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9.10 VIBRATION ISOLATION MATERIALS
There are several commonly used materials for vibration isolation applications,
including felt, cork, rubber or elastomers, and metal springs. The
characteristics of these materials are presented in this section.
9.10.1 Cork and Felt Resilient Materials
Cork and felt materials are used in vibration isolation applications in which
relatively low surface pressures (less than about 200 kPa or 30 psi) can be
attained, and applications where the required static deflection is in the range
of 0.25–1.8mm (0.01–0.07 in). In applications where cork and felt materials
are used, care should be exercised to prevent contaminants, such as oil and
water, from coming into contact with the material. Exposure to oil and
water over an extended period of time will cause the material to deteriorate.
Cork and felt materials are well-suited for applications in which high-frequency
vibration must be damped.
Felt may be composed of wool fibers, and synthetic fibers may also be
mixed with the wool fibers. Because the felt has no strong binder for the
fibers, felt material should be used only in the form of pads loaded in
compression.
When felt is used as a vibration isolation material, the lowest transmissibility
is obtained by using a small surface area, a soft felt (one having a
low density), and a large thickness. Felt material generally has a damping
ratio around _ ј 0:060. The uncompressed thickness of the felt pad is
usually selected in practice as 30 or more times the static deflection. The
surface pressure for felt should generally be limited to values less than about
140–200 kPa (20–30 psi).
An empirical relationship (Crede, 1951) between the natural frequency
and the surface pressure for a felt pad is as follows:
fnfHzg ј C1 ю
14:93Ѕ1 _ р0:148Ю=р_f=_ref Ю_
рPs=Pref Ю
_ _
рhЮ (9-149)
446 Chapter 9
Copyright © 2003 Marcel Dekker, Inc.
The quantity Ps is the surface pressure on the pad, and the quantity C1 is a
function of the density of the felt _f :
C1 ј
14:29
1 _ 1:068р_f=_ref Ю
(9-150)
The quantity рhЮ is a function of the unloaded thickness of the felt pad h:
рhЮ ј 1 _ 0:25 lnрh=href Ю (9-151)
The reference surface pressure Pref ј 101:325 kPa (14.696 psi), the reference
density _ref ј 1000 kg=m3 (62.43 lbm=ft3), and the reference thickness is
href ј 25:4mm (1.00 in). The empirical relationship is valid for the following
range of the variables:
180 kg=m3 _ _ _ 400 kg=m3
30 kPa _ Ps _ 250 kPa
5mm _ h _ 200mm
Cork is one of the oldest materials used for vibration isolation. It is
made from the elastic outer layer of bark of the cork oak tree. Like felt, cork
is used only in the form of pads under compressive loading. The slope of the
load–deflection curve for cork is not constant, but tends to increase as the
load on the cork is increased. Typical thickness of cork used for pads is
between about 25mm (1 in) and 150mm (6 in). The allowable static deflection
for cork should be in the range from 0.4mm (0.015 in) to 2.0mm
(0.079 in). The damping ratio for cork is approximately _ ј 0:075.
An empirical expression (Crede, 1951) for the undamped natural frequency
as a function of the surface pressure for cork material is as follows:
fnfHzg ј C2рPs=Pref Ю_1=3 рhЮ (9-152)
It is recommended that the design natural frequency be selected as about 1.5
times the value given by Eq. (9-152). The quantity Ps is the surface pressure
on the pad and the quantity C2 is a function of the density of the cork _c:
C2 ј
19:75
1 _ 1:091р_c=_ref Ю
(9-153)
The quantity рhЮ is given by Eq. (9-151), and the reference surface pressure,
density, and thickness are the same as those used in the expressions for the
natural frequency for felt.
The recommended maximum allowable surface pressure for cork is a
function of the density of the cork:
Pmax=Pref ј 13р_c=_ref Ю (9-154)
Vibration Isolation for Noise Control 447
Copyright © 2003 Marcel Dekker, Inc.
The empirical relationships are valid for the following range of the
variables:
200 kg=m3 _ _ _ 350 kg=m3
70 kPa _ Ps _ 400 kPa
25mm _ h _ 150mm
Cork material is sometimes sold in units of board feet, where 1 board
foot is equal to 144 in3 ј 1
12 ft3 ј 2:36dm3. The conversion factor for density
expressed in units of lbm=bd ft is as follows:
192:22 рkg=m3Ю=рlbm=bd ftЮ ј 12:00 рlbm=ft3Ю=рlbm=bd ftЮ
Example 9-9. A machine having a total weight of 1800N (405 lbf ) operates
at a speed of 6000 rpm or 100 Hz. Four felt pads (damping ratio, _ ј 0:060)
are used to support the machine. The pads have a density of 260 kg=m3
(16.2 lbm=ft3) and a thickness of 40mm (1.57 in). The maximum transmissibility
for the system is to be 0.100. Determine the size of the felt pads.
