot-shoe brakes, and slip clutches are also discussed.

Machine elements in this chapter: Brakes and clutches of all kinds.
Typical applications: Automobiles, aircraft, vehicles of all kinds; shafts on any machinery; transmissions.
Competing machine elements: Shaft couplings (Ch. 11), springs (Ch. 17).

519

520 Chapter 18 Brakes and Clutches

Symbols same angular velocity, to act as a coupling without slip or
A area, m2; constant loss of speed in the driving shaft. A brake is a device used
to bring a moving system to rest, to slow its speed, or to con-
b cone or face width, m trol its speed to a certain value. The function of the brake
is to turn mechanical energy into heat. The design of brakes
C cost and clutches is subjected to uncertainties in the value of the
Cp specific heat, J/(kg◦C) coefficient of friction that must necessarily be used. Material
selection topics from Sections 3.5 and 3.7, as well as friction
c constant and wear covered in Sections 8.8 and 8.9 will be used in this
chapter.
D largest diameter of cone, m
Figure 18.1 illustrates selected brakes and clutches that
d smallest diameter of cone, m are covered in this chapter. These include a rim type that has
internal expanding and external contracting shoes, a band
d1-d10 distances used for brakes, m brake, a thrust disk, and a cone disk. This figure also shows
the actuating forces being applied to each brake or clutch. Ta-
F friction force, N ble 18.1 summarizes the types of brakes and clutches as well
as some typical applications.
F1 pin reaction force, N
Brakes and clutches are similar, but different from other
F2 actuating force, N machinery elements in that they are tribological systems
where friction is intended to be high. Therefore, much effort
hp work or energy conversion rate, W has been directed toward identifying and developing materi-
als that result in simultaneous high coefficients of friction and
M moment, N-m low wear so that a reasonable combination of performance
and service life can be achieved. In previous years, brake
ma mass, kg and clutch materials were asbestos-fiber-containing compos-
ites, but the wear particles associated with these materials
N number of sets of disks resulted in excessive health hazards to maintenance person-
nel. Modern brakes and clutches use “semimetallic” mate-
ns safety factor rials (i.e., metals produced using powder metallurgy tech-
niques) in the tribological interface, even though longer life
P normal force, N could be obtained by using the older asbestos-based linings.
This substitution is a good example of multidisciplinary de-
p contact pressure, Pa sign, in that a consideration totally outside of mechanical en-
gineering has eliminated a class of materials from considera-
Q energy, J tion.

po uniform pressure, Pa Typical brake and clutch design also involves selecting
components of sufficient size and capacity to attain reason-
R reaction force, N able service life. Many of the problems are solid mechanics
oriented; the associated theory is covered in Chapters 4 to
r radius, m 7 and is not repeated here. Because this chapter is mainly
concerned with the performance of brake and clutch sys-
T torque, N-m tems, the focus here is on the actuating forces and resultant
torques. Components of brake systems, such as springs, riv-
T¯ dimensionless torque, T ets, etc., are considered elsewhere and will not be repeated
(2µP ro) here. However, it should be noted that a brake or clutch con-
sists of a number of components integrated into a system.
tm temperature, ◦C
18.2 Thermal Considerations
u sliding velocity, m/s
A critical consideration in the design of brake and clutch com-
W actuating force, N ponents is temperature. Whenever brakes or clutches are ac-
tivated, one high-friction material slides over another under
α half-cone angle, deg a large normal force. The associated energy is converted into
ri heat which always results in elevated temperatures in the lin-
β radius ratio, ro ing material. While all brakes encounter wear, thermal ef-
fects can lead to accelerated wear and can compromise per-
βo optimum radius ratio formance and life. Obviously, such circumstances can only
be discovered through periodic inspection. Therefore, regu-
γ extent of brake pad lar maintenance of brake and clutch systems is essential, and
their service lives are often much lower than those of other
θ circumferential coordinate, deg machinery elements.

θa angle where p = pmax, deg As mentioned above, thermal effects are important in
braking and clutch systems. If temperatures become too high,
θ1 location where shoe begins, deg damage to components could result, which could compro-
mise the useful life or performance of brake and clutch sys-
θ2 location where shoe ends, deg tems. Thermally induced damage often takes the following
forms:
µ coefficient of friction

φ wrap angle, deg

ω angular velocity, rad/s

Subscripts

c conduction
d deenergizing
F friction force
f friction
h convection
i inner
m mean
o outer
P normal force
p uniform pressure
s self-energizing, storage
w uniform wear

18.1 Introduction

Brakes and clutches are examples of machine elements that
use friction in a useful way. Clutches are required when
shafts must be frequently connected and disconnected. The
function of a clutch is twofold: first, to provide a gradual in-
crease in the angular velocity of the driven shaft, so that its
speed can be brought up to that of the driving shaft with-
out shock; second, when the two shafts are rotating at the

Thermal Considerations 521

P PP

PP P P

P PP
P

(a) (b) (c) (d) (e)

Figure 18.1: Five types of brake and clutch. (a) Internal, expanding rim type; (b) external contracting rim type; (c) band brake;
(d) thrust disk; (e) cone disk.

Table 18.1: Typical applications of clutches and brakes.

Type Application notes
Thrust pad (disc)
Extremely common and versatile arrangement; can be wet or dry; wide variety of materials including
Cone carbon-carbon composites for aircraft brakes; preferred for front axles of vehicles because of superior
convective cooling; cannot self-lock.
Block or short-shoe Higher pressure and torque for the same sized clutch compared to thrust pad due to wedging action of
Long-shoe (drum) cone; common for lower speed applications with little sliding such as washing machines or extractors,
Pivot-shoe or high-performance applications such as vehicle racing.
Band brakes Available in a wide variety of configurations and capacities; commonly applied to roller coasters, in-
Slip clutches dustrial equipment and positioning devices.
Widely applied in vehicles on rear axles; self-locking promotes “parking brake” function; economical
and reliable; limited heat dissipation capability.
Used for low-torque applications in architecture, fishing equipement; higher torque applications in-
clude hoists and cranes; difficult to properly locate pivot.
Simple, compact, and rugged, widely applied to chain saws, go-karts, motorcycles, and some bicycles;
susceptible to chatter or grabbing.
Used to prevent excessive torque transfer to machinery; available in a wide variety of sizes and ca-
pacities; applied to machinery to prevent overload, some garage door operators, cranes as an anti-two
blocking device; torque is difficult to control.

