Page 305 - Mechanical Behavior of Materials

306 Chapter 7 Yielding and Fracture under Combined Stresses

Table E7.7

(a) Given Stresses (b) Calculated Values
σ 3 σ 1 = σ 2 y x Center
MPa MPa |σ 1 − σ 3 | σ 1 + σ 3 (σ 1 + σ 3 )/2
3 0 — — 1.5
−100 0 100 −100 −50
−700 −100 600 −800 −400
−1230 −200 1030 −1430 −715
Source: Data in [Jaeger 69] pp. 75 and 156.

Solution (a) Values of y and x are calculated from Eq. 7.53(a), as given in Table E7.7(b). A
linear least squares fit of these values, with the simple tension test not being included, yields

a =−0.6995, b = 33.37 MPa

Equation 7.53(b) then gives
b 33.37 MPa
m =−a = 0.6995, τ i = √ = √ = 23.35 MPa Ans.
2 1 − m 2 2 1 − 0.6995 2
The additional values desired can then be calculated from Eqs. 7.47 and 7.43:
90 − φ
◦
−1
◦
φ = sin m = 44.39 , μ = tan φ = 0.9789, θ c = = 22.81 ◦ Ans.
2
(b) The failure envelope line is then given by substituting the constants obtained into
Eq. 7.42:

|τ| + 0.9789σ = 23.35 MPa Ans.

This line is plotted in Figs. E7.7(a) and (b), where the latter shows the region near the origin in
more detail. Also plotted are the largest Mohr’s circles from each test, where the centers of each
are calculated in Table E7.7(b) as a convenience. The line is in reasonable agreement with the
circles for the three tests in compression, but it is far above the circle for the simple tension test.
(c) The values of the strengths in simple compression and tension expected from the fitted

envelope, σ and σ , are obtained by substituting m and τ i from the previous fit into Eqs. 7.48
uc ut
and 7.49:

σ uc =−111.1, σ ut = 19.63 MPa Ans.

The value of σ uc is about 10% larger than σ uc =−100 MPa, which is perhaps within statistical
scatter. But σ is drastically larger than σ ut = 3 MPa, and the fitted envelope obviously does not

ut
agree with the tension test data.