Spring design

Notation: $F$ = axial load, $D$ = mean coil diameter, $d$ = wire diameter, $C = D/d$ (spring index), $n$ = number of active coils, $G$ = shear modulus.

Stress

A section of the wire is subjected to a direct shear force $F$ and a torque $T = FD/2$.

        F
        |
    ---(  )---   wire section: shear F + torque T = F D/2
       D/2

Torsional shear stress: $\tau_t = \dfrac{16T}{\pi d^3} = \dfrac{8FD}{\pi d^3}$

Direct shear stress: $\tau_d = \dfrac{F}{\pi d^2/4} = \dfrac{4F}{\pi d^2}$

Maximum stress (inner fibre) by simple addition:

$$\tau = \frac{8FD}{\pi d^3}\left(1 + \frac{d}{2D}\right) = K_s\,\frac{8FD}{\pi d^3}, \quad K_s = 1 + \frac{1}{2C}$$

Wahl correction factor

Because the wire is curved, the stress is higher at the inner side of the coil, and the direct shear is not uniform. A. M. Wahl included the curvature effect and direct shear in one factor:

$$K = \frac{4C-1}{4C-4} + \frac{0.615}{C}$$ $$\tau_{max} = K\,\frac{8FD}{\pi d^3} = K\,\frac{8FC}{\pi d^2}$$

For $C = 6$: $K = 1.2525$. For small $C$ (tight coils) $K$ is large; for static loads of ductile wire $K_s$ may be used, but for fatigue $K$ is used.

Deflection (strain energy method)

Strain energy due to torsion of the wire (length $l = \pi D n$, polar moment $J = \pi d^4/32$):

$$U = \frac{T^2 l}{2GJ} = \frac{(FD/2)^2(\pi D n)}{2G(\pi d^4/32)} = \frac{4F^2D^3 n}{G d^4}$$

Work done by load $= \tfrac{1}{2}F\delta$. Equating, $\tfrac12 F\delta = U$:

$$\delta = \frac{8FD^3 n}{G d^4} = \frac{8FC^3 n}{G d}$$

Stiffness

$$k = \frac{F}{\delta} = \frac{G d^4}{8D^3 n} = \frac{G d}{8C^3 n}$$

Stiffness increases with the fourth power of the wire diameter and falls with the cube of coil diameter.

Data

$F = 800$ N, $\delta = 30$ mm, $C = 6$, $\tau = 450$ MPa, $G = 81.4$ GPa $= 81\,400$ N/mm².

Wire diameter

Wahl factor:

$$K = \frac{4C-1}{4C-4} + \frac{0.615}{C} = \frac{23}{20} + 0.1025 = 1.2525$$

From $\tau = K\dfrac{8FC}{\pi d^2}$:

$$d^2 = \frac{8KFC}{\pi\tau} = \frac{8\times1.2525\times800\times6}{\pi\times450} = 34.02 \;\Rightarrow\; d = 5.83\ \text{mm}$$

Select standard wire $d = 6$ mm. Check: $\tau = 1.2525\times\dfrac{8\times800\times6}{\pi\times 36} = 425\ \text{MPa} < 450$ MPa (safe).

Mean diameter

$$D = C d = 6\times6 = 36\ \text{mm}, \qquad D_o = D + d = 42\ \text{mm}$$

Active turns

From $\delta = \dfrac{8FC^3 n}{Gd}$:

$$n = \frac{G d\,\delta}{8FC^3} = \frac{81400\times6\times30}{8\times800\times216} = 10.6$$

Take $n = 11$ active turns. The actual stiffness is $k = \dfrac{Gd}{8C^3n} = \dfrac{81400\times6}{8\times216\times11} = 25.7$ N/mm, so the actual deflection is $800/25.7 = 31.1$ mm.

Total turns and lengths

Squared and ground ends add 2 inactive turns: $n_t = 11 + 2 = 13$.

