1. Introduction
Springs are widely used in mechanisms to exert a force. In many critical applications, vehicle suspension springs and engine valve springs, they are subjected to rapid changes in length and their mass must be kept to a minimum to reduce undesirable dynamic effects. This inevitably means that such springs are working at high stress levels which has implications concerning the choice of materials and manufacturing processes. This section will concentrate on coil springs working in compression, dealing with design, stress analysis, choice of materials and manufacturing processes.

2. Forces Acting
The equations below assume that the spring coil is loaded centrally along the spring axis, which remains straight and the ends are free to rotate about the spring axis relative to one another. In practice these conditions are often not met, end fixings commonly do not distribute the load centrally and in some car suspension systems the coil axis assumes a curved shape that varies depending upon the deflection. The design factor used must take account of any departures from the ideal assumptions.

The loads in the wire can be deduced from the layout diagram on the left and the FBD on the right of a part of the coil and are:

A force F parallel to the spring axis acting nearly transverse to the wire axis and through its centre and
A torque T = FD/2 about the axis of the wire.

Both of these generate shear stresses in the wire.

Ss = ± FDd/4J + F/A where

J is the polar moment of area of the wire cross section, D is the mean diameter of the coil and d the wire diameter.

The first term is due to torsion and the second term is due to the transverse shear force and this expression ignores the curvature of the wire in the spring.

As J = 3.14159 d4/32 ..and.. A = 3.14159 d2/4

The equation for shear stress can be written as:

Ss = [8 F D/d3 + 4 F/d2]/3.14159

The spring index C is defined as D/d and for many springs is in the range 6-12. The two terms in the previous expression for shear stress can be combined by introducing a shear stress correction factor:

Ks =(2 C + 1)/2C ... which gives

Ss = Ks8 F D/3.14159d3

3. Curvature Effect
As the wire is coiled, the rotation of the wire on the inner side of the coil occurs over a shorter distance than rotation on the outer side of the coil. This means that the shear stress on the inner side of the coil must be greater than that at the outside of the coil. This also means that the centre of rotation of the wire must be displaced away from the wire axis towards the centre of the coil, although the actual displacement is quite small.

Two factors have been proposed to include the effects of both the curvature and the transverse shear:

Kw, the Wahl factor ..and.. KB, the Bergstrasser factor.
The results from these two methods usually differ by less than 1%.

The curvature correction factor can be determined by cancelling the effect of transverse shear, done below for the Bergstrasser factor:

When fatigue is likely to be a factor (or the spring material must be considered to be brittle) Kc is used as a stress concentration factor. Normally in fatigue calculations the stress concentration factor would be corrected to Kf because of notch sensitivity, but for high strength steels the notch sensitivity is close to 1, so the full value of Kc (or KB or KW in some procedures) is used.

4. Equations Used in Design
The relationship between load and deflection is given by:

F = G d4 ð/8 D3 Na ... where
G is the shear modulus, for spring steel a value of 80 GPa can be used.
ð is the deflection and
Na is the number of active turns, in determining this account must be taken of the ways that the ends of the coil are finished and appropriate corrections made.

The linear spring rate is then:

R = F / ð = G d4/8 D3 Na

The basic stress is given by: Ss = 8 D F/3.14159 d3 .. and the corrected stress:

Ssc = 8 D F KW/3.14159 d3

Design methods used to be based on nomograms, however spreadsheets are now used. The Society of Automotive Engineers (SAE) publish a 'Spring design Manual' that contains information about design, appropriate design methodology, reliability and materials.

Normally a design will involve some constraints about the space available, (governing D) required spring rate, limits of motion, availability of wire diameter, material, maximum allowable stress when the spring is solid. Some iterations will probably be needed to reach the best solution. Fatigue testing is commonly carried out on new designs of springs destined for critical applications.

5. Materials and Manufacture
'Music wire', AISI 1085 steel is used in diameters up to 3 mm for the highest quality springs.

For diameters up to 12 mm, AISI 1065 may be used in the hardened and tempered condition or cold drawn.

For larger wire diameter, or for highly stressed applications, low alloy steels containing chrome - vanadium, chrome - silicon and silicon - manganese, hot rolled, hardened and tempered (to 50 - 53 Rc hardness, equivalent to about 1600 to 1700 MPa UTS) are used.

For most spring materials increasing the wire diameter reduces the UTS, and if the UTS is plotted against the wire diameter on semi - log graph paper, the line is often nearly straight for many of the metals used for springs.

Research suggests that the yield shear stress of most metals lies in the range of 0.35 to 0.55 times the UTS, in the absence of specific information, a value of 0.5 time the UTS can be used for hardened and tempered carbon and low alloy steels.

The fatigue strength of springs can be increased by cold setting and particularly by shot peening, which can increase the fatigue endurance strength by as much as 50%. Both processes are routinely carried out on highly loaded automotive suspension springs.

Titanium Springs
Springs made from titanium alloys are 50 - 70% lighter than equivalent steel springs and occupy about 25% less space. Titanium alloys have a lower modulii of elasticity than steels.

Commercially pure titanium has been used as has Ti-6Al-4V. However for heavier duty applications high strength beta alloys such as Ti-13V-11Cr-3Al, Ti-15V-3Cr-3Sn-3Al and Ti-8V-6Cr-4Mo-4Zr-3Al have been used. These highly alloyed beta alloys are costly and their use has been limited mainly to aerospace applications. Less costly beta alloys, with fewer alloying additions have been developed, Ti-6.8 Mo-4.5 Fe-1.5 Al-0.15 O and been shown to properties competitive with more expensive alloys.

6. The Effective Mass of a Spring
Springs are normally used to provide a force in a dynamic system, this frequently means that one end of the spring will be moving and the other end will either be stationary or moving at a different velocity. This section looks at the derivation of the effective mass of a spring with one end fixed and the other moving at a velocity Ve, see diagram below:

It is assumed that the spring is uniform along its length, L and that the total mass of the spring is ms, so the mass of a small element of the spring of length dx at x from the fixed end is given by:

dm = msdx/L

The velocity of the element x from the fixed end is given by:

vx = vex/L The kinetic energy of the element, mass(velocity)2/2, is given by:


Rearranging and integrating gives:

The KE of the spring is one third that of a mass equal to that of the spring moving at a velocity of ve showing that the equivalent mass to be used in such circumstances is 1/3 of the mass of the spring.
The effective mass being equal to 1/3 of the actual mass of the spring is valid whether the spring is a conventional coil or a uniform bar.

7. Surge in Springs
If one end of a helical spring is held fixed and the other end suddenly compressed and then held fixed, a compression wave is formed that travels along the spring, is reflected from fixed ends (now both ends) and the motion continues until it is damped out. This phenomena is referred to as 'surge' and the natural frequencies of this wave are important in spring design as it is undesirable that the frequency of a disturbance coincides with any of the spring's natural frequencies.

The wave equation for the longitudinal vibration of rods is modified so it can be applied to springs. The derivation of this is given in ref. 10.1 and the result of this is:

omegan = 3.14159 n (k/ms)0.5 rad/s.

fn = (n/2)(k g/Ws)0.5 Hz .. where n = 1, 2, 3 ...

where k is the spring stiffness, Ws is the total weight of the spring.

Dr David J Grieve. Revised: 26th October 2010