Chapter 5, Problem 1P

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states:

(a) σ_x = 100 MPa, σ_y = 100 MPa

(b) σ_x = 100 MPa, σ_y = 50 MPa

(c) σ_x = 100 MPa, τ_xy = −75 MPa

(d) σ_x = −50 MPa, σ_y = −75 MPa τ_xy = −50 MPa

(e) σ_x = 100 MPa, σ_y = 20 MPa τ_xy = −20 MPa

To determine

The factor of the safety using maximum shear stress theory.

The factor of the safety using maximum distortion energy theory.

If ${\sigma }_x=100\text{Mpa, }{\sigma }_y=100\text{MPa}$.

Answer to Problem 1P

The factor of the safety using maximum shear stress theory is $3.5$.

The factor of the safety using maximum distortion energy theory is $3.5$.

Explanation of Solution

Write the expression for the maximum principle stress.

    ${\sigma }_1=\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)+\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$ (I)

Here, the normal stress in x direction is ${\sigma }_x$, the normal stress in y direction is ${\sigma }_y$ and shear stress is ${\tau }_{xy}$.

Write the expression for the minimum principle stress.

    ${\sigma }_2=\left(\frac{{\sigma }_x+{\sigma }_y}{2}\right)-\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$ (II)

Here, the normal stress in x direction is ${\sigma }_x$, the normal stress in y direction is ${\sigma }_y$ and shear stress is ${\tau }_{xy}$.

Write the expression for the factor of the safety for MSST.

    $n=\frac{S_{yt}}{\max \text{of}\left[\left|\left({\sigma }_1-{\sigma }_2\right),{\sigma }_2,{\sigma }_1\right|\right]}$ (III)

Here, the yield strength of the material in tension is $S_{yt}$ , the maximum principle stress is ${\sigma }_1$ and minimum principle stress is ${\sigma }_2$.

Write the expression for the factor of the safety for MDET.

    $n=\frac{S_{yt}}{\sqrt{{\sigma }_1{}^2-{\sigma }_1{\sigma }_2+{\sigma }_2{}^2}}$ (IV)

Here, the yield strength of the material in tension is $S_{yt}$ , the maximum principle stress is ${\sigma }_1$ and minimum principle stress is ${\sigma }_2$.

Conclusion:

Substitute $100\text{MPa}$ for ${\sigma }_x$, $100\text{MPa}$ for ${\sigma }_y$, and $0\text{MPa}$ for ${\tau }_{xy}$ in Equation (I).

    $\begin{array}{c}{\sigma }_1=\left(\frac{100\text{MPa}+100\text{MPa}}{2}\right)+\sqrt{{\left(\frac{100\text{MPa}-100\text{MPa}}{2}\right)}^2+{\left(0\text{MPa}\right)}^2}\\ =\frac{200\text{MPa}}{2}\\ =100\text{MPa}\end{array}$

Substitute $100\text{MPa}$ for ${\sigma }_x$, $100\text{MPa}$ for ${\sigma }_y$, and $0\text{MPa}$ for ${\tau }_{xy}$ in Equation (II).

    $\begin{array}{c}{\sigma }_2=\left(\frac{100\text{MPa}+100\text{MPa}}{2}\right)-\sqrt{{\left(\frac{100\text{MPa}-100\text{MPa}}{2}\right)}^2+{\left(0\text{MPa}\right)}^2}\\ =\frac{200\text{MPa}}{2}\\ =100\text{MPa}\end{array}$

The maximum value from $\left({\sigma }_1-{\sigma }_2\right)$, ${\sigma }_2$, and ${\sigma }_1$ is ${\sigma }_1$.

Substitute $100\text{MPa}$ for ${\sigma }_1$ and $350\text{MPa}$ for $S_{yt}$ in Equation (III).

    $\begin{array}{c}n=\frac{350\text{MPa}}{100\text{MPa}}\\ =3.5\end{array}$

Thus, the factor of the safety using maximum shear stress theory is $3.5$.

Substitute $100\text{MPa}$ for ${\sigma }_1$, $100\text{MPa}$ for ${\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (IV).

