426 RICE ART. K 



7X2). For equilibrium the sum of this and Fx, the external sur- 

 face force-component per unit area, must be zero. 



It was stated that the stress at a point was determined by six 

 independent quantities, but so far we seem to have reduced it 

 to a representation by nine. So we shall now turn our atten- 

 tion to three relations which exist between these nine constit- 

 uents, and which are given in [375] and [376], proving these, 

 however, by a more direct and more easily grasped method than 

 that employed by Gibbs. To this end let us once more give 

 our attention to the conditions controlling the equilibrium of the 

 parallelopiped (Fig. 6), and recall the fact that not only must 

 the total resultant force on the parallelopiped vanish, but also 

 the total couple as well. This couple is obtained by taking 

 moments about the point P, and has three components, one 

 around the local axis of x through P, one around the local axis 

 of y', and one around the local axis of z. Consider the contri- 

 butions made by each influence on the parallelopiped to the 

 component of the total couple round the local axis of x. The 

 pulls across the faces involving the constituents Xx, Yy, Zz are 

 symmetrical with regard to P and contribute nothing to the 

 couple. On the other hand the individual shearing forces 

 obviously tend to produce twists. Those that tend to twist the 

 element around the local axis of x are the shearing forces parallel 

 to the local axes of y and z, and they are the following four: 



/ dZy \ 



4f ^ across the face containing R, 



( dZr \ 



~ [Zy — T~ V ) 4f ^ across the face containing V, 



dYz \ „ 



Yz + ~r~ r ) 4^77 across the face containing S, 

 az J 



( 



/ dYz \ 



— [Yz — "r~ r ) 4^77 across the face containing W . 



The moment of the first about the local axis of x is 



/ dZy \ 



