Chapter 4.10, Problem 166P

(a)

To determine

Find the error in the computation of maximum stress by assuming the bar as straight.

Answer to Problem 166P

The error in the computation of maximum stress by assuming the bar as straight is $\underset{\_}{34.36\%}$.

Explanation of Solution

Given information:

The value of h is $40\text{mm}$.

The inner $\left(r_1\right)$ and outer radius $\left(r_2\right)$ of the curved bar is $20\text{mm}$ and $60\text{mm}$.

The width and depth of the bar are $b=60\text{mm}$ and $h=40\text{mm}$.

The moment (M) is $120\text{N}\cdot \text{m}$.

Calculation:

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

Substitute $60\text{mm}$ for $b$ and $40\text{mm}$ for $h$.

$\begin{array}{c}A=bh\\ =60\times 40\\ =2,400{\text{mm}}^2\end{array}$

Calculate the moment of inertia (I) of the cross-section of the bar using the relation:

$I=\frac{1}{12}b{h}^3$ (1)

Substitute $60\text{mm}$ for $b$ and $40\text{mm}$ for $h$ in Equation (1).

$\begin{array}{c}I=\frac{1}{12}\times 60\times {40}^3\\ =320,000{\text{mm}}^4\end{array}$

Calculate the distance (c) between the neutral axis and the extreme fiber using the relation:

$c=\frac{h}{2}$ (2)

Substitute $40\text{mm}$ for $h$ in Equation (2).

$\begin{array}{c}c=\frac{40}{2}\\ =20\text{mm}\end{array}$

Calculate the stress $\left({\sigma }_s\right)$ using the relation:

${\sigma }_s=-\frac{Mc}{I}$ (3)

Substitute $120\text{N}\cdot \text{m}$ for M, $20\text{mm}$ for c, and $320,000{\text{mm}}^4$ for I in Equation (3).

$\begin{array}{c}{\sigma }_s=-\frac{120\text{N}\cdot \text{m}\times 20\text{mm}\times \left(\frac{1\text{m}}{1000\text{mm}}\right)}{320,000{\text{mm}}^4\times \left(\frac{\text{1}{\text{m}}^4}{{10}^{12}{\text{mm}}^4}\right)}\\ =-\frac{2.4}{0.32\times {10}^{-6}}\\ =-7.5\times {10}^6\text{Pa}\times \left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =-7.5\text{MPa}\end{array}$

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

$R=\frac{h}{\ln \frac{r_2}{r_1}}$ (4)

Substitute $40\text{mm}$ for $h$, $20\text{mm}$ for $r_1$, and $60\text{mm}$ for $r_2$ in Equation (4).

$\begin{array}{c}R=\frac{40}{\ln \left(\frac{60}{20}\right)}\\ =\frac{40}{\ln \left(3\right)}\\ =\frac{40}{1.09861}\\ =36.4096\text{mm}\end{array}$

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

$\overline{r}=\frac{1}{2}\left(r_1+r_2\right)$ (5)

Substitute $20\text{mm}$ for $r_1$ and $60\text{mm}$ for $r_2$ in Equation (5).

$\begin{array}{c}\overline{r}=\frac{1}{2}\left(20+60\right)\\ =40\text{mm}\end{array}$

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

$e=\overline{r}-R$ (6)

Substitute $40\text{mm}$ for $\overline{r}$ and $36.4096\text{mm}$ for R in Equation (6).

$\begin{array}{c}e=40-36.4096\text{mm}\\ =3.5904\text{mm}\end{array}$

Calculate the actual stress using the relation:

${\sigma }_a=\frac{M\left(r_1-R\right)}{Ae{r}_1}$ (7)

Substitute $120\text{N}\cdot \text{m}$ for M, $36.4096\text{mm}$ for $R$, $2,400{\text{mm}}^2$ for A, $3.5904\text{mm}$ for e, $20\text{mm}$ for $r_1$ in Equation (7).