    $\begin{array}{c}n=\frac{350\text{MPa}}{\sqrt{{\left(100\text{MPa}\right)}^2-\left(100\text{MPa}\times 100\text{MPa}\right)+{\left(100\text{MPa}\right)}^2}}\\ =\frac{350\text{MPa}}{100\text{MPa}}\\ =3.5\end{array}$

Thus the factor of the safety using maximum distortion energy theory is $3.5$.

To determine

The factor of the safety using maximum shear stress theory.

The factor of the safety using maximum distortion energy theory.

If ${\sigma }_x=100\text{Mpa, }{\sigma }_y=50\text{MPa}$.

Answer to Problem 1P

The factor of the safety using maximum shear stress theory is $3.5$.

The factor of the safety using maximum distortion energy theory is $4.041$.

Explanation of Solution

Conclusion:

Substitute $100\text{MPa}$ for ${\sigma }_x$, $50\text{MPa}$ for ${\sigma }_y$, and $0$ for ${\tau }_{xy}$ in Equation (I).

    $\begin{array}{c}{\sigma }_1=\left(\frac{100\text{MPa}+50\text{MPa}}{2}\right)+\sqrt{{\left(\frac{100\text{MPa}-50\text{MPa}}{2}\right)}^2+{\left(0\right)}^2}\\ =\frac{150\text{MPa}}{2}+25\text{MPa}\\ =100\text{MPa}\end{array}$

Substitute $100\text{MPa}$ for ${\sigma }_x$, $50\text{MPa}$ for ${\sigma }_y$, and $0\text{kpsi}$ for ${\tau }_{xy}$ in Equation (II).

    $\begin{array}{c}{\sigma }_2=\left(\frac{100\text{MPa}+50\text{MPa}}{2}\right)-\sqrt{{\left(\frac{100\text{MPa}-50\text{MPa}}{2}\right)}^2+{\left(0\right)}^2}\\ =\frac{150\text{MPa}}{2}-25\text{MPa}\\ =50\text{MPa}\end{array}$

The maximum value between $\left({\sigma }_1-{\sigma }_2\right)$, ${\sigma }_2$, and ${\sigma }_1$ is ${\sigma }_1$.

Substitute $100\text{MPa}$ for ${\sigma }_1$, and $350\text{MPa}$ for $S_{yt}$ in Equation (III).

    $\begin{array}{c}n=\frac{350\text{MPa}}{100\text{MPa}}\\ =\frac{350\text{MPa}}{100\text{MPa}}\\ =3.5\end{array}$

The factor of the safety using maximum shear stress theory is $3.5$.

Substitute $100\text{MPa}$ for ${\sigma }_1$, $50\text{MPa}$ for ${\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (IV).

    $\begin{array}{c}n=\frac{350\text{MPa}}{\sqrt{{\left(100\text{MPa}\right)}^2-\left(100\text{MPa}\times 50\text{MPa}\right)+{\left(50\text{MPa}\right)}^2}}\\ =\frac{350\text{MPa}}{50\sqrt{3}\text{MPa}}\\ =4.041\end{array}$

Thus, the factor of the safety using maximum distortion energy theory is $4.041$.

To determine

The factor of the safety using maximum shear stress theory.

The factor of the safety using maximum distortion energy theory.

If ${\sigma }_x=100\text{Mpa, }{\tau }_{xy}=-75\text{MPa}$.

Answer to Problem 1P

The factor of the safety using maximum shear stress theory is $1.941$.

The factor of the safety using maximum distortion energy theory is $2.135$.

Explanation of Solution

Conclusion:

Substitute $100\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$, and $-75\text{MPa}$ for ${\tau }_{xy}$ in Equation (I).

    $\begin{array}{c}{\sigma }_1=\left(\frac{100\text{MPa}+0}{2}\right)+\sqrt{{\left(\frac{100\text{MPa}-0}{2}\right)}^2+{\left(-75\text{MPa}\right)}^2}\\ =50\text{MPa}+25\sqrt{13}\text{MPa}\\ =140.138\text{MPa}\end{array}$

Substitute $100\text{MPa}$ for ${\sigma }_x$, $0$ for ${\sigma }_y$, and $-75\text{MPa}$ for ${\tau }_{xy}$ in Equation (II).

