Chapter 3, Problem 15P

For each of the plane stress states listed below, draw a Mohr’s circle diagram properly labeled, find the principal normal and shear stresses, and determine the angle from the x axis to σ₁. Draw stress elements as in Fig. 3–11c and d and label all details.

(a)    σ_x = 20 kpsi, σ_y = 210 kpsi, τ_xy = 8 kpsi cw

(b)    σ_x = 16 kpsi, σ_y = 9 kpsi, τ_xy = 5 kpsi ccw

(c)    σ_x = 10 kpsi, σ_y = 24 kpsi, τ_xy = 6 kpsi ccw

(d)    σ_x = 212 kpsi, σ_y = 22 kpsi, τ_xy = 12 kpsi cw

To determine

The principle normal stress.

The shear stress.

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

Answer to Problem 15P

The principle normal stress ${\sigma }_1$ is $22\text{kpsi}$ and ${\sigma }_2$ is $-12\text{kpsi}$.

The shear stress is $17\text{kpsi}$.

The angle from ${\sigma }_1$ to x-axis is $14.04°$.

Explanation of Solution

The Figure (1) shows the state of stress on the element for ${\sigma }_x=20\text{kpsi}$, ${\sigma }_y=-10\text{kpsi}$ and ${\tau }^{cw}=8\text{kpsi}$ clockwise.

Figure (1)

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 ${\sigma }_y$, 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.

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

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

Write the expression for the angle of maximum shear plane.

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

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

Conclusion:

Substitute $20\text{kpsi}$ for ${\sigma }_x$, $-10\text{kpsi}$ for ${\sigma }_y$,$8\text{kpsi}$ for ${\tau }^{cw}$ and $-8\text{kpsi}$ for ${\tau }^{ccw}$ in Equation (I) and Equation (II).

    $\begin{array}{c}\text{A}=\left(20,8\right)\\ \text{B}=\left(-10,-8\right)\end{array}$

Draw the Mohr’s circle diagram.

The Figure (2) shows the Mohr’s circle diagram.

Figure (2)

Substitute the value $20\text{kpsi}$ for ${\sigma }_x$, $-10\text{kpsi}$ for ${\sigma }_y$ in Equation (III).

    $\begin{array}{c}C=\frac{20\text{kpsi}+10\text{kpsi}}{2}\\ =\frac{30\text{kpsi}}{2}\\ =15\text{kpsi}\end{array}$

Substitute the value $20\text{kpsi}$ for ${\sigma }_x$, $-10\text{kpsi}$ for ${\sigma }_y$,$8\text{kpsi}$ for ${\tau }^{cw}$ in Equation (IV).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{2\times 8\text{kpsi}}{20\text{kpsi}-\left(-10\text{kpsi}\right)}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{8\text{kpsi}}{15\text{kpsi}}\right)\\ =14.04°\end{array}$

Substitute $20\text{kpsi}$ for ${\sigma }_x$, $-10\text{kpsi}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}R=\sqrt{{\left(\frac{20\text{kpsi}+10\text{kpsi}}{2}\right)}^2+{\left(8\text{kpsi}\right)}^2}\\ =\sqrt{{15}^2+8^2}\text{kpsi}\\ =\sqrt{225+64}\text{kpsi}\\ =17\text{kpsi}\end{array}$

Substitute $5\text{kpsi}$ for $C$ and $15\text{kpsi}$ for $R$ in and Equation (VII).

    $\begin{array}{l}{\sigma }_1=5\text{kpsi}+17\text{kpsi}\\ {\sigma }_1=22\text{kpsi}\\ {\sigma }_2=5\text{kpsi}-17\text{kpsi}\\ {\sigma }_2=-12\text{kpsi}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $22\text{kpsi}$ and ${\sigma }_2$ is $-12\text{kpsi}$.

Substitute $20\text{kpsi}$ for ${\sigma }_x$, $-10\text{kpsi}$ for ${\sigma }_y$ in Equation (VIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{20\text{kpsi}+10\text{kpsi}}{2}\right)}^2+8^2}\\ =\sqrt{{15}^2+8^2}\text{kpsi}\\ =\sqrt{225+64}\text{kpsi}\\ =17\text{kpsi}\end{array}$

Thus, the maximum shear stress is $17\text{kpsi}$.

