Transformation of stress and strain

Sign convention

Tensile normal stress positive. Shear stress $\tau_{xy}$ is positive when it acts in the $+y$ direction on the face whose outward normal is $+x$. Angle $\theta$ is measured anticlockwise from the $x$ axis to the normal of the inclined plane.

        sy (top face)
      ^   ^
   +-----------+
   |           |  txy on the +x face points up (+y)
 <-|           |->  sx
   |           |
   +-----------+
      sy (bottom face)

Equilibrium of a wedge

Take a triangular wedge with the inclined face of area $dA$, normal at $\theta$. The areas of the vertical face and horizontal face are $dA\cos\theta$ and $dA\sin\theta$.

Forces along the normal $n$ ($\sum F_n = 0$):

$$\sigma_\theta\,dA = \sigma_x\cos^2\theta\,dA + \sigma_y\sin^2\theta\,dA + 2\tau_{xy}\sin\theta\cos\theta\,dA$$

Forces along the plane ($\sum F_t = 0$):

$$\tau_\theta\,dA = -(\sigma_x-\sigma_y)\sin\theta\cos\theta\,dA + \tau_{xy}(\cos^2\theta-\sin^2\theta)\,dA$$

Using double-angle identities:

$$\boxed{\sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta}$$ $$\boxed{\tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta}$$

Principal stresses

Principal planes have $\tau_\theta = 0$:

$$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y}$$

This gives two values of $2\theta_p$ that differ by $180^\circ$, so the two principal planes are $90^\circ$ apart. Substituting back:

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$$

(Equivalently $d\sigma_\theta/d\theta = 0$.)

Maximum shear stress

Setting $d\tau_\theta/d\theta = 0$:

$$\tan2\theta_s = -\frac{\sigma_x-\sigma_y}{2\tau_{xy}}$$

so $\theta_s = \theta_p \pm 45^\circ$ (planes of maximum shear are at $45^\circ$ to the principal planes).

$$\tau_{max} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = \frac{\sigma_1-\sigma_2}{2}$$

On these planes the normal stress is $\sigma_{avg} = (\sigma_x+\sigma_y)/2$.

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

$$\frac{\sigma_x+\sigma_y}{2} = 20\ \text{MPa}, \qquad \frac{\sigma_x-\sigma_y}{2} = 60\ \text{MPa}$$

(a) Stresses on the 30 degree plane

$2\theta = 60^\circ$, $\cos60^\circ = 0.5$, $\sin60^\circ = 0.866$.

$$\sigma_\theta = 20 + 60(0.5) + 30(0.866) = 20 + 30 + 25.98 = 76.0\ \text{MPa}$$ $$\tau_\theta = -60(0.866) + 30(0.5) = -51.96 + 15 = -37.0\ \text{MPa}$$

The negative sign means the shear acts opposite to the positive direction (clockwise sense on this face).

Check on the perpendicular plane ($\theta = 120^\circ$): $\sigma = 20 - 30 - 25.98 = -36.0$ MPa. The sum $76.0 - 36.0 = 40 = \sigma_x+\sigma_y$ confirms the calculation.

(b) Principal stresses

$$R = \sqrt{60^2 + 30^2} = \sqrt{4500} = 67.08\ \text{MPa}$$ $$\sigma_1 = 20 + 67.08 = 87.1\ \text{MPa (tension)}, \qquad \sigma_2 = 20 - 67.08 = -47.1\ \text{MPa (compression)}$$

Principal planes:

$$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{60}{120} = 0.5 \;\Rightarrow\; 2\theta_p = 26.57^\circ,\ \theta_p = 13.28^\circ$$

The plane at $13.28^\circ$ anticlockwise from the $x$ axis carries $\sigma_1 = 87.1$ MPa (check: $20+60\cos26.57^\circ+30\sin26.57^\circ = 87.1$). The other principal plane is at $13.28^\circ+90^\circ = 103.28^\circ$ and carries $\sigma_2$.

(c) Maximum shear stress

$$\tau_{max} = R = 67.1\ \text{MPa}$$

acting on planes at $\theta_p \pm 45^\circ$, i.e. $58.3^\circ$ and $148.3^\circ$, with normal stress $20$ MPa.

Answer: (a) $\sigma_\theta = 76.0$ MPa, $\tau_\theta = -37.0$ MPa; (b) $\sigma_1 = 87.1$ MPa, $\sigma_2 = -47.1$ MPa at $13.28^\circ$; (c) $\tau_{max} = 67.1$ MPa.

Mohr's circle construction

Convention: tension positive to the right on the $\sigma$ axis; a shear stress that rotates the element clockwise is plotted upward.

1. Plot X $(\sigma_x,\ \tau) = (100,\ +40)$ for the x face.
2. Plot Y $(\sigma_y,\ \tau) = (20,\ -40)$ for the y face.
3. Join XY; it cuts the $\sigma$ axis at the centre $C$.
4. Draw the circle with centre $C$ and diameter XY.

$$\sigma_{avg} = \frac{100+20}{2} = 60\ \text{MPa}, \qquad R = \sqrt{40^2 + 40^2} = 56.57\ \text{MPa}$$

  tau (cw +)
     |        X(100,40)
     |      .-*-.
     |    .'  |  '.
  ---+---*----C----*---- sigma
     0   B   60    A
     |    '.  |  .'
     |      '-*-'
     |        Y(20,-40)

Principal stresses (points A and B)

$$\sigma_1 = 60 + 56.57 = 116.6\ \text{MPa}, \qquad \sigma_2 = 60 - 56.57 = 3.43\ \text{MPa}$$

Maximum in-plane shear stress

Top and bottom points of the circle:

$$\tau_{max,\,in\text{-}plane} = R = 56.6\ \text{MPa}, \quad \text{with normal stress } 60\ \text{MPa}$$

Because both principal stresses are tensile, the absolute maximum shear stress (including the third principal stress, $\sigma_3=0$) is $\sigma_1/2 = 58.3$ MPa.

