THE MECHANICAL INTERPRETATION OF JOINTS 3 
to 72,73. Similarly resolve the stresses p, into three components: 
parallel to r,, parallel to 7;, and normal to 7,, 7;. Of these com- 
ponents, only the first named of each set, namely, 
Px COS (oz) P2 COS (p.7:), 
have moments about 7;. Hence for equilibrium as regards rotation 
about 7;, we have 
2dA +p; COS (pr 2) -PC,=2dA * p2 COS (p27:)° PC, ; 
or, since PC, = PC., 
Px COS (172) = pz COS (p21). (x) 
The angles indicated are those between the stresses and outwardly 
directed radii. The fundamental equation (1) may be stated as 
follows: 
Tf px, p2 are unit stresses at the surface elements r;, fo, the 
projection of p; upon r, is equal to the projection of p. wpon 1;. 
We proceed to put equation (1) in a more usable form. Let us 
regard angles in the plane of 7;, r. as positive when clockwise; and 
let s,, s, denote radii in this plane 90° ahead of r,, 7, respectively. 
Now resolve p,; into components as follows: (1) o:=p: Cos (p:/s), 
parallel to 7:; (2) t=: cos (p:5:), parallel to s,; (3) parallel to 73. 
Also resolve p2 into the components: (1) ¢.=p. cos (p2r2), parallel 
tO 72; (2) T2=p2 COS (p52), parallel to s,; (3) parallel to 7; The 
normal stresses o;, 02 (i.e., normal to their surface elements) are 
positive when directed outward from P, or when they are fensions. 
The components 7;, 72 of the shearing stresses are positive when 
they produce clockwise moments about 7;. We now transform the 
cosines in (1) by means of the formula for the angle between two 
directions in space: 
cos (p:2) =Cos (prs) COS (721) + cos (px52) COS (7251) COS (pr) + COS (7275) 
=cos (pyr) Cos (7172) 0s (px5:) sin (r12), 
cos (p,7:) =cos (p.72) Cos (712) + COS (p252) COS (7:52) Cos (p273) COS (7773) 
=cos (p,r2) cos (7:72) — COs (p22) sin (7272). 
