Elasticity_ Barber

16 1 Introduction
e yy = ∂u y
∂y
; e zz = ∂u z
∂z . (1.37)
Notice how easy the problem of Figure 1.6 becomes when we use these
definitions. We get
from (1.34, 1.36) and hence
∂u x
∂x
=
ρg(L − x)
E
u x = ρg(2Lx − x2 )
2E
, (1.38)
+ A , (1.39)
where A is an arbitrary constant of integration which expresses the fact that
our knowledge of the stresses and hence the strains in the body is not sufficient
to determine its position in space. In fact, A represents an arbitrary rigid-body
displacement. In this case we need to use the fact that the top of the bar is
joined to a supposedly rigid ceiling — i.e. u x (0) = 0 and hence A = 0 from
(1.39).
1.2.2 Rotation and shear strain
Noting that the two x’s in e xx correspond to those in its definition ∂u x /∂x,
it is natural to seek a connection between the shear strain e xy and one or
both of the derivatives ∂u x /∂y, ∂u y /∂x. As a first step, we shall discuss the
geometrical interpretation of these derivatives.
Figure 1.8: Rotation of a line segment.
Figure 1.8 shows a line segment P Q of length δx, aligned with the x-axis,
the two ends of which are displaced in the y-direction. Clearly if u y (Q) ≠
u y (P ), the line P Q will be rotated by these displacements and if the angle of
rotation is small it can be written
φ = u y(x + δx) − u y (x)
δx
, (1.40)