Eight people step into a lift on the ground floor of a tall building. The car hangs from a steel cable 30 m long. As the people step in, the floor of the car sinks by a few millimetres, though nobody notices. When they step out at the top, the car rises back by the same amount. The steel cable looks perfectly rigid, yet it has stretched and returned.
Push, and the solid pushes back
In the chapters on Dynamics, strings did not stretch and blocks did not squash. That was a useful simplification. A real solid always gives a little when a force acts on it. A force that changes the size or the shape of a body is called a deforming force. Squeeze a rubber ball, pull a steel wire or bend a ruler, and in each case the body changes.
The body does not simply give way. As it deforms, it develops an internal force that opposes the change. This is the restoring force. Pull a wire, and the wire pulls back on your hand. For a body at rest under a load, the restoring force grows until it balances the deforming force.
Elastic, plastic and rigid
Remove the deforming force, and bodies behave in two ways. A rubber band, a steel spring or a steel wire under a small load returns to its original size and shape. This property is elasticity, and such a body is elastic. A lump of wet clay or a ball of dough stays flattened. This is plasticity, and such a body is plastic.
Elasticity is the property of a body by which it returns to its original size and shape when the deforming force is removed. Plasticity is the property of keeping the deformed shape.
No real body is perfectly elastic or perfectly plastic. Steel is close to perfectly elastic under small loads, and putty is close to perfectly plastic. A rigid body, which does not deform at all, is an idealisation. In Rotational Mechanics that idealisation was useful. In this chapter the small deformation itself is the subject.
A steel wire and a strip of chewing gum are both stretched a little and then released. The wire returns to its length and the gum stays longer. Which statement is right?
A solid is atoms held by springs
Why does a solid push back? A solid is made of atoms held at definite distances by the forces between them. Pull two neighbouring atoms apart, and the force between them becomes an attraction that pulls them together. Push them closer, and it becomes a repulsion. So each bond between two atoms behaves like a small spring.
Watch Fig 1.1 closely. The same force acts through every bond of a row, so every bond stretches by the same amount. The row of 6 bonds therefore stretches twice as far as the row of 3. The stretch per bond, or the fraction \(\Delta L/L\), is the same for both rows. When the pull eases off, every bond returns to its natural length, and the whole row comes back.
Think of a solid as a mattress of balls joined by springs in all three directions. Press it anywhere, and the springs near your hand squeeze and push back. Take your hand away, and the springs return every ball to its place. That is elasticity, seen from inside.
Stress: the restoring force per area
Hang a load of weight \(F\) from a wire, and imagine the wire cut across at any point. The part below the cut is at rest, so the part above must pull it up with a force \(F\). This internal force is spread over the cross-section of area \(A\). A thick wire shares the same force among more atoms, so each bond carries less.
So the measure of how hard the material is working is the force per area. The restoring force per unit area of a cross-section is called stress. At rest it equals the deforming force divided by the area, \(F/A\). Its SI unit is the newton per square metre, called the pascal (Pa). A megapascal is \(1\,\text{MPa} = 10^6\,\text{Pa}\).
Stress is the internal restoring force per unit area of a cross-section of a deformed body. For a body at rest under a force \(F\) spread over an area \(A\), the stress is \(F/A\).
Three kinds of stress
The force on a face can point along the normal to the face or along the face itself. A normal force that pulls produces tensile stress, as in a lift cable. A normal force that pushes produces compressive stress, as in the pillar of a bridge. A force along the face, a tangential force, produces shearing stress, as in a rivet that holds two plates sliding past each other.
Look at the shape of the block in each chip of Fig 1.2. Under tensile stress the block grows longer and a little thinner. Under compressive stress it grows shorter and a little fatter. Under shearing stress its length and height stay the same, but the top face slides over the bottom face, so the square becomes slanted.
Stress is not the pressure you apply from outside. It is the force that the material develops inside, per area of a cut. At rest the two have the same size, which is why \(F/A\) gives the stress. Also convert the area to square metres before dividing: \(1\,\text{mm}^2 = 10^{-6}\,\text{m}^2\), not \(10^{-3}\,\text{m}^2\).
(a) A ceiling fan of mass \(6\,\text{kg}\) hangs from a solid steel rod of diameter \(1\,\text{cm}\). Take \(g = 10\,\text{m/s}^2\). Find the stress in the rod. (b) A pillar of a Metro viaduct has a cross-section of \(2\,\text{m}^2\) and carries a load of \(4 \times 10^6\,\text{N}\). Find the stress in it, and name its kind.
