mechanical springs shigley ch 10 lecture 21 · mechanical springs why do we need springs? •...

Mechanical Springs

Shigley Ch 10

Lecture 21

Mechanical Springs Why do we need springs?

• Flexibility of structure

• Storing and releasing of energy

Standard spring types:

• Wire springs

- Helical springs of round or square wire, made to deflect under Compression, tension or torsion

- Can be used as a straight piece of wire for stiffer spring types (torsion bar suspensions springs)

• Flat springs

- Cantilever, elliptical, wound clock type (spiral), Flat Spring Washers (usually called Belleville springs), leaf springs, etc

• Special shape springs also exist for custom cases

� Most springs can be Linear or non linear, depending on geometry

Helical compression spring

Flat-leaf spring

Belleville spring

Torsion spring

Stresses in Helical Spring

23

max

48

d

F

d

FD

A

F

J

Tr

ππτ

τ

+=

+=

Torsion Shear force

Maximum shear equation

We define a term called the Spring index as

C-Usually ranges between 4 and 12

d

DC =

3

8

d

FDK s

πτ =

Next define a term Ks=shear stress correction factor

Then the stress equation becomes

C

CKs

2

12 +=

Spring design consideration

• Use round spring wire where possible, if space

is limited use nested round springs

• Springs with round wire are made

in large quantities, so for cost avoid

square/other shape sections

Valve spring in action, showing nested round springs

(Grey, red and white)

The Curvature effect

Helical springs are not straight but curved

Taking the curvature into account replace Ks

with KB (Bergsträsser factor)

Thus the largest shear stress is now given by

34

24

−

+=

C

CKB

3

8

d

FDK B

πτ =

C

CKs

2

12 +=

Deflection of Helical Springs

Strain energy for a helical spring is given by adding

the torsional-and shear-energy components

Substituting T, L, A and J gives

Where N=Na is the number of active coils

AG

lF

GJ

lTU

22

22

+=

Gd

DNF

Gd

NDFU

2

2

4

32 24+=

From Castigliano’s theorem the total deflection

Spring rate or also called “scale” of the spring is

given by

Gd

NFD

CGd

NFD

Gd

FDN

Gd

NFD

F

Uy

4

3

24

3

24

3

8

2

11

8

48

≅

+=

+=∂

∂=

ND

Gd

y

Fk

3

4

8≅=

Compression Springs, end conditions

End bent down to form 0° helix angle End ground flat to form flat

mounting surface for spring

End Conditions may differ check with manufacturer

or count and measure them

Stability

As long columns, long springs can buckle with

high loads

Where

Effective slenderness ratio

α End condition constant from table 10-2

Elastic constants

−−=

21

2

'

2'

1 11eff

ocr

CCLy

λ

D

L oeff

αλ =

)(2

'

1GE

EC

−=

EG

GEC

+

−=

2

)(2 2'

2

π

Table 10-2

End condition Constant α

Spring supported between flat parallel

surfaces

0.5

One end on flat surface perpendicular to

axis (fixed) other end hinged

0.707

Both ends hinged 1

One end clamped other end free 2

Absolute stability is given by

For steels

Squared and ground ends thus

21

02

)(2

+

−<

EG

GEDL

α

π

α

DL 63.20 <

5.0=α

DL 26.50 <

Spring Materials

Hot- or cold-working processes, depending on

size, spring index and properties desired

In general Pre Hardened material should not to

be used if

D/d < 4 or

d > 6mm

Table 10-3 shows most commonly used spring

materials

Table 10-3

Spring Materials

Graph of tensile strength vs. wire diameter is a

straight line when plotted on log-log paper

using the following equation:

Table 10-4 gives values for m and A

mutd

AS =

Table 10-4

Spring Calculations

Torsional yield strength is needed in designing springs butbecause tensile strength is easy to determine the torsionalyield strength is usually estimated from that

From DET the torsional yield strength is given by

The yield strength is between 60 and 90 percent of the ultimate tensile strength for steel, this approach results in the range

Table 10-5 gives the range of Ssy and table 10-6 gives the maximum percentage of tensile strength

ysy SS 577.0=

utsyut SSS 52.035.0 ≤≤

Table 10-5

• Set removal or presetting

– Make spring longer than required, close/press to

solid length to introduce residual stresses (10 to

30 percent of length is reduced this way)

– Stress at solid height is between 1.1 and 1.3 the

torsional yield strength

– Make spring stronger by inducing residual stresses

opposite those induced in service

– Not good for fatigue applications