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CHAPTER 10

Mechanical Springs

Flexibility is sometimes needed and is often provided by metal bodies with cleverly

controlled geometry

Such flexibility can be linear or nonlinear in relating deflection to load.

These devices allow controlled application of force or torque; the storing and release of

energy can be another purpose

In general, spring may be classified as:

1. 

Wire springs such as helical springs of round or square wire, made to resist anddeflect under tensile, compressive, and torsional loads.

2.  Flat springs which includes cantilever and elliptical types, wound motor-or clock-type power springs, a flat spring washers, usually called Belleville springs.

3.  Special-shaped springs

10.1 Stresses in Helical Springs

A round-wire helical compression spring loaded by the axial force F is shown in figure10-1a.

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Two important parameters in spring design:

1.  The mean coil diameter D 2.  The wire diameter d .

If the spring is cut at some point, the effect of the removed portion replaced by the netinternal reactions. See figure 10-1b.

Using the equation of equilibrium, the cut portion would contain a direct shear force  F  and a torsion T = FD/2

The maximum stress in the wire may be computed by superposition of the direct shear

stress and the torsional shear stress

 A

 F 

 J 

Tr +=maxτ  

Replace T  = FD/2, r =d /2, J =πd 4/32, A=πd 2/4, τmax=τ, we have

23

48

d 

 F 

d 

 FD

ππτ   +=   (10-1)

Defined the spring index as:

d 

 DC  =   (10-2)

Equation (10-2) measures the coil curvature.

Substitute equation (10-2) into equation (10-1), we have:

3

8

d 

 FD K S 

πτ  =   (10-3)

Where K S  is a shear-stress correction factor and is defined by the equation:

C 

C  K S 

2

12   +=   (10-4)

For most springs the range for C  is: 6 ≤ C  ≤ 12.

 Note that equation (10-3) is a general formula and is applied for both static and fatigue

loads.

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10.2 The Curvature Effect

Equation (10-1) is based on the wire being straight. But, the wire has a curvature.

The curvature of the wire increases the stress on the inside of the spring but decreases it

only slightly on the outside.

In static load: these stresses can normally be neglected because it will be relieved by local

yielding with first application of a load.

In fatigue load: It is important to include the curvature stress because of its high

localized.

To include the curvature in equation (10-3), the factor K S  is need to be modified.

Two different curvature factors known as:

Wahl factor:

C C 

C  K W 

615.0

44

14+

−−

=   (10-5)

Bergstrasser factor:

34

24

−+

=C 

C  K  B   (10-6)

The different between the two equations is less than one percent which makes equation

(10-6) is preferred to used

The curvature correction factor can now be obtained by canceling out the effect of the

direct shear form K  B, thus

( )

( )( )1234

242

+−+

==C C 

C C 

 K 

 K  K 

S 

 BC    (10-7)

To predict the largest shear stress we will use the equation:

3

8

d 

 FD K  B πτ  =  

10.3 Deflection of Helical Springs 

Using the strain energy method to include both the torsional and shear components, thus

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 AG

l  F 

GJ 

l T U 

22

22

+=  

Substituting T  = FD/2, l =π DN , J =πd 4/32, A=πd 2/4, in the previous equation we have:

Gd 

 DN  F 

Gd 

 N  D F U 

2

2

4

32 24 +=  

where N = N a = number of active coils

The total deflection y can now be calculated by:

Gd 

 FDN 

Gd 

 N  FD

 F 

U  y

24

3 48+=

∂∂

=  

Since, C = D/d , the deflection y becomes:

Gd 

 N  FD y

4

38≅   (10-8)

The spring rate can be calculated by

 N  D

Gd 

 y

 F k 

3

4

8==   (10-9)

10.4 Compression Springs

There are four standard types on helical compression springs. They are plain end, squared

end, plain-ground end, and squared-ground as illustrated in figure 10-2.

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A spring with plain ends has a noninterrupted helicoids; the ends are the same as if a long

spring had been cut into sections.

A spring with plain ends that are squared or closed is obtained by deforming the ends to a

zero-degree helix angle.

Springs should always be both squared and ground for important applications, because a

 better transfer of the load is obtained.

A spring with squared and ground ends compressed between rigid plates can be

considered to have fixed ends.

The type of end used affects the number of active coils  N a  and the solid height of thespring.

Square ends effectively decrease the number of total coils N t  by approximately two:

 N t = N a+2

Table 10-1 shows how the type of end used affects the number of coils and the spring length.

