V
151
CONTACT OF ELASTIC SOLIDS
differential equations given for E, n, 5, and the displacements
For the pressure components we find
vanish at infinity.
(221
20 ap
Xx
X₁ =
-
2K
+
да K(1+20) Oz j'
X=-2K
a²11
даду
(021
- 2K-
Y₁ =
-
20 ӘР)
X,
Z₂
Y₂
= -
+
dy2 K(1+20) dz
21 ӘР) ӘР
2K +9
дідж Jx
= 2K
-
= -
= 2z
dxdz
(ӘП, 2(2 + 30) 0P)
+
0-2 K(1+20) dz'
ap
ӘР
=2z
ləyəz + I
dy дудъ
The last two formulæ show that for the given system the
stress-components perpendicular to the z-axis vanish through-
out the plane z = 0. We determine the displacement and
the normal pressure Z, at the plane z = 0, and find
<=JP, Z₁ = -2
Ꮡ
ӘР
2oz
The density of the electricity producing the potential P is
-(1/2)(P/2), hence we have the following theorem. The
displacement in the surface, which corresponds to the normal
pressure Z, is equal to 9/47 times the potential due to an
electrical density numerically equal to the pressure Z,
We now consider again both bodies: we imagine the
electricity whose potential is P to be distributed over a finite
portion only of the plane z = 0; we make II, and II, equal to
the expressions derived from the given expression for II by
giving to the symbols K and @ the index 1 and 2, and put
$2
&
=
әп2
дх
әп,
дх
71=
=
әп,
მყ
"
n2
----
ап2
dy
"
52
=
әп,
Əz
+29,P,
2.9,P;
a12
Əz
-