Stress Tensor

Stress at a point

Cut the body by a plane through point $P$. If a small area $\Delta A$ around $P$ carries a resultant force $\Delta \mathbf{F}$, the stress vector is

$$\mathbf{t} = \lim_{\Delta A \to 0} \frac{\Delta \mathbf{F}}{\Delta A}$$

It depends on the orientation of the plane. The state of stress at $P$ is fully defined by the stresses on three mutually perpendicular planes, i.e. by the stress tensor.

Notation and sign convention

$\sigma_{ij}$: first subscript $i$ = direction of the normal to the face; second subscript $j$ = direction of the stress component. Normal stresses are written $\sigma_x, \sigma_y, \sigma_z$, shear stresses $\tau_{xy}$, etc.

$$[\sigma] = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_y & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_z \end{bmatrix}$$

• A face is positive if its outward normal points in a positive axis direction.
• A stress component is positive if it acts in the positive axis direction on a positive face, or in the negative direction on a negative face.
• So tension is positive and compression negative; positive shear points the same way as the axis on a positive face.

Proof of $\tau_{xy}=\tau_{yx}$

Take an element $dx \times dy \times dz$ in equilibrium under plane stress (no body moments).

• Right and left faces (area $dy\,dz$): shear $\tau_{xy}$ acts up on the right face and down on the left face, forming a couple $(\tau_{xy}\,dy\,dz)\,dx$ (anticlockwise).
• Top and bottom faces (area $dx\,dz$): shear $\tau_{yx}$ acts to the right on the top face and to the left on the bottom face, forming a couple $(\tau_{yx}\,dx\,dz)\,dy$ (clockwise).

Moments about the z-axis through the centre:

$$\begin{aligned} \sum M_z &= 0 \\ (\tau_{xy}\, dy\, dz)\, dx - (\tau_{yx}\, dx\, dz)\, dy &= 0 \\ \tau_{xy} &= \tau_{yx} \end{aligned}$$

In the same way $\tau_{yz}=\tau_{zy}$ and $\tau_{zx}=\tau_{xz}$. Hence the stress tensor is symmetric and has only six independent components, not nine.

Use the plane-stress transformation equations (Rajput, Timoshenko).

Given

$\sigma_x = 80$, $\sigma_y = -40$, $\tau_{xy} = 30$ MPa.

(a) Principal stresses

$$\begin{aligned} \sigma_{avg} &= \frac{\sigma_x+\sigma_y}{2} = \frac{80-40}{2} = 20.0\ \text{MPa} \\ R &= \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{60^2+30^2} = 67.08\ \text{MPa} \\ \sigma_{1,2} &= \sigma_{avg} \pm R = 87.08,\ -47.08\ \text{MPa} \end{aligned}$$

Orientation:

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{60}{120} = 0.5 \quad\Rightarrow\quad 2\theta_p = 26.57^\circ,\ \theta_p = 13.28^\circ$$

Substituting $\theta = 13.28^\circ$ in the $\sigma_n$ equation gives $87.08$ MPa, so this plane carries $\sigma_1$. The other principal plane is at $13.28 + 90 = 103.28^\circ$. Shear stress is zero on both.

(b) Maximum shear stress

$$\tau_{max} = R = 67.08\ \text{MPa}$$

It acts on planes at $\theta_s = \theta_p + 45^\circ = 58.28^\circ$, with a normal stress $\sigma_{avg} = 20$ MPa on those planes. (The out-of-plane principal stress is zero; $(\sigma_1 - 0)/2 = 43.5$ MPa is smaller, so the in-plane value governs.)

(c) Plane at $\theta = 30^\circ$

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 20 + 60\cos 60^\circ + 30 \sin 60^\circ = 20 + 30.00 + 25.98 = 75.98\ \text{MPa} \\[4pt] \tau_{n} &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -60\sin 60^\circ + 30\cos 60^\circ = -51.96 + 15.00 = -36.96\ \text{MPa} \end{aligned}$$

Check: the normal stress on the perpendicular plane ($120^\circ$) is $-35.98$ MPa, and $75.98 + (-35.98) = 40 = \sigma_x + \sigma_y$ (first invariant preserved).

