Chapter 4.3, Problem 19P

4.19 and 4.20 Knowing that for the extruded beam shown the allowable stress is 120 Mpa in tension and 150 Mpa in compression, determine the largest couple M that can be applied.

Fig. P4.19

To determine

Find the largest couple moment (M) that can be applied to the beam.

Answer to Problem 19P

The largest couple moment (M) applied to the beam is $\underset{\_}{3.79\text{kN}\cdot \text{m}}$.

Explanation of Solution

Given information:

The allowable stress in tension is ${\sigma }_{\text{bottom}}=120\text{MPa}$.

The allowable stress in compression is ${\sigma }_{\text{top}}=150\text{MPa}$.

Calculation:

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

Refer Figure 1.

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

$\begin{array}{c}{\overline{y}}_0=\frac{\left(A_1\right)y_1+\left(A_2\right)y_2}{A_1+A_2}\\ =\frac{\left(b_1{h}_1\right)y_1+\left(\frac{1}{2}\times b_2{h}_2\right)y_2}{b_1{h}_1+\frac{1}{2}\times b_2{h}_2}\end{array}$ (1)

Substitute $40\text{mm}$ for $b_1$, $54\text{mm}$ for $h_1$, $40\text{mm}$ for $b_2$, $54\text{mm}$ for $h_2$, $27\text{mm}$ for $y_1$, and $36\text{mm}$ for $y_2$ in Equation (1).

$\begin{array}{c}{\overline{y}}_0=\frac{\left(40\times 54\right)\times 27+0.5\left(40\times 54\right)\times 36}{40\times 54+0.5\times 40\times 54}\\ =\frac{97,200}{3,240}\\ =30\text{mm}\end{array}$

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

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

Substitute $40\text{mm}$ for $b_1$, $54\text{mm}$ for $h_1$, $27\text{mm}$ for $y_1$ and $30\text{mm}$ for ${\overline{y}}_0$ in Equation (2).

$\begin{array}{c}{I}_1=\left(\frac{1}{12}\times 40\times {54}^3\right)+40\times 54\times {\left(30-27\right)}^2\\ =\left(\frac{1}{12}\times 40\times {54}^3\right)+\left(40\times 54\times 9\right)\\ =524.88\times {10}^3+19.44\times {10}^3\\ =544.32\times {10}^3{\text{mm}}^4\end{array}$

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

$\begin{array}{c}{I}_2=\frac{1}{36}{b}_2{h}_2^3+A_2{\left({\overline{y}}_0-y_2\right)}^2\\ =\frac{1}{36}{b}_2{h}_2^3+\frac{1}{2}\times \left(b_2{h}_2\right){\left({\overline{y}}_0-y_2\right)}^2\end{array}$ (3)

Substitute $40\text{mm}$ for $b_2$, $54\text{mm}$ for $h_2$, $36\text{mm}$ for $y_2$ and $30\text{mm}$ for ${\overline{y}}_0$ in Equation (3).

$\begin{array}{c}{I}_2=\frac{1}{36}\times 40\times {54}^3+\frac{1}{2}\times 40\times 54\times {\left(36-30\right)}^2\\ =174.96\times {10}^3+38.88\times {10}^3\\ =213.84\times {10}^3{\text{mm}}^4\end{array}$

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

$I=I_1+I_2$ (4)

Substitute $544.32\times {10}^3{\text{mm}}^4$ for $I_1$, and $213.84\times {10}^3{\text{mm}}^4$ for $I_2$ in Equation (4).

$\begin{array}{c}I=544.32\times {10}^3+213.84\times {10}^3\\ =758.16\times {10}^3{\text{mm}}^4\end{array}$

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:

$\begin{array}{c}{y}_{\text{bottom}}=-{\overline{y}}_0\\ =-30\text{mm}\end{array}$

$\begin{array}{c}{y}_{\text{top}}=54-{\overline{y}}_0\\ =54\text{mm}-30\text{mm}\\ =24\text{mm}\end{array}$

Calculate the maximum moment $\left(M_{\text{top}}\right)$ at the top fiber as follows:

$\left|{\sigma }_{\text{top}}\right|=\left|\frac{M_{\text{top}}{y}_{\text{top}}}{I}\right|$ (5)

Substitute $120\text{MPa}$ for ${\sigma }_{\text{top}}$, $24\text{mm}$ for $y_{\text{top}}$, and $146.383\times {10}^3{\text{mm}}^4$ for I in Equation (5).

$\begin{array}{c}\left|120\text{MPa}\times \left(\frac{{10}^6\frac{\text{N}}{{\text{m}}^2}}{1\text{MPa}}\right)\right|=\left|\frac{M_{\text{top}}\times 24\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{758.16\times {10}^3{\text{mm}}^4\times \left(\frac{1{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)}\right|\\ 120\times {10}^6=\frac{0.024M_{\text{top}}}{758.16\times {10}^{-9}}\\ M_{\text{top}}=\frac{120\times {10}^6\times 758.16\times {10}^{-9}}{0.024}\\ =\frac{90.979}{0.024}\end{array}$

$M_{\text{top}}=3.7908\times {10}^3\text{N}\cdot \text{m}$

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

$\left|{\sigma }_{\text{bottom}}\right|=\left|\frac{M_{\text{bottom}}{y}_{\text{bottom}}}{I}\right|$ (6)

Substitute $150\text{MPa}$ for ${\sigma }_{\text{bottom}}$, $30\text{mm}$ for $y_{\text{bottom}}$, and $758.16\times {10}^3{\text{mm}}^4$ for I in Equation (6).

$\begin{array}{c}\left|150\text{MPa}\times \left(\frac{{10}^6\frac{\text{N}}{{\text{m}}^2}}{1\text{MPa}}\right)\right|=\left|\frac{M_{\text{bottom}}\times 30\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{758.16\times {10}^3{\text{mm}}^4\times \left(\frac{1{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)}\right|\\ 150\times {10}^6=\frac{0.03M_{\text{bottom}}}{758.16\times {10}^{-9}}\\ M_{\text{bottom}}=\frac{150\times {10}^6\times 758.16\times {10}^{-9}}{0.03}\\ =\frac{113.724}{0.02341}\end{array}$

$\begin{array}{c}{M}_{\text{bottom}}=3.7908\times {10}^3\text{N}\cdot \text{m}\\ =3.7908\times {10}^3\text{N}\cdot \text{m}\times \frac{1\text{kN}}{{10}^3\text{N}}\\ =3.79\text{kN}\cdot \text{m}\end{array}$

Compare $M_{\text{bottom}}=3.7908\times {10}^3\text{N}\cdot \text{m}$ and $M_{\text{top}}=3.7908\times {10}^3\text{N}\cdot \text{m}$.

Get the largest moment applied on the beam as $M=3.7908\times {10}^3\text{N}\cdot \text{m}$.

Thus, the largest couple moment (M) applied to the beam is $\underset{\_}{3.79\text{kN}\cdot \text{m}}$.