VI
167
ON HARDNESS
2
ap
K₂(1+20₂) dz
In the first place, this system satisfies the equations of
equilibrium, for we have
▼² + (1 + 20₁)
до,
дүгП,
2 22P
+
=
0,
მა
d.x
K₁ Əzəx
1
and similar equations hold for 2, 71, 72; for the 's we get the
same result, remembering that 2P = 0. For the tangential
stress components at the surface (0) we find, leaving out
the indices:-
=
0,
=
X, K
-
૦૬ × ૦૬
a
+ǝz
дх
ƏºII
ӘР
72P
=
K2.
+29
=
2z
дъл
Jx
Əxəz
=3
=
-
K(2
a²II ӘР
+29
ǝyǝz да
a²P
= 22
=
дудъ
Y--K(+)-
Əy
as the second condition requires.
0,
It is more troublesome to prove that the third condition
is satisfied. We again omit indices, as the calculation applies
equally to both bodies. We have generally
2₂
= -
მ
2K (+8) -
Θσ
-
=
- 2K
2z
2P
მ-2
Ə²²
21 2(2+30 ap
+
K(1+20) Əz
ap
მა
Əz² - 2 azi
ap
therefore at the surface Z.-2. Now, using the equa-
=
tion for u, we have generally
ӘР
Əz
=
3p
2
00
αλ
8π² ) λ√(a²+λ)(b² +λ)λ'
น
ӘР
дъ
and therefore at the surface vanishes, as it must do, and
with it Z₂, at any rate outside the surface of pressure.
In
the compressed surface, where u = 0, the expression takes the