First, let us determine the frequency ratio. The _ parameter may be
calculated from Eq. (9-109):
_ ј 1 ю р2Юр0:060Ю2р1 _ 0:102Ю
р0:10Ю2 ј 1 ю 0:7128 ј 1:7128
The frequency ratio is found from Eq. (9-108):
r4 _ р2Юр1:7128Юr2 _
1 _ 0:102
0:102 ј 0
r2 ј 1:7128 ю Ѕр1:7128Ю2 ю 99_1=2 ј 11:809
r ј 3:436 ј f =fn
The undamped natural frequency for the system is as follows:
fn ј р100Ю=р3:436Ю ј 29:1Hz
Let us check the static deflection for the system, using Eq. (9-17):
d ј
g
4_2f 2
n ј р9:806Ю
р4_2Юр29:1Ю2 ј 0:293 _ 10_3 m ј 0:293mm р0:0115 inЮ
h=d ј р40Ю=р0:293Ю ј 137 > 30
Thus, the static deflection is satisfactory in this case.
The surface pressure may now be calculated. The coefficient C1 may be
calculated from Eq. (9-150):
448 Chapter 9
Copyright © 2003 Marcel Dekker, Inc.
C1 ј р14:29Ю
1 _ р1:068Юр0:260Ю ј 19:78
The thickness parameter is found from Eq. (9-151):
рhЮ ј 1 _ р0:25Ю lnр40=25:4Ю ј 0:886
The surface pressure may be found from Eq. (9-149):
fn ј 29:1Hz ј 19:78 ю р14:93ЮЅ1 _ р0:148Ю=р0:260Ю_
рPs=Pref Ю
_ _
р0:886Ю
Ps=Pref ј р6:431Ю
Ѕр29:1Ю=р0:886Ю_ _ 19:78 ј 0:4941
Ps ј р0:4941Юр101:325Ю ј 50:0 kPa р7:25 psiЮ
The total required load-bearing area for the pads is as follows:
Sf ј р1800Ю
р50:0Юр103Ю ј 0:0360m2 ј 360 cm2 р55:8 in2Ю
The area per pad (four pads are used) is as follows:
1
4 Sf ј р360Ю=р4Ю ј 90:0 cm2 ј 9000mm2 р13:95 in2Ю
If the cross-sectional area of the pad is square, the edge dimensions of the
pad are as follows:
a ј р9000Ю1=2 ј 94:9mm р3:72 inЮ
Let us determine the size of cork pads having a density of 250 kg/m3
(15.61 lbm=ft3 or 1.30 lbm=bd ft) and a thickness of 50mm (1.969 in) for the
same application. The damping ratio for cork is _ ј 0:075.
Let us determine the design natural frequency:
_ ј 1 ю р2Юр0:075Ю2р1 _ 0:102Ю
р0:10Ю2 ј 1 ю 1:1138 ј 2:1138
The frequency ratio is found from Eq. (9-108):
r4 _ р2Юр2:1138Юr2 _
1 _ 0:102
0:102 ј 0
r2 ј 2:1138 ю Ѕр2:1138Ю2 ю 99_1=2 ј 12:286
r ј 3:505 ј f =fn
The design undamped natural frequency for the system is as follows:
fnрdesignЮ ј р100Ю=р3:505Ю ј 28:53 Hz
Vibration Isolation for Noise Control 449
Copyright © 2003 Marcel Dekker, Inc.
