Engineering Rock Mechanics

64 Strain und the theory of elasticity

We invert these equations to find the strains sx, sy and yxy as
-1
cos2 OP sin2 OP sin ~p cos OP

cos2 OR sin2 sin OR COS OR

Inverting the matrix of trigonometrical functions used to be difficult,

and is the reason why strain gauge rosettes of specific geometry (e.g.

rectangular and delta rosettes) were developed: the inverse of the matrix

for these special geometries has a particularly simple form. However, it

is now easier to perform the inversion of a general matrix on modern

spreadsheets and scientific calculators, and so this constraint on rosette

geometry is diminishing.

The solution to the problem is found using the matrix equation given

above. In this case we have Op = 20°, Op = 80" and OR = 140°, and so the

matrix equation is as follows:

0.883 0.117 0.321

0.884 -0.293 0.449

0.857 0.456 -1.313

43.0 x lop6

7.8 x low6

17.0 x

43.6 x 41.6 x

[ 7.9 x lo-.] = [ 3.6 x lo-.].

17.0 x 18.1 x 10-6

The principal strains and their orientations are then computed from sx, sy,

and yxy. In this case we find e1 = 43.7 x with the

angle between the x-direction and the major principal strain being 12.7".

To compute the stress state from the strain state we use the stress-

strain relations for an isotropic material, i.e.

and s2 = 1.52 x

[::I;$" -v 1 : I.[:]

YXY 0 2(1+ v) txy

which when inverted gives

From these we find that ax = 7.04 MPa, ay = 2.65 MPa and txy =

1.04 MPa. Computing the principal stresses and their orientations from

these values gives c1 = 7.28 MPa and a2 = 2.41 MPa, with the angle

between the x-direction and the major principal stress being 12.7".

Notice that because this is an isotropic material, the orientations of the

principal stresses and the principal strains are identical. Note also that

we can perform the complement of this calculation, i.e. determine the