Forces change the state of motion of an object (post 16.13). But, in post 16.50 on springs, we saw that forces could also change the dimensions of objects. A spring can be stretched by a force (a *tensile force*) or compressed by a force (a *compressive force*). This effect is not restricted to springs – a force can change the dimensions of any object, although for a very stiff object this change may be difficult to detect. Forces can also change the shape of an object, for example when a pot is made from clay by pushing and pulling it into the desired shape.

Change in dimensions and shape of an object are called *deformations*; deformations that involve only a change in dimensions are the subject of this post.

Let’s think about pulling a long, thin wire. The wire gets longer (and thinner) but the change in shape is negligible. So, the force acting on the wire can be considered to stretch it without changing its shape – it is (almost) subject to pure *tension*.

We can define the stiffness of the wire in exactly the same way as we define the stiffness of a spring; if a force, *F*, changes the length, *L*, of a wire by Δ*L*, its *stiffness*, *k*, is given by

*k* = *F*/Δ*L*

(post 16.50). For tension Δ*L* is defined to be positive; for compression, it is defined to be negative.

Now suppose we wanted to compare the behaviour of a steel wire and a copper wire. We could compare stiffness of the two wires by comparing values of *k*. But if we wanted to compare the stiffness of the materials steel and copper, we would have a problem, unless the two wires had exactly the same dimensions – because a thin wire will stretch more easily than a thick wire.

To compare the stiffness of materials we define the concepts of *stress* and *strain*. Be careful – in everyday speech “stress” and “strain” mean (almost) the same thing – in physics and engineering they don’t! (see post 17.14).

If our wire has a cross-sectional area *A*, stress is defined by

*σ* = *F*/*A*.

If the stress is tensile, it is defined to be positive; for a compressive stress, it is negative. But we need to be careful; when the wire stretches, it gets thinner – so *A* decreases. If we use the original value of *A* to calculate *σ*, the result is called *engineering stress*; if we use the true value of *A* at each stage of extension, the result is called *true stress*. In practice, the change in *A* is usually so small that engineering stress and true stress are almost the same; we usually measure engineering stress because it is much easier to measure *A* before the wire stretches. Note that stress is a force divided by an area so is measured in N.m^{-2}; this unit has a special name – the pascal (abbreviated to Pa), the unit of pressure (all this is explained in post 17.5). But, although strain and pressure are measured in the same units, they are not the same, because pressure is a scalar (see below).

Also, if we have two wires, made of the same material and with identical cross-sectional areas, it is easier to stretch a wire whose length is 1.0 m by 0.1 mm than it is to stretch a wire of length 10 cm by the same amount. So, if we are comparing properties of materials, like steel and copper, we need to allow for the effect of the length of the wires. We do this by calculating the fractional increase in length, called *strain*, by

*ε* = Δ*L*/*L*.

This equation defines what is called *engineering strain*; this is the definition of strain that is most often used; *true strain* is defined in the appendix. A tensile strain is defined to be positive; a compressive strain is defined to be negative. Because stress is defined to be a length divided by another length, it is a number with no units – it is dimensionless (post 17.41). But, because strain values are so small, we sometimes talk about *percentage strain* – the strain value multiplied by 100.

We define the stiffness of a material by

*E *= *σ*/*ε*.

*E* is usually called *Young’s modulus*, after the British scientist Thomas Young (1773-1829). There is an assumption hidden in this simple definition of *E* (called Hooke’s law) that we will think about in the next post. Since *E* is stress (measured in Pa) divided by strain (no units), it is measured in the same units as strain (Pa).

The table above shows some approximate values of *E* for different materials. The prefix G in the units used here (GPa) is explained in post 16.12; the way in which numbers, like 2 × 10^{2} are written, is explained in post 16.7.

How can we use this table? Here is an example. Suppose we have a vertical bar of steel, length 1 m, with a cross-section that is a square of side 1 mm. A mass of 100 kg hangs from its end. How much does the bar stretch? The gravitational force acting on the mass is 1000 N (post 16.16) and the cross-sectional area is 10^{-3} × 10^{-3} = 10^{-6} m^{2}. So

*σ* = 1000/10^{-6} = 10^{9} Pa.

From the table

*E* = 2 × 10^{2} GPa = 2 × 10^{2} × 10^{9} = 2 × 10^{11} Pa.

Δ*L* = *εL* = (*σ*/*E*)*L* = 10^{9}/(2 × 10^{11}) × 1 = ½ × 10^{-2} m = 5 mm.

Now let’s look at the relationship between *k*, the stiffness of the wire, and *E*, the stiffness of the stuff it is made from. We have seen that

*E* = *σ*/*ε* = (*F*/*A*)/(Δ*L*/*L*) = (*F*/Δ*L*)/(*AL*) = *k*/*V*,

where *V* is the volume of the wire. So we calculate the stiffness of the material by dividing the stiffness of the wire by its volume.

So far we have ignored something important about force – it is a vector (post 16.50). But if force is a vector, is stress a vector (a force divided by an area) or a scalar (the magnitude of a force divided by an area)? (See post 17.2 for more information on vectors and scalars). It’s neither! A stress makes a wire longer but it also makes it thinner – it has an effect perpendicular to the direction in which it acts; a vector has no effect in a perpendicular direction (post 16.50). And because the wire gets thinner, as it gets longer, the strain is one direction is associated with a strain in a perpendicular direction. Stress and strain are examples of *tensors* – numbers that have direction as well as magnitude but whose components are not orthogonal (post 16.50). I hope to explain more about tensors in a later post.

Although strain and pressure are measured in the same units (pascals), they are not the same – strain is a tensor and pressure is a scalar (post 17.5).

In conclusion, an extension (or compression) produced by a force is used to define the stiffness of an object. The extension (or compression) produced by a stress is used to define the stiffness of a material (its Young’s modulus).

__Related posts__

16.49 Springs

16.37 Solids, liquids and gases

Follow-up posts

20.3 Hooke’s law

20.5 Poisson’s ratio

20.6 Elasticity

20.9 Stiffness and strength

20.11 Elasticity of rubber

20.13 Fracture

20.19 Deformation of objects – changes in shape

20.20 Deformation of objects – isometric compression

21.17 Beams

__Appendix__

The purpose of this appendix is to explain the idea of *true strain*.

Let’s suppose that wire is stretched, by an infinitesimal amount (post 17.4) *δL*. Then its strain is

*δε*’ = *δL*/*L*.

Integrating (post 17.19) this result, for a wire stretched from length *L* to *L* + Δ*L* gives

To find out how the integral was evaluated see post 18.15.

To find out about logarithms (like log_{e}*L*), and how I have manipulated them here, see post 18.3.

You might think that true stain, *ε*’, and engineering strain, *ε*, are very different. But, most of the time, we see strains whose values are small; a typical value might be 0.01 (a 1% strain). The table below shows that the values of *ε*’ and *ε* are then almost identical.