Types of Springs Ends

Term Plain Plain and groundSquared or

ClosedSquared and

Ground

End coils, N e  0 1 2 2

Total coils, N t    N a   N a+1  N a+2  N a+2

Free length, L0   pN a+d    p( N a+1)  pN a+3d    pN a+2d  

Solid length, L s  d ( N t +1) dN t   d ( N t +1) dN t  

Pitch, p  ( L0-d )/ N a   L0/ ( N a+1) ( L0-3d )/ N a  ( L0-2d )/ N a 

Forys gives an expression for calculating the solid length of squared and ground ends

 L s=( N t - a)d  

Where a varies, with an average of 0.75 which means in this case that the entry dN t   in

table 10-1may be overstated.

The way to check these variations is to take a spring and count the wire diameters in thesolid stack

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Set removal or presetting:

A process used to induce (make) useful residual stresses.It is done by making the spring longer than needed and then compressing it to its solid

height L s.

This operation sets the spring to the required final free length L0 and, since the torsional

yield strength has been exceeded, induces (makes) residual stresses opposite in the

direction to those induced in service.

Set removed increases the strength of the spring and so is especially useful when the

spring is used for energy-storage purposes. But, set removal should not be used when

springs are subjects to fatigue

10.5 Stability

Compression coil springs may buckle when the deflection becomes too large.

The critical deflection is given by the equation:

   

  

 −−=

2/1

2

'

2'

10 11eff 

cr 

C C  L y

λ  (10-10)

where ycr  is the deflection corresponding to the onest of instability

λeff  is the effective slenderness ratio and is given by:

 D

 Leff 

0αλ   =   (10-11)

α the end condition constant

'

2

'

1 &C C   are elastic constants defined by the equations:

( )

( ) E G

G E C 

G E 

 E C 

+−=

−=

2

2

2

2'

2

'

1

π  

The end condition constant α depends upon how the ends of the spring are support.

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Table 10-2 gives values of α for usual end conditions.

End condition Constant α Spring supported between flat parallel surfaces (fixed ends) 0.5

One end supported by flat surface perpendicular to spring axis (fixed);

other end pivoted (hinged)

0.707

Both ends pivoted (hinged) 1

One end clamped; other end free 2

Absolute stability occurs when equation the term 2'2 / eff C    λ  in equation (10-10) is greater

than unity. Thus, for stability we have:

( )2/1

02

2

+−

< E G

G E  D L

α

π  (10-12)

For steels, equation 10-12 becomes: L0 < 2.63 D/α For squared and ground ends α =0.5 and L0 < 5.26 D 

10.6 Spring Materials

A great variety of spring materials are available to the designer, including plain carbon

steels, alloy steels, and corrosion-resisting steels, as well as nonferrous materials such as

spring brass, and various nickel alloys.

Most common used steels are found in table 10-3

The UNS steel listed in appendix A should be used in designing hot-worked, heavy-coilsprings, as well as flat springs, leaf springs, and torsion bars.

Spring materials may be compared by an examination of their tensile strength; these vary

so much with wire size that they cannot be specified until the wire size is known

The graph of tensile strength versus wire diameter is almost a straight line for some

materials

This can be expressed as a linear function which can be written as:

mut  d 

 AS    =   (10-14)

where A and m are constants and are given in table 10-4.

 Note:

If d  measured in mm, then the unit of A is MPa.mmm.

If d  measured in inches, then the unit of A is kpsi.inm.

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A very rough estimate of the torsional yield strength can be obtained by assuming that the

tensile yield strength is between 60 and 90 percent of the tensile strength.

ut  yut    S S S  9.06.0  

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 F 

C 

C  K 

 D K 

S d  F 

d 

 FD K S 

 B

 B

 sy

 B sy

734.25

124.13)82.10(4

2)82.10(4

34

24;

8

8

3

3

=⇒

=−+

=−+

==⇒

==

π

πτ

 

c)  the scale of the spring:

a N  D

Gd k 

3

4

8=  

From table 10-5, d  = 0.94mm = 0.037in ⇒ G=80GPa N a for both ends squared: N t  = N a +2 ⇒  N a = 12.5-2=10.5 turns

m N  N  D

Gd k 

a

/7078 3

4

==  

d) 

The deflection that would be caused by the load in part (b):

mmmk 

 F  y 4.360364.0

707

734.25====  

e)  The solid length of the spring, from table 10-1:

mm N d  L t  s 69.12)15.12(94.0)1(   =+=+=  

f)  The free length of the spring suppose to have to be sure when it is compressed and

then released, there will be no permanent change in it is free length:

mm y L L  s 09.494.3669.120   =+=+=  

g)  To Check if buckling is possible:

