Learn about elasticity and how to determine the force exerted by a spring. 

What is a spring?

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.
There is nothing particularly magical about the shape of a coil spring that makes it behave like a spring. The 'springiness', or more correctly, the elasticity is a fundamental property of the wire that the spring is made from. A long straight metal wire also has the ability to ‘spring back’ following a stretching or twisting action. Winding the wire into a spring just allows us to exploit the properties of a long piece of wire in a small space. This is much more convenient for building mechanical devices.
The response of a metal wire to stretching (axial load) and twisting (torsion) is governed by different physics and the design of a particular spring can exploit one kind of deformation over another. In addition, the elastic properties of metals depend strongly on their grain microstructure. This can be changed by stress and a controlled heating/cooling process known as annealing. If a metal wire was formed from a straight section into a coil then it is likely that it would need to be re-annealed to restore its original elastic properties.

What happens when a material is deformed?

When a force is placed on a material, the material stretches or compresses in response to the force. We are all familiar with materials like rubber which stretch very easily.
In mechanics, the force applied per unit area is what is important, this is called the stress (symbol σ\sigma). The extent of the stretching/compression produced as the material responds to stress in called the strain (symbol ϵ\epsilon). Strain is measured by the ratio of the difference in length ΔL\Delta L to original length L0L_0 along the direction of the stress, i.e. ϵ=ΔL/L0\epsilon=\Delta L/L_0.
Every material responds differently to stress and the details of the response are important to engineers who must select materials for their structures and machines that behave predictably under expected stresses.
For most materials, the strain experienced when a small stress is applied depends on the tightness of the chemical bonds within the material. The stiffness of the material is directly related to the chemical structure of the material and the type of chemical bonds present. What happens when the stress is removed depends on how far the atoms have been moved. There are broadly two types of deformation:
  1. Elastic deformation. When the stress is removed the material returns to the dimension it had before the load was applied. The deformation is reversible, non-permanent.
  2. Plastic deformation. This occurs when a large stress is applied to a material. The stress is so large that when removed, the material does not spring back to its previous dimension. There is a permanent, irreversible deformation. The minimal value of the stress which produces plastic deformation is known as the elastic limit for the material.
Any spring should be designed and specified such that it only ever experiences elastic deformation when built into a machine under normal operation.

