312 REPORTS ON THE STATE OF SCIENCE, ETC. 
Thus we canzot, in this case, take E= E, and y=y, and the function E (and therefore 
also the stresses) must involve the elastic constants. 
8. Let us now return to our plate of ideal material and introduce in it a simple 
dislocation with respect to any given hole corresponding to a unit relative translation 
of the faces of the cut parallel to 2 (The unit, of course, must in this case be 
a small length, so that the squares of the strains introduced are negligible.) No 
forces are to be applied to any boundary. 
Corresponding to this dislocation of the ideal plate, there will be an internal 
strain and stress defined by the stress function E,, and let ~, be the corresponding 
value of ~ Owing to the absence of external forces, E, and its differential 
coefficients are acyclic throughout. The corresponding displacements U,, V, are 
cyclic for a circuit enclosing the hole considered. Also 
Cy(U,)=1,  Cy(V,)=0. 
Hence, applying Cy to (15). 
“on (BY (ay 
Qu, =1— 1 0O=1-— } ee 5 é A 
y= 1~0,Cy( “ 0s (3) (18) 
Similarly if E,, y, belong to a simple dislocation with respect to the same hole 
corresponding to a unit relative translation of the faces of the cut parallel to y 
0=1—0,Cy (3) 2m =1—oyCs( Ne). ooh bse eto 
In like manner, unit translational dislocations with reference to the other holes 
will lead to stress functions E,,E,.. . E,,_,,E,, and corresponding wy functions 
Why, Ee i Von—pYon- 
Build up a solution 
E=E,+4,B,+0,B,+0,B,+a,By+ ©. . + a9n—1Boy—y + Oy Ban § 
WHY y + Wy + Oo t+ Ogg + Oy + . 6. + Gxn—Won—1 + GonWony 
where a, . . . a, are constants to be determined. 
If this solution is to lead to acyclic values for the displacements U, V, we must 
have, for every irreducible circuit 
e(B8) =n) 
sE 2) 
Or(3,)= Co) e>(2) 
If we take Cy to refer to an irreducible circuit enclosing the hole to which solu- 
tions 1 and 2 refer, but none of the others, then Cy(E,)=Cy(E,) = Cy(E,) = Cy(H,) 
=.... =Cy(Es,=0, and also, since the displacements U,, V,, say, are acyclic except 
for a circuit enclosing that hole to which solutions 3 and 4 refer, 
Cy a) 5 Cy 3) = Dyer VES.Oy 
by ba 
Eqnations (20) then become, on substitution, 
dE dy dy, (5¥. 
\ °)= — | ed) — J BC — Cy( — 
CO (3!) (1—c) Cy( 5") +a,(1 7) Cys) ac ov) “ae) 
0y(',.") =(l-o) ox) +a (1- 7) cy) +a.(1 —oycy(*) 
wherce, using (18) and (19), 
] 6h OW 2 My, 
ew OL) (ig ae) a | oo) = on! 
l—o be ni Rs, l-c¢ 
1 5B a) Quyd, [ 
_ Cy A) PEER 6 at \ pea es 
L307; "(5,') (en) =] 
Since the cyclic functions on the left-hand side are known to reduce to mere 
constants, equations (21) determine a, and a,, 
Similarly, taking irreducible circuits about the other holes, we obtain ay, a,, etc., 
and we have finally an acyclic solution which holds good for the actual plate, but 
39 (n+ 1) een nea 
| 
(21) 
