A circular steel wire 2.00 m is to stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of wire. What minimum diameter is required for the wire?

Answer 1

area = pi d^2/4
length = 2 m
change in length = .0025 m

change in length/length = .0025/2

= tensile stress/E = 400/(pi d^2/4 E)
so
.00125 = 400/ (pi d^2/4 E)