According to the recommended design procedure, the design frequency
should be about 1.50 times the frequency given by Eq. (9-152):
fnЅEq. (9-152)] ј р28:53Ю=р1:50Ю ј 19:02 Hz
The pad surface pressure may be determined from Eq. (9-152). The
coefficient C2 is found from Eq. (9-153):
C2 ј р19:75Ю
1 _ р1:091Юр0:250Ю ј 27:157
The thickness function is found from Eq. (9-151):
рhЮ ј 1 _ р0:25Ю lnр50=25:4Ю ј 0:8307
fn ј 19:02 Hz ј р27:157Юр0:8307Ю
рPs=Pref Ю1=3
Ps=Pref ј р1:1861Ю3 ј 1:668
Ps ј р1:668Юр101:325Ю ј 169:1 kPa р24:5 psiЮ
The maximum recommended surface pressure for cork is found from Eq. (9-
154):
Pmax=Pref ј р13Юр0:250Ю ј 3:25
Pmax ј р3:25Юр101:325Ю ј 329 kPa > Ps ј 169 kPa
The required load-bearing area of the cork is as follows:
Sc ј р1800Ю
р169:1Юр103Ю ј 0:01065m2 ј 106:5 cm2 р16:51 in2Ю
The area per pad (four pads are used) is as follows:
1
4 Sc ј р106:5Ю=р4Ю ј 26:6 cm2 ј 2660mm2 р4:13 in2Ю
Suppose the cork pad is cylindrical. The diameter of one pad may be found
as follows:
Dc ј Ѕр4=_Юр2660_1=2 ј 58:2mm р2:29 inЮ
It is noted that both the felt pads and the cork pads provide a practical
solution to the vibration isolation problem in this case.
9.10.2 Rubber and Elastomer Vibration Isolators
Rubber and elastomer (such as neoprene) materials are used in vibration
isolation applications in which the static deflection on the order of 10–
15mm (0.40–0.60 in) is required and where moderate surface pressures are
encountered. Some of the properties of rubber materials are given in Table
450 Chapter 9
Copyright © 2003 Marcel Dekker, Inc.
9-5. It is noted that the physical properties of rubber are strongly dependent
on the hardness of the material.
The design of rubber isolators is best carried out by consulting the
data from manufacturers’ catalogs, because there are a wide variety of
shapes and sizes of mount available. For compression loading, the static
deflection should be limited to about 10–15%of the unloaded thickness. For
shear loading, the static deflection should be limited to about 20–30%. The
design or maximumoperating compressive stress for rubber isolators should
generally not exceed about 600 kPa (90 psi), and the design shear stress for
shear-loaded isolators should be limited to about 200–275 kPa (30–40 psi).
The manfacturers’ catalog data generally states the maximum allowable
load for the isolator. The normal operating temperatures for rubber mounts
is usually below 608C (1408F), although temperatures as high as 758C
(1708F) may be used for some mounts without seriously deteriorating service
performance.
The spring constant for a rubber mounting loaded in compression is
strongly dependent on the friction between the mount and the support for
unbonded mounts. An approximate empirical relationship for the spring
constant for compression-loaded rubber isolators, similar to that shown in
Fig. 9-12, is as follows:
KS ј
SEc
h
S1=2
рhoh2a2=b2Ю1=3
" #
(9-155)
The quantity a is the length (longer dimension, a b), b is the width, and h
is the thickness of the rubber slab. The surface area S ј ab, and the quantity
Vibration Isolation for Noise Control 451
TABLE 9-5 Properties of Rubber at 218C (708F)
Durometer
hardness,
Shore A
Density,
kg/m3
Static modulus, MPaa Damping
ratio,
Compression Tension Shear Bulk _
30 1,010 1.28 1.21 0.345 2,030 0.015
40 1,060 1.86 1.59 0.483 2,220 0.020
50 1,110 2.59 2.10 0.655 2,380 0.045
60 1,180 3.79 3.10 0.965 2,550 0.075
70 1,250 5.17 4.21 1.34 2,870 0.105
80 1,310 8.27 7.07 1.76 3,130 0.140
aTo convert (MPa) to (psi), the conversion factor is 145.038 psi/MPa.
Source: U.S. Rubber (1941).
Copyright © 2003 Marcel Dekker, Inc.
ho ј 25:4mm (1.00 in). The quantity Ec is Young’s modulus in compression
for the rubber.
For the case of a mount loaded in shear, consisting of two rubber pads
bonded between three steel plates, as shown in Fig. 9-13, the following
expression applies for the spring constant:
KS ј
2SG
h
(9-156)
452 Chapter 9
FIGURE 9-12 Rubber or elastomer vibration isolator loaded in compression. The
dimension a is the larger dimension, i.e., a b.
FIGURE 9-13 Rubber or elastomer vibration isolator loaded in shear.
Copyright © 2003 Marcel Dekker, Inc.
The quantity S is the surface area of the pad on one side, G is the modulus of
elasticity in shear, and h is the thickness of one rubber pad. If a single side is
used (one rubber pad between two steel plates), the factor 2 is omitted from
Eq. (9-156).
For a shear-loaded mount, consisting of a cylinder of rubber mounted
between two cylinders of steel, as shown in Fig. 9-14, the spring constant is
given by the following expression:
KS ј
2_hG
lnрDo=DiЮ
(9-157)
The quantities Do and Di are the outside diameter and inside diameter of the
rubber, respectively, and h is the length of the cylinder. The quantity G is the
modulus of elasticity in shear for the rubber pad.
