Page:Elementary Text-book of Physics (Anthony, 1897).djvu/129

§ 101] traction. It may be considered as made up of three equal tractions $$\frac{P}{3}$$. Apply to each of the four other faces of the cube two opposite stresses, each equal to $$\frac{P}{3}$$. Two of these pairs of stresses are represented in the figure. These stresses on the sides of the cube, being equal and opposite, are equivalent to no stress. It is evident that the combination of stresses here described is equivalent to a tension $$\frac{P}{3}$$ applied to each face of the cube, to a shearing stress $$\frac{P}{3}$$ acting in the plane of the figure, and to a shearing stress $$\frac{P}{3}$$ acting in the plane at right angles to the plane of the figure. Thus the longitudinal traction may be resolved into a tension uniform in all directions and two shearing stresses, all of the same numerical value.

The uniform tension just employed is an example of a hydrostatic stress. More generally, a hydrostatic stress is a stress which is normal to any surface element drawn in a body, whatever be its direction. The numerical value of a hydrostatic stress is the same in whatever direction the surface be drawn to which it is applied. To show this, we examine the relations of the pressures on the faces of the tetrahedron formed bypassing a plane through the points $$ABC$$ taken infinitely near the point (Fig. 37) on lines drawn through that point in the directions of the three coordinate axes. Let $$l, m, n$$ represent the direction cosines of the normal to the face $$ABC$$, and let a represent the area of this face; the areas of the other faces are respectively equal to $$al, am, an$$. Let $$X, P, Q, R$$ represent the pressures on the faces in the order mentioned: the forces acting on the faces are then $$Xa, Pal, Qam$$, and $$Ran$$. By the definition of hydrostatic