    $\begin{array}{c}{\sigma }_2=\left(\frac{100\text{MPa}+0}{2}\right)-\sqrt{{\left(\frac{100\text{MPa}-0}{2}\right)}^2+{\left(-75\text{MPa}\right)}^2}\\ =50\text{MPa-}25\sqrt{13}\text{MPa}\\ =-40.138\text{MPa}\end{array}$

The maximum value between $\left({\sigma }_1-{\sigma }_2\right)$, ${\sigma }_2$, and ${\sigma }_1$ is $\left({\sigma }_1-{\sigma }_2\right)$.

Substitute $140.138\text{MPa}$ for ${\sigma }_1$, $-40.138\text{MPa}$ for ${\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (III).

    $\begin{array}{c}n=\frac{350\text{MPa}}{140.138\text{MPa}-\left(-40.138\text{MPa}\right)}\\ =\frac{350\text{MPa}}{180.276\text{MPa}}\\ =1.941\end{array}$

Thus, the factor of the safety using maximum shear stress theory is $1.941$.

Substitute $140.138\text{MPa}$ for ${\sigma }_1$, $-40.138\text{MPa}$ for ${\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (IV).

    $\begin{array}{c}n=\frac{350\text{MPa}}{\sqrt{{\left(140.138\text{MPa}\right)}^2-\left(140.138\text{MPa}\times -40.138\text{MPa}\right)+{\left(-40.138\text{MPa}\right)}^2}}\\ =\frac{350\text{MPa}}{163.93\text{MPa}}\\ =2.135\end{array}$

Thus, the factor of the safety using maximum distortion energy theory is $2.135$.

To determine

The factor of the safety using maximum shear stress theory.

The factor of the safety using maximum distortion energy theory.

If ${\sigma }_x=-50\text{Mpa, }{\sigma }_y=-75\text{MPa }{\tau }_{xy}=-50\text{MPa}$

Answer to Problem 1P

The factor of the safety using maximum shear stress theory is $3.4$.

The factor of the safety using maximum distortion energy theory is $3.21$.

Explanation of Solution

Conclusion:

Substitute $-50\text{MPa}$ for ${\sigma }_x$, $-75\text{MPa}$ for ${\sigma }_y$, and $-50\text{MPa}$ for ${\tau }_{xy}$ in Equation (I).

    $\begin{array}{c}{\sigma }_1=\left(\frac{-50\text{MPa}-75\text{MPa}}{2}\right)+\sqrt{{\left(\frac{-50\text{MPa+}75\text{MPa}}{2}\right)}^2+{\left(-50\text{MPa}\right)}^2}\\ =-62.5\text{MPa}+51.54\text{MPa}\\ =-10.96\text{MPa}\end{array}$

Substitute $-50\text{MPa}$ for ${\sigma }_x$, $-75\text{MPa}$ for ${\sigma }_y$, and $-50\text{MPa}$ for ${\tau }_{xy}$ in Equation (II).

    $\begin{array}{c}{\sigma }_2=\left(\frac{-50\text{MPa}-75\text{MPa}}{2}\right)-\sqrt{{\left(\frac{-50\text{MPa+}75\text{MPa}}{2}\right)}^2+{\left(-50\text{MPa}\right)}^2}\\ =-62.5\text{MPa}-51.54\text{MPa}\\ =-114.04\text{MPa}\end{array}$

Calculate ${\sigma }_1-{\sigma }_2$.

    $\begin{array}{c}{\sigma }_1-{\sigma }_2=\left(-10.96\text{MPa}\right)-\left(-114.04\text{MPa}\right)\\ =103.08\text{MPa}\end{array}$

The maximum values among $\left({\sigma }_1-{\sigma }_2\right)$, ${\sigma }_2$, and ${\sigma }_1$ is ${\sigma }_1-{\sigma }_2$.

Substitute $103.08\text{MPa}$ for ${\sigma }_1-{\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (III).

