弹性波动力学笔记(八) 应力张量极值点分析

3.8 Stationary values and directions of the normal and shearing stress vectors

Given a stress tensor \(\tau_{ij}\) and a point P within the stressed medium, there are an infinite number of stress vector vectors, each of them associate with one of the infinitely many surface elements through \(P\). Each surface element has an associated normal vector and shear vector \(\mathbf{n}\) . and now the question we want to ask is whether there are special directions along which the absolute values of the normal and shearing stress vectors have maximum and or minimum values. Therefore, the problem is to find the vector or vectors \(\mathbf{n}\) that render \(\mathbf{T}^{\mathbf{N}}\) and \(\mathbf{T}^{\mathbf{S}}\) stationary. The only restrictions on \(\mathbf{n}\) is that absolute its absolute value is equal to \(1\) . There will be no restrictions on \(\tau_{ij}\) except for two special cases discussed at the end of this section. The analysis will be carried out in the principal axes system, as in this system all the equations are simpler. Working in this preferred system does not introduce any restriction; it is always possible to go back to the original coordinate system by applying appropriate rotations. This section is based on Sokolnikoff (1956).

Let assume direction unit vector in principal stress-axes system of \(\tau_{ij}\)

\[\mathbf{n}=n_1\mathbf{a}_1+n_2\mathbf{a}_2+n_3\mathbf{a}_3 \tag{3.8.1} \]

be the unit normal to a plane through some point \(P\) inside the medium. Then using equation (3.7.1) and (3.5.3)

\[\mathbf{T}\cdot\mathbf{n}=\tau_{N}=(n_1\tau_1,n_2\tau_2,n_3\tau_3)\cdot(n_1,n_2,n_3)=\tau_1n_1^2+\tau_2n_2^2+\tau_3n_3^2 \tag{3.8.2} \]

Now use the relation

\[|\mathbf{n}|^2=n_1^2+n_2^2+n_3^2=1 \tag{3.8.3} \]

to solve for \(n_1^2\) ,which, after substitution in equation (3.8.2) gives

\[ \tau_N=\tau_1(1-n_1^2-n_2^2)+\tau_2n_2^2+\tau_3n_3^2 \tag{3.8.4} \]

We are interested in determining the maximum or minimum values of \(\tau_{N}\) . To do that we will determine the value of \(n_2\) and \(n_3\) for which \(\tau_N\) is stationary. In other words, we are interested in the values of \(n_2\) and \(n_3\) that satisfy the following conditions:

\[ \frac{\part{\tau_N}}{\part{n_2}}=0,\space \frac{\part{\tau_N}}{\part{n_3}}=0 \tag{3.8.5} \]

This gives

\[\begin{equation} \begin{aligned} &\frac{\part{\tau_{N}}}{\part{n_2}}=-2n_2\tau_1+2n_2\tau_2=0\\ &\frac{\part{\tau_{N}}}{\part{n_1}}=-2n_3\tau_1+2n_3\tau_3=0 \end{aligned} \tag{3.8.6} \end{equation} \]

Equations (3.8.6) imply \(n_2=n_3=0\), so that

\[\mathbf{n}=(\pm1,0,0); \space \tau_N=\tau_1 \tag{3.8.7} \]

using (3.8.2) and (3.8.3) for \(n_2\) and \(n_3\), and repeating the previous steps gives

\[\mathbf{n}=(0,\pm1,0); \space \tau_N=\tau_2 \tag{3.8.8} \]

and

\[\mathbf{n}=(0,0,\pm1);\space \tau_N=\tau_3 \tag{3.8.9} \]

Equations (3.8.7)-(3.8.9) show that the principal directions are those along which normal stresses are stationary.

Now we will determine the stationary values of the shearing stresses. From equation(3.7.3) (stress vector decomposition into \(\tau_N\) and \(\tau_S\)) we know that

\[|\mathbf{T}|^2=|\mathbf{T^N}|^2+|\mathbf{T^S}|^2 \tag{3.8.10} \]

Therefore, using Cauchy's Stress Law and equation (3.8.2), we can gain

\[|\mathbf{T^S}|^2=\tau_S^2=|\mathbf{T}|^2-|\mathbf{T^N}|^2=\tau_1^2n_1^2+\tau_2^2n_2^2+\tau_3^2n_3^2-(\tau_1n_1^2+\tau_2n_2^2+\tau_3n_3^2)^2 \tag{3.8.11} \]