Shear-loaded rubber mounts are usually used in more applications
than compression-loaded mounts because the spring constant for the
shear-loaded mounts is generally much smaller than that for compressionloaded
mounts. The smaller spring constant results in a smaller undamped
natural frequency and a larger frequency ratio, which produces a smaller
transmissibility.
Vibration Isolation for Noise Control 453
FIGURE 9-14 Cylindrical rubber or elastomer vibration isolator loaded in shear.
The load is applied to the ends of the inner and outer steel cylinders.
Copyright © 2003 Marcel Dekker, Inc.
Compression-loaded rubber mounts are used in applications in which
greater load-carrying capacity per unit volume of rubber is required. Rubber
in compression is also used where large deflections are not required.
Example 9-10. A machine having a total mass of 1400 kg (3086 lbm) is
supported by four rubber isolators loaded in compression. The rubber has
a hardness of 50 Durometer. The dimensions of each pad are 76.2mm
(3.00in) _ 76.2mm _ 25.4mm (1.00in) thick. The effective dynamic force
on the system has an amplitude of 1.60 kN (360 lbf) and a frequency of
45 Hz. Determine the maximum amplitude of motion of the machine and
the force transmitted to the foundation.
The surface area of the pad is as follows:
S ј ab ј р0:0762Юр0:0762Ю ј 58:06_10_4 m2 ј 58:06cm2 р9:00in2Ю
The Young’s modulus in compression for the rubber is found from Table 9-
5 to be Ec ј 2:59MPa. The spring constant for one compression-loaded pad
may be found from Eq. (9-155):
KS ј р58:06Юр10_4Юр2:59Юр106Ю
р0:0254Ю
Ѕр58:06Юр10_4Ю_1=2
Ѕр0:0254Юр0:0254Ю2р1Ю2_1=3
( )
KS ј р0:5920Юр106Юр3:00Ю ј 1:776 _ 106 N=m
ј 1:776MN=m р10,140 lbf=inЮ
The spring constant for the system (four pads) is as follows:
4Ks ј р4Юр1:776Ю ј 7:104MN=m
We may check the static deflection at this point. The total supported
weight of the machine is as follows:
Mg ј р1400Юр9:806Ю ј 13,730N ј 13:73kN р3086 lbf Ю
The static deflection is found from Eq. (9-16):
d ј р13,730Ю
р7:104Юр106Ю ј 0:001932m ј 1:932mm р0:076 inЮ
The static deflection is р1:832=25:4Ю ј 0:072 ј 7:2% of the unloaded thickness,
which is satisfactory (maximum allowable is between 10 and 15%).
The surface pressure due to the static load (weight of the machine) is as
follows:
Ps ј р13,730=р4Юр58:06 _ 10_4Ю ј 591:2 _ 103 Pa
ј 591:2 kPa р85:7 psiЮ
454 Chapter 9
Copyright © 2003 Marcel Dekker, Inc.
The value is satisfactory, because the surface pressure should be limited to
about 600 kPa (90 psi).
The undamped natural frequency for the system may be found from
Eq. (9-15):
fn ј Ѕр7:104Юр106Ю=р1400Ю_1=2
р2_Ю ј 11:34Hz
The frequency ratio r ј f =fn is as follows:
r ј р45Ю=р11:34Ю ј 3:969
From Table 9-5, we find the damping ratio for the rubber to be
_ ј 0:045. The magnification factor is found from Eq. (9-81):
MF ј
1
fЅ1_р3:969Ю2_2 юЅр2Юр0:045Юр3:969Ю_2g1=2 ј 0:06776 ј
KSymax
Fo
The maximum amplitude of vibration of the machine is as follows:
ymax ј р0:06776Юр1600Ю=р7:104Юр106Ю ј 0:0153_10_3m ј 0:0153mm
The transmissibility may be calculated from Eq. (9-103):
Tr ј
1юЅр2Юр0:045Юр3:969Ю_2
Ѕ1_р3:969Ю2_2 юЅр2Юр0:045Юр3:969Ю_2
( )1=2
ј 0:07195
The transmissibility level is found from Eq. (9-104):
LTr ј 20log10р0:07195Ю ј _22:9 dB
The force transmitted to the foundation is as follows:
FT ј FoTr ј р1600Юр0:07195Ю ј 115:1N р25:9 lbf Ю
Example 9-11. An instrument package having a total mass of 20kg
(44.1 lbm) is connected to a structure through four single-shear rubber isolator
mounts, as shown in Fig. 9-15. The hardness of the rubber isolator
material is 40 Durometer, and the diameter of the cylindrical isolator is
25mm (0.984 in). The structure vibrates with a maximum amplitude of
3.6mm (0.142 in) with a frequency of 62 Hz. Determine the thickness of
the isolator such that the maximum amplitude of vibration for the instrument
package is limited to 0.06mm (0.0024 in).
