# Technical Information on spring design

This page was made to help you with your spring design. First let us explain some spring design terminology then we can get to business.

O.D. = Outside diameter of spring

I.D. = Inside diameter of spring

d = Wire diameter in inches

D = Mean diameter(O.D. - d) in inches

N = Number of total coils

n = Number of active coils

FL = Free length

F = Deflection

P = Applied load

R = Spring rate in load per inch

θ = Angular position

This is not all of the design terminology, but enough for this brief web page.

Let's begin, to us the first design requirement in spring design is the material type. Where will this spring be used, what type of environment will this spring be in, what stress will be on this spring, all effected by the material type. Next would be design tolerances, allowing ample tolerances will make the spring more economical to make. Normal load tolerances are ±10% for precision springs and ±16% for commerical springs. Then the type of ends you need for your spring. The following examples are the minimum information we need to help design your spring. Now to the types of springs.

**Compression springs:** First would be to determine the installed height(L1) and load required(P1) at this height of the spring, then the load(P2) required at the deflected compressed height(L2). If you are looking for a long life spring, keep the total deflection of the spring between 20% and 80% of FL. Compression spring O.D. will grow as the spring is compressed so remember this in your design if space is tight.

F = L1 - L2.

P = P2 - P1.

Using these two numbers we can determine the spring rate required, so we take P / F = R. Then we can determine the free length of the spring by using P2 / R + L2 = FL.**Example:**

2.8LBS at .800" installed height, 5.6LBS at .600" compressed height, must fit in .250 hole, must fit over .156 shaft, use music wire, ends closed.

.800 - .600 = .200"

5.6 - 2.8 = 2.8LBS.

2.8 / .200 = 14LBS spring rate.

5.6 / 14 + .600 = 1.000" FL

Using our spring design software, this example would result in the following parameters:

.240" ± .004" O.D.

.176" I.D.

1.000" ± .046" FL

.032" diameter wire(d)

13.9 coils

.479" solid height

Spring is not above allowable design stresses when compressed so this spring will have a long life.

**Extension springs:** First determine the installed extension length(L1) and load(P1), then the extended length(L2) and load(P2). The spring rate is determined by R = (P2 - P1) / (L2 - L1). **Example:**

.25LBS at 1.200" installed extension, .5LBS at 1.450" extended length, must hook on .125 diameter shaft, use stainless steel, machine hooks, minimum 1,000,000 cycles.

(.5 - .25) / (1.450 - 1.200) = 1LBS spring rate.

Using our spring design software, this example would result in the following parameters:

.187" ± .004" O.D.

.149" I.D.

.970" ± .020" FL inside hooks

.019" diameter wire(d)

302 stainless steel

Cycle life greater than 1,000,000

**Torsion springs:** Loads used for torsion springs are determined by angular position(θ), not by length. Measure angles by having one leg horizontal at 0°, then to the other leg in the opposite direction the spring is wound. First determine the load(M1) at the installed deflection angle(θ1), then the load(M2) at final deflection angle(θ2). R = (M2 - M1) / (θ2 - θ1) * 360.**Example:** .18LBS at 43.6 degrees, .76LBS at 23 degrees, use stainless steel, must fit around .236" diameter pin, straight torsion legs .750" long, maximum spring length .100".

(.76 - .18) / (43.6 - 23) * 360 = 10.13 spring rate.

Using our spring design software, this example would result in the following parameters:

.330" OD

.256" ID ± .003"

1.64 active coils(n)

.037" diameter wire(d)

50° free angle ± 3.5°

.098" spring height

302 stainless steel

If you have any questions, contact us at info@badgerspring.com and we will try to help in your spring design.