outline the function(s) of springs · the function(s) of springs most fundamentally: ... **see...

1

OutlineSpring Functions & TypesHelical Springs

CompressionExtensionTorsional

The Function(s) of Springs

Most fundamentally: to STORE ENERGY

Many springs can also: pushpulltwist

Some ReviewF

y

k

linear springs: k=F/y

nonlinear springs:dydFk =

Parallel

ktotal=k1+k2+k3

Series

321

1111kkkktotal

++=

Types of SpringsHelical:

Compression

Extension Torsion

More Springs

Washer Springs:

Beams:Power springs:

Helical Compression Springs

d diameter of wireD mean coil diameterLf free lengthp pitchNt Total coils

may also need:Do and Di

2

Length Terminology

Free Length AssembledLength

Max WorkingLoad

Bottomed Out

minimum of 10-15%clash allowance

Lf La Lm Ls

End ConditionsPlain

Square

Plain Ground

Square Ground

Na=Active Coils

F

F

F

F

Stresses in Helical SpringsSpring Index C=D/d

CCKwhere

d

FDK ss 212,8

3max+

==π

τ

FF

T

T

Curvature Stress

under static loading, local yielding eliminates stress concentration, so use Ks

under dynamic loading, failure happens below Sy: use Ks for mean, Kw for alternating

Inner part of spring is a stress concentration(see Chapter 4)

Kw includes both the direct shear factor and the stress concentration factor

CCCKwhere

d

FDK ww615.0

4414,8

3max +−−

==π

τ

Spring Deflection

Gd

NFDy a4

38≈

Spring Rate

Gd

NFDy a4

38≈

aND

Gdk 3

4

8≈

k=F/y

3

Helical SpringsCompression

NomenclatureStressDeflection and Spring ConstantStatic DesignFatigue Design

ExtensionTorsion

Static Spring DesignInherently iterative

Some values must be set to calculate stresses, deflections, etc.

Truly Designthere is not one “correct” answermust synthesize (a little bit) in addition to analyze

Material PropertiesSut ultimate tensile strength

Figure 13-3Table 13-4 with Sut=Adb

Sys torsional yield strengthTable 13-6 – a function of Sut and set

Spring/Material TreatmentsSetting

overstress material in same direction as applied load

» increase static load capacity 45-65%» increase energy storage by 100%

use Ks, not Kw (stress concentration relieved)Load Reversal with SpringsShot Peening

What type of failure would this be most effective against?

What are You Designing?Given

F, yk, y

Find

kF

Such that:Safety factor is > 1Spring will not buckleSpring will fit in hole, over pin, within vertical space

d, C, D*, Lf*, Na

*, clash allowance (α)**, material**

design variables

+

* - often can calculate from given** - often given/defined

Static Spring Flow Chart

STRESSES

Ns=Sys/τ

for shut spring if possibleif not, for max working load

if GIVEN F,y, then find k; If GIVEN k, y, then find F

D, Ks, Kw

material strengths

DEFLECTION

Lf, yshut, Fshut

Three things to know:• effect of d• shortcut to finding d• how to check buckling

ITERATE?

d, C

material

Na, α

CHECK

buckling, Nshut, Di, Do

Nshut=Sys/τshut

4

Static Design: Wire Diameter


3max8

dFDKsπ

τ =Gd

NFDy a4

38≈

**see Example 13-3A on Norton CD*maintain units (in. or mm) for A, b

use Table 13-2 to select standard d near calculated d

( ) ( ) ( )[ ] )2(115.08 b

minitialworks

AKFFCNd

+

⎭⎬⎫

⎩⎨⎧ −++

=π

αα

Based on Ns=Ssy/τ and above equation for τ:

Km=Sys/Sut

Buckling

f

workinginit

f

Lyy

y

DL

RS

+=′

=..




ExtensionTorsion

Material PropertiesSus ultimate shear strength

Sus≈0.67 Sut

Sfw´ torsional fatigue strengthTable 13-7 -- function of Sut, # of cyclesrepeated, room temp, 50% reliability, no corrosion

Sew´ torsional endurance limitfor steel, d < 10mmsee Eq. 13-12 (=45 ksi if unpeened, =67.5 ksi if peened)repeated, room temp., 50% reliability, no corrosion

Modified Goodman for SpringsSfw, Sew are for torsional strengths, so von Mises not used

0.5 Sfw

0.5 Sfw

τa

τm

Repea

ted

SfsC

B

Sus

A

( )fwus

usfwfs SS

SSS

5.05.0

−=

Fatigue Safety Factor

0.5 Sfw

0.5 Sfw

τa

τm

Sfs

Sus

load

line

Saτa

τi τm

mload

mgoodaa

fsSNτ

=

τa,load = τa,good at intersection( )

