T − C ±
(T + C )2+ 12 (C − T ) p
S =
TC
3 + (C − T ) p
Of course, the Mises criterion is unaffected by pressure.
The pressure testing data, also from Pai [7.1], are at 1 kBar pressure = 101.4 MPa, which is about twice the compressive strength of the material.
The data and predictions for the strengths under pressure are shown in Table 7.1.
It should be noted that the original data were given in psi units to two significant figures. The Mises criterion is calibrated by the tensile strength, as is conventional. The total square errors for the properties predictions compared with the data are given by
Failure criterion 5.5% error Mises criterion 45.5% error
It is seen that the worst of the three predictions by the Mises criterion is for uniaxial compression, and it differs from the data by a factor of 2.0. Incidentally, the total square error for the predictions from the C–M criterion is 15.4%.
7.2 Brittle Polymers
This example is similar to that of the example in Section 7.1, but for a dif-ferent materials type that moves a little more toward the brittle end of the scale. Specifically, the materials type is typified by that of PMMA, which
is a transparent, rather brittle thermoplastic, widely used as an inexpens-ive replacement for glass, and often used in model experiments, especially for dynamic crack propagation. Although as a plastic it is considered as brittle, its strength ratio T/C ∼= 1/2 places it as intermediate on the over-all scale for over-all materials. The uniaxial and shear failure criterion under pressure are the same as those given in the example in Section 7.1, and the strengths for the C–M criterion under pressure are given by
C–M criterion
It is interesting to note that the C–M forms give a linear dependence on pressure p, while the failure criterion forms in the example in Section 7.1 give a square-root dependence on p. The strengths at T/C = 1/2 and for pressure p = 2C (which is of about the same pressure ratio in the example in Section 7.1) are given in Table 7.2.
The greatest difference between these is for uniaxial compressive strength at a factor of
η = 1.40
Thus even at a reasonable pressure of p = 2C these polymers show a wide difference between the two predictions. The differences widen
Table 7.2 Brittle polymer strengths under pressure σ11T
Failure criterion 1.64 –2.14 1.08
C–M criterion 3/2 –3 1
Glasses 91 even further at larger pressures for all three strengths. Furthermore, the effect of the pressure has a very strengthening effect on the strengths themselves.
7.3 Glasses
The problem is that of pushing a curved indenter into an elastic but brittle material such as a glass or ceramic. This is the classical Hertz contact problem.
If the material were a ductile metal, the yielding would begin straight under the indenter, but at some depth. The interest here is with glass, and finding the location of its initial failure, and finding the load at which failure occurs. Take the indenter to be rigid and spherical of radius R (Fig. 7.1). It is found that for the properties (Poisson’s ratio) of the brittle material of the half space, both failure criteria maximize at the edge of the contact zone. At this point the surface tractions vanish, and the in-plane stresses are
σr = (1− 2ν) 3 q0 σθ =−(1− 2ν)
3 q0
where q0is the maximum pressure immediately under the indenter and on its center line. The present failure criterion is controlled by the fracture form, and the analysis of the problem gives the force P acting on the indenter at failure as
Fig. 7.1 Spherical indenter on deformable medium.
Fracture criterion
For T/C = 1/5, which corresponds to a high-quality glass such as S-glass, the two predictions of the load at failure differ by a factor of
η = 1.73
Timoshenko and Goodier [7.2] note that many indentation experiments on glass have verified this location for the failure. In fact, it even goes back to the original experiments of Hertz.
7.4 Ceramics
In this example the effect of a two-dimensional pressure upon the shear strength will be examined. The material of application will be for a typi-cal ceramic. Ceramics function poorly under tensile stress states, though they are very useful under combinations of shear stress and compression.
Consider a thin layer of a ceramic as a heat shield and subject to pres-sure in its plane along with shear stress. In contrast to the example in Section 7.2, here both the failure criterion and the fracture criterion will be found to play a role. The two-dimensional pressure σ11 =σ22 = –p2D, σ33 = 0 will be taken in the same plane as the shear stress σ12, having strength S. The governing criteria are
Failure criterion
3S2 = TC − p2D2 + 2(C − T ) p2D
Fracture criterion
S = T + p2D
Minerals 93
2 1
1
Fracture Mode
Failure Mode
C-M S
C
T C
1 7
p2D
=
Fig. 7.2 Shear strength dependence on two-dimensional pressure.
C–M criterion
For p2D ≤ C 2
S = TC + (C − T ) p2D
T + C For p2D ≥ C
2 S = C − p2D
At T/C = 1/7, the failure envelopes are as shown in Fig. 7.2.
The shortcoming of the C–M criterion in part relates to the fact that it does not account for the intermediate principal stress, σ2, in its formulation.
7.5 Minerals
This problem is motivated by drilling through hard rock in exploring for water or hydrocarbons. The drill bit causes an intense local state of stress that is primarily composed of shear stress under pressure. The difference between this example and the example in Section 7.4 is that here the pressure is three-dimensional, p, rather than two-dimensional pressure.
The governing failure criteria are
Failure criterion
S =
TC
3 + (C − T ) p
Fracture criterion
S = T + p
C–M criterion
S = TC + (C − T ) p T + C
The plot of S versus p is given in Fig. 7.3 at T/C = 1/15.
There is seen to be a major difference in behavior for three-dimensional pressure from that which occurred in the example in Section 7.4 with two-dimensional pressure. In this example the C–M criterion is mostly unconservative relative to the present failure criterion, whereas in the example in Section 7.4 it is conservative. The present example is much
5 5
Fracture Mode
Failure Mode C-M S
C T
C 1 15
p
=
Fig. 7.3 Shear strength dependence on three-dimensional pressure.
Geo-Materials 95 like that in Section 6.3, where the test data for dolomite were found to bifurcate away from the fracture mode and follow the failure mode as pressure increases, rather than following the straight line projection of Coulomb–Mohr.