The shear force on each isolator is as follows:
1
4Mg ј р1
4Юр20Юр9:806Ю ј 49:03N р11:02 lbf Ю
The shear area is as follows:
Vibration Isolation for Noise Control 455
Copyright © 2003 Marcel Dekker, Inc.
S ј р_=4Юр0:025Ю2 ј 4:909_10_4 m2 ј 4:909 cm2 р0:0761 in2Ю
The shear stress on each isolator is as follows:
ss ј р49:03Ю=р4:909_10_4Ю ј 99:9 kPa р14:5psiЮ
This value is satisfactory, because the shear stress for shear-loaded isolators
should be limited to about 200–275 kPa (30–40 psi).
The required transmissibility for the support system is determined
from Eq. (9-131):
Tr ј j y2mj
j y1mj ј
0:06
3:60 ј 0:01667
The damping ratio for 40 Durometer rubber is _ ј 0:020, as given in Table
9-5. The _-parameter is found from Eq. (9-109):
_ ј 1 ю р2Юр0:020Ю2р1_0:016672Ю
р0:01667Ю2 ј 1ю2:8792 ј 3:8792
The frequency ratio is found from Eq. (9-108):
r4 _р2Юр3:8792Юr2 _ р1_0:016672Ю
р0:01667Ю2 ј 0
r2 ј 3:8792юЅр3:8792Ю2 ю3599_1=2 ј 64:00
r ј 8:00 ј f =fn
The required undamped natural frequency for the systemis as follows:
fn ј р62Ю
р8:00Ю ј 7:75Hz ј рKS=MЮ1=2
2_
456 Chapter 9
FIGURE9-15 Diagramfor the single-shear rubber isolator mount in Example 9-11.
Copyright © 2003 Marcel Dekker, Inc.
The spring constant for the system is as follows:
KS ј р4_2Юр7:75Ю2р20:0Ю ј 47,420N=m
The spring constant for one of the four individual isolators is as follows:
1
4KS ј р1
4Юр47,420Ю ј 11,860N=m ј 11:86 kN=m р67:7 lbf=inЮ
The shear modulus for the rubber material is G ј 0:483MPa from
Table 9-5. The thickness for the isolator is as follows:
h ј
SG
р1
4KSЮ ј р4:909Юр10_4Юр0:483Юр106Ю
р11,860Ю ј 0:0200m
ј 20:0mm р0:787 inЮ
Let us check the static deflection for each isolator:
d ј
Mg
KS ј р1
4MЮg
р1
4KSЮ ј р20Юр9:806Ю
р47,420Ю ј 0:00414m ј 4:14mm р0:163 inЮ
This value is satisfactory, because the static deflection is р4:14=20:0Ю ј 0:207 ј 20:7% of the unloaded thickness. This is within the range of
20–30%. The maximum dynamic deflection of the isolator is as follows:
_ ј d юр y2m юy1mЮ ј 4:14юр3:60ю0:06Ю ј 7:80mm р0:307 inЮ
9.10.3 Metal Spring Isolators
Metal springs are commonly used elements in vibration isolation, especially
for applications in which the required undamped natural frequency is less
than 5Hz and large (up to 125mm or 5 in) static deflections are encountered.
Metal springs have been used (Beranek, 1971) to isolate small delicate
instrument packages and have been used to isolate masses as large as
400Mg (400 metric tons or 900,000lbm). Metal springs have the advantage
that spring materials that are not adversely affected by oil and water can be
selected.
In many cases, pads of neoprene or other elastomers are mounted in
series with the spring (between the spring and the supporting structure, as
shown in Fig. 9-16) to prevent high-frequency waves fromtraveling through
the spring into the support structure.
Metal springs have been constructed of several materials, including
spring steel, 304 stainless steel, spring brass, phosphor bronze, and beryllium
copper. The pertinent physical properties of these materials are listed in
Table 9-6. The standard size (wire gauge) for ferrous wire, excluding
music wire, is the Washburn and Moen gauge (W&M). The Music Wire
Vibration Isolation for Noise Control 457
Copyright © 2003 Marcel Dekker, Inc.
gauge is used for music wire sizes. For non-ferrous metals, the Brown and
Sharp gauge (B&S) or the American wire gauge (AWG) are used (Avallone
and Baumeister, 1987).