Substitute the value $14.04°$ for ${\varphi }_p$.

    $\begin{array}{c}{\varphi }_m=45°-14.04°\\ =30.96°\end{array}$

Thus, the angle from ${\sigma }_1$ to x-axis is $14.04°$.

The Figure (3) shows the maximum in plane normal stress distribution about the plane.

Figure (3)

The Figure (4) shows stress distribution at maximum shear stress plane.

Figure (4)

To determine

The principle normal stress.

The shear stress.

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

Answer to Problem 15P

The principle normal stress ${\sigma }_1$ is $18.6\text{kpsi}$ and ${\sigma }_2$ is $6.4\text{kpsi}$.

The shear stress is $6.10\text{kpsi}$.

The angle from ${\sigma }_1$ to x-axis is $27.5°$.

Explanation of Solution

The Figure (5) shows the state of stress on the element for ${\sigma }_x=16\text{kpsi}$, ${\sigma }_y=9\text{kpsi}$ and ${\tau }^{ccw}=5\text{kpsi}$ counterclockwise.

Figure (5)

Conclusion:

Substitute the value $16\text{kpsi}$ for ${\sigma }_x$, $9\text{kpsi}$ for ${\sigma }_y$,$5\text{kpsi}$ for ${\tau }^{cw}$ and $-5\text{kpsi}$ for ${\tau }^{ccw}$ in Equation (I) and Equation (II).

    $\begin{array}{c}\text{A}=\left(16,-5\right)\\ \text{B}=\left(9,5\right)\end{array}$

Draw the Mohr’s circle diagram.

The Figure (2) shows the Mohr’s circle diagram.

Figure (6)

Substitute the value $16\text{kpsi}$ for ${\sigma }_x$, $9\text{kpsi}$ for ${\sigma }_y$ in Equation (III).

    $\begin{array}{c}C=\frac{9\text{kpsi}+16\text{lbf}}{2}\\ =\frac{25\text{kpsi}}{2}\\ =12.5\text{kpsi}\end{array}$

Substitute the value $16\text{kpsi}$ for ${\sigma }_x$, $9\text{kpsi}$ for ${\sigma }_y$,$5\text{kpsi}$ for ${\tau }^{cw}$ in Equation (IV).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{2\times 5\text{kpsi}}{16\text{kpsi}-9\text{kpsi}}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{10\text{kpsi}}{7\text{kpsi}}\right)\\ =27.5°\end{array}$

Substitute the value $16\text{kpsi}$ for ${\sigma }_x$, $9\text{kpsi}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}R=\sqrt{{\left(\frac{16\text{kpsi}-9\text{kpsi}}{2}\right)}^2+{\left(5\text{kpsi}\right)}^2}\\ =\sqrt{{3.5}^2+8^2}\text{kpsi}\\ =6.10\text{kpsi}\end{array}$

Substitute the value $12.5\text{kpsi}$ for $C$ and $6.10\text{kpsi}$ for $R$ in and Equation (VI) Equation (VII).

    $\begin{array}{l}{\sigma }_1=12.5\text{kpsi}+6.1\text{kpsi}\\ {\sigma }_1=18.6\text{kpsi}\\ {\sigma }_2=12.5\text{kpsi}-6.1\text{kpsi}\\ {\sigma }_2=6.4\text{kpsi}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $18.6\text{kpsi}$ and ${\sigma }_2$ is $6.4\text{kpsi}$.

Substitute the value $16\text{kpsi}$ for ${\sigma }_x$, $9\text{kpsi}$ for ${\sigma }_y$ in Equation (VIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{16\text{kpsi}-9\text{kpsi}}{2}\right)}^2+{\left(5\text{kpsi}\right)}^2}\\ =\sqrt{{\left(3.5\text{kpsi}\right)}^2+{\left(5\text{kpsi}\right)}^2}\\ =6.1\text{kpsi}\end{array}$

Thus, the maximum shear stress is $6.10\text{kpsi}$.

Substitute $27.5°$ for $\varphi$ in Equation (IX).

    $\begin{array}{c}{\varphi }_m=45°-27.5°\\ =17.5°\end{array}$

Thus, the angle from ${\sigma }_1$ to x-axis is $27.5°$.

