Peter's Physics Pages

An introduction to Electricity and Strength of Materials with Peter Eyland

Lecture 10 (Elasticity)

In this lecture the following are introduced:
•Normal Stress, Shear Stress and Bulk Stress
•Normal Strain, Shear Strain and Bulk Strain
•Young's Modulus, Shear/Rigidity Modulus and Bulk Modulus
•Poisson's ratio
•Young’s Modulus and the Atomic Spring Constant

Elastic Solids

An external force on a solid will deform it. An elastic solid will return to its original size and shape when the force is removed. Elastic responses only happen with small forces. If the force is large enough the solid will be permamently deformed. The elastic limit is the largest applied force that gives elastic behaviour. To deal with the various shapes and sizes of solids and the angles of the forces on them, some definitions are needed.

Stress

Stress is generally, a force applied over an area of a solid. It is often more useful than force because it gives an idea of the force on a single bond. The symbol for stress is often the Greek letter "sigma". i.e. σ. Stress has the units of force over area i.e. N/m2 which is given the special name Pascal.

There are three main types of stress, defined by the way the force interacts with the area over which it is applied.

 Normal stress  Shear stress  Bulk stress  Strain

Strain is the result of stress and is generally a change in a dimension over an original dimension. This makes it independent of the size of the solid. The symbol for strain is often the Greek letter "epsilon". i.e. ε.

As there are three stresses, so there are three strains.

 Normal strain  Since strain is m/m it is dimensionless. Shear strain Note 1: the volume of the solid is not changed by shear strain. Note 2: the angle is is radians, not degrees. Bulk strain  Elastic Modulus

Stress produces a strain, but how much strain is produced depends on the solid itself. The solid is then characterised by an elastic modulus that relates strain to stress. There are three elastic moduli, related to the three stresses and strains.

 Young's modulus  Rigidity modulus  Bulk modulus  Elastic Moduli of some common materials.

 Material Y GPa G GPa K GPa Poisson’s ratio* aluminium, Al 70 24 75 0.34 copper, Cu 123 45 131 0.34 gold, Au 80 28 166 0.42 lead, Pb 15 5.4 50 0.45 steel, Fe(C) 206 89 181 0.33 iron, Fe 206 brass (70% Cu, 30% Zn) 110 quartz, SiO2 310 30 0.37 glass (silicate) 70 30 50 0.25 alumina, Al2O3 355 MgO 207 SiC 414 vulcanised rubber 3.4 1.2 5.7 0.4 polystyrene 3.8 1.0 4.7 0.4 polyethylene 0.14 0.05 0.35 0.4

Young’s Modulus and the Interatomic Spring Constant

To find the connection between the macroscopic stress, strain, modulus and the microscopic picture, we need to return to the interatomic bond.
 A typical interatomic potential and force function is: The force graph has the form. The important thing to notice is that very close to r0, the curve looks like a straight line. We can find the slope of this straight line from: This says that very close to r0, there is a restoring force that is proportional to the distance from r0, and the constant can be found from the interatomic potential.

Hooke's law for springs says the same thing, so near r0, interatomic forces can be modelled as little springs, with k as the interatomic spring constant.

Given that the interatomic force can be modelled as as a spring, the interatomic spring constant, k, can be related to Young's modulus. Define dF as the force between two atoms. It will also be the force along a column of atoms, as shown in the diagram. From the definitions of stress and strain Using these relationships This result says that the interatomic spring constant equals Young's modulus times the equilibrium spacing (k=Er0). It is important because it connects the macroscopic modulus with a microscopic "spring" from the interatomic potential.

From the typical interatomic potential above: This shows that Young's Modulus is directly proportional to the strength of the attractive potential and varies in some inverse way with the interatomic spacing.

*Poisson's ratio

The three stresses and strains do not operate independently. Stresses produce strains in lateral directions as the solid tries to retain its original volume. Poisson's ratio is a measure of how successful this is.  The minus sign gives a positive value to the material parameter since positive longitudinal strains usually produce negative lateral strain. As seen from the table, Poisson's ratio is usually about 0.3. It can be shown that a Poisson's ratio of 0.5 is needed for the volume to stay the same as it was originally.

Note:
Engineering Stress doesn't take the reduction in cross-sectional area into account.
True stress includes this reduction in area.