Answer: $\sigma_1 = 87.1$ MPa, $\sigma_2 = -47.1$ MPa at $\theta_p = 13.3^\circ$; $\tau_{max} = 67.1$ MPa; on the $30^\circ$ plane $\sigma_n = 76.0$ MPa and $\tau = -37.0$ MPa.

Derivation

Consider a wedge cut from a plane-stress element. Let the inclined plane have its normal $n$ at angle $\theta$ (anticlockwise) from the x-axis and area $dA$. The faces on the x and y sides then have areas $dA\cos\theta$ and $dA\sin\theta$.

        y
        |   /  n (normal at angle theta)
   sigma_y  /
  +-----+/
  |     /|
  |    / |-- tau_xy
  +---+--+---- x

Resolve forces along the normal $n$:

$$\sigma_n\, dA = \sigma_x (dA\cos\theta)\cos\theta + \sigma_y (dA\sin\theta)\sin\theta + \tau_{xy}(dA\cos\theta)\sin\theta + \tau_{yx}(dA\sin\theta)\cos\theta$$

Using $\tau_{yx}=\tau_{xy}$ and double-angle identities:

$$\sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta$$

Resolving along the plane gives

$$\tau_{nt} = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta$$

Principal planes: $\tau_{nt}=0$ gives $\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$, and $\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$.

Mohr's circle

Squaring and adding the two equations eliminates $\theta$:

$$\left(\sigma_n - \sigma_{avg}\right)^2 + \tau_{nt}^2 = R^2$$

This is a circle with centre $(\sigma_{avg}, 0)$ and radius $R$ in the $\sigma$-$\tau$ plane.

Construction

1. Take $\sigma$ (tension positive) to the right and $\tau$ downward for clockwise-positive shear (or upward with the opposite convention; state the one used).
2. Plot $X(\sigma_x, \tau_{xy})$ and $Y(\sigma_y, -\tau_{xy})$.
3. Join $XY$; it cuts the $\sigma$-axis at the centre $C$.
4. Draw the circle with diameter $XY$.

   tau
    |        . S (tau max)
    |     .     .
    |   .         .
 ---+--+----C------+----- sigma
    | sigma2       sigma1
    |   .         .
    |     .     .
    |        . S'

Reading results

• Points where the circle cuts the $\sigma$-axis give $\sigma_1 = OC + R$ and $\sigma_2 = OC - R$.
• The highest and lowest points give $\tau_{max} = \pm R$, with normal stress $\sigma_{avg}$.
• The angle $XC$ to the $\sigma$-axis is $2\theta_p$: angles on the circle are twice the physical angles, and rotate in the same sense as the plane rotates.
• A point on the circle at $2\theta$ from $X$ gives $(\sigma_n, \tau_{nt})$ for the plane at $\theta$.

(a) Invariants and principal stresses

$$[\sigma] = \begin{bmatrix} 100 & 40 & 0 \\ 40 & -20 & 30 \\ 0 & 30 & 60 \end{bmatrix}\ \text{MPa}$$

The characteristic equation is $\sigma^3 - I_1\sigma^2 + I_2\sigma - I_3 = 0$.

$$\begin{aligned} I_1 &= \sigma_x+\sigma_y+\sigma_z = 100 - 20 + 60 = 140 \\ I_2 &= (\sigma_x\sigma_y - \tau_{xy}^2) + (\sigma_y\sigma_z - \tau_{yz}^2) + (\sigma_z\sigma_x - \tau_{zx}^2) \\ &= (-2000-1600) + (-1200-900) + (6000-0) = 300 \\ I_3 &= \det[\sigma] = 100(-1200-900) - 40(2400-0) + 0 = -306000 \end{aligned}$$

So $\sigma^3 - 140\sigma^2 + 300\sigma + 306000 = 0$.

Solving (trigonometric method or trial and error):

$$\sigma_1 = 113.68,\quad \sigma_2 = 66.68,\quad \sigma_3 = -40.37\ \text{MPa}$$

Check: sum $= 140.0 = I_1$; product $= -306000 = I_3$.