$\begin{array}{c}{\sigma }_a=\frac{120\text{N}\cdot \text{m}\times \left(20-36.4096\right)\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{2400{\text{mm}}^2\times \left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\times 3.5904\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)\times 20\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}\\ =\frac{120\times \left(-16.4096\right)\times {10}^{-3}}{2400\times {10}^{-6}\times 3.5904\times {10}^{-3}\times 20\times {10}^{-3}}\\ =-\frac{1.9691}{0.172339\times {10}^{-6}}\\ \approx -11.426\times {10}^6\text{Pa}\times \frac{1\text{MPa}}{{10}^6\text{Pa}}\end{array}$

${\sigma }_a=-\text{11}\text{.426}\text{MPa}$

Calculate the error in the computation of maximum stress by assuming the bar as straight using the relation:

$\%\text{error}=\left(\frac{{\sigma }_a-{\sigma }_s}{{\sigma }_a}\right)\times 100$ (8)

Substitute $-\text{11}\text{.426}\text{MPa}$ for ${\sigma }_a$ and $-7.5\text{MPa}$ for ${\sigma }_s$ in Equation (8).

$\begin{array}{c}\%\text{error}=\left(\frac{-\text{11}\text{.426}-\left(-7.5\right)}{-\text{11}\text{.426}}\right)\times 100\\ =\frac{-3.926}{-11.426}\times 100\\ =34.36\%\end{array}$

Thus, the in the computation of maximum stress by assuming the bar as straight is $\underset{\_}{34.36\%}$.

(b)

To determine

Find the error in the computation of maximum stress by assuming the bar as straight.

Answer to Problem 166P

The error in the computation of maximum stress by assuming the bar as straight is $\underset{\_}{6\%}$.

Explanation of Solution

Given information:

The value of h is $40\text{mm}$.

The inner $\left(r_1\right)$ and outer radius $\left(r_2\right)$ of the curved bar is $200\text{mm}$ and $240\text{mm}$.

The width and depth of the bar are $b=60\text{mm}$ and $h=40\text{mm}$.

The moment (M) is $120\text{N}\cdot \text{m}$.

Calculation:

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

Substitute $40\text{mm}$ for $h$, $200\text{mm}$ for $r_1$, and $240\text{mm}$ for $r_2$ in Equation (4).

$\begin{array}{c}R=\frac{40}{\ln \left(\frac{240}{200}\right)}\\ =\frac{40}{\ln \left(1.2\right)}\\ =\frac{40}{0.182321}\\ =219.39326\text{mm}\end{array}$

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

Substitute $200\text{mm}$ for $r_1$ and $240\text{mm}$ for $r_2$ in Equation (5).

$\begin{array}{c}\overline{r}=\frac{1}{2}\left(200+240\right)\\ =220\text{mm}\end{array}$

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

Substitute $220\text{mm}$ for $\overline{r}$ and $219.39326\text{mm}$ for R in Equation (6).

$\begin{array}{c}e=220-219.39326\text{mm}\\ =0.60674\text{mm}\end{array}$

Calculate the actual stress using the relation:

Substitute $120\text{N}\cdot \text{m}$ for M, $219.39326\text{mm}$ for $R$, $2,400{\text{mm}}^2$ for A, $0.60674\text{mm}$ for e, $200\text{mm}$ for $r_1$ in Equation (7).

$\begin{array}{c}{\sigma }_a=\frac{120\text{N}\cdot \text{m}\times \left(200-219.39326\right)\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{2400{\text{mm}}^2\times \left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\times 0.60674\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)\times 200\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}\\ =\frac{120\times \left(-19.39326\right)\times {10}^{-3}}{2400\times {10}^{-6}\times 0.60674\times {10}^{-3}\times 200\times {10}^{-3}}\\ =-\frac{2.327}{0.291235\times {10}^{-6}}\\ \approx -7.990\times {10}^6\text{Pa}\times \frac{1\text{MPa}}{{10}^6\text{Pa}}\end{array}$

${\sigma }_a=-7.990\text{MPa}$

Calculate the error in the computation of maximum stress by assuming the bar as straight using the relation:

Substitute $-7.99\text{MPa}$ for ${\sigma }_a$ and $-7.5\text{MPa}$ for ${\sigma }_s$ in Equation (8).