For steel and squared ground: L0 < 5.26 D ⇒  L0 < 53.94mm, since L0 = 49.9mm,therefore there will be no buckling. 

h)  The pitch of the body coil is:

( )   ( )41.4

5.10

94.0(309.4930 =−

=−

=a N 

d  L p  

10.7 Helical Compression Spring Design for Static Service

The design of a new spring involves the following considerations:

•  Space into which the spring must fit and operate•  Values of working forces and deflections•  Accuracy and reliability needed•  Tolerances and permissible variations in specifications

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•  Environmental conditions such as temperature and presence of a corrosiveatmosphere

•  Cost and quantities needed

Designers use these factors to select a material and specify suitable values for wire size,

the number of turns, the diameter and free length, the type of ends, and the spring rateneeded to satisfy the working force-deflection requirements.

Some important limits:

•  The range index is 4 ≤ C  ≤ 12, Lower indexes being more difficult to form (because

of the danger of surface cracking). Higher indexes tending to tangle (knot) often

enough to require individual packing.

•  The recommended range of active turns is 3 ≤  N a ≤ 15.

• 

To maintain linearity when a spring is about to close, it is necessary to avoid thegradual touching of coils.

•  The characteristic of the force-deflection for the helical coil spring is nearly linear.

•  The spring force is not reproducible for very small deflections and near closure, ⇒ nonlinear behavior begins as the number of active turns diminishes as coils begin totouch.

•  The designer confines (limits) the spring’s operating point to the central 75% of thecurve between no load F =0 and closure (end)  F = F S  

Thus, the maximum operating force

S  F  F 8

7max ≤  

Defining the fractional overrun to closure as ξ, where

( ) max1   F  F S    ξ+=   (10-17)

From the pervious two equations:

S S   F 

 F 

8

7

1≤

+ ξ 

Solve for ξ gets:

7

1≥ξ  

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Summery of the recommended design conditions:

124   ≤≤ C    (10-18)

153   ≤≤   a N    (10-19)

7

1≥ξ   (10-20)

2.1≥ sn   (10-21)

•  Figure of merit:

1.  The cost of wire from which the spring is wound (coiled)2.  Help in making the decision for the optimal spring design.

3. 

Formula: ( )4

22

 D N d  RMC  fom   t γπ−=   (10-22)

where RMC  is the relative material cost.

•  Spring design is an open-ended process•  There are many decisions to be made, and may possible solution paths as well as

solutions.

•  One possible approach for design spring coil:

1. 

Make a priori decisions, with hard-drawn steel wire the first choice (relativematerial cost =1)

2.  Choose a wire size d

3.  generate a column of parameters: d , D, OD or ID, N a, L s, Lo, ( Lo)cr , ns, and fom

By incrementing wire sizes available, a table of parameters will be generated. Then, thedesign recommendation conditions are applied to choose the good design.

After wire sizes are eliminated, choose the spring design with the highest figure of merit.

This will give the optimal design despite the presence of a discrete design variable d andaggregation of equality and inequality constraints.

The column vector of information can be generated by using the flowchart displayed infigure 10-3.

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As wound and set-removed springs, operating over a rod, or in a hole free of rod or hole.

In as-wound springs the controlling equation must be solved for the spring index as

follows:

From equation 10-3, with: τ =S  sy/n s, C = D/d , K  B 

3

8

d 

 D F  K 

n

S S 

 B

 s

 sy

π=  

With

34

24

−+

=C 

C  K  B  and ( ) max1   F  F S    ξ+=  

Therefore:

( )

  +

−+

=2

max18

34

24

d 

C  F 

C 

C 

n

S 

 s

 sy

π

ξ 

Let

( )2

max18&d 

 F 

ns

S  sy

π

ξβα

  +==  

Substituting α and β into the torsion equation and simplifying yields a quadraticequations in C . The larger of the two solutions will yield the spring index

2/12

4

3

4

2

4

2

  

 

 

 

 −  

 

 

 

    −+

−=

β

α

β

βα

β

βαC    (10-23)

Example

An A228 wire helical compression spring is needed to support a 20-Ibf load after beingcompressed 2 in. Because of assembly considerations the solid height cannot exceed 1 in

and the free length cannot be more than 4 in. Design the spring.