Absolute maximum shear stress:

$$\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{113.68 - (-40.37)}{2} = 77.03\ \text{MPa}$$

(b) Plane with $n = (2/3, 1/3, 2/3)$

Cauchy's formula $\mathbf{t} = [\sigma]\mathbf{n}$:

$$\begin{aligned} t_x &= 100\left(\tfrac23\right) + 40\left(\tfrac13\right) + 0 = 80.00 \\ t_y &= 40\left(\tfrac23\right) - 20\left(\tfrac13\right) + 30\left(\tfrac23\right) = 40.00 \\ t_z &= 0 + 30\left(\tfrac13\right) + 60\left(\tfrac23\right) = 50.00 \end{aligned}$$ $$\begin{aligned} \sigma_n &= \mathbf{t}\cdot\mathbf{n} = 80\left(\tfrac23\right) + 40\left(\tfrac13\right) + 50\left(\tfrac23\right) = 100.00\ \text{MPa} \\ |\mathbf{t}| &= \sqrt{80^2+40^2+50^2} = 102.47\ \text{MPa} \\ \tau &= \sqrt{|\mathbf{t}|^2 - \sigma_n^2} = \sqrt{10500 - 10000} = 22.36\ \text{MPa} \end{aligned}$$

Answer: $I_1 = 140$, $I_2 = 300$, $I_3 = -306000$ (MPa units); $\sigma_1 = 113.7$, $\sigma_2 = 66.7$, $\sigma_3 = -40.4$ MPa; $\tau_{max} = 77.0$ MPa; on the given plane $\sigma_n = 100.0$ MPa and $\tau = 22.4$ MPa.

Cauchy's formula

Take a tetrahedron at a point with three faces on the coordinate planes and a fourth inclined face of area $dA$ and unit normal $\mathbf{n} = (l, m, n)$ (direction cosines). The coordinate-plane faces then have areas

$$dA_x = l\,dA,\quad dA_y = m\,dA,\quad dA_z = n\,dA$$

Let $\mathbf{t} = (t_x, t_y, t_z)$ be the traction (stress vector) on the inclined face. The faces on the coordinate planes have outward normals along $-x, -y, -z$, so their stresses act opposite to the positive components.

        z
        |  C
        | /|\
        |/ | \
        A--+--B  (inclined face ABC)
       /  O
      y       x

Force equilibrium in the x-direction (body force and inertia are of higher order, since volume $\propto h^3$ while area $\propto h^2$ as the tetrahedron shrinks):

$$\begin{aligned} t_x\, dA - \sigma_x (l\,dA) - \tau_{yx}(m\,dA) - \tau_{zx}(n\,dA) &= 0 \\ t_x &= \sigma_x l + \tau_{yx} m + \tau_{zx} n \end{aligned}$$

Similarly for $y$ and $z$. In matrix form (symmetric tensor):

$$\begin{Bmatrix} t_x \\ t_y \\ t_z \end{Bmatrix} = \begin{bmatrix} \sigma_x & \tau_{xy} & \tau_{xz} \\ \tau_{xy} & \sigma_y & \tau_{yz} \\ \tau_{xz} & \tau_{yz} & \sigma_z \end{bmatrix} \begin{Bmatrix} l \\ m \\ n \end{Bmatrix}$$

or $t_i = \sigma_{ij} n_j$. The normal and shear stress on the plane are then

$$\sigma_n = \mathbf{t}\cdot\mathbf{n},\qquad \tau = \sqrt{|\mathbf{t}|^2 - \sigma_n^2}$$

So the six components of $\sigma_{ij}$ determine the traction on every plane through the point.

Equilibrium equations

For a body force per unit volume $(X, Y, Z)$:

$$\begin{aligned} \frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau_{xy}}{\partial y} + \frac{\partial \tau_{xz}}{\partial z} + X &= 0 \\ \frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_y}{\partial y} + \frac{\partial \tau_{yz}}{\partial z} + Y &= 0 \\ \frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_z}{\partial z} + Z &= 0 \end{aligned}$$

Moment equilibrium gives the symmetry $\tau_{ij}=\tau_{ji}$. At the boundary, the traction must satisfy $t_i = \sigma_{ij}n_j$ equal to the applied surface load.