Problem 5.44 The rod has a diameter of 18 mm and weight of 20 N/m (Figure...

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Answer 1

Answer #1

torsion due to self weight=20*0.3 + 20*1*1/2=16 Nm=16000 Nmm

diameter of shaft= 18 mm

shear stress due to torsion in circular section is given by = Tr/J

Max shear stress will be at extreme fiber

r=18/2=9 mm

J=pi*r⁴/2=10306 mm⁴

shear stress=16000*9/10306=13.97 N/mm²

Answer 2

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