Solution:

For A228 wire helical spring: From table, A = 201 kpsi-inm and m = 0.145

From table 10-5, E = 28.5Mpsi, G =11.75Mpsi (choosing d  > 0.064 in)

Ends squared and groundFunction: F max=20 Ibf, ymax=2 in

Safety: use design factor at solid height of (n s)d = 1.2

Select robust linearity: ξ= 0.15Use as wound spring, S sy=0.45S ut from table 10-6

Decision variable: d  = 0.08 in.

From fig. 10-3 and table 10-6

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( )

( )

417.04/6.2

43.4/63.2

264.3)1(

964.0

05.12205.10

05.10)8/(

923.0

2.1/

7.108)/()1(8

128.1)34/()24(

8424.0

53.104

3

4

2

4

2

15.918

713.108

46.13045.0

22

max

max

3

max

4

3

max

2/12

2max

−=−=

===++=+=

===+=

===+=

===+=

=−+===

=  

 

 

 

 −  

 

  

    −+

−=

=+=

==

==

 D N d  fom

in D L

in y L y L L

indN  L

turns N 

turns F  D yGd  N 

ind  DOD

S n

kpsid  D F  K 

C C  K 

Cd  D

C 

kpsid  F 

kpsin

S 

kpsid 

 AS 

t 

cr o

 s s so

t  s

t 

a

 s sy s

 B s

 B

 s

 sy

m sy

π

α

ξ

τ

πξτ

β

α

β

βα

β

βα

πξβ

α

 

Repeat the above analysis for other diameters and form a table to select the best spring

design:d 0.063 0.067 0.071 0.075 0.08 0.085 0.09 0.095

D 0.391 0.479 0.578 0.688 0.843 1.017 1.211 1.427

C 6.205 7.153 8.143 9.178 10.53 11.96 13.46 15.02

OD 0.454 0.546 0.649 0.763 0.923 1.102 1.301 1.522

N a   39.1 26.9 19.3 14.2 10.1 7.3 5.4 4.1

L s   2.587 1.936 1.513 1.219 0.964 0.790 0.668 0.581

L o   4.887 4.236 3.813 3.519 3.264 3.090 2.968 2.881

(L o )cr  2.06 2.52 3.04 3.62 4.43 5.35 6.37 7.51

n s   1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2

fom-0.409 -0.399 -0.398 -0.404 -0.417 -0.438 -0.467 -0.505

Examine the table and perform the adequacy assessment.

The constraint 3≤  N a ≤ 15 cancel wire diameters less than 0.08 inThe constraint 4≤ C ≤ 12 cancel diameters larger than 0.085 in.The constraint L s > 1 in cancel diameters less than 0.080 in.

The constraint Lo > 4 in cancel diameters less than 0.071 in.

The buckling criterion cancel free length longer than (Lo)cr , which cancel diameters lessthan0.075in.

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The result is that there are only two springs in the feasible domain.

The figure of merit decides that the wire diameter is 0.08in.

10.8 Critical Frequency of Helical Springs 

When helical springs are used in applications requiring a rapid reciprocating motion, the

designer must be certain that the physical dimensions of the spring are not such as to

create a natural vibratory frequency close to the frequency of the applied force; otherwiseresonance may occur, resulting in damaging stresses.

The governing equation for the translational vibration of a spring is the wave equation:

2

2

22

2

t 

u

kgl 

W 

 x

u

∂∂

=∂∂

  (10-24)

wherek  = spring rate

g = acceleration due to gravityl = length of spring

W = weight of spring

 x = coordinate along length of springu = motion of any particle at distance x 

Equation (10-24) has a harmonic solution and it depends on:

1.  Given physical properties

2. 

End conditions of the spring

The natural frequencies for a spring placed between two flat and parallel plates:

,...3,2,1==   mW 

kg mπω  

Since =2π f , thus

,...3,2,12

1==   m

W 

kg 

m f   

If m =1, it is called the fundamental frequency and it is equal to:

W 

kg  f 

2

1=   (10-25)

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For spring has one end against a flat plate and other end free, the frequency is

W 

kg  f 

4

1=   (10-26)

W  can be calculated as:

( )   γ πγ ππ

γ    aa   DN d  DN d 

 ALW  222

4

1

4===  

where  is the specific weight

To avoid resonance with the harmonic it is required that the fundamental critical

frequency is 15~20 ! the frequency of the force or motion of the spring.

