A steel shaft is 500 mm long and has 50 mm diameter for 200 mm of its length and 30 mm diameter for the remaining 300 mm length. A tensile load F is applied to the rod so that the... A steel shaft is 500 mm long and has 50 mm diameter for 200 mm of its length and 30 mm diameter for the remaining 300 mm length. A tensile load F is applied to the rod so that the maximum stress induced in the material is 150 N/mm². Given: E = 200GN/m². (a) Show that the magnitude of the tensile load F is 106 kN. (b) Calculate the stress in the 50 mm diameter section and hence find the total extension of the rod. (c) A hole is to be drilled through the section with 50 mm diameter so that it has the same stress as the section with 30 mm diameter. Calculate the inner diameter required. Read more

Steps to Solve

1. Calculate the area of the smaller diameter section

The smaller diameter section has the maximum stress. Calculate the area $A_2$ using the formula for the area of a circle: $A = \pi r^2 = \pi (\frac{d}{2})^2$ Given $d_2 = 30 \text{ mm}$, $A_2 = \pi (\frac{30}{2})^2 = \pi (15)^2 = 225\pi \text{ mm}^2$

2. Calculate the tensile load $F$

We know that stress $\sigma = \frac{F}{A}$. Therefore, $F = \sigma \times A$. The maximum stress $\sigma_{max} = 150 \text{ N/mm}^2$ occurs in the smaller diameter section. $F = \sigma_{max} \times A_2 = 150 \text{ N/mm}^2 \times 225\pi \text{ mm}^2 = 33750\pi \text{ N} \approx 106028.75 \text{ N}$ Convert Newtons to kilonewtons: $F \approx 106.03 \text{ kN}$ This confirms that the tensile load $F$ is approximately $106 \text{ kN}$.

3. Calculate the stress in the larger diameter section

Calculate the area of the larger diameter section. Given $d_1 = 50 \text{ mm}$, $A_1 = \pi (\frac{50}{2})^2 = \pi (25)^2 = 625\pi \text{ mm}^2$ Calculate the stress in the $50 \text{ mm}$ diameter section: $\sigma_1 = \frac{F}{A_1} = \frac{33750\pi}{625\pi} = \frac{33750}{625} = 54 \text{ N/mm}^2$

4. Calculate the extension of the rod in the larger diameter section

Use the formula for extension, $\Delta L = \frac{FL}{EA}$. For the larger diameter section: $F = 33750\pi \text{ N}$, $L_1 = 200 \text{ mm}$, $E = 200 \text{ GN/m}^2 = 200 \times 10^3 \text{ N/mm}^2$, and $A_1 = 625\pi \text{ mm}^2$ $\Delta L_1 = \frac{33750\pi \times 200}{200 \times 10^3 \times 625\pi} = \frac{33750 \times 200}{200 \times 10^3 \times 625} = \frac{6750000}{125000000} = 0.054 \text{ mm}$

5. Calculate the extension of the rod in the smaller diameter section

For the smaller diameter section: $F = 33750\pi \text{ N}$, $L_2 = 300 \text{ mm}$, $E = 200 \times 10^3 \text{ N/mm}^2$, and $A_2 = 225\pi \text{ mm}^2$ $\Delta L_2 = \frac{33750\pi \times 300}{200 \times 10^3 \times 225\pi} = \frac{33750 \times 300}{200 \times 10^3 \times 225} = \frac{10125000}{45000000} = 0.225 \text{ mm}$

6. Calculate the total extension

The total extension is the sum of the extensions in both sections: $\Delta L_{total} = \Delta L_1 + \Delta L_2 = 0.054 + 0.225 = 0.279 \text{ mm}$

7. Calculate the inner diameter required

The stress in the section with the hole should be the same as the stress in the 30 mm diameter section, which is $150 \text{ N/mm}^2$. Let $d$ be the inner diameter of the hole. The area of the section with the hole is $A_{hole} = \pi (\frac{50}{2})^2 - \pi (\frac{d}{2})^2 = \pi (25^2 - (\frac{d}{2})^2) = \pi (625 - \frac{d^2}{4})$ Since the stress is the same, we have: $\sigma = \frac{F}{A_{hole}} = 150 \text{ N/mm}^2$ $150 = \frac{33750\pi}{\pi (625 - \frac{d^2}{4})} = \frac{33750}{625 - \frac{d^2}{4}}$ $150 (625 - \frac{d^2}{4}) = 33750$ $625 - \frac{d^2}{4} = \frac{33750}{150} = 225$ $\frac{d^2}{4} = 625 - 225 = 400$ $d^2 = 400 \times 4 = 1600$ $d = \sqrt{1600} = 40 \text{ mm}$