• Warped components, such as out-of-round drums or 10 mm
non-planar rotors, result from excessive heating and are Figure 18.2: Brake drum surface showing a high level of heat
often associated with recovery of residual stresses. Such checking. Source: Courtesy Webb Wheel Products, Inc.
conditions can also result from improper machining of
components during service or repair, or from mishan- Hard spot
dling or improper installation.
10 mm
• Heat checks, shown in Fig. 18.2, are small cracks caused Figure 18.3: Hard spot on a brake drum. Source: Courtesy
by repeated heating and cooling of brake surfaces, and Webb Wheel Products, Inc.
could result in high wear rates and compromised perfor-
mance. Light heat checking is a normal condition, and
the small cracks form and are worn away during normal
operation; the component can be machined to remove
the damaged surface during service, as long as manufac-
turer’s tolerances are maintained. Excessive heat checks
can be caused by an operator using the brakes exces-
sively, damaged brake components (such as a worn re-
turn spring or bushing) or by out-of round or warped
drums or rotors.

• Glazed lining surfaces are associated with excessive
heat that can be attributable to a number of causes, and
result in lower friction and diminished performance.
Unless the lining is significantly worn, the glazed sur-
faces can be removed with an abrasive such as emery
cloth to re-establish proper performance.

• Hard spots in the brake surface (Fig. 18.3) are caused by
highly localized heating and cooling cycles, and result
in chatter or pulsation during brake action, with the ulti-
mate result of compromised performance and life. Hard
spots are often associated with warped or out-of-round
components.

522 Chapter 18 Brakes and Clutches

Predicting temperatures of brake and clutch systems is Table 18.2: Product of contact pressure and sliding velocity
extremely difficult in practice because they are operated un-
der widely varying conditions. Neglecting radiation, the first for brakes and clutches. Source: Adapted from Juvinall, R.C.,
law of thermodynamics requires that
and Marshek, K.M. [2006].

pu,

Operating condition (kPa)(m/s)

Qf = Qc + Qh + Qs, (18.1) Continuous: poor heat dissipation 1050
Occasional: poor heat dissipation 2100
Continuous: good heat dissipation 3000

as in oil bath

where 18.3 Thrust Pad Clutches and
Qf = energy input into brake or clutch system from Brakes
friction between sliding elements
Qc = heat transferred by conduction through A thrust disk has its axis of rotation perpendicular to the con-
machinery elements tacting surfaces, as shown in Fig. 18.1d; Fig. 18.4 illustrates
Qh = heat transferred by convection to surrounding the components of an automotive disk brake. Basically, a ro-
environment tor or disk is attached to the vehicle’s axle, and a caliper that
Qs = energy stored in brake and clutch components, is mounted on the automobile body contains two brake pads.
resulting in temperature increase The pads consist of a friction material supported by a back-
ing plate. Pressurized brake fluid hydraulically actuates the
If conduction and convection are negligible, the temperature brake cylinder, causing brake pads to bear against the rotor
rise in the brake or clutch material is given by on opposite sides. The pressure applied determines the con-
tacting pressure, friction, and torque, as will be shown below.
∆tm = Qf , (18.2)
Cpma Figure 18.5 shows the various radii of the thrust disk
clutch or brake. A typical design task is to obtain the ax-
where Cp is the specific heat of the material, and ma is the ial force, P , needed to produce a certain torque, T , and the
resulting contact pressure, p, and wear depth, δ. For some
mass. This equation is useful for determining the instanta- elemental area,
neous temperature rise in a brake or clutch pad, since the
frictional energy is dissipated directly on the contacting sur- dA = (r dθ) dr, (18.3)
faces and does not have time to be conducted or conveyed
away. Brake pads and clutches usually have an area in con- the normal force and the torque can be expressed as
tact that then moves out of contact (i.e., heat is conveyed)
and can cool. The actual maximum operating temperature dP = p dA = pr dθ dr, (18.4)
is therefore a complicated function of heat input and cooling
rates. and

The main difficulty in predicting brake system tempera- T = r dF = µr dP = µpr2 dr dθ. (18.5)
tures is that the heat conducted and the heat conveyed de-
pend on the machine ambient temperature and the brake Only a single set of disks will be analyzed in the follow-
or clutch geometry, and can vary widely. In previous cir- ing sections, but the torque for a single set of disks is multi-
cumstances in this text, a worst-case analysis would be per- plied by N to get the torque for N sets of disks.
formed, which in this case quickly reduces to circumstances
where brakes and clutches become obviously overheated. 18.3.1 Uniform Pressure Model
This result is not incorrect: most brake and clutch systems
are overheated when abused and can sustain serious damage For new, accurately flat, and aligned disks, the pressure will
as a result. The alternative is to use such massive brake and
clutch systems as to make the economic burden unbearable to be uniform, or p = po. Substituting this into Eqs. (18.4) and
responsible users. It is far more reasonable to use brake sys-
tems that require periodic maintenance and can be damaged (18.5) gives
through abuse than to incur the economic costs of surviv-
ing worst-case analyses. This differs from previous circum- Pp = πpo ro2 − ri2 , (18.6)
stances, where a worst-case analysis still resulted in a reason-
able product. Tp = 2πµpo ro3 − ri3 = 2µPp ro3 − ri3 . (18.7)
3 3 (ro2 − ri2)
Some clutches are intended to be used with a fluid (wet
clutches) to aid in cooling the clutch. Similarly, some pads or Thus, expressions for the normal load and torque can be de-
shoes will include grooves so that air or fluid can be better termined from the uniform pressure, the geometry (ro and
entrained, and increased flow and convective heat transfer ri), and the coefficient of friction, µ.
result. Predicting wet clutch temperatures is a complex prob-
lem and requires numerical, usually finite element, methods. 18.3.2 Uniform Wear Model

Obviously, the proper size of brake components is ex- If the mating surfaces of the clutch are sufficiently rigid, it
tremely difficult to determine with certainty. For the pur- can be assumed that uniform wear will occur. This assump-
poses of this text the values given by Juvinall and Marshek tion generally holds true after some initial running in. For
[2006] for the product of brake shoe or pad contact pressure example, consider the Archard wear law of Eq. (8.39). For a
and sliding velocity, pu, can be used to estimate component thrust disk clutch, the sliding distance per revolution is pro-
sizes (Table 18.2). As discussed in Section 18.3.2, pu is pro- portional to the radius; that is, the outside of the disk sees a
portional to the power dissipated. Most manufacturers rely
heavily on experimental verification of designs; the applica-
tion of these numbers in the absence of experimental verifi-
cation requires extreme caution, but is useful for evaluating
designs and estimating component sizes.

Thrust Pad Clutches and Brakes 523

Rotor (disk) Brake pads larger sliding distance than the inner radius. If the pressure
is uniform, and the hardness of a brake pad or clutch disk
Caliper is constant, then more wear will occur on the outside of the
disc. Of course, this results in a redistribution of pressure, so
Studs that after initial wear has taken place, uniform wear becomes
possible. Thus, for the disk shown in Fig. 18.5 the wear is
constant over the surface area ri ≤ r ≤ ro and around the
circumference of the disk.

