Tài liệu Handbook of Mechanical Engineering Calculations P23 pdf

23.1
SECTION 23
SPRING SELECTION AND
ANALYSIS
Proportioning Helical Springs by
Minimum Weight 23.1
Determining Safe Torsional Stress for a
Helical Spring 23.3
Designing a Spring with a Specific
Number of Coils 23.4
Determining Spring Dimensions Before
and After Load Applications 23.5
Analysis and Design of Flat Metal
Springs 23.6
Sizing Torsional Leaf Springs 23.7
Designing a Spring to Fire a Projectile
23.12
Spring Support of Machinery to Control
Vibration Forces 23.13
Spring Selection for a Known Load and
Deflection 23.14
Spring Wire Length and Weight 23.15
Helical Compression and Tension Spring
Analysis 23.16
Selection of Helical Compression and
Tension Springs 23.17
Sizing Helical Springs for Optimum
Dimensions and Weight 23.19
Selection of Square- and Rectangular-
Wire Helical Springs 23.20
Curved Spring Design Analysis 23.21

Round- and Square-Wire Helical Torsion-
Spring Selection 23.23
Torsion-Bar Spring Analysis 23.26
Multirate Helical Spring Analysis 23.27
Belleville Spring Analysis for Smallest
Diameter 23.28
Belleville Spring Computations for Disk
Deflection, Load, and Number 23.30
Ring-Spring Design Analysis 23.32
Liquid-Spring Selection 23.34
Selection of Air-Snubber Dashpot
Dimensions 23.37
Design Analysis of Flat Reinforced-
Plastic Springs 23.39
Life of Cyclically Loaded Mechanical
Springs 23.42
Shock-Mount Deflection and Spring
Rate 23.43
PROPORTIONING HELICAL SPRINGS BY
MINIMUM WEIGHT
The detent helical spring in Fig. 1 is to be designed so that force P
1
for the extended
condition is to be 20 lb (88.9 N). After an additional compression of 0.625 in (1.59
cm) the shearing stress in the spring is to be 75,000 lb/in
2
(517 MPa). The spring
index, c, is approximately 8, based on past experience, and the modulus of elasticity
in shear is 11,500,000 lb/in
2

(79.2 GPa). Find the diameter of the wire, helix radius,
and number of active coils for the smallest amount of material in the spring.
Calculation Procedure:
1. Find the diameter of the spring wire
A spring which will contain the smallest possible amount of material will have
design parameters selected so that the maximum force, stress, and deflection are
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Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS
23.2 DESIGN ENGINEERING
FIGURE 1 Spring-loaded de-
tent. (Product Engineering.)
exactly twice the minimum force, stress, and deflection, respectively. With such
parameters, the spring wire diameter,
16Pc
1
d ϭ
Ί
␲
(S )
s max
where d ϭ spring wire diameter, in (cm); P
1
ϭ force in the extended condition, lb
(N); c
ϭ spring index, dimensionless;
␲
ϭ 3.1416; (S
s

) ϭ shear stress in the
max
spring wire, lb/in
2
(kPa). Substituting, d ϭ (16 ϫ 20 ϫ 8 / 75,000
␲
) ϭ 0.1042 in
0.5
(0.265 cm). Referring to a spring wire table, choose No. 12 wire with d ϭ 0.1055
in (0.268 cm) as the nearest commercially available wire size.
2. Compute the mean radius of the spring helix
The helix mean radius,
3
␲
d (S )
s max
R ϭ
32P
1
where R ϭ mean radius of the helix, in (cm); other symbols as defined earlier.
Substituting, R
ϭ (
␲
ϫ 0.1055
3
ϫ 75,000)/32 ϫ 20 ϭ 0.4323 in (1.09 cm).
3. Determine the number of active coils in the spring
Use the relation, N
ϭ number of active coils in the spring,
(

␦
Ϫ
␦
) dG
21
N ϭ
2
2
␲
R (S )
s max
where,
␦
2
ϭ maximum compression of spring, in (cm);
␦
1
ϭ minimum compression
of the spring, in (cm); G
ϭ modulus of elasticity in shear; other symbols as before.
Substituting, N
ϭ (0.625 ϫ 0.1055 ϫ 11,500,000) / (2
␲
ϫ 0.4323
2
ϫ 75,000 ϭ 8.61
active coils.
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.3
4. Find the volume of the spring
Volume of a helical spring is given by V
ϭ 0.5 ϫ
␲
2
ϫ d
2
ϫ RN, where the
symbols are as given earlier. Substituting, V
ϭ 0.5 ϫ 3.14
2
ϫ 0.1055
2
ϫ 0.4323
ϫ 8.61 ϭ 2.044 in
3
(30.0335 dm
3
).
Related Calculations. Because of the many variables involved, the proportion-
ing of a helical spring is usually done by trial and error. An unlimited number of
springs can be obtained that will satisfy a given set of equations for stress and
deflection. However, all springs found this way will contain more material than
necessary unless the conditions for the mathematical minimum are fulfilled.
The functioning of a helical spring usually occurs when the spring is extended,
and is least able to exert a force. For example, in the indexing mechanism shown
in Fig. 1, when the ball is in the detent, the holding power of the mechanism
depends on the force exerted by the spring. After relative motion between the parts

occurs, the ball is out of the detent and the additional force from the increased
compression serves no useful purpose and may be a disadvantage. In this condition
the shearing stress in the wire is usually the governing factor.
Many similar applications arise in which the design is controlled by (a) the
spring force in the extended condition and (b) the shearing stress for the most
compressed condition. It has been proved for this instance, that the spring will
contain the smallest possible amount of material when the design parameters are
selected so that the maximum force, stress, and deflection are exactly twice the
minimum force, stress, and deflection, respectively.
Although not directly specified, the designer can usually estimate a suitable value
for the spring index, c. Taking the maximum load as twice the minimum load, the
wire-diameter, d, is computed using the equation given in step 1, above. Next the
helix radius, R, is computed using the d value from step 1. Then the number of
active coils is computed, step 3. Lastly, the minimum volume of the spring is
determined, as in step 4.
This procedure is the work of M. F. Spotts, Northwestern Technological Institute,
as reported in Product Engineering magazine.
DETERMINING SAFE TORSIONAL STRESS FOR A
HELICAL SPRING
The load on a helical spring is 1600 lb (726.4 kg) and the corresponding deflection
is to be 4 in (10.2 cm). Rigidity modulus is 11
ϫ 10
6
lb/in
2
(75.8 GPa) and the
maximum intensity of safe torsional stress is 60,000 lb/in
2
(413.4 MPa). Design
the spring for the total number of turns if the wire is circular in cross section with

a diameter of 0.625 in (1.5875 cm) and a centerline radius of 1.5 in (3.81 cm).
Calculation Procedure:
1. Find the number of active coils in the spring
Use the relation y
ϭ N(64)( P)(r
3
)/G(d
4
), where y ϭ total deflection of the spring,
in (cm); N
ϭ number of active coils in the spring; P ϭ load on spring, lb (N); r
ϭ radius as axis to centerline of wires, in (cm); G ϭ modulus of elasticity of spring
material in shear, lb/in
2
(kPa); d ϭ spring wire diameter, in (cm). Solving for the
number of active coils, N
ϭ y(G)(d
4
)/64(P)(r
3
). Or N ϭ 4(11 ϫ 11
6
)(0.625
4
)/
64(1600)(1.5
3
) ϭ 19.4 active coils.
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SPRING SELECTION AND ANALYSIS
23.4 DESIGN ENGINEERING
2. Check for the safe limit of torsional stress in the spring
Use the relation S,
ϭ 16(P)(r)/
␲
(d
3
). Solving, S ϭ 16(1600)(1.5)/
␲
(0.625
3
) ϭ
50,066 lb/in
2
(344.95 MPa). This torsional stress is less than the safe limit of 60,000
lb/in
2
(413.4 MPa) chosen for this spring. Hence, the spring is acceptable.
The total number spring turns for two inactive coils (one at each end) is N
ϩ 2
ϭ 19.4 ϩ 2 ϭ 21.4, or 22 coils.
Related Calculations. Use this general approach for any helical spring you
wish to analyze for safe torsional stress.
DESIGNING A SPRING WITH A SPECIFIC
NUMBER OF COILS
A coiled spring with an outside diameter of 1.75 in (4.45 cm) is required to work
under a load of 140 lb (622.7 N) and have seven active coils with the ends closed
and round. Determine unit deflection, total number of coils, and length of spring

