Given that extension = ∆l and force applied = F, we assume that the wire’s length is L.
Stress is directly proportional to strain if the elastic limit is not exceeded.
where stress (σ = F/A) is the force applied per unit area.
Additionally, strain is defined as extension per length unit (ε = ∆l/l).
Therefore,
Energy stored = ½ x stress x strain x volume
Energy stored = ½ x (F/A) x (∆l / L) x A x L
Energy stored = ½ F∆l
Given that extension = ∆l and force applied = F, we assume that the wire’s length is L.
Stress is directly proportional to strain if the elastic limit is not exceeded.
where stress (σ = F/A) is the force applied per unit area.
Additionally, strain is defined as extension per length unit (ε = ∆l/l).
Therefore,
Energy stored = ½ x stress x strain x volume
Energy stored = ½ x (F/A) x (∆l / L) x A x L
Energy stored = ½ F∆l