Chapter 3, Problem 18P

For each of the stress states listed below, find all three principal normal and shear stresses. Draw a complete Mohr’s three-circle diagram and label all points of interest.

(a)    σx = –80 MPa, σ_y = 230 MPa, τ_xy = 20 MPa cw

(b)    σx = 30 MPa, σ_y = 260 MPa, τ_xy = 30 MPa cw

(c)    σ_x = 40 MPa, σ_z = –30 MPa, τ_xy = 20 MPa ccw

(d)    σ_x = 50 MPa, σ_z = –20 MPa, τ_xy = 30 MPa cw

To determine

The principal normal stresses.

The principal shear stresses.

The Mohr circle diagram.

If ${\sigma }_x=-80\text{MPa, }{\sigma }_y=-30\text{MPa,}{\tau }_{xy}=20\text{MPa}\text{cW}$.

Answer to Problem 18P

The principal normal stresses are $0$, $-23\text{MPa}$ and $-87\text{MPa}$.

The principal shear stresses are $11.25\text{MPa}$, $32\text{MPa}$ and $43.5\text{MPa}$.

The Mohr circle diagram is shown in Figure-(1).

Explanation of Solution

Write the expression for principal stress.

    $\left[\begin{array}{l}{\sigma }^3-\left({\sigma }_x+{\sigma }_y+{\sigma }_z\right){\sigma }^2+\left({\sigma }_x{\sigma }_y+{\sigma }_x{\sigma }_z+{\sigma }_y{\sigma }_z-{\tau }_{xy}^2-{\tau }_{yz}^2-{\tau }_{zx}^2\right)\sigma \\ -\left({\sigma }_x{\sigma }_y{\sigma }_z+2{\tau }_{xy}{\tau }_{yz}{\tau }_{zx}-{\sigma }_x{\tau }_{yz}^2-{\sigma }_y{\tau }_{yz}^2-{\sigma }_z{\tau }_{xy}^2\right)\end{array}\right]=0$ (I)

Here, normal stresses in x, y and z directions are ${\sigma }_x$,${\sigma }_y$ and ${\sigma }_z$ and shear stresses are ${\tau }_{xy}$,${\tau }_{yz}$ and ${\tau }_{zx}$.

Write the expression for first principal shear stress.

    ${\tau }_{1/2}=\frac{{\sigma }_1-{\sigma }_2}{2}$                                                                                           (II)

Here, first principal shear stress is ${\tau }_{1/2}$.

Write the expression for second principal shear stress.

    ${\tau }_{2/3}=\frac{{\sigma }_2-{\sigma }_3}{2}$                                                                                         (III)

Here, second principal shear stress is ${\tau }_{2/3}$.

Write the expression for third principal shear stress.

    ${\tau }_{1/3}=\frac{{\sigma }_1-{\sigma }_3}{2}$                                                                                          (IV)

Here, third principal shear stress is ${\tau }_{1/3}$.

Write the expression for the distance of center of the Mohr’s circle from origin.

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

Here, the distance of center of the Mohr’s circle from origin is $C$.

Write the expression for radius of circle.

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

Here, radius of the circle is $R$.

Write the expression for distance $CD$.

    $CD=\frac{{\sigma }_x-{\sigma }_y}{2}$                                                                                        (VII)

Here, the distance between point C and D is $CD$.

Conclusion:

Substitute $80\text{MPa}$ for ${\sigma }_x$, $-30\text{MPa}$ for ${\sigma }_y$, $0$ for ${\sigma }_z$, $20\text{MPa}$ for ${\tau }_{xy}$, $0$ for ${\tau }_{yz}$ and $0$ for ${\tau }_{zx}$ in Equation (I).

