Chapter 4, Problem 192RP

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. P4.192

To determine

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

Answer to Problem 192RP

The maximum tensile and compressive stress in the portion BC of the beam are $\underset{\_}{{\sigma }_{top}=-81.8\text{MPa}}$ and $\underset{\_}{{\sigma }_{bottom}=67.8\text{MPa}}$

Explanation of Solution

Given information:

The load P acting on the beam is $4\text{kN}$.

Consider the radius of the semi-circular region is $r=25\text{mm}$.

Calculation:

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

Refer Figure 1.

The cross-section of the beam consist of a semi-circle 1 and a rectangle 2.

Calculate the area of the semi-circle 1 and a rectangle 2 as follows:

$\begin{array}{c}{A}_1=\frac{\pi }{2}{r}^2\\ =\frac{\pi }{2}{\left(25\right)}^2\\ =981.7{\text{mm}}^2\end{array}$

$\begin{array}{c}{A}_2=bh\\ =50\times 25\\ =1,250{\text{mm}}^2\end{array}$

Consider the distance of the centroid of the region 1 and 2 from their bases are denoted by ${\overline{y}}_1$ and ${\overline{y}}_2$.

Calculate the value of the distances ${\overline{y}}_1$ as follows:

${\overline{y}}_1=\frac{4r}{3\pi }$

Substitute $25\text{mm}$ for $r$.

$\begin{array}{c}{\overline{y}}_1=\frac{4\left(25\right)}{3\pi }\\ =10.610\text{mm}\end{array}$

Calculate the value of the distances ${\overline{y}}_2$ as follows:

${\overline{y}}_2=-\frac{h}{2}$

Substitute $25\text{mm}$ for $h$.

$\begin{array}{c}{\overline{y}}_2=-\frac{25}{2}\\ =-12.5\text{mm}\end{array}$

Calculate the distance $\left(\overline{y}\right)$ of the neutral axis of the entire cross-section using the relation:

$\overline{y}=\frac{A_1{\overline{y}}_1+A_1{\overline{y}}_1}{A_1+A_2}$ (1)

Substitute $981.7{\text{mm}}^2$ for $A_1$, $10.610\text{mm}$ for ${\overline{y}}_1$, $1,250{\text{mm}}^2$ for $A_2$, $-12.5\text{mm}$ for ${\overline{y}}_2$ in Equation (1).

$\begin{array}{c}\overline{y}=\frac{981.7\times 10.61+1250\times \left(-12.5\right)}{1250+981.7}\\ =-2.334\text{in}\text{.}\end{array}$

Calculate the total moment of inertia of the cross-section (I) using the rerlation:

$\begin{array}{c}I=I_1+I_2\\ =\left({\overline{I}}_1+A_1{d}_1^2\right)+\left({\overline{I}}_2+A_2{d}_2^2\right)\\ =\left(\left[I_{x_1}-A_1{\overline{y}}_1^2\right]+A_1{d}_1^2\right)+\left({\overline{I}}_2+A_2{d}_2^2\right)\\ =\left\{\left(I_{x_1}-A_1{\overline{y}}_1^2\right)+A_1{\left({\overline{y}}_1-\overline{y}\right)}^2\right\}+\left\{{\overline{I}}_2+A_2{\left({\overline{y}}_2-\overline{y}\right)}^2\right\}\end{array}$

$I=\left\{\frac{b_1{h}_1^3}{12}-A_1{\overline{y}}_1^2+A_1{\left(\overline{y}-{\overline{y}}_1\right)}^2\right\}+\left\{\frac{\pi }{8}{r}^4+A_2{\left(\overline{y}-{\overline{y}}_2\right)}^2\right\}$

Substitute $50\text{mm}$ for $b_1$, $25\text{mm}$ for $h_1$, $981.7{\text{mm}}^2$ for $A_1$, $25\text{mm}$ for $r$, $1,250{\text{mm}}^2$ for $A_2$, $-12.5\text{mm}$ for ${\overline{y}}_2$, $10.610\text{mm}$ for ${\overline{y}}_1$, and $-2.334\text{mm}$ for $\overline{y}$.

$\begin{array}{c}I=\left\{\frac{50\times {25}^3}{12}-981.7\times {10.610}^2+981.7{\left(10.610+2.334\right)}^2\right\}+\left\{\frac{\pi }{8}{\left(25\right)}^4+1250{\left(-12.5\text{+}2.334\right)}^2\right\}\\ =\left\{-45,407.86+164,481.02\right\}+\left\{282,582.52\right\}\\ =401.6\times {10}^3{\text{mm}}^4\end{array}$

Show the forces acting on the beam as shown in Figure 2.

Calculate the value of moment M as follows:

$\begin{array}{c}M=Pa\\ =4\text{kN}\times \text{300}\text{mm}\\ \text{=1200}\text{kN}\cdot \text{mm}\end{array}$

Calculate the stress at the top fiber as follows:

${\sigma }_{top}=\frac{-M{y}_{top}}{I}$ (2)

Here, $y_{top}$ is the distance of the neutral axis to the top fiber.

Calculate the value of $y_{top}$ as follows:

$\begin{array}{c}{y}_{top}=25+2.334\\ =27.334\text{mm}\end{array}$

Substitute $\text{1200}\text{kN}\cdot \text{mm}$ for M, $27.334\text{mm}$ for $y_{top}$ and $401.6\times {10}^3{\text{mm}}^4$ for I in Equation (2).

$\begin{array}{c}{\sigma }_{top}=\frac{-\text{1200}\times 27.334}{401.6\times {10}^3}\\ =-81.76\times {10}^{-3}\frac{\text{kN}}{{\text{mm}}^2}\times \left(\frac{1,000\text{MPa}}{1\frac{\text{kN}}{{\text{mm}}^2}}\right)\\ \approx -81.8\text{MPa}\end{array}$

Calculate the stress at the top fiber as follows:

${\sigma }_{bottom}=\frac{-M{y}_{bottom}}{I}$ (3)

Here, $y_{bottom}$ is the distance of the neutral axis to the bottom fiber.

Calculate the value of $y_{top}$ as follows:

$\begin{array}{c}{y}_{bottom}=-25+2.334\\ =22.666\text{mm}\end{array}$

Substitute $\text{1200}\text{kN}\cdot \text{mm}$ for M, $-22.666$ for $y_{bottom}$ and $401.6\times {10}^3{\text{mm}}^4$ for I in Equation (2).

$\begin{array}{c}{\sigma }_{bottom}=\frac{-\text{1200}\times \left(-22.666\right)}{401.6\times {10}^3}\\ =67.8\times {10}^{-3}\frac{\text{kN}}{{\text{mm}}^2}\times \left(\frac{1,000\text{MPa}}{1\frac{\text{kN}}{{\text{mm}}^2}}\right)\\ \approx 67.8\text{MPa}\end{array}$

Thus, the maximum tensile and compressive stress in the portion BC of the beam are $\underset{\_}{{\sigma }_{top}=-81.8\text{MPa}}$ and $\underset{\_}{{\sigma }_{bottom}=67.8\text{MPa}}$.