ME3180
ME 3180: Machine Design
Helical Torsion Springs
Lecture Notes
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ME3180
Helical Torsion Springs
Can be loaded in torsion instead of compression or
tension
Ends are extended tangentially to provide lever arms
on which to apply moment load
Ends come in variety of shapes to suit application
Coils are close wound like extension springs (but do
not have any initial tension), but in few cases are
wound with spacing like compression spring (this will
avoid friction between coils)
Applied moment should always be arranged to close
coils rather than open them because residual stresses
from coil-winding are favorable against a closing
moment (i.e., residual stresses oppose working
stresses).
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ME3180
Helical Torsion Springs
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ME3180
Helical Torsion Springs – Cont’d
Dynamic loading should be repeated or fluctuating with stress ratio R  0
Applied moment should never be reversed in service
Normal stresses are produced in torsion springs
Load should be defined at angle  between tangent ends in loaded
position rather than at deflection from free position
Rectangular wire is more efficient (because load is in bending) in terms of
stiffness per unit volume (larger I for same dimension)
However, most helical torsion springs are made from round wire
because of its lower cost and larger variety of available sizes and materials
Torsion springs are used in door hinges, rat traps, automobile starters,
finger exercisers, garage doors and etc
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ME3180
Helical Torsion Springs – Cont’d
Number of Coils in Torsion Springs
For straight ends, the contribution to equation 13.26b can be expressed as
an equivalent number of coils Ne:
Active coils:
(13.26a)
L L
Ne 
1
3D
2
N a  Nb  N e
(13.26b)
Where Nb is number of coils in spring body
Deflection
Angular deflections of coil-end is normally expressed in radians, but is
often converted to revolutions. Revolutions will be used.
θ rev
1
1 ML w

θ rad 
2
2 EI
(13.27a)
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ME3180
Where: M is applied moment
Lw is length of wire
E is Young’s modulus
I is second moment of area for wire cross section about neutral axis
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ME3180
Helical Torsion Springs – Cont’d
In specifying torsion spring, ends must be located relative to each
other. Commercial tolerances on these relative positions are listed
in Table 10-9.
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ME3180
Helical Torsion Springs – Cont’d
Simplest scheme for expressing initial unloaded location of one end with
respect to the other is in terms of angle  defining partial turn present in
coil body as N P   / 360, as shown in Fig. 10-10. For analysis purpose
nomenclature of Fig. 10-10 can be used. Communication with springmaker is often in terms of the back-angle .
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ME3180
Helical Torsion Springs – Cont’d
Number of body turns N b is number of turns in free spring body
by count.
Body-turn count is related to the initial position angle  by
N b  integer 

360 
 integer  N p
where N p is number of partial turns.
The above equation means that N b takes on non-integer, discrete
values such as 5.3, 6.3, 7.3,…, with successive differences of 1 as
possibilities in designing a specific spring. This consideration will
be discussed later.
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Helical Torsion Springs – Cont’d
ME3180
Bending Stress
Torsion spring has bending induced in coils, rather than torsion.
Means that residual stresses built in during winding are in same
direction but of opposite sign to working stresses that occur during use.
Strain-strengthening locks in residual stresses opposing working
stresses provided load is always applied in winding sense.
Torsion springs can operate at bending stresses exceeding yield
strength of wire from which it was wound.
Bending stress can be obtained from curved-beam theory expressed in
form shown below:
Mc
 K
I
where K is stress-correction factor.
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ME3180
Helical Torsion Springs – Cont’d
Value of K depends on shape of wire cross section and whether
stress is sought at inner or outer fiber. Wahl analytically
determined values of K to be, for round wire,
4C 2  C  1
Ki 
4C (C  1)
4C 2  C  1
Ko 
4C (C  1)
(10-43)
where C is spring index and subscript i and o refer to inner and
outer fibers, respectively.
In view of fact that Ko is always less than unity, we shall use Ki to
estimate the stresses. When bending moment is M = Fr and
3
section modulus I / C  d / 32 , we express bending equations as
  Ki
32Fr
 d3
(10-44)
which gives the bending stress for a round-wire torsion spring.
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Helical Torsion Springs – Cont’d
ME3180
Note: Next two slides are from Norton
Maximum compressive bending stress at inside coil diameter of round
wire helical torsion spring, loaded to close its coils is:
i
max
 K bi
M maxc
32 M max
 K bi
I
d 3
(13.32a)
Tensile bending stresses at the outside of the coil:
32M min
 omin  K bo
;
3
d
 omax   omin
 omean 
;
2
32M max
 omax  K bo
d 3
 omax   omin
 oalt 
2
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(13.32b)
(13.32c)
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ME3180
Helical Torsion Spring - Cont’d
For static failure (yielding) of torsion spring loaded to close its coils,
compressive stress σimax at inside of coil is of most concern
For fatigue failure, which is a tensile-state phenomenon σomax at outside
of coils is of concern
Alternating and mean stresses are calculated at outside of coil
Since closely spaced coils prevent shot from impacting inside diameter
of coil, shot peening may not be effective in torsion springs
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ME3180
Helical Torsion Springs – Cont’d
Deflection and Spring Rate
For torsion springs, angular deflection can be expressed in radians
or revolutions (turns). If term contains revolution units, term will
be expressed with a prime sign.
The spring rate K  is expressed in units of torque/revolution (lbf.
in/rev or N. mm/rev) and moment is proportional to angle  ,
expressed in turns rather than radians.
Spring rate, if linear, can be expressed as
M1 M 2 M 2  M1
k 


