Answer:
True Strain at failure = 1.386
Explanation:
For the question, ductility is given in reduction In the area to be 0.75
Let the initial Cross sectional Area of wire be Ao
And the final cross sectional Area of wire of cross sectional Area of wire at fracture be A
(Ao - A)/Ao = ductility = 0.75
Ao - A = 0.75Ao
A = Ao - 0.75Ao
A = 0.25 Ao
True Strain = In (Lf/Lo)
To obtain the ratio of the lengths of wire,
The volume of the wire stays constant, that is, Vo = Vf; Vo = Ao × Lo and Vf = A × Lf
AoLo = ALf
Lf/Lo = Ao/A
In (Lf/Lo) = In (Ao/A) = In (Ao/0.25Ao) = In 4 = 1.386
True Strain = 1.386