The Figure (3) shows the maximum in plane normal stress distribution about the plane.

Figure (7)

The Figure (4) shows stress distribution at maximum shear stress plane.

Figure (8)

To determine

The principle normal stress.

The shear stress.

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

Answer to Problem 15P

The principle normal stress ${\sigma }_1$ is $26.22\text{kpsi}$ and ${\sigma }_2$ is $7.78\text{kpsi}$.

The shear stress is $9.22\text{kpsi}$.

The angle from ${\sigma }_1$ to x-axis is $69.7°$.

Explanation of Solution

The Figure (5) shows the state of stress on the element for ${\sigma }_x=16\text{kpsi}$, ${\sigma }_y=9\text{kpsi}$ and ${\tau }^{ccw}=5\text{kpsi}$ counterclockwise.

Figure (9)

Conclusion:

Substitute the value $24\text{kpsi}$ for ${\sigma }_x$, $10\text{kpsi}$ for ${\sigma }_y$,$6\text{kpsi}$ for ${\tau }^{cw}$ and $-6\text{kpsi}$ for ${\tau }^{ccw}$ in Equation (I) and Equation (II).

    $\begin{array}{c}\text{A}=\left(24,6\right)\\ \text{B}=\left(10,-6\right)\end{array}$

Draw the Mohr’s circle diagram.

The Figure (2) shows the Mohr’s circle diagram.

Figure (10)

Substitute the value $24\text{kpsi}$ for ${\sigma }_x$, $10\text{kpsi}$ for ${\sigma }_y$ in Equation (III).

    $\begin{array}{c}C=\frac{24\text{kpsi}+10\text{kpsi}}{2}\\ =\frac{34\text{kpsi}}{2}\\ =17\text{kpsi}\end{array}$

Substitute the value $24\text{kpsi}$ for ${\sigma }_x$, $10\text{kpsi}$ for ${\sigma }_y$,$6\text{kpsi}$ for ${\tau }^{cw}$ in Equation (IV).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{2\times 6\text{kpsi}}{24\text{kpsi}-10\text{kpsi}}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{12\text{kpsi}}{14\text{kpsi}}\right)\\ =24.7°\end{array}$

Write the value of ${\varphi }_p$ in clockwise direction.

    $\begin{array}{c}{\varphi }_p=45°+24.7°\\ =69.7°\end{array}$

Substitute the value $24\text{kpsi}$ for ${\sigma }_x$, $10\text{kpsi}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}R=\sqrt{{\left(\frac{24\text{kpsi}-10\text{kpsi}}{2}\right)}^2+6^2}\\ =\sqrt{{\left(7\text{kpsi}\right)}^2+{\left(6\text{kpsi}\right)}^2}\\ =9.22\text{kpsi}\end{array}$

Substitute the value $17\text{kpsi}$ for $C$ and $9.22\text{kpsi}$ for $R$ in and Equation (VII).

    $\begin{array}{l}{\sigma }_1=17\text{kpsi}+9.22\text{kpsi}\\ {\sigma }_1=26.22\text{kpsi}\\ {\sigma }_2=17\text{kpsi}-9.22\text{kpsi}\\ {\sigma }_2=7.78\text{kpsi}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $26.22\text{kpsi}$ and ${\sigma }_2$ is $7.78\text{kpsi}$.

Substitute $24\text{kpsi}$ for ${\sigma }_x$, $10\text{kpsi}$ for ${\sigma }_y$ in Equation (VIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{24\text{kpsi}-10\text{kpsi}}{2}\right)}^2+{\left(6\text{kpsi}\right)}^2}\\ =\sqrt{7^2+6^2}\text{kpsi}\\ =9.22\text{kpsi}\end{array}$

Thus, the maximum shear stress is $9.22\text{kpsi}$.

Substitute the value $27.5°$ for $\varphi$.

    $\begin{array}{c}{\varphi }_m=69.7°-45°\\ =24.7°\end{array}$

Thus, the angle from ${\sigma }_1$ to x-axis is $69.7°$.

The Figure (3) shows the maximum in plane normal stress distribution about the plane.

Figure (11)

The Figure (4) shows stress distribution at maximum shear stress plane.

Figure (12)

To determine

The principle normal stress.

The shear stress.

