Calculation of true thickness of the bed

116. An earthquake occurring at every 100 years, leads to a displacement of 2 m over a 40 km long fault. The corresponding shear strain accumulation rate is 

Solution:

Stress: Force per unit area.

Strain: change in length / original length.

Shear Strain: Which are changes in the angular relationships between parts of a body.

Here i'm using Shear strain and strain are both are same. 

$Shear Strain=\frac {change in length} {Original length}$

From above problem,

time = 100 years

displacement (change in length) =2 m

Original length = 40 km

                        = 40000m

$Shear Strain=\frac{ 2}{40000}$

Shear Strain = 0.00005

Therefore,

$rate=\frac{ Shear Strain}{time}$

$rate=\frac{ 0.00005}{100}$

$rate=0.5*10^{-6}$