Chapter 4.3, Problem 10P

4.9 through 4.11 Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.

Fig. P4.10

To determine

The maximum tensile and compressive stresses in portion BC of the beam.

Answer to Problem 10P

The maximum compressive and tensile stress in the in the section BC are $\underset{\_}{-10.38\text{ksi}}$ and $\underset{\_}{15.40\text{ksi}}$.

Explanation of Solution

Given information:

Calculation:

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

Refer to Figure 1.

Calculate the value of ${\overline{y}}_0$ using the relation:

$\begin{array}{c}{\overline{y}}_0=\frac{\left(A_1\right)y_1+\left(A_2\right)y_2+\left(A_3\right)y_3}{A_1+A_2+A_3}\\ =\frac{\left(b_1{h}_1\right)y_1+\left(b_2{h}_2\right)y_2+\left(b_3{h}_3\right)y_3}{b_1{h}_1+b_2{h}_2+b_3{h}_3}\end{array}$

Substitute $8\text{in}\text{.}$ for $b_1$, $1\text{in}\text{.}$ for $h_1$, $1\text{in}\text{.}$ for $b_2$, $6\text{in}\text{.}$ for $h_2$, $4\text{in}\text{.}$ for $b_3$, $1\text{in}\text{.}$ for $h_3$, $7.5\text{in}\text{.}$ for $y_1$, $4\text{in}\text{.}$ for $y_2$, and $0.5\text{in}\text{.}$ for $y_3$.

$\begin{array}{c}{\overline{y}}_0=\frac{\left(8\times 1\right)\times 7.5+\left(1\times 6\right)\times 4+\left(4\times 1\right)\times 0.5}{8\times 1+1\times 6+4\times 1}\\ =\frac{86}{18}\\ =4.778\text{in}\text{.}\end{array}$

Calculate the moment of inertia $\left(I_1\right)$ of the rectangle 1 as follows:

$\begin{array}{c}{I}_1=\frac{1}{12}{b}_1{h}_1^3+A_1{\left(y_1-{\overline{y}}_0\right)}^2\\ I_1=\frac{1}{12}{b}_1{h}_1^3+\left(b_1{h}_1\right){\left(y_1-{\overline{y}}_0\right)}^2\end{array}$ (1)

Substitute $8\text{in}\text{.}$ for $b_1$, $1\text{in}\text{.}$ for $h_1$, $7.5\text{in}\text{.}$ for $y_1$ and $4.778\text{in}\text{.}$ for ${\overline{y}}_0$ in Equation (1).

$\begin{array}{c}{I}_1=\frac{1}{12}\times 8\times 1^3+\left[8\times 1\times {\left(7.5-4.778\right)}^2\right]\\ =\frac{1}{12}\times 8\times 1^3+\left(8\times 1\times {2.722}^2\right)\\ =0.666+59.274\\ =59.94\text{in}{\text{.}}^4\end{array}$

Calculate the moment of inertia $\left(I_2\right)$ of the rectangle 2 as follows:

$\begin{array}{c}{I}_2=\frac{1}{12}{b}_2{h}_2^3+A_2{\left({\overline{y}}_0-y_1\right)}^2\\ I_2=\frac{1}{12}{b}_2{h}_2^3+\left(b_2{h}_2\right){\left({\overline{y}}_0-y_1\right)}^2\end{array}$ (2)

Substitute $1\text{in}\text{.}$ for $b_2$, $6\text{in}\text{.}$ for $h_2$, $4\text{in}\text{.}$ for $y_2$ and $4.778\text{in}\text{.}$ for ${\overline{y}}_0$ in Equation (2).

