The basic elastic constants

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Problem 2.1a

Show that when the only nonzero applied stress is , Hooke’s law requires that the normal strains , and that Poisson’s ratio , defined as , satisfies the equation


(2.1a)

Background

Stress is force/unit area and is denoted by , etc., where a force in the -direction acts upon a surface perpendicular to the -axis. The stresses and are, respectively, a normal stress and a shearing stress.

Stresses produce strains (changes in size and/or shape). If the stresses cause a point to have displacements along the coordinate axes, the basic strains are given by derivatives of these displacements as follows:


(2.1b)


(2.1c)

The vector displacement is


(2.1d)

where are unit vectors in the -directions (see Sheriff and Geldart, 1995, problem 15.3). The dilatation is the change in volume per unit volume, i.e.,


(2.1e)

A pressure produces a decrease in volume, the proportionality constant being the bulk modulus :


(2.1f)

Sometimes the compressibility, , is used instead of .

In addition to creating strains, stresses cause rotation of the medium, the vector rotation being equal to


(2.1g)

where , ,

For small strains and an isotropic medium (where properties are the same regard-less of the direction of measurement), Hooke’s law relates the stresses to the strains as follows:


(2.1h)


(2.1i)

where and are Lamé’s constants ( is usually called the modulus of rigidity or the shear modulus).

Solution

Subtracting equation (2.1h) for from the same equation for gives .

Dividing equation (2.1h) for by gives

so

Problem 2.1b

2.1b Show that Young’s modulus , defined as , is given by the equation


(2.1j)

Solution

Adding the three equations (2.1h) for , , and , and recalling that , we get

Dividing both sides by gives

Using equation (2.1a) we get

Problem 2.1c

A pressure is equivalent to stresses . Derive the following result for the bulk modulus :


(2.1k)

Solution

Since , we add equation (2.1h) for each of the three values , obtaining , so from equation (2.1f),

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