Acylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the bar axis. determine the force that will produce an elastic reduction of 2.7 ×10-3 mm (1.063 ×10-4 in.) in the diameter. for steel, values for the elastic modulus (e) and poisson's ratio (ν) are, respectively, 207 gpa and 0.30.

14778.29 N

Explanation:

Diameter, d=10.1mm=

Since stress, where F is force, A is area and since specimen is cylindrical,  

A=  Therefore,

Also strain where L is length

Poison’s ratio,v is the ratio of lateral strain to longitudinal strain hence

V=

From Hooke’s law,

Conclusively,

F=

F= = 14778.29N

Therefore, required force F is 14778.29 N

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