Principal planes

$$\tan2\theta_p = \frac{2\times40}{100-20} = 1 \;\Rightarrow\; 2\theta_p = 45^\circ$$

On the circle the angle XCA is $45^\circ$ measured clockwise, so the plane carrying $\sigma_1$ is $22.5^\circ$ clockwise from the x axis; $\sigma_2$ acts on the plane $90^\circ$ from it.

Stresses on the plane at 20 degrees anticlockwise

Rotate the radius CX by $2\theta = 40^\circ$ anticlockwise. The radius makes $45^\circ + 40^\circ = 85^\circ$ with the $\sigma$ axis:

$$\sigma = 60 + 56.57\cos85^\circ = 64.9\ \text{MPa}$$ $$\tau = 56.57\sin85^\circ = 56.4\ \text{MPa (clockwise sense)}$$

Check by formula: $\sigma = 60 + 40\cos40^\circ - 40\sin40^\circ = 64.9$ MPa.

Answer: $\sigma_1 = 116.6$ MPa, $\sigma_2 = 3.43$ MPa; $\tau_{max} = 56.6$ MPa; principal plane $22.5^\circ$ clockwise from x; on the 20 degree plane $\sigma = 64.9$ MPa, $\tau = 56.4$ MPa.

All three are special cases of the plane stress equations of transformation:

$$\sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta, \quad \tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta$$

Uniaxial stress ($\sigma_y = \tau_{xy} = 0$)

A bar under a single axial stress $\sigma$.

$$\sigma_\theta = \frac{\sigma}{2}(1+\cos2\theta) = \sigma\cos^2\theta, \qquad \tau_\theta = -\frac{\sigma}{2}\sin2\theta$$

$\tau_{max} = \sigma/2$ at $\theta = 45^\circ$. Principal stresses: $\sigma$ and $0$.

Biaxial stress ($\tau_{xy} = 0$)

Normal stresses $\sigma_x$ and $\sigma_y$ in two perpendicular directions, e.g. a thin plate or shell.

$$\sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta, \qquad \tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta$$

$\tau_{max} = \dfrac{\sigma_x-\sigma_y}{2}$ at $45^\circ$. The x and y planes are principal planes.

Pure shear ($\sigma_x = \sigma_y = 0$, $\tau_{xy} = \tau$)

Occurs in a twisted circular shaft.

$$\sigma_\theta = \tau\sin2\theta, \qquad \tau_\theta = \tau\cos2\theta$$

Principal planes: $\tan2\theta_p = \infty$, so $\theta_p = 45^\circ$ and $135^\circ$.

$$\sigma_1 = \tau\ (\text{at }45^\circ), \qquad \sigma_2 = -\tau\ (\text{at }135^\circ)$$

So a state of pure shear $\tau$ is the same as tension $\tau$ and compression $\tau$ on planes inclined at $45^\circ$, with no shear on those planes. Maximum shear stress $= \tau$. This is why brittle materials fail along a $45^\circ$ helix in torsion (tension failure), while ductile materials fail by shear on the cross-section.

Stresses at the surface point

$$\sigma_x = \frac{32M}{\pi d^3} = \frac{32\times1.5\times10^6}{\pi\times60^3} = 70.7\ \text{MPa (tensile)}$$ $$\tau_{xy} = \frac{16T}{\pi d^3} = \frac{16\times2.0\times10^6}{\pi\times60^3} = 47.2\ \text{MPa}$$

and $\sigma_y = 0$ (no stress transverse to the axis).

Principal stresses

$$\frac{\sigma_x}{2} = 35.4\ \text{MPa}, \qquad R = \sqrt{35.4^2 + 47.2^2} = 58.9\ \text{MPa}$$ $$\sigma_1 = 35.4 + 58.9 = 94.3\ \text{MPa (tension)}$$ $$\sigma_2 = 35.4 - 58.9 = -23.6\ \text{MPa (compression)}$$

Maximum shear stress

$$\tau_{max} = R = 58.9\ \text{MPa}$$

(The same value follows from $\frac{16}{\pi d^3}\sqrt{M^2+T^2} = \frac{16\times2.5\times10^6}{\pi\times60^3} = 58.9$ MPa, which is the basis of the equivalent torque $T_e = \sqrt{M^2+T^2} = 2.5$ kN m.)

Principal planes

$$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x} = \frac{2\times47.2}{70.7} = 1.335 \;\Rightarrow\; 2\theta_p = 53.1^\circ,\ \theta_p = 26.6^\circ$$

The plane carrying $\sigma_1$ is inclined at $26.6^\circ$ to the plane perpendicular to the shaft axis (the plane normal is at $26.6^\circ$ to the axis); the plane carrying $\sigma_2$ is $90^\circ$ from it.

Answer: $\sigma_1 = 94.3$ MPa, $\sigma_2 = -23.6$ MPa, $\tau_{max} = 58.9$ MPa, $\theta_p = 26.6^\circ$.