Chapter 3, Problem 16P

Repeat Prob. 3–15 for:

(a)    σ_x = 28 MPa, σ_y = 7 MPa, τ_{xy =} 6 MPa cw

(b)    σ_x = 9 MPa, σ_y = –6 MPa, τ_{xy =} 3 MPa cw

(c)    σ_x = –4 MPa, σ_y = 12 MPa, τ_{xy =} 7 MPa ccw

(d)    σ_x = 6 MPa, σ_y = –5 MPa, τ_{xy =} 8 MPa ccw

To determine

The principle normal stress.

The shear stress.

The angle from ${\sigma }_1$ to $x$-axis.

Answer to Problem 16P

The principle normal stress ${\sigma }_1$ is $9.10\text{MPa}$ and ${\sigma }_2$ is $-10.1\text{MPa}$.

The shear stress is $9.60\text{MPa}$.

The angle from ${\sigma }_1$ to x-axis is $64.33°$.

Explanation of Solution

Write the coordinates of the points through which the Mohr’s circle pass.

    $\text{A}=\left({\sigma }_x,{\tau }^{cw}\right)$                                                                           (I)

    $\text{B}=\left({\sigma }_y,{\tau }^{ccw}\right)$                                                                          (II)

Here, the stress along x face is ${\sigma }_x$, stress along y face is ${\sigma }_y$, shear stress above the $\sigma$ axis is ${\tau }^{cw}$, shear stress below the $\sigma$ axis is ${\tau }^{ccw}$, and the points in $\sigma -\tau$ plane is A,B.

Draw the $\sigma$ axis and $\tau$ axis and locate the points A and B. Join the line AB. It forms the diameter for the circle and intersects the $\sigma$ axis at the center.

Write the formula for the center point.

    $C=\frac{{\sigma }_x+{\sigma }_y}{2}$                                                                            (III)

Here, the center point is $C$.

Write the expression for the angle between the line joining points A and B with $\sigma$ axis.

    ${\varphi }_p=\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{2{\tau }^{cw}}{{\sigma }_x-{\sigma }_y}\right)\right]$                                                (IV)

Here, the angle made by the diameter with positive x-axis in the counterclockwise direction is ${\varphi }_p$.

Write the expression of the radius of circle.

    $R=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left({\tau }^{cw}\right)}^2}$                                                               (V)

Write the expression maximum in plane normal stress.

    ${\sigma }_1=C+R$                                                                               (VI)

    ${\sigma }_2=C-R$                                                                              (VII)

Here, the maximum in plane normal stress are ${\sigma }_1$ and ${\sigma }_2$.

Write the expression of maximum in plane shear stress.

    ${\tau }_{\max }=\frac{{\sigma }_x-{\sigma }_y}{2}$                                                                         (VIII)

Here, the maximum shear stress is ${\tau }_{\max }$.

Write the expression for the angle of maximum shear plane.

    ${\varphi }_m={\varphi }_p-45°$                                                                               (IX)

Here, the angle is ${\varphi }_m$.

Conclusion:

Substitute the value $7\text{MPa}$ for ${\sigma }_x$, $-8\text{MPa}$ for ${\sigma }_y$,$6\text{MPa}$ for ${\tau }^{cw}$ and $-6\text{MPa}$ for ${\tau }^{ccw}$ in Equation (I) and Equation (II).

    $\begin{array}{c}\text{A}=\left(7,6^{ccw}\right)\\ \text{B}=\left(-8,6^{cw}\right)\end{array}$

Draw the Mohr’s circle diagram.

The Figure (1) shows the Mohr’s circle diagram.

Figure (1)

Substitute the value $7\text{MPa}$ for ${\sigma }_x$, $-8\text{MPa}$ for ${\sigma }_y$ in Equation (III).

