174 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
given to a second approximation, and each of the stresses, except z, to a first approxi- — 
mation. The result is obviously the same as that found by integrating the permanent 
terms (65) of the original source solution. 
If we write 
then the displacements for Z force of intensity Z(a, y, 2’) per unit area on z=2’ are 
o= > 
lz 3 | . /atd 4-3 9, +1 a—3,., 
4 “Jamal! aa ( 5 ete Zz ae at ne!) 
v= > 
dy — 
i (a+ DF + vR(So8 24 So 802 _ any tt pp (72 
w= ici a Vv 5) 5 ‘ : : ) 
The corresponding results for a volume distribution of force, Z(x, y, z) per unit 
volume, are found by integrating these with respect to 2’ from —/ to +h. 
In order to calculate the stress z; from displacements, we should need the value of w — 
to a third approximation. It is therefore easier to find z directly from (71) and ( 2) 
On dividing by 2u(a+1), we find, corresponding to (72), 
3h'2 — 
ese Sie ey . GB 
When the force is Z(a, y, z) per unit volume, this leads to 
hz — 23 
if a ae ae q 
os ie Za, y, ddd —3[° Ble yp eae! + 5 [ Bla, nid... 
=~ 
a 
We can now find the stresses xz, xy, yy to a second approximation. 
For 
me PE av dw 
ma op ¢ 
Be ee ‘ ie 
A du dw 
Fe Fae at ar ae 
Peay bee ae ra 222) Viie= 
dx dy/ X+ Sie! 
and similarly 
~ 4u(A+ Mo = r 
WS Oe oan ay Nea, 
~ (oe a 
a dy dx 
We have now only to put in the values of u, v from (72) and the value of x from (73). 
zz, Where o= aM + ft). 
Also 