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

Answer to Problem 15P

The principle normal stress ${\sigma }_1$ is $25.81\text{kpsi}$ and ${\sigma }_2$ is. $-15.81\text{kpsi}$.

The shear stress is $20.81\text{kpsi}$.

The angle from ${\sigma }_1$ to x-axis is $72.39°$.

Explanation of Solution

The Figure (5) shows the state of stress on the element for ${\sigma }_x=16\text{kpsi}$, ${\sigma }_y=9\text{kpsi}$ and ${\tau }^{ccw}=5\text{kpsi}$ counterclockwise.

Figure (13)

Conclusion:

Substitute the value $-12\text{kpsi}$ for ${\sigma }_x$, $22\text{kpsi}$ for ${\sigma }_y$,$12\text{kpsi}$ for ${\tau }^{cw}$ and $-12\text{kpsi}$ for ${\tau }^{ccw}$ in Equation (I) and Equation (II).

    $\begin{array}{c}\text{A}=\left(-12,12\right)\\ \text{B}=\left(22,-12\right)\end{array}$

Draw the Mohr’s circle diagram.

The Figure (2) shows the Mohr’s circle diagram.

Figure (14)

Substitute the value for ${\sigma }_x$, $22\text{kpsi}$ for ${\sigma }_y$ in Equation (III).

    $\begin{array}{c}C=\frac{-12\text{kpsi}+22\text{kpsi}}{2}\\ =\frac{10\text{kpsi}}{2}\\ =5\text{kpsi}\end{array}$

Substitute the value $-12\text{kpsi}$ for ${\sigma }_x$, $22\text{kpsi}$ for ${\sigma }_y$,$6\text{kpsi}$ for ${\tau }^{cw}$ in Equation (IV).

    $\begin{array}{c}{\varphi }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{2\times 12\text{kpsi}}{-12\text{kpsi}+22\text{kpsi}}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{12\text{kpsi}}{5\text{kpsi}}\right)\\ =27.39°\end{array}$

Write the value of ${\varphi }_p$ in clockwise direction.

    $\begin{array}{c}{\varphi }_p=45°+27.39°\\ =72.39°\end{array}$

Substitute the value $-12\text{kpsi}$ for ${\sigma }_x$, $22\text{kpsi}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}R=\sqrt{{\left(\frac{-12\text{kpsi}-22\text{kpsi}}{2}\right)}^2+{\left(12\text{kpsi}\right)}^2}\\ =\sqrt{{\left(-17\text{kpsi}\right)}^2+{\left(12\text{kpsi}\right)}^2}\\ =20.81\text{kpsi}\end{array}$

Substitute the value $5\text{kpsi}$ for $C$ and $20.81\text{kpsi}$ for $R$ in and Equation (VII).

    $\begin{array}{l}{\sigma }_1=5\text{kpsi}+20.81\text{kpsi}\\ {\sigma }_1=25.81\text{kpsi}\\ {\sigma }_2=5\text{kpsi}-20.81\text{kpsi}\\ {\sigma }_2=-15.81\text{kpsi}\end{array}$

Thus, the principle normal stress ${\sigma }_1$ is $25.81\text{kpsi}$ and ${\sigma }_2$ is. $-15.81\text{kpsi}$.

Substitute the value $-12\text{kpsi}$ for ${\sigma }_x$, $22\text{kpsi}$ for ${\sigma }_y$ in Equation (VIII).

    $\begin{array}{c}{\tau }_{\max }=\sqrt{{\left(\frac{-12\text{kpsi}-22\text{kpsi}}{2}\right)}^2+{\left(12\text{kpsi}\right)}^2}\\ =\sqrt{{\left(-17\right)}^2+{12}^2}\text{kpsi}\\ =20.81\text{kpsi}\end{array}$

Thus, the shear stress is $20.81\text{kpsi}$.

Substitute the value $72.39°$ for ${\varphi }_p$.

    $\begin{array}{c}{\varphi }_m=72.39°-45°\\ =27.39°\end{array}$

Thus, the angle from ${\sigma }_1$ to x-axis is $72.39°$.

The Figure (3) shows the maximum in plane normal stress distribution about the plane.

Figure (15)

The Figure (4) shows stress distribution at maximum shear stress plane.

Figure (16)