Chapter 4.10, Problem 161P

For the curved bar shown, determine the stress at point A when (a) h = 50 mm, (b) h = 60 mm.

Fig. P4.161 and P4.162

To determine

The stress at point A.

Answer to Problem 161P

The stress at A is $\underset{\_}{\text{-77}\text{.3}\text{MPa}}$.

Explanation of Solution

Given information:

The value of h is $50\text{mm}$.

The inner $\left(r_1\right)$ and outer radius $\left(r_2\right)$ of the curved bar is $50\text{mm}$ and $110\text{mm}$.

The width and depth of the bar are $b=24\text{mm}$ and $h=50\text{mm}$.

The moment (M) is $600\text{N}\cdot \text{m}$.

Calculation:

Calculate the cross-section area (A) of the bar as follows:

$\begin{array}{c}A=bh\\ =24\times 50\\ =1200{\text{mm}}^2\end{array}$

Calculate the radius (R) of the neutral surface using the relation:

$R=\frac{h}{\ln \frac{r_2}{r_1}}$ (1)

Substitute $50\text{mm}$ for $h$, $50\text{mm}$ for $r_1$, and $100\text{mm}$ for $r_2$ in Equation (1).

$\begin{array}{c}R=\frac{50}{\ln \left(\frac{100}{50}\right)}\\ =\frac{50}{\ln \left(2\right)}\\ =\frac{50}{0.69314}\\ =72.135\text{mm}\end{array}$

Calculate the mean radius $\left(\overline{r}\right)$ of the curved bar using the relation:

$\overline{r}=\frac{1}{2}\left(r_1+r_2\right)$ (2)

Substitute $50\text{mm}$ for $r_1$ and $100\text{mm}$ for $r_2$ in Equation (2).

$\begin{array}{c}\overline{r}=\frac{1}{2}\left(50+100\right)\\ =75\text{mm}\end{array}$

The distance (e) between the neutral axis and the centroid of the cross-section using the relation:

$e=\overline{r}-R$ (3)

Substitute $75\text{mm}$ for $\overline{r}$ and $72.135\text{mm}$ for R in Equation (3).

$\begin{array}{c}e=75-72.135\text{mm}\\ =2.865\text{mm}\end{array}$

Calculate the value of $y_A$ using the relation:

$\begin{array}{c}{y}_A=R-r_1\\ =72.135\text{mm-50}\text{mm}\\ =22.135\text{mm}\end{array}$

The distance $\left(r_A\right)$ between the point A and the point C is $50\text{mm}$.

Calculate the stress at point A using the relation:

${\sigma }_A=-\frac{M{y}_A}{Ae{r}_A}$ (4)

Substitute $600\text{N}\cdot \text{m}$ for M, $22.135\text{mm}$ for $y_A$, $1200{\text{mm}}^2$ for A, $2.865\text{mm}$ for e, $50\text{mm}$ for $\left(r_A\right)$ in Equation (4).

$\begin{array}{c}{\sigma }_A=-\frac{600\text{N}\cdot \text{m}\times 22.135\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{1200{\text{mm}}^2\times \left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\times 2.865\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)\times 50\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}\\ =-\frac{600\times 22.135\times {10}^{-3}}{1200\times {10}^{-6}\times 2.865\times {10}^{-3}\times 50\times {10}^{-3}}\\ =-\frac{13.281}{0.171\times {10}^{-6}}\\ \approx -77.3\times {10}^6\text{Pa}\end{array}$

${\sigma }_A\text{=}-\text{77}\text{.3}\text{MPa}$

Thus, the stress at point A is $\underset{\_}{\text{-77}\text{.3}\text{MPa}}$.

To determine

The stress at point A.

Answer to Problem 161P

The stress at A is $\underset{\_}{-\text{55}\text{.7}\text{MPa}}$.

Explanation of Solution

Given information:

The value of h is $60\text{mm}$.

The inner $\left(r_1\right)$ and outer radius $\left(r_2\right)$ of the curved bar is $50\text{mm}$ and $110\text{mm}$.

The width and depth of the bar are $b=24\text{mm}$ and $h=60\text{mm}$.

The moment (M) is $600\text{N}\cdot \text{m}$.

Calculation:

Calculate the cross-section area (A) of the bar as follows:

$\begin{array}{c}A=bh\\ =24\times 60\\ =1,440{\text{mm}}^2\end{array}$

Calculate the radius (R) of the neutral surface using the relation:

$R=\frac{h}{\ln \frac{r_2}{r_1}}$ (5)

Substitute $60\text{mm}$ for $h$, $50\text{mm}$ for $r_1$, and $110\text{mm}$ for $r_2$ in Equation (5).

$\begin{array}{c}R=\frac{60}{\ln \left(\frac{110}{50}\right)}\\ =\frac{60}{\ln \left(2.2\right)}\\ =\frac{60}{0.69314}\\ =76.09796\text{mm}\end{array}$

Calculate the mean radius $\left(\overline{r}\right)$ of the curved bar using the relation:

$\overline{r}=\frac{1}{2}\left(r_1+r_2\right)$ (6)

Substitute $50\text{mm}$ for $r_1$ and $110\text{mm}$ for $r_2$ in Equation (6).

$\begin{array}{c}\overline{r}=\frac{1}{2}\left(50+110\right)\\ =80\text{mm}\end{array}$

The distance (e) between the neutral axis and the centroid of the cross-section using the relation:

$e=\overline{r}-R$ (7)

Substitute $80\text{mm}$ for $\overline{r}$ and $76.09796\text{mm}$ for R in Equation (7).

$\begin{array}{c}e=80-76.09796\text{mm}\\ =3.90204\text{mm}\end{array}$

Calculate the value of $y_A$ using the relation:

$\begin{array}{c}{y}_A=R-r_1\\ =76.09796\text{mm-50}\text{mm}\\ =26.09796\text{mm}\end{array}$

The distance $\left(r_A\right)$ between the point A and the point C is $50\text{mm}$.

Calculate the stress at point A using the relation:

${\sigma }_A=-\frac{M{y}_A}{Ae{r}_A}$ (8)

Substitute $600\text{N}\cdot \text{m}$ for M, $26.09796\text{mm}$ for $y_A$, $1,440{\text{mm}}^2$ for A, $3.90204\text{mm}$ for e, $50\text{mm}$ for $\left(r_A\right)$ in Equation (8).

$\begin{array}{c}{\sigma }_A=-\frac{600\text{N}\cdot \text{m}\times 26.09796\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{1,440{\text{mm}}^2\times \left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\times 3.90204\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)\times 50\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}\\ =-\frac{600\times 26.09796\times {10}^{-3}}{1,440\times {10}^{-6}\times 3.90204\times {10}^{-3}\times 50\times {10}^{-3}}\\ =-\frac{15.658}{0.280946\times {10}^{-6}}\\ \approx -55.7\times {10}^6\text{Pa}\end{array}$

${\sigma }_A\text{=}-\text{55}\text{.7}\text{MPa}$

Thus, the stress at point A is $\underset{\_}{-\text{55}\text{.7}\text{MPa}}$.