| sign | Meaning of the sign | |
|---|---|---|
| d | wire diameter(φ) |
  |
| D1 | inside diameter of coil (㎜) | |
| D2 | outside diameter of coil(㎜) | |
| D | mean diameter of coil （D1+D2）/2 |
|
| Na | number of active coils | |
| Nt | total number of turns | |
| L | free length（㎜） | |
| P | load（N） | |
| δ | spring travel（㎜） | |
| k | spring constant（N/㎜） | |
| G | modulus of transverse elasticity（N/㎟） | |
| c | spring index（D/d） |
| material | modulus of transverse elasticity（N/㎜） |
| hard steel wire | ７８５００ |
| piano wire | ７８５００ |
| oil tempered wires | ７８５００ |
| stainless steel | ６８５００ |
Calculate the weight of the spring
Sample⇒piano wireφ2.0 active coils 5 coil diameter φ15.0
1.Find the Mass of the spring
cross section × The length of the spring ＝ Mass of the spring
formula ⇒ （1.0×1.0×3.14）×（15.0×3.14×7）＝3.14×329.7＝1035.258㎣
2.Find the weight of the spring
mass × specific gravity ＝ weight of the spring
formula ⇒ 1035.258㎣ × 0.00784g/㎥ ＝ 8.116ｇ
Calculate the fixed number of the spring
Sample⇒piano wireφ2.0 active coils5 coil diameteφ15.0
spring constant ＝ （modulus of transverse elasticity×wire diameter to the 4）÷（８×number of active coils×mean diameter to the 3）
formula ⇒ （78,500×2.0＾4）÷（8×5×15.0＾3）＝9.304N/㎜
Calculate the load
Spring properties mentioned above. free length 30㎜ Installation length 25㎜.
load ＝ spring travel × spring constant │ spring travel ＝ free length － installation length
formula ⇒ （30-25）×9.304 ＝ 46.52N
Calculate the stress of the spring
When assume it spring properties mentioned above.
spring index（ｃ）＝mean diameter of coil（D）÷wire diameter（ｄ）
Waal correction coefficient（K）＝{（４×spring index）－１}÷{４×spring index－４}＋（０．６１５÷spring index）
stress of the spring（T)＝（８×Waal correction coefficient×mean diameter）÷（Pi×wire diameter to the 3）×load
spring index ⇒ 15.0÷2＝7.5
Waal correction coefficient ⇒ {（4×7.5-1）/（4×7.5-4）}＋（0.615÷7.5）＝1.1974
stress of the spring ⇒ {（8×1.1974×15.0）÷（3.14×2.0＾3）}×46.52＝266.097N/㎟