Question

a load of 20kg is applied to the ends of wire 4m along and produce an extension of 0.24mm if the diameter of the wire is 2mm find the stress on the wire and strain producers​

Answer 1

Explanation: Force = mg as the wire is fixed at one point and the load is inserted at one end.

g=9.8m/s²

F= mg

F= 20×9.8= 19.6 newton

Strain= Δl/L

where Δl= change in length

and L= original length

Strain = 0.24×0.001 m/4m

Strain=0.00006

Stress= F/A

where A = cross sectional area

therefore A = π× (diametre/2)²

A=3.14×1²≈3.14 mm²= 0.000000314m²

Stress = 19.6/0.000000314=62420382 Pascal