Calculation example of combined spring

3.9 Calculation example of combined spring

Table 12-2-23

Items

Unit

Formulas and Data

Original condition

Minimum working load P₁
Maximum working load P_n
Working range h
Load properties
Spring Type
End Type
Spring Material

N
N
mm

P₁=340
P_n=900
h=10
Impact load
Class II
Tighten and grind both ends together
Carbon Spring Wire Grade C

Parameter calculation

Maximum working load of outer and inner springs P_n1,P_n2

P_n1=(5/7)P_n=5/7*900=643
P_n2=P_n-P_n1=900-643=257

Minimum working load of outer and inner springs P₁₁,P₁₂

P₁₁=(5/7)P₁=5/7*340=243
P₁₂=P₁-P₁₁=340-243=97

Required stiffness of outer and inner springs p ‘

N/mm

P₁’=(P_n1-P₁₁)/h=(643-243)10=40

P₂’=(P_n2-P₁₂)/h=(257-97)10=16

P’=P₁’+P₂’=40+16=56

Required working limit load P_j

P_j1=1.25P_n1=1.25*643=803.75

P_j2=1.25P_n2=1.25*257=321.25

Spring material diameter d and spring diameter D

According to P, and P value, select from table 12-2-19, and the relevant parameters are as follows:

Spring parting	d	D	P_j	f_j	P’_d
Outer spring	5	35	850	5.903	144
Inner spring	3	20	339.76	3.398	100 1

Radial clearance of outer and inner springs δr

δr=(D₁₁-D₀₂)/2≥（d₁-d₂)/2
δr=[(35-5)-(20+3)]/2≥（5-3)/2=3.5＞1

Deformation under maximum working load F_n

F_n=P_n*h/(P_n-P₁)=16
F_n1=F_n2=F_n=16

Maximum working load of the selected spring P_n

P_n1≤0.8xP_j≤0.8×850≤680
P_n2≤0.8xP_j≤0.8×339.76≤272

Minimum working load of the selected spring P₁

P₁₁=P_n1(F_n1-h)/F_n1=680(16-10)/16=255

P₁₂=P_n2(F_n2-h)/F_n2=272(16-10)/16=102

Check working load P

Maximum working load (P_n=900)
P_n1+P_n2=952＞900
Minimum working load (P₁=900)
P₁₁+P₁₂=255+102=357＞340

Single coil deformation under maximum working load f_n

f_n1=0.8f_j=0.8*5.903=4.72

f_n2=0.8f_j=0.8*3.398=2.72

Effective coils n

coil

n₀₁=F_n1/f_n1=16/4.72=3.38,take n=3.5
n₀₂=F_n2/f_n2=16/2.72=5.58,take n=6

Total coils n₁

coil

inner n₁=n₀₁+2=3.5+2=5.5

outer n₁=n₀₂+2=6+2=8

Actual deformation under maximum working load F_n

F_n1=nf_n1=3.54*4.72=16.52

F_n2=nf_n2=6*2.72=16.32

Actual deformation under minimum working load F₁

F₁₁=F_n1*P_11/P_n1=16.52*255/680=6.19

F₁₂=F_n2*P_12/P_n2=16.32*102/272=6.12

Deformation under ultimate working load F_j

F_j1=nf_j=3.5*5.903=20.65

F_j2=nf_j=6*3.398=20.40

Pitch t

t₁=d+f_j=5+5.903=10.9

t₂=d+f_j=3+3.398=6.4

Free height H₀

H₀₁=nt₁+1.5d=3.5*10.9+1.5*5=46

H₀₂=nt₂+1.5d=6*6.4+1.5*3=38

Inner spring needs to be padded, thickness =46-38=8mm

Actual spring stiffness p ‘

P’₁=P’_d/n₀₁=144/3.5=41

P’₂=P’_d/n₀₂=100/6=16

P’₁+P’₂=40+16=57≈P’(56)

Winding ratio C

C₁=D/d=35/5=7

C₂=D/d=20/3=6.7

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3.9 Calculation example of combined spring

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