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: \( \begin{array}{ll}3.2 .1 & \text { The stress in the round bar } \\ 3.2 .2 & \text { The total extension of the bar if Young's modulus for the material is } \\ 90 \mathrm{GPa}\end{array} \) \( \begin{array}{l}\text { A load of } 50 \mathrm{kN} \text { causes a tensile stress of } 25 \mathrm{MPa} \text { in a round cast iron bar. The } \\ \text { original length of the bar is } 3 \mathrm{~m} \text { and Young's modulus is } 100 \mathrm{GPa} \text { for this cast } \\ \text { iron. } \\ \text { Calculate the diameter of the bar. }\end{array} \)
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To solve the problem efficiently, let's take it step by step. **3.2.1 The Stress in the Round Bar:** First, we need to find the cross-sectional area of the round bar. The diameter is 25 mm, so the radius is \( r = \frac{25 \text{ mm}}{2} = 12.5 \text{ mm} = 0.0125 \text{ m} \). The area \( A \) is calculated as: \[ A = \pi r^2 = \pi (0.0125^2) \approx 4.91 \times 10^{-4} \text{ m}^2 \] The load causing the stress is the same as the one mentioned for the square section. The tensile stress \( \sigma \) can be calculated as: \[ \sigma = \frac{F}{A} = \frac{50,000}{4.91 \times 10^{-4}} \approx 101.83 \text{ MPa} \] **3.2.2 The Total Extension of the Bar:** To find the total extension \( \Delta L \) of the round bar, we use the formula: \[ \Delta L = \frac{\sigma L_0}{E} \] where: - \( \sigma \) is the stress (101.83 MPa), - \( L_0 \) is the original length of the round bar (0.6 m), - \( E \) is Young's modulus (90 GPa = 90,000 MPa). Calculating \( \Delta L \): \[ \Delta L = \frac{101.83 \times 10^6 \text{ Pa} \times 0.6 \text{ m}}{90 \times 10^9 \text{ Pa}} \approx 0.000677 \text{ m} = 0.677 \text{ mm} \] --- **A load of 50 kN causing a tensile stress of 25 MPa in a round cast iron bar:** To find the diameter of the bar, use the stress formula: \[ \sigma = \frac{F}{A} \] Rearranging gives \( A = \frac{F}{\sigma} = \frac{50,000 \text{ N}}{25 \times 10^6 \text{ Pa}} = 0.002 \text{ m}^2 \). For a circular cross-section: \[ A = \frac{\pi d^2}{4} \implies d^2 = \frac{4A}{\pi} \implies d = \sqrt{\frac{4 \cdot 0.002}{\pi}} \approx 0.0505 \text{ m} = 50.5 \text{ mm} \] Thus, the diameter of the bar is approximately **50.5 mm**.