Example S2

A glass cube (Poisson's ratio 0.25 and sides 250mm) has a 500N normal force applied outwards across opposite faces producing a extension of 30nm between those faces. Find
(i) the normal stress in the cube,
(ii) the normal strain in the cube
(iii) Young's modulus for the glass, and
(iv) the lateral displacement of the sides of the glass block.

Answer S2

 Normal stress Normal strain Young's modulus Poisson's ratio Example S3

Copper has a rigidity modulus 45 GPa. A cube of copper with 500mm sides is subjected to a shear force of 2250 N. Find
(a) the shear stress on the cube,
(b) the shear strain on the cube,
(c) the shear displacement of the upper face, and
(d) the angle (in degrees) the side makes with the normal.

Answer S3 Example S4

A solid sphere is immersed in a liquid so that the hydrostatic pressure is 6.15 GPa. This pressure causes a contraction of 1% in the diameter of the sphere. Find the bulk modulus.

Answer S4 From the table, this material could be iron or steel.

Example S5

A cube of isotropic material experiences a uniform hydrostatic pressure, P. Find the resultant strain at right angles to any cube face, in terms of the pressure, Young's modulus (E) and Poisson's ratio (ν).

Answer S5

The total strain the x direction is due to stress in the x, y and z directions.

 Take the x direction first: Define σx as the stress applied in the x direction, and εxx as the strain in the x direction due to the x stress. Now consider the y stress: Define σy as the stress applied in the y direction, and εyy as the strain in the y direction due to the y stress. Taking lateral directions: Define εxy as the strain in the x direction due to the y stress. εxy can be found from εyy through Poisson's ratio Starting with Young's modulus in the y direction, the y strain can be replaced with the corresponding x strain through Poisson's ratio. Since there is no difference in principle between the z direction and the y direction, εxz the strain in the x direction due to the z stress is: The total strain in the x direction, due to stresses in the x, y and z directions is: Example S6
Starting from a uniform hydrostatic pressure on a cube of side :
(a) find the new volume,
(b) find the fractional change in volume, and
(c) show that from the bulk modulus definition: .

Answer S6

For a uniform hydrostatic pressure on a cube of side .

 From example S5, there is a strain in each direction given by: The new length of each side is given from the definition of strain. The new lengths are each The new volume is: The fractional change in volume is: From the definition of the bulk modulus: Example S7 (Poisson's ratio example with a single Tensile stress)
Find the value of Poisson's ratio, for no volume change to occur under a small tensile stress in the x direction.

Answer S7
Take the example of a cube of side , under a tensile stress σx:
 The new x length is: Poisson's ratio gives the new y & z lengths: The new volume is: For no change in volume: (1 - 2ν) = 0, i.e. ν ⇒ 0.5

Example S8 (Isotropic solid with mixed tensile and compressive loads)
A steel bar is 100 mm long, and has a 20 mm by 10 mm cross-sectional area. It has Young's modulus of 207 GPa, Poisson's ratio of 0.26 and subjected to

• a tensile load of 104 N along its length,
• a compressive load of 105 N on its cross-sectional area, and
• a tensile load of 105 N on its 100 mm 10 mm faces.

(a) Find the change in volume of the bar.
(b) Find what the compressive load has to be raised to, so that there is no change from the original volume.

Answer S8
(a) For the volume change.

 The volume change comes from the definition of Bulk Modulus. The total Bulk Stress is needed and comes from the stresses in the three directions. The Bulk Modulus is needed and comes from Young's modulus and Poisson's ratio. Combining these results gives the ratio: The volume change is: (b) For no change in volume.

 The total stress must equal zero The compressive stress and force for this is: Summarising:

An elastic solid will return to its original size and shape when the deforming force is removed.
Normal Stress is Normal Force/Area,
Shear Stress is Tangential Force/Area, and
Bulk Stress is the Pressure.
Normal Strain is Extension (force direction)/Length (force direction),
Shear Strain is Extension (force direction)/Length (at right angles to force direction), and
Bulk Stress is the Change in Volume/Unstressed Volume.
An elastic modulus shows how much strain results from the stress.
Young's Modulus is Normal Stress/Normal Strain,
Shear Modulus is Shear Stress/Shear Strain, and
Bulk Modulus is the Pressure/Bulk Strain.
The restoring force between atoms can be represented as springs if the applied force is small.
The interatomic spring constant equals Young's modulus times the equilibrium spacing. email me a note if you found this useful
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