    $\begin{array}{l}\left[\begin{array}{l}{\sigma }^3-\left(-80\text{MPa}+\left(-30\text{MPa}\right)+0\right){\sigma }^2\\ +\left(\begin{array}{l}\left(-80\text{MPa}\right)\left(-30\text{MPa}\right)+\\ \left(-80\text{MPa}\right)0+\left(-30\text{MPa}\right)0\\ -0-0-0\end{array}\right)\sigma \\ -\left(\left(-80\text{MPa}\right)\left(-30\text{MPa}\right)0+2\times 0-0-\left(-30\text{MPa}\right)0-0\times 0\right)\end{array}\right]=0\\ {\sigma }^3+110{\sigma }^2+2000\sigma =0\\ \sigma \left({\sigma }^2+110\sigma +2000\right)=0\end{array}$ (VIII)

Solve Equation (VIII) to obtain roots of Equation as $0$ for ${\sigma }_1$, $-23\text{MPa}$ for ${\sigma }_2$ and $-87\text{MPa}$ for ${\sigma }_3$.

Substitute $0$ for ${\sigma }_1$, $-23\text{MPa}$ for ${\sigma }_2$ in Equation (II).

    $\begin{array}{c}{\tau }_{1/2}=\frac{0-\left(-23\text{MPa}\right)}{2}\\ =11.25\text{MPa}\end{array}$

Substitute $-23\text{MPa}$ for ${\sigma }_2$ and $-87\text{MPa}$ for ${\sigma }_3$ in Equation (III).

    $\begin{array}{c}{\tau }_{2/3}=\frac{-23\text{MPa}-\left(-87\text{MPa}\right)}{2}\\ =\frac{64\text{MPa}}{2}\\ =32\text{MPa}\end{array}$

Substitute $0$ for ${\sigma }_1$, ${\sigma }_2$ and $-87\text{MPa}$ for ${\sigma }_3$ in Equation (IV).

    $\begin{array}{c}{\tau }_{1/3}=\frac{0-\left(-87\text{MPa}\right)}{2}\\ =43.5\text{MPa}\end{array}$

Thus, the principal shear stresses are $11.25\text{MPa}$, $32\text{MPa}$ and $43.5\text{MPa}$.

Substitute $-80\text{MPa}$ for ${\sigma }_x$, $-30\text{MPa}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}C=\frac{-80\text{MPa}+\left(-30\text{MPa}\right)}{2}\\ =\frac{-110\text{MPa}}{2}\\ =-55\text{MPa}\end{array}$

Substitute $-80\text{MPa}$ for ${\sigma }_x$, $-30\text{MPa}$ for ${\sigma }_y$ and $20\text{MPa}$ for ${\tau }_{xy}$ in Equation (VI).

    $\begin{array}{c}R=\sqrt{{\left(\frac{-80\text{MPa}-\left(-30\text{MPa}\right)}{2}\right)}^2+{\left(20\text{MPa}\right)}^2}\\ =\sqrt{{\left(25\text{MPa}\right)}^2+{\left(20\text{MPa}\right)}^2}\\ \approx 32.02\text{MPa}\end{array}$

Substitute $-80\text{MPa}$ for ${\sigma }_x$, $-30\text{MPa}$ for ${\sigma }_y$ in Equation (VII).

    $\begin{array}{c}CD=\frac{-80\text{MPa}-\left(-30\text{MPa}\right)}{2}\\ =\frac{-50\text{MPa}}{2}\\ =-25\text{MPa}\end{array}$

The procedure to draw the Mohr’s circle is given below.

• Draw $\sigma$ and $\tau$ axes.
• Mark ${\sigma }_1$, ${\sigma }_2$ and ${\sigma }_3$ on $\sigma$ axis.
• Draw three circles with shear stresses as radius.

The figure below shows the Mohr’s Circle.

Figure-(1)

To determine

The principal normal stresses.

The principal shear stresses.

The Mohr circle diagram.

If ${\sigma }_x=30\text{MPa, }{\sigma }_y=-60\text{MPa,}{\tau }_{xy}=30\text{MPa}\text{cW}$.

Answer to Problem 18P

The principal normal stresses are $39.08\text{MPa}$, $0\text{MPa}$ and $-69.08\text{MPa}$.

The principal shear stresses are $19.54\text{MPa}$, $34.54\text{MPa}$ and $54.08\text{MPa}$.

The Mohr circle diagram is shown in Figure-(2).

Explanation of Solution

Conclusion:

Substitute $30\text{MPa}$ for ${\sigma }_x$, $-60\text{MPa}$ for ${\sigma }_y$, $0$ for ${\sigma }_z$, $30\text{MPa}$ for ${\tau }_{xy}$, $0$ for ${\tau }_{yz}$ and $0$ for ${\tau }_{zx}$ in Equation (I).

