Chapter 4.6, Problem 90P

(a)

To determine

Find the residual stress at $y=4.5\text{in}\text{.}$

Answer to Problem 90P

The residual stress is $\underset{\_}{{\sigma }_{\text{res}}=-22.19\text{ksi}}$.

Explanation of Solution

Given information:

The yield stress for the beam is ${\sigma }_Y=42\text{ksi}$.

The Young’s modulus of steel is $E=29\times {10}^6\text{psi}$.

Calculation:

Show the cross-section of the beam as shown in Figure 1.

Refer Figure 1.

Calculate the area of the cross section $\left(A\right)$ as shown below.

$A=bd$ (1)

Here, b is the width of the cross section and d is the depth of the cross section.

Calculate the area of the portion (1) $\left(A_1\right)$ as shown below.

Substitute $6\text{in}\text{.}$ for b and $3\text{in}\text{.}$ for d in Equation (1).

$\begin{array}{c}{A}_1=6\times 3\\ =18\text{in}{\text{.}}^2\end{array}$

Calculate the area of the portion (2) $\left(A_2\right)$ as shown below.

Substitute $3\text{in}\text{.}$ for b and $3\text{in}\text{.}$ for d in Equation (1).

$\begin{array}{c}{A}_2=3\times 3\\ =9\text{in}{\text{.}}^2\end{array}$

Calculate the moment of inertia $\left(I\right)$ as shown below.

$I=\frac{b{d}^3}{12}+A{h}^2$ (2)

Calculate the moment of inertia of portion (1) $\left(I_1\right)$ as shown below.

Substitute $6\text{in}\text{.}$ for b, $3\text{in}\text{.}$ for d, $18\text{in}{\text{.}}^2$ for A and $3\text{in}\text{.}$ for h in Equation (2).

$\begin{array}{c}{I}_1=\frac{6\times 3^3}{12}+\left(18\times 3^2\right)\\ =175.5\text{in}{\text{.}}^4\end{array}$

Calculate the moment of inertia of portion (2) $\left(I_2\right)$ as shown below.

Substitute $3\text{in}\text{.}$ for b, $3\text{in}\text{.}$ for d, $9\text{in}{\text{.}}^2$ for A and $0$ for h in Equation (2).

$\begin{array}{c}{I}_2=\frac{3\times 3^3}{12}+0\\ =6.75\text{in}{\text{.}}^4\end{array}$

Calculate the total moment of inertia $\left(I\right)$ as shown below.

$I=2I_1+I_2$

Substitute $175.5\text{in}{\text{.}}^4$ for $I_1$ and $6.75\text{in}{\text{.}}^4$ for $I_2$.

$\begin{array}{c}I=2\times 175.5+6.75\\ =357.75\text{in}{\text{.}}^4\end{array}$

Calculate the centroid (c) as shown below.

$c=\frac{h}{2}$

Substitute $9\text{in}\text{.}$ for h.

$\begin{array}{c}c=\frac{9}{2}\\ =4.5\text{in}\text{.}\end{array}$

Sketch the stress acting on the cross-section of the beam as shown in Figure 2.

Refer Figure 2.

Calculate the area of the portion (2) $\left(A_2\right)$ as shown below.

$A_2=\frac{1}{2}bh$

Substitute $3\text{in}\text{.}$ for b and $1.5\text{in}\text{.}$ for d.

$\begin{array}{c}{A}_2=\frac{1}{2}\times 1.5\times 3\\ =2.25\text{in}{\text{.}}^2\end{array}$

Calculate the reaction applied to portion (1) $\left(R_1\right)$ as shown below.

$R_1={\sigma }_Y{A}_1$

Substitute $42\text{ksi}$ for ${\sigma }_Y$ and $18\text{in}{\text{.}}^2$ for $A_1$.

$\begin{array}{c}{R}_1=42\times 18\\ =756\text{kips}\end{array}$

Calculate the reaction applied to portion (2) $\left(R_2\right)$ as shown below.

$R_2={\sigma }_Y{A}_2$

Substitute $42\text{ksi}$ for ${\sigma }_Y$ and $2.25\text{in}{\text{.}}^2$ for $A_2$.

$\begin{array}{c}{R}_2=42\times 2.25\\ =94.5\text{kips}\end{array}$

Calculate the moment $\left(M\right)$ as shown below.

$M=2\left(R_1{y}_1+R_2{y}_2\right)$

Substitute $756\text{kips}$ for $R_1$, $\left(1.5+\frac{3}{2}\right)\text{in}\text{.}$ for $y_1$, $94.5\text{kips}$ for $R_2$, and $\left(\frac{2}{3}1.5\right)\text{in}\text{.}$ for $y_2$.

$\begin{array}{c}M=2\left(756\times \left(1.5+\frac{3}{2}\right)+94.5\times \left(\frac{2}{3}\times 1.5\right)\right)\\ =4,725\text{kip}\cdot \text{in}\text{.}\end{array}$

Calculate the stress $\left({\sigma }^{\prime }\right)$ as shown below.