Equation (3.8.11) is subject to the condition that \(|\mathbf{n}^2|=n_in_i=1\) . To find stationary values of \(\tau_S^2\) the method of Lagrange multipliers will be used. To do that let

\[\phi=n_in_i-1\tag{3.8.12} \]

and determine the stationary values of the new objective function

\[F=\tau_S^2+\lambda\phi \tag{3.8.13} \]

where \(\lambda\) is an unknown scalar to be determined. Now set \({\part{F}}/{\part{n_i}}=0\) and use \(|\mathbf{n}|=1\). This gives

\[\frac{\part{F}}{\part{n_i}}=2n_i\left[ \tau_i^2-2\left(\tau_1n_1^2+\tau_2n_2^2+\tau_2n_3^2\right)+\lambda\right]=0\tag{3.8.14} \]

We will solve equation (3.8.14) by examining a number of possibilities. To do that note that (3.8.14) can be written as

\[n_if(n_i,\tau_i,\lambda)=0 \tag{3.8.15} \]

with \(f\) representing the expression within brackets. An obvious solution is \(\mathbf{n}=(0,0,0)\) , but it does not satisfy the constraint \(\mathbf{n^2}=1\) .

Let us investigate the possibility that two of the \(n_i\) are equal to zero. Assume that \(\mathbf{n}=(\pm1,0,0)\). Replacing this expression in equation (3.8.15) and (3.8.12) gives \(\lambda=\tau_1^2\) and \(\tau_S^2=0\). The latter is an expected result because \(\mathbf{n}=(\pm1,0,0)\) is normal to one of the principal planes of stress, on which \(\tau_S\) is known to be zero.

Similar results are obtained when \(\mathbf{n}=(0,\pm1,0)\) or \(\mathbf{n}=(0,0,\pm1)\) .

Another possibility is that only one of the \(n_i\) is equal to zero. Let us assume that \(n_1=0\). In this case we find

\[\begin{equation} \begin{aligned} n_2^2+n_3^2&=1\\ \tau_2^2-2(\tau_2n_2^2+\tau_3n_3^2)+\lambda&=0\\ \tau_3^2-2(\tau_2n_2^2+\tau_3n_3^2)+\lambda&=0 \end{aligned} \tag{3.8.16} \end{equation} \]

Equation (3.8.16) represent a system of three equations and three unknowns variables (\(n_2^2,n_3^2,\lambda\)). The Solution is

\[\begin{aligned} n_3^2=n_2^2&=\frac{1}{2};\\ \lambda&=\tau_2\tau_3 \end{aligned} \tag{3.8.17} \]

Then we can obtain vector \(\mathbf{n}\)

\[\mathbf{n}=\left(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\right) \tag{3.8.18} \]

and from equation (3.8.11)

\[\begin{equation} \begin{aligned} \tau_S^2&=\frac{1}{2}(\tau_2^2+\tau_{2}{3})-\frac{1}{4}(\tau_2+\tau_3)^2 \\&=\frac{1}{4}\tau_2^2+\frac{1}{4}\tau_3^2-\frac{1}{2}\tau_2\tau_3\\ &=\frac{1}{2}(\tau_2-\tau_3)^2 \end{aligned} \tag{3.8.19} \end{equation} \]

so that

\[\tau_S=\pm\frac{1}{2}(\tau_2-\tau_3) \tag{3.8.20} \]

Since \(\tau_{S}\) and \(\tau_S^2\) are not functions of \(\mathbf{n}\), their derivatives with respect to \(n_i\) are equal to zero. Therefore, \(\mathbf{n}\) given by (3.8.18) represents a stationary direction.

If instead of \(n_1=0\) we had chosen \(n_2=0\) or \(n_3=0\), the corresponding results would have been:

\[\mathbf{n}=\left(\pm\frac{1}{\sqrt{2}},0,\pm\frac{1}{\sqrt{2}}\right); |\tau_S|=\frac{1}{2}|\tau_1-\tau_3| \tag{3.8.21} \]

\[\mathbf{n}=\left(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}},0\right); |\tau_S|=\frac{1}{2}|\tau_1-\tau_2| \tag{3.8.22} \]

If now we assume that \(\tau_1>\tau_2>\tau_3\), then \((\tau_1-\tau_3)/2\) is the greatest shearing stress at \(P\). This stress is one-half of the difference between the greatest and least principal stresses. It acts on plane that contain \(\mathbf{a}_2\) and that bisect the right angles between the directions of two principal stresses.