Excluding hooks.

〔equivalent to(a),(b)and(c)in Fig.2〕

SWOSC-V*,SWOSM,SWOSC-BSWO,SWO-V,SWOC-V,

9260, 5155, 5160, 6150, 51B60, 4161

78×103 {8×103 }

}3{8×10378×10

X1and X2：indicate each number of coils of both ends.

X

(a)When only the tip of the coil touches the next free coil,

－1.5

N a＝ tN

N t＝ aN

N a＝ t－2N

X 1＝ X 2＝ 0.75therefore

(b)When coil end dose not touch the next coil, 　 length of the 　grinding part 〔equivalent to(e)and(f)in Fig. 2〕

therefore1＝ 2X＝ 1X

X(N )2＋ 1－ t＝ aN

(2)　Tension Springs Effective turns of tension springs are as below.

(1)　In case of compression spring

At spring design, the number of active turns should be fixed equal to the number of free coils as follows.

1.3.2　Number of Active Coils

χ

(1)

c＝ dD

SUS 316

SUS 631 J1

SUS 304

SUS 304N1（S30451）

SUS 302

SWP Equivalent to ASTM A228（　　　　）

SW40C, SW60C

SUS　302

SUS　304

SUS　316

SUS　631　J1 }3{7.5×10374×10

69×103 {7×103 }Stainless steel wire

* SWOSC-V : Steel Wire Oil-temper Silicon For Valves/ASTM A401

78×103 {8×103 }

}3{8×10378×10

Music wire

Oil tempered steel wire

Hard steel wire

Spring steel

SymbolValue N/mm2(kgf/mm2)Material

Table 2. Modulus of Rigidity（G）

Fix the value of modulus of rigidity G in designing spring with Table 2 as a rule.

　 1.3　Considerable matters in designing springs　1.3.1　Modulus of rigidity

δ )iP＋P

i)P－P

iτ ＋ ……( 4′

i

)

)

)

…( 7′

)

)

)

)

)

…( 2′

)iP－P

　1.2.2 In case of tension spring with initial tension

＝ …( 1′ 8N aD3(Gd

k ＝ P－Pδ ＝

Gd8N aD3

4

τ 8DPπ 3 ……………( 3′ ＝ 0

d

0＝ τ 2π Gdδ N aD

＝ τ ……………………( 5′ χ τ 0

……( 6′ d ＝ 8DP ＝ π τ 03

τ π χ 8 DP

N ＝ a8D3(

Gd4δ ＝ Gd8

4

D3k

……………( 8′ U ＝ 2

(

2

2kδ ＝ Pδ

2＝ U ……………( 8 )

k3D

4

8Gd＝ δ 4Gd

P3D8a …………( 7 )＝ N

DP8π τ

330τ π ＝

DP8＝ d ……( 6 )

0τ χ ……………………( 5 )τ ＝

DaNGdδ π 2 …………( 4 )τ ＝ 0

d0＝ ……………( 3 )3π

DP8τ

4

3DaN8Gd＝ δ

P＝ ………( 2 )k

4GdP3DaN8 …………( 1 )δ

　1.2.1 In case of tension spring without compression spring and initial tension1.2 Basic formula used in designing springs

－2d)2

1D

D

For reference：L＝Na・P＋1.5d

L

d P

2D

For reference：L＝Na・d＋2(D

1

D

Dd

2D

L

Note：（1）Gravitational acceleration is rounded off as 9800mm/s2 in designing springs though prescribed as 9806.65mm/s（2）in the regulation of measure.

i

0

i

t

a

s

g

W

ω

U

f

χ

τ

τ

τ

k

δ

P

G

c

P

p

H

L

N

N

D

D

D

d

2

N・mm{kgf・mm}

}2{kgf/mm2

mm/s

N{kgf}

－

Hz

N/mm{kgf/mm}

mm

N/mm

N{kgf}

N{kgf}

－

mm

mm

mm

－

－

mm

mm

mm

mm

Energy saved in spring

Gravitational acceleration

Weight of spring moving part

Material weight of a volume unit

Frequency

Stress correction factors

Torsional stress by initial tension

Torsional amending stress

Torsional stress

Spring constant

Deflection of a spring

Load on a spring

Modulus of rigidity

Spring index

Initial tension

Pitch

Solid height

Number of free coils

Number of active coils

Total number of coils

Average diameter of a coil＝　　　

Outside diameter of a coil

Inside diameter of a coil

Diameter of material

Symbol UnitMeaning of Symbols

Symbols used in design formula for spring are shown in Table 1.

Table 1. Meaning of Symbols

1.1 Symbols used in design formula for spring　 1.　Calculation

100c

In this case, torsional stress caused by initial tension is ranged inside the shaded portion in Fig. 2 as a rule.

However, the solid height of the compression spring shall not be specified by the purchaser, generally.

　In calculating the design of springs, the following matters also have to be considered.

……………………………………………………(12)＝　　 i

ds ＋ p＝

NL－H　

a…………………………………………………(14 )

(2)

(1)

When both ends are free or fixed type：

Torsional stress by initial tension ×0.8（The value of 0.8 is reduced by 20％ through low-temperature annealing.）

Initial tension

Initial tension

×0.75（The value of 0.75 is reduced by 25％ through low-temperature annealing.） Torsional stress by initial tensionτ i＝ G

(2)　When low-temperature annealing is performed after forming, the value of the steel wire 　　　such as piano wire, hard steel wire, etc. is reduced by 20～35％, and the value of 　　　stainless steel wire is reduced by 15～25％ from the value obtained in above case.