    $\begin{array}{c}n=\frac{350\text{MPa}}{\left|103.08\text{MPa}\right|}\\ =\frac{350}{103.08}\\ =3.395\\ \approx 3.4\end{array}$

Thus, the factor of the safety using maximum shear stress theory is $3.4$.

Substitute $-10.96\text{MPa}$ for ${\sigma }_1$, $-114.04\text{MPa}$ for ${\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (IV).

    $\begin{array}{c}n=\frac{350\text{MPa}}{\sqrt{{\left(-10.96\text{MPa}\right)}^2-\left(-10.96\text{MPa}\right)\left(-114.04\text{MPa}\right)+{\left(-114.04\text{MPa}\right)}^2}}\\ =\frac{350\text{MPa}}{108.97\text{MPa}}\\ =3.21\end{array}$

Thus, the factor of the safety using maximum distortion energy theory is $3.21$.

To determine

The factor of the safety using maximum shear stress theory.

The factor of the safety using maximum distortion energy theory.

If ${\sigma }_x=100\text{Mpa, }{\sigma }_y=20\text{MPa,}{\tau }_{xy}=-20\text{MPa}$.

Answer to Problem 1P

The factor of the safety using maximum shear stress theory is $3.34$.

The factor of the safety using maximum distortion energy theory is $3.57$.

Explanation of Solution

Conclusion:

Substitute $100\text{MPa}$ for ${\sigma }_x$, $20\text{MPa}$ for ${\sigma }_y$, and $-20\text{MPa}$ for ${\tau }_{xy}$ in Equation (I).

    $\begin{array}{c}{\sigma }_1=\left(\frac{100\text{MPa}+20\text{MPa}}{2}\right)+\sqrt{{\left(\frac{100\text{MPa}-20\text{MPa}}{2}\right)}^2+{\left(-20\text{MPa}\right)}^2}\\ =60\text{MPa}+44.7213\text{MPa}\\ =104.7213\text{MPa}\\ \simeq 104.7\text{MPa}\end{array}$

Substitute $100\text{MPa}$ for ${\sigma }_x$, $20\text{MPa}$ for ${\sigma }_y$, and $-20\text{MPa}$ for ${\tau }_{xy}$ in Equation (II).

    $\begin{array}{c}{\sigma }_2=\left(\frac{100\text{MPa}+20\text{MPa}}{2}\right)-\sqrt{{\left(\frac{100\text{MPa}-20\text{MPa}}{2}\right)}^2+{\left(-20\text{MPa}\right)}^2}\\ =60\text{MPa}-44.7213\text{MPa}\\ =15.2787\text{MPa}\\ \simeq 15.3\text{MPa}\end{array}$

Calculate ${\sigma }_1-{\sigma }_2$.

    $\begin{array}{c}{\sigma }_1-{\sigma }_2=\left(104.7\text{MPa}\right)-\left(15.3\text{MPa}\right)\\ =89.4\text{MPa}\end{array}$

The maximum values among $\left({\sigma }_1-{\sigma }_2\right)$, ${\sigma }_2$, and ${\sigma }_1$ is ${\sigma }_1$.

Substitute $104.7\text{MPa}$ for $\left({\sigma }_1-{\sigma }_2\right)$, and $350\text{MPa}$ for $S_{yt}$ in Equation (III).

    $\begin{array}{c}n=\frac{350\text{MPa}}{104.7\text{MPa}}\\ =3.3428\\ \approx 3.34\end{array}$

Thus, the factor of the safety using maximum shear stress theory is $3.34$.

Substitute $103.25\text{MPa}$ for ${\sigma }_1$, $-23.25\text{MPa}$ for ${\sigma }_2$ and $350\text{MPa}$ for $S_{yt}$ in Equation (IV).

    $\begin{array}{c}n=\frac{350\text{MPa}}{\sqrt{{\left(104.7\text{MPa}\right)}^2-\left(104.7\text{MPa}\times 15.3\text{MPa}\right)+{\left(15.3\text{MPa}\right)}^2}}\\ =\frac{350\text{MPa}}{97.9503\text{MPa}}\\ =3.573\\ \simeq 3.57\end{array}$

Thus, the factor of the safety using maximum distortion energy theory is $3.57$.