A thin strip of rubber has an unstretched length of 15 in If

A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5 in., determine the average normal strain in the strip.

Solution

L = 15 in

Circumference of pipe = pi*diameter

Circumference = 3.141*5 = 15.708 in

Change in length (dL) = circumference- original length

dL = 15.708-15 = 0.708 in

Average strain = dL/L

= 0.708/15

Average Strain = 0.04719