at the point. These are called the components of strain at (x, y, z). The first three are stretches parallel to the axes, the other three are shearing strains. We may get an idea of the nature of these strains from two simple typical cases. 1. Let u = ex, v = 0, w = 0. This makes du/dx = e, and the other five strains zero. But we see that the displacement of every particle is perpendicular to the yz plane, and proportional to its distance from that plane. Every line parallel to the axis of x is therefore elongated by a definite fraction of its original length, the value of this fraction being e, which is du/dx. The strains du/dx, dv/dy, dw/dz are therefore stretches parallel to the axes. 2. Let u = 0, v = cz, w = 0. This gives dv/dz + du/dy = c, and the other five strains zero. The displacement of every particle is parallel to the axis of y, and proportional to its distance from the plane xy. The strain is therefore a slide, or shear of planes parallel to xy in the direction of the axis of y.
Stresses.—To specify the stress round a point (x, y, z), consider a small plane area through the point. The material on one side of this acts on the material on the other side with a certain force whose components parallel to the axes are F, G, H, say, per unit area. If we know F, G, H for every orientation of the small plane area, the state of stress round (x, y, z) is defined. But it is easy to show, as below, that we can find F, G, H for every area if we know them for areas parallel to the co-ordinate planes. We are thus led to the specification of the stress round (x, y, z) by the six components of stress, xx, yy, zz, xy, xz, yz; where xy, for example, means the force per unit area parallel to Ox exerted on a plane perpendicular to Oy by the material on the positive side of that plane on the material on its negative side. Thus e.g. the components of the force per unit area exerted across the yz plane through (x, y, z) by the
material on the positive side of that plane are xx, xy, xz. It is important to note that xy = yx. This is easily proved by considering the equilibrium of a small rectangular volume of the material round (x, y, z) as centre, with its edges parallel to the axes; if xy were not equal to yx, there would be a residual couple in the plane xy.
Relations between the Strains and the Stresses.—When the strains are known, the stresses can be found from a generalized Hooke's law, which can be deduced from the principle of energy, combined with consideration of symmetry. If the solid is isotropic, i.e. is symmetrical in its elastic properties in all directions round a point, the relations between stress and strain are of the form
| xx = λ | ( | du | + | dv | + | dw | ) | + 2μ | du | , | ||
| dx | dy | dz | dx |
| yz = μ | ( | dw | + | dv | ) | ; | ||
| dy | dz |
the values of the other four components of strain can be written down from symmetry. Here λ and μ are constants, each being a modulus of elasticity. In particular μ is the shape modulus or the rigidity, already referred to. The Young's modulus and the bulk modulus can easily be found in terms of λ and μ.
Equations of Equilibrium in Terms of the Stresses.—A rectangular element dx, dy, dz, of the body is held in equilibrium by the body force, and the tractions on its faces arising from the stress. The tractions per unit area, parallel to the axis of x, on the six faces, are: on the plane x, -xx; on the plane x + dx, xx + (d/dx xx)dx; on the plane y, -xy; on the plane y + dy, xy + (d/dy xy)dy; and similarly for the plane z. Let the force acting on the mass of the body, such as its weight, be (X, Y, Z) per unit mass, and let ρ be the density. By equating the sum of the x components of all the forces to zero, we get
| dxx | + | dxy | + | dxz | + ρX = 0; | |
| dx | dy | dz |
| Similarly | dxy | + | dyy | + | dyz | + ρY = 0, |
| dx | dy | dz |