Principal Stresses. Maximum Shear Stress

In stress analysis problems one is interested in finding the maximum and the minimum normal stress and the maximum shearing stress at a point. In the following discussion, the point $\mathbf{x}$ in the present configuration is kept fixed.

At a point $\mathbf{x}$ in the present configuration of the body, consider a plane whose outer unit normal is $\mathbf{n}$. The stress vector $\mathbf{f}$ on this plane given by equation (4.1.12) is $T_{ij}n_j$. The normal stress on this plane is given by

$$T_{\rm normal} = \mathbf{n} \cdot \mathbf{f} = n_iT_{ij}n_j.$$ (4.5.1)

Thus the problem of finding the maximum or the minimum normal stress at the point $\mathbf{x}$ reduces to finding the unit vector $\mathbf{n}$ for which $T_{\rm normal}$ is maximum or minimum. Using the method of Lagrange multiplier, the problem becomes that of finding extreme values of the function

$$F(\mathbf{n}) = n_iT_{ij}n_j - \lambda (n_in_i - 1)$$ (4.5.2)

in which $\lambda$ is an arbitrary scalar. The vector $\mathbf{n}$ which makes $F$ assume a maximum or minimum value is given by

$$\frac{\partial F}{\partial n_i} = 0,\ \frac{\partial F}{\partial\lambda} = 0.$$ (4.5.3)

That is

$$(T_{ij} - \lambda \delta_{ij}) n_j = 0,\ n_in_i - 1 = 0.$$ (4.5.4)

A nontrivial solution of eqn. (4.5.4)$_1$ exists if and only if

$${\rm det}\ [T_{ij} - \lambda\delta_{ij}] = 0$$ (4.5.5)

which gives the cubic equation

$$\lambda^3 - I_T\lambda^2 + II_T\lambda - III_T = 0.$$ (4.5.6)

Here $I_T,\ II_T$ and $III_T$ are the principal invariants of $T_{ij}$. Since $T_{ij}$ is symmetric, therefore, eqn. (4.5.6) has three real eigenvalues $\lambda^{(1)},\lambda^{(2)},\lambda^{(3)}$. Corresponding to each of these eigenvalues we can find a unit vector $\mathbf{n}$ from eqn. (4.5.4). For example, $\mathbf{n}^{(1)}$ corresponding to $\lambda^{(1)}$ is given by

$$(T_{ij} - \lambda^{(1)}\delta_{ij})n^{(1)}_j = 0,\ n^{(1)}_j n^{(1)}_j = 1.$$ (4.5.7)

$\mathbf{n}^{(1)},\ \mathbf{n}^{(2)},\ \mathbf{n}^{(3)}$ are normals to planes on which $T_{\rm normal}$ assumes extreme values. Since

$$T^{(1)}_{\rm normal} = n^{(1)}_iT_{ij}n^{(1)}_j = n^{(1)}_i\lambda^{(1)}\delta_{ij}n^{(1)}_j = \lambda^{(1)},$$ (4.5.8)

therefore $\lambda^{(1)},\ \lambda^{(2)}$ and $\lambda^{(3)}$ are extreme values of normal stresses. Also the stress vector on the plane with outer unit normal $\mathbf{n}^{(1)}$ is given by

$$f_i = T_{ij}n^{(1)}_j = \lambda^{(1)}\delta_{ij}n^{(1)}_j = \lambda^{(1)}n^{(1)}_i.$$ (4.5.9)

Let $\mathbf{e}$ a unit vector in this plane. Then the shear stress in the direction of $\mathbf{e}$ on the $\mathbf{n}^{(1)}$ plane is given by

$$T_{\rm shear} = \mathbf{f}\cdot \mathbf{e} = \lambda^{(1)} e_in^{(1)}_i = 0.$$ (4.5.10)

The plane on which the shear stress is zero is called a principal plane and the normal stress on a principal plane is called a principal stress. Thus $\lambda^{(1)},\ \lambda^{(2)}$ and $\lambda^{(3)}$ are principal stresses and $\mathbf{n}^{(1)},\ \mathbf{n}^{(2)}$ and $\mathbf{n}^{(3)}$ are normals to principal planes. $\mathbf{n}^{(1)},\ \mathbf{n}^{(2)}$ and $\mathbf{n}^{(3)}$ are called principal axes of the stress.

Whenever $\lambda^{(1)}\ne \lambda^{(2)}\ne \lambda^{(3)},\ \mathbf{n}^{(1)},\ \mathbf{n}^{(2)}$ and $\mathbf{n}^{(3)}$ are uniquely determined and are mutually orthogonal. However, when any two or all three of the principal stresses are equal, then $\mathbf{n}^{(1)},\ \mathbf{n}^{(2)}$ and $\mathbf{n}^{(3)}$ are not uniquely determined but can still be found so that they are mutually orthogonal. Henceforth, in this section, we will assume that the principal axes of the stress tensor are mutually orthogonal.

