Consider the effect of a partial strain all ay other displacements being zero. There will be no loss in generality if we take a square stead of a rectangle in Fig. Z because the strain components are independent of the mag A stress will be denoted by a capi tal letter to indicate its direction, with a sub script to show the normal to plane on which it acts. Thus Xx is parallel to X on a plane (YZ) normal to X and is a normal stress; is a shear parallel to Y on the XY plane. Fig. 3 shows an element under coplanar stress, nitudes of dx, dy, dz. From Fig. 2 dy= tan (45de) ay au Now dy is small compared with dx because ay u is by assumption small compared with x; then as dy= dx, the equation reduces to d8=4-- au In the same way 47 is the counterclockwise rotation of the diagonal due to shearing of the right side of the element. The resultant tive (X toward Y) rotation about the Z-axis is ax atv av likewise 'ix= (3) (au aw - as ax ox and being obtained by cyclic permuta tion. These are the component rotations; when they vanish the strain is irrotational or pure. There are always at least three orthogo nal lines whose directions remain unaltered by strain; they are called the principal axes and the planes normal to them the' principal planes. If u v, w are eliminated from equa tions (1) and (2) there will be three equations of the form ay'axay azay (4) and three of the form 2 me.. a ce., 84, &LA (4)) = ayas ax as ax ay/ the others being written by permuting x, y, z.
These are the equations of compatibility and must be satisfied by every solution of a prob lem in elasticity. Many of the formulas de rived in the strength of materials are not com patible with theory although they may be rea sonably in accord with experiment.
all stresses parallel to Z being zero. It will be seen from the theorem about to be derived that there can be no shear on planes parallel to the paper if there is none normal to the paper. Taking moments of the forces about the upper right-hand corner (edge) we find, after rejecting terms which vanish in the limit, Xy= Yx; it is to be observed that the weight and the moment of inertia of the element are vanish ingly small. The diagram shows that the
shears on two orthogonal faces both point away from, or both toward the edge, hence the Theorem: Shear stress on any plane is accompanied by equal shear stress on a per. pendicular plane, both acting away from or to ward the edge of intersection and both being normal to it.
This is due to Cauchy. For the other shears Z2Xs The translations of the element in Fig. 1 are u, v, w, whence the axial accelerations are die derivatives are partial be ats ' at'ap cause they must denote only time-changes and not space-changes. From the general free body of which Fig. 3 is a special case we get by resolving the forces axially ax a x =___ a ax as ais a Yx a a v. alp . . . . (5) + + + a Y = a ax az ap az. , az, az. , 1- Ts -I- where a is the density and X, Y, Z are the components of the applied forces (e.g. weight) per unit mass.
At the surface the internal stresses Xy, , .... must be in equilibrium with the external or applied stresses.
Relations between stress and strain. A ma terial is elastically isotropic when it resists stress with equal intensity in all directions. Crystals, fibrous materials, and metals which have been heavily rolled or otherwise worked are unequally strong in different directions; they are eolotropic. It has been found by ex penment that for many bodies stress= C X strain; C is called the modulus of elasticity. It is con stant for a given isotropic material but de pends upon the kind of stress; an eolotropic body has several moduli for each type of stress. The law takes the following special forms.
(1) For normal stress p and linear strain e in the direction of p p= Ee; E is called Young's modulus after Thomas Young who introduced it in 1807. For granu lar materials like cast iron and stone a more accurate form is p----Een where n lies between / and /./; we shall assume that n=/. (See