when under load when G
ϭ 12 ϫ 10
6
lb/in
2
(82.7 GPa) and mean spring diameter
(outside diameter
Ϫ wire diameter) is to be 0.779 in (1.98 cm).
Calculation Procedure:
1. Find the safe shearing stress in the spring
Use the relation, S
t
ϭ 8PD/
␲
d
3
, where S ϭ fiber stress in shear, lb/in
2
(kPa); P
ϭ axial load on spring, lb (N); D ϭ mean diameter of spring as defined above, in
(cm); d
ϭ diameter of spring wire, in (cm). Substituting, S ϭ 9(140)(0.779)(2)/
␲
(0.192
3
) ϭ 78,475 lb /in
2
(540.7 MPa).
2. Determine the deflection of the spring
Use the relation, y

ϭ 4(
␲
)(N)(r
2
)(S)/Gd, where y ϭ total deflection, in (cm); N ϭ
number of active coils; G ϭ rigidity modulus of the spring material in shear, lb /
in
2
(kPa); other symbols as defined earlier. Substituting, y ϭ 4(
␲
)(7)(0.779
2
)
(78,475)/12
ϫ 10
6
(0.192) ϭ 1.818 in (4.6 cm), or about 1
13
⁄
16
in.
3. Compute the total number of coils in the spring
For the ends of a spring to be closed and ground smooth, 1.5 coils should be taken
as inactive. In compression springs the number of active coils depends on the style
of ends, as follows: open ends, not ground—all coils are active; open ends,
ground—0.5 coil inactive; closed ends, not ground—1 coil inactive; closed ends,
ground—1.5 coils inactive; squared ends, ground—2 coils inactive. Since this
spring is to have ends that are closed and round; i.e., closed and ground, use 1.5
inactive coils. Hence, the total number of coils will be 7
ϩ 1.5 ϭ 8.5, say 9 coils.

4. Find the length of the spring when under load
The solid height of the spring when it is entirely compressed is 9 coils
ϫ 0.192
ϭ 1.728 in (4.39 cm), say 1
3
⁄
4
in.
The total deflection under load is 1
13
⁄
16
in (4.6 cm). Using a total free space
between coils of 1 in (2.54 cm), the free length of the spring will be 1
3
⁄
4
ϩ 1
13
⁄
16
ϩ 1 ϭ 4
9
⁄
16
in (11.59 cm). Hence, the length of the spring when under load ϭ 4
9
⁄
16
Ϫ 1

13
⁄
16
ϭ 2
3
⁄
4
in (6.98 cm).
Related Calculations. The procedure given here is valid for sizing any spring
when it must have a specified number of coils.
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.5
DETERMINING SPRING DIMENSIONS BEFORE
AND AFTER LOAD APPLICATION
A cylindrical helical spring of circular cross-section wire is to be designed to safely
carry an axial compressive load of 1200 lb (5338 N) at a maximum stress of
110,000 lb/in
2
(757.9 MPa). The spring is to have a deflection scale of approxi-
mately 150 lb/in (262 N/cm). Spring proportions are to be: Mean diameter of
coil/diameter of wire
ϭ 6 to 8; spring length when closed/mean diameter of coil
ϭ 1.7 to 2.3. Determine (a) mean diameter of coil; (b) diameter of wire; (c) length
of coil when closed, and (d ) length of coil before load application. Use G
ϭ rigidity
modulus of steel in shear
ϭ 11.5 ϫ 10

6
(79,235 MPa).
Calculation Procedure:
1. Find the trial wire size
In this trial design we can assume the ratio of D / d to be 6:8, as given. Then d
ϭ
D/7, using the midpoint ratio, where D ϭ mean diameter of spring coil/diameter
of spring wire. Use the relation, D
ϭ S(
␲
)(d
3
)/8(P), where S ϭ spring stress, lb/
in
2
(kPa); D ϭ mean diameter of the spring ϭ outside diameter Ϫ wire diameter,
in (cm); d
ϭ diameter of spring wire, in (cm). Substituting, D ϭ
(110,000)(3.1416)[(D/7)
3
]/8(1200); D ϭ 4.45 in (2.02 cm).
The wire diameter is d
ϭ 4.45/7 ϭ 0.635 in (1.6 cm). The nearest wire standard
wire size is 0.6666 in (1.69 cm). Solving for D / d
ϭ 4.45 / 0.6666 ϭ 6.68. This lies
within the limits of 6 to 8 set for this spring for D /d.
2. Determine the spring scale
Use the relation P/y
ϭ G(d
4

)/N(64)(r
3
), where y ϭ total deflection of the spring,
in (cm); N
ϭ number of coils; other symbols as before. For this spring, P / y ϭ
150. Substituting, 150 ϭ (11.5 ϫ 10
6
)(0.625
4
)/N(64)(2.38
3
). Solving, N ϭ 13.6
active coils.
Total turns for this spring, assuming closed ends ground
ϭ 13.6 ϩ 1.5 ϭ 15.1.
3. Evaluate the length closed/mean diameter ratio
This ratio must be between 1.7 and 2.3 if the spring is to meet its design require-
ments. Using the relation, (d)(total turns) / D, we have (15.5)(0.6666)/4.45
ϭ 2.26,
which is within the required limits. Then, the length of the closed coil
ϭ 15.1 ϫ
0.6666 ϭ 10.065 in (25.56 cm).
4. Find the coil length before load application
To find the length of the spring coil before the load is applied we add the closed
length of the coil to the total deflection of the spring. To find the total deflection,
use the relation, y
ϭ N(64)(P)(2.38
3
)/11.5 ϫ 10
6

(0.6666
4
); or y ϭ
13.6(64)(1200)(2.38
3
)/11.5 ϫ 10
6
(0.6666
4
) ϭ 6.2 in(15.75 cm). Then, the length
of the coil before load application
ϭ 10 ϩ 6.2 ϭ 16.2 in (41.1 cm).
Related Calculations. The steps in this procedure are useful when designing
a spring to meet certain preset requirements. Using the given steps you can develop
a spring meeting any of several dimensional or stress parameters.
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SPRING SELECTION AND ANALYSIS
23.6 DESIGN ENGINEERING
ANALYSIS AND DESIGN OF FLAT METAL
SPRINGS
Determine the width and thickness of the leaves of a six-leaf steel cantilever spring
13-in (33-cm) long to carry a load of 375 lb (1668 N) with a deflection of 1.25 in
(3.2 cm). The maximum stress allowed in this spring is 50,000 lb / in
2
(344.5 MPa);
the modulus of elasticity of the steel spring material is 30
ϫ 10
6

lb/in
2
(206,700
MPa).
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.7
Calculation Procedure:
1. Find the leaf thickness for this steel spring
See the list of equations below. Knowing the deflection, we can use the equation,
F
ϭ S(l
2
)/E(t), where F ϭ deflection, in (cm); S ϭ safe tensile stress in the spring
metal, lb / in
2
(kPa); l ϭ spring length, in (cm); E ϭ modulus of elasticity as given
above; t
ϭ spring leaf thickness in (cm). substituting, 1.25 ϭ 50,000(13
2
)/30 ϫ
10
6
(t); t ϭ 0.225 in (0.57 cm).
W
ϭ save load or pull, lb (N)
F
ϭ deflection at point of application, in (cm)

