Page 46 - Rock Mechanics For Underground Mining

STRESS AND INFINITESIMAL STRAIN

components of stress ( xx , yy , xy ) and the antiplane components ( zz , yz , zx ). In
the particular case where the z direction is a principal axis, the antiplane shear stress
components vanish. The plane geometric problem can then be analysed in terms of
the plane components of stress, since the zz , component is frequently neglected. A
state of biaxial (or two-dimensional) stress at any point in the medium is defined by
three components, in this case xx , yy , xy .
The stress transformation equations related to ll , mm , lm in equation 2.22, for
the biaxial state of stress, may be recast in the form
1
1
ll = ( xx + yy ) + ( xx − yy ) cos 2 + xy sin 2
2 2
1
1
mm = ( xx + yy ) − ( xx − yy ) cos 2 − xy sin 2 (2.23)
2 2
1
lm = xy cos 2 − ( xx − yy ) sin 2
2
In establishing these equations, the x, y and l, m axes are taken to have the same
sense of ‘handedness’, and the angle is measured from the x to the l axis, in a sense
that corresponds to the ‘handedness’ of the transformation. There is no inference
of clockwise or anticlockwise rotation of axes in establishing these transformation
equations. However, the way in which the order of the terms is specified in the
equations, and related to the sense of measurement of the rotation angle , should be
examined closely.
Consider now the determination of the magnitudes and orientations of the plane
principal stresses for a plane problem in the x, y plane. In this case, the zz , yz , zx
stress components vanish, the third stress invariant vanishes, and the characteristic
equation, 2.18, becomes

2
− ( xx + yy ) p + xx yy − 2 = 0
p xy
Solution of this quadratic equation yields the magnitudes of the plane principal
stresses as

1/2
1 1 2 2
1,2 = ( xx + yy ) ± ( xx − yy ) + (2.24a)
2 4 xy
The orientations of the respective principal stress axes are obtained by establishing
the direction of the outward normal to a plane which is free of shear stress. Suppose
ab, shown in Figure 2.5, represents such a plane. The outward normal to ab is Ol, and
therefore defines the direction of a principal stress, p . Considering static equilibrium
of the element aOb under forces operating in the x direction:

p ab cos − xx ab cos − xy ab sin = 0
or

p − xx
tan =
xy
i.e.
Figure 2.5 Problem geometry for
determination of plane principal = tan −1 p − xx (2.24b)
stresses and their orientations. xy
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