(1)　In the case of the stainless steel witre, reduce 15％ from the initial stress of the steel wire.

（value prior to low-temperature annealing, formed from steel wire） iτ

Further, the initial stress !!14!!, when it has been formed into solid-wound from the steelHowever, the value of initial stress read from the range of the oblique lines of Fig. 3 shall be corrected as given in the following, according to the properties of material other than steel wires and the process of low temperature annealing.

P

(g)　Open end 　 end turns（unground）

…

＝　

mm )2　(kgf/

　 (4)　Pitch　The pitch should be 0.5D or less because the pitch exceeding 0.5D generally makes the coil diameter change as the deflection（load）increases, which requires compensation of the deflection and tortional stress obtained from the basic formula. The pitch of the spring shall generally be calculated from the following approximate calculation formula.

(3)　Number of active coil　The number of active coils should be 3 or more because the spring property tends to be unstable if the number is less than 3.

(2)　Aspect ratio　The aspect ratio of a compression spring（ratio between free height and mean diameter of coil）has to be 0.8 or more in order to secure effective turns, generally a ratio within the range 0.8 and 4 should be chosen to prevent buckling.

(1)　Spring index　When the spring index is small, local stress becomes excessive. If the spring index is too large or small, workability becomes a problem.Therefore, a spring index of between 4 and 15 should be chosen in the case of hot forming, and one of between 4 and 22 when cold forming is used.

1.3.7　Other considerations

)2Da

dN

5＝3.56×101f

When w＝ 76.93×10－6N/mm3{7.85×10－6kgf/mm3} and both ends of the spring are free or fixed, primary natural frequency of the spring is obtainable with the following formula.

G of steel ＝ 78×103N/mm2{8×103kgf/mm2},

i＝1,2,3－1i

a＝　　 ：In case one end fixed and the other free4

2

Here,

……………………………………………(13)ω Gd70

2DaNπ kgW

＝a　　＝af

Natural frequency of spring is

To avoid surging, natural frequency of the spring should be selected not to resonate to all the vibrations which affect the spring.

1.3.6　Surging

2323 }〕 kgf/mm{7×10N/mm〔G＝69×10In case of stainless steel wire

}〕 2kgf/mm3{8×102N/mm3〔G＝78×10In case of hard steel wire and piano wireThen the initial tension is as follows.

cG

100＝ iτ

Instead of reading out the value of initial stress before low-temperature annealing from Fig. 3, it can be calculated using the following empirical formula.

For reference

Fig 3 Initial stress：

3 4 5 6 7 8 9 121110 13141516171819202122

Dd

Spring index　c

iτ

(8)(6)(4)(2)(0)

(10)(12)(14)(16)(18)(20)

020406080

100120140160180200220

τ 3

Dπd8 i

1.3.5　Initial tension of tension springSolid coiled cold formed extension springs have initial tension Pi.

Fig. 2 Coil tip

　Pig tail end（unground）

　Tangent tail end（unground）

　Open end（unground） 　Closed end（tapered）

　Open end（tapered） 　Open end（ground）

　Closed end（ground） 　Closed end（unground）

(f)(e)

(i)　

(h)

(d)(c)

(b)(a)

Note;（t1+t2）：Sum of the both ends thickness of the coil.

)t2 …………………………………(10)＋ 1t＋(d－1)s N t＝(H

In the case of compression springs having shape（b）,（c）,（e）, or（f）shown in Fig. 2 at its both ends and a specific solid height needs to be specified, specify the value obtained by the following formula as the maximum solid height. However, please take note

max×dtN＝ sH

Note; d（max）：Diameter of max. allowance of d.

The solid height of the spring shall generally be calculated from the following approximate calculation formula.

1.3.4　Solid height

……………………………………………(11)

dD＝　 c　 Spring index

Fig. 1 Stress correction factors：χ

４ ３ ５ ６ ７ ８ ９ 10 11 12 13 14 15 16 17 18 19 20 21 221.0

1.3

1.2

1.1

1.4

1.5

1.6

………………………………( 9 )0.615c

－1－4c

c44χ＝　　　＋

Stress correction factors against the value c of spring index should be fixed to the formula below or Fig. 1.

1.3.3　Stress correction factors

………………………………………………(13′

（But P＞Pi）

4

SUS　304N1（S30451）

４ ３

2＋ 2D1D

3

100cG＝ iτ

＝ iPπd 3

D8τ i＝

255D

4Gd2×0.8

D

422 d22

d＝ 216 4

D

i

2a＝　　

D

4229＝ d2 2

d24 4

D×0.752

Gd 4

D255＝ iτ

8D

3πdPi＝

3／4

}2{kgf/mm2N/mm

}2{kgf/mm2N/mm

}2{kgf/mm2N/mm

}3{kgf/mm3N/mm

1

2

δ

＝

χ

Str

ess

corr

ectio

n fa

ctor

s

mm 2iτ

N/

Torsi

onal

stres

s by i

nitial

tens

ion

Excerpt from JIS B 2704（1999） Calculation of Springs〔Technical Data〕

13621361