The Archard wear law suggests that the rate of wear is
proportional to the product of force and velocity, as can be
seen by taking the derivative with respect to time of Eq. (8.39).
Since the product of force and speed is power,

hp = F u = µP u = µpAu, (18.8)

Dust cap Ventilation slots where
Hub
F = friction force, N

µ = coefficient of friction

P = normal force, N

u = velocity, m/s
A = area, m2

(a) If the brake cylinder actuates linearly, and tolerances are rea-
Caliper
Piston sonably tight, any initial uniform pressure will need to adjust.

Where the surfaces wear the most, the pressure decreases the

most. Thus, multiplying pressure and velocity will produce a

Pad constant work or energy conversion, implying that the wear

Backing should be uniform at any radius. Then, µpAu remains con-
plate
stant; and if µA is constant, p is inversely proportional to u,

and for any radius, r, c
p= .
(18.9)
r

Substituting Eq. (18.9) into Eq. (18.4) gives

Brake fluid Pw = 2πc (ro − ri) . (18.10)

Since p = pmax at r = ri, it follows from Eq. (18.9) that

c = pmaxri. (18.11)

Hub Rotor/disc Substituting Eq. (18.11) into Eq. (18.10) results in

(b) Pw = 2πpmaxri (ro − ri) , (18.12)

Figure 18.4: Thrust brake terminology and operation. (a) Il- and inserting Eq. (18.9) into Eq. (18.5) produces
lustration of a thrust brake, with wheel removed for clarity.
Note that the caliper shown has a window to allow obser- Tw = µc 2πµc ro2 − ri2 . (18.13)
vation of the brake pad thickness, a feature that is not al- r dr dθ =
ways present. (b) Section view of the disk brake, showing
the caliper and brake cylinder. 2

Substituting Eq. (18.11) into Eq. (18.13) gives

Tw = πµripmax ro2 − ri2 . (18.14)
Substituting Eq. (18.12) into Eq. (18.14) gives

Tw = Fw rm = µPw (ro + ri) . (18.15)
2

θ dr By coincidence, Eq. (18.15) gives the same result as if the
ro r torque was obtained by multiplying the mean radius rm =
(ro + ri)/2 by the friction force, F .
ri
Equations (18.7) and (18.15) can be expressed as the di-
mensionless torque for uniform pressure, T¯p, and for uniform
wear, T¯w, by the following equations:

T¯w = Tw = (1 + β) , (18.16)
2µPw ro 4 (18.17)

Figure 18.5: Thrust disk clutch surface with various radii. T¯p Tp (1 − β3)
2µPpro 3(1 − β2) ,
= =

524 Chapter 18 Brakes and Clutches

Table 18.3: Representative properties of contacting materials operating dry, when rubbing against smooth cast iron or steel.

Maximum contact Maximum bulk
Friction Coefficient of pressure,a pmax temperature, tm,
material friction, µ ◦C max
kPa

Molded 0.25–0.45 1030–2070 204–260
Woven 0.25–0.45 345–690 204–260
Sintered metal 0.15–0.45 204–677
Cork 0.30–0.50 1030–2070
Wood 0.20–0.30 55–95 82
Cast iron; hard steel 0.15–0.25 345–620 93
260
690–1720

a Use of lower values will give longer life.

T 0.6 Table 18.4: Coefficient of friction for contacting materials op-
2µPro
erating in oil when rubbing against steel or cast iron.

Friction material Coefficient of friction, µ

–T–= 0.5 Molded 0.06–0.09
Woven 0.08–0.10
torque, 0.4 Uniform pressure Sintered metal 0.05–0.08
Paper 0.10–0.14
Dimensionless 0.3 Uniform wear Graphitic 0.12 (avg.)
Polymeric 0.11 (avg.)
Cork 0.15–0.25
Wood 0.12–0.16
Cast iron; hard steel 0.03–0.16

0.2 0 0.2 0.4 0.6 0.8 1.0
Radius ratio, β = ri/ro
18.3.3 Partial Contact
Figure 18.6: Effect of radius ratio on dimensionless torque for
uniform pressure and uniform wear models. It should be noted that many brakes do not use a full circle
of contact, as shown in Fig. 18.4. A common practice is to
where β = ri . assume that a pad can be approximated as a sector or wedge,
ro so that it covers a fraction of the rotor or disc. This has to be
(18.18) accommodated in the equations for the actuating force and
torque developed for the uniform pressure and uniform wear
Figure 18.6 shows the effect of radius ratio, β, on dimen- models. This can be easily accomplished by defining γ as
sionless torque for the uniform pressure and uniform wear the percentage of the rotor covered by the pad. Then, for the
models. The largest difference between these models occurs uniform pressure model, Eqs. (18.6) and (18.7) become
at a radius ratio of zero, and the smallest difference occurs at a
radius ratio of 1. Also, for the same dimensionless torque the Pp = γπpo ro2 − ri2 , (18.19)
uniform wear model requires a larger radius ratio than does
the uniform pressure model. This larger radius ratio implies Tp = 2γπµpo ro3 − ri3 , (18.20)
that a smaller area is predicted by the uniform wear model. 3
Thus, the uniform wear model may be viewed as the safer
approach, although the two approaches yield similar torque and for the uniform wear model, Eq. (18.12) and (18.14) be-
predictions. Usually a brake or clutch is analyzed using both come
conditions to make sure a design is robust over its life. The
rationale is that, when new, the pressure can be uniform, but Pw = 2πγpmaxri (ro − ri) , (18.21)
as the brake wears, the pressure will adjust until wear is uni-
form across the pad area. Tw = πγµripmax ro2 − ri2 . (18.22)

Table 18.3 gives the coefficient of friction for several ma- Example 18.1: Optimum Size of a
terials rubbing against smooth cast iron or steel under dry Thrust Disk Clutch
conditions. It also gives the maximum contact pressure and
the maximum bulk temperature for these materials. Table Given: A single set of thrust disk clutches is to be designed
18.4 gives the coefficient of friction for several materials, in- for use in an engine with a maximum torque of 150 N-m.
cluding those in Table 18.3, rubbing against smooth cast iron A woven fiber reinforced polymer will contact steel in a dry
or steel in oil. As would be expected, the coefficients of fric- environment. A safety factor of 1.5 is assumed in order to
tion are much smaller in oil than under dry conditions. account for slippage at full engine torque. The outside diam-
eter should be as small as possible.
Equations (18.6) and (18.7) for the uniform pressure
model and Eqs. (18.12) and (18.15) for the uniform wear Find: Determine the appropriate values for ro, ri, and P .
model, which are applicable for thrust disk clutches, are also
applicable for thrust disk brakes provided that the disk shape
is similar to that shown in Fig. 18.5. A detailed analysis
of disk brakes gives equations that result in slightly larger
torques than those resulting from the clutch equations. This
text assumes that the brake and clutch equations are identi-
cal.

