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