Question

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.9 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 376 MPa causes the specimen to plastically elongate to a length of 53.5 mm. If it is known that the strain-hardening exponent for this alloy is 0.1, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 49.9 mm to a length of 58.2 mm.

Answer 1

The true stress required = 407 MPa

Step-by-step 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.