STRENGTH OF MATERIALS).
The longitudinal strain e is always accom panied by a strain e' of the transverse dimen sions; for a given material e'=cre where a is constant for isotropy. Thus if a rod receives a longitudinal strain e the strain of any straight line in a cross-section is ae: at is not an areal strain. In engineering. m is used for a. a is called Poisson's ratio after Poisson, who in 1828 calculated it to be for all materials; experiment has not verified his prediction.
(2) For shear stress q and strain q-=-14; p is the modulus of shear or of rigidity. Young was the first to point out that resistance to (detrusion,p as he called shear, was different from resistance to stretching. But he did not introduce the shear modulus; this was done by Navier in 1833. The symbol p was first used by Lame in 1852; in books on the strength of materials N and G are used for p.
.(3) A constant normal stress p over the entire surface of a body produces a volumetric strain A, called the dilatation, where k being the bulk modulus or modulus of com pression.
For isoteopic bodies there are thus four constants of elasticity: E, p, a, k. It will be shown later that only two of them are inde pendent. Stokes in 1845 ((Mathematical and Physical Papers,' Vol. I, p. 75) pointed out that p and k are of basic importance in theo retical work; in engineering E and p are more convenient.
For eolotropic substances we may still as sume the stress to be a linear function of the strain, in which case the generalized forms of Hooke's law are Cyy+ Cu ess+Cu Gay+ Cis els+ esz Cn eyy + eXtl+ CIS eyx+ Cs" and so on for Zs, Xy, Z., the 36 coeffi cients, Cu to Cu, being the elastic constants. Green in 1837 proved that for conservative systems Css. Cam whereby the constants re duce to 21 for eolotropy. For isotropy they reduce to 2. To express the stresses in terrns of strains for isotropy consider the effect of a single tension X.. 13y Hooke's law it pro duces a strain Xi/E in its own direction, and according to Poisson's ratio, a lateral com pressive strain aX./ E. Hence if a parallele piped is acted on by tensions X., and Zs, the resultant axial strains will be given by E,e.. X — a ( Yy Zz) Eeyy = Yy — agz -1-Xx) . . . (6) Efts = Z.— a(Xx Ifv); exx, , est are called the equivalent simple strains, and when multiplied by E, the equiva lent simple stresses. By addition eyy ett)= (X.+ Yy Zs) (i— 2c7) ; the first parenthesis is the volumetric strain because the change of volume is (I -I- e..) dx (1 -I- eind dy (I -I- est) ds—dx dy dz =(azz+ eins dx dy dm if the strains are so small that terms above the first order are negligible. The volumetric strain or dilatation A is then A Cyy es: Now it is obvious from Fig. 2 that the infinites • au imal shear does not change the area of aY the face dx dy; if one does not, none vvIll, so that shear does not change areas. In the" same
way it follows that infinitesimal shear will not alter volumes; this is why the dilatation contains only linear strains.
If the stresses in the equation following (6) are all equal to X.
3(1 — 2a) X.= EA whence, as p.---kA k —2a) If a> 5/2, k is negative whence volumes would be increased by compression and decreased by tension; as this does not occur a cannot ex ceed %. Furthermore a negative a would mean lateral expansion under tension; this is not true of isotropic materials. Therefore a is a positive fraction not larger than Vs; ex periment verifies this.
If the first equation of (6) is written Xx=Eezz±a(Xx+ +Z.)-0X.
and the parenthesis eliminated by means of the equation below (6), there results Xs aa 2ut..; similarly Yy= RA ± Zuevy (7) Zs= RA + 2pes.
where etE -Fe).
0 +00-2a) It will be proved ldter that p is the modulus of shear as defined above; hence Xy=pess , Ys=pess , Z.=---pezx , (8) Navier, Poisson, and Cauchy, the founders of the theory of elastidty, derived their equations from a hypothesis of intermolecular actions the consequences of which demanded that X=0; then contrary to experiment. They belonged to what Pearson has called the rari-constant, as opposed to the multi-constant, school of elasticians. The weight of evidence is in favor of the necessity of two constants for specifying the elastic properties of isotro pic materials. To interpret p in the equations just found, consider a cubical element under tension p on one pair of faces and compres sion p on a perpendicular pair, as on the full line square in Fig. 4. By taking as a free-body the shaded corner cut off by a 45-degree plane and resolving the forces (stressXarea) par allel and normal to the oblique surface, we find that a stress q=-p on the oblique face is neces sary and sufficient for equilibrium. That is, orthogonal, equal, unlike normal stresses pro duce pure shear of equal magnitude on any plane at 45 degrees. Take now the dash-line square as a free-body: it is in pure shear of magnitude q=p. Since the change of a right angle is the shear strain 0, the change of ik=45°) is 0/2. As the sides of the inner square are not altered in length, D is constant in dl D=1 cos whence = &kin° dl But and dB= 0 •• = A By eq (6) — E and P E 2(1 + a) For the purposes of integration it is con venient to eliminate the internal stresses from equations (5). Substituting equations (7) and (8) in (5), replacing the strains by their values in (1) and (2) and using the symbolic abbrevia tion — the Lapacian operator — a. a. as , axt az? we find atu (X — Vls pX = p at.