${\sigma }^{\prime }=\frac{Mc}{I}$

Substitute $4,725\text{kip}\cdot \text{in}\text{.}$ for M, $4.5\text{in}\text{.}$ for c, and $357.75\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}{\sigma }^{\prime }=\frac{4,725\times 4.5}{357.75}\\ =59.434\text{ksi}\end{array}$

Calculate the stress $\left({\sigma }^{″}\right)$ as shown below.

${\sigma }^{″}=\frac{My}{I}$

Substitute $4,725\text{kip}\cdot \text{in}\text{.}$ for M, $1.5\text{in}\text{.}$ for y, and $357.75\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}{\sigma }^{″}=\frac{4,725\times 1.5}{357.75}\\ =19.811\text{ksi}\end{array}$

Calculate the residual stress at $y=c$ as shown below.

${\sigma }_{\text{res}}={\sigma }^{\prime }-{\sigma }_Y$

Substitute $59.434\text{ksi}$ for ${\sigma }^{\prime }$ and $42\text{ksi}$ for ${\sigma }_Y$.

$\begin{array}{c}{\sigma }_{\text{res}}=59.434-42\\ =17.434\text{ksi}\end{array}$

Calculate the residual stress at $y=y_Y$ as shown below.

${\sigma }_{\text{res}}={\sigma }^{″}-{\sigma }_Y$

Substitute $19.811\text{ksi}$ for ${\sigma }^{\prime }$ and $42\text{ksi}$ for ${\sigma }_Y$.

$\begin{array}{c}{\sigma }_{\text{res}}=19.811-42\\ =-22.189\text{ksi}\end{array}$

Sketch the stress distribution as shown in Figure 3.

Hence, the residual stress is $\underset{\_}{{\sigma }_{\text{res}}=-22.19\text{ksi}}$.

(b)

To determine

Find the point where the residual stress is zero.

Answer to Problem 90P

The point where the residual stress is zero is $\underset{\_}{y_0=3.18\text{in}\text{.}}$.

Explanation of Solution

Given information:

The yield stress for the beam is ${\sigma }_Y=42\text{ksi}$.

The Young’s modulus of steel is $E=29\times {10}^6\text{psi}$.

Calculation:

Consider that the residual stress ${\sigma }_{\text{res}}=0$.

Calculate the yield stress $\left({\sigma }^{\prime }\right)$ as shown below.

${\sigma }^{\prime }=\frac{M{y}_0}{I}$

Calculate the point where the residual stress is zero as shown below.

${\sigma }_{\text{res}}={\sigma }^{\prime }-{\sigma }_Y$

Substitute $\frac{M{y}_0}{I}$ for ${\sigma }^{\prime }$ and $0$ for ${\sigma }_{\text{res}}$.

$\begin{array}{l}0=\frac{M{y}_0}{I}-{\sigma }_Y\\ {\sigma }_Y=\frac{M{y}_0}{I}\\ y_0=\frac{{\sigma }_{Y}I}{M}\end{array}$

Substitute $42\text{ksi}$ for ${\sigma }_Y$, $4,725\text{kip}\cdot \text{in}\text{.}$ for M, and $357.75\text{in}{\text{.}}^{{}^4}$ for I.

$\begin{array}{c}{y}_0=\frac{42\times 357.75}{4,725}\\ =3.18\text{in}\text{.}\end{array}$

Therefore, the point where the residual stress is zero is $\underset{\_}{y_0=3.18\text{in}\text{.}}$.

(c)

To determine

Find the radius of curvature corresponding to the permanent deformation of the bar.

Answer to Problem 90P

The radius of curvature is $\underset{\_}{\rho =163.4\text{ft}}$.

Explanation of Solution

Given information:

The yield stress for the beam is ${\sigma }_Y=42\text{ksi}$.

The Young’s modulus of steel is $E=29\times {10}^6\text{psi}$.

Calculation:

Refer to part (a).

The residual stress ${\sigma }_{\text{res}}=-22.19\text{ksi}$.

Calculate the radius of curvature $\left(\rho \right)$ as shown below.

${\sigma }^{\prime }=\frac{M{y}_0}{I}$

Calculate the point where the residual stress is zero as shown below.

$\rho =-\frac{Ey}{\sigma }$

Substitute $29\times {10}^6\text{psi}$ for E, $-22.19\text{ksi}$ for $\sigma$ and $1.5\text{in}\text{.}$ for $y$.

$\begin{array}{c}\rho =-\frac{29\times {10}^6\text{psi}\times \frac{1\text{ksi}}{1,000\text{psi}}\times 1.5\text{in}\text{.}}{-22.19\text{ksi}}\\ =1,960.34\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =163.4\text{ft}\end{array}$

Therefore, the radius of curvature is $\underset{\_}{\rho =163.4\text{ft}}$.