Stress and Strain Analysis

Reading of the figure: $\sigma_x = 80$ MPa on the vertical faces (horizontal direction), $\sigma_y = 50$ MPa on the horizontal faces (vertical direction), $\tau_{xy} = 40$ MPa; the plane is at $20^\circ$ to the vertical, so its normal is $20^\circ$ from the $x$-axis.

Stresses on the inclined plane ($\theta = 20^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 65.00 + (15.00)\cos 40^\circ + (40.00)\sin 40^\circ = 102.20\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(15.00)\sin 40^\circ + (40.00)\cos 40^\circ = 21.00\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{102.20^2+21.00^2} = 104.34\ \text{MPa}$, inclined at $11.61^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 65.00 \pm \sqrt{(15.00)^2+(40.00)^2} = 65.00 \pm 42.72$$

$\sigma_1 = 107.72\ \text{MPa}$ and $\sigma_2 = 22.28\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(40.00)}{30.00} \Rightarrow 2\theta_p = 69.44^\circ$$

So $\theta_p = 34.72^\circ$ (normal of the plane carrying $\sigma_1$) and $124.72^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 42.72\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 79.72^\circ$ and $169.72^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 65.00\ \text{MPa}$.

Check with Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(80, 40)$ and $Y(50, -40)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (65.00, 0)$.
• Draw the circle with centre C and radius $CX = 42.72$.
• It cuts the $\sigma$-axis at $\sigma_1 = 107.72$ and $\sigma_2 = 22.28$; $2\theta_p = \angle XCP_1 = 69.44^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm42.72$ with $\sigma = 65.00$.
• Turn CX through $2\theta = 40^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 102.20$ and $\tau \approx 21.00$, which match the analytical values.

Answer: $\sigma_n = 102.20$ MPa, $\tau = 21.00$ MPa; $\sigma_1 = 107.72$ MPa, $\sigma_2 = 22.28$ MPa at $\theta_p = 34.72^\circ$ and $124.72^\circ$; $\tau_{max} = 42.72$ MPa.

Reading of the figure: $\sigma_x = 60$ MPa (tensile) on the vertical faces, $\sigma_y = 30$ MPa (tensile) on the horizontal faces, $\tau_{xy} = 50$ MPa. The plane AB is at $25^\circ$ to the vertical, so its normal is $25^\circ$ from the $x$-axis.

Take tensile stress as positive, compressive as negative. Here $\sigma_x = 60\ \text{MPa}$ (on the vertical faces), $\sigma_y = 30\ \text{MPa}$ (on the horizontal faces) and $\tau_{xy} = 50\ \text{MPa}$. The angle $\theta$ is the angle of the plane normal from the $x$-axis, measured anticlockwise.

Stresses on the inclined plane ($\theta = 25^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 45.00 + (15.00)\cos 50^\circ + (50.00)\sin 50^\circ = 92.94\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(15.00)\sin 50^\circ + (50.00)\cos 50^\circ = 20.65\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{92.94^2+20.65^2} = 95.21\ \text{MPa}$, inclined at $12.53^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 45.00 \pm \sqrt{(15.00)^2+(50.00)^2} = 45.00 \pm 52.20$$

$\sigma_1 = 97.20\ \text{MPa}$ and $\sigma_2 = -7.20\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{30.00} \Rightarrow 2\theta_p = 73.30^\circ$$

So $\theta_p = 36.65^\circ$ (normal of the plane carrying $\sigma_1$) and $126.65^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 52.20\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 81.65^\circ$ and $171.65^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 45.00\ \text{MPa}$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(60, 50)$ and $Y(30, -50)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (45.00, 0)$.
• Draw the circle with centre C and radius $CX = 52.20$.
• It cuts the $\sigma$-axis at $\sigma_1 = 97.20$ and $\sigma_2 = -7.20$; $2\theta_p = \angle XCP_1 = 73.30^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm52.20$ with $\sigma = 45.00$.
• Turn CX through $2\theta = 50^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 92.94$ and $\tau \approx 20.65$, which match the analytical values.