If it is not enough, the spring should be redesign to increase k  or decrease W  

10.9 Fatigue Loading of Helical Compression Springs

Helical springs are never used as both compression and extension springs. This is becausethey are usually assembled with a preload so that the working load is additional. Thus,the spring application fall under the condition of fluctuating loads. Thus,

am

a

 F  F  F  F 

 F 

 F  F  F 

+=+

=

−=

minminmax

minmax

2

2 (10-31)

Therefore, the amplitude and midrange shear stresses respectively can be written as:

3

3

8

8

d 

 D F  K 

d 

 D F  K 

m Bm

a Ba

πτ

πτ

=

=  (10-32,33)

Endurance limits for infinite life were found to be for unpeened and peened springs:

Unpeened: S  sa=35.0 kpsi (241MPa) S  sm=55.0kpsi (379MPa)Peened: S  sa=57.5 kpsi (398MPa) S  sm=77.5kpsi (534MPa)

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Then S  se can be calculated using Goodman theory:

( ) su sm

 sa

 seS S 

S S 

/1−=   where S  su=0.67S ut  

Example

A helical compression spring, made of A228 wire, has a wire size of 0.092 in, an outside

diameter of 0.5625 in, a free length of 4.125 in, 21 active coils, and both ends squared

and ground. The spring is to be assembled with a preload of 5 lb and will operate to amaximum load of 35 lb during use. Knowing that the spring is unpeened type.

1.  Find the factor of safety guarding against a fatigue failure.

2.  Find the critical operating frequency

Solution

Given OD=0.5625 in, d =0.092 in, N a=21

A228 spring type material, G=11.75GPsi

 F max=35lb, F min=5lbBoth ends squared and ground, and unpeened.

1.  The fatigue factor of safety:

( )

( )

kpsid 

 D F  K 

kpsid  D F  K 

lb F 

lb F 

C C  K 

d  DC 

ind OD D

m

 Bm

a Ba

m

a

 B

8.338

7.298

202/535

152/535

287.1)34/()24(

11.5/

4705.0

3

3

==

==

=+==−=

=−+===

=−=

πτ

πτ

 

From table 10-4, A=201kpsi, m=0.145, therefore

kpsiS S 

kpsid 

 AS 

ut  su

mut 

347.19067.0

1.284

==

== 

For unpeened spring:

( )  kpsi

S S 

S S 

kpsiS kpsiS 

 su sm

 sa se

 sm sa

22.49/1

5535

=−

=

== 

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  118

The factor of safety guarding against failure to be

28.11

=+

=⇒=+ sem sua

 su se

 f 

 f  su

m

 se

a

S S 

S S n

nS S    ττ

ττ 

2. 

The critical frequency:

 Hz W 

kg  f 

lbf W 

inlb DN d W 

inlb N  D

Gd k 

a

a

281586.0

)386(1.48

2

1

2

1

0586.0

/284.0;4

1

/1.488

322

3

4

===

=∴

==

==

γ γ π 

Check for operating frequency:

For good design 1.1420

20   ≤⇒≤⇒≥   opopop

 f  f 

 f  f 

 f  

10.10 Helical Compression Spring Design for Fatigue Loading

Example

A helical compression spring, made of A228 wire, with infinite life is needed to resist adynamic load that varies from 5 to 20 lbf at 5 Hz while the end deflection varies from 0.5

to 2 in. Because of assembly considerations, the solid height cannot exceed 1.2 in and the

free length cannot be more than 4 in. The springmaker has the following wire sizes instock: 0.069, 0.071, 0.080, 0.085, 0.090, 0.095, 0.105, and 0.112 in.

SolutionFrom table 10-4 for A228: A=201 kpsi.in

m, m=0.145, G =11.75Mpsi, relative cost of

wire=2.6

Surface treatment: unpeended

End treatment: squqred and ground

Select robust linearity: ξ= 0.15 and f op = 5HzFatigue safety: n f =1.5 using the Sines-Zimmerli fatigue-failure criterion

Use as wound spring, S  sy = 0.45S ut  from table 10-6

 F min = 5lbf, F max= 20lbf, ymin= 0.5 in, ymax = 2 in, spring operates free ( no rod or hole)

Decision variable: wire size d  

For d  = 0.112 in

8/21/2019 Me 301 Chapter 10

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  119

kpsiS S 

kpsiS S 

kpsiS 

inlbf  y

 F k 

lbf  F lbf  F 

ut  sy

ut  su

ut 

ma

2.12445.0

0.18567.0

1.276112.0201

/102

20

5.122

5205.7

2

520

145.0

max

max

====

==

===

=+

==−

=

 

For Unpeened spring form equation (10-28): S  sa=35.0 kpsi S  sm=55.0kpsi

In Sines failure criterion, the terms S  sm ignores, thus

( )  kpsi

S S 

S S 

 su sm

 sa

 se 3501

35

/1=

−=

−=  

To find C , we replace S  sy by S  se, n s by n f , and (1+ξ) F max by F a, thus:

