420 



Lord Kelvin on the Elasticity 



N is the number of points per unit volume, SW becomes the 

 w of § 3, we find 



w 



+ r<l>'(r)Q(e,f t g, a, b, c) 



+ i^~P {e% 2 +ff+gz*+ayz + bzx + cxy) 2 \ (14). 



§ 14. Let us now suppose, for simplicity, the whole 

 assemblage, in its unstrained condition, to be a cube of 

 unit edge, and let P be the sum of the normal components 

 of the extraneous forces applied to the points of the surface- 

 layer in one of the faces of the cube. The equilibrium of the 

 cube, as a whole, requires an equal and opposite normal com- 

 ponent P in the opposite face of the cube. Similarly, let Q 

 and R denote the sums of the normal components of extraneous 

 force on the two other pairs of faces of the cube. Let T be 

 the sum of tangential components, parallel to OZ, of the 

 extraneous forces on either of the YZ faces. The equilibrium 

 of the cube as a whole requires four such forces on the four 

 faces parallel to OY, constituting two balancing couples, as 

 shown in the accompanying diagram. Similarly, we must 



Fig. 1. 



Z 



S- 



o 



,Y 



T. 



T 



have four balancing tangential forces S on the four faces 

 parallel to OX, and four tangential forces U on the four 

 faces parallel to OZ. 



§ 15. Considering now an infinitely small change of strain 

 in the cube from (e, f, g, a, h, c) to (e + de, f+df, g + dg, 

 a -{-da, b + db, c + dc) ; the work required to produce it, 

 as we see by considering the definitions of the displacements 

 e, /, g, a, b, c, explained above in § 8, is as follows : — ■ 



dw = Tde -F'Qd/+ Bdg + Sda + Tdb + JJdc . (15). 