There are several types of metal springs, including helical springs, leaf
springs, Belleville springs (coned disk springs), and torsion springs. In this
section, we will concentrate on helical compression springs.
Helical compression springs may be used as freestanding springs
(unrestrained springs) or as housed or restrained springs. For freestanding
springs, care must be taken to avoid sideways (lateral) instability or buckling.
The unrestrained compression spring will always be stable if the following
condition is valid:
_ _
1 ю 2
2 ю
_ _
_D
Ho
_ _2
1 (9-158)
For a value of Poisson’s ratio ј 0:3, Eq. (9-158) reduces to the following:
D=Ho 0:382 (9-159)
If the value of the _ ratio is less than 1, the spring will be stable if the ratio of
the total deflection _y to the free height Ho (spring height when unloaded)
meets the following criterion (Timoshenko and Gere, 1961):
458 Chapter 9
FIGURE 9-16 Metal spring support with damping pad or neoprene or other elastomer.
Copyright © 2003 Marcel Dekker, Inc.
_y
Ho
<
1 ю
1 ю 2
_ _
1 _ 1 _
1 ю 2
2 ю
_ _
_D
Ho
_ _2
" #1=2
8<
:
9=
; (9-160)
The quantity is Poisson’s ratio, D is the mean coil diameter for the spring,
and Ho is the free height for a spring that is not clamped at the ends. If both
ends of the spring are clamped, use Ho ј р2 _ free height for springЮ. If
Poisson’s ratio for the spring material is ј 0:3, Eq. (9-160) reduces to
the following:
р_y=HoЮ < 0:8125f1 _ Ѕ1 _ 6:87рD=HoЮ2_1=2g (9-161)
The spring constant for axial compression of a helical spring is given
by the following expression:
KS ј
Gd4w
8D3Nc
(9-162)
Vibration Isolation for Noise Control 459
TABLE 9-6 Properties of Metal Spring Materials
Material
Density, ,
kg/m3
Young’s
modulus,
E,
GPa
Shear
modulus,
G,
GPa
Poisson’s
ratio,
_
Shear yield
strength,
sys,
MPa
Spring steel 7,830 203.4 79.3 0.287 a
304 stainless steel 7,820 190.3 73.1 0.305 179
Spring brass 8,550 106.0 40.1 0.324 200
Phosphor bronze 8,800 111.0 41.4 0.349 315
Beryllium copper 8,230 124.0 48.3 0.285 675
aNote that the strength properties are strongly dependent on the heat treatment, cold
working, etc. The shear yield strength of spring steels is also dependent on the wire size.
The shear yield strength may be approximated by the following expression for small sizes:
sys ј sys1рdref=dwЮn
where dref ј 1mm and sys1 and n are as follows:
Size range, mm sys1, MPa Exponent, n
Music wire 0.10–6.5 940 0.146
>6.5 715 0
Oil-tempered wire 0.50–12 814 0.186
>12 513 0
Hard-drawn wire 0.70–12 758 0.192
>12 470 0
Copyright © 2003 Marcel Dekker, Inc.
The quantity G is the shear modulus, dw is the wire diameter, D is the mean
diameter of the wire coil, and Nc is the number of active coils in the spring.
The ratio D=dw is called the spring index, and usually has values in the range
between 6 and 12 (Shigley and Mischke, 1989).
The number of active coils for a spring depends on the treatment of the
ends of the spring wire. A spring with plain ends has no special treatment of
the ends; the ends are the same as if a spring had been cut to make two
shorter springs. For the case of plain and ground ends, the last coil on the end
of the spring has the wire ground with a flat surface so that approximately
half of the coil is in direct contact with the supporting surface. For a squared
or closed end, the end coil is deformed to a zero degree helix angle such that
the entire coil touches the supporting surface. For a squared and ground end,
the end coil is squared, then the wire is ground with a flat surface such that
practically all of the end coil is in direct contact with the supporting surface.
The number of active coils for the various end treatments is summarized in
Table 9-7. Unless other factors indicate otherwise, the ends of the springs
should be both squared and ground because better transfer of the load on
the spring is achieved for this end treatment.
It is obvious that the spring should not be compressed solid (i.e., with
the coils in contact with the adjacent coils) during operation of the spring.
The expressions for the solid height of spring with various end treatments
are also given in Table 9-7. For a helical compression spring, the spring
height under maximum deflection conditions should not be less than about
1.20 times the solid height.
