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Chapter 19

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Chapter 19
Springs
Chapter 19: Springs
Springs Characterized By:
пЃ®
пЃ®
пЃ®
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?
пЃ®
пЃ®
пЃ®
Shock/Vibration protection
Store/Release PE
Offer resisting force or reaction force for
mechanism
пЃ· Example:
пЃ®
пЃ®
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 Compression
Springs
пЃ®
пЃ®
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
Ff – zero force
Ls – solid length
Fs – solid force
Li – installed length
Fi – min. operating force
Lo – operating length
Fo – max. operating force
Chapter 19: Springs - Terminology
k – spring rate
C – spring index = Dm/Dw
N – number of coils
Na – number of active coils (careful Na may be different from N
– depends on end condition – see slide 9)
p – pitch
λ – pitch angle
cc – coils clearance
K – Wahl factor
f – linear deflection
G – shear modulus
t – torsional shear stress
to – stress under operation
ts – stress at solid length
td – design stress
Spring Rate
Chapter 19: Springs
Chapter 19: Springs – Analysis Process:
Key Equations:
Determine: K, П„o, П„s to determine if OK for your application
1.) Calc spring rate :
2.) Based on geometry:
Shear mod table 19-4
C = spring index = Dm/Dw
Wire dia. Table 19-2
**Need to know k to get spring forces to get spring stresses**
3.) Shear stress in spring:
(accounts for curvature of spring)
Show slide
**Compare τo & τs to material allowable (Figure 19-8 – 19-13)**
Chapter 19: Springs – Analysis Process:
4) Buckling Analysis – usually final analysis done to
make sure there’s no stability issue. If so, may be
as simple as supporting the spring through id or od
•Calculate Lf/Dm
•From Fig 19-15, get fo/Lf
•fo = buckling deflection – you
want your maximum deflection
to 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!!
Spring Analysis Example:
Given: Spring- 34ga Music Wire
Dm = 1.0 “
Lf = 3.0”
Li = 2.5”
Lo = 2.0”
Na = 15 (squared and ground end)
Find:
Spring rate, П„o, П„s
Dw = .100” (table 19-2)
Spring Analysis Example:
Fo = k О”L = 9.875 lb/in (1 in) = 9.875 lb
operating force
Lf - Lo
3” – 2”
o
IS this stress ok? See figure 19-9
(severe or average service)
Spring Analysis Example:
Squared and ground
(Max force)
CHECK FOR
BUCKLING!!!
IS this stress ok? See figure 19-9
(light service since only happens
once!!)
Spring Design Example:
Given: Lo = 2.0 in
Fo = 90lb
Li = 2.5 in
Fi = 30 lb
Find:
k пЂЅ
пЃ„L пЂЅ
Generally all that’s
given based on
application!!
Severe service
Suitable compression spring
пЃ„F
пЃ„L
пЃ„F
k
пЂЅ
60 lb
пЂЅ 120 lb / in .
. 5 in
пЂЅ
30 lb
120 lb / in .
пЂЅ . 25 in . пЂЅ L f пЂ­ L i
L f пЂЅ L i пЂ« пЃ„ L пЂЅ 2 . 5 in пЂ« . 25 in пЂЅ 2 . 75 in
Looks OK compared
to ~ 3 in. length
Spring Design Example:
Try: Cr – Si Alloy, A-401
Guess: Dm = .75 in.
Guess: П„allow = 115 ksi (Figure 19-12)
1/ 3
Dw
пѓ¦ 3 . 06 ( Fo)( D m ) пѓ¶
пѓ·
пЂЅ пѓ§пѓ§
пѓ·
t allow
пѓЁ
пѓё
Dw
пѓ¦ 3 . 06 ( 90 lb )(. 75 in ) пѓ¶
пЂЅ пѓ§пѓ§
пѓ·пѓ·
2
115
,
000
lb
/
in
пѓЁ
пѓё
1/ 3
пЂЅ . 1216 in .
Try: 11 gage = .1205 in. (Table 19-2)
Spring Design Example:
FROM APPENDIX 19-5:
П„severe allow = 122 ksi (operating max.)
П„light allow = 177 ksi (solid max.)
C пЂЅ
Dm
Dw
K пЂЅ
пЂЅ
. 75 in .
пЂЅ 6 . 224
. 1205 in .
4C пЂ­ 1
4C пЂ­ 4
пЂ«
. 615
C
пЂЅ 1 . 242
( > 5, so OK )
Spring Design Example:
to пЂЅ
8 KFC
o
пЃ°D
2
w
пЂЅ
8 (1 . 242 )( 90 lb )( 6 . 224 )
пЃ° (. 1205 in )
2
3
f пЂЅ
8 FC Na
п‚®
GD q
fGD w
8 FC
6
Na пЂЅ
пЂЅ 112 , 048 psi
3
OK < 122 ksi
пЂЅ Na
2
. 75 in (11 . 2 x10 lb / in )(. 1205 in )
8 ( 90 lb )( 6 . 224 )
3
пЂЅ 5 . 83
Spring Design Example:
L s пЂЅ D w ( Na пЂ« 2 ) пЂЅ . 1205 in ( 5 . 83 пЂ« 2 ) пЂЅ . 942 in
(squared and ground ends assumed)
F s пЂЅ k пЃ„ L пЂЅ k ( L f пЂ­ L s ) пЂЅ 120 lb / in ( 2 . 75 in пЂ­ . 942 in ) пЂЅ 217 lb
2
ts пЂЅ
8 KF s C
пЃ°D
2
w
пЂЅ
8 (1 . 242 )( 217 lb )( 6 . 224 )
пЃ° (. 1205 in )
2
пЂЅ 294 ,129 psi
294 ksi > 177 ksi -пѓ WILL YIELD, NOT ACCEPTABLE
Spring Design Example:
How to check buckling:
Lf
Dm
пЂЅ
2 . 75 in
пЂЅ 3 . 667
Critical ratio = ?
. 75 in
For any fo/Lf This spring is below fixed end line and
fixed-pinned. If pinned-pinned critical ratio = .23
fo
Lf
пЂЅ
2 . 75 in пЂ­ 2 . 0 in
пЂЅ . 273
2 . 75 in
.273 > .23: So it would buckle
Spring Design Example:
Trials
1
2
3
4
5
6
Dm
1.00
1.25
0.75
1.00
1.00
1.10
Dw
0.1350
0.1483
0.1483
0.1483
0.1620
0.1620
C
7.41
8.43
5.06
6.74
6.17
6.79
Select one of these
Na
3.88
2.89
13.38
5.64
8.04
6.04
П„s
292
280
Ls > Lo
185
101
140
Side Info: How determine initial estimate for Dw
t пЂЅ
8 KFD
Equation for shear stress
where geometry is known
m
пЃ°D w
3
But…. Τallow and K depend on Dw
So……Guess K is mid-range, about 1.2
Then:
8K
пЃ°
пЂЅ
8 (1 . 2 )
пЃ°
пЂЅ 3 . 06
пѓ¦ 3 . 06 ( F 0 )( D m ) пѓ¶
пѓ·
D
пЂЅ
Re-arrange…. w 
пѓ·
t
allow
пѓЁ
пѓё
1/ 3
This Dw is about where to start for spring design. Both K and П„allow
may be found for selected Dw.
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