    $\begin{array}{c}C=\frac{-8\text{MPa}+7\text{MPa}}{2}\\ =\frac{-1\text{MPa}}{2}\\ =-0.5\text{MPa}\end{array}$

Substitute the value $7\text{MPa}$ for ${\sigma }_x$, $-8\text{MPa}$ for ${\sigma }_y$,$6\text{MPa}$ for ${\tau }^{cw}$ in Equation (IV).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{2\times 6\text{MPa}}{7\text{MPa}-\left(-8\text{MPa}\right)}\right)\right]\\ =\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{12\text{MPa}}{15\text{MPa}}\right)\right]\\ =\frac{1}{2}\left(90°+38.66°\right)\\ =64.33°\end{array}$

Thus, the angle from ${\sigma }_1$ to x-axis is $64.33°$.

Substitute the value $7\text{MPa}$ for ${\sigma }_x$, $-8\text{MPa}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}R=\sqrt{{\left(\frac{7\text{MPa}+8\text{MPa}}{2}\right)}^2+{\left(6\text{MPa}\right)}^2}\\ =\sqrt{\left({7.5}^2+6^2\right){\text{MPa}}^2}\\ =\sqrt{92.25{\text{MPa}}^2}\\ \approx 9.6\text{MPa}\end{array}$

Substitute the value $-0.5\text{MPa}$ for $C$ and $9.6\text{MPa}$ for $R$ in Equation (VI).    $\begin{array}{c}{\sigma }_1=9.6\text{MPa}-0.5\text{MPa}\\ =9.10\text{MPa}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $9.10\text{MPa}$.

Substitute the value $-0.5\text{MPa}$ for $C$ and $9.6\text{MPa}$ for $R$ in Equation (VII).

    $\begin{array}{c}{\sigma }_2=-0.5\text{MPa}-9.6\text{MPa}\\ =-10.1\text{MPa}\end{array}$

Thus, the principle normal stress ${\sigma }_2$ is $-10.1\text{MPa}$.

Substitute the value $7\text{MPa}$ for ${\sigma }_x$, $-8\text{MPa}$ for ${\sigma }_y$ in Equation (VIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{8-\left(-7\right)}{2}\right)}^2+8^2}\\ =\sqrt{{7.5}^2+8^2}\\ =9.60\text{MPa}\end{array}$

Thus, the shear stress is $9.60\text{MPa}$.

Substitute the value $70.67°$ for ${\varphi }_p$ in Equation (X).

    $\begin{array}{c}{\varphi }_m=45°-70.67°\\ =-25.67°\text{ccw}\\ =25.67°\text{cw}\end{array}$

The Figure (2) shows the maximum in plane normal stress distribution about the plane.

Figure (2)

The Figure (3) shows stress distribution at maximum shear stress plane.

Figure (3)

To determine

The principle normal stress.

The shear stress.

The angle from ${\sigma }_1$ to $x$-axis.

Answer to Problem 16P

The principle normal stress ${\sigma }_1$ is $9.58\text{MPa}$ and ${\sigma }_2$ is $-6.58\text{MPa}$.

The shear stress is $8.078\text{MPa}$.

The angle from ${\sigma }_1$ to x-axis is $10.9°$.

Explanation of Solution

Write the coordinates of the points through which the Mohr’s circle pass.

    $\text{A}=\left({\sigma }_x,{\tau }^{cw}\right)$                                                                              (X)

    $\text{B}=\left({\sigma }_y,{\tau }^{ccw}\right)$                                                                             (XI)

Draw the $\sigma$ axis and $\tau$ axis and locate the points A and B. Join the line AB. It forms the diameter for the circle and intersects the $\sigma$ axis at the center.

Write the formula for the center point.

    $C=\frac{{\sigma }_x+{\sigma }_y}{2}$                                                                             (XII)

Write the expression for the angle between the line joining points A and B with $\sigma$ axis.

    ${\varphi }_p=\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{2{\tau }^{cw}}{{\sigma }_x-{\sigma }_y}\right)\right]$                                                 (XIII)

Write the expression of the radius of circle.

    $R=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left({\tau }^{cw}\right)}^2}$                                                       (XIV)

Write the expression maximum in plane normal stress.

    ${\sigma }_1=C+R$                                                                              (XV)

    ${\sigma }_2=C-R$                                                                             (XVI)

Write the expression of maximum in plane shear stress.

    ${\tau }_{\max }=\frac{{\sigma }_x-{\sigma }_y}{2}$                                                                        (XVII)

Write the expression for the angle of maximum shear plane.

