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Discover the suitable diameter of the shaft:
A shaft is required to fail while the torque on the shaft is 20 kN-m. Discover the suitable diameter of the shaft of the modulus of rupture of torsion is 200 mN/m2.
Solution
T = 200 kN-m = 20 × 103 N-m
τmax = 200 MN/m2 = 200 × 106 N/m2
d = ?
τmax = 16T /πd3
⇒200 × 106 = 16 × 20 × 10/ πd 3
∴ d = 7.99 × 10- 2 m = 80 mm
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