$\begin{array}{c}{I}_2=\left(\frac{1}{12}\times 1\times 6^3\right)+\left[1\times 6\times {\left(4.778-4\right)}^2\right]\\ =18+3.63\\ =21.63\text{in}{\text{.}}^4\end{array}$

Calculate the moment of inertia $\left(I_3\right)$ of the rectangle 3 as follows:

$\begin{array}{c}{I}_3=\frac{1}{12}{b}_3{h}_3^3+A_3{\left({\overline{y}}_0-y_3\right)}^2\\ I_3=\frac{1}{12}{b}_3{h}_3^3+\left(b_3{h}_3\right){\left({\overline{y}}_0-y_3\right)}^2\end{array}$ (3)

Substitute $4\text{in}\text{.}$ for $b_3$, $1\text{in}\text{.}$ for $h_3$, $0.5\text{in}\text{.}$ for $y_3$ and $4.778\text{in}\text{.}$ for ${\overline{y}}_0$ in Equation (3).

$\begin{array}{c}{I}_3=\left(\frac{1}{12}\times 4\times 1^3\right)+\left[4\times 1\times {\left(4.778-0.5\right)}^2\right]\\ =0.333+73.20\\ =73.54\text{in}{\text{.}}^4\end{array}$

Calculate the total moment of inertia (I) of the cross-section as follows:

$I=I_1+I_2+I_3$ (4)

Substitute $59.94\text{in}{\text{.}}^4$ for $I_1$, $21.63\text{in}{\text{.}}^4$ for $I_2$ and $73.54\text{in}{\text{.}}^4$ for $I_3$ in Equation (4).

$\begin{array}{c}I=59.94+21.63+73.54\\ =155.11\text{in}{\text{.}}^4\end{array}$

Refer to Figure 1.

Consider the distance between the neutral axis and the top fiber and bottom fiber of the beam is $y_{\text{top}}$ and $y_{\text{bottom}}$.

Calculate $y_{\text{top}}$ and $y_{\text{bottom}}$ as follows:

$\begin{array}{c}{y}_{\text{bottom}}=-{\overline{y}}_0\\ =-4.778\text{in}\text{.}\end{array}$

$\begin{array}{c}{y}_{\text{top}}=8\text{in}\text{.}-{\overline{y}}_0\\ =8\text{in}\text{.}-4.778\text{in}\text{.}\\ =\text{3}\text{.222}\text{in}\text{.}\end{array}$

Show the section of the beam left of C as shown in Figure 2.

Refer to Figure 2.

Calculate the moment M using the relation:

$M=Pa$ (5)

Substitute $25\text{kip}$ for P, and $20\text{in}\text{.}$ for a in Equation (5).

$\begin{array}{c}M=25\times 20\\ =500\text{kip}\cdot \text{in}\text{.}\end{array}$

Calculate the value of stress $\left({\sigma }_{\text{top}}\right)$ at the top fiber as follows:

${\sigma }_{\text{top}}=-\frac{M{y}_{\text{top}}}{I}$ (6)

Substitute $500\text{kip}\cdot \text{in}\text{.}$ for M, $\text{3}\text{.222}\text{in}\text{.}$ for $y_{\text{top}}$, and $155.11\text{in}{\text{.}}^4$ for I in Equation (6).

$\begin{array}{c}{\sigma }_{\text{top}}=-\frac{500\times \text{3}\text{.222}}{155.11}\\ =-\frac{1,611}{155.11}\\ =-10.38\text{ksi}\\ =10.38\text{ksi}\left(\text{compressive}\right)\end{array}$

Calculate the value of stress $\left({\sigma }_{\text{bottom}}\right)$ at the top fiber as follows:

${\sigma }_{\text{bottom}}=-\frac{M{y}_{\text{bottom}}}{I}$ (7)

Substitute $500\text{kip}\cdot \text{in}\text{.}$ for M, $\text{-4}\text{.778}\text{in}\text{.}$ for $y_{\text{bottom}}$, and $155.11\text{in}{\text{.}}^4$ for I in Equation (7).

$\begin{array}{c}{\sigma }_{\text{bottom}}=-\frac{500\times \left(-4.778\right)}{155.11}\\ =\frac{2,389}{155.11}\\ =15.40\text{ksi}\left(\text{tensile}\right)\end{array}$

Thus, the maximum compressive and tensile stress in the in the section BC are $\underset{\_}{-10.38\text{ksi}}$ and $\underset{\_}{15.40\text{ksi}}$.