Answer: $\sigma_n = 92.94$ MPa, $\tau = 20.65$ MPa on AB; $\sigma_1 = 97.20$ MPa, $\sigma_2 = -7.20$ MPa; $\tau_{max} = 52.20$ MPa.

Reading of the figure: $\sigma_x = 50$ MPa on the left/right faces, $\sigma_y = 0$ and a shear stress $\tau_{xy} = 30$ MPa on all four faces. The plane is at $30^\circ$ to the vertical, so its normal is at $30^\circ$ to the $x$-axis.

Stresses on the inclined plane ($\theta = 30^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 25.00 + (25.00)\cos 60^\circ + (30.00)\sin 60^\circ = 63.48\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(25.00)\sin 60^\circ + (30.00)\cos 60^\circ = -6.65\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{63.48^2+-6.65^2} = 63.83\ \text{MPa}$, inclined at $5.98^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 25.00 \pm \sqrt{(25.00)^2+(30.00)^2} = 25.00 \pm 39.05$$

$\sigma_1 = 64.05\ \text{MPa}$ and $\sigma_2 = -14.05\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{50.00} \Rightarrow 2\theta_p = 50.19^\circ$$

So $\theta_p = 25.10^\circ$ (normal of the plane carrying $\sigma_1$) and $115.10^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 39.05\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 70.10^\circ$ and $160.10^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 25.00\ \text{MPa}$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(50, 30)$ and $Y(0, -30)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (25.00, 0)$.
• Draw the circle with centre C and radius $CX = 39.05$.
• It cuts the $\sigma$-axis at $\sigma_1 = 64.05$ and $\sigma_2 = -14.05$; $2\theta_p = \angle XCP_1 = 50.19^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm39.05$ with $\sigma = 25.00$.
• Turn CX through $2\theta = 60^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 63.48$ and $\tau \approx -6.65$, which match the analytical values.

Answer: $\sigma_n = 63.48$, $\tau = -6.65$ on the inclined plane; $\sigma_1 = 64.05$, $\sigma_2 = -14.05$ at $\theta_p = 25.10^\circ$ and $115.10^\circ$; $\tau_{max} = 39.05$ with normal stress $25.00$ (MPa).

Mohr's circle is a graphical method that represents the state of plane stress at a point; every point on the circle gives the normal and shear stress on one plane.

Given: $\sigma_x$, $\sigma_y$ and $\tau_{xy}$ on an element, and a plane whose normal is at angle $\theta$ to the $x$-axis.

Procedure

1. Draw the $\sigma$-axis horizontally (tension positive to the right of O) and the $\tau$-axis vertically through O. Select a suitable scale.
2. Plot point $X(\sigma_x, \tau_{xy})$ for the face on which $\sigma_x$ acts (shear that turns the element clockwise is plotted downward, anticlockwise upward; adopt one rule consistently).
3. Plot point $Y(\sigma_y, -\tau_{xy})$ for the perpendicular face.
4. Join X and Y. The line cuts the $\sigma$-axis at $C$, the centre, where $OC = \dfrac{\sigma_x+\sigma_y}{2}$.
5. With C as centre and $CX$ as radius, draw the circle. Radius $R = \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$.
6. The circle cuts the $\sigma$-axis at $P_1$ and $P_2$: $\sigma_1 = OC + R$, $\sigma_2 = OC - R$. The angle $XCP_1 = 2\theta_p$ gives the principal plane.
7. For the stresses on the given inclined plane, rotate the radius CX through $2\theta$ about C in the same sense as the plane is rotated, reaching point $Q$. The abscissa of Q is the normal stress $\sigma_n$ and the ordinate is the shear stress $\tau_\theta$. The diametrically opposite point $Q'$ gives the stresses on the perpendicular plane.
8. The resultant stress is $OQ$, and its obliquity is $\angle QOC$ measured from the $\sigma$-axis.
9. The top and bottom points of the circle give the maximum shear stress $\tau_{max} = R$.