    ${\varphi }_m={\varphi }_p-45°$                                                                          (XVIII)

Write the expression for the angle between the line joining points A and B with $\sigma$ axis.

    ${\varphi }_p=\frac{1}{2}\left[{\tan }^{-1}\left(\frac{2\tau }{{\sigma }_x-{\sigma }_y}\right)\right]$                                                           (XIX)

Conclusion:

Substitute $9\text{MPa}$ for ${\sigma }_x$ and $3\text{MPa}$ for ${\tau }^{cw}$ in Equation (X).

    $\text{A}=\left(9,3^{cw}\right)$

Substitute $-6\text{MPa}$ for ${\sigma }_y$ and $-3\text{MPa}$ for ${\tau }^{ccw}$ in Equation (XI).

    $\text{B}=\left(-6,3^{ccw}\right)$

Draw the Mohr’s circle diagram.

The Figure (4) shows the Mohr’s circle diagram.

Figure (4)

Substitute the value $9\text{MPa}$ for ${\sigma }_x$, $-6\text{MPa}$ for ${\sigma }_y$ in Equation (XII).

    $\begin{array}{c}C=\frac{9\text{MPa}-6\text{MPa}}{2}\\ =\frac{3\text{MPa}}{2}\\ =1.5\text{MPa}\end{array}$

Substitute the value $9\text{MPa}$ for ${\sigma }_x$, $-6\text{MPa}$ for ${\sigma }_y$,$3\text{MPa}$ for ${\tau }^{cw}$ in Equation (XX).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}\left[{\tan }^{-1}\left(\frac{2\times 3\text{MPa}}{9\text{MPa}-\left(-6\text{MPa}\right)}\right)\right]\\ =\frac{1}{2}\left[{\tan }^{-1}\left(\frac{6}{15}\right)\right]\\ =0.5\times \left(21.80°\right)\\ \approx 10.9°\text{cw}\end{array}$

Thus, the angle from ${\sigma }_1$ to x-axis is $10.9°$.

Substitute the value $9\text{MPa}$ for ${\sigma }_x$, $-6\text{MPa}$ for ${\sigma }_y$,$3\text{MPa}$ for ${\tau }^{cw}$ in Equation (XIV).

    $\begin{array}{c}R=\sqrt{{\left(\frac{9\text{MPa}+6\text{MPa}}{2}\right)}^2+{\left(3\text{MPa}\right)}^2}\\ =\sqrt{{\left(7.5\text{MPa}\right)}^2+{\left(3\text{MPa}\right)}^2}\\ =\sqrt{65.25{\text{MPa}}^2}\\ =8.078\text{MPa}\end{array}$

Substitute $1.5\text{MPa}$ for $C$ and $8.078\text{MPa}$ for $R$ in Equation (XV).

    $\begin{array}{c}{\sigma }_1=1.5\text{MPa}+8.078\text{MPa}\\ =9.58\text{MPa}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $9.58\text{MPa}$.

Substitute $1.5\text{MPa}$ for $C$ and $8.078\text{MPa}$ for $R$ in Equation (XVI).

    $\begin{array}{c}{\sigma }_2=1.5\text{MPa}-8.078\text{MPa}\\ =-6.58\text{MPa}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $9.58\text{MPa}$.

Substitute the value $9\text{MPa}$ for ${\sigma }_x$, $-6\text{MPa}$ for ${\sigma }_y$ in Equation (XVII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{9\text{MPa}-\left(-6\text{MPa}\right)}{2}\right)}^2+{\left(3\text{MPa}\right)}^2}\\ =\sqrt{{\left(7.5\text{MPa}\right)}^2+{\left(3\text{MPa}\right)}^2}\\ =\sqrt{65.24{\text{MPa}}^2}\\ =8.078\text{MPa}\end{array}$

Thus, the shear stress is $8.078\text{MPa}$.

Substitute the value $10.9°$ for ${\varphi }_p$ in Equation (XVIII).

    $\begin{array}{c}{\varphi }_m=45°-10.9°\\ =34.1°\text{ccw}\end{array}$

The Figure (5) shows the maximum in plane normal stress distribution about the plane.