Cone Clutches and Brakes dr 525
sin α dθ θ
Solution: For a woven material in contact with steel in a
dry environment, Table 18.3 gives the coefficient of friction α dr dA rdθ
as µ = 0.35 and the maximum contact pressure as pmax = 345 dP
kPa = 0.345 MPa. The average coefficient of friction has been
used to estimate average performance; however, the smallest
pressure has been selected for a long life. Making use of the
above and Eq. (18.14) gives

ri ro2 − ri2 = nsTw dw
πµpmax d
(1.5)(150) r
D
=
π(0.35) (0.345 × 106)

= 5.931 × 10−4 m3.

Solving for the outside radius, ro,

ro = 5.931 × 10−4 + ri2. (a) b
ri Figure 18.7: Forces acting on elements of a cone clutch.

The minimum outside radius is obtained by taking the
derivative of the outside radius with respect to the inside ra-
dius and setting it equal to zero:

10−4 dP = p dA. (18.24)

dro = 0.5 − 5.931 × + 2ri = 0, The actuating force is the thrust component, dW , of the nor-
dri ri2 mal force, dP , or
5.931 × 10−4 + ri2
ri

which is numerically solved as ri = 66.69 mm. Therefore, dW = dP sin α = p dA sin α = pr drdθ.
Eq. (a) gives
Using Eq. (18.4) gives the actuating force as

ro = 5.931 × 10−4 + 0.066692 = 0.1155 m = 115.5 mm. D/2
0.06669
W= pr dr dθ = 2π pr dr. (18.25)

d/2

The radius ratio is Similarly, Eq. (18.5) gives the torque as

β = ri = 66.69 = 0.5774. T= µr dP = 2π D/2 (18.26)
ro 115.5
µpr2 dr.
sin α d/2
The radius ratio required to maximize the torque capacity is

the same as the radius ratio required to minimize the outside

radius for a given torque capacity. Thus, the radius ratio for 18.4.1 Uniform Pressure Model

maximizing the torque capacity or for minimizing the out-

side radius is As was discussed in Section 18.3.1, the pressure for a new
thrust disk clutch is assumed to be uniform over the surfaces,
βo = 1 or p = po. Using this uniform pressure model for a cone
= 0.5774. clutch gives the actuating force as
3

From Eq. (18.15), the maximum normal force that can be ap- W = πpo D2 − d2 . (18.27)
plied to the clutch without exceeding the pad pressure con- 4
straint is

P = 2nsTw = 2(1.5)(150) = 7057 N. Similarly, the torque is

µ(ro + ri) (0.35)(0.1155 + 0.06669)

T = 2πpoµ 1 D3 − d3 = πpoµ D3 − d3 . (18.28)
3 sin α 8 12 sin α

18.4 Cone Clutches and Brakes Making use of Eq. (18.27) enables Eq. (18.28) to be rewritten

as µW D3 − d3

Cone clutches use wedging action to increase the normal T = 3 sin α (D2 − d2) . (18.29)
force on the lining, thus increasing the friction force and the
torque. Usually the half cone angle, α, is above 4◦ to avoid 18.4.2 Uniform Wear Model
jamming, and is usually between 6◦ and 8◦. Figure 18.7
shows a conical surface with forces acting on an element. The Substituting Eq. (18.9) into Eq. (18.25) gives the actuating
area of the element and the normal force on the element are

force as D/2

dA = (r dθ) dr , (18.23) W = 2πc dr = πc (D − d) . (18.30)
sin α
d/2

526 Chapter 18 Brakes and Clutches
d4 W
Similarly, substituting Eq. (18.9) into Eq. (18.26) gives the
torque as d3
C
T = 2πµc D/2 πµc D2 − d2 . (18.31) D µP B d1
r dr = P
sin α d/2 4 sin α

d2

Making use of Eq. (18.30) enables Eq. (18.31) to be rewritten ω
as

T = µW (D + d) . (18.32) r
4 sin α

Example 18.2: Cone Clutch Figure 18.8: Block, or short-shoe brake, with two configura-
tions.
Given: A cone clutch similar to that shown in Fig. 18.7 has
the following dimensions: D = 330 mm, d = 306 mm, and From Eq. (18.27), the uniform pressure required is
b = 60 mm. The clutch uses sintered metal on steel, with a
coefficient of friction of 0.26, and the torque transmitted is 4W
200 N-m. pmax = po = π (D2 − d2)

Find: Determine the minimum required actuating force and 4(948.4)
the associated contact pressure by using the uniform pres- = π(0.3302 − 0.3062) Pa
sure and uniform wear models. = 79.11 kPa.

Solution: 18.5 Block or Short-Shoe Brakes

1. Uniform wear: From Fig. 18.7, the half-cone angle of the A block, or short-shoe, brake can be configured to move
cone clutch is radially against a cylindrical drum or plate, as shown in
Fig. 18.8, although other configurations exist. A normal force,
tan α = D − d = 165 − 153 = 12 = 0.200, P , develops a friction force, F = µP , on the drum or plate,
2b 60 60 where µ is the coefficient of friction. The actuating force, W ,
is also shown in Fig. 18.8 along with critical hinge pin dimen-
α = 11.31◦. sions d1, d2, d3, and d4. The normal force, P , and the fric-
tion force, µP , are the forces acting on the brake. For block
The pressure is a maximum when r = d/2. Thus, mak- or short-shoe brakes, a constant pressure is assumed over the
ing use of Eqs. (18.31) and (18.9) gives pad surface. As long as the pad is short relative to the circum-
ference of the drum, this assumption is reasonably accurate.
T = πµdpmax D2 − d2 ,
8 sin α A brake is considered to be self-energizing if the friction
moment assists the actuating moment in applying the brake.
and This implies that the signs of the friction and actuating mo-
ments are the same. Deenergizing effects occur if the friction
8T sin α moment counteracts the actuating moment in applying the
pmax = πµd (D2 − d2) brake. Figure 18.8 can be used to illustrate self-energizing
and deenergizing effects. Note that merely changing the
8(200) sin 11.31◦ cylinder’s direction of rotation will change a self-energizing
= π(0.26)(0.306)(0.3302 − 0.3062) shoe into a deenergizing shoe.
= 82.25 kPa.
Summing the moments about the hinge at C (see
From Eq. (18.32), the actuating force can be written as Fig. 18.8) and setting equal to zero results in

W= 4T sin α = 4(200) sin 11.31◦ = 948.8 N.

