Chapter 11.3, Problem 39P

(a)

To determine

The factor of safety associated with the yield strength ${\sigma }_z=+16\text{ksi}$.

Answer to Problem 39P

The factor of safety associated with the yield strength ${\sigma }_z=+16\text{ksi}$ is $\underset{\_}{F.S.=2.33}$.

Explanation of Solution

Given information:

The stress component along x direction is ${\sigma }_x=8\text{ksi}$.

The stress component along z direction is ${\sigma }_z=0$.

The shear stress component is ${\tau }_{xz}=14\text{ksi}$.

The yield stress is ${\sigma }_Y=65\text{ksi}$.

The principal stress is ${\sigma }_z=+16\text{ksi}$.

Calculation:

Sketch the state of stress in a machine component as shown in Figure 1.

Refer to Figure 1.

$\begin{array}{l}{\sigma }_x=8\text{ksi}\\ {\sigma }_z=0\\ {\tau }_{xz}=14\text{ksi}\end{array}$

Apply the procedure to construct the Mohr’s circle as shown below.

• Find the centre of the circle C located ${\sigma }_{\text{avg}}=\frac{{\sigma }_x+{\sigma }_z}{2}$ from the origin.
• Plot the reference points A having coordinates $A\left({\sigma }_x,{\tau }_A\right)$.
• Connect the point A with C and from the shaded triangle find the radius R of the circle.
• Sketch the circle once R has been determined.

Construct the Mohr’s circle as shown below.

Calculate the centre of the circle $\left({\sigma }_{\text{avg}}\right)$ using average normal stress as shown below.

${\sigma }_{\text{avg}}=\frac{{\sigma }_x+{\sigma }_z}{2}$

Substitute $8\text{ksi}$ for ${\sigma }_x$ and $0$ for ${\sigma }_z$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{8+0}{2}\\ =4\text{ksi}\end{array}$

The centre of the circle is $C=4\text{ksi}$.

Coordinates of the reference point z.

$z=\left({\sigma }_z,-{\tau }_{xz}\right)$

Substitute $0$ for ${\sigma }_z$ and $14\text{ksi}$ for ${\tau }_{xz}$.

$z=\left(\text{0, }-14\text{ ksi}\right)$

Coordinates of the reference point x.

$x=\left({\sigma }_x,{\tau }_{xz}\right)$

Substitute $8\text{ksi}$ for ${\sigma }_x$ and $14\text{ksi}$ for ${\tau }_{xz}$.

$x=\left(8\text{ksi},14\text{ksi}\right)$

Calculate the radius (R) of the circle as shown below.

$R=\sqrt{{\left({\sigma }_x-{\sigma }_{\text{avg}}\right)}^2+{\left({\tau }_{xz}\right)}^2}$

Substitute $8\text{ksi}$ for ${\sigma }_x$, $4\text{ksi}$ for ${\sigma }_{\text{avg}}$, and $14\text{ksi}$ for ${\tau }_{xz}$.

$\begin{array}{c}R=\sqrt{{\left(8-4\right)}^2+{\left(14\right)}^2}\\ =\sqrt{212}\\ =14.56\text{ksi}\end{array}$

Sketch the Mohr’s circle as shown in Figure 2.

Refer to Figure 2.

Calculate the principal stresses $\left({\sigma }_{\max }\text{and }{\sigma }_{\min }\right)$ as shown below.

${\sigma }_{\max ,\min }={\sigma }_{\text{avg}}\pm R$

Substitute $4\text{ksi}$ for ${\sigma }_{\text{avg}}$ and $14.56\text{ksi}$ for $R$.

$\begin{array}{l}{\sigma }_{\max ,\min }=4\pm 14.56\\ {\sigma }_a=18.56\text{ksi}\\ {\sigma }_b=-10.56\text{ksi}\end{array}$

Also the principal stress ${\sigma }_c={\sigma }_y$.

Apply maximum distortion energy criterion as shown below.

${\left({\sigma }_a-{\sigma }_b\right)}^2+{\left({\sigma }_b-{\sigma }_c\right)}^2+{\left({\sigma }_c-{\sigma }_a\right)}^2=2{\left(\frac{{\sigma }_Y}{F.S.}\right)}^2$ (1)

Substitute $18.56\text{ksi}$ for ${\sigma }_a$, $-10.56\text{ksi}$ for ${\sigma }_b$, $+16\text{ksi}$ for ${\sigma }_c$, $2.2$ for $F.S.$, and $65\text{ksi}$ for ${\sigma }_Y$ in Equation (1).

$\begin{array}{l}{\left(18.56-\left(-10.56\right)\right)}^2+{\left(\left(-10.56\right)-16\right)}^2+{\left(16-18.56\right)}^2=2\times {\left(\frac{65}{F.S.}\right)}^2\\ 847.9744+705.4336+6.5536=\frac{8,450}{{\left(F.S.\right)}^2}\\ {\left(F.S.\right)}^2=5.4168\\ F.S.=2.33\end{array}$

Hence, the factor of safety associated with the yield strength ${\sigma }_z=+16\text{ksi}$ is $\underset{\_}{F.S.=2.33}$.

(b)

To determine

The factor of safety associated with the yield strength ${\sigma }_z=-16\text{ksi}$.

Answer to Problem 39P

The factor of safety associated with the yield strength ${\sigma }_z=-16\text{ksi}$ is $\underset{\_}{F.S.=2.02}$.

Explanation of Solution

Given information:

The stress component along x direction is ${\sigma }_x=8\text{ksi}$.

The stress component along z direction is ${\sigma }_z=0$.

The shear stress component is ${\tau }_{xz}=14\text{ksi}$.

The yield stress is ${\sigma }_Y=65\text{ksi}$.

The principal stress is ${\sigma }_z=-16\text{ksi}$.

Calculation:

Refer to part (a).

The principal stresses ${\sigma }_a=18.56\text{ksi and }{\sigma }_b=-10.56\text{MPa}$.

Apply maximum distortion energy criterion as shown below.

Substitute $18.56\text{ksi}$ for ${\sigma }_a$, $-10.56\text{ksi}$ for ${\sigma }_b$, $-16\text{ksi}$ for ${\sigma }_c$, $2.2$ for $F.S.$, and $65\text{ksi}$ for ${\sigma }_Y$ in Equation (1).

$\begin{array}{l}{\left(18.56-\left(-10.56\right)\right)}^2+{\left(\left(-10.56\right)-\left(-16\right)\right)}^2+{\left(\left(-16\right)-18.56\right)}^2=2\times {\left(\frac{65}{F.S.}\right)}^2\\ 847.9744+29.5936+1,194.3936=\frac{8,450}{{\left(F.S.\right)}^2}\\ {\left(F.S.\right)}^2=4.0783\\ F.S.=2.02\end{array}$

Therefore, the factor of safety associated with the yield strength ${\sigma }_z=-16\text{ksi}$ is $\underset{\_}{F.S.=2.02}$.