Answer:
The true stress required = 407 MPa
Explanation:
True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,
σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent
True strain is given by
Epsilon t =㏑ (l/l₀)
Substitute㏑(l/l₀) for ε(t)
σ(t) = K(㏑(l/l₀))ⁿ
Given values l₀ = 49.9mm, l =53.5mm , n =0.1 , σ(t) =376Mpa
376 x 10⁶ = K (㏑(53.5/49.9))^0.1
K = 376 x 10⁶/(㏑(53.5/49.9))^0.1
K = 490.78 MPa
Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.
σ(t) = K(㏑(l/l₀))ⁿ
l₀ = 49.9mm, l = 58.2mm, n = 0.1, K = 490.78Mpa
σ(t) = 490.78 x 10⁶ x (㏑(58.2/49.9))^0.1
σ(t) = 407 MPa
The true stress necessary to plastically elongate the specimen is 407 MPa.
