3.2 A round bar of 25 mm diameter and a length of $$ 0,6 \mathrm{~m} $$ is machined to have a square section of 12 mm sides over the length of 0.4 m . If the stress in the square section is not to exceed 250 MPa , calculate the following: $$ \begin{array}{ll}3.2 .1 & ext { The stress in the round bar } \ 3.2 .2 & ext { The total extension of the bar if Young's modulus for the material is } \ 90 \mathrm{GPa}\end{array} $$ $$ \begin{array}{l} ext { A load of } 50 \mathrm{kN} ext { causes a tensile stress of } 25 \mathrm{MPa} ext { in a round cast iron bar. The } \ ext { original length of the bar is } 3 \mathrm{~m} ext { and Young's modulus is } 100 \mathrm{GPa} ext { for this cast } \ ext { iron. } \ ext { Calculate the diameter of the bar. }\end{array} $$

Question

3.2 A round bar of 25 mm diameter and a length of $$ 0,6 \mathrm{~m} $$ is machined to have a square section of 12 mm sides over the length of 0.4 m . If the stress in the square section is not to exceed 250 MPa , calculate the following: $$ \begin{array}{ll}3.2 .1 & 	ext { The stress in the round bar } \ 3.2 .2 & 	ext { The total extension of the bar if Young's modulus for the material is } \ 90 \mathrm{GPa}\end{array} $$ $$ \begin{array}{l}	ext { A load of } 50 \mathrm{kN} 	ext { causes a tensile stress of } 25 \mathrm{MPa} 	ext { in a round cast iron bar. The } \ 	ext { original length of the bar is } 3 \mathrm{~m} 	ext { and Young's modulus is } 100 \mathrm{GPa} 	ext { for this cast } \ 	ext { iron. } \ 	ext { Calculate the diameter of the bar. }\end{array} $$

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$$
\textbf{3.2.1. Maximum load from the square section:}
$$

The square section has sides of 12 mm, so its cross‐sectional area is
$$
A_{\text{square}} = 12\,\text{mm} \times 12\,\text{mm} = 144\,\text{mm}^2.
$$
To keep the stress in the square section below 250 MPa, the maximum load $$P$$ is
$$
P = \sigma_{\text{square}} \, A_{\text{square}} = 250\,\text{MPa} \times 144\,\text{mm}^2.
$$
Remember that $$1\,\text{MPa} = 1\,\text{N/mm}^2$$, so
$$
P = 250\,\frac{\text{N}}{\text{mm}^2} \times 144\,\text{mm}^2 = 36\,000\,\text{N}.
$$

$$
\textbf{Stress in the round bar (3.2.1):}
$$

The original round bar has a diameter of 25 mm, so its area is
$$
A_{\text{round}} = \frac{\pi}{4} \times (25\,\text{mm})^2 = \frac{\pi \times 625}{4} \approx 490.9\,\text{mm}^2.
$$
Since the same load $$P = 36\,000\,\text{N}$$ is carried by the round section, the stress in the round bar, $$\sigma_{\text{round}}$$, is
$$
\sigma_{\text{round}} = \frac{P}{A_{\text{round}}} = \frac{36\,000\,\text{N}}{490.9\,\text{mm}^2} \approx 73.3\,\text{MPa}.
$$

$$
\textbf{Extension of the bar (3.2.2):}
$$

The bar consists of two regions:
- A machined square-section region of length $$L_1 = 0.4\,\text{m}$$ and area $$A_1 = 144\,\text{mm}^2 = 144 \times 10^{-6}\,\text{m}^2$$.
- A remaining round-section region of length $$L_2 = 0.6\,\text{m} - 0.4\,\text{m} = 0.2\,\text{m}$$ and area $$A_2 = 490.9\,\text{mm}^2 = 490.9 \times 10^{-6}\,\text{m}^2$$.

With Young’s modulus $$E = 90\,\text{GPa} = 90 \times 10^9\,\text{Pa}$$, the extension $$\delta$$ in each region is given by
$$
\delta = \frac{P\,L}{A\,E}.
$$

For the square section:
$$
\delta_1 = \frac{36\,000\,\text{N} \times 0.4\,\text{m}}{144 \times 10^{-6}\,\text{m}^2 \times 90 \times 10^9\,\text{Pa}}.
$$
Calculate the denominator first:
$$
144 \times 10^{-6} \times 90 \times 10^9 = 144 \times 90 \times 10^3 = 12\,960\,000.
$$
Then,
$$
\delta_1 = \frac{14\,400}{12\,960\,000} \approx 0.001111\,\text{m} \approx 1.11\,\text{mm}.
$$

For the round section:
$$
\delta_2 = \frac{36\,000\,\text{N} \times 0.2\,\text{m}}{490.9 \times 10^{-6}\,\text{m}^2 \times 90 \times 10^9\,\text{Pa}}.
$$
The denominator is:
$$
490.9 \times 10^{-6}\,\text{m}^2 \times 90 \times 10^9\,\text{Pa} = 490.9 \times 90 \times 10^3 \approx 44\,181\,000.
$$
Thus,
$$
\delta_2 = \frac{7\,200}{44\,181\,000} \approx 0.000163\,\text{m} \approx 0.163\,\text{mm}.
$$

The total extension of the bar is
$$
\delta_{\text{total}} = \delta_1 + \delta_2 \approx 1.11\,\text{mm} + 0.163\,\text{mm} \approx 1.27\,\text{mm}.
$$

$$
\textbf{Cast iron bar: Determination of the diameter:}
$$

For the cast iron bar, the given data are:
$$
P = 50\,\text{kN} = 50\,000\,\text{N}, \quad \sigma = 25\,\text{MPa} = 25 \times 10^6\,\text{Pa}.
$$
The cross‐sectional area required to produce this stress under the applied load is
$$
A = \frac{P}{\sigma} = \frac{50\,000\,\text{N}}{25 \times 10^6\,\text{Pa}} = 0.002\,\text{m}^2.
$$
For a round bar, the area is related to the diameter $$d$$ by
$$
A = \frac{\pi}{4} d^2.
$$
Thus,
$$
d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 0.002}{\pi}}.
$$
Calculate the numerical value:
$$
\frac{4 \times 0.002}{\pi} = \frac{0.008}{\pi} \approx \frac{0.008}{3.1416} \approx 0.0025465,
$$
so
$$
d = \sqrt{0.0025465} \approx 0.0505\,\text{m} \approx 50.5\,\text{mm}.
$$

$$
\textbf{Summary of Answers:}
$$
$$
\begin{array}{rl}
\text{3.2.1:} & \sigma_{\text{round}} \approx 73.3\,\text{MPa} \
\text{3.2.2:} & \delta_{\text{total}} \approx 1.27\,\text{mm} \
\text{Cast iron bar:} & d \approx 50.5\,\text{mm}
\end{array}
$$
**3.2.1. Stress in the Round Bar:** The stress in the round bar is approximately 73.3 MPa. **3.2.2. Total Extension of the Bar:** The total extension of the bar is approximately 1.27 mm. **Cast Iron Bar:** The diameter of the cast iron bar is approximately 50.5 mm.

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