Basis of Calculation
Symbols
d
D
De
Dd
Dh
Lo
L1
L2
Ln
Lc
S1
S2
Sn
Sc
V
G
Sa
Sk
n
nt
R
R1
F1
F2
Fn
Fc
t1
t2
tn
tc
tzul
K
d
D
De
Dd
Dh
Lo
L1
L2
Ln
Lc
S1
S2
Sn
Sc
V
G
Sa
Sk
n
nt
R
R1
F1
F2
Fn
Fc
t1
t2
tn
tc
tzul
K
Unit
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
N/mm
N/mm
N
N
N
N
N/mm2
N/mm2
N/mm2
N/mm2
N/mm2
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
mm
N/mm
N/mm
N
N
N
N
N/mm2
N/mm2
N/mm2
N/mm2
N/mm2
Description
Wire diameter
Mean coil diameter
Outer coil diameter
Arbor diameter
Bush diameter
Unstressed length
Pre-stressed length
Stressed length
Max. stressed length
Block length
Pre-stressed spring deflection
Stressed spring deflection
Max. stressed spring deflection
Spring deflection to block
Bearing coefficient
G-Module (see table of materials)
Total minimum distance
Spring deflection bend
Active coil number
Total coil number
Spring rate
Spring rate for one coil
Pre-stressed force
Stressed force
Max. spring force
Theoretical block force
Shear with F1
Shear with F2
Shear with Fn
Shear with Fc
Permitted shear
Shear coefficient
Wire diameter
Mean coil diameter
Outer coil diameter
Arbor diameter
Bush diameter
Unstressed length
Pre-stressed length
Stressed length
Max. stressed length
Block length
Pre-stressed spring deflection
Stressed spring deflection
Max. stressed spring deflection
Spring deflection to block
Bearing coefficient
G-Module (see table of materials)
Total minimum distance
Spring deflection bend
Active coil number
Total coil number
Spring rate
Spring rate for one coil
Pre-stressed force
Stressed force
Max. spring force
Theoretical block force
Shear with F1
Shear with F2
Shear with Fn
Shear with Fc
Permitted shear
Shear coefficient
Formulae
Wire diameter
Spring deflection
Spring defelction to bend
Number of spring coils
Total minimum distance
Spring rate
Spring force
Springing
Shear static
Shear coeffiecient
Shear dynamic
Examples for Calculation of a compression spring
1. Statically Loadable Compression Spring
Given:
Needed:
F1=8 N; F2=24 N; Ds=16 mm; Di=10 mm; spring rigidity tolerance TR=±5%
Compression spring data (d, De, D, Di, n, R, s1, s2, sn, LC, Sa, Ln, L1, L2, w, k, spring material)
Solution:
a.) Dimensions
Assumptions:
Wire diameter:
Chosen:
F2=0,9 · Fn; D=14 mm; patented steel spring wire DIN 17223/01 with Rmmin=1600 N/mm² and G=80000 N/mm²; tzul=0,5 · Rmmin =800 N/mm²
d=1,1 mm
Required spring rigidity:
Rerf = (F2-F1)/Ds = (24 N-8 N)/16 mm = 1 N/mm
with the tolerances
Rmin = 0,95 N/mm < Rerf < Rmax = 1,05 N/mm
coil number:
n = (G · d4) / (8 · D3 · Rerf)
n = (80000 N/mm² · 1,14 mm4) / (8 · 14³ mm³ · 1 N/mm) = 5,34
Chosen:
n=5,5
b) recalculation:
Rvorh = (G · d4) / (8 · D³ · n) = (80000 N/mm² · 1,14 mm4) / (8 · 14³ mm³ · 5,5) = 0,970 N/mm
The condition Rmin < Rvorh = 0,97 N/mm < Rmax is fulfilled.
tvorh = (8 · Fn · D) / (p · d³) = (8 · 24 N · 14 mm) / (0,9 · 3,14 · 1,1³ mm³) = 714,6 N/mm²
Class C steel spring wire may be selected with Rmmin = 1690 N/mm² (tzul = 845 N/mm²). The condition tvorh = 714,6 N/mm² < tzul = 845 N/mm² is fulfilled, also according to the tension increase by factor k according to the formula:
w = D/d = 14/1,1 mm = 12,73
k = (w + 0,5) / (w - 0,75) = (12,73 + 0,5) / (12,73 - 0,75) = 1,104
tk = ktvorh = 1,104 · 714,6 N/mm² = 789 N/mm².