• Solid length: $L_s = n_t d = 13\times6 = 78$ mm
• Clearance: $0.15\times31.1 \approx 4.7$ mm
• Free length: $L_f = L_s + \delta + \text{clearance} = 78 + 31.1 + 4.7 = 113.8$ mm

Pitch $p \approx \dfrac{L_f - 2d}{n} = \dfrac{113.8-12}{11} = 9.26$ mm.

Check for buckling: $L_f/D = 3.16 < 4$ for squared ends, so no guide is needed (for ground ends buckling is unlikely at this ratio).

Answer: $d = 6$ mm, $D = 36$ mm, $n = 11$, total 13 turns, $L_s = 78$ mm, $L_f \approx 114$ mm.

Loads

$$F_m = \frac{450+150}{2} = 300\ \text{N}, \qquad F_a = \frac{450-150}{2} = 150\ \text{N}$$

Spring index $C = 6$:

$$K_w = \frac{23}{20}+\frac{0.615}{6} = 1.2525, \qquad K_s = 1 + \frac{0.5}{6} = 1.0833$$

Stresses (with $D = Cd$)

$$\tau_a = K_w\frac{8F_aC}{\pi d^2}, \qquad \tau_m = K_s\frac{8F_mC}{\pi d^2}$$

Soderberg line

$$\frac{1}{n} = \frac{\tau_a}{S_{se}} + \frac{\tau_m}{S_{sy}}$$

Substitute and solve for $d^2$:

$$d^2 = \frac{8C\,n}{\pi}\left(\frac{K_wF_a}{S_{se}} + \frac{K_sF_m}{S_{sy}}\right)$$ $$d^2 = \frac{8\times6\times1.4}{\pi}\left(\frac{1.2525\times150}{300} + \frac{1.0833\times300}{680}\right) = 21.39\,(0.6263 + 0.4779) = 23.62$$ $$d = 4.86\ \text{mm} \;\Rightarrow\; \text{use } d = 5\ \text{mm}$$

Check with $d = 5$ mm

$D = 30$ mm.

$$\tau_a = 1.2525\times\frac{8\times150\times30}{\pi\times125} = 114.8\ \text{MPa}, \quad \tau_m = 1.0833\times\frac{8\times300\times30}{\pi\times125} = 198.6\ \text{MPa}$$ $$\frac{1}{n} = \frac{114.8}{300} + \frac{198.6}{680} = 0.3827 + 0.2921 = 0.6748 \;\Rightarrow\; n = 1.48 \ (>1.4)$$

Maximum stress $= 1.2525\times\dfrac{8\times450\times30}{\pi\times125} = 344$ MPa, well below the yield 680 MPa.

Number of active turns

Stiffness required: $k = \dfrac{450-150}{12} = 25$ N/mm.

$$n = \frac{G d}{8C^3 k} = \frac{79300\times5}{8\times216\times25} = 9.18 \;\Rightarrow\; 9\ \text{turns}$$

With $n = 9$: $k = 25.5$ N/mm and the deflection between loads is $300/25.5 = 11.8$ mm.

Answer: $d = 5$ mm ($D = 30$ mm), safety factor 1.48, 9 active turns.

(a) Surge

Surge is the resonance of the spring coils in longitudinal vibration. When the frequency of the applied load (for example, valve motion from a cam, including its harmonics) equals the natural frequency of the spring, a wave of compression runs along the coils, the stress rises sharply and the spring may fail and the valve may bounce.

Reduction:

• Make the natural frequency high: increase $d$, reduce $D$ and active turns, use light wire.
• Use dual (concentric) springs with different frequencies, whose friction damps surge.
• Use springs with unequal (variable) pitch (progressive), damping or dampers.