    $\begin{array}{l}\left[\begin{array}{l}{\sigma }^3-\left(30\text{MPa}+\left(-60\text{MPa}\right)+0\right){\sigma }^2\\ +\left(\left(30\text{MPa}\right)\left(-60\text{MPa}\right)+0+0-0-{\left(-30\text{MPa}\right)}^2-0-0\right)\sigma \\ -\left(0+0-0-0-0\times 0\right)\end{array}\right]=0\\ {\sigma }^3+30{\sigma }^2-2700\sigma =0\\ \sigma \left({\sigma }^2+30\sigma -2700\right)=0\end{array}$ (IX)

Solve Equation (IX) to obtain roots of Equation as $39.08\text{MPa}$ for ${\sigma }_1$, $0$ for ${\sigma }_2$ and $-69.08\text{MPa}$ for ${\sigma }_3$.

Substitute $39.08\text{MPa}$ for ${\sigma }_1$,$0$ for ${\sigma }_2$ in Equation (II).

    $\begin{array}{c}{\tau }_{1/2}=\frac{39.08\text{MPa}-\left(0\right)}{2}\\ =19.54\text{MPa}\end{array}$

Substitute $0$ for ${\sigma }_2$ and $-69.08\text{MPa}$ for ${\sigma }_3$ in Equation (III).

    $\begin{array}{c}{\tau }_{2/3}=\frac{0-\left(-69.08\text{MPa}\right)}{2}\\ =34.54\text{MPa}\end{array}$

Substitute $39.08\text{MPa}$ for ${\sigma }_1$, and $-69.08\text{MPa}$ for ${\sigma }_3$ in Equation (IV).

    $\begin{array}{c}{\tau }_{1/3}=\frac{39.08\text{MPa}-\left(-69.08\text{MPa}\right)}{2}\\ =\frac{108.16\text{MPa}}{2}\\ =54.08\text{MPa}\end{array}$

Thus, the principal shear stresses are $19.54\text{MPa}$, $34.54\text{MPa}$ and $54.08\text{MPa}$.

Substitute $30\text{MPa}$ for ${\sigma }_x$, $-60\text{MPa}$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}C=\frac{30\text{MPa}+\left(-60\text{MPa}\right)}{2}\\ =\frac{-30\text{MPa}}{2}\\ =-15\text{MPa}\end{array}$

Substitute $30\text{MPa}$ for ${\sigma }_x$, $-60\text{MPa}$ for ${\sigma }_y$ and $30\text{MPa}$ for ${\tau }_{xy}$ in Equation (VI).

    $\begin{array}{c}R=\sqrt{{\left(\frac{30\text{MPa}-\left(-60\text{MPa}\right)}{2}\right)}^2+{\left(30\text{MPa}\right)}^2}\\ =\sqrt{{\left(45\text{MPa}\right)}^2+{\left(30\text{MPa}\right)}^2}\\ =54.1\text{MPa}\end{array}$

Substitute $30\text{MPa}$ for ${\sigma }_x$, $-60\text{MPa}$ for ${\sigma }_y$ in Equation (VII).

    $\begin{array}{c}CD=\frac{30\text{MPa}-\left(-60\text{MPa}\right)}{2}\\ =\frac{+90\text{MPa}}{2}\\ =45\text{MPa}\end{array}$

The procedure to draw the Mohr’s circle is given below.

• Draw $\sigma$ and $\tau$ axes.
• Mark ${\sigma }_1$, ${\sigma }_2$ and ${\sigma }_3$ on $\sigma$ axis.
• Draw three circles with shear stresses as radius.

The figure below shows the Mohr’s Circle.

Figure-(2)

To determine

The principal normal stresses.

The principal shear stresses.

The Mohr circle diagram.

If ${\sigma }_x=40\text{MPa, }{\sigma }_y=-30\text{MPa,}{\tau }_{xy}=20\text{MPa}\text{ccw}$.