Sketch

        tau
         |        T (tau max = R)
         |     .--*--.
         |   /    |    \
         |  *Q    |     * X(sx, txy)
 --------+-P2-----C------P1------- sigma
         O  \     |    /
         |   *Y   |   /
         |     '--*--'
         |        T'
 P2 = sigma_2, P1 = sigma_1, angle XCP1 = 2*theta_p

Reading of the figure: both normal stresses are tensile: $\sigma_x = 50$ MPa (left/right faces), $\sigma_y = 30$ MPa (top/bottom faces), shear stress $\tau_{xy} = 30$ MPa. The plane is at $40^\circ$ to the vertical, so its normal is at $40^\circ$ to the $x$-axis.

Stresses on the inclined plane ($\theta = 40^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 40.00 + (10.00)\cos 80^\circ + (30.00)\sin 80^\circ = 71.28\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(10.00)\sin 80^\circ + (30.00)\cos 80^\circ = -4.64\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{71.28^2+-4.64^2} = 71.43\ \text{MPa}$, inclined at $3.72^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 40.00 \pm \sqrt{(10.00)^2+(30.00)^2} = 40.00 \pm 31.62$$

$\sigma_1 = 71.62\ \text{MPa}$ and $\sigma_2 = 8.38\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{20.00} \Rightarrow 2\theta_p = 71.57^\circ$$

So $\theta_p = 35.78^\circ$ (normal of the plane carrying $\sigma_1$) and $125.78^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress (extra)

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 31.62\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 80.78^\circ$ and $170.78^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 40.00\ \text{MPa}$.

Answer: $\sigma_n = 71.28$, $\tau = -4.64$ on the inclined plane; $\sigma_1 = 71.62$, $\sigma_2 = 8.38$ at $\theta_p = 35.78^\circ$ and $125.78^\circ$; $\tau_{max} = 31.62$ with normal stress $40.00$ (MPa).

Reading of the figure: $\sigma_x = 80$ MPa (left/right faces), $\sigma_y = 60$ MPa (top/bottom faces), $\tau_{xy} = 50$ MPa.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 70.00 \pm \sqrt{(10.00)^2+(50.00)^2} = 70.00 \pm 50.99$$

$\sigma_1 = 120.99\ \text{MPa}$ and $\sigma_2 = 19.01\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{20.00} \Rightarrow 2\theta_p = 78.69^\circ$$

So $\theta_p = 39.35^\circ$ (normal of the plane carrying $\sigma_1$) and $129.35^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 50.99\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 84.35^\circ$ and $174.35^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 70.00\ \text{MPa}$.

Answer: $\sigma_1 = 120.99$, $\sigma_2 = 19.01$ at $\theta_p = 39.35^\circ$ and $129.35^\circ$; $\tau_{max} = 50.99$ with normal stress $70.00$ (MPa).

Reading of the figure: shear stresses on perpendicular faces must be equal, so the labels are read as $\sigma_x = 60$ MPa (left/right faces), $\sigma_y = 15$ MPa (top/bottom faces) and $\tau_{xy} = 30$ MPa.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 37.50 \pm \sqrt{(22.50)^2+(30.00)^2} = 37.50 \pm 37.50$$

$\sigma_1 = 75.00\ \text{MPa}$ and $\sigma_2 = 0.00\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{45.00} \Rightarrow 2\theta_p = 53.13^\circ$$

So $\theta_p = 26.57^\circ$ (normal of the plane carrying $\sigma_1$) and $116.57^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 37.50\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 71.57^\circ$ and $161.57^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 37.50\ \text{MPa}$.

Answer: $\sigma_1 = 75.00$, $\sigma_2 = 0.00$ at $\theta_p = 26.57^\circ$ and $116.57^\circ$; $\tau_{max} = 37.50$ with normal stress $37.50$ (MPa).

