force required to reduce the diameter

samedi 31 août 2013

1. The problem statement, all variables and given/known data



A mild steel bar 40 mm diameter and 100 mm long is subjected to a tensile force along its axis.

Young’s modulus of elasticity for mild steel = 200 GN m–2.

Poisson’s ratio is 0.3.



Calculate the force (F) required to reduce the diameter to 39.99 mm.



Use the x–y coordinate system as shown above.



2. Relevant equations



Poissson's Ratio = - (transverse strain / axial strain)



force = Stress x Area



3. The attempt at a solution



transverse strain = (39.99 - 40) / 40 = -0.25x10^-3



axial strain = - (-0.25x10^-3 / 0.3) = 833.333 x 10^-6



axial stress = (833.333 x 10^-6) x (200 x 10^9) = 166.666 x 10^6



force (F) = 166.666x10^6 x (0.25∏ x 0.04^2) = 209439.51 N






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