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28. The upper end of a wire 1 m long and 4 mm radius isclamped. The lower end is twisted by an angle of 300. Theangle of shear is:(á) 12(b) 1.20 (c) 0.12 (d) 0.012° |
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Answer» angle of shear = ɸ = (rɸ /L)r = radius = 4mm = 4 × 10–3mθ = 30° = 30 × (π / 180) = (π/6) rad L = 1 m henceɸ = {(4 × 10–3× {π/6}) / 1} = 2.09 × 10–3radi.e.ɸ = 2.09 × 10–3rad= {2.09 × 10–3× (180 / π)}°= 0.12° |
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