( ) ausimfs

iusfsfs SS

SSN

ττττ+−

−=

Eq. 13-17a

Fi=Fmin

Fa=(Fmax-Fmin)/2Fm=(Fmax+Fmin)/2

5

What are you Designing?Given

Fmax,Fmin, ∆yk, ∆ y

Find

kF

Such that:Fatigue Safety Factor is > 1Shut Static Safety Factor is > 1Spring will not buckleSpring is well below natural frequencySpring will fit in hole, over pin, within vertical space

d, C, D*, Lf*, Na

*, clash allowance (α)**, material**

design variables

+

* - often can calculate from Given** - often given/defined

Fatigue Spring Design Strategy

d, C

if GIVEN F,y, then find k; If GIVEN k, y, then find F

Nshut=Sys/τshut

CHECKbuckling, frequency,Nshut, Di, Do

D, Ks, Kw

material strengths

DEFLECTION

Lf, yshut, Fshut

Two things to know:• shortcut to finding d• how to check frequency

ITERATE?

material

Na, α

STRESSES( )

( ) ausimfs

iusfsfs SS

SSN

ττττ+−

−=

Fatigue Design:Wire Diameteras before, you can iterate to find d, or you can use an equation

derived from relationships that we already know:

)2(1

min 134.11

67.08

b

awfw

bs

fs

fsms

fs FKSAdFK

NN

FKA

CNd

+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−−=

π

use Table 13-2 to select standard d near calculated d


**see Example 13-4A on Norton CD*maintain units (in. or mm) for A, b

Natural Frequency: Surge


Surge == longitudinal resonance

for fixed/fixed end conditions:

an W

kgf21

= (Hz)

ideally, fn will be at least 13x more than fforcing… it should definitely be multiple times bigger

Review of Design Strategy

Find LoadingSelect C, d

Find stressesDetermine material propertiesFind safety factor

Solve for d, pick standard dFind stressesDetermine material propertiesCheck safety factor

Find LoadingSelect C, safety factor

ITERATIVE USING d EQUATION

Strategy Review ContinuedFind spring constant, Na, Nt

Find FSHUT (must find lengths and y’s to do this)Find static shut shear stress and safety factor

Check Buckling

Check Surge

Check Di, Do if pin to fit over, hole to fit in

6

Consider the Following: Helical SpringsCompression


ExtensionTorsion

Extension SpringsAs before, 4 < C < 12

However, no peening, no setting,no concern about buckling

aws forKusedFDK τπ

τ ,83max =

surge check is same as before

Lb=d(Na+1)

Difference 1: Initial Force

deflection y

force F

Fi

“preloading”

F=Fi+kya

iND

Gdy

FFk 3

4

8=

−=

Difference 1a: Deflection

Gd

NDFFy ai4

3)(8 −≈

Difference 2: Initial Stress

take initial stress as the average stress between these lines, then find Fi

7

Difference 3: Ends!: Bending

23416dF

dDFKba

ππσ +=

CdD

dRC

CCCC

Kb

===

−

−−=

222

)1(414

11

11

121

67.0

)()(

minmin

ese

altutmeaneute

fb

SS

SSSSN

=

+−−

=σσσ

σ

standardend

Difference 3a: Ends: Torsion

4414,8

22

3max 22 −−

==CCK

dFDK wwπ

τ

C2=2R2/d

pick a value >4

Materials

Sut – SameSys, Sfw, Sew – same for bodySys, Sfw, Sew – see Tables 13-10 and 13-11 for ends

Strategysimilar to compression + end stresses - buckling



ExtensionTorsion

Torsion Springs

• close-wound, always load to close

Deflection & Spring Rate

Ed

MDN

DNwireoflengthLEI

ML

aroundwirerev

aww

rev

4, 8.10

,21

=

===

θ

ππ

θ

rev

Mkθ

=

8

Stresses

Compressive is Max – Use for Static – Inside of Coil

3maxmax 32

max d

MKI

cMKii bbi

πσ ==

)1(414 2

−−−

=CC

CCKib

For Fatigue – Slightly lower Outside Tensile Stress – Outside of Coil

)1(414 2

+−+

=CC

CCKob

3max32

max d

MKobo

πσ =

3min32

min d

MKobo

πσ =

Materialssee Tables 13-13 and 13-14

follow book on Sewb=Sew/0.577

Strategy

M K

θ

• fit over pin (if there is one)• don’t exceed stresses

Select C, dHelical Springs

CompressionNomenclatureStressDeflection and Spring ConstantStatic DesignFatigue Design

ExtensionTorsion

outline the function(s) of springs · the function(s) of springs most fundamentally: ... **see...

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