Taking $x_1,x_2,x_3$ axes along the principal axes of the stress, we see that with respect to these axes $T_{ij}$ has the form

$$[T_{ij}] = \left[\begin{array}{ccc} \lambda^{(1)} & 0 & 0\\ 0 & \lambda^{(2)} & 0\\ 0 & 0 & \lambda^{(3)}\end{array}\right].$$ (4.5.11)

We now find the plane of the maximum shear stress. Let an outer unit normal to this plane be $\mathbf{n}$. Then, taking principal axes of stress as the coordinate axes,

$$f_i = T_{ij}n_j = \lambda^{(1)}n_1\delta_{i1} + \lambda^{(2)}n_2\delta_{i2} + \lambda^{(3)}n_3\delta_{i3}$$ (4.5.12)

gives the traction on this plane. Therefore the shear stress on this plane is given by

$$(T_{\rm shear})^2 =\ f_if_i - (f_in_i)^2,$$

$$=\ \lambda^{(1)^2}n^2_1 + \lambda^{(2)^2}n^2_2 + \lambda^{(3)^2} n^2_3 - (\lambda^{(1)}n^2_1 + \lambda^{(2)}n^2_2 + \lambda^{(3)}n^2_3)^2.$$ (4.5.13)

Thus the problem of determining maximum $T_{\rm shear}$ reduces to that of finding extreme values of the function

$$G(\mathbf{n}) = T^2_{\rm shear} + \mu (n_in_i - 1)$$ (4.5.14)

in which $\mu$ is a Lagrange multiplier. The necessary conditions are

$$\frac{\partial G}{\partial n_i} = 0,\ \frac{\partial G}{\partial \mu} = 0,$$

which are equivalent to

$$\lambda^{(i)^2}n_i - (\lambda^{(1)}n^2_1 + \lambda^{(2)}n^2_2 + \lambda^{(3)}n^2_3) (2\lambda^{(i)}_i) n_i + \mu n_i = 0\ \ \ ( \ i)\ ,$$ (4.5.15)

$$n_in_i = 1\ .$$ (4.5.16)

Three equations given by (4.5.15) for $i = 1,2$ and $3$ and the eqn. (4.5.16) determine $n_i$ and $\mu$. A solution of these equations for which two out of $n_1,\ n_2$ and $n_3$ are zero is not interesting since this corresponds to finding $T_{\rm shear}$ on a principal plane which is zero. The solution of (4.5.15) and (4.5.16) for which $n_1 \ne n_2\ne n_3\ne 0$ exists only if $\lambda^{(1)} = \lambda^{(2)} = \lambda^{(3)}$. In this case too, eqn. (4.5.13) gives $T_{\rm shear} = 0$.

The remaining possibility is that one out of $n_1,\ n_2$ and $n_3$ be zero, say $n_1 = 0$, $n_2\ne 0\ne n_3$. For this case, eqn. (4.5.15) for $i = 1$ is satisfied identically. For $i = 2$ and $3$ we obtain

$$\lambda^{(2)^2} n_2 - (\lambda^{(2)}n^2_2 + \lambda^{(3)} n^2_3)2\lambda^{(2)} n_2 + \mu n_2 = 0,$$

$$\lambda^{(3)^2} n_3 - (\lambda^{(2)} n^2_2 + \lambda^{(3)} n^2_3)2\lambda^{(3)}n_3 + \mu n_3 = 0,$$

and, of course,

$$n^2_2 + n^2_3 = 1.$$

A solution of these is

$$n_2 = n_3 = \pm \frac{1}{\sqrt 2}\ .$$ (4.5.17)

For the values of $n_2$ and $n_3$ given by (4.5.17) and $n_1 = 0$, eqn. (4.5.13) gives

$$T^2_{\rm shear} = \frac{1}{4} (\lambda^{(2)} - \lambda^{(3)})^2,$$

and hence

$$T_{\rm shear} = \pm \frac{1}{2}(\lambda^{(2)} - \lambda^{(3)}).$$ (4.5.18)

Similarly for $n_2 = 0$, a solution of (4.5.15) and (4.5.16) is $n_1 = n_3 = \pm \frac{1}{\sqrt 2}$ and the shear stress on this plane is

$$T_{\rm shear} = \pm \frac{1}{2}(\lambda^{(1)} - \lambda^{(3)}).$$ (4.5.19)

For $n_3 = 0$, a solution of (4.5.15) and (4.5.16) is $n_1 = n_2 = \pm \frac{1}{\sqrt 2}$ and the shear stress on this plane is

$$T_{\rm shear} = \pm \frac{1}{2}(\lambda^{(1)} - \lambda^{(2)}).$$ (4.5.20)

Thus the extreme values of the shear stress are given by (4.5.18), (4.5.19) and (4.5.20). Note that on planes of maximum shear stress, the normal stress need not be zero. In fact, the normal stress on the plane for which

$$\mathbf{n} = \left( 0,\ \frac{1}{\sqrt 2},\ \frac{1}{\sqrt 2}\right) \ {\rm is}\ \frac{1}{2} (\lambda^{(2)} + \lambda^{(3)} ).$$