Figure (5)

The Figure (6) shows stress distribution at maximum shear stress plane.

Figure (6)

Thus, the principle normal stress ${\sigma }_1$ is $9.58\text{MPa}$ and ${\sigma }_2$ is $-6.58\text{MPa}$. The shear stress is $8.078\text{MPa}$, and the angle from ${\sigma }_1$ to x-axis is $10.9°$.

To determine

The principle normal stress.

The shear stress.

The angle from ${\sigma }_1$ to $x$-axis.

Answer to Problem 16P

The principle normal stress ${\sigma }_1$ is $14.63\text{MPa}$ and ${\sigma }_2$ is $-6.63\text{MPa}$.

The shear stress is $10.63\text{MPa}$.

The angle from ${\sigma }_1$ to x-axis is $69.4°$.

Explanation of Solution

Write the coordinates of the points through which the Mohr’s circle pass.

    $\text{A}=\left({\sigma }_x,{\tau }^{cw}\right)$                                                                           (XX)

    $\text{B}=\left({\sigma }_y,{\tau }^{ccw}\right)$                                                                          (XXI)

Draw the $\sigma$ axis and $\tau$ axis and locate the points A and B. Join the line AB. It forms the diameter for the circle and intersects the $\sigma$ axis at the center.

Write the formula for the center point.

    $C=\frac{{\sigma }_x+{\sigma }_y}{2}$                                                                            (XXII)

Write the expression for the angle between the line joining points A and B with $\sigma$ axis.

    ${\varphi }_p=\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{2{\tau }^{cw}}{{\sigma }_x-{\sigma }_y}\right)\right]$                                                (XXIII)

Write the expression of the radius of circle.

    $R=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left({\tau }^{cw}\right)}^2}$                                                              (XXIV)

Write the expression maximum in plane normal stress.

    ${\sigma }_1=C+R$                                                                              (XXV)

    ${\sigma }_2=C-R$                                                                              (XXVI)

Write the expression of maximum in plane shear stress.

    ${\tau }_{\max }=\frac{{\sigma }_x-{\sigma }_y}{2}$                                                                         (XXVII)

Write the expression for the angle of maximum shear plane.

    ${\varphi }_m={\varphi }_p-45°$                                                                               (XXVIII)

Conclusion:

Substitute the value $12\text{MPa}$ for ${\sigma }_x$ and $7\text{MPa}$ for ${\tau }^{cw}$ in Equation (XX).

    $\text{A}=\left(12,7^{cw}\right)$

Substitute $-4\text{MPa}$ for ${\sigma }_y$ and $-7\text{MPa}$ for ${\tau }^{ccw}$ in Equation (XXI).

    $\text{B}=\left(-4,7^{ccw}\right)$

Draw the Mohr’s circle diagram.

Figure (7) shows the Mohr’s circle diagram.

Figure (7)

Substitute the value $12\text{MPa}$ for ${\sigma }_x$, $-4\text{MPa}$ for ${\sigma }_y$,$7\text{MPa}$ for ${\tau }^{cw}$ and $-7\text{MPa}$ for ${\tau }^{ccw}$ in Equation (XXII).

    $\begin{array}{c}C=\frac{-4\text{MPa}+12\text{MPa}}{2}\\ =\frac{8\text{MPa}}{2}\\ =4\text{MPa}\end{array}$

Substitute the value $12\text{MPa}$ for ${\sigma }_x$, $-4\text{MPa}$ for ${\sigma }_y$ and $7\text{MPa}$ for ${\tau }^{cw}$ in Equation (XXIII).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{2\left(7\text{MPa}\right)}{12\text{MPa}-\left(-4\text{MPa}\right)}\right)\right]\\ =\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{14}{16}\right)\right]\\ =\frac{1}{2}\left(90°+41.185°\right)\\ \approx 65.6°\text{ccw}\end{array}$

Substitute the value $12\text{MPa}$ for ${\sigma }_x$ and $-4\text{MPa}$ for ${\sigma }_y$ in Equation (XXIV).

