Consider a bar of uniform cross section under longitudinal tension as in Figure 1. Assume an imaginary break a-a normal to the longitudinal axis of the bar. For equilibrium to be maintained, there must be internal forces between the two adjoining parts of the bar with a resultant equal to the externally applied force. In the simplest case, when the external force is uniformly distributed over the end, the internal forces are also uniformly distributed over any cross section a-a. The stress is defined as the intensity of the internal force, i.e., the total tensile force P divided by the cross sectional-area A.
In the most general case, consider a solid under a systems of external forces Pi as in Figure 2. On a plane section a-a, defined by its normal direction r and area A, the resultant of the internal forces or stress resultant, R, is a vector which can be regarded as the sum of a force N normal to the surface and another force along the surface, S. The normal stress resultant is defined as the ratio between N and A. The shear stress is the ratio S/A. The usual symbol for normal stress is σ and for the shear stress it is τ. It should be emphasized that a stress is not a vector but the intensity of internal force acting on a surface with a given orientation. This can be best understood by isolating from the solid an elemental cube of unit side oriented along the x,y,z coordinate axes, Figure 3. On the face ABCD, normal to the x axis, the normal stress is σx. The shear stress can be resolved along the y and z directions as τxy and τxz. Similarly, stresses σy, τyx , τyz and σz, τzx, τzy are defined. The stresses acting on faces such as A'B'C'D' may be deduced from these by considering the equilibrium of the element. It can be shown that
and that the state of stress at a point O which may be regarded as in the center of the cube is characterized by the six stress components σx σy σz τxy τxz τyz. These may be grouped in the form of a tensor symmetrical about the principal diagonal,