(b) Numerical

Spring mass $m = \rho\cdot\dfrac{\pi d^2}{4}\cdot\pi D n$ and stiffness $k = \dfrac{Gd^4}{8D^3n}$. Fundamental frequency with both ends fixed:

$$f = \frac{1}{2}\sqrt{\frac{k}{m}} = \frac{d}{2\pi D^2 n}\sqrt{\frac{G}{2\rho}}$$

Substituting $d = 0.004$ m, $D = 0.032$ m, $n = 7$:

$$\sqrt{\frac{G}{2\rho}} = \sqrt{\frac{79.3\times10^9}{2\times7850}} = 2247.6\ \text{m/s}$$ $$f = \frac{0.004}{2\pi\,(0.032)^2\,7}\times2247.6 = 0.08883\times2247.6 = 199.6\ \text{Hz}$$

Cam frequency at 3000 rpm $= 50$ Hz. The ratio $f/f_{cam} = 199.6/50 = 3.99 \approx 4$, so the 4th harmonic of the cam excites the spring at maximum speed (resonance at about 2993 rpm, almost the top speed).

Cam speed at which the 13th harmonic resonates:

$$N = \frac{60 f}{13} = \frac{60\times199.6}{13} = 921\ \text{rpm}$$

(The fundamental would resonate at $60f = 11\,976$ rpm.)

Acceptance: required natural frequency $\ge 13\times50 = 650$ Hz, but $f = 199.6$ Hz. The spring is not acceptable; it would need a much stiffer, lighter design (larger $d/D^2$, fewer turns) or dual springs.

Answer: $f = 199.6$ Hz; 4th harmonic resonates near maximum cam speed; 13th harmonic at 921 rpm; spring is unsafe against surge.

Common spring materials

Selection factors: strength and fatigue limit, stress-relaxation and sag resistance, operating temperature, corrosion, electrical or magnetic needs, wire availability and cost, and size (thin wire is stronger).

Estimation of strength

Tensile strength of drawn wire depends on diameter (thin wire is more worked and stronger):

$$S_{ut} = \frac{A}{d^{\,m}}$$

where $A$ and $m$ are constants for the material and $d$ is in mm. Springs fail in torsion, so the torsional yield strength is taken as a fraction of $S_{ut}$:

• cold-drawn carbon steels (music wire, hard-drawn): $S_{sy} \approx 0.45\,S_{ut}$
• hardened and tempered steels (oil-tempered, alloy wires): $S_{sy} \approx 0.50\,S_{ut}$
• the shear ultimate $S_{us} \approx 0.67\,S_{ut}$ is used for design in static strength.

The allowable design shear stress is $S_{sy}$ divided by the factor of safety (often 1.2 to 1.5).

Example: music wire, $d = 2$ mm

$$S_{ut} = \frac{2211}{2^{0.145}} = \frac{2211}{1.1057} = 2000\ \text{MPa}$$ $$S_{sy} = 0.45\times2000 = 900\ \text{MPa}$$

Answer: $S_{ut} \approx 2000$ MPa and $S_{sy} \approx 900$ MPa.

Compression spring ends

Squared and ground ends seat properly and give better load transfer and less buckling.

Extension spring ends

An extension spring works in tension, so its ends need hooks or loops (full loop, half loop, cross-over loop, extended hook, threaded plug). The stresses concentrate at the bend of the hook (bending and torsion), so a hook of larger radius lowers the stress and the hook is designed separately from the body. The body coils are wound tight (touching), unlike a compression spring where space is left between coils for deflection.

Initial tension

Extension springs are wound with coils pressed together, which stores a pretension $F_i$. No extension takes place until the applied load exceeds $F_i$. Then

$$F = F_i + k\,y$$

Initial tension keeps the spring tight, gives a definite starting load and reduces the length needed. It depends on spring index and manufacturing, and is usually limited so that the stress due to $F_i$ stays within the recommended band for the spring index.

Springs in combination

Series (same load, deflections add): $\dfrac{1}{k_s} = \dfrac{1}{k_1}+\dfrac{1}{k_2} = \dfrac{1}{20}+\dfrac{1}{30}$, so $k_s = 12$ N/mm.

Parallel (same deflection, loads add): $k_p = k_1+k_2 = 20 + 30 = 50$ N/mm.

Extension of the series combination under 300 N: $y = 300/12 = 25$ mm.

Answer: $k_{series} = 12$ N/mm, $k_{parallel} = 50$ N/mm, extension $= 25$ mm.