ch19 springs

8/13/2019 CH19 Springs

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Chapter 19

Springs



Chapter 19: Springs

Springs Characterized By:

Ability to deform significantly without

failure Ability to store/release PE over large

deflections

Provides an elastic force for useful purpose



Chapter 19: Springs

How used in mechanical design? Shock/Vibration protection

Store/Release PE Offer resisting force or reaction force for

mechanism Example:

Valve spring pushes rocker arm so lifter follows cam VCR lid – torsion springs keeps door closed



Types of springs



Chapter 19: Springs

Our focus will be on Helical CompressionSprings

Standard round wire wrapped into cylinder,usually constant pitch

We will cover design process and analysis



Chapter 19: Springs - Terminology

Helical Compression Springs:ID –inside diameter of helix

OD – outside diameter of helix

Dm – mean diameter of helix

Lf – free length

Ls – solid length

Li – installed length

Lo – operating length

Ff – zero force

Fs – solid force

Fi – min. operating force

Fo – max. operating force



k – spring rateC – spring index = Dm/DwN – number of coilsNa – number of active coils (careful Na may be different from N – depends on end condition – see slide 9)p –

pitchλ – pitch anglecc – coils clearanceK – Wahl factorf – linear deflection

G –

shear modulust – torsional shear stressto – stress under operation

ts – stress at solid length

td – design stress

Chapter 19: Springs - Terminology



Spring Rate



Chapter 19: Springs



Chapter 19: Springs – Analysis Process:

4) Buckling Analysis – usually final analysis done tomake sure there’s no stability issue. If so, may beas simple as supporting the spring through id or od

•Calculate Lf/Dm

•From Fig 19-15, get fo/Lf

•fo = buckling deflection – youwant your maximum deflectionto be less than this!!

Buckling procedure



2 Categores:

Spring Analysis – spring already exists –

verify design requirements are met (namely,stiffness, buckling and stress is acceptable)

Spring Design – design spring from scratch – involves iterations!!



Fo = k ΔL = 9.875 lb/in (1 in) = 9.875 lb

operating force Lf - Lo 3” – 2”

IS this stress ok? See figure 19-9(severe or average service)

Spring Analysis Example:

o



Squared and ground

(Max force)

Spring Analysis Example:

IS this stress ok? See figure 19-9(light service since only happensonce!!)CHECK FOR

BUCKLING!!!



Given: Lo = 2.0 in Fo = 90lb

Li = 2.5 in Fi = 30 lb

Severe service

Find: Suitable compression spring

./1205.

60inlb

in

lb

L

F k

i f L Lininlb

lb

k

F L

.25.

./120

30

ininin L L L i f 75.225.5.2 Looks OK compared

to ~ 3 in. length

Spring Design Example:

Generally all that’s

given based onapplication!!



Guess: Dm = .75 in. Try: Cr – Si Alloy, A-401

Guess: τallow = 115 ksi (Figure 19-12)

3/1

))((06.3

allow

mw

D F D

t

.1216./000,115

)75)(.90(06.3 3/1

2 in

inlb

inlb Dw

Try: 11 gage = .1205 in. (Table 19-2)


o



FROM APPENDIX 19-5:

τsevere allow = 122 ksi (operating max.)

τlight allow = 177 ksi (solid max.)

224.6.1205.

.75.

in

in

D

DC

w

m

242.1615.

44

14

C C

C K

( > 5, so OK )




psiin

lb

D

KFC

w

o 048,112)1205(.

)224.6)(90)(242.1(8822

t OK < 122 ksi

Na FC

fGD

GD

Na FC f w

q

3

3

8

8

83.5)224.6)(90(8

)1205)(./102.11(75.3

26

lb

ininlb xin Na


o



inin Na D L w s 942.)283.5(1205.)2(

(squared and ground ends assumed)

lbinininlb L Lk Lk F s f s 217)942.75.2(/120)( 2

psiin

lb

D

C KF

w

s s 129,294

)1205(.

)224.6)(217)(242.1(8822

t

294 ksi > 177 ksi - WILL YIELD, NOT ACCEPTABLE




How to check buckling:

667.375.

75.2

in

in

D

L

m

f Critical ratio = ?

For any f o/Lf This spring is below fixed end line and

fixed-pinned. If pinned-pinned critical ratio = .23

273.75.2

0.275.2

in

inin

L

f

f

o

.273 > .23: So it would buckle




TrialsDm Dw C Na τs

1 1.00 0.1350 7.41 3.88 292

2 1.25 0.1483 8.43 2.89 280

3 0.75 0.1483 5.06 13.38 Ls > Lo

4 1.00 0.1483 6.74 5.64 185

5 1.00 0.1620 6.17 8.04 101

6 1.10 0.1620 6.79 6.04 140

Select one of these




But…. Τallow and K depend on Dw

So……Guess K is mid-range, about 1.2

Then:

Side Info: How determine initial estimate for Dw

3

8

w

m

D

KFD

t Equation for shear stress

where geometry is known

Re-arrange….

3/1

0 ))((06.3

allow

m

w

D F D

t

06.3)2.1(88

K

This Dw is about where to start for spring design. Both K and τallow

may be found for selected Dw.

ch19 springs

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