S
ϭ safe tensile stress of material, lb /in
2
(kPa)
E
ϭ modulus of elasticity, 30 ϫ 10
6
for steel (kPa)
Other symbols as shown above.
2. Determine the width of each leaf in the spring
Use the relation, W
ϭ S(N )(b)(t
2
)/6(l ), where the symbols are as defined above
and W
ϭ safe load or pull, lb (N). Substituting, 375 ϭ 50,000(6)(b)(.225
2
)/6(13);
b
ϭ 1.93 in (4.9 cm).
Related Calculations. Use the spring layout and equations above to design any
of the three types of metal springs shown. Flat metal springs find wide use in a
variety of mechanical applications. The three types shown above—flat parallel, flat
triangular, and flat leaf—are the most popular today. Using the data given here,
engineers can design a multitude of flat springs for many different uses.
SIZING TORSIONAL LEAF SPRINGS
Size a torsional leaf spring, Fig. 2, for a 38-in ϫ 38-in (96.5-cm ϫ 96.5-cm) work
platform weighing 145 lb (65.8 kg) which requires an assist spring to reduce the
lifting force that must be exerted by a human being. A summation of the moments
about a line through both platform pivots, Fig. 3, shows that a lifting force of only

23 lb (102.3 N) is needed if a torsional spring provides an assist moment of M
t
ϭ
1880 lb ⅐ in (212.4 N ⅐ m). An assist moment is required to raise the platform from
either the stowed or the operating position. As the platform is rotated from one
position to the other, the assist spring undergoes a load reversal. The required spring
wind-up angle,
␪
ϭ
␲
/2 radians; the spring length ϭ 36 in (91.4 cm). AISI 4130
alloy steel heat-treated to 31 Rc minimum with an allowable shear stress of 68,000
lb/in
2
(468.5 MPa) and a shear modulus of G ϭ 11.5 ϫ 10
6
lb/in
2
(79,235 MPa)
is to be used for this spring. Size the spring, determining the required dimensions
for blade thickness, number of blades, aspect ratio, and blade width.
Calculation Procedure:
1. Estimate the blade thickness
Use the equation
b
␶
␤
ϭϭ ,
l 2G
␪

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SPRING SELECTION AND ANALYSIS
23.8 DESIGN ENGINEERING
FIGURE 2 Torsional leaf springs normally are twisted to a specific wind-up angle. This
two-blade spring is wound at right angles. (Machine Design.)
FIGURE 3 Work platform requires an assist spring to reduce required lifting force. (Machine
Design.)
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.9
where the symbols are as given below. Solving for the blade thickness using the
given data, b
ϭ (68.000)(36) /2(11.5 ϫ 10
6
)(
␲
/2) ϭ 0.068 in (0.172 cm). Note that
this is the blade half-thickness.
To keep manufacturing costs low, a stock plate thickness would be chosen. The
nearest stock plate thickness, 2b
ƒ
, is 0.125 in (0.3175 cm). So b
ƒ
ϭ 0.125/2 ϭ
0.0625 in (0.159 cm), which is less than b ϭ 0.068 in (0.172 cm) the computed
required blade half-thickness.

Nomenclature
a ϭ blade half-width, in (cm)
a
ƒ
ϭ selected blade half-width, in (cm)
a
r
ϭ required blade half-width, in (cm)
b
ϭ blade half-thickness, in (cm)
b
ƒ
ϭ selected blade half-thickness, in (cm)
F
ϭ torsional stress factor, 1 / in
3
(1/cm
3
)
ƒ
ϭ normalized torsional stress factor
G ϭ shear modulus, lb / in
2
(kPa) R
0
ϭ estimated aspect ratio
K
ϭ torsional stiffness factor, in
4
(cm

4
) R
r
ϭ required aspect ratio
k
ϭ normalized torsional stiffness s ϭ normalized shear stress
factor
␣
ϭ normalized half-width
l
ϭ spring length, in (cm)
␤
ϭ normalized half-thickness
M
t
ϭ applied moment, lb ⅐ in (N ⅐ m)
␪
ϭ spring wind-up angle, rad
m
t
ϭ normalized applied moment
␶
ϭ torsional stress, lb / in
2
(kPa)
n
ϭ number of blades
␶
a
ϭ allowable torsional stress, lb / in

2
(kPa)
n
f
ϭ selected number of blades
R
ϭ aspect ratio
2. Select the number of blades for this torsion leaf spring
Use the equation,
3l(M /
␪
)
t
nR ϭ
4
16Gb
ƒ
where the symbols are as given above. Substituting, nR ϭ 3(36)[1880 / (
␲
/2)]/
16(11.5
ϫ 10
6
)(0.0626
4
) ϭ 46.
Next, use the equation,
nR
4 Ͻ R ϭϽ15,
0

n
f
letting n
f
ϭ 5; then R
0
ϭ 46/5 ϭ 9.2.
3. Choose the aspect ratio
The required aspect ratio, R
r
, is established by employing the computed value for
R
0
in step 2 as an initial estimate in the equation
lM
t
ͩͪͩͪ
G
␪
R ϭ
i
ϩ
1
16 3.36 1
4
nb Ϫ 1 Ϫ
ͫͩͪͬ
ƒƒ
4
3 R 12R

ii
where i ϭ 0,1,2,
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SPRING SELECTION AND ANALYSIS
23.10 DESIGN ENGINEERING
Usually, after three or so iterations, computed values R converge to a single value
i
ϩ
1
which is the required aspect ratio, R
r
. Substituting, using R
0
ϭ 9.2, as computed in
step 2, for R
1
,
36 1880
ͩͪͩͪ
6
11.5 ϫ 10
␲
/2
R ϭ
1
16 3.36 1
4
5(0.0625 ) Ϫ 1 Ϫ

ͫͩ ͪͬ
4
3 9.2 12(9.2 )
49.108
ϭ
16 3.36 1
Ϫ 1 Ϫ
ͩͪ
4
3 9.2 12(9.2 )
ϭ 9.88
Using the R
1
value of 9.88 found above, substitute it in the equation again,
49.108
R ϭ
2
16 3.36 1
Ϫ 1 Ϫ
ͩͪ
4
3 9.88 12(9.88 )
ϭ 9.83 ϭ R
r
Further iterations show that the aspect ratio converges to a value of 9.83. Therefore,
let R
r
ϭ 9.83.
4. Determine the blade width
Given R

r
ϭ 9.83 and 2b
ƒ
ϭ 0.125 in (0.3175 cm), the required blade width, 2a,is
found from 2a
r
ϭ R
r
(2b
ƒ
). Or, 2a
r
ϭ (9.83)(0.125) ϭ 0.123 in (0.31 cm). For
manufacturing simplicity, the blades will be cut to a 1.25-in (3.18-cm) width from
stock 0.125-in (3.18-cm) thick plates. Therefore, let 2a
ƒ
ϭ 1.25 in (3.18 cm).
5. Check the assist moment and stress
The assist spring has been sized to have five 36-in (91.4-cm)-long blades with a
0.125-in (3.18-cm) wide by 0.125-in (3.18-cm) thick cross section. The assist mo-
ment provided by such a spring stack is determined from
KG
␪
M ϭ
t
l
4
nG b b
16
ƒƒƒ

3
ϭ ab Ϫ 3.36 1 Ϫ
␪
ͫͩͪͬ
ƒƒ
4
l 3 a 12a
ƒƒ
Ϸ M
t required
Substituting,
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.11
6
5(11.5 ϫ 10 )
3
M ϭ (0.625)(0.0625 )
t
36
4
16 0.0625 0.0625
␲
ϫϪ3.36 1 Ϫ
ͫͩͪͩ ͪͬ
4
3 0.625 12(0.625 ) 2
ϭ 1913 lb ⅐ in

Ϸ M ϭ 1880 lb ⅐ in (212.4 N ⅐ m)
t required
Because the actual assist moment of 1913 lb ⅐ in (216.2-N ⅐ m) is slightly greater
than the required moment of 1880 lb
⅐ in (212.4 N ⅐ m), the lifting force needed will
be less than 23 lb (102.3 N).
6. Find the shear stress in each blade
Use the equation,
␶
ϭ FM
t
3a ϩ 1.8b
1
ƒƒ
ϭ M
ͩͪ
t
22
n 8ab
ƒƒƒ
Ͻ
␶
a
to determine the shear stress in each blade. Substituting,
1 3(0.625)
ϩ 1.8(0.0625)
␶
ϭ (1913)
ͫͬ
22