µ(D + d) (0.26)(0.330 + 0.306)

2. Uniform pressure: From Eq. (18.29), the actuating force d4W + µP d1 − d3P = 0.
can be expressed as
Since the signs of the friction and actuating moments are the
3T sin α D2 − d2 same, the brake hinged at C is self-energizing. Solving for the
W= normal force gives

µ (D3 − d3) P = d4W . (18.33)
3(200) sin 11.31◦ 0.3302 − 0.3062 d3 − µd1
= (0.26)(0.3303 − 0.3063)
= 948.4 N. The braking torque at C is

T = Fr = µrP = µrd4W , (18.34)
d3 − µd1

Block or Short-Shoe Brakes 527

350 W Example 18.3: Short-Shoe Brake
37.5 P
Given: A 350-mm-radius brake drum contacts a single short
350 shoe, as shown in Fig. 18.9, and sustains 225 Nm of torque at
900 500 rpm. Assume that the coefficient of friction for the drum
and shoe combination is µ = 0.30.
Figure 18.9: Short-shoe brake used in Example 18.3. Dimen-
sions in millimeters. Find: Determine the following:

(a) Normal force acting on the shoe

(b) Required actuating force, W , when the drum has clock-
wise rotation

(c) Required actuating force, W , when the drum has coun-
terclockwise rotation

(d) Required change in the 37.5 mm dimension (Fig. 18.9)
for self-locking to occur if the other dimensions do not
change

where r is the radius of the brake drum. Summing the mo- Solution:
ments about the hinge at D (see Fig. 18.8) and setting the sum (a) The torque of the brake is
equal to zero results in

T = rF = rµP,

−W d4 + µP d2 + d3P = 0. where

Since the signs of the friction and actuating moments are op- P = T = 225 = 2140 N,
posite, the brake hinged at D is deenergizing. Solving for the µr (0.3)(0.350)
normal force gives

P = W d4 . (18.35) and
d3 + µd2 µP = (0.3)(2140) = 643 N.

The torque of the brake hinged at D in Fig. 18.8 is (b) For clockwise rotation, summing the moments about
the hinge pin and setting the sum equal to zero yields

(0.0375)(643) + (0.900)W − (0.350)(2140) = 0, (a)

T = µd4rW . (18.36) which is solved as W = 805 N. Since the signs of the
d3 + µd2 friction and actuating moments are the same, the brake
is self-energizing.
A brake is considered to be self-locking if the actuating force
(W in Fig. 18.8) equals zero for a non-zero torque. Self- (c) For counterclockwise rotation, summing the moments
locking brake geometries are avoided, since they seize or about the hinge pin and setting the sum equal to zero
grab, thus operating unsatisfactorily or even dangerously. gives

Block brakes such as shown in Fig. 18.8 reduce an angu- (0.0375)(643) − (0.900)W + (0.350)(2140) = 0,
lar velocity, and in vehicle applications require a high friction
force to exist between a wheel and road to affect braking. This or W = 859 N. Since the signs of the friction and actu-
does not exist for some conveyor designs, or for vehicles such ating moments are not the same, the brake is deenergiz-
as roller coasters that ride on rails where the friction between ing.
the wheel and rail is very low. In such circumstances, a com-
mon brake uses a vehicle-mounted rectangular pad or fin that (d) If, in Eq. (a), W = 0,
bears against a flange or fin on the moving component. This
is discussed in Case Study 18.1. x(643) + (0.900)W − (0.350)(2140) = 0,

With such a block brake, when a plate or fin approaches or, solving for x,
the brake system, the fins bear against the brake pads and
thereby develop a normal force. Not surprisingly, the lead- x = (0.350)(2140) = 1165 mm
ing edge of the brake pad will wear more than other sec- 643
tions of the brake. Clearly, the brake does not encounter
uniform pressure. However, from a tribological standpoint, Therefore, self-locking will occur if the distance of 37.5
there is little difference between a concentrated load and a mm in Fig. 18.9 is changed to 1.165 m. Since self-locking
distributed load in generating a friction force. As long as is not a desirable effect in a brake and 37.5 mm is quite
the real area of contact between the pad and the fin remains different from 1.165 m, we would not expect the brake
small, a linear relationship exists between the friction and the to self-lock.
normal force. For all practical purposes, this approximation
is reasonable for the entire time of brake-fin contact.

528 Chapter 18 Brakes and Clutches

Brake drum Brake shoe Brake shoe Backing plate
Lining hold-down pin

Wheel cylinder

Return Bleeder valve
springs Brake line

Hold-down spring Brake adjuster

Figure 18.10: A typical automotive long-shoe, internal, expanding rim brake, commonly called a drum brake.

18.6 Long-Shoe, Internal, WW Rotation
Expanding Rim Brakes

Figure 18.10 shows a long-shoe, internal, expanding rim d6 Shoe
brake with two pads or shoes, often referred to as drum θ2
brakes. They are widely used, especially in automobiles, Drum
where they are commonly applied for rear axle braking. d7 θ Lining
As mentioned previously, disc brakes are commonly used
for front brakes because they promote more convective heat θ1
transfer than drum designs, and the front axle encounters rA
more air flow over the brake or disc. This can be accentu-
ated with grooves or holes in the rotor, as shown in this chap- d5 d5 Hinge pin
ter’s opening illustration. Drum brakes are used because they
have a self-energizing shoe, which is very easy to incorporate Figure 18.11: Long-shoe, internal, expanding rim brake with
into a parking brake feature; disc brakes do not have self- two shoes.
energizing shoes, and when all-wheel disc brakes are used, it
is common that a drum brake is built into the rear disc brakes hinge pin. Also, the shoe lining does not begin at θ = 0◦, but
to provide a parking brake feature. at some θ1 and extends until θ2. At any angle, θ, of the lining,
the differential normal force dP is
Figure 18.11 illustrates the shoes and the drum, and de-
fines important dimensions. The hinge pin for the right shoe dP = pbr dθ, (18.38)
is at A in Fig. 18.11. The heel of the pad is the region closest to
the hinge pin, and the toe is the region closest to the actuating where b is the face width of the shoe (the distance perpen-
force, W . A major difference between a short shoe (Fig. 18.8)
and a long shoe (Fig. 18.11) is that the pressure can be con- dicular to the paper). Substituting Eq. (18.37) into Eq. (18.38)
sidered constant for a short shoe but not for a long shoe. In
a long shoe, little if any pressure is applied at the heel, and
the pressure increases toward the toe. This sort of pressure
variation suggests that the pressure may vary sinusoidally.
Thus, a relationship of the contact pressure, p, in terms of the
maximum pressure, pmax, may be written as

sin θ gives
p = pmax sin θa ,
(18.37) dP = pmaxbr sin θ dθ . (18.39)
sin θa

where θa is the angle where pressure is at a maximum value. From Eq. (18.39), the normal force has a moment arm of
Observe from Eq. (18.37) that p = pmax when θ = 90◦. If the d7 sin θ so that its associated moment is
shoe has an angular extent less than 90◦ (such as is the case
in Fig. 18.11 if θ2 was less than 90◦), p = pmax when θa = θ2. MP = d7 sin θ dP = d7brpmax θ2
sin θa
Observe also in Fig. 18.11 that the distance d6 is perpen- sin2 θ dθ
dicular to the actuating force W . Figure 18.12 shows the
forces and critical dimensions of a long-shoe, internal, ex- θ1
panding rim brake. In Fig. 18.12, the θ-coordinate begins with
a line drawn from the center of the drum to the center of the = brd7pmax 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1 ,
4 sin θ1 180◦