Answer to Problem 18P

The principal normal stresses are $48.3\text{MPa}$, $-8.3\text{MPa}$ and $-30\text{MPa}$.

The principal shear stresses are $28.3\text{MPa}$, $10.85\text{MPa}$ and $39.15\text{MPa}$.

The Mohr circle diagram is shown in Figure-(3).

Explanation of Solution

Conclusion:

Substitute $40\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$, $-30\text{MPa}$ for ${\sigma }_z$,$20\text{MPa}$ for ${\tau }_{xy}$, $0$ for ${\tau }_{yz}$ and $0$ for ${\tau }_{zx}$ in Equation (I).

    $\begin{array}{l}\left[\begin{array}{l}{\sigma }^3-\left(40\text{MPa}+\left(-30\text{MPa}\right)+0\right){\sigma }^2\\ +\left(\left(0\right)+\left(40\text{MPa}\right)\left(-30\text{MPa}\right)0+0-0-{\left(20\text{MPa}\right)}^2-0-0\right)\sigma \\ -\left(0+0-0-0-\left(-30\text{MPa}\right)\left(20\text{MPa}\right)\right)\end{array}\right]=0\\ {\sigma }^3-10{\sigma }^2-1600\sigma -12000=0\end{array}$ (X).

Solve Equation (X) to obtain roots of Equation as $48.3\text{MPa}$ for ${\sigma }_1$, $-8.3\text{MPa}$ for ${\sigma }_2$ and $-30\text{MPa}$ for ${\sigma }_3$.

Substitute $48.3\text{MPa}$ for ${\sigma }_1$,$-8.3\text{MPa}$ for ${\sigma }_2$ in Equation (II).

    $\begin{array}{c}{\tau }_{1/2}=\frac{48.3\text{MPa}-\left(-8.3\text{MPa}\right)}{2}\\ =\frac{56.6\text{MPa}}{2}\\ =28.3\text{MPa}\end{array}$

Substitute $-8.3\text{MPa}$ for ${\sigma }_2$ and $-30\text{MPa}$ for ${\sigma }_3$ in Equation (III).

    $\begin{array}{c}{\tau }_{2/3}=\frac{-8.3\text{MPa}-\left(-30\text{MPa}\right)}{2}\\ =\frac{21.7\text{MPa}}{2}\\ =10.85\text{MPa}\end{array}$

Substitute $48.3\text{MPa}$ for ${\sigma }_1$, and $-30\text{MPa}$ for ${\sigma }_3$ in Equation (IV).

    $\begin{array}{c}{\tau }_{1/3}=\frac{48.3\text{MPa}-\left(-30\text{MPa}\right)}{2}\\ =\frac{78.3\text{MPa}}{2}\\ =39.15\text{MPa}\end{array}$

Thus, the principal shear stresses are $28.3\text{MPa}$, $10.85\text{MPa}$ and $39.15\text{MPa}$.

Substitute $40\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}C=\frac{40\text{MPa}+\left(0\right)}{2}\\ =20\text{MPa}\end{array}$

Substitute $40\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$ and $20\text{MPa}$ for ${\tau }_{xy}$ in Equation (VI).

    $\begin{array}{c}R=\sqrt{{\left(\frac{40\text{MPa}-\left(0\right)}{2}\right)}^2+{\left(20\text{MPa}\right)}^2}\\ =\sqrt{{\left(20\text{MPa}\right)}^2+{\left(20\text{MPa}\right)}^2}\\ =28.3\text{MPa}\end{array}$

Substitute $40\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$ in Equation (VII).

    $\begin{array}{c}CD=\frac{40\text{MPa}-\left(0\right)}{2}\\ =20\text{MPa}\end{array}$

The procedure to draw the Mohr’s circle is given below.

• Draw $\sigma$ and $\tau$ axes.
• Mark ${\sigma }_1$, ${\sigma }_2$ and ${\sigma }_3$ on $\sigma$ axis.
• Draw three circles with shear stresses as radius.

The figure below shows the Mohr’s Circle.

Figure-(3)

To determine

The principal normal stresses.

The principal shear stresses.

The Mohr circle diagram.

If ${\sigma }_x=50\text{MPa, }{\sigma }_y=-20\text{MPa,}{\tau }_{xy}=30\text{MPa}\text{cw}$.

