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WUMERICALS11. A weight of 1.0 kg is suspended from the lower endof a wire of cross-section 10 mm2. Find the magnitudeand direction of the stress produced in it.(g = 9.8 m s.)Ans. 9.8 × 105 N m. upwards2 |
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Answer» Stress = Force/ Area. Force= m x a . Assuming a=g= 9.81 m/s^2. Force= 9.81 N Area of a circle= (3.14 * diameter^2)/4. = 0.785 * (.01^2) = 0.0000785 m^2. Stress= 9.81/0.0000785 = 124970 N/m^2. = 124970 Pa = 124.9 kPa. |
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