(18.40)

Long-Shoe, Internal, Expanding Rim Brakes 529

y d7 sin θ Figure 18.12 shows the reaction forces as well as the friction
dP sin θ force and the normal force. Summing the forces in the x-
µdP dP dP cos θ direction and setting the sum equal to zero results in
θθ
µdP cos θ r – d7 cos θ Rxs + Wx − cos θ dP + µ sin θ dP = 0. (18.45)
Wx x

W µdP sin θ Ry Substituting Eq. (18.39) into Eq. (18.45) gives the reaction
Wy θ2 force in the x-direction for a self-energizing shoe as

θ Rxs = pmax,s br θ2

sin θ cos θ dθ
sin θa θ1
θ1 A
d6 Rx − µpmax,sbr θ2
sin θa
d7 sin2 θ dθ − Wx,
r
θ1

or = −Wx + pmax,s br 2 sin2 θ2 − sin2 θ1
Rxs 4 sin θa

− µpmaxbr 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1 ,
4 sin θa 180◦

(18.46)

Rotation where θ1 and θ2 are in degrees. From Fig. 18.12, summing
the forces in the y-direction and setting the sum equal to zero
Figure 18.12: Forces and dimensions of long-shoe, internal gives
expanding rim brake.

Rys + Wy − µ dP cos θ − dP sin θ = 0, (18.47)

where θ1 and θ2 are in degrees. From Eq. (18.39), the friction and
force has a moment arm of r − d7 cos θ, so that its moment is

MF = (r − d7 cos θ) µ dP Rys = −Wy + 2µpmax,sbr sin2 θ2 − sin2 θ1
4 sin θa

= µpmaxbr θ2 + pmax,sbr 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1 .
sin θ1 4 sin θa 180◦
(r − d7 cos θ) sin θ dθ,

θ1

or (18.48)

MF = − µpmaxbr r (cos θ2 + cos θ1) − d7 sin2 θ2 − sin2 θ1 In Eqs. (18.45) to (18.48), the reference system has its origin
sin θa 2 at the center of the drum. The positive x-axis is taken to be
through the hinge pin. The positive y-axis is in the direction
(18.41) of the shoe.

Long-shoe brakes can be designed so that all, some, or none

of the shoes are self-energizing. Each type of shoe needs to be

analyzed separately. 18.6.2 Deenergizing Shoe

18.6.1 Self-Energizing Shoe If, in Fig. 18.12, the direction of rotation is changed from
clockwise to counterclockwise, the friction forces change di-
Setting the sum of the moments about the hinge pin equal to rection. Thus, summing the moments about the hinge pin
zero results in and setting the sum equal to zero,

−W d6 − MF + MP = 0. (18.42) −W d6 + MF + MP = 0. (18.49)

Since the actuating and friction moments have the same sign The only difference between Eq. (18.42) and Eq. (18.49) is the
in Eq. (18.42), the shoe shown in Fig. 18.12 is self-energizing.
It can also be concluded just from Fig. 18.12 that the shoe is sign of the friction moment. Solving for the actuating force in
self-energizing because Wx and µ dP sin θ are in the same di-
rection. Solving for the actuating force in Eq. (18.42) gives Eq. (18.49) gives

W = MP + MF . (18.50)
d6
W = MP − MF .
d6 (18.43) For a deenergizing shoe, the only changes from the equations
derived for the self-energizing shoe are that in Eqs. (18.46)
From Eqs. (8.5) and (18.39), the braking torque is and (18.48) a sign change occurs for terms containing the co-
efficient of friction, µ, resulting in the following:

T = rµ dP

µpmaxbr2 θ2 Rxd = −Wx + pmax,d br 2 sin2 θ2 − sin2 θ1
4 sin θa
= sin θ dθ
sin θa θ1 + µpmax,dbr π
4 sin θa 2 (θ2 − θ1) 180◦ − sin 2θ2 + sin 2θ1 ,
= µpmaxbr2 (cos θ1 − cos θ2) .
sin θa (18.44) (18.51)

530 Chapter 18 Brakes and Clutches
y
Ryd = −Wy − µpmax,dbr sin2 θ2 − sin2 θ1
2 sin θa 15° 15°

+ pmax,dbr 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1 .
4 sin θa 180◦

(18.52) WW

The maximum contact pressure used in evaluating a self- dd
energizing shoe is taken from Table 18.3. The maximum b
contact pressure used in evaluating a deenergizing shoe is
less than that for the self-energizing shoe, since the actuating 10° a a 10° x
force is the same for both types of shoe.
10° A B 10°
Design Procedure 18.1: Long-Shoe,
Internal, Expanding Brake Analysis rb
dd
It will be assumed that the application has a known veloc-
ity, and that the physical dimensions of the brake are known. WW
This Design Procedure outlines the method used to obtain ω
the maximum allowable brake force (which can be controlled
by design of the hydraulic or pneumatic actuators) and brak- 15° 15°
ing torque.
Figure 18.13: Four-long-shoe, internal expanding rim brake
1. Select a brake material. A reasonable starting point is to used in Example 18.4.
assume the drum is made of steel, using sintered metal
lining material. Table 18.2 then allows estimation of 7. Equations (18.46) and (18.48) can be used to obtain the
maximum allowable contact pressure and friction co- hinge pin reaction forces.
efficient. Table 18.3 also recommends a maximum pres-
sure, but based on thermal conditions. The lower of the 8. In most brakes, the force applied to the self-energizing
two contact pressures should be used for further analy- and deenergizing shoes are the same. However, the
sis. maximum pressure on the deenergizing shoe will be
lower than the self-energizing one. Thus, the applied
2. Draw a free body diagram of the brake shoes, pay- force and pressure can be taken from the self-energizing
ing special attention to the force that acts on the shoes shoe analysis, as this will reflect the higher stress.
due to friction. Identify which of the shoes, if any,
are self-energizing or deenergizing. In a self-energizing 9. Equation (18.50) allows calculation of the maximum
shoe, the moment due to frictional force applied to the pressure on the deenergizing shoe.
shoe will have the same sign as the moment due to
the applied force. If it is not clear that a shoe is self- 10. The torque can be obtained from Eq. (18.45) using the
energizing or deenergizing, then assume the brake is maximum pressure for the deenergizing shoe.
self-energizing in order to be conservative regarding
maximum shoe pressure. In any case, if the friction 11. Equations (18.51) and (18.52) allow calculation of the
moment is close to zero, then the braking torque will hinge pin reaction forces.
be similar whether the brake was assumed to be self-
energizing or deenergizing. Example 18.4: Long-Shoe Internal
Brake
3. Evaluate MP and MF from Eqs. (18.40) and (18.41), re-
spectively. Note that one or more terms may be un- Given: Figure 18.13 shows four long shoes in an internal,
known, but they can be treated as variables. expanding rim brake. The brake drum has a 400-mm inner
diameter. Each hinge pin (A and B) supports a pair of shoes.
4. Consider the self-energizing shoe first. The self- The actuating mechanism is to be arranged to produce the
energizing shoe will encounter a higher pressure than same actuating force W on each shoe. The shoe face width
the deenergizing shoe, so that the limiting pressure de- is 75 mm. The material of the shoe and drum produces a co-
termined above can be used to evaluate MP and MF . efficient of friction of 0.24 and a maximum contact pressure
of 1 MPa. Additional dimensions for use in Fig. 18.13 are as
5. Equation (18.43) can be used to determine the maxi- follows: d = 50 mm, b = 165 mm, and a = 150 mm.
mum braking force. Note that a lower braking force can
be applied, but a higher braking force would exceed the
allowable stress of the lining material, leading to plastic
deformation or compromised brake life. If the braking
force was prescribed, then Eq. (18.43) can be used to ob-
tain the pressure in the shoe, which can be compared to
the maximum allowable pressure obtained previously.