(a) The rod carries the weight \(F = 60\,\text{N}\). Its area is \(A = \pi r^2 = \pi \times (0.005)^2 = 7.85 \times 10^{-5}\,\text{m}^2\). The stress is \(60/(7.85 \times 10^{-5}) = 7.6 \times 10^5\,\text{Pa}\), a tensile stress.
(b) The stress is \(4 \times 10^6/2 = 2 \times 10^6\,\text{Pa}\), or \(2\,\text{MPa}\). The load pushes down on the pillar, so the stress is compressive.
A load of \(20\,\text{kg}\) hangs from a wire of cross-section \(2\,\text{mm}^2\). Take \(g = 10\,\text{m/s}^2\). What is the tensile stress in the wire, in MPa?
Two wooden planks overlap and are glued face to face. They are pulled in opposite directions along their length. What kind of stress acts in the layer of glue?
Strain: how much, as a fraction
Fig 1.1 showed why a long wire stretches more than a short one under the same load. Every bond stretches by the same amount, so the stretch grows with the number of bonds, that is, with the length. The honest measure of the deformation is therefore the change divided by the original size. This fraction is called strain.
There is one strain for each kind of stress. A change of length \(\Delta L\) in a length \(L\) gives the longitudinal strain \(\Delta L/L\). A change of volume \(\Delta V\) in a volume \(V\) gives the volume strain \(\Delta V/V\). A shear that moves the top face by \(x\) over a height \(L\) gives the shear strain \(\theta = x/L\). It is the small angle through which the side faces turn.
A strain is a length divided by a length, or a volume divided by a volume. The units cancel, so strain is a pure number with no unit. It is often written as a percentage. A steel wire in everyday use rarely reaches a strain of \(0.1\) percent.
Find the strain in each of three cases, (a) to (c). A wire \(2\,\text{m}\) long stretches by \(1\,\text{mm}\). One litre of water is squeezed to \(995\,\text{mL}\). The top face of a rubber block \(10\,\text{cm}\) high shifts sideways by \(1\,\text{mm}\).
(a) \(\Delta L/L = 10^{-3}/2 = 5 \times 10^{-4}\), or \(0.05\) percent.
(b) \(\Delta V/V = (995 - 1000)/1000 = -5 \times 10^{-3}\). The minus sign says the volume decreased.
(c) \(\theta = x/L = 0.1\,\text{cm}/10\,\text{cm} = 0.01\,\text{rad}\).
What is the SI unit of strain?
A wire \(3\,\text{m}\) long stretches by \(1.5\,\text{mm}\) under a load. What is its longitudinal strain, as a percentage?
A person of weight \(600\,\text{N}\) stands on two legs. Each thigh bone has a cross-section of about \(3\,\text{cm}^2\) of solid bone. Estimate the compressive stress in each thigh bone, in pascals.
Now return to the lift. The car of \(1000\,\text{kg}\) and eight people of \(60\,\text{kg}\) weigh \(14\,800\,\text{N}\). The cable has a metal cross-section of \(2\,\text{cm}^2\), so the stress in it is \(14\,800/(2 \times 10^{-4}) = 7.4 \times 10^7\,\text{Pa}\). When the people step in, the stress rises by \(2.4 \times 10^7\,\text{Pa}\). Every bond in the steel must stretch to supply that extra restoring force, so the cable must lengthen.
In the lesson on Young's modulus you will find how much: about \(3.6\,\text{mm}\) on a \(30\,\text{m}\) cable. That is the few millimetres by which the car sinks.
Concrete bears compressive stress very well but cracks under a small tensile stress. The beams and pillars of a Metro viaduct therefore carry steel bars inside the concrete. Where a beam bends, its lower side is stretched. The steel bars there take the tensile stress, and the concrete above them takes the compressive stress. Each material does the job it is good at.
- Using an area in mm² or cm² directly in \(F/A\).Convert first: \(1\,\text{mm}^2 = 10^{-6}\,\text{m}^2\) and \(1\,\text{cm}^2 = 10^{-4}\,\text{m}^2\).
- Giving strain a unit, such as metres.Strain is a change divided by the original size, a pure number.
- Calling a body "more elastic" because it stretches easily.Elasticity is the power to return. A rubber band stretches easily; steel returns more exactly.
- Mixing up shearing stress with tensile stress.Ask where the force points relative to the face: along the normal is tensile or compressive, along the face is shearing.