Hooke's law

When studying springs and elasticity, the 17ᵗʰ century physicist Robert Hooke noticed that the stress vs strain curve for many materials has a linear region. Within certain limits, the force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring. This is known as Hooke's law and commonly written:
Where FF is the force, xx is the length of extension/compression and kk is a constant of proportionality known as the spring constant which is usually given in N/m\mathrm{N/m}.
Though we have not explicitly established the direction of the force here, the negative sign is customarily added. This is to signify that the restoring force due to the spring is in the opposite direction to the force which caused the displacement. Pulling down on a spring will cause an extension of the spring downward, which will in turn result in an upward force due to the spring.
It is always important to make sure that the direction of the restoring force is specified consistently when approaching mechanics problems involving elasticity. For simple problems we can often interpret the extension xx as a 1-dimensional vector; in this case the resulting force will also be a 1-dimensional vector and the negative sign in Hooke’s law will give the correct direction of the force.
When calculating xx, it is important to remember that the spring itself will also have some nominal length L0L_0. The total length LL of a spring under extension is equal to the nominal length plus the extension, L=L0+xL=L_0 + x. For a spring under compression, it would be L=L0xL=L_0-x.
Exercise 1: A 75 kg person stands on a compression spring with spring constant 5000 N/m5000~\mathrm{N/m} and nominal length 0.25 m0.25~\mathrm{m}. What is the total length of the loaded spring?
Using Hooke's law we find the extension,
x=Fk=mgk=(75 kg)(9.81 m/s2)5000 N/m0.15 m\begin{aligned} x &= \frac{F}{k} \\ &= \frac{mg}{k} \\ &= \frac{(75~\mathrm{kg})\cdot(9.81~\mathrm{m/s^2})}{5000~\mathrm{N/m}}\\&\simeq 0.15~\mathrm{m} \end{aligned}
We now subtract this from the nominal length of the spring:
L=L0x=0.250.15 m=0.1 m\begin{aligned} L &= L_0-x \\ &= 0.25-0.15~\mathrm{m}\\&= 0.1~\mathrm{m}\end{aligned}
Exercise 2a: You are designing a mount for moving a 1 kg camera smoothly over a vertical distance of 50 mm. The design calls for the camera to slide on a pair of rails, with a spring supporting the camera and pulling it up against the tip of an adjustment screw as shown in Figure 1. The nominal length of the spring is L0=50 mmL_0=50~\mathrm{mm}. What is the minimum spring constant required for this design?
Why not connect the camera directly to the screw? While this would work, it would not result in a mechanism with a smooth and repeatable movement as the screw is adjusted. This is because there is backlash in any mechanism where a screw turns in a nut or threaded section. This is due to the finite spacing of the threads. A common engineering solution to the problem of backlash is to use a screw pushing against a spring, as in this example.
Figure 1: Camera height adjustment mechanism (exercise 2).
Figure 1: Camera height adjustment mechanism (exercise 2).
The spring must be elastic enough to supply enough force to pull the camera against the screw tip at all times. The force will be weakest when the spring is at its minimum extension, i.e. when the distance between the top and bottom of the spring is 100 mm.
Because the spring is specified to have nominal length of 50 mm, the spring will have a minimum extension x=100 mm50 mm=50 mmx=100~\mathrm{mm}-50~\mathrm{mm} = 50~\mathrm{mm}. The spring force must oppose the force due to gravity on the camera of mg=(1 kg)(9.81 m/s2)=9.81 Nmg = (1~\mathrm{kg})\cdot(9.81~\mathrm{m/s^2}) = 9.81~\mathrm{N}.
Using Hooke's law we find the required spring constant:
k=Fx=9.81 N50103 m196 N/m\begin{aligned} k &= \frac{F}{x} \\ &= \frac{9.81~\mathrm{N}}{50\cdot 10^{-3}~\mathrm{m}} \\ &\simeq 196~\mathrm{N/m}\end{aligned}
Exercise 2b: What is the minimum elastic limit required by your spring?
The spring must be robust enough to not break or exceed its elastic limit when the force on it is at the maximum permitted by the design. The spring force is maximized when the extension of the spring is maximized.
We know that the maximum extension is x=150 mm50 mm=100 mmx=150~\mathrm{mm}-50~\mathrm{mm} = 100~\mathrm{mm} and we assume we have selected a spring with spring constant of 196 N/m196~\mathrm{N/m}.
We can use Hooke's law to find the maximum extension force. This corresponds to the minimum elastic limit that we need our spring to have.
F=kx=(196 N/m)(100103 m)=19.6 N\begin{aligned} F &= kx \\ &= (196~\mathrm{N/m}) \cdot (100\cdot 10^{-3}~\mathrm{m}) \\ &= 19.6~\mathrm{N}\end{aligned}

Young's modulus and combining springs

Young's modulus (also known as the elastic modulus) is a number that measures the resistance of a material to being elastically deformed. It is named after the 17ᵗʰ century physicist Thomas Young. The stiffer a material, the higher its Young's modulus.
Young's modulus is usually given the symbol EE, and is defined as:
E=σϵ=StressStrainE = \frac{\sigma}{\epsilon} = \frac{\mathrm{Stress}}{\mathrm{Strain}}
Young's modulus can be defined at any strain, but where Hooke's law is obeyed it is a constant. We can directly obtain the spring constant kk from the Young's modulus of the material, the area AA over which the force is applied (since stress depends on the area) and nominal length of the material LL.
This treatment holds reasonably well for a simple elastic material, for example a block of rubber. A metal coil spring is an example of a relatively complicated structure which exploits both axial and torsional deformation. In this case further analysis is required to determine the correct value for the spring constant based on the properties of the bulk metal. However, the general form of the relationship between the spring constant and the springs geometry remains the same.
k=EALk = E \frac{A}{L}
This is a very useful relationship to understand when thinking about the properties of combinations of springs. Consider the case of two similar ideal springs with spring constant kk which can be arranged to support a weight end-to-end (series) or jointly (parallel) as shown in Figure 2. What is the effective spring constant for the combination in each case?
Figure 2: Series and parallel combinations of two similar springs.
Figure 2: Series and parallel combinations of two similar springs.
In the series configuration, we can see that the combined spring is equivalent to one spring with double the length. The spring constant in this case must therefore be half that of an individual spring, keffective=k/2k_\mathrm{effective}=k/2.
In the parallel configuration, the length remains the same but the force is distributed over twice the area of material. This doubles the effective spring constant of the combination, keffective=2kk_\mathrm{effective}=2 k.
Does this relationship show up elsewhere in physics? It turns out that the behavior of spring constants for series and parallel combinations exactly mirrors the series and parallel combinations of capacitors in electrical circuits.
In series,
1keffective=1k1+1k2+\frac{1}{k_\mathrm{effective}} = \frac{1}{k_1} + \frac{1}{k_2} + \ldots.
In parallel,
keffective=k1+k2+k_\mathrm{effective} = k_1 + k_2 + \ldots.

