DUE TO THE WEIGHT OP CONTINENTS. 
199 
In the case considered in the preceding sections S and U vanish, and the cubic 
reduces to the quadratic 
(P-\)(R-X)-T 3 =:0 
of which the solution is 
2X=P+R±a/(P-R) 2 +4T 3 
m is obviously zero and l, n are determinable from 
P(P-X)=n 2 (R-X) 
Let 
Then it is easily proved that 
l— cos 3-, n— sin 3- 
cot 23 — 
P-E 
’ 2T 
(27) 
This equation gives the directions of the principal stress-axes. 
The two principal stresses N 1? N 3 are the two values of X, so that 
N 1 =i(P+E)+i v / (P-E) 3 +4T^ 
N 3 =!(P+R)-iy(P-R)H4'P 
(28) 
and the third principal stress, which we suppose intermediate in value between 
and N 8 , is of course Q. 
When an elastic solid is in a state of stress it is supposed, in all probability with 
justice, that the tendency of the solid to rupture at any point is to be estimated by 
the form of the stress quadric. At any rate the hypothesis is here adopted that the 
tendency to break is to be estimated by the difference between the greatest and least 
principal stresses. For the sake of brevity I shall refer to the difference between the 
greatest and least principal stresses as “the stress-difference.” This quantity I shall 
find it convenient to indicate by A. 
We may also look at the subject from another point of view:—It is a well-known 
theorem in the theory of elastic solids that the greatest shearing stress at any point is 
equal to a half of the stress-difference. It is difficult to conceive any mode in which 
an elastic solid can rupture except by shearing, and hence it appears that the greatest 
shearing stress is a proper measure of the tendency to break. This measure of ten¬ 
dency to break is exactly one-half of the stress-difference, and it is therefore a matter 
of indifference whether we take greatest shearing stress or stress-difference. For the 
sake of comparison with experimental results as to the stresses under which wires and 
rods of various materials will break and crush, I have found it more convenient to use 
stress-difference throughout; but the results may all be reduced to shearing stresses 
by merely halving the numbers given. 
From (28) we have then 
A = v / (P-E) 2 +4T 3 . 
and the greatest shearing stress at the same point is ^ A. 
(29) 
