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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

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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