Page 316 - Mechanical Behavior of Materials

Section 7.8 Modified Mohr Fracture Criterion 317

Example 7.9
A block of the granite rock of Table 7.1 is subjected to a confining pressure on all sides of
p = 150 MPa, due to the weight of rock above, as well as a shear stress τ xy ,asshown in
Fig. E7.9(a).

(a) What value of shear stress τ xy will cause the block to fracture?
(b) What is the largest value of τ xy that can be allowed if a safety factor of 2.0 against
fracture is desired?

Solution (a) Since a fit was done to obtain the constants in Table 7.1 for this material, it
is preferable to employ σ uc from Eq. 7.48, rather than the tabulated value of σ uc from a simple

compression test. Using m = 0.824 and τ i = 19.42 MPa from Table 7.1, we find that the value is

! !
1 + m 1 + 0.824

σ uc =−2τ i =−2(19.42 MPa) =−125.0MPa
1 − m 1 − 0.824
The given state of stress is σ x = σ y = σ z =−p =−150 MPa, and unknown τ xy , with τ yz =
τ zx = 0 MPa. This is a state of generalized plane stress, so that one principal normal stress is
σ 3 = σ z =−150 MPa, and the other two are

2

σ x + σ y σ x − σ y
2
σ 1 ,σ 2 = ± + τ xy =−150 ± τ xy MPa
2 2
Since σ 1 and σ 2 are determined by adding and subtracting the same value from σ 3 =−150 MPa,
the three Mohr’s circles must be configured as in Fig. E7.9(b), with the circle formed by σ 1 and
σ 2 being the largest. Hence, C 12 is the largest and controlling value for Eqs. 7.58 and 7.59, so
C 23 and C 31 can be disregarded. Assuming for the present that the C–M component controls, we
have

1 " #
σ
uc
C 12 = |σ 1 − σ 2 | + m(σ 1 + σ 2 ) =¯σ CM =
1 − m X CM

(−150 + τ xy ) − (−150 − τ xy ) + 0.824(−300) = (1 − 0.824)(125.0)/1.00 MPa

p μ τ
(b)
(a)
τ xy τ i
p σ
ut
σ
σ σ σ 0
p 2 3 1
Figure E7.9