1  2
 2  1
(10-45)
where the moment M can be expressed as Fl or Fr .
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ME3180
Helical Torsion Springs – Cont’d
Total angular deflection in radian is:
t 
64 MDN b 64 ML1 64 ML 2 64 MD
L1  L2



(
N

)
b
4
4
4
4
d E
3d E 3d E
d E
3D
(10-47)
Equivalent number of active turns Na is expressed as
L1  L2
N a  Nb 
3D
(10-48)
Spring rate k in torque per radian is
k 
Fr
t

M
t
d 4E

64DNa
(10-49)
Spring rate may also be expressed as torque per turn. Expression for
this is obtained by multiplying Eq. (10-49) by 2 rad/turn. Thus
spring rate k ' (units torque/turn) is
2d 4 E
d 4E
k' 

64DNa
10.2 DNa
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(10-50)
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ME3180
Helical Torsion Springs – Cont’d
Tests show that effect of friction between coils and arbor is such
that constant 10.2 should be increased to 10.8. The equation above
becomes
d 4E
k' 
(10-51)
10.8DNa
(unit torque per turn). Equation(10-51)gives better results. Also Eq.
(10-47) becomes
10 .8MD
l1  l2
(10-52)
 t' 
(
N

)
b
4
d E
3D
Torsion springs are frequently used over round bar or pin. When
load is applied to torsion spring, spring winds up, causing decrease
in inside diameter of coil body.
It is necessary to ensure that inside diameter of coil never becomes
equal to or less than diameter of pin, in which case loss of spring
function would ensure.
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ME3180
Helical Torsion Springs – Cont’d
Helix diameter of coil D ' becomes
Nb D
D' 
N b   c'
(10-53)
where c is angular deflection of body of coil in number of turns,
given by
10 .8MDN b
'
(10-54)
c 
'
d 4E
New inside diameter Di'  D'  d makes diametral  clearance
between body coil and pin of diameter D p equal to
Nb D
  D  d  Dp 
 d  Dp
'
Nb  c
'
(10-55)
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ME3180
Helical Torsion Springs – Cont’d
Equation(10-55) solved for N b is
Nb 
 c' (  d  D p )
D    d  Dp
(10-56)
which gives the number of body turns corresponding to a specified
diametral clearance of arbor.
This angle may not be in agreement with necessary partial-turn
reminder. Thus, diametral clearance may be exceeded but not equaled
Static Strength
First column entries in Table 10-6 can be divided by 0.577 (from
distortion-energy theory) to give
0.7 8S ut

S y  0.8 7S ut
0.6 1S
ut

Music wire and cold-drawn carbon steels
(10-57)
QQ&T (hardened & tempered) carbon and low-alloy steels
Austenitic stainless steel and nonferrous alloys
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ME3180
Helical Torsion Springs – Cont’d
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ME3180
Helical Torsion Springs – Cont’d
Fatigue Strength
Since spring wire is in bending, Sines equation is not applicable.
The Sines model is in the presence of pure torsion. Since
Zimmerli’s results were for compression springs (wire in pure
torsion), we will use the repeated bending stress (R = 0) values
provided by Associated Spring in Table 10-10.
As in Eq. (10-40) we will use the Gerber fatigue-failure criterion
incorporating the Associated Spring R = 0 fatigue strength Sr :
Sr / 2
Se 
S /2
1  ( r )2
S ut
(10-58)
Value of S r (and Se ) has been corrected for size, surface
condition, and type of loading, but not for temperature or
miscellaneous effects.
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ME3180
Helical Torsion Springs – Cont’d
Gerber fatigue criterion is now defined. Strength-amplitude component
is given by Table 6-7, p. 307, as
r 2 Sut2
Sa 
[1 
2S e
2Se 2
1 (
) ] (10-59)
rS ut
where slope of load line is r  M a / M m . Load line is radial through
origin of designer’s fatigue diagram. Factor of safety guarding against
fatigue failure is
S
nf  a
(10-60)

a
Alternatively, we can find n f directly by using Table 6-7, p. 307:
 m Se 2
1  a Sut 2
nf 
( ) [1  1  (2
) ]
2 Se  m
Sut  a
(10-61)
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