5 8(0.625 )(0.0625 )
22
ϭ 62,300 lb / in Ͻ
␶
ϭ 68,000 lb/in
a
Since the blade shear stress of 62,300 lb / in
2
(429.2 MPa) is less than the allowable
shear stress of 68,000 lb / in
2
(468.5 MPa), this leaf spring can be used as sized
here.
Related Calculations. Leaf springs, in most applications where they transmit
forces, are subjected primarily to bending loads. Design of such springs normally
is based upon cantilever beam theory, which has been well-documented in the
literature on structures and machine elements.
It is not widely known that leaf springs also can be designed to twist. When so
used, leaf springs can generate high torques while occupying a small space. Such
torsional leaf springs are most commonly used as assist devices on hatches, doors,
and folding platforms or walkways. Typical applications are on a variety of vehicles
in which space is restricted. The procedure presented here applies to blades having
identical dimensions, the usual configuration.
A manual available from the Society of Automotive Engineers provides design
techniques for leaf springs for vehicular applications. The SAE manual describes a
design procedure for leaf springs consisting of blades with either the same or dif-
ferent widths and thicknesses. The procedure given here is said, by its developer
(see below), to be a more straightforward method.
Stresses in torsional leaf springs can be reduced significantly by increasing either
the number of blades, the aspect ratio, or both. Single torsional springs made of

square, round, or rectangular bar materials have higher stresses than do torsional
leaf springs, given the same assist moment and angular deflection.
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SPRING SELECTION AND ANALYSIS
23.12 DESIGN ENGINEERING
SI Values
8 in. (20.3 cm)
26 in. (66 cm)
FIGURE 4 Spring
for firing a projectile.
According to Timoshenko and Goddier, a local irregularity in the stress distri-
bution occurs near the attachment points when the aspect ratio of the spring is
large. Because no specific information is generally available, experience and good
judgment must be used to determine how large the aspect ratio can be before local
attachment stresses become significant. Generally, the aspect ratio should be greater
than 4 but less than 15. Also, the number of blades should be limited so that the
spring stack height does not exceed the spring width.
This procedure is the work of Russel Lilliston, Staff Engineer, Martin Marietta
Aerospace, as reported in Machine Design magazine.
DESIGNING A SPRING TO FIRE A PROJECTILE
A 22-coil squared and ground spring is designed to fire a 10-lb (4.54-kg) projectile
into the air. The spring has a 6-in (15.24-cm) diameter coil with 0.75-in (1.9-cm)
diameter wire. Free length of the spring, Fig. 4, is 26 in (66 cm); it is compressed
to 8 in when loaded or set. The shear elastic limit for the spring material is 85,000
lb/in
2
(585.65 MPa). The spring constant Wahl factor is 1.18. Shear modulus of
elasticity of the spring material is 12

ϫ 10
6
lb/in
2
(82,680 MPa). Determine (a) the
height to which the projectile will be fired; (b) the safety factor for this spring.
Calculation Procedure:
1. Find the actual stress in the spring
Use the relation, actual stress, S
a
ϭ (L
c
)(G)(d)(W)/(4
␲
)(n)(r
2
), where S
c
ϭ actual
stress, lb/in
2
(kPa); L ϭ compressed length of spring, in (cm); G ϭ shear modulus
of elasticity, lb/in
2
(kPa); d ϭ spring-wire diameter, in (cm); W ϭ Wahl factor; n
ϭ number of coils; r ϭ spring radius ϭ diameter/2, in (cm). Substituting, S
c
ϭ
(8)(12 ϫ 10
6

)(0.75)(1.18)/(4
␲
)(22)(3
2
) ϭ 34,146 lb/in
2
(235.2 MPa).
2. Compute the force to which the spring is stressed when loaded
Use the relation, P ϭ 0.1963(d
3
)(S
c
)/(r)(W ), where P ϭ loaded force, lb (N); other
symbols as given earlier. Substituting, P
ϭ (0.1963)(0.75
3
)(34,146)/(3)(1.18) ϭ
798.8 lb (3553 N).
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.13
3. Calculate the distance the projectile will be fired
Use the relation (weight, lb)(distance fired, s)
ϭ ( P)(deflection) / 2. Or s ϭ
(798.8)(8)/(2)(10) ϭ 319.5 in (811.6 cm).
4. Determine the factor of safety for the spring
The factor of safety
ϭ (stress at elastic limit)/(actual stress) ϭ 85,000 / 34,146 ϭ

2.487.
Related Calculations. Use this general procedure to determine the height to
which a projectile can be fired for toys, military devices, naval applications, etc.
SPRING SUPPORT OF MACHINERY TO CONTROL
VIBRATION FORCES
A machine weighing 2 tons (1816 kg) creates a disturbing force of 1200 lb (5338
N) at a frequency of 2000 cpm. The machine is to be supported on six springs,
each taking an equal share of the load in such a way that the force transmitted to
the building housing the machine is not to exceed 20 lb (89 N). Guides limit the
machine motion to just the vertical direction. What is the required scale, lb / in
(kg/cm), of each spring?
Calculation Procedure:
1. Set up the frequency and amplitude equations for this machine
The natural frequency of free vibration of this system is given by
1 kg
ƒ ϭ
Ί
2
␲
W
where k
ϭ spring modulus, lb / in (kg / cm); g ϭ gravitational constant, in/s
2
(cm/
s
2
); W ϭ weight of each spring, lb (kg).
The amplitude of the forced vibration of this machine, A, is given by
1
A ϭ x

s
2
1 Ϫ (ƒ /ƒ)
1
where x
s
ϭ displacement of the spring, in (cm), caused by steady-state vibration of
the machine; ƒ
1
ϭ W /2
␲
, or frequency of the exciting force, cps; ƒ ϭ natural
frequency of the system, cps.
2. Set up the magnification factor
The ratio of ƒ is called the frequency ratio, and the ratio of 1 / (1
Ϫ [ƒ ]
2
) may
1/ƒ 1/ƒ
be interpreted as the magnification factor. Now, because of proportionality factors,
we can say
20 1
ϭ
2
1200
(2000/60)
1 Ϫ
ͫͬ
2
ƒ

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SPRING SELECTION AND ANALYSIS
23.14 DESIGN ENGINEERING
For the reduction in the disturbing force which we seek, the ratio A / x
s
is neg-
ative. Solving for ƒ in the above equation gives ƒ
ϭ 4.339 cps.
3. Find the spring modulus
Substitute the value found for ƒ in the first equation above for natural frequency
and we find ƒ
ϭ 4.339 ϭ (1 / 6.28)(k ϫ 386.4/[4000/6] ); k ϭ 128.1 lb/in (2243
0.5
N/cm).
Related Calculations. Isolating the vibration caused by machinery is an im-
portant design challenge, especially in heavily populated structures. Springs are
widely used to isolate the vibrations produced by reciprocating engines and com-
pressors, pumps, presses, and similar machinery. This general procedure gives a
useful approach to analyzing springs for such applications.
SPRING SELECTION FOR A KNOWN LOAD AND
DEFLECTION
Give the steps in choosing a spring for a known load and an allowable deflection.
Show how the type and size of spring are determined.
Calculation Procedure:
1. Determine the load that must be handled
A spring may be required to absorb the force produced by a falling load or the
recoil of a mass, to mitigate a mechanical shock load, to apply a force or torque,
to isolate vibration, to support moving masses, or to indicate or control a load or

torque. Analyze the load to determine the magnitude of the force that is acting and
the distance through which it acts.
Once the magnitude of the force is known, determine how it might be
absorbed—by compression or extension (tension) of a spring. In some applications,
either compression or extension of the spring is acceptable.
2. Determine the distance through which the load acts
The load member usually moves when it applies a force to the spring. This move-
ment can be in a vertical, horizontal, or angular direction, or it may be a rotation.
With the first type of movement, a compression, or tension, spring is generally
chosen. With a torsional movement, a torsion-type spring is usually selected. Note
that the movement in either case may be negligible (i.e., the spring applies a large
restraining force), or the movement may be large, with the spring exerting only a
nominal force compared with the load.
3. Make a tentative choice of spring type
Refer to Table 1, entering at the type of load. Based on the information known
about the load, make a tentative choice of the type of spring to use.
4. Compute the spring size and stress
Use the methods given in the following calculation procedures to determine the
spring dimensions, stress, and deflection.
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.15
TABLE 1 Metal Spring Selection Guide
5. Check the suitability of the spring
Determine (a) whether the spring will fit in the allowable space, (b) the probable
spring life, (c) the spring cost, and (d) the spring reliability. Based on these findings,
use the spring chosen, if it is satisfactory. If the spring is unsatisfactory, choose
another type of spring from Table 1 and repeat the study.

