## 3

Note that J1, J2, and J3 can also be expressed in terms of I1, I2, and I3, or the three maximum deviatoric stresses, as follows:

J2 + 3^1 - 2VsP1 + sP2 + sP3) J3 — h + 31112 + 2713 — 1(sp1 + sp2 + SP3) — SP1SP2SP3

### 1.1.6 Maximum Shear Stresses

If the principal stresses are known, the maximum shear stresses that act on each of the three orthogonal planes, which bisect the angle between the principal planes with direction indices (n1 = n2 = ±1/P2, n3 = 0), (n1 = 0, n2 = ±1/P2, n3 = ±1/^2), (n1 = ±1/p2, n2 = 0, n3 = ±1/P2) with respect to the principal axes, are given by

Note that the planes (called principal shear planes) on which these stresses act are not pure shear planes. The corresponding normal stresses that act on these principal shear planes are (sP1 + sP2)/2, (sP2 + sP3)/2, and (sP1 + sP3)/2, respectively.

1.2 Strains

### 1.2.1 Strain Components

Corresponding to the six stress components described in the preceding section are six strain components. With reference to a Cartesian coordinate system with axes labeled 1, 2, and 3 as in Figure 1.1, these strains are denoted as e11, e22, e33, g12 = 2e12, g23 = 2e23, and g31 = 2e31. e11, e22, and e33 are called normal strains and g12, g23, and g31 are called shear strains. Using the definitions for engineering strains (Bathe 1982), normal strain is defined as the ratio of the change in length to the original length of a straight line element, and shear strain is defined as the change in angle (when the element is in a strained state) from an originally right angle (when the element is in an unstrained state).

### 1.2.2 Strain-Displacement Relationships

If we denote u, v, and w as the translational displacements in the 1, 2, and 3 (or x, y, and z) directions, respectively, then according to the small-displacement theory the six engineering strain components can be written in terms of these displacements as

0U 0V 0W

e11 = o"" , e22 = 3"" , 633 = o- , 0x 0y 0z

0u 0v 0v 0w 0w 0u g12 =2612 = qy+qx, g23 = 2623 = qz+ay, g31 = 2631 = gx+0z

1.2.3 Strain Analysis

Like stresses, strains can be transformed from one Cartesian coordinate system to another. This can be done by replacing a by e in Equation 1.9. In addition, one can calculate the three principal strains (eP1, eP2, and eP3) and the maximum shear strains (emax1, emax2, and emax3) acting on the three orthogonal planes by following the same procedure outlined in the preceding section for stresses simply by replacing all occurrences of a by e in Equations 1.12 and 1.24. However, it should be noted that except for isotropic elastic materials, principal planes for stresses and principal planes for strains do not necessarily coincide, nor do the planes of maximum shear stresses and maximum shear strains.

### 1.3 Equilibrium and Compatibility

By using an infinitesimal parallelepiped element subject to a system of positive three-dimensional stresses, equilibrium of the element requires that the following three equations relating the stresses be satisfied (Wang 1953; Timoshenko and Goodier 1970):

0aii 0a12 Qa13

9ff12 9ff22 9ff23

0x 0y 0z

oa13 oa23 oa33

H + ^ + B2 = 0 ox oy oz where Bx, By, and Bz are the body forces per unit volume acting in the 1, 2, and 3 (or x, y, and z) directions, respectively.

According to Equations 1.25, the six strain components can be expressed in terms of just three displacement variables (u, v, and w). To obtain a unique solution for the displacements for a given loading condition, these strains must be related. By manipulating Equations 1.25, it can be shown (Wang 1953; Timoshenko and Goodier 1970) that the strains are related by the following compatibility equations:

Q2en 82e22 _ 2 82e12

oy2 ox2 oxoy

Q2e22 Q2e33 _ 2 Q2e23 Qz2 Qy2 QyQz

Q2e11 Q^=2 Q2e13 Qz2 Qx2 QxQz

QyQz Qx2 QxQy QxQz

Q2e22 _ Q2e23 Q2e13 Q2e12 QxQz QxQy Qy2 QyQz

0^33 _ Q2e23 Q2e13 Q2e12 QxQy QxQz QyQz Qz2

Since Equations 1.27 were derived from Equations 1.25, they should not be regarded as an independent set of equations. The six stress components (a11, a22, a33, a12, a23, and a13), the six strain components (e11, e22, e33, e12, e23, and e13), and the three displacement components (u, v, and w) constitute a total of 15 unknowns, which cannot be solved using the three equilibrium equations (Equations 1.26) and the six compatibility equations (Equations 1.27). To do so, six additional equations are needed. These equations, which relate stresses with strains, are described in the next section.