A helical compression spring

A spring is three constraints at once: it has to give the right force at the right deflection, it has to survive the shear stress that produces, and it has to fit — solid length, free length and buckling all decide whether the rate that came out of the first constraint can be built. This one returns a valve: about 50 N at 4 mm of working deflection, in a bore 24 mm across.

use steel

Four significant figures.

digits 4

d_{wire}

3 mm

D_{mean}

20 mm

n_{active}

8

Spring steel is a little stiffer in shear than structural steel; this is the conventional design value for cold-drawn wire.

G_{spring}

79.3 GPa

The spring index

The index is the single most useful number about a spring. Below four the wire cannot be coiled without cracking; above twelve the coil tangles in handling and the spring is slack in its bore. Six to eight is comfortable.

C_{index}

\frac{D_{mean}}{d_{wire}}

\frac{20 mm}{3 mm}

6.667

check

C_{index} \geq 4 and C_{index} \leq 12

6.667 \geq 4 and 6.667 \leq 12

pass

Rate and deflection

k_{rate}

\frac{G_{spring} \cdot {d_{wire}}^{4}}{8 \cdot {D_{mean}}^{3} \cdot n_{active}}

\frac{79.3 GPa \cdot {(3 mm)}^{4}}{8 \cdot {(20 mm)}^{3} \cdot 8}

12.55 N/mm

F_{work}

50 N

x_{work}

\frac{F_{work}}{k_{rate}}

\frac{50 N}{12.55 N/mm}

3.985 mm

The requirement was a force at a deflection, so that is what gets checked — the rate is only the means. A spring is wound to a tolerance on its rate, so this is a band rather than a number; ±10% is ordinary for a commercial spring.

x_{required}

4 mm

F_{at_required}

k_{rate} \cdot x_{required}

12.55 N/mm \cdot 4 mm

50.18 N

check

F_{at_required} \geq 0.9 \cdot F_{work} and F_{at_required} \leq 1.1 \cdot F_{work}

50.18 N \geq 0.9 \cdot 50 N and 50.18 N \leq 1.1 \cdot 50 N

pass

Shear stress, with the curvature correction

The direct shear formula understates the stress on the inside of the coil, where the wire is tighter and the shear from the load adds to the torsional shear. The Wahl factor corrects for both, and at this index it is worth a quarter of the answer.

K_{wahl}

\frac{4 \cdot C_{index} - 1}{4 \cdot C_{index} - 4} + \frac{0.615}{C_{index}}

\frac{4 \cdot 6.667 - 1}{4 \cdot 6.667 - 4} + \frac{0.615}{6.667}

1.225

τ_{work}

\frac{K_{wahl} \cdot 8 \cdot F_{work} \cdot D_{mean}}{π \cdot {d_{wire}}^{3}}

\frac{1.225 \cdot 8 \cdot 50 N \cdot 20 mm}{π \cdot {(3 mm)}^{3}}

115.5 MPa

The allowable shear stress belongs to the wire and its diameter — a drawn wire is stronger in a thin section than a thick one — so it is taken from the wire's specification rather than computed. This is a music wire value.

τ_{allow}

700 MPa

check

τ_{work} \leq τ_{allow}

115.5 MPa \leq 700 MPa

pass

Whether it fits

Solid length is where the coils touch and the spring stops being a spring. Two dead coils at the ends, squared and ground.

n_{total}

n_{active} + 2

8 + 2

10

L_{solid}

n_{total} \cdot d_{wire}

10 \cdot 3 mm

30 mm

L_{free}

40 mm

The deflection available before the coils shut, and the working deflection against it. Working past about 80% of the way to solid is how a spring takes a set.

x_{solid}

L_{free} - L_{solid}

40 mm - 30 mm

10 mm

check

x_{work} \leq 0.8 \cdot x_{solid}

3.985 mm \leq 0.8 \cdot 10 mm

pass

Buckling, which a compression spring does exactly as a column does. For squared and ground ends held between flat parallel surfaces, the free length must stay under about 2.6 times the mean diameter.

check

L_{free} \leq 2.6 \cdot D_{mean}

40 mm \leq 2.6 \cdot 20 mm

pass

And it has to go in the bore, with clearance for the coils to spread as it compresses.

bore

24 mm

check

D_{mean} + d_{wire} \leq bore - 1 mm

20 mm + 3 mm \leq 24 mm - 1 mm

pass

Force against deflection

Straight, which is the whole idea of a helical spring — and drawn only as far as the coils allow. Past solid length there is no spring left to plot: the stack becomes a short steel column, the force rises without bound, and a line continued through that point would be a wrong picture of a right equation.

fn force defined

plot (force, 0, 10)