CH19_Springs.ppt

8/9/2019 CH19_Springs.ppt

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Chapter 19

Springs


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Chapter 19: Springs

Springs Characterized By: Ability to deform signicantly without

failure Ability to store/release ! o"er large

de#ections

ro"ides an elastic force for usefulpurpose


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Chapter 19: Springs

$ow used in mechanical design% Shoc&/'ibration protection

Store/(elease ! )*er resisting force or reaction force

for mechanism!+ample:

'al"e spring pushes roc&er arm so lifterfollows cam

'C( lid torsion springs &eeps door closed


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Types of springs


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Chapter 19: Springs

)ur focus will be on $elicalCompression Springs Standard round wire wrapped into cylinder,

usually constant pitch

-e will co"er design process and analysis


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Chapter 19: Springs .erminology$elical Compression Springs:0 2inside diameter of heli+) 2 outside diameter ofheli+

m2 mean diameter of heli+3f2 free length

3s2 solid length

3i2 installed length

3o2 operating length

4f2 zero force

4s2 solid force

4i2 min5 operating force

4o2 ma+5 operating force


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& spring rateC spring inde+ 6 m/w7 number of coils7anumber of acti"e coils 8careful 7amay be di*erent

from 7 2 depends on end condition 2 see slide 9p pitch pitch anglecc coils clearance; -ahl factorf linear de#ection< shear modulustorsional shear stressostress under operationsstress at solid lengthddesign stress

Chapter 19: Springs .erminology


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Spring (ate


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Chapter 19: Springs


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Chapter 19: Springs Analysisrocess:

etermine: ;, =o, =sto determine if ); for your

application15 Calc springrate :

>5 Based ongeometry:

Shear mod table19.?

-ire dia5 able 19.>

C 6 spring inde+ 6m/w

@@7eed to &now & to get spring forces to getspring stresses@@

5 Shear stress in

spring: 8accounts for cur"ature of spring

@@Compare =o

=s

to material allowable 84igure 19. 2 19.1@@

;ey !Duations:

Show slide


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Chapter 19: Springs Analysisrocess:

? Buc&ling Analysis 2 usually nal analysisdone to ma&e sure thereEs no stability issue50f so, may be as simple as supporting thespring through id or od

Calculate 3f/m

4rom 4ig 19.1F, get fo/3f

fo 6 buc&ling de#ection 2you want your ma+imumde#ection to be less thanthisGG

Buc&ling procedure


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> Categores:

Spring Analysis 2 spring already e+ists 2

"erify design reDuirements are met8namely, sti*ness, buc&ling and stressis acceptable

Spring esign 2 design spring fromscratch 2 in"ol"es iterationsGG


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Given: Spring. ?ga Husic -irem6 15I J

3f6 5IK 3i6 >5FK

3o6 >5IK

7a 6 1F 8sDuared and ground endFind: Spring rate, =o, =s

w6 51IIK 8table 19.>

Spring Analysis !+ample:


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4o 6 & L3 6 95MF lb/in 81 in 695MF lb

operatingforce

3f. 3o K 2>K

0S this stress o&% See gure19.9 8se"ere or a"erageser"ice


o


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SDuared andground

8Ha+force


0S this stress o&% See gure19.9 8light ser"ice since onlyhappens onceGGC$!C; 4)(

BNC;307


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Given: Lo= 2.0 in Fo= 90lb

Li= 2.5 in Fi= 30 lb

Severe serviceFind: Suitable compression spring

./1205.

60inlb

in

lb

L

Fk ==

=

if LLininlb

lb

k

FL ===

= .25.

./120

30

inininLLL if 75.225.5.2 =+=+=Loos !" compare#

to $ 3 in. lengt%

Spring esign !+ample:


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&uess' (m= .)5 in. Try' *r + Si ,lloy- ,/0

&uess' =allow6 11F &si 84igure 19.

1>3/1

))((06.3

=

allow

m

w

DFD

.1216./000,115

)75)(.90(06.3 3/1

2 in

inlb

inlbDw =

=

Try' gage = .205 in. 1Table 92


o


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F!4 ,67(8 95'

=se"ere allow6 1>> &si 8operating ma+5

=light allow6 1MM &si 8solid ma+5

224.6.1205.

.75.===

in

in

D

DC

w

m

242.1615.

44

14=+

=

CC

CK

1 : 5- so !"



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ininNaDL ws 942.)283.5(1205.)2( =+=+=1s86L(- 7!T ,**6T,?L6



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@oA to c%ec bucling'

667.375.

75.2==

in

in

D

L

m

f *ritical ratio = B

For any foCLfT%is spring is beloA fiDe# en# line an#

fiDe#pinne#. 8f pinne#pinne# critical ratio = .23

273.75.2

0.275.2=

=

in

inin

L

f

f

o

.2)3 : .23' So it Aoul# bucle



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Trials

Select one of t%ese



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?utE.Oallowand ; depend on w

SoPP

hen:

Side 0nfo: $ow determine initial estimate for w

3

8

w

m

D

KFD

= 6


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CH19_Springs.ppt

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