The potential energy of a strained body comes out to be ½ x stress x strain x volume.
Can you derive this relation?
The potential energy of a strained body comes out to be ½ x stress x strain x volume.
Can you derive this relation?
When a wire is stretched, some work is done against the internal restoring forces acting between particles of the wire. This work done appears as elastic potential energy in the wire.
Consider a wire of length l and area of cross section a. Let F be the stretching forces applied on the wire and Δl be the increase in length of the wire.
Initially, the internal restoring force was zero but when length is increased by Δl ,
the average internal force for an increase in length Δl of the wire
= (0 + F)/2 = F/2
Hence, work done on the wire, w = average force × increase in length = [F/2 ] × Δl
This is stored as elastic potential energy U in the wire.
∴ U = (1/2) × F × Δl
multiplying a×l on numerator & denominator
we have
U= (1/2) × F/a × Δl/l × al
we know F/a= force/area=stress
=strain
a×l=area × length=volume of the wire
thus,
U= (1/2)(stress) × (strain) × volume of the wire
Hope you understand
When a wire is stretched, some work is done against the internal restoring forces acting between particles of the wire. This work done appears as elastic potential energy in the wire.
Consider a wire of length l and area of cross section a. Let F be the stretching forces applied on the wire and Δl be the increase in length of the wire.
Initially, the internal restoring force was zero but when length is increased by Δl ,
the average internal force for an increase in length Δl of the wire
= (0 + F)/2 = F/2
Hence, work done on the wire, w = average force × increase in length = [F/2 ] × Δl
This is stored as elastic potential energy U in the wire.
∴ U = (1/2) × F × Δl
multiplying a×l on numerator & denominator
we have
U= (1/2) × F/a × Δl/l × al
we know F/a= force/area=stress
=strain
a×l=area × length=volume of the wire
thus,
U= (1/2)(stress) × (strain) × volume of the wire
Hope you understand
