Rock Mechanics.pdf - Mining and Blasting

Figure 2.8 Displacement components
produced by pure longitudinal
strain.
Figure 2.9 Displacement produced
by pure shear strain.
DISPLACEMENT AND STRAIN
and
dux =−z dy
du y = z dx
(2.28)
The total displacement due to the various rigid-body rotations is obtained by addition
of equations 2.26, 2.27 and 2.28, i.e.
dux =−z dy + y dz
du y = z dx − x dz
duz =−y dx + x dy
These equations may be written in the form
⎡ ⎤ ⎡
dux
⎣ du y ⎦ = ⎣
0
z
−z
0
⎤ ⎡ ⎤
y dx
−x⎦
⎣ dy ⎦ (2.29a)
duz −z x 0 dz
or
[d ′ ] = [Ω][dr] (2.29b)
The contribution of deformation to the relative displacement [d] is determined
by considering elongation and distortion of the element. Figure 2.8 represents the
elongation of the block in the x direction. The element of length dx is assumed to be
homogeneously strained, in extension, and the normal strain component is therefore
defined by
εxx = dux
dx
Considering the y and z components of elongation of the element in a similar way,
gives the components of relative displacement due to normal strain as
dux = εxx dx
du y = εyy dy
duz = εzz dz
(2.30)
The components of relative displacement arising from distortion of the element
are derived by considering an element subject to various modes of pure shear strain.
Figure 2.9 shows such an element strained in the x, y plane. Since the angle is
small, pure shear of the element results in the displacement components
dux = dy
du y = dx
Since shear strain magnitude is defined by
31
xy =
− = 2
2