$\begin{array}{c}\%\text{error}=\left(\frac{-7.990-\left(-7.5\right)}{-7.990}\right)\times 100\\ =\frac{-0.490}{-7.990}\times 100\\ \approx 6\%\end{array}$

Thus, the in the computation of maximum stress by assuming the bar as straight is $\underset{\_}{34.36\%}$.

(c)

To determine

Find the error in the computation of maximum stress by assuming the bar as straight.

Answer to Problem 166P

The error in the computation of maximum stress by assuming the bar as straight is $\underset{\_}{0.6\%}$.

Explanation of Solution

Given information:

The value of h is $40\text{mm}$.

The inner radius $\left(r_1\right)$ is $2\text{m}$.

Show the unit conversion of inner radius as follows:

$\begin{array}{c}{r}_1=2\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ =2000\text{mm}\end{array}$

The inner $\left(r_1\right)$ and outer radius $\left(r_2\right)$ of the curved bar is $2,000\text{mm}$ and $2,040\text{mm}$.

The width and depth of the bar are $b=60\text{mm}$ and $h=40\text{mm}$.

The moment (M) is $120\text{N}\cdot \text{m}$.

Calculation:

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

Substitute $40\text{mm}$ for $h$, $2,000\text{mm}$ for $r_1$, and $2,040\text{mm}$ for $r_2$ in Equation (4).

$\begin{array}{c}R=\frac{40}{\ln \left(\frac{2,040}{2,000}\right)}\\ =\frac{40}{\ln \left(1.02\right)}\\ =\frac{40}{0.01980262}\\ =2019.934\text{mm}\end{array}$

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

Substitute $2000\text{mm}$ for $r_1$ and $2040\text{mm}$ for $r_2$ in Equation (5).

$\begin{array}{c}\overline{r}=\frac{1}{2}\left(2000+2040\right)\\ =2020\text{mm}\end{array}$

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

Substitute $2020\text{mm}$ for $\overline{r}$ and $2019.934\text{mm}$ for R in Equation (6).

$\begin{array}{c}e=2020-2019.934\text{mm}\\ =0.066\text{mm}\end{array}$

Calculate the actual stress using the relation:

Substitute $120\text{N}\cdot \text{m}$ for M, $2019.9\text{mm}$ for $R$, $2,400{\text{mm}}^2$ for A, $0.1\text{mm}$ for e, $200\text{mm}$ for $r_1$ in Equation (7).

$\begin{array}{c}{\sigma }_a=\frac{120\text{N}\cdot \text{m}\times \left(2,000-2019.934\right)\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}{2400{\text{mm}}^2\times \left(\frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\right)\times 0.1\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)\times 2,000\text{mm}\times \left(\frac{1\text{m}}{1,000\text{mm}}\right)}\\ =\frac{120\times \left(-19.934\right)\times {10}^{-3}}{2400\times {10}^{-6}\times 0.066\times {10}^{-3}\times 2,000\times {10}^{-3}}\\ =-\frac{2.39208}{0.3168\times {10}^{-6}}\\ \approx -7.550\times {10}^6\text{Pa}\times \frac{1\text{MPa}}{{10}^6\text{Pa}}\end{array}$

${\sigma }_a=-7.550\text{MPa}$

Calculate the error in the computation of maximum stress by assuming the bar as straight using the relation:

Substitute $-7.55\text{MPa}$ for ${\sigma }_a$ and $-7.5\text{MPa}$ for ${\sigma }_s$ in Equation (8).

$\begin{array}{c}\%\text{error}=\left(\frac{-7.550-\left(-7.5\right)}{-7.550}\right)\times 100\\ =\frac{-0.050}{-7.550}\times 100\\ \approx 0.6\%\end{array}$

Thus, the in the computation of maximum stress by assuming the bar as straight is $\underset{\_}{0.6\%}$.