- 1.1 Introduction
- 1.2 Scope of Treatment
- 1.3 Analysis and Design
- 1.4 Conditions of Equilibrium
- 1.5 Definition and Components of Stress
- 1.6 Internal Force-Resultant and Stress Relations
- 1.7 Stresses on Inclined Sections
- 1.8 Variation of Stress Within a Body
- 1.9 Plane-Stress Transformation
- 1.10 Principal Stresses and Maximum in-plane Shear Stress
- 1.11 Mohr's Circle for Two-Dimensional Stress
- 1.12 Three-Dimensional Stress Transformation
- 1.13 Principal Stresses in Three Dimensions
- 1.14 Normal and Shear Stresses on an Oblique Plane
- 1.15 Mohr's Circles in Three Dimensions
- 1.16 Boundary Conditions in Terms of Surface Forces
- 1.17 Indicial Notation
1.13 Principal Stresses in Three Dimensions
For the three-dimensional case, it is now demonstrated that three planes of zero shear stress exist, that these planes are mutually perpendicular, and that on these planes the normal stresses have maximum or minimum values. As has been discussed, these normal stresses are referred to as principal stresses, usually denoted s 1, s 2, and s 3. The algebraically largest stress is represented by s 1, and the smallest by s 3: s 1 > s 2 > s 3.
We begin by again considering an oblique x' plane. The normal stress acting on this plane is given by Eq. (1.28a):
The problem at hand is the determination of extreme or stationary values of s x' . To accomplish this, we examine the variation of s x' relative to the direction cosines. Inasmuch as l, m, and n are not independent, but connected by l 2 + m 2 + n 2 = 1, only l and m may be regarded as independent variables. Thus,
Differentiating Eq. (a) as indicated by Eqs. (b) in terms of the quantities in Eq. (1.26), we obtain
From n 2 = 1 – l 2 – m 2, we have n/l = –l/n and n/m = –m/n. Introducing these into Eq. (c), the following relationships between the components of p and n are determined:
These proportionalities indicate that the stress resultant must be parallel to the unit normal and therefore contains no shear component. It is concluded that, on a plane for which s x' has an extreme or principal value, a principal plane, the shearing stress vanishes.
It is now shown that three principal stresses and three principal planes exist. Denoting the principal stresses by s p , Eq. (d) may be written as
These expressions, together with Eq. (1.26), lead to
A nontrivial solution for the direction cosines requires that the characteristic determinant vanish:
Expanding Eq. (1.32) leads to
The three roots of the stress cubic equation (1.33) are the principal stresses, corresponding to which are three sets of direction cosines, which establish the relationship of the principal planes to the origin of the nonprincipal axes. The principal stresses are the characteristic values or eigenvalues of the stress tensor t ij . Since the stress tensor is a symmetric tensor whose elements are all real, it has real eigenvalues. That is, the three principal stresses are real [Refs. 1.8 and 1.9]. The direction cosines l, m, and n are the eigenvectors of t ij .
It is clear that the principal stresses are independent of the orientation of the original coordinate system. It follows from Eq. (1.33) that the coefficients I 1, I 2, and I 3 must likewise be independent of x, y, and z, since otherwise the principal stresses would change. For example, we can demonstrate that adding the expressions for s x' , s y' , and s z' given by Eq. (1.28) and making use of Eq. (1.30a) leads to I 1 = s x' + s y' + s z' = s x + s y + s z . Thus, the coefficients I 1, I 2, and I 3 represent three invariants of the stress tensor in three dimensions or, briefly, the stress invariants. For plane stress, it is a simple matter to show that the following quantities are invariant (Prob. 1.27):
Equations (1.34) and (1.35) are particularly helpful in checking the results of a stress transformation, as illustrated in Example 1.7.
If now one of the principal stresses, say s 1 obtained from Eq. (1.33), is substituted into Eq. (1.31), the resulting expressions, together with l 2 + m 2 + n 2 = 1, provide enough information to solve for the direction cosines, thus specifying the orientation of s 1 relative to the xyz system. The direction cosines of s 2 and s 3 are similarly obtained. A convenient way of determining the roots of the stress cubic equation and solving for the direction cosines is presented in Appendix B, where a related computer program is also included (see Table B.1).
Example 1.6. Three-Dimensional Stress in a Hub
A steel shaft is to be force fitted into a fixed-ended cast-iron hub. The shaft is subjected to a bending moment M, a torque T, and a vertical force P, Fig. 1.20a. Suppose that at a point Q in the hub, the stress field is as shown in Fig. 1.20b, represented by the matrix
Figure 1.20 Example 1.6. (a) Hub-shaft assembly. (b) Element in three-dimensional stress.
Determine the principal stresses and their orientation with respect to the original coordinate system.
Substituting the given stresses into Eq. (1.33) we obtain from Eqs. (B.2)
s 1 = 11.618 MPa,
s 2 = –9.001 MPa,
s 3 = –25.316 MPa
Successive introduction of these values into Eq. (1.31), together with Eq. (1.30a), or application of Eqs. (B.6) yields the direction cosines that define the orientation of the planes on which s 1, s 2, and s 3 act:
l 1 = 0.0266,
l 2 = –0.6209,
l 3 = 0.7834
m 1 = –0.8638,
m 2 = 0.3802,
m 3 = 0.3306
n 1 = –0.5031,
n 2 = –0.6855,
n 3 = –0.5262
Note that the directions of the principal stresses are seldom required for purposes of predicting the behavior of structural members.
Example 1.7. Three-Dimensional Stress in a Machine Component
The stress tensor at a point in a machine element with respect to a Cartesian coordinate system is given by the following array:
Determine the state of stress and I 1, I 2, and I 3 for an x', y', z' coordinate system defined by rotating x, y through an angle of q = 45° counterclockwise about the z axis (Fig. 1.21a).
Figure 1.21 Example 1.7. Direction cosines for q = 45°.
The direction cosines corresponding to the prescribed rotation of axes are given in Fig. 1.21b. Thus, through the use of Eq. (1.28) we obtain
It is seen that the arrays (f) and (g), when substituted into Eq. (1.34), both yield I 1 = 100 MPa, I 2 = 1400 (MPa)2, and I 3 = –53,000 (MPa)3, and the invariance of I 1, I 2, and I 3 under the orthogonal transformation is confirmed.