Page 259 - Mechanical Behavior of Materials

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

Combining these results with Eq. 6.30 gives

2
2
2
(2.414m 1 ) + m + 0 = 1, m 1 = 0.383
1
The value of l 1 is then easily obtained, so that the three values are

l 1 = 0.924, m 1 = 0.383, n 1 = 0 Ans.

Similar solutions for σ 2 =−93.1 and for σ 3 = 40 MPa give the direction cosines for the other
two principal axes:

l 2 =−0.383, m 2 = 0.924, n 2 = 0
Ans.
l 3 = 0, m 3 = 0, n 3 = 1

The right-handedness of the 1-2-3 system of the direction cosines needs to be checked by
using Eq. 6.31. We have

i j k i j k

l 1 m 1
n 1 = l 3 i + m 3 j + n 3 k, 0.924 0.383 0 = 1.000k

l 2 m 2 n 2
−0.383 0.924 0
where i, j, k are unit vectors for the x, y, z directions, respectively, and the cross product is done
in determinate form. The l 3 , m 3 , n 3 direction cosines from the preceding analysis are confirmed,
and no sign changes are needed.
Comments From Fig. E6.3(b), the angles between the principal axes and the x, y, z axes are
as follows:
◦
1-axis: θ x = 22.5 , θ y = 67.5 , θ z = 90 ◦
◦
◦
2-axis: θ x = 112.5 ,θ y = 22.5 , θ z = 90 ◦
◦
◦
◦
3-axis: θ x = 90 , θ y = 90 , θ z = 0 ◦
The cosines of these angles are seen to agree with the values of the direction cosines that
have just been found. For a state of stress with no zero components in the original x -y -z system,
◦
there would be no 90 or 0 angles, so none of the direction cosines would be zero or unity.
◦
6.5 STRESSES ON THE OCTAHEDRAL PLANES
Consider an oblique plane oriented relative to the 1-2-3 principal axes, as shown in Fig. 6.15(a). A
normal stress σ and a shear stress τ act on this plane. The direction of the normal to the oblique
plane is specified by the angles α, β, and γ to the principal axes.