( )

turns N 

turnsk  DGd  N 

lbf  F  F 

inCd  D

C 

kpsid 

 F 

kpsin

S 

t 

a

 s

a

 f 

 se

98.7298.5

98.5)8/(

231

569.1

005.144

3

4

2

4

2

523.18

333.23

34

max

2/12

2

=+===

=+=

==

=  

 

 

 

 −  

 

  

    −+

−=

==

==

ξ

β

α

β

βα

β

βα

πβ

α

 

( )

( )

 Hz W 

k  f 

lbf  DN d 

W 

in D L

in L L y

ind  DOD

ind  D ID

ink  F  L y L L

indN  L

n

a

cr o

 so s

 s s s so

t  s

108386

5.0

0825.04

253.8/63.2

3.2

681.1

457.1

194.3/

894.0

22

==

==

===−==+==−=

=+=+===

γ π

α 

8/21/2019 Me 301 Chapter 10

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  120

01.14/6.2

74.1/

5.1/

56.718

89.388

334.238

094.1)34/()24(

22

3

3

3

−=−=

==

==

==

==

==

=−+=

 D N d  fom

S n

S n

kpsid 

 D F  K 

kpsi

d 

 D F  K 

kpsid 

 D F  K 

C C  K 

t 

 s sy s

a sa f 

 s B s

m Bm

a

 Ba

 B

π

τ

τ

πτ

π

τ

πτ

 

Repeat the above analysis for other wire diameters and form a table to select the best

spring design:

d 0.069 0.071 0.080 0.085 0.090 0.095 0.105 0.112

D 0.297 0.332 0.512 0.632 0.767 0.919 1.274 1.569

ID 0.228 0.261 0.432 0.547 0.677 0.824 1.169 1.457

OD 0.366 0.403 0.592 0.717 0.857 1.014 1.379 1.681

C 4.33 4.67 6.40 7.44 8.53 9.67 12.14 14.00

N a   127.2 102.4 44.8 30.5 21.3 15.4 8.63 6.0

L s   8.916 7.414 3.74 2.75 2.10 1.655 1.116 0.895

L o   11.216 9.714 6.040 5.05 4.40 3.955 3.416 3.195

(L o )cr  1.562 1.744 2.964 3.325 4.036 4.833 6.703 8.25

n f   1.50 1.50 1.50 1.50 1.50 1.50 1.50 1.50

n s   1.86 1.85 1.82 1.81 1.79 1.78 1.75 1.74

f n  

87.5 89.7 96.9 99.7 101.9 103.8 106.6 108fom -1.17 -1.12 -0.983 -0.948 -0.930 -0.927 -0.958 -1.01

Examine the table and perform the adequacy assessment.

•  General Constraints:o  The constraint 3≤  N a ≤ 15 cancel wire diameters less than 0.105 ino  The constraint 4≤ C ≤ 12 cancel diameters larger than 0.105 in.

•  Problem Constraints:o  The constraint L s ≤ 1.2 in cancel diameters less than 0.1050 ino

 

The constraint Lo ≤ 4 in cancel diameters less than 0.095 ino   f n ! 20 f op ⇒  f n ! 100Hz cancel diameters less than 0.090 ino  The buckling criterion cancel free length longer than ( Lo)cr , which cancel

diameters less than0.075in.

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  121

10.11 Extension Springs

Most of the preceding discussion of compression springs applied equally to helicalextension springs

The natural frequency of a helical extension spring with both ends fixed against axialdeflection is the same as that for a helical spring in compression

But, In extension spring, the coils are usually close wound so that there is an initialtension or so termed preload  F . Therefore, no deflection occurs until the initial tension

 built into the spring is overcome. That is: the applied load F  becomes larger than initial

tension ( F  > F i)

The load transfer can be done with: a threaded plug or a swivel hook

Figure 10-6 shows types of ends used on extension springs

Stresses in the body of the extension spring are handled the same as compression springs.

In designing a spring with a hook end, bending and torsion in the hook must be included

in the analysis.

Figure 10-7 shows two common used method of designing the end. c and d shows an

improved design due to a reduced coil diameter, not to elimination of stress

concentration. The reduced coil diameter results in a lower stress because of the shortermoment arm.