SPRING WIRE LENGTH AND WEIGHT
How long a wire is needed to make a helical spring having a mean coil diameter
of 0.820 in (20.8 mm) if there are five coils in the spring? What will this spring
weigh if it is made of oil-tempered spring steel 0.055 in (1.40 mm) in diameter?
Calculation Procedure:
1. Compute the spring wire length
Find the spring length from l
ϭ
␲
nd
m
, where l ϭ wire length, in; n ϭ number of
coils in the spring, in. Thus, for this spring, l
ϭ
␲
(5)(0.820) ϭ 12.9 in (32.8 cm).
2. Compute the weight of the spring
Find the spring weight from w
ϭ 0.224ld
2
, where w ϭ spring weight, lb; d ϭ
spring wire diameter, in. For this spring, w ϭ 0.224(12.9)(0.055)
2
ϭ 0.0087 lb
(0.0387 kg).
Related Calculations. The weight equation in step 2 is valid for springs made
of oil-tempered steel, chrome vanadium steel, silica-manganese steel, and silicon-
chromium steel. For stainless steels, use a constant of 0.228, in place of 0.224, in
the equation. The relation given in this procedure is valid for any spring having a
continuous coil—helical, spiral, etc. Where a number of springs are to be made,

simply multiply the length and weight of each by the number to be made to de-
termine the total wire length required and the weight of the wire.
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SPRING SELECTION AND ANALYSIS
23.16 DESIGN ENGINEERING
HELICAL COMPRESSION AND TENSION SPRING
ANALYSIS
Determine the dimensions of a helical compression spring to carry a 5000-lb
(22,241.1-N) load if it is made of hard-drawn steel wire having an allowable shear
stress of 65,000 lb /in
2
(448,175.0 kPa). The spring must fit in a 2-in (5.1-cm)
diameter hole. What is the deflection of the spring? The spring operates at atmos-
pheric temperature, and the shear modulus of elasticity is 5
ϫ 10
6
lb/in
2
(34.5 ϫ
10
9
Pa).
Calculation Procedure:
1. Choose the tentative dimensions of the spring
Since the spring must fit inside a 2-in (5.1-cm) diameter hole, the mean diameter
of the coil should not exceed about 1.75 in (4.5 cm). Use this as a trial mean
diameter, and compute the wire diameter from d
ϭ [8Ld

m
k/(
␲
s
s
)] , where d ϭ
1/3
spring wire diameter, in; L ϭ load on spring, lb; D
m
ϭ mean diameter of coil, in;
k
ϭ spring curvature correction factor ϭ (4c Ϫ 1)/(4c Ϫ 4) ϩ 0.615 / c for heavily
coiled springs, where c
ϭ 2r
m
/d ϭ d
m
/d; s
s
ϭ allowable shear stress material, lb/
in
2
. For lightly coiled springs, k ϭ 1.0. Thus, d ϭ [8 ϫ 5000 ϫ 1.75 ϫ 1.0/(
␲
ϫ
65,000)] ϭ 0.70 in (1.8 cm). So the outside diameter d
o
of the spring will be d
o
1/3

ϭ d
m
ϩ 2(d /2) ϭ d
m
ϩ d ϭ 1.75 ϩ 0.70 ϭ 2.45 in (6.2 cm). But the spring must
fit a 2-in (5.1-cm) diameter hole. Hence, a smaller value of d
m
must be tried.
Using d
m
ϭ 1.5 in (3.8 cm) and following the same procedure, we find d ϭ [8
ϫ 5000 ϫ 1.50 ϫ 1.0/(
␲
ϫ 65,000)] ϭ 0.665 in (1.7 cm). Then d
o
ϭ 1.5 ϩ
1/3
0.665 ϭ 2.165 in (5.5 cm), which is still too large.
Using d
m
ϭ 1.25 in (3.2 cm), we get d ϭ [8 ϫ 1.25 ϫ 1.0/(
␲
ϫ 65,000)] ϭ
1/3
0.625 in (1.6 cm). Then d
o
ϭ 1.25 ϩ 0.625 ϭ 1.875 in (4.8 cm). Since this is
nearly 2 in (5.1 cm), the spring probably will be acceptable. However, the value of
k should be checked.
Thus, c

ϭ 2r
m
/d ϭ 2(1.25 / 2)/0.625 ϭ 2.0. Note that r
m
ϭ d
m
/2 ϭ 1.25/2 in
this calculation for the value of c. Then k
ϭ [(4 ϫ 2) Ϫ 1]/[(4 ϫ 2) Ϫ 4] ϩ
0.615/2 ϭ 2.0575. Hence, the assumed value of k ϭ 1.0 was inaccurate for this
spring. Recalculating, d
ϭ [8 ϫ 5000 ϫ 1.25 ϫ 2.0575 / (
␲
ϫ 65,000)] ϭ 0.796
1/3
in (2.0 cm). Now, 1.25 ϩ 0.796 ϭ 2.046 in (5.2 cm), which is still too large.
Using d
m
ϭ 1.20 in (3.1 cm) and assuming k ϭ 2.0575, then d ϭ [8 ϫ 5000 ϫ
1.20 ϫ 2.0575/(
␲
ϫ 65,000)] ϭ 0.785 in (2.0 cm) and d
o
ϭ 1.20 ϩ 0.785 ϭ
1/3
1.985 in (5.0 cm). Checking the value of k gives c ϭ 1.20/0.785 ϭ 1.529 and k
ϭ [(4 ϫ 1.529) Ϫ 1] / [(4 ϫ 1.529) Ϫ 4] ϩ 0.615/1.529 ϭ 2.820. Recalculating
again, d
ϭ [8 ϫ 5000 ϫ 1.20 ϫ 2.820/(
␲

ϫ 65,000)] ϭ 0.872 in (2.2 cm). Then
1/3
d
o
ϭ 1.20 ϩ 0.872 ϭ 2.072 in (5.3 cm), which is worse than when d
m
ϭ 1.25 in
(3.2 cm) was used.
It is now obvious that a practical trade-off must be utilized so that a spring can
be designed to carry a 5000-lb (22,241.1-N) load and fit in a 2-in (5.1-cm) diameter
hole, d
h
, with suitable clearance. Such a trade-off is to use hard-drawn steel wire
of greater strength at higher cost. This trade-off would not be necessary if d
h
could
be increased to, say, 2.13 in (5.4 cm).
Using d
m
ϭ 1.25 in (3.2 cm) and a clearance of, say, 0.08 in (0.20 cm), then d
ϭ d
h
Ϫ d
m
Ϫ 0.08 ϭ 2.00 Ϫ 1.25 Ϫ 0.08 ϭ 0.67 in (1.7 cm). Also, c ϭ d
m
/d ϭ
1.25/0.67 ϭ 1.866 and k ϭ [(4 ϫ 1.866) Ϫ 1] / [(4 ϫ 1.866) Ϫ 4] ϩ 0.615/1.866
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.17
ϭ 2.196. The new allowable shear stress can now be found from s
s
ϭ 8Ld
m
k/
␲
d
3
ϭ 8 ϫ 5000 ϫ 1.25 ϫ 2.196 /(
␲
ϫ 0.67
3
) ϭ 116,206 lb/in
2
(801,212.1 kPa). Hard-
drawn spring wire (ASTM A-227-47) is available with s
s
ϭ 117,000 lb / in
2
(806,686.6 kPa) and G ϭ 11.5 ϫ 10
6
lb/in
2
(79.29 ϫ 10
9
Pa).
Hence, d