Springs with mass

Consider the setup shown in Figure 3. A spring supports a 1 kg mass horizontally via a pulley (which can be assumed to be frictionless) and an identical spring supports the same mass vertically. Suppose the spring has mass of 50 g, spring constant k=200 N/m. What is the extension of the spring in each case?
Figure 3: Comparison of a spring used horizontally and vertically.
Figure 3: Comparison of a spring used horizontally and vertically.
In both cases, the force on the spring due to the mass has the same magnitude, mgmg. So we might first assume that the extension is the same in both cases. It turns out that for a real spring this is not true.
The complication here is that the spring itself has mass. In the vertical case, the force of gravity acts on the spring in the same direction as the force due to the mass. So the mass of the spring adds to that of the weight. The extended spring is supporting a total weight of 1.05 kg which causes an extension of
Although the mass of the spring msm_s is distributed along its length, this situation is equivalent to a massless spring with constant kk supporting a point mass of msm_s.
Imagine dividing a massless spring in half and putting a mass msm_s in the mid point. Here we have two springs in series, so the spring constant of each half-spring will be k/2k/2. The extension of this top spring will be mg/(k/2)mg/(k/2), or double the extension of the original spring. The bottom spring however is now unloaded and has no extension. The overall result is the same extension as a single spring supporting mass msm_s.
1.05 kg9.81 m/s2200 N/m=51.5 mm\frac{1.05~\mathrm{kg} \cdot 9.81~\mathrm{m/s^2}}{200~\mathrm{N/m}}=51.5~\mathrm{mm}
In the horizontal case, the pulley has changed the direction of the force. The force due to the 1 kg weight acting on the spring is now orthogonal to the force of gravity acting on the spring. So the extension of the spring is supporting only 1 kg. It therefore extends
1 kg9.81 m/s2200 N/m=49 mm\frac{1~\mathrm{kg} \cdot 9.81~\mathrm{m/s^2}}{200~\mathrm{N/m}}=49~\mathrm{mm}
This difference can be quite significant and if not taken into account it can lead to incorrect results in the laboratory. In physics teaching laboratories, we often use spring balances to measure force. A spring balance (Figure 4) is simply a spring with a pointer attached and a scale from which the force can be read.
Figure 4: A common spring balance.
Figure 4: A common spring balance.
Because the manufacturers of spring balances expect their product to be used vertically (for example, by a fisherman measuring the mass of his fish) the scale is calibrated to take into account the mass of the spring and hook. It will give an incorrect absolute result if used to measure a horizontal force. However, Hooke's law tells us that there is a linear relationship between force and extension. Because of this we can still rely on the scale for relative measurements when used horizontally. Some spring balances have an adjustment screw which allows the zero point to be calibrated, eliminating this problem.
What do we mean by absolute and relative? An absolute measurement is a measurement of a quantity that has a defined reference point where the quantity is zero. A measurement of length by a ruler with a printed scale is an example of an absolute measurement. A ruler without printed numbers (only tick marks) would only be able to make a relative measurement.
To convert a relative measurement to an absolute one we need to choose our own reference point. In the case of a horizontal spring balance we would probably choose to establish the reference point by finding the force measured when the spring is unloaded. We would then subtract this from the value shown on the scale of the balance when making a measurement.