Answer to Problem 18P

The principal normal stresses are $64.1\text{MPa}$, $-14.1\text{MPa}$ and $-20\text{MPa}$.

The principal shear stresses are $39.1\text{MPa}$, $2.95\text{MPa}$ and $42.05\text{MPa}$.

The Mohr circle diagram is shown in Figure-(4).

Explanation of Solution

Conclusion:

Substitute $50\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$, $-20\text{MPa}$ for ${\sigma }_z$,$30\text{MPa}$ for ${\tau }_{xy}$, $0$ for ${\tau }_{yz}$ and $0$ for ${\tau }_{zx}$ in Equation (I).

    $\begin{array}{l}\left[\begin{array}{l}{\sigma }^3-\left(50\text{MPa}+\left(-20\text{MPa}\right)+0\right){\sigma }^2\\ +\left(\left(0\right)+\left(50\text{MPa}\right)\left(-20\text{MPa}\right)0+0-0-{\left(30\text{MPa}\right)}^2-0-0\right)\sigma \\ -\left(0+0-0-0-\left(-20\text{MPa}\right)\left(30\text{MPa}\right)\right)\end{array}\right]=0\\ {\sigma }^3-30{\sigma }^2-1900\sigma -18000=0\end{array}$ (XI).

Solve Equation (XI) to obtain roots of Equation as $64.1\text{MPa}$ for ${\sigma }_1$, $-14.1\text{MPa}$ for ${\sigma }_2$ and $-20\text{MPa}$ for ${\sigma }_3$.

Substitute $64.1\text{MPa}$ for ${\sigma }_1$, $-14.1\text{MPa}$ for ${\sigma }_2$ in Equation (II).

    $\begin{array}{c}{\tau }_{1/2}=\frac{64.1\text{MPa}-\left(-14.1\text{MPa}\right)}{2}\\ =\frac{78.2\text{MPa}}{2}\\ =39.1\text{MPa}\end{array}$

Substitute $-14.1\text{MPa}$ for ${\sigma }_2$ and $-20\text{MPa}$ for ${\sigma }_3$ in Equation (III).

    $\begin{array}{c}{\tau }_{2/3}=\frac{-14.1\text{MPa}-\left(-20\text{MPa}\right)}{2}\\ =\frac{5.9\text{MPa}}{2}\\ =2.95\text{MPa}\end{array}$

Substitute $64.1\text{MPa}$ for ${\sigma }_1$, and $-20\text{MPa}$ for ${\sigma }_3$ in Equation (IV).

    $\begin{array}{c}{\tau }_{1/3}=\frac{64.1\text{MPa}-\left(-20\text{MPa}\right)}{2}\\ =\frac{84.1\text{MPa}}{2}\\ =42.05\text{MPa}\end{array}$

Thus, the principal shear stresses are $39.1\text{MPa}$, $2.95\text{MPa}$ and $42.05\text{MPa}$.

Substitute $50\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$ in Equation (V).

    $\begin{array}{c}C=\frac{50\text{MPa}+\left(0\right)}{2}\\ =25\text{MPa}\end{array}$

Substitute $50\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$ and $30\text{MPa}$ for ${\tau }_{xy}$ in Equation (VI).

    $\begin{array}{c}R=\sqrt{{\left(\frac{50\text{MPa}-\left(0\right)}{2}\right)}^2+{\left(30\text{MPa}\right)}^2}\\ =\sqrt{{\left(25\text{MPa}\right)}^2+{\left(30\text{MPa}\right)}^2}\\ =39.1\text{MPa}\end{array}$

Substitute $50\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$ in Equation (VII).

    $\begin{array}{c}CD=\frac{50\text{MPa}-\left(0\right)}{2}\\ =25\text{MPa}\end{array}$

The procedure to draw the Mohr’s circle is given below.

• Draw $\sigma$ and $\tau$ axes.
• Mark ${\sigma }_1$, ${\sigma }_2$ and ${\sigma }_3$ on $\sigma$ axis.
• Draw three circles with shear stresses as radius.

The figure below shows the Mohr’s Circle.

Figure-(4)