Expression for stresses on an inclined plane (bi-axial loading)

Consider a small element subjected to tensile stresses $\sigma_x$ and $\sigma_y$ on two perpendicular faces (no shear stress). Cut it by a plane AB whose normal makes angle $\theta$ with the $x$-axis (the plane makes $\theta$ with the vertical face). Let the area of AB be $dA$ (unit thickness).

        sy
     B  |  |  |
     |\ v  v  v
     | \
  sx |  \ AB (area dA)
 ----> |___\ A
       (BA vertical face = dA cos(theta))

Areas: vertical face $= dA\cos\theta$; horizontal face $= dA\sin\theta$.

Forces on the wedge: $\sigma_x\,dA\cos\theta$ (horizontal) and $\sigma_y\,dA\sin\theta$ (vertical).

Normal stress $\sigma_n$ (resolve along the normal to AB):

$$\sigma_n\,dA = \sigma_x\,dA\cos\theta\cdot\cos\theta + \sigma_y\,dA\sin\theta\cdot\sin\theta$$ $$\sigma_n = \sigma_x\cos^2\theta + \sigma_y\sin^2\theta = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta$$

Shear (tangential) stress $\tau$ (resolve along AB):

$$\tau\,dA = \sigma_x\,dA\cos\theta\sin\theta - \sigma_y\,dA\sin\theta\cos\theta$$ $$\tau = \frac{\sigma_x-\sigma_y}{2}\sin 2\theta$$

Resultant stress: $\sigma_R = \sqrt{\sigma_n^2 + \tau^2}$, at obliquity $\tan\phi = \tau/\sigma_n$ to the normal.

Maximum $\sigma_n = \sigma_x$ at $\theta = 0^\circ$; maximum shear $= \dfrac{\sigma_x-\sigma_y}{2}$ at $\theta = 45^\circ$.

Mohr's graphical method

• Draw a horizontal axis for normal stress. Mark $OA = \sigma_y$ and $OB = \sigma_x$ to scale.
• Bisect AB at C and draw a circle with diameter AB (centre C, radius $\frac{\sigma_x-\sigma_y}{2}$).
• From C draw a line making angle $2\theta$ with CB, cutting the circle at Q.
• The horizontal distance of Q from O is $\sigma_n$ and the vertical distance of Q from the axis is $\tau$, because $OC + CQ\cos2\theta = \sigma_n$ and $CQ\sin2\theta = \tau$.
• The line OQ is the resultant stress; $\angle QOC$ is its obliquity.

Assumption: the 30 MN/m² on plane GE is the normal (tensile) stress on it, as in the figure. Shear stresses on two perpendicular planes are equal in magnitude, so the shear on GE equals the shear on EF.

Components on plane EF

The 50 MN/m² resultant is at $30^\circ$ to the normal of EF:

$$\sigma_{EF} = 50\cos 30^\circ = 43.30\ \text{MN/m}^2, \qquad \tau_{EF} = 50\sin 30^\circ = 25.00\ \text{MN/m}^2$$

On the perpendicular plane GE: $\sigma_{GE} = 30$ and $\tau_{GE} = 25.00$ MN/m².

So $\sigma_x = 43.30$, $\sigma_y = 30$, $\tau_{xy} = 25.00$ (all in MN/m²).

(i) Resultant stress on plane GE

$$\sigma_R = \sqrt{30^2 + 25.00^2} = 39.05\ \text{MN/m}^2, \qquad \tan\alpha = \frac{25.00}{30} \Rightarrow \alpha = 39.81^\circ\ \text{to the normal of GE}$$

(ii) Principal stresses and planes

$$\sigma_{1,2} = \frac{43.30+30}{2} \pm \sqrt{\left(\frac{43.30-30}{2}\right)^2 + 25.00^2} = 36.65 \pm 25.87$$ $$\sigma_1 = 62.52\ \text{MN/m}^2, \qquad \sigma_2 = 10.78\ \text{MN/m}^2$$ $$\tan 2\theta_p = \frac{2(25)}{43.30-30} \Rightarrow 2\theta_p = 75.10^\circ,\ \theta_p = 37.55^\circ$$

$\sigma_1$ acts on the plane whose normal is at $37.55^\circ$ to the normal of EF; $\sigma_2$ acts on the plane at $127.55^\circ$ (both planes are at right angles).