    $\begin{array}{c}R=\sqrt{{\left(\frac{12\text{MPa}-\left(-4\text{MPa}\right)}{2}\right)}^2+{\left(7\text{MPa}\right)}^2}\\ =\sqrt{{\left(8\text{MPa}\right)}^2+{\left(7\text{MPa}\right)}^2}\\ =\sqrt{113{\text{MPa}}^2}\\ =10.63\text{MPa}\end{array}$

Substitute the value $4\text{MPa}$ for $C$ and $10.63\text{MPa}$ for $R$ in Equation (XXV).

    $\begin{array}{c}{\sigma }_1=4\text{MPa}+10.63\text{MPa}\\ =14.63\text{MPa}\end{array}$

Substitute the value $4\text{MPa}$ for $C$ and $10.63\text{MPa}$ for $R$ in Equation (XXVI).

    $\begin{array}{c}{\sigma }_2=4\text{MPa}-10.63\text{MPa}\\ =-6.63\text{MPa}\end{array}$

Substitute the value $12\text{MPa}$ for ${\sigma }_x$, $-4\text{MPa}$ for ${\sigma }_y$ and $7\text{MPa}$ for ${\tau }^{cw}$ in Equation (XXVII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{12\text{MPa}-\left(-4\text{MPa}\right)}{2}\right)}^2+{\left(7\text{MPa}\right)}^2}\\ =\sqrt{{\left(8\text{MPa}\right)}^2+{\left(7\text{MPa}\right)}^2}\\ =\sqrt{113{\text{MPa}}^2}\\ =10.63\text{MPa}\end{array}$

Substitute the value $69.4°\text{ccw}$ for $\varphi$.

    $\begin{array}{c}{\varphi }_m=69.4°-45°\\ =24.4°\text{ccw}\end{array}$

The Figure (8) shows the maximum in plane normal stress distribution about the plane.

Figure (8)

The Figure (9) shows stress distribution at maximum shear stress plane.

Figure (9)

Thus, the he principle normal stress ${\sigma }_1$ is $14.63\text{MPa}$ and ${\sigma }_2$ is $-6.63\text{MPa}$. The shear stress is $10.63\text{MPa}$. The angle from ${\sigma }_1$ to x-axis is $69.4°$.

To determine

The principle normal stress.

The shear stress.

The angle from ${\sigma }_1$ to $x$-axis.

Answer to Problem 16P

The principle normal stress ${\sigma }_1$ is $10.21\text{MPa}$ and ${\sigma }_2$ is $-9.21\text{MPa}$.

The shear stress is $9.71\text{MPa}$.

The angle from ${\sigma }_1$ to x-axis is $27.75°$.

Explanation of Solution

Write the coordinates of the points through which the Mohr’s circle pass.

    $\text{A}=\left({\sigma }_x,{\tau }^{cw}\right)$                                                                            (XXIX)

    $\text{B}=\left({\sigma }_y,{\tau }^{ccw}\right)$                                                                         (XXX)

Draw the $\sigma$ axis and $\tau$ axis and locate the points A and B. Join the line AB. It forms the diameter for the circle and intersects the $\sigma$ axis at the center.

Write the formula for the center point.

    $C=\frac{{\sigma }_x+{\sigma }_y}{2}$                                                                           (XXXI)

Write the expression for the angle between the line joining points A and B with $\sigma$ axis.

    ${\varphi }_p=\frac{1}{2}\left[90°+{\tan }^{-1}\left(\frac{2{\tau }^{cw}}{{\sigma }_x-{\sigma }_y}\right)\right]$                                                   (XXXII)

Write the expression of the radius of circle.

    $R=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\left({\tau }^{cw}\right)}^2}$                                                          (XXXIII)

Write the expression maximum in plane normal stress.

    ${\sigma }_1=C+R$                                                                                  (XXXIV)

    ${\sigma }_2=C-R$                                                                                   (XXXV)

Write the expression of maximum in plane shear stress.

    ${\tau }_{\max }=\frac{{\sigma }_x-{\sigma }_y}{2}$                                                                            (XXXVI)

Write the expression for the angle of maximum shear plane.

    ${\varphi }_m={\varphi }_p-45°$                                                                             (XXXVII)

Write the expression for the angle between the line joining points A and B with $\sigma$ axis.

