Elasticity_ Barber

12.6 Problems with two deformable bodies 185
The surface displacement u y1 (x, 0) of body 1 is simply (12.26) generalized
to include the effect of the tangential traction p x (see equation (12.25)) — i.e.
∫
(κ + 1) a
u y1 (x, 0) = − p y (ξ) ln |x − ξ|dξ
4πµ
−
(κ − 1)
8µ
−a
∫ a
−a
p x (ξ)sgn(x − ξ)dξ (12.64)
and hence 6
du y1
dx
= −(κ
+ 1)
4πµ
∫ a
−a
p y (ξ)dξ (κ − 1)
−
(x − ξ) 4µ p x(x) . (12.65)
We also record the corresponding expression for the tangential displacement
u x1 , which is
du x1
dx
= −(κ
+ 1)
4πµ
∫ a
−a
p x (ξ)dξ (κ − 1)
+
(x − ξ) 4µ p y(x) , (12.66)
from (12.24).
Comparing Figure 12.11 with Figure 12.3, we see that equations (12.65,
12.66) can be used for the displacements u x , u y of body 1, but for the corresponding
displacements of body 2 we have to take account of the fact that
the tractions p x , p y are reversed and that the y-axis is now directed into the
body. It is easily verified that this can be achieved by changing the signs in
the expressions involving u y , p x , whilst leaving those expressions with u x , p y
unchanged. It then follows that
where
d
dx (u y1 − u y2 ) = − A 4π
d
dx (u x1 − u x2 ) = − A 4π
∫ a
−a
∫ a
−a
p y (ξ)dξ
(x − ξ) − B 4 p x(x) (12.67)
p x (ξ)dξ
(x − ξ) + B 4 p y(x) , (12.68)
A = (κ 1 + 1)
µ 1
+ (κ 2 + 1)
µ 2
(12.69)
B = (κ 1 − 1)
µ 1
− (κ 2 − 1)
µ 2
. (12.70)
6 Notice that an alternative representation of sgn(x) is 2H(x)−1, where H(x) is
the Heaviside step function. It follows that the derivative of sgn(x) is 2δ(x) and
that the derivative of the integral in the second term of (12.64) is simply twice
the value of p x(ξ) at the point ξ =x.