| and | dxz | + | dyz | + | dzz | + ρZ = 0. |
| dx | dy | dz |
The Surface Tractions in Terms of the Stresses.—Draw a small tetrahedron round (x, y, z) with its faces perpendicular to the axes and to the direction (l, m, n), and consider the equilibrium of this small body. Let F, G, H be the components per unit area of the force on the plane whose direction cosines, drawn outwards, are l, m, n. If A be the area of the face perpendicular to (l, m, n), we get, by resolving parallel to Ox,
F·A = xx·lA + xy·mA + xz·nA.
Hence F = lxx + mxy + nxz,
and similarly G = lxy + myy + nyz,
H = lxz + myz + nzz.
If (x, y, z) is a point at the surface of the body, and l, m, n are the direction cosines of the outward normal at that point, these values of F, G, H are the component of the force that must be applied from outside to the surface at (x, y, z) to maintain the state of stress.
The Equations connecting the Displacements and the Applied Forces.—By using, in the equations of equilibrium, the values of the stresses in terms of the strains, we find the body equations of equilibrium in terms of displacements,
| μ | ( | d2u | + | d2u | + | d2u | ) | + (λ + μ) | d | ( | du | + | dv | + | dw | ) | + Xρ = 0, | |||||
| dx2 | dy2 | dz2 | dx | dx | dy | dz |
with two similar equations.