Other materials show characteristically dif ferent curves. With increase of carbon, nickel, or chromium the serrated line in the strain di agram of steel smooths out and plastic ductility sets in immediately after the yield point; the curve from A onward is different enough in each case to indicate quite closely the nature of the metal under test. Neither cast iron (in tension or compression) not concrete (in compression) follows Hooke's law. Some writers infer from this that these materials are never perfectly elastic. This is unjustifiable because there is at present no known connec tion between perfect elasticity and the pro portionality of stress and strain. Conclusions drawn from experiments on cast iron are un satisfactory on account of the existence of initial stress due to sudden cooling. Rolled aluminium does not obey Hooke's law but is de cidedly elastic. For materials such as these Bach and Schiile have proposed the empirical law pn = Es where n is a constant equal in the case of cast iron in tension to 1.07.

Stress Relations.— In most engineering problems the stresses at any point in a body are all parallel to one plane and are, therefore, called plane stresses; we shall limit the dis cussion to this case. Stress on one plane at any point produces stress on every plane at that point. Fig. 5 shows an infinitesimal ele ment in the most general state of plane stress, there being by assumption no stresses per pendicular to the paper. The stresses on any face, for example the left, are /it normal, and q1 tangential; the corresponding forces are p,dy and gady if the thickness is unity; the other stresses are assumed to act as in the diagram. As the element is in equilibrium un less vibrations occur the sum of the moments of all forces — not stresses — about a central axis normal to the paper must vanish; al though the stress may vary from point to point it is constant over an infinitesimal area. Hence 4(q, d= dxdy =-- dydx.

In the limit el =q, and 0= es q,= q, or: shear stress on any plane through a point is always accompanied by equal shear stress on a per pendicular plane through that point, both stresses acting away from or toward the edge of intersection and normal to it. This is Cauchy's theorem.

If the stresses on two planes X and Y at a point P are known as in Fig. 6, the stress on any other plane through thatpoint can be found. Pass any plane at an infinitesimal dis tance from P and take the wedge thus formed as a free body. There will in general be nor mal and shear stresses p' and q on the oblique plane. If the hypothenuse of the wedge has an

area dA the forces on it will be p'dA and q'dA ; the horizontal force on the vertical face is and so on for the other forces. By resolving along p' and q' respectively and replacing functions of 8 by functions of 20 we get Pr—Pvcos ze-q sin 20 ... (1) 2 2 q' = sin 20 —q cos 20 (2) 2 which remain unchanged when dA becomes zero. They are the stresses at P on any oblique plane of angle O..

When tan 2B= Pz 2q(3) which, being satisfied by the two values and 0+90', shows that there are always two orthogonal planes free from shear stress. They are called principal planes and the pure normal stresses on them principal stresses. If (3) is substituted into (1) the principal stresses, say p, and pi, are found to be Pi,th= Px Pi'± }NAN —Pv)° 44° • • • (4) By putting dy/d0 from (1) equal to zero we get the important fact that the principal stresses are the largest and smallest normal stresses in the body at P.

While there are always two pure normal stresses, pure shear exists only under special conditions. Take, instead of X and Y, the principal planes as the planes of reference. Then in (1) and (2) q drops out, ps and ps, must be replaced by P. and P,, and 8 stands for the angle between the oblique plane and one of the principal planes; P' + P' — cos 20 . . . 2 2, — sin 20 . . . (6) q 2 The condition for pure shear on the oblique plane is p'.= 0 from which 28 = 4 — this is impossible unless p. and pt are opposite in sign or unlike in sense. Hence pure shear exists only when the principal stresses are un like. If they are unlike but numerically equal, But pure shears are not in general the greatest or least shears. It is evident from (6) that maximum and minimum shears occur when 0....45°; in other words, the planes of maximum and minimum (greatest numerical) shear lie at 45° to the principal planes. The largest shear stress is, therefore, r - - pi) - 11/ (p. - + The effect of a stress p is a strain p/E in the direction of P and, by Poisson's law, a lateral strain thisE perpendicular to P; they are opposite in sign. Consequently when a block is acted on by tensions on one pair of opposite faces and tensions p, on a pair per pendicular to the first, the resultant strain E. along is Pi P• E mE A single stress equal to Ee, acting alone in the direction of p. will produce the same strain in this direction as both p. and p.; in this re spect it is equivalent to them and is called the equivalent simple stress.