6. Equation (18.44) can be used to obtain the torque for the
self-energizing shoe.

Long-Shoe, Internal, Expanding Rim Brakes 531

Find: MP d = brd7pmax,d
4 sin θa
(a) Which shoes are self-energizing and which are deener-
gizing? × 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1
180◦
(b) What are the actuating forces and total torques for the
four shoes? = (0.075)(0.2)(0.15)pmax,d
4 sin 75◦
(c) What are the hinge pin reactions as well as the resultant
reaction? × 2 (75◦ − 10◦) π − sin 150◦ + sin 20◦
180◦
Solution:
= 0.001229pmax,d,
(a) With the drum rotating in the clockwise direction
(Fig. 18.13) the top-right and bottom-left shoes have MF d = µpmax,dbr [−r (cos θ2 − cos θ1)
their actuating and friction moments acting in the same sin θa
direction. Thus, they are self-energizing shoes. The top-
left and bottom-right shoes have their actuating and − d7 sin2 θ2 − sin2 θ1
friction moments acting in opposite directions. Thus, 2
they are deenergizing shoes.
= 0.0002888pmax,d.
(b) The dimensions given in Fig. 18.13 correspond to the
dimensions given in Figs. 18.11 and 18.12 as d5 = 50 mm, From Eq. (18.50) the actuating load for the deenergizing
d6 = 165 mm, and d7 = 150 mm. Also, since θ2 < 90◦, shoes is
then θ2 = θa. Because the hinge pins in Fig. 18.13 are at
A and B, θ1 = 10◦ and θ2 = θa = 75◦. Wd = Ws = 5698 N = MP d + MF d
d6
Self-energizing shoes: Making use of the above and
Eq. (18.40) gives the normal force moment as 0.001229 + 0.0002888

= 0.165 pmax,d .

MP s = brd7pmax,s
4 sin θa
Solving for the maximum pressure yields pmax,d =
× 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1 0.6194 MPa. The braking torque for the deenergizing
180◦ shoes is

(0.075)(0.2)(0.15) 1 × 106 Td = Ts pmax,d = 541.2 0.6194 = 335.2 N-m.
= 4 sin 75◦ pmax,s 1.000

× 2 (75 − 10) π − sin 150◦ + sin 20◦
180◦

= 1229 N-m. The total braking torque of the four shoes, two of which
are self-energizing and two of which are deenergizing,
From Eq. (18.41), the friction moment is is

− µpmax,sbr T = 2 (Ts + Td) = 2(541.2 + 335.2) = 1753 N-m.
sin θa
MF s = [r (cos θ2 − cos θ1)

+ d7 sin2 θ2 − sin2 θ1 (c) The hinge reactions are obtained from Eqs. (18.46) and
2 (18.48) for the self-energizing and deenergizing shoes,
respectively. First, for the self-energizing shoes:
= 288.8 N-m.
From Eq. (18.46),

From Eq. (18.43), the actuating force for both the self- Rxs = −Wx + pmax,s br 2 sin2 θ2 − sin2 θ1
energizing and deenergizing shoes is 4 sin θa

Ws = Wd = MP − MF = 1229 − 288.8 = 5698 N. −µ µpmaxbr
d6 0.165 4 sin θa

From Eq. (18.44), the braking torque for each self- × 2 (θ2 − θ1) π − sin 2θ2 + sin 2θ1 ,
energizing shoe is 180◦

= −654.6 N

Ts = µpmax,sbr2 (cos θ1 − cos θ2) From Eq. (18.48)
sin θa
(1 × 106)(0.075)(0.2)
(0.24) 0.075 × 106 (0.2)2(cos 10◦ − cos 75◦) Rys = 4 sin 75◦
= sin 75◦
π
= 541.2 N-m. × 2(75◦ − 10◦) 180◦ − sin 150◦ + sin 20◦

Deenergizing shoes: The only change in the calculation +2(0.24) sin2 75◦ − sin2 10◦
of the normal and friction moments for the deenergiz-
ing shoes is the maximum pressure. This is unknown, = 9878 N.
but using Eqs. (18.40), (18.41) and (18.50) will allow its
determination. MP d and MF d are obtained as Similarly, Rxd = −137.5 N.

532 Chapter 18 Brakes and Clutches
Wx
mation of Eq. (18.42), the actuating and friction moments
Wy W have the same sign. The external brake shoe in Fig. 18.14 is
y deenergizing for clockwise rotation. Summing the moments
and equating the sum to zero

µdP sinθ d6

µdP θ µdP cosθ

dP W d6 − MF − MP = 0. (18.53)

θ2 θ dP sinθ
dP cosθ

θ1 Rx x The actuating and friction moments have opposite signs and
θ thus the external brake shoe shown in Fig. 18.14 is deenergiz-
A ing.
d7
r Ry If, in Figs. 18.12 and 18.14, the direction of rotation were
changed from clockwise to counterclockwise, the friction mo-
ments in Eqs. (18.42) and (18.53) would have opposite signs.
Therefore, the internal brake shoe would be deenergizing and
the external brake shoe would be self-energizing.