ϭ [8 ϫ 5000 ϫ 1.25 ϫ 2.196/(
␲
ϫ 117,000)] ϭ 0.668 in (1.7 cm).
1/3
Thus, d
o
ϭ 1.25 ϩ 0.668 ϭ 1.918 in (4.9 cm). Now, the spring is acceptable because
further recalculations would show that d
o
will remain less than 2 in (5.1 cm),
regardless.
2. Compute the spring deflection
The deflection of a helical compression spring is given by ƒ
ϭ 64nr L/(d
4
G) ϭ
3
m
4
␲
nr s
s
/(dGk), where ƒ ϭ spring deflection, in; n ϭ number of coils in this spring;
2
m
r
m
ϭ mean radius of spring coil, in; L, d, G, s
s
, and k are as determined for the

acceptable spring.
Assuming n
ϭ 10 coils, we find ƒ ϭ 64 ϫ 10 ϫ (1.25 /2)
3
ϫ 5000/(0.668
4
ϫ
11.5 ϫ 10
6
) ϭ 0.341 in (0.9 cm). Or, ƒ ϭ 4
␲
ϫ 10 ϫ (1.25/2)
2
ϫ 117,000 /(0.668
ϫ 11.5 ϫ 10
6
ϫ 2.196) ϭ 0.340 (0.9 cm), a close agreement.
The number of coils n assumed for this spring is based on past experience with
similar springs. However, where past experience does not exist, several trial values
of n can be used until a spring of suitable deflection and length is obtained.
Related Calculations. Use this general procedure to analyze helical coil com-
pression or tension springs. As a general guide, the outside diameter of a spring of
this type is taken as (0.96)(hole diameter). The active solid height of a compression-
type spring, i.e., the height of the spring when fully closed by the load, usually is
nd, or (0.9) (final height when compressed by the design load).
SELECTION OF HELICAL COMPRESSION AND
TENSION SPRINGS
Choose a helical compression spring to carry a 90-lb (400.3-N) load with a stress
of 50,000 lb/in
2

(344,750.0 kPa) and a deflection of about 2.0 in (5.1 cm). The
spring should fit in a 3.375-in (8.6-cm) diameter hole. The spring operates at about
70
ЊF (21.1ЊC). How many coils will the spring have? What will the free length of
the spring be?
Calculation Procedure:
1. Determine the spring outside diameter
Using the usual relation between spring outside diameter and hole diameter, we get
d
o
ϭ 0.96d
h
, where d
h
ϭ hole diameter, in. Thus, d
o
ϭ 0.96(3.375) ϭ 3.24 in, say
3.25 in (8.3 cm).
2. Determine the required wire diameter
he equations in the previous calculation procedure can be used to determine the
required wire diameter, if desired. However, the usual practice is to select the wire
diameter by using precomputed tabulations of spring properties, charts of spring
properties, or a special slide rule available from some spring manufacturers. The
tabular solution will be used here because it is one of the most popular methods.
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SPRING SELECTION AND ANALYSIS
23.18 DESIGN ENGINEERING
TABLE 2 Load and Spring Rates for Helical Compression and Tension

Springs*
Table 2 shows typical loads and spring rates for springs of various outside di-
ameters and wire diameters based on a corrected shear stress of 100,000 lb/in
2
(689,500.0 kPa) and a shear modulus of G ϭ 11.5 ϫ 10
6
lb/in
2
(79.3 ϫ 10
9
Pa).
Before Table 2 can be used, the actual load must be corrected for the tabulated
stress. Do this by taking the product of (actual load, lb)(table stress, lb / in
2
)/(allow-
able spring stress, lb / in
2
). For this spring, tabular load, lb ϭ (90)(100,000 / 50,000)
ϭ 180 lb (800.7 N). This means that a 90-lb (400.3-N) load at a 50,000-lb/in
2
(344,750.0-kPa) stress corresponds to a 180-lb (800.7-N) load at 100,000-lb/in
2
(689,500-kPa) stress.
Enter Table 2 at the spring outside diameter, 3.25 in (8.3 cm), and project ver-
tically downward in this column until a load of approximately 180 lb (800.7 N) is
intersected. At the left read the wire diameter. Thus, with a 3.25-in (8.3-cm) outside
diameter and 183-lb (814.0-N) load, the required wire diameter is 0.250 in (0.635
cm).
3. Determine the number of coils required
The allowable spring deflection is 2.0 in (5.1 cm), and the spring rate per single

coil, Table 2, is 208 lb/in (364.3 N/cm) at a tabular stress of 100,000 lb/in
2
(689,500 kPa). We use the relation, deflection ƒ, in ϭ load, lb/desired spring rate,
lb/in, S
R
; or, 2.0 ϭ 90 /S
R
; S
R
ϭ 90/20 ϭ 45 lb / in (78.8 N/cm).
4. Compute the number of coils in the spring
The number of active coils in a spring is n ϭ (tabular spring rate, lb/in)/(desired
spring rate, lb /in). For this spring, n
ϭ 208/45 ϭ 4.62, say 5 coils.
5. Determine the spring free length
Find the approximate length of the spring in its free, expanded condition from l in
ϭ (n ϩ i)d ϩ ƒ, where l ϭ approximate free length of spring, in; i ϭ number of
inactive coils in the spring; other symbols as before. Assuming two inactive coils
for this spring, we get l
ϭ (5 ϩ 2)(0.25) ϩ 2 ϭ 3.75 in (9.5 cm).
Related Calculations. Similar design tables are available for torsion springs,
spiral springs, coned-disk (Belleville) springs, ring springs, and rubber springs.
These design tables can be found in engineering handbooks and in spring manu-
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.19
facturers’ engineering data. Likewise, spring design charts are available from many
of these same sources. Spring design slide rules are generally available free of

charge to design engineers from spring manufacturers.
SIZING HELICAL SPRINGS FOR OPTIMUM
DIMENSIONS AND WEIGHT
Determine the dimensions of a helical spring having the minimum material volume
if the initial, suddenly applied, load on the spring is 15 lb (66.7 N), the mean coil
diameter is 1.02 in (2.6 cm), the spring stroke is 1.16 in (2.9 cm), the final spring
stress is 100,000 lb / in
2
(689,500 kPa), and the spring modulus of torsion is 11.5
ϫ 10
6
lb/in
2
(79.3 ϫ 10
9
Pa).
Calculation Procedure:
1. Compute the minimum spring volume
Use the relation
v
m
ϭ 8ƒLG/s , where v
m
ϭ minimum volume of spring, in
3
;ƒϭ
2
ƒ
spring stroke, in
3

; L ϭ initial load on spring, lb; G ϭ modulus of torsion of spring
material, lb/in
2
; s
ƒ
ϭ final stress in spring, lb/in
2
. For this spring, v
m
ϭ
8(1.16)(15)(11.5 ϫ 10
6
)/(100,000)
2
ϭ 0.16 in
3
(2.6 cm
3
). Note: s
ƒ
ϭ 2s
s
, where s
s
ϭ shear stress due to a static, or gradually applied, load.
2. Compute the required spring wire diameter
Find the wire diameter from d ϭ [16Ld
m
/(
␲

s
ƒ
)] , where d ϭ wire diameter, in;
1/3
d
m
ϭ mean diameter of spring, in; other symbols as before. For this spring, d ϭ
[16 ϫ 15 ϫ 1.02/(
␲
ϫ 100,000)] ϭ 0.092 in (2.3 mm).
1/3
3. Find the number of active coils in the spring
se the relation n
ϭ 4v
m
/(
␲
2
d
2
d
m
), where n ϭ number of active coils; other symbols
as before. Thus, n
ϭ 4(0.16)/[
␲
2
(0.092)
2
(1.02)] ϭ 7.5 coils.