(iii) Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = 25.87\ \text{MN/m}^2$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $82.55^\circ$ and $172.55^\circ$ from the normal of EF, with a normal stress of $36.65$ MN/m² on them.

Given: $\sigma_x = 120$ and $\sigma_y = 60\ \text{MN/m}^2$ (tensile), no shear stress ($\tau_{xy} = 0$), $\nu = 1/4$.

Principal stresses and directions

Since there is no shear on the X and Y planes, these planes are already principal planes.

$$\sigma_{1,2} = \frac{120+60}{2} \pm \sqrt{\left(\frac{120-60}{2}\right)^2 + 0} = 90 \pm 30$$ $$\sigma_1 = 120\ \text{MN/m}^2\ \text{(on the plane normal to the X-axis, }\theta_p = 0^\circ), \qquad \sigma_2 = 60\ \text{MN/m}^2\ \text{(}\theta_p = 90^\circ)$$

$\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y} = 0 \Rightarrow \theta_p = 0^\circ$ and $90^\circ$.

Stress invariant

The sum of normal stresses on any two perpendicular planes is constant (an invariant of plane stress). Check with a plane at, say, $\theta = 30^\circ$ and its perpendicular ($120^\circ$):

$$\sigma_{30} = 90 + 30\cos60^\circ = 105, \qquad \sigma_{120} = 90 + 30\cos240^\circ = 75$$ $$\sigma_x + \sigma_y = 120 + 60 = 180, \qquad \sigma_1 + \sigma_2 = 120 + 60 = 180, \qquad \sigma_{30} + \sigma_{120} = 105 + 75 = 180$$

So $\sigma_x + \sigma_y = \sigma_1 + \sigma_2 = 180\ \text{MN/m}^2$ is verified.

Equivalent stress for the same maximum strain

Maximum strain (in the direction of $\sigma_1$):

$$\varepsilon_{max} = \frac{1}{E}(\sigma_1 - \nu\sigma_2) = \frac{1}{E}(120 - 0.25\times60) = \frac{105}{E}$$

A single stress $\sigma_e$ acting alone gives the strain $\sigma_e/E$. Equating:

$$\sigma_e = 105\ \text{MN/m}^2$$

Answer: $\sigma_1 = 120$, $\sigma_2 = 60$ MN/m² (at $0^\circ$ and $90^\circ$); the invariant is $180$ MN/m²; equivalent stress $= 105$ MN/m².

Reading of the figure: $\sigma_x = 75$ N/mm² tensile (left and right faces), $\sigma_y = -30$ N/mm² (compressive, top and bottom faces), $\tau_{xy} = 18$ N/mm².

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 22.50 \pm \sqrt{(52.50)^2+(18.00)^2} = 22.50 \pm 55.50$$

$\sigma_1 = 78.00\ \text{N/mm}^2$ and $\sigma_2 = -33.00\ \text{N/mm}^2$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(18.00)}{105.00} \Rightarrow 2\theta_p = 18.92^\circ$$

So $\theta_p = 9.46^\circ$ (normal of the plane carrying $\sigma_1$) and $99.46^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 55.50\ \text{N/mm}^2$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 54.46^\circ$ and $144.46^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 22.50\ \text{N/mm}^2$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(75, 18)$ and $Y(-30, -18)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (22.50, 0)$.
• Draw the circle with centre C and radius $CX = 55.50$.
• It cuts the $\sigma$-axis at $\sigma_1 = 78.00$ and $\sigma_2 = -33.00$; $2\theta_p = \angle XCP_1 = 18.92^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm55.50$ with $\sigma = 22.50$.

Answer: $\sigma_1 = 78.00$, $\sigma_2 = -33.00$ at $\theta_p = 9.46^\circ$ and $99.46^\circ$; $\tau_{max} = 55.50$ with normal stress $22.50$ (N/mm²).

Reading: the maximum (principal) stress must not exceed 150 MPa. Let $\sigma_x = 100$ MPa (tension), $\sigma_y = -60$ MPa (compression) and $\tau$ the unknown shear stress.