The shear stress in a helical compression spring is a function of the
force applied F, which includes both the supported weight and the dynamic
force, and the dimensions of the spring:
ss ј
8FDksh
_d3w
(9-163)
460 Chapter 9
TABLE 9-7 Characteristics of Helical Coil Springs
End treatment
Active coils,
Nc
Free height,
Ho
Solid height,
Hs
Spring pitch,
ps
b
Plain Nt
a psNt
ю dw dw
рNt
ю 1Ю рHo
_ dw
Ю=Nt
Plain and ground Nt
_ 1 psNt dwNt Ho=Nt
Squared Nt
_ 2 psNc
ю 3dw dw
рNt
ю 1Ю рHo
_ 3dw
Ю=Nc
Squared and ground Nt
_ 2 psNc
ю 2dw dwNt
рHo
_ 2dw
Ю=Nc
aNt is the total number of coils for the spring.
bps is the spring pitch (reciprocal of the number of coils per unit height of the spring).
Copyright © 2003 Marcel Dekker, Inc.
The quantity ksh is a shear-stress correction factor, given by the following
expression:
ksh ј
2рD=dwЮ ю 1
2рD=dwЮ
(9-164)
Springs that support machinery are often subjected to loads in the
lateral direction (perpendicular to the axis of the spring). The spring constant
in the lateral direction Klat is related to the spring constant in the axial
direction KS by the following expression:
Klat
KS ј
2р1 ю Ю
1 ю 4р2 ю ЮрHo=DЮ2 (9-165)
For the special case of Poisson’s ratio ј 0:3, Eq. (9-165) reduces to the
following:
Klat
KS ј
2:60
1 ю 9:20рHo=DЮ2 (9-166)
Another factor that must be considered in spring design is the problem
of spring surge. If one end of a helical spring is forced to oscillate, a wave
will travel from the moving end to the fixed end of the spring, where the
wave will be reflected back to the other end. The critical or surge frequency
for a spring that has one end against a flat plate and the other end driven by
an oscillatory force is given by (Wolford and Smith, 1976):
fs ј
dwрG=2_Ю1=2
2_D2Nc
(9-167)
The quantity dw is the diameter of the spring wire, G is the shear modulus, _
is the density of the spring material, D is the mean diameter of the spring
coil, and Nc is the number of active coils for the spring. The surge frequency
for the spring should be at least 15 times the forcing frequency for the
system to avoid problems with resonance in the spring. The surge frequency
may be increased by using a larger spring wire diameter or a smaller spring
coil diameter (or a smaller spring index, D=dw).
Example 9-12. A machine having a mass of 80 kg (176.4 lbm) is to be
supported by four metal springs. The springs are to be made of harddrawn
steel wire and have squared and ground ends. The damping ratio
for the springs is _ ј 0:050, and the required transmissibility is 0.05 or
_26 dB. The driving force for the machine has a maximum amplitude of
5.00kN and a frequency of 36 Hz. Determine the dimensions of the spring.
First, let us determine the required frequency ratio. The parameter _ is
found from Eq. (9-109):
Vibration Isolation for Noise Control 461
Copyright © 2003 Marcel Dekker, Inc.
_ ј 1 ю р2Юр0:050Ю2р1 _ 0:052Ю
р0:05Ю2 ј 1 ю 1:995 ј 2:995
The frequency ratio is as follows:
r4 _ р2Юр2:995Юr2 _ р1 _ 0:052Ю
р0:05Ю2 ј 0
r2 ј 2:995 ю р2:9952 ю 399Ю1=2 ј 23:193
r ј 4:816 ј f =fn
The undamped natural frequency for the system is as follows:
fn ј р36Ю=р4:816Ю ј 7:475 Hz
The required spring constant for one spring, supporting a mass of
р1
4MЮ ј 20 kg may now be found:
KS ј р4_2Юр7:475Ю2р20Ю ј 44:12 _ 103 N=m ј 44:12kN=m р252 lbf=inЮ
Let us try a spring with a spring index рD=dwЮ _ 6 and Nc ј 5 active coils.
The spring wire diameter may be found from Eq. (9-162) with a shear
modulus of 79.3 GPa:
dw ј р8Юр44:12Юр103Юр6Ю3р5Ю
р79:3Юр109Ю ј 4:807 _ 10_3 m ј 4:807mm р0:1893 inЮ
The next larger standard gauge is #6 W&M gauge wire, with a diameter of
dw ј 0:1920 in ј 4:877 mm. Let us try this size wire for the spring.