4. Determine the active solid height of the spring
The solid height H
s
ϭ (n ϩ 1)d, in, or H
s
ϭ (7.5 ϩ 1)(0.092) ϭ 0.782 in (2.0 cm).
For a practical design, allow 10 percent clearance between the solid height and the
minimum compressed height H
c
. Thus, H
c
ϭ 1.1H
s
ϭ 1.1(0.782) ϭ 0.860 in (2.2
cm). The assembled height H
a
ϭ H
c
ϩ ƒ ϭ 0.860 ϩ 1.16 ϭ 2.020 in (5.13 cm).
5. Compute the spring load-deflection rate
The load-deflection rate R
ϭ Gd
4
/(8dn), where R ϭ load-deflection rate, lb/in;
3
m
other symbols as before. Thus, R ϭ (11.5 ϫ 10
6
)(0.092)
4

/[8(1.02)
3
(7.5)] ϭ 12.9
lb/in (2259.1 N /m).
The initial deflection of the spring is ƒ
i
ϭ L/R in, or ƒ
i
ϭ 15/12.9 ϭ 1.163 in
(3.0 cm). Since the free height of a spring H
ƒ
ϭ H
a
ϩ ƒ
i
, the free height of this
spring is H
ƒ
ϭ 2.020 ϩ 1.163 ϭ 3.183 in (8.1 cm).
Related Calculations. The above procedure for determining the minimum
spring volume can be used to find the minimum spring weight by relating the spring
weight W lb to the density of the spring material
␳
lb/in
3
in the following manner:
For the required initial load L
1
lb, W ϭ
␳

(8ƒL
1
G/s
ƒ
2
). For the required energy
min
capacity E in ⅐ lb, W ϭ
␳
(4ED/s
ƒ
2
). For the required final load L
2
lb, W ϭ
min min
␳
(2ƒ
2
L
2
G/s ).
2
ƒ
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SPRING SELECTION AND ANALYSIS
23.20 DESIGN ENGINEERING
The above procedure assumes the spring ends are open and are not ground. For

other types of end conditions, the minimum spring volume will be greater by the
following amount: For squared (closed) ends,
v
m
ϭ 0.5
␲
2
d
2
d
m
. For ground ends,
v
m
ϭ 0.25
␲
2
d
2
d
m
. The methods presented here were developed by Henry Swies-
kowski and reported in Producted Engineering.
SELECTION OF SQUARE- AND RECTANGULAR-
WIRE HELICAL SPRINGS
Choose a square-wire spring to support a load of 500 lb (2224.1 N) with a deflection
of not more than 1.0 in (2.5 cm). The spring must fit in a 4.25-in (10.8-cm) diameter
hole. The modulus of rigidity for the spring material is G
ϭ 11.5 ϫ 10
6

lb/in
2
(79.3 ϫ 10
9
Pa). What is the shear stress in the spring? Determine the corrected
shear stress for this spring.
Calculation Procedure:
1. Determine the spring dimensions
Assume that a 4-in (10.2-cm) diameter square-bar spring is used. Such a spring
will fit the 4.25-in (10.8-cm) hole with a small amount of room to spare.
As a trial, assume that the width of the spring wire
ϭ 0.5 in (1.3 cm) ϭ a. Since
the spring is square, the height of the spring wire
ϭ 0.5 in (1.3 cm) ϭ b.
With a 4-in (10.2-cm) outside diameter and a spring wire width of 0.5 in (1.3
cm), the mean radius of the spring coil r
m
ϭ 1.75 in (4.4 cm). This is the radius
from the center of the spring to the center of the spring wire coil.
2. Compute the spring deflection
The deflection of a square-wire tension spring is ƒ
ϭ 45Lr n /(Ga
4
), where ƒ ϭ
3
m
spring deflection, in; L ϭ load on spring, lb; n ϭ number of coils in spring; other
symbols as before. To solve this equation, the number of coils must be known.
Assume, as a trial value, five coils. Then ƒ
ϭ 45(500)(1.75)

3
(5)/[(11.5 ϫ 10
6
)(0.5)
4
]
ϭ 0.838 in (2.1 cm). Since a deflection of not more than 1.0 in (2.5 cm) is per-
mitted, this spring is probably acceptable.
3. Compute the shear stress in the spring
Find the shear stress in a square-bar spring from S
s
ϭ 4.8Lr
m
/a
3
, where S
s
ϭ
spring shear stress, lb/in
2
; other symbols as before. For this spring, S
s
ϭ
(4.8)(500)(1.75)/(0.5)
3
ϭ 33,600 lb / in
2
(231,663.8 kPa). This is within the allow-
able limits for usual spring steel.
4. Determine the corrected shear stress

Find the shear stress in a square-bar spring from S
s
ϭ 4.8Lr
m
/a
3
, where s
s
ϭ spring
correction factor k
ϭ 1 ϩ 1.2/c ϩ 0.56/c
2
ϩ 0.5/c
3
, where c ϭ 2r
m
/a. For this
spring, c
ϭ (2 ϫ 1.75)/0.5 ϭ 7.0. Then k ϭ 1 ϩ 1.2/7 ϩ 0.56/7
2
ϩ 0.5/7
3
ϭ 1.184.
Hence, the corrected shear stress is S
ϭ ks
s
,orS ϭ (1.184)(33,600) ϭ 39,800ЈЈ
ss
lb/in
2

(274,411.3 kPa). This is still within the limits for usual spring steel.
Related Calculations. Use a similar procedure to select rectangular-wire
springs. Once the dimensions are selected, compute the spring deflection from
ƒ
ϭ 19.6Lr n /[Gb
2
(a Ϫ 0.566)], where all the symbols are as given earlier in this
3
m
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.21
FIGURE 5 Typical curved spring. (Product
Engineering.)
calculation procedure. Compute the uncorrected shear stress from S
s
ϭ Lr
m
(3a
ϩ 1.8b)/(a
2
b
2
). To correct the stress, use the Liesecke correction factor given in
Wahl—Mechanical Springs. For most selection purposes, the uncorrected stress is
satisfactory.
CURVED SPRING DESIGN ANALYSIS
Find the maximum load P, maximum deflection F, and spring constant C for the

curved rectangular wire spring shown in Fig. 5 if the spring variables expressed in
metric units are E
ϭ 14,500 kg / mm
2
, S
b
ϭ 55 kg / mm
2
, b ϭ 1.20 mm, h ϭ 0.30
mm, r
1
ϭ 0.65 mm, r
2
ϭ 1.75 mm, L ϭ 9.7 mm, u
1
ϭ 1.7 mm, and u
2
ϭ 5.6 mm.
Calculation Procedure:
1. Divide the spring into analyzable components
Using Fig. 6, developed by J. Palm and K. Thomas of West Germany, as a guide,
divide the spring to be analyzed into two or more analyzable components, Fig. 5.
Thus, the given spring can be divided into two springs—a type D (Fig. 6), called
system I, and a type A (Fig. 6), called system II.
2. Compute the spring force
The spring force P
ϭ P
I
ϭ P
II

. Since (u
2
ϩ r
2
) Ͼ (u
1
ϩ r
1
), the spring in system
II exerts a larger force. From Fig. 6 for
␤
ϭ 90Њ, P ϭ S
␴
/(u
2
ϩ r
2
), where S ϭ
max
section modulus, mm
3
, of the spring wire. Since S ϭ bh
2
/6 for a rectangle, P ϭ
bh
2
␴
/[6(u
2
ϩ r

