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2 April, 10:58

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 379 MPa causes the specimen to plastically elongate to a length of 51.7 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 49.7 mm to a length of 57.1 mm.

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  1. 2 April, 11:06
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    Answer:i ea
  2. 2 April, 13:41
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    The true stress required = 379 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 = 49.7mm, l = 51.7mm, n = 0.2, σ (t) = 379Mpa

    379 x 10⁶ = K (㏑ (51.7/49.7)) ^0.2

    K = 379 x 10⁶ / (㏑ (51.7/49.7)) ^0.2

    K = 723.48 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 = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

    σ (t) = 723.48 x 106 x (㏑ (51.7/49.7)) ^0.2

    σ (t) = 379 MPa

    The true stress necessary to plastically elongate the specimen is 379 MPa.
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