 No stress-concentration factor is needed for the axial component of the load

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  122

The maximum tensile stress at A, due to bending and axial loading, is given by:

( )

+=

23

416

d d 

 D K  F   A A

ππσ   (10-34)

Where ( K ) A is a bending stress correction factor for curvature, given by:

( )( )   d 

r C 

C C 

C C  K   A

11

11

1

2

1 2

14

14=

−−−

=   (10-35)

The maximum torsional stress at point B is given by:

( )3

8

d 

 FD K   B B

πτ   =   (10-36)

where ( K ) B is the stress correction factor for curvature and it is given by:

( )d 

r C 

C 

C  K   B

2

2

2

2 2

44

14=

−−

=   (10-37)

8/21/2019 Me 301 Chapter 10

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  123

If coils in contact with one another, the springs is known as close-wound.

To have a more accurate free length, it is preferred that some initial tension in closewound springs.

Figure 10-8a shows the load-deflection curve, where y  is the extension beyond the freelength Lo and F i is the initial tension in the spring that must be exceeded before the spring

deflects

Thus, the load-deflection relation can be written as:

 F = F i+ky 

Where k  is the spring rate

The free length Lo of a spring measured inside the end loops or hooks as shown in figure

10-8b can be expressed as

 Lo= 2( D-d ) + ( N 

b+1)d  = (2C -1+ N 

b)d  

 D mean coil diameter

 N b number of body coils

C  the spring index

The equivalent number of active helical turns N a for use in equation 10-9 is

 N a = N b + G/ E  

where G and E  are the shear the tensile modulus of elasticity

8/21/2019 Me 301 Chapter 10

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  124

The amount of initial tension that a springmaker can routinely incorporate is as shown in

figure 10-8c

The preferred range can be expressed in terms of the uncorrected torsional stress τi as

( )  psi

C 

C i    

  

     −−±=

5.6

341000

105.0exp

33500τ   (10-41)

The maximum allowable corrected stresses ( K W  or K  B) for static applications of extension

springs are given in table 10-7

Percent of Tensile Strength

Materials body end endPatented, cold-drawn or hardened and tempered

carbon and low-alloy steels

45-50 40 75

Austenitic stainless steel and nonferrous alloys 35 30 55

In Torsion In Bending

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  125

Example

A hard drawn steel wire extension spring has a wire diameter of 0.035 in, an outside coil

diameter of 0.248 in, hook radii of r 1= 0.106 in and r 2 = 0.089in, and an initial tension of1.19 lbf. The number of body turns is 12.17. From the given information:

1.  Determine the physical parameters of the spring( D, C , K  B, N a, k , Lo, ymax)

2. 

Check the initial preload stress conditions3.  Find the factors of safety under a static 5.25 lbf load.

Solution

d  = 0.035 in, OD = 0.248 in, r 1 = 0.106 in, r 2 = 0.089 in, N b = 12.17

 F i = 1.19 lbf, F max=5.25

HD steel ⇒ Form table 10-5, with 0.033 < d  < 0.063 ⇒  E = 28.7Mpsi, G = 11.6MpsiFrom table 10-4, A = 140kpsi-in

m, m = 0.19

1.  The physical parameters

( )   ind  N C  L

inlbf  N  D

Gd k 

turns E G N  N 

C 

C  K 

d  DC d OD D

bo

a

 Ba

 B

817.012

/76.178

57.127.28/6.1117.12/

234.134

24

086.6/213.0035.0248.0

3

4

=+−=

==

=+=+=

=−+

=

== =−=−=

 

The deflection under the service load is:( )

ink 

 F  F  y   i 229.0

76.17

19.125.5maxmax   =

−=

−=  

Therefore the maximum spring length is:

 L = Lo + ymax =0.817+0.229 = 1.046 in

2.  Initial preload stress condition:

The uncorrected initial stress is given by equation (10-3) without the correction factor:

( )   kpsid 

 D F iuncorr i

1.158

3  ==

πτ  

The preferred range is given by equation (10-41):

( )( )

  kpsior kpsiC 

C  pref i

2.212.143525176815.6

341000

105.0exp

33500=±= 

  

     −−±=τ  

Thus, the initial tension of 15.1 kpsi is in the preferred range.

