Transformation of Stress Tensor under Rotation of Axes

Let us consider two sets of co-ordinate axes $x_i$ and $x^\prime_i$ related by eqn. (2.7.4), that is,

$$\mathbf{e}^\prime_i = a_{ij}\mathbf{e}_j.$$ (2.7.4)

Consider a plane the outer normal to which points in the positive $x^\prime_1$ direction. Therefore, a unit normal $\mathbf{n}$ to this plane can be written as

$$\mathbf{n} = \mathbf{e}^\prime_1 = a_{1j}\mathbf{e}_j$$ (4.4.1)

or

$$n_i = a_{1i}.$$ (4.4.2)

The stress vector $\mathbf{f}^\prime$ on this plane is given by

$$f^\prime_i = T_{ij}n_j = T_{ij}a_{1j}.$$ (4.4.3)

Recalling the interpretations of various components $T_{11},\ T_{12}$ etc. of the stress tensor given after eqn. (4.1.3), we note that $T^\prime_{11},\ T^\prime_{21},\ T^\prime_{31}$ equal respectively, the components of $\mathbf{f}^\prime$ in the positive $x^\prime_1,\ x^\prime_2$, and $x^\prime_3$ directions. Thus

$$T^\prime_{11} = \mathbf{f}^\prime \cdot \mathbf{e}^\prime_1 = a_{1i} T_{ij}a_{1j},$$

$$T^\prime_{21} = \mathbf{f}^\prime \cdot \mathbf{e}^\prime_2 = a_{2i}T_{ij}a_{1j},$$ (4.4.4)

$$T^\prime_{31} = \mathbf{f}^\prime \cdot \mathbf{e}^\prime_3 = a_{3i}T_{ij}a_{1j}.$$

These three equations can be written as

$$T^\prime_{k1} = a_{ki}T_{ij}a_{1j}.$$ (4.4.5)

Similarly, now consider planes whose outer normals point in the positive $x^\prime_2,\ x^\prime_3$ directions and proceed the way it was done to arrive at (4.4.5). The result is

$$T^\prime_{k2} = a_{ki}T_{ij}a_{2j},$$ (4.4.6)

$$T^\prime_{k3} = a_{ki}T_{ij}a_{3j}.$$ (4.4.7)

Equations (4.4.5), (4.4.6) and (4.4.7) can be collectively written as

$$T^\prime_{kp} = a_{ki}T_{ij}a_{pj};\ [\mathbf{T}^\prime] = [\mathbf{a}][\mathbf{T}][\mathbf{a}]^T.$$ (4.4.8)

A comparison of this equation with eqn. (2.8.4) reveals that $\mathbf{T}$ is indeed a second order tensor.

An application of eqn. (4.4.8) is the problem of finding the normal and the shear stress on an oblique plane when the stress-state on horizontal and vertical planes in the plane-stress problem is known.

For the axes shown,

$$[\mathbf{a}] = \left[\begin{array}{ccc} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{array}\right].$$

Therefore

$$\left[\begin{array}{ccc}T^\prime_{11} & T^\prime_{12} & T^\prime_... ...& -\sin\theta & 0\\ \sin\theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right]$$

$$= \left[\begin{array}{ccc}T_{11}\cos^2\theta + T_{22}\sin^2 \thet... ...22}\cos^2\theta - 2T_{12}\sin\theta\cos\theta & 0\\ 0 & 0 & 0\end{array}\right]$$

Thus one can find $T^\prime_{11},\ T^\prime_{12}$, the normal and shear stresses on the oblique plane. Equations relating $T^\prime_{11},\ T^\prime_{12}$ to $T_{11},\ T_{12}$ are called ``stress transformation equations'' in the Mechanics of Deformable Bodies course.

Exercise: The stress matrix $\mathbf{T}$ referred to $x_1x_2x_3$-axes is shown below at the left (in ksi). New $x^\prime_1x^\prime_2x^\prime_3$-axes are chosen by rotating the unprimed axes as shown to the right.

$$[\mathbf{T}]= \left[\begin{array}{ccc} 50 & 37.5 & 0\\ 37.5 & ... ...-.1in} \multicolumn{4}{l}{\underline{\hspace{1.5in}}} \end{array}$$

a)

Determine the traction vectors on each of the new coordinate $(x^\prime_i)$ planes in terms of components referred to the old $(x_i)$ axes. For example, determine $\mathbf{f}$ on the $x^\prime_1$ plane in the form $\mathbf{f} = (-)\mathbf{e}_1 + (-)\mathbf{e}_2 + (-)\mathbf{e}_3$.

(b)

Now project each of the vectors obtained in (a) onto the three new coordinate axes, and verify that the nine new components thus obtained for the stress matrix $\mathbf{T}^\prime$ are the same as those given by the formulae for the transformation of a second-order tensor under the rotation of axes.

Exercise: The work $W$ done by external forces during the deformation of a continuous body is given by

$$W = \int\mathbf{f} \cdot \mathbf{u} da + \int\rho \mathbf{g} \cdot\mathbf{u} dv,$$

By substituting for $\mathbf{f}$ from eqn. (4.1.12), using the divergence theorem and assuming that the body is deformed quasi-statically (that is, the inertia force $\rho a_i$ is negligible), show that

$$W = \int T_{ij} u_{i,j} dv.$$

Hence show that for small deformations

$$W = \int T_{ij} e_{ij} dv.$$

Exercise: The power $P$ of external forces is defined as

$$P = \int \mathbf{f} \cdot\mathbf{v} da + \int \rho\mathbf{g} \cdot \mathbf{v} dv.$$

By substituting for $\mathbf{f}$ from eqn. (4.1.12), using the divergence theorem and Cauchy's first law of motion, show that

$$P = \dot K + \int T_{ij} D_{ij} dv$$

where

$$K = \frac{1}{2}\int \rho v_i v_i dv$$

is the kinetic energy of the body.