Rotation Example 18.5: External Long-Shoe
Brake
Figure 18.14: Forces and dimensions of long-shoe, external,
contracting rim brake. Given: An external, long-shoe rim brake is to be cost opti-
mized. Three lining geometries are being considered, cover-
Deenergizing shoes: From Eq. (18.52), ing the entire 90◦ of the shoe, covering only 45◦ of the central
portion of the shoe, or covering only 22.5◦ of the central por-
Ryd = (0.6194)(1 × 106)(0.075)(0.2) tion of the shoe. The braking torque must be the same for all
4 sin 75◦ three geometries, and the cost of changing any of the brake
linings is half of the cost of a 22.5◦ lining. The cost of the lin-
× 2(75◦ − 10◦ ) π − sin 150◦ + sin 20◦ ing material is proportional to the wrap angle. The wear rate
180◦ is proportional to the pressure. The input parameters for the
90◦ lining are d7 = 100 mm, r = 80 mm, b = 25 mm, θ1 = 0◦,
−2(0.24)(sin2 75◦ − sin2 10◦) θ2 = 90◦, µ = 0.27, and T = 125 N-m.

= 4034 N. Find: Which of the wrap angles (90◦, 45◦, or 22.5◦) would be
the most economical?
The resultant forces of the reactions in the hinge pin in the
horizontal and vertical directions are Solution: The braking torque is given by Eq. (18.44) and is
the same for all three geometries. For the 90◦ wrap angle
(θ1 = 0◦, θ2 = 90◦, and θa = 90◦):

RH = −654.6 − 137.5 = −792.1 N, T sin θa
µbr2(cos θ1 − cos θ2)
RV = 9878 − 4034 = 5844 N. (pmax )90◦ =
The resultant force at the hinge pin is
(125) sin 90◦
= (0.27)(0.025)(0.08)2(cos 0◦ − cos 90◦)

R = RH2 + RV2 = (−792.1)2 + (5844)2 = 5897 N. = 2.894 × 106 Pa

= 2.894 MPa.

18.7 Long-Shoe, External, For the 45◦ wrap angle (θ1 = 22.5◦, θ2 = 67.5◦, and θa =
67.5◦):
Contracting Rim Brakes
(125) sin 67.5◦
Figure 18.14 shows the forces and dimensions of a long-shoe, (pmax)45◦ = (0.27)(0.025)(0.08)2(cos 22.5◦ − cos 67.5◦)
external, contracting rim brake. In Fig. 18.12, the brake is in-
ternal to the drum, whereas in Fig. 18.14 the brake is exter- = 4.940 × 106 Pa
nal to the drum. The symbols used in these figures are simi- = 4.940 MPa.
lar. The equations developed in Section 18.6 for internal shoe
brakes are exactly the same as those for external shoe brakes For the 22.5◦ wrap angle (θ1 = 33.75◦, θ2 = 56.25◦, and
as long as one properly identifies whether the brake is self- θa = 56.25◦):
energizing or deenergizing.
(125) sin 56.25◦
The internal brake shoe in Fig. 18.12 was found to be self- (pmax)22.5◦ = (0.27)(0.025)(0.08)2(cos 33.7◦ − cos 56.2◦)
energizing for clockwise rotation, since in the moment sum-
which is solved as 8.720 MPa.

Symmetrically Loaded Pivot-Shoe Brakes 533

The cost of changing the lining is C. The lining costs are 2C y
for a 22.5◦ lining, 4C for a 45◦ lining, and 8C for a 90◦ lining. µdP sin θ
The wear rate is proportional to the pressure, or the time it
takes for the shoe to wear out is inversely proportional to the Rotation µdP µdP cos θ
pressure. Thus, the times it takes for the shoe to wear out for r
the three geometries are d7 cos θ – r
Rx
t90◦ = A = A = 3.455 × 10−7A, dP dP sin θ
(pmax )90◦ (2.894 × 106) Ry
θ
where A is a constant independent of geometry. Similarly, θ2 dP cos θ x

t45◦ = A = A = 2.024 × 10−7A, θ1
(pmax )45◦ (4.940 × 106) r cos θ

t22.65◦ = A = A = 1.147 × 10−7A.
(pmax )22.5◦ (8.720 × 106)

The costs per unit time for the three geometries are:

90◦ wrap angle: d7
Figure 18.15: Symmetrically loaded pivot-shoe brake.
8C + C 26.05 × 106 C.
(3.455 × 10−7) A = A

45◦ wrap angle: the friction moment, when set equal to zero, is

4C + C 24.70 × 106 C. θ2 (18.56)
(2.024 × 10−7) A = A
MF = 2 µ dP (d7 cos θ − r) = 0.
22.5◦ wrap angle:
0

Substituting Eq. (18.55) into Eq. (18.56) gives

2C + C 26.16 × 106 C. θ2 d7 cos2 θ − r cos θ = 0.
(1.147 × 10−7) A = A
2µpmaxbr

0

The 45◦-wrap-angle shoe gives the lowest cost, 5.6% lower This reduces to
than the 22.5◦-wrap-angle shoe and 5.2% lower than the 90◦-
wrap-angle shoe. d7 = 4r sin θ2 . (18.57)
π
2θ2 180◦ + sin 2θ2

18.8 Symmetrically Loaded Pivot- This value of d7 produces a friction moment equal to zero
Shoe Brakes (MF = 0). The braking torque is

θ2

T = 2 rµ dP

Figure 18.15 shows a symmetrically loaded pivot-shoe brake. 0
The major difference between the internal and external rim
brakes considered previously and the symmetrically loaded = 2µr2bpmax θ2
pivot-shoe brake shown in Fig. 18.15 is the pressure distribu-
tion around the shoe. Recall from Eq. (18.37) for the internal cos θ dθ
rim brake that the maximum pressure was at θ = 90◦ and the
pressure distribution from the heel to the top of the brake was 0
sinusoidal. For the symmetrically loaded pivot-shoe brake
(Fig. 18.15), the maximum pressure is at θ = 0◦, which sug- = 2µr2bpmax sin θ2. (18.58)
gests the pressure variation is
Note from Fig. 18.15 that, for any x, the horizontal friction
p = pmax cos θ = pmax cos θ. (18.54) force components of the upper half of the shoe are equal and
cos θa opposite in direction to the horizontal friction force compo-
nents of the lower half of the shoe. For a fixed x, the horizon-
tal normal components of both halves of the shoe are equal
and in the same direction, so that the horizontal reaction force
is

For any angular position from the pivot, θ, a differential nor- Rx = 2 θ2 dP cos θ = pmaxbr 2θ2 π + sin 2θ2 .
mal force dP acts with a magnitude of 02 180◦ (18.59)

dP = pbr dθ = pmaxbr cos θ dθ. (18.55) Making use of Eq. (18.57) gives

The design of a symmetrically loaded pivot-shoe brake is Rx = 2br2pmax sin θ2 . (18.60)
such that the distance d7, which is measured from the center d7
of the drum to the pivot, is chosen so that the resulting fric-
tion moment acting on the brake shoe is zero. From Fig. 18.15, For a fixed y, the vertical friction force components of the up-
per half of the shoe are equal and in the same direction as