Answer:
0.05%
Explanation:
From the question, we have;
The yield strength of the mild steel, = 43.5 ksi
Young's modulus of elasticity, ∈ = 29,000 ksi
The total strain, = 0.2% = 0.002
The inelatic strain is given as follows;
=
-
/∈
Therefore, we have;
= 0.002 - 43.5/(29,000) = 0.0005
Therefore, the inelastic strain, = 0.0005 = 0.05%
Taking the inelastic strain as the residual strain, we have;
The residual strain = 0.05%
