142 APPLIED MECHANICS 
Let BE be a plane section, upon which there is pure normal stress of 
intensity 7. It is required to find r, and 9, the inclination of BE to CD, 
- in terms of p, g, and f. 
Let the edge of the cube be denoted by x. Consider the aiydlitierats 
of the prism BCE. i 
P the resultant of the stress p on BC = pa? ; 
Q ” ” ” qd » CE = gz? cot 6. 
R is RL ete r , BE=ra?/ sin 0. 
F ” ” ” Ba ” BC = fu. 
8 ” ” ” I ” CE ED bo cot 0. 
Resolving vertically. Resolving horizontally. 
R cos 0= F+Q. R sin 0=P+8. 
ra? cot 0=fa? + qa? cot 0. rx" = px? + fx? cot 0. 
rcot0=f+qeot0. . (1) r=p+foot@. . . (2) 
Solving equations (1) and (2), 7=43{p+q+ /(p—q)?+4f"! 
tan 20 7F or tan §2.5 Soe J(p~ 9) + 4p? 
q-P ios dhe: * af 
There are two values of 0 which satisfy the above equations, and 
these values differ by 90°. Corresponding to the two values of @ there 
are two values of 7. 
The two values of 7 acting in directions at right angles to one another 
are called principal stresses, and two lines parallel to the directions of the 
principal stresses are called principal axes up stress, 
As a numerical example, let p=6, g=3, and f=2, all in tons per 
643+ (6-3)? +4 x 2? 
2 
square inch. Then r= 
=7, or 2 tons per square 
inch, tan alee aed (3S or pL see 
2 2 a 
To show the positions of the principal axes of stress, draw HK (Fig. 
191) parallel to the direction of p, and make it equal to unity on any 
convenient scale. Draw KL at right angles to KH, and make it=2. 
Join HL, then the angle LHK is one value of 6. Produce KH to 
M, making HM=2. Draw MN at right angles to MH, and make 
it=1. Join NH, then the angle NHK is the other value of 6, and it 
is easily seen that the angle NHL is a right angle. Through any 
point O in the original cube ABCD draw OX and OY perpendicular to 
HL and HN respectively. OX and OY are principal axes of stress, and 
if a rectangular element be taken at O, with its faces parallel to OX 
and OY, the stresses on the faces of this element will be entirely normal, 
the stress in the direction OX being a tension of 7 tons per square 
inch, and the stress in the direction OY a tension of 2 tons per square 
inch. 
In Fig. 190, p eae gq are shown as tensile stresses. If p or g, or both, 
