Page 248 - Mechanical Behavior of Materials

Section 6.3 Principal Stresses and the Maximum Shear Stress 249

From Fig. E6.3(a), the directions for these stresses are given by a rotation θ n relative to the
original x-y axes:

◦
◦
tan 2θ n = 80/80 = 1.00, 2θ n = 45 , θ n = 22.5 (CCW)
This rotation gives the 1-2 principal axes. Since the third principal stress σ 3 is coincident with
σ z , the, 1-2-3 principal axes are as shown in Fig. E6.3(b).
The maximum normal stress is the largest of σ 1 , σ 2 , σ 3 , so that σ max = 133.1MPa (Ans.).
The points (σ 1 , 0), (σ 2 , 0), and (σ 3 , 0) now fix the circles for the three principal planes, as
shown in Fig. E6.5(c). The radii of these circles are the principal shear stresses:

|σ 2 − σ 3 | |−93.1 − 40|
τ 1 = = = 66.6MPa Ans.
2 2

|σ 1 − σ 3 | |133.1 − 40|
τ 2 = = = 46.6MPa Ans.
2 2

|σ 1 − σ 2 | |133.1 − (−93.1)|
τ 3 = = = 113.1MPa Ans.
2 2
The largest of these is τ max = 113.1MPa. Ans.
Comments The principal normal stresses and directions could have been determined from
Eqs. 6.6 and 6.7 rather than from Mohr’s circle. If desired, the three circles could then still be
plotted from the values of σ 1 , σ 2 , and σ 3 , with half circles as in (c) being sufficient to allow τ 1 ,
τ 2 , and τ 3 to be visualized.

6.3.2 Plane Stress Revisited
Consider plane stress in the x-y plane, so that σ z = τ yz = τ zx = 0. If the principal normal stresses in
the x-y plane are σ 1 and σ 2 , then the third principal normal stress σ 3 is zero. Even for this situation,
shear stresses are in general present on all of the principal shear planes of Fig. 6.8. From Eq. 6.18,
for the case of σ z = σ 3 = 0, the principal shear stresses are

|σ 2 − σ 3 | |σ 2 | |σ 1 − σ 3 | |σ 1 | |σ 1 − σ 2 |
τ 1 = = , τ 2 = = , τ 3 = (6.21)
2 2 2 2 2

Thus, in a sense, there is no such thing as a state of plane stress, as stresses occur on planes
associated with choices of coordinate axes that are not in the x-y plane. Furthermore, it is hazardous