2
)] where b ϭ spring wire width, mm; h ϭ spring wire height,
max
mm;
␴
ϭ maximum bending stress in the spring, kg/mm
3
; other symbols as
max
given in Fig. 5. Then P ϭ (1.20)(0.30)
2
ϫ (55)/[6(5.6 ϩ 1.75)] ϭ 0.135 kg.
3. Compute the spring deflection
The total deflection of the springs is F
ϭ 2F
I
ϩ F
II
, where F ϭ spring deflection,
mm, and the subscripts refer to each spring system. Taking the sum of the deflec-
tions as given in Fig. 6, we get F
ϭ [2P /(3EI)][2K
1
r (m
1
ϩ
␤
1
/2)
2

ϩ (v
1
Ϫ u
1
)
33
1
ϩ K
2
r (m
2
ϩ
␤
2
)
3
], where E ϭ Young’s modulus, kg/mm
2
; I ϭ spring wire moment
3
2
of inertia, mm
4
; K ϭ correction factor for the spring from Fig. 7, where the sub-
scripts refer to the radius being considered in the relation u/r; m
ϭ u/r;
␤
ϭ angle
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SPRING SELECTION AND ANALYSIS
23.22 DESIGN ENGINEERING
FIGURE 6 Deflection, force, and stress relations for curved springs. (Product Engineering.)
FIGURE 7 Correction factors for curved springs. (Product Engineering.)
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SPRING SELECTION AND ANALYSIS
SPRING SELECTION AND ANALYSIS 23.23
FIGURE 8 Typical torsion spring. (Product Engineering.)
of spring curvature, rad. Where the subscripts 1 and 2 are used in this equa-
tion, they refer to the respective radius identified by this subscript. Since I
ϭ
bh
3
/12 for a rectangle, or I ϭ (1.20)(0.30)
3
/12 ϭ 0.0027 mm
4
, F ϭ {[2(0.135)/
[(3)(14,500)(0.0027)]}[2(0.92)(0.65)(2.62
ϩ 1.57)
3
ϩ 0 ϩ 0.94(1.75)
3
(3.2 ϩ
1.57)
3
] ϭ 1.34 mm.

4. Compute the spring constant
The spring constant C ϭ P / F ϭ 0.135 ϭ 0.135 / 1.34 ϭ 0.101 kg / mm.
Related Calculations. The relations given here can also be used for round-
wire springs. For accurate results, h /r for flat springs and d
o
/r for round-wire
springs should be less than 0.6. The various symbols used in this calculation pro-
cedure are defined in the text and illustrations. Since the equations given here
analyze the springs and do not contain any empirical constants, the equations can
be used, as presented, for both metric and English units. Where a round spring is
analyzed, h
ϭ b ϭ d
o
, where d
o
ϭ spring outside diameter, mm or in.
ROUND- AND SQUARE-WIRE HELICAL TORSION-
SPRING SELECTION
Choose a round-music-wire torsion spring to handle a moment load of 15.0 lb ⅐ in
(1.7 N
⅐ m) through a deflection angle of 250Њ. The mean diameter of the spring
should be about 1.0 in (2.5 cm) to satisfy the space requirements of the design.
Determine the required diameter of the spring wire, the stress in the wire, and the
number of turns required in the spring. What is the maximum moment and angular
deflection the spring can handle? What is the maximum moment and deflection
without permanent set?
Calculation Procedure:
1. Select a suitable wire diameter
To reduce the manufacturing cost of a spring, a wire of standard diameter should
be used, whenever possible, for the spring, Fig. 8. Usual torsion-spring wire di-

ameters and the side of square-wire springs range from 0.02 to 0.60 in (0.05 to
1.52 cm), depending on the moment the spring must carry and the angular deflec-
tion.
Assume a wire diameter of 0.10 in (0.25 cm) and a bending stress of 150,000
lb/in
2
(1.03 ϫ 10
9
Pa) as trial values for this spring. [Typical round-wire and square-
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SPRING SELECTION AND ANALYSIS
23.24 DESIGN ENGINEERING
wire torsion-spring bending stresses range from 100,000 to 200,000 lb/in
2
(689.5
ϫ 10
6
to 1.38 ϫ 10
9
Pa), depending on the material used in the spring.]
Compute the twisting moment corresponding to the assumed stress from M
i
ϭ
␲
d
3
S
b

/32, where M
i
ϭ twisting moment load, lb ⅐ in; d ϭ spring wire diameter,
in; S
b
ϭ bending stress in spring, lb /in
2
. Thus, M
i
ϭ
␲
(0.10)
3
(150,000)/32 ϭ 14.7
lb
⅐ in (1.66 N ⅐ m). This is very close to the actual moment load of 15.0 lb ⅐ in (1.7
N
⅐ m). Therefore, the assumed spring diameter and bending stress are acceptable,
thus far.
2. Compute the actual spring stress
Use the following relation to find the actual bending stress S
b
lb/in
2
in the spring:
S
b
ϭ (actual spring moment lb ⅐ in/computed spring moment, lb ⅐ in)(assumed
stress, lb / in
2

); S
b
ϭ (15.0/14.7)(150,000) ϭ 153,000 lb / in
2
(1.05 ϫ 10
9
Pa).
3. Check the actual vs. recommended spring stress
Enter Fig. 9 at the wire diameter of 0.10 in (0.25 cm), and project vertically upward
to the music-wire curve to read the recommended bending stress for music wire as
159,000 lb / in
2
(1.10 ϫ 10
9
Pa). Since the actual stress, 153,000 lb/in
2
(1.05 ϫ 10
9
Pa), is less than but reasonably close to the recommended stress, the selected wire
diameter is acceptable for the planned load on the spring. This chart and calculation
procedure were developed by H. F. Ross and reported in Product Engineering.
4. Determine the angular deflection per spring coil
Compute the angular deflection per coil from
␾
ϭ 360S
b
d
m
/(Ed ), where
␾

ϭ
angular deflection per spring coil, degrees; d
m
ϭ mean diameter of spring, in; E ϭ
Young’s modulus for spring material ϭ 30 ϫ 10
6
lb/in
2
(206.9 ϫ 10
9
Pa) for spring
steel; other symbols as before. Thus, by using the assumed bending stress in the
spring,
␾
ϭ 360(150,000)(1.0) / [(30 ϫ 10
6
)(0.1)] ϭ 18Њ. This value is the maximum
safe deflection per coil for the spring.
5. Compute the number of coils required
The number of coils n required in a helical torsion spring is n
ϭ
␾
t
(assumed stress,
lb/in
2
)/
␾
(actual stress, lb / in
2

), where
␾
t
ϭ total angular deflection of spring, de-
grees;
␾
ϭ maximum safe deflection per coil, degrees. Thus, n ϭ 250(150,000)/
[(18)(153,000)]
ϭ 13.6 coils; use 14 coils.
6. Determine the maximum moment the spring can handle
On the basis of the maximum recommended stress, the moment can be increased
to M
i
ϭ [(maximum recommended stress, lb / in
2
)/(assumed stress, lb/in
2
)](actual
moment, lb
⅐ in). Read the maximum recommended stress from Fig. 9 as 159,000
lb/in
2
(1.10 ϫ 10
9
Pa) for 0.1-in (0.25-cm) diameter music wire, as in step 3. Thus,
M
i
ϭ (159,000/150,000)(14.7) ϭ 15.6 lb ⅐ in (1.8 N ⅐ m).
7. Compute the maximum angular deflection
The maximum angular deflection per coil is

␾
ϭ [(maximum recommended stress,
lb/in
2
)/(assumed stress, lb / in
2
)](computed angular deflection per coil, degrees)
ϭ 159,000/150,000 ϫ 18 ϭ 19.1Њ per coil.
8. Determine the special-case moment and deflection
The maximum moment M and deflection
␾
, without permanent set, can be
max max
one-third greater than in steps 6 and 7, or M ϭ 15.6(1.33) ϭ 20.8 lb ⅐ in (2.4 N
max
⅐ m), and
␾
ϭ 19.1(1.33) ϭ 25.5Њ per coil. These maximum values allow for
max
overloads on the spring.
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SPRING SELECTION AND ANALYSIS
23.25
FIGURE 9 Recommended bending stresses for torsion springs. (Product Engineering.)
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SPRING SELECTION AND ANALYSIS