Page 247 - Mechanical Behavior of Materials

248 Chapter 6 Review of Complex and Principal States of Stress and Strain

of the principal normal stresses:

σ 3 = σ z = 40 MPa Ans.

Mohr’s circle may then be employed for the x-y plane just as for a two-dimensional problem.
The two ends of a diameter are

σ x , −τ xy = (100, −80), σ y ,τ xy = (−60, 80) MPa

The resulting circle is shown in Fig. E6.3(a).
Simple geometry, as in Ex. 6.2, is next needed to locate the ends of the horizontal diameter.
In particular, the center of the circle is located at a σ value of

σ x + σ y 100 − 60
a = = = 20 MPa
2 2
From the cross-hatched triangle, the radius of the circle is

2
2
r = 80 + 80 = 113.1MPa
This gives the two remaining principal normal stresses:
σ 1 ,σ 2 = a ± r = 133.1, −93.1MPa Ans.

y
2
τ, MPa
x-y plane
(also 1-2 plane) (b) 22.5° 1
x
(–60, 80)
z, 3
2θ n
0 80 σ, MPa τ
(–93.1, 0) (133.1, 0) (c)
20 80 113.1
(a) 113.1 46.6
(100, –80) 66.6
σ
(–93.1, 0) 0 (40, 0) (133.1, 0)

Figure E6.3 Mohr’s circle, principal axes, and principal stresses for the three-dimensional
state of stress example.