Chapter 4.3, Problem 11P

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. P4.11

To determine

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

Answer to Problem 11P

The maximum compressive and tensile stress in the in the section BC are $\underset{\_}{-102.4\text{MPa}}$ and $\underset{\_}{73.2\text{MPa}}$.

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 $10\text{mm}$ for $b_1$, $60\text{mm}$ for $h_1$, $10\text{mm}$ for $b_2$, $60\text{mm}$ for $h_2$, $30\text{mm}$ for $b_3$, $10\text{mm}$ for $h_3$, $30\text{mm}$ for $y_1$, $30\text{mm}$ for $y_2$, and $5\text{mm}$ for $y_3$.

$\begin{array}{c}{\overline{y}}_0=\frac{\left(10\times 60\right)\times 30+\left(10\times 60\right)\times 30+\left(30\times 10\right)\times 5}{10\times 60+10\times 60+30\times 10}\\ =\frac{37,500}{1,500}\\ =25\text{mm}\end{array}$

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 $10\text{mm}$ for $b_1$, $60\text{mm}$ for $h_1$, $30\text{mm}$ for $y_1$ and $25\text{mm}$ for ${\overline{y}}_0$ in Equation (1).

$\begin{array}{c}{I}_1=\frac{1}{12}\times 10\times {60}^3+10\times 60\times {\left(30-25\right)}^2\\ =\frac{1}{12}\times 10\times {60}^3+10\times 60\times 25\\ =180\times {10}^3+15\times {10}^3\\ =195\times {10}^3{\text{mm}}^4\end{array}$

The moment of inertia of rectangle 2 is same as the moment of inertia of rectangle 1. Then,

$\begin{array}{c}{I}_2=I_1\\ =195\times {10}^3{\text{mm}}^4\end{array}$

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

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

Substitute $30\text{mm}$ for $b_3$, $10\text{mm}$ for $h_3$, $5\text{mm}$ for $y_3$ and $\text{25}\text{mm}$ for ${\overline{y}}_0$ in Equation (2).

$\begin{array}{c}{I}_3=\frac{1}{12}\times 30\times {10}^3+30\times 10\times {\left(25-5\right)}^2\\ =2.5\times {10}^3+120\times {10}^3\\ =122.5\times {10}^3{\text{mm}}^4\end{array}$

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

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

Substitute $195\times {10}^3{\text{mm}}^4$ for $I_1$, $195\times {10}^3{\text{mm}}^4$ for $I_2$ and $122.5\times {10}^3{\text{mm}}^4$ for $I_3$ in Equation (4).

$\begin{array}{c}I=195\times {10}^3+195\times {10}^3+122.5\times {10}^3\\ =512.5\times {10}^3{\text{mm}}^4\end{array}$

Refer 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:

$\begin{array}{c}{y}_{\text{bottom}}=-{\overline{y}}_0\\ =-25\text{mm}\end{array}$

$\begin{array}{c}{y}_{\text{top}}=60-{\overline{y}}_0\\ =60\text{mm}-25\text{mm}\\ =\text{35}\text{mm}\end{array}$

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 $10\text{kN}$ for P, and $150\text{mm}$ for a in Equation (5).

$\begin{array}{c}M=10\text{kN}\times \left(\frac{1,000\text{N}}{1\text{kN}}\right)\times 150\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)\\ =1,500\text{N}\cdot \text{m}\end{array}$

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 $1,500\text{N}\cdot \text{m}$ for M, $\text{35}\text{mm}$ for $y_{\text{top}}$, and $512.5\times {10}^3{\text{mm}}^4$ for I in Equation (6).

$\begin{array}{c}{\sigma }_{\text{top}}=-\frac{1,500\text{N}\cdot \text{m}\times \text{35}\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{512.5\times {10}^3{\text{mm}}^4\times \left(\frac{1{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)}\\ =-\frac{52.5}{512.5\times {10}^{-9}}\\ =-102.4\times {10}^6\frac{\text{N}}{{\text{m}}^2}\times \frac{1\text{MPa}}{{10}^6\frac{\text{N}}{{\text{m}}^2}}\\ =102.4\text{MPa}\left(\text{compressive}\right)\end{array}$

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 $1,500\text{N}\cdot \text{m}$ for M, $\text{-25}\text{mm}$ for $y_{\text{bottom}}$, and $512.5\times {10}^3{\text{mm}}^4$ for I in Equation (6).

$\begin{array}{c}{\sigma }_{\text{bottom}}=-\frac{1,500\text{N}\cdot \text{m}\times \left(\text{-25}\text{mm}\right)\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{512.5\times {10}^3{\text{mm}}^4\times \left(\frac{1{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)}\\ =\frac{37.5}{512.5\times {10}^{-9}}\\ =73.2\times {10}^6\frac{\text{N}}{{\text{m}}^2}\times \frac{1\text{MPa}}{{10}^6\frac{\text{N}}{{\text{m}}^2}}\\ =73.2\text{MPa}\left(\text{tensile}\right)\end{array}$

Thus, the maximum compressive and tensile stress in the in the section BC are $\underset{\_}{-102.4\text{MPa}}$ and $\underset{\_}{73.2\text{MPa}}$.