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  126

3.  The factor of safety under static load

We need to check three positions:

•  The shear stress under the service load•  The bending at the end hook which is represented by point A 

• 

The torsion at the end hook which is represented by point B 

1.  For the shear stress under the service load

( )( )

45.182

7.26445.0

82035.0

)213.0)(25.5(8234.1

7.264035.0

140

8&45.0;

3max

19.0

3

max

max

max

==

==

===

===

n

kpsi

kpsid 

 AS 

d 

 D F  K S S 

S n

mut 

 But  sy

 sy

πτ

πτ

τ

 

2.  The bending at the end hook which is represented by point A 

( )

( ) ( )( )

( )

27.19.156

5.198

9.15614.1

057.6/214

14;

416

5.1987.26475.075.0;

11

11

1

2

1

23max

==

=⇒=

==−−−

=

+=

====

 A

 A A

 A A A

ut  y

 A

 y

 A

n

kpsi K 

d r C C C 

C C  K 

d d 

 D K  F 

kpsiS S S 

n

σ

ππσ

σ

 

3.  The torsion at the end hook which is represented by point B 

( )

( ) ( )

( )

35.14.78

88.105

4.7818.1

086.5/244

14;

8

88.1057.2644.04.0;

22

2

2

3

max

==

=⇒=

==−−

==

====

 A

 B B

 B B B

ut  sy

 B

 sy

 B

n

kpsi K 

d r C C 

C  K 

d 

 D F  K 

kpsiS S S 

n

τ

πτ

τ

 

 Note that S  sy=0.4Sut from table 10-7 under torsion for the end part.

From all three calculations, the yield will first occur due to the bending of the end hook.

8/21/2019 Me 301 Chapter 10

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  127

S  sy S  sm 

S  sy 

S  sa 

r

1=+ sy

 sm

 sy

 sa

S 

S 

S 

S  

τi 

Example: Fatigue

The helical extension spring of the pervious example is subjected to a dynamic loading

from 1.5 lbf to 5 lbf. Estimate the factors of safety using Goodman failure criterion for1.  The coil fatigue

2.  The coil yield

3. 

The end hook bending fatigue at point A 4.  The end hook torsion fatigue at point B 

Solution

d  = 0.035 in, D = 0.213 in, r 1= 0.106 in, r 2 = 0.089 in, N b = 12.17

 F i = 1.19 lbf, F min=1.5, F max=5,

From the pervious example we have:

C =6.086, Lo=0.817, k =17.76 lbf/in

 K B=1.234, ( K )A=1.14, ( K )B=1.18, (τi)uncorr =15.1 kpsiS ut = 264.7kpsi, S  su = 0.67S ut = 177.3 kpsi, S  y =198.5 kpsi, S  sy =119.1 kpsi (shear in body)

1. 

The fatigue in the body coil:( )

( )

kpsid 

 D F  K 

kpsid 

 D F  K 

lbf  F 

lbf  F 

m Bm

a

 Ba

m

a

7.508

3.278

25.32/5.15

75.12/5.15

3

3

==

==

=+==−=

πτ

πτ  

For unpeened spring:

( )  kpsi

S S 

S S 

kpsiS kpsiS 

 su sm

 sa

 se

 sm sa

74.50/1

5535

=−

=

== 

The factor of safety guarding against failure to be

214.11

=+

=⇒=+ sem sua

 su se

 f 

 f  su

m

 se

a

S S 

S S n

nS S    ττ

ττ 

2. 

The coil yield:

a

 sa

 y

S n

τ

=  

To find the values of S  sa:The load line:

im

ar ττ

τ

−= (1)

1=+ sy

 sm

 sy

 sa

S 

S 

S 

S (2)

8/21/2019 Me 301 Chapter 10

28/28

The intersection between 1 & 2 gives:

( )i sy sa   S r 

r S    τ−

+=

1 

Therefore:

69.13.27

2.46

:

2.46

85.06.187.50

3.27

6.183.2775.1

19.1

==

=

=−

=

=   

  =  

  

  =⇒=

 y

 sa

a

a

i

i

a

a

i

i

n

Thus

kpsiS 

r 

kpsi F 

 F 

 F  F ττττ

 

3. 

The end hook bending fatigue at point A ( )

( )

( ) 87.0

1.67577.0

1.97

;3.52416

;///1

23

=⇒

==

=   

  

 =

=

+=

+=

 A f 

 se

e

a

a

m

m

 Aaa

ut mea f 

n

kpsiS 

S 

kpsi F 

 F 

kpsid d 

 D K  F 

S S n

σσ

ππσ

σσ

 

4.  The end hook torsion fatigue at point B 

( )

( ) ( )

( ) ( )

( ) 27.1

5.488

1.268

///1

3

3

=⇒

==

==

+=

 B f 

m

 B Bm

a

 B Ba

 su sm se sa f 

n

kpsid 

 D F  K 

kpsid 

 D F  K 

S S n

πτ

πτ

ττ