Page 258 - Mechanical Behavior of Materials

Section 6.4 Three-Dimensional States of Stress 259

obtained as follows:

σ 1 = 133.1, σ 2 =−93.1, σ 3 = 40.0MPa Ans.

These are numbered consistently with the Ex. 6.3 solution. The maximum normal stress is
the largest of σ 1 , σ 2 , σ 3 , so that σ max = 133.1 MPa. Principal shear stresses τ 1 , τ 2 , τ 3 and the
maximum shear stress, τ max , can then be calculated as already done in Ex. 6.3.
Comments Note that we did not take advantage of the fact that this is a state of generalized
plane stress. Hence, the foregoing procedure can be applied for any state of stress, as for cases
where there are no zero stress components for the original x-y-z system. Where there are zero
components, the determinate form of the cubic, Eq. 6.26, may be useful. In this particular case,
Eq. 6.26 gives

(100 − σ) 80 0

80 (−60 − σ) 0 = 0

0 0 (40 − σ)
Using the last column to expand yields
2
(40 − σ)[(100 − σ)(−60 − σ) − 6400] = 0, (σ − 40)(σ − 40σ − 12,400) = 0
Hence, from the (σ − 40) factor, it is evident that one root is σ 3 = 40.0 MPa. The remaining two
can be found by applying the quadratic formula to the equation that is the other factor.

Example 6.7
Find the direction cosines for each principal normal stress axis for the stress state of Ex. 6.6.
Solution The principal stresses as already determined in Ex. 6.6 are needed:

σ 1 = 133.1, σ 2 =−93.1, σ 3 = 40 MPa
Equation 6.25 must now be applied for each of these stresses. Substituting for σ 1 , and also for
σ x , σ y , σ z , τ xy , τ yz , and τ zx on the original coordinate axes, we have

(100 − 133.1)l 1 + 80m 1 = 0
80l 1 + (−60 − 133.1)m 1 = 0

(40 − 133.1)n 1 = 0

The last of these is satisfied only by n 1 = 0, and the first two both give the same result:

l 1 = 2.414m 1