The actual mean diameter of the spring may be found from Eq. (9-
162):
D ј р79:3Юр109Юр0:004877Ю4
р8Юр44:12Юр103Юр5Ю
" #1=3
ј 0:02940m ј 29:40mm р1:157 inЮ
The actual spring index is as follows:
D=dw ј р29:4Ю=р4:877Ю ј 6:029
The outside diameter of the spring is as follows:
Do ј D ю dw ј 29:40 ю 4:877 ј 34:28mm р1:349 inЮ
Let us check the static shear stress in the spring. The shear correction
factor is found from Eq. (9-164):
ksh ј р2Юр6:029Ю ю 1
р2Юр6:029Ю ј 1:041
462 Chapter 9
Copyright © 2003 Marcel Dekker, Inc.
The static shear stress for the spring is found from Eq. (9-163):
ss ј р8Юр1:041Юр20Юр9:806Юр0:02940Ю
р_Юр0:004877Ю3 ј 131:8_106 Pa
ј 131:8MPa р19,120 psiЮ
The shear yield strength for a hard-drawn wire with a diameter of 4.877mm
is found from the data in Table 9-6.
sys ј р758Юр1=4:877Ю0:192 ј 559:2MPa р81,100 psiЮ
The static factor of safety for the spring is as follows:
FS ј
sys
ss ј
559:2
131:8 ј 4:24 > 3
The static shear stress level is satisfactory.
The total number of coils for the spring with squared and ground ends
is as follows:
Nt ј Nc ю 2 ј 5 ю 2 ј 7 coils total
The solid height of the spring is as follows:
Hs ј р4:877Юр7Ю ј 34:14mm р1:344 inЮ
The static deflection for the spring is found from Eq. (9-16):
d ј р20Юр9:806Ю
р44,120Ю ј 0:004445m ј 4:445mm р0:175 inЮ
The magnification factor is found from Eq. (9-81):
MF ј
1
fЅ1 _ р4:816Ю2_2 ю Ѕр2Юр0:050Юр4:816Ю_2g1=2 ј 0:04505 ј
KSymax
Fo
The maximum amplitude of vibration for the system is as follows:
ymax ј р0:04505Юр5000Ю
р44,120Ю ј 0:005105m ј 5:105mm р0:201 inЮ
The maximum deflection of the spring is the sum of the static and dynamic
displacements:
dmax ј d ю ymax ј 4:445 ю 5:105 ј 9:55mm р0:376 inЮ
To ensure that the spring will not be compressed solid, let us take the design
maximum deflection as follows:
dmax(design) ј 1:25dmax ј р1:25Юр9:55Ю ј 11:94mm р0:470 inЮ
Vibration Isolation for Noise Control 463
Copyright © 2003 Marcel Dekker, Inc.
The design free height of the spring may now be determined as the sum
of the solid height and the design maximum deflection:
Ho ј Hs юdmax(designЮ ј 34:14ю11:94 ј 46:08mm р1:814 inЮ
The pitch of the spring may be determined from the data in Table 9-7.
ps ј
Ho _ 2dw
Nc ј
46:08 _ р2Юр4:877Ю
р5Ю ј 7:27mm р0:286 inЮ
The pitch is the center-to-center spacing of the wire in adjacent coils of the
spring. There are р1=7:27Ю ј 0:1376 coils/mm ј 1:376 coils/cm or 3.50 coils/
inch height of the spring.
Let us check the buckling stability of the spring. The _ parameter may
be found from Eq. (9-158):
_ ј Ѕ1 ю р2Юр0:287Ю_Ѕр_Юр29:40Ю=р46:08Ю_2
р2 ю 0:287Ю ј 2:765 > 1
The spring is quite stable and buckling will not be a problem.
Finally, let us check the surge frequency from Eq. (9-167):
fs ј р0:00487Ю
р2_Юр0:02940Ю2р5Ю
р79:3Юр109Ю
р2Юр7830Ю
" #1=2
ј 404 Hz
The ratio of the surge frequency to the driving force frequency is as follows:
fs=f ј р404Ю=р36Ю ј 11:2
Although the surge frequency is not greater than 15 times the forcing frequency,
surging probably would to be a serious problem, in this case,
because of the damping in the support system.
A summary of the spring characteristics is as follows:
spring wire diameter, dw 4.877mm (0.1920 in), #6 W&M
gauge wire
spring mean diameter, D 29.40mm (1.157 in)
spring outside diameter, Do 34.28mm (1.349 in)
number of coils 7 total coils; 5 active coils
spring pitch, ps 7.27mm (0.286 in)
free height, Ho 46.08mm (1.814 in)
solid height, Hs 34.14mm (1.344 in)
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