To find the smallest diameter bolt that can be used in a clevis given the shearing strength of the bolt and the applied force, we can follow these steps:
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Calculate the Shear Force: The force \( F \) acting on the bolt is given as \( 380 \text{ kN} \).
\[ F = 380 \text{ kN} = 380,000 \text{ N} \]
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Use Shear Strength: The shearing strength of the bolt \( \tau \) is given as \( 400 \text{ MPa} \).
\[ \tau = 400 \text{ MPa} = 400 \times 10^6 \text{ Pa} = 400 \times 10^6 \text{ N/m}^2 \]
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Calculate the Required Area: The shear strength is related to the shear force and the area over which that force is applied. The area \( A \) needed can be calculated using the formula:
\[ \tau = \frac{F}{A} \]
Rearranging the equation to find the area, we have:
\[ A = \frac{F}{\tau} \]
Substitute the values into the equation:
\[ A = \frac{380,000 , \text{N}}{400 \times 10^6 , \text{N/m}^2} \]
\[ A = \frac{380,000}{400,000,000} = 0.00095 , m^2 \]
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Relate Area to Diameter: The shear area for a circular bolt is given by the formula:
\[ A = \frac{\pi d^2}{4} \]
Where \( d \) is the diameter of the bolt. Setting the expressions for area equal gives us:
\[ \frac{\pi d^2}{4} = 0.00095 \]
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Solve for Diameter: To find \( d \), rearrange the equation:
\[ d^2 = \frac{4 \times 0.00095}{\pi} \]
\[ d^2 \approx \frac{0.0038}{3.14159} \approx 0.00121 \]
\[ d \approx \sqrt{0.00121} \approx 0.0348 \text{ m} = 34.8 \text{ mm} \]
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Final Diameter: Since bolts come in standard sizes, you'll want to round up to the nearest standard bolt size. The smallest diameter that can be used for this application is:
\[ d = 35 \text{ mm} \]
Conclusion: The smallest diameter bolt that can be used in the clevis, considering the shear force and the shearing strength of the bolt, is 35 mm.