    ${\varphi }_p=\frac{1}{2}\left[{\tan }^{-1}\left(\frac{2{\tau }^{cw}}{{\sigma }_x-{\sigma }_y}\right)\right]$                                                            (XXXVIII)

Conclusion:

Substitute $5\text{MPa}$ for ${\sigma }_x$ and $8\text{MPa}$ for ${\tau }^{cw}$ in Equation (XXIX).

    $\text{A}=\left(6\text{MPa},-8\text{MPa}\right)$

Substitute $6\text{MPa}$ for ${\sigma }_y$ and $-8\text{MPa}$ for ${\tau }^{ccw}$ in Equation (XXX).

    $\text{B}=\left(-5\text{MPa},8\text{MPa}\right)$

Draw the Mohr’s circle diagram.

The Figure (10) shows the Mohr’s circle diagram.

Figure (10)

Substitute the value $5\text{MPa}$ for ${\sigma }_x$, $6\text{MPa}$ for ${\sigma }_y$ in Equation (XXXI).

    $\begin{array}{c}C=\frac{-5\text{MPa}+6\text{MPa}}{2}\\ =\frac{1\text{MPa}}{2}\\ =0.5\text{MPa}\end{array}$

Substitute the value $5\text{MPa}$ for ${\sigma }_x$, $6\text{MPa}$ for ${\sigma }_y$,$8\text{MPa}$ for ${\tau }^{cw}$ in Equation (XXXII).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{2\times 8\text{MPa}}{6\text{MPa}+5\text{MPa}}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{16\text{MPa}}{11\text{MPa}}\right)\\ =\left(0.5\right)\left(55.49°\right)\\ \approx 27.75°\text{ccw}\end{array}$

Substitute the value $5\text{MPa}$ for ${\sigma }_x$, $6\text{MPa}$ for ${\sigma }_y$ and $8\text{MPa}$ for ${\tau }^{cw}$ in Equation (XXXIII).

    $\begin{array}{c}R=\sqrt{{\left(\frac{6\text{MPa}+5\text{MPa}}{2}\right)}^2+{\left(8\text{MPa}\right)}^2}\\ =\sqrt{{\left(5.5\text{MPa}\right)}^2+{\left(8\text{MPa}\right)}^2}\\ =\sqrt{94.25{\text{MPa}}^2}\\ =9.71\text{MPa}\end{array}$

Substitute the value $0.5\text{MPa}$ for $C$ and $9.71\text{MPa}$ for $R$ in Equation (XXXIV).

    $\begin{array}{c}{\sigma }_1=0.5\text{MPa}+9.71\text{MPa}\\ =10.21\text{MPa}\end{array}$

Substitute the value $0.5\text{MPa}$ for $C$ and $9.71\text{MPa}$ for $R$ in Equation (XXXV).

    $\begin{array}{c}{\sigma }_2=0.5\text{MPa}-9.71\text{MPa}\\ =-9.21\text{MPa}\end{array}$

Substitute the value $5\text{MPa}$ for ${\sigma }_x$, $6\text{MPa}$ for ${\sigma }_y$ and $8\text{MPa}$ for ${\tau }^{cw}$ in Equation (XXXVI).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{6\text{MPa}+5\text{MPa}}{2}\right)}^2+{\left(8\text{MPa}\right)}^2}\\ =\sqrt{{\left(5.5\text{MPa}\right)}^2+{\left(8\text{MPa}\right)}^2}\\ =\sqrt{94.25{\text{MPa}}^2}\\ =9.71\text{MPa}\end{array}$

Substitute the value $72.39°$ for ${\varphi }_p$ in Equation (XXXVII)

    $\begin{array}{c}{\varphi }_m=45°-27.75°\\ =17.25°\text{cw}\end{array}$

The Figure (11) shows the maximum in plane normal stress distribution about the plane.

Figure (11)

The Figure (12) shows stress distribution at maximum shear stress plane.

Figure (12)

Thus, the principle normal stress ${\sigma }_1$ is $10.21\text{MPa}$ and ${\sigma }_2$ is. $-9.21\text{MPa}$. The shear stress is $9.71\text{MPa}$. The angle from ${\sigma }_1$ to x-axis is $27.75°$.