Shear stress that can be applied

$$\frac{\sigma_x+\sigma_y}{2} = \frac{100-60}{2} = 20\ \text{MPa}, \qquad \frac{\sigma_x-\sigma_y}{2} = 80\ \text{MPa}$$

The maximum principal stress is

$$\sigma_1 = \frac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} = 150$$ $$\sqrt{80^2 + \tau^2} = 150 - 20 = 130 \Rightarrow \tau^2 = 130^2 - 80^2 = 10500$$ $$\tau = 102.47\ \text{MPa}$$

Maximum shear stress in the material

$$\tau_{max} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} = 130\ \text{MPa}$$

Principal stresses and planes

$$\sigma_1 = 20 + 130 = 150\ \text{MPa}, \qquad \sigma_2 = 20 - 130 = -110\ \text{MPa}$$ $$\tan 2\theta_p = \frac{2\tau}{\sigma_x-\sigma_y} = \frac{2(102.47)}{160} = 1.2809 \Rightarrow 2\theta_p = 52.02^\circ,\ \theta_p = 26.01^\circ$$

$\sigma_1$ acts on the plane whose normal is at $26.01^\circ$ to the direction of the 100 MPa stress, and $\sigma_2$ on the plane at $116.01^\circ$ (anticlockwise, with $\tau$ taken positive as in the standard sign convention).

Answer: shear stress $= 102.47$ MPa; $\tau_{max} = 130$ MPa; $\sigma_1 = 150$ MPa, $\sigma_2 = -110$ MPa at $26.01^\circ$ and $116.01^\circ$.

Reading of the figure: $\sigma_x = 80$ MPa (horizontal, on left/right faces), $\sigma_y = 60$ MPa (vertical, on top/bottom faces), shear $\tau_{xy} = 20$ MPa; all tensile.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 70.00 \pm \sqrt{(10.00)^2+(20.00)^2} = 70.00 \pm 22.36$$

$\sigma_1 = 92.36\ \text{MPa}$ and $\sigma_2 = 47.64\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(20.00)}{20.00} \Rightarrow 2\theta_p = 63.43^\circ$$

So $\theta_p = 31.72^\circ$ (normal of the plane carrying $\sigma_1$) and $121.72^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 22.36\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 76.72^\circ$ and $166.72^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 70.00\ \text{MPa}$.

Answer: $\sigma_1 = 92.36$, $\sigma_2 = 47.64$ at $\theta_p = 31.72^\circ$ and $121.72^\circ$; $\tau_{max} = 22.36$ with normal stress $70.00$ (MPa).

When the two principal stresses are known, the plane is located by the angle $\theta$ between its normal and the major principal stress ($\sigma_1$) axis. Here $\sigma_1 = 100$ MPa, $\sigma_2 = -40$ MPa and $\theta = 60^\circ$.

Normal stress

$$\begin{aligned} \sigma_n &= \frac{\sigma_1+\sigma_2}{2} + \frac{\sigma_1-\sigma_2}{2}\cos 2\theta \\ &= \frac{100-40}{2} + \frac{100+40}{2}\cos 120^\circ = 30 + 70(-0.5) = -5.00\ \text{MPa} \end{aligned}$$

(negative: compressive).

Shear stress

$$\tau = \frac{\sigma_1-\sigma_2}{2}\sin 2\theta = 70\sin 120^\circ = 60.62\ \text{MPa}$$

Resultant stress and its direction

$$\sigma_R = \sqrt{\sigma_n^2+\tau^2} = \sqrt{(-5.00)^2+(60.62)^2} = 60.83\ \text{MPa}$$ $$\tan\phi = \frac{|\sigma_n|}{\tau} = \frac{5.00}{60.62} \Rightarrow \phi = 4.72^\circ$$

The resultant makes only $4.72^\circ$ with the plane itself (it is almost along the plane, because $\sigma_n$ is small); from the normal the angle is $94.72^\circ$, measured on the compressive side.

Answer: $\sigma_n = -5.00$ MPa, $\tau = 60.62$ MPa, resultant $= 60.83$ MPa.