The functional and stress analysis conditions are fulfilled by the calculated spring dimensions.
c) Calculation of further data
Spring length:
LC < (n · d) = (n + 2) · d = (5,5 + 2) · 1,1 mm = 8,25 mm
Sa = (x · d · n) = (0,4 · 1,1 mm · 5,5) = 2,42 mm
Ln = LC + Sa = 8,25 mm + 2,42 mm = 10,67 mm
Sn = Fn / Rvorh = F2 / (0,9 · Rvorh) = 24 N / (0,9 · 0,97
N/mm) = 27,49 mm
L0 = Ln + sn = 10,67 mm + 27,49 mm = 38,16 mm
SW = [(L0 - d) · (nt - n)] / n = 6,55 mm
spring defelection and installation lengths:
s1 = F1 / Rvorh = 8 N / 0,97 N/mm = 8,25 mm
s2 = F2 / Rvorh = 24 N / 0,97 N/mm = 24,74 mm
L1 = L0 - s1 = 38,2 mm - 8,25 mm = 29,95 mm
L2 = L0 - s2 = 38,2 mm - 24,74 mm = 13,46 mm
It is possible to round up the unstressed spring length by choosing greater coil spacings.
Spring data
d=1,1 mm; De=15,1 mm; Di=12,9 mm; D=14mm; L0=38,2 mm; n=5,5; class C spring steel wire according to DIN 17223; production compensation via L0.
d) Bending safety
h = (100% · s2) / L0 = (100% · 24,74 mm) / 38,2 mm = 64,76%
l = L0 / D = 38,2 mm / 14 mm = 2,73
l=2,73 is valid for springing required for bending herf=50% for compression springs with changeable bearing and herf=68% for compression springs guided mounting and parallel ground spring bearing area. The spring is not bend-proof for changing bearing. Re-calculation is possible.
The spring is not bend-proof under the assumed mounting conditions (KL=1). Parallel mounting and guidance of spring ends is necessary.
If you are having problems with the mounting of your springs or looking for the perfect springs for narrow mountings please call our development department!
Examples for the Calculation of a Tension Spring
1. Statically Loadable Tension Spring
Given:
Needed:
F1 = 40N; F2 = 100N; Hub-sh = 30mm; Da <= 11mm
Dimensions of spring with minimum installation length
Dimensions of spring with minimum installation length
Solution:
a) Dimensions
Assumptions:
D = 8,5mm; tzul = 900N/mm² for patented steel spring wiret according to DIN 17223
wire diameter:
chosen:
d=1,4mm
coil diameter: For class D patented spring steel wire with d=1,4mm, Rm=2110 N/mm² is valid. The permissible shear stress may be calculated as follows:
The coil diameter can thus be increased to the maximum possible value Dm = 9,6m
Spring rate:
R = (F2 - F1) / Sh = (100N - 40N) / 30mm = 2N/mm
Number of coils:
n = (G · d4) / (8 · D3 · R) = (81400 N/mm² · 1,44 mm4) / (8 · 9,6³ mm³ · 2 N/mm) = 22,09
gewählt:
n=22.
b) Minimum unstressed length L1
The unstressed length decreases with the spring deflection s1. s1 is dependant on the magnitude of the initial tension F0. For a tension spring without included initial tension: ist
s1 = F1 / R = 40N / 2N/mm = 20mm. for the current coiling w = Dm / d = 9,6 / 1,4 = 6,86 an included initial tension t0 = 0,1 · tzul = 0,1·972N/mm² = 97,2N/mm² is permissible.
This leads to an initial tension of F0=(p · t0 · d³) / (8 · D) = (3,14 · 97,2N/mm² · 1,4³mm³) / (8 · 9,6mm) = 10,9N
The unstressed length decreases with the spring defelction s1. s1 is dependant on the magnitude of the initial tension F0. For a tension spring without included initial tension:
s1 = F1 / R = 40N / 2N/mm = 20mm for the current coiling. w = Dm / d = 9,6 / 1,4 = 6,86 ian included initial tension of t0 = 0,1 · tzul = 0,1·972N/mm² = 97,2N/mm² is permissible.
This leads to an initial tension F0=(p · t0 · d³) / (8 · D) = (3,14 · 97,2N/mm² · 1,4³mm³) / (8 · 9,6mm) = 10,9N
Thus, the unstressed lengths are:
L1 = L0 + s1 = 45,4mm + 14,6mm = 60mm
L2 = Ln = L1 + sh = 60mm + 30mm = 90mm
A smaller value for the unstressed length may be obtained by a larger value for the included initial tension F0. This is only possible by the usage of specialised techniques.
c) Tension Re-Calculation
t = (8 · D · Fn) / (p · d³) = (8 · 9,6mm · 100N) / (3,14 ·1,4³mm³)
t = 915,5N/mm² < tzul = 949,5N/mm²