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Axial stress in connecting rod AB: - Force P = 30 kN - Allowable stress = 100 MPa - Required area of rod: A = P/σ = 30,000 N / 100,000,000 Pa = 0.3 cm2 - Rod is circular, so: A = πr2 0.3 cm2 = πr2 r = √(0.3/π) cm = 1 cm Shearing stress in pin D: - Shear force = P = 30 kN - Area of pin = π(0.02)2 cm2 - Shear stress = P/A =
- Author
- prince janeo
- Language
- EN