Calculation Of Springs

〔Technical Data〕

Excerpt from JIS B 2704（1999）

Calculation of Springs

Table 1. Meaning of Symbols Symbol d D1

Meaning of Symbols

Unit

Diameter of material

mm

Inside diameter of a coil

mm

D2

Outside diameter of a coil

mm

D

D1＋D2 Average diameter of a coil＝    2

mm

Nt

Total number of coils

−

Na

Number of active coils

−

L

Number of free coils

mm

Hs

Solid height

mm

p

Pitch

Pi

Initial tension

c

D Spring index c＝ d

−

G

Modulus of rigidity

N/mm2{kgf/mm2 }

P

Load on a spring

δ

Deflection of a spring

mm N{kgf}

N{kgf}

8N aD3 P …………( 1 ) Gd4

τ ＝ χτ0 ……………………( 5 )

k ＝

P Gd ………( 2 ) ＝ δ 8N aD3

d ＝3

τ0＝

8DP πd3

N a＝

4

k

Spring constant

N/mm{kgf/mm}

Torsional stress

N/mm2{kgf/mm2 }

τ

Torsional amending stress

N/mm2{kgf/mm2 }

τi

Torsional stress by initial tension

N/mm2{kgf/mm2 }

−

Stress correction factors

Hz

f

Frequency

U

Energy saved in spring

N・mm{kgf・mm}

ω

Material weight of a volume unit

N/mm3{kgf/mm3 }

W

Weight of spring moving part

N{kgf}

g

Gravitational acceleration

mm/s 2

(1)

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.

……………( 3 )

Gdδ …………( 4 ) τ0＝ πN aD2

8N aD 3 ( P−P i ) …( 1′) Gd4 4 Gd P−P i ＝ k ＝ …( 2′) 8N aD 3 δ

δ＝

Gd4 δ Gd4 …………( 7 ) ＝ 8D3 P 8D3 k

Pδ kδ 2 U ＝ ……………( 8 ) ＝ 2 2

τ0＝

8DP ……………( 3′) πd3

τ0＝

Gdδ ＋ τi ……( 4′) πN aD 2

) τ ＝ χτ0 ……………………( 5′ d ＝3

8DP 8χDP ) ……( 6′ ＝3 πτ πτ0

Gd4 δ Gd4 …( 7′) ＝ 8D3 k 8D 3 ( P−P i ) ( P＋P i ) δ U ＝ ……………( 8′) 2 N a＝

1.3 Considerable matters in designing springs   Fix the value of modulus of rigidity G in  1.3.1 Modulus of rigidity

designing spring with Table 2 as a rule.

Table 2. Modulus of Rigidity（G） Value N/mm2(kgf/mm2)

Symbol

Spring steel

78×10 3 {8×10 3 }

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

Hard steel wire

78×10 3 {8×10 3 }

SW40C, SW60C

Music wire

78×10 3 {8×10 3 }

SWP ASTM A228

78×10 3 {8×10 3 }

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

Material

Oil tempered steel wire

（    ） Equivalent to

SUS 302

SUS 302

SUS 304

SUS 304

3 3 Stainless steel SUS 304N1 69×10 {7×10 } （S30451） wire

SUS 316 SUS 631 J1

SUS 304N1（S30451） SUS 316

74×103

{7.5×103 }

1.3.2 Number of Active Coils At spring design, the number of active turns should

1.5

1.3 1.2

χ 1.1

1.0 ３ ４ ５ ６ ７ ８ ９ 10 11 12 13 14 15 16 17 18 19 20 21 22 D Spring index c ＝  d

1.3.4 Solid height The solid height of the spring shall generally be calculated from the following approximate calculation formula. However, the solid height of the compression spring shall not be specified by the purchaser, generally. H s＝( N t −1) d ＋( t 1＋t 2 )…………………………………(10) Note;（t1+t2）：Sum of the both ends thickness of the coil. 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 m aximum solid height. However, please take note H s＝N t ×d max ……………………………………………(11) Note; d（max）：Diameter of max. allowance of d.

D2

P

D1

L

D

L

D2

X 1＝X 2＝0.75 therefore N a＝N t−1.5

d For reference：L＝Na・d＋2(D 2−2d)

(2) Tension Springs Effective turns of tension springs are as below. D1 D

Excluding hooks.

N a＝N t

220 (20) 200 (18) 180 (16) 160 (14) 140 (12) 120 τi (10)  (kgf/mm 2 ) 100 (8) 80 (6) 60 (4) 40 (2) 20 (0) 0 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

D Spring index c ＝  d (1) In the case of the stainless steel witre, reduce 15％ from the initial stress of the steel wire. (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. For reference 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. G τi＝ 100 c Then the initial tension is as follows. (1) In case of hard steel wire and piano wire〔G＝78×10 3N/mm 2 {8×10 3 kgf/mm 2}〕 G ×0.75（The value of 0.75 is reduced by 25％ through low-temperature annealing.） Torsional stress by initial tension τi＝ 100c πd 3 Gd 4 229 d 4 24 d 4 τi＝ Initial tension P i＝ ×0.75 ＝ 2 8D 255 D 2 D2 D (2) In case of stainless steel wire〔G＝69×10 3N/mm 2 {7×10 3 kgf/mm 2 }〕 G Torsional stress by initial tension τi＝ ×0.8（The value of 0.8 is reduced by 20％ through low-temperature annealing.） 100c πd 3 Gd 4 216 d 4 22 d 4 τi＝ Initial tension P i＝ ×0.8 ＝ 2 255D 2 8D D2 D

1.3.6 Surging Fig. 2 Coil tip (a)  Closed end（unground）

(b)  Closed end（ground）

(c)  Closed end（tapered）

(d)  Open end（unground）

(e)  Open end（ground）

(f)  Open end（tapered）

(g)  Open end3／  4 end turns（unground） (h) Tangent tail end（unground）

To avoid surging, natural frequency of the spring should be selected not to resonate to all the vibrations which affect the spring. Natural frequency of spring is kg 70 d G f ＝a   ＝a ……………………………………………(13) W πN aD 2 ω i Here, When both ends are free or fixed type： a＝   2 2 i−1 a＝   ：In case one end fixed and the other free i＝1,2,3 … 4 G of steel ＝ 78×103N/mm2{8×103kgf/mm2}, 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. f 1＝3.56×105

d ………………………………………………(13′) N aD 2

1.3.7 Other considerations

be fixed equal to the number of free coils as follows.

N a＝N t−(X 1＋X 2) X1and X2：indicate each number of coils of both ends. (a)When only the tip of the coil touches the next free coil, 〔equivalent to(a),(b)and(c)in Fig.2〕 X 1＝X 2＝1 therefore N a＝N t−2 ３ (b)When coil end dose not touch the next coil,   ４ length of the  grinding part 〔equivalent to(e)and(f)in Fig. 2〕

Fig 3 Initial stress：τi（value prior to low-temperature annealing, formed from steel wire）

τi N/ mm 2

1.4

(1) In case of compression spring

For reference：L＝Na・P＋1.5d

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1.6

SUS 631 J1

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

d

Fig. 1 Stress correction factors：χ

8DP 8χDP ＝3 ……( 6 ) πτ πτ0

 1.2.2 In case of tension spring with initial tension（But P＞Pi）

mm

τ0

χ

δ＝

Stress correction factors against the value c of spring index should be fixed to the formula below or Fig. 1. 4 c −1 0.615 ………………………………( 9 ) χ＝   ＋ 4 c −4 c

Torsional stress by initial tension

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

1.3.3 Stress correction factors

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

Stress correction factors

1. Calculation 1.1 Symbols used in design formula for spring 

 In calculating the design of springs, the following matters also have to be considered. (i)   Pig tail end（unground）

1.3.5 Initial tension of tension spring Solid coiled cold formed extension springs have initial tension Pi. In this case, torsional stress caused by initial tension is ranged inside the shaded portion in Fig. 2 as a rule. πd 3 P i ＝  τi ……………………………………………………(12) 8D 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.

(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. (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. (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. (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. p＝

L−H  s ＋d …………………………………………………(14 ) Na

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