Numerical - Original length of bar, given modulus of elasticity - Basic Mechanical Engineering

Q.  The elongation of a bar is 0.5 mm, when a tensile stress of 200 N/mm2 acts on it. Determine original length of a bar if modulus of elasticity is 150 x 103.
- Published on 01 Sep 15

a. 375.93 mm
b. 300 mm
c. 360 mm
d. None of the above

ANSWER: 375.93 mm

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    Discussion

  • MIDHUN R   -Posted on 21 Jul 20
    There is a typing mistake , that the answer given is not for change in length. The answer 375.93mm is original length.
  • Suri   -Posted on 15 Feb 20
    In answer,
    that is is original length,
    not the change in length
  • Sravanthi   -Posted on 06 Nov 15
    Given: Modulus of elasticity = 150 x 103, tensile stress = 200 N/mm2, elongation of bar = 0.5 mm

    Formula: According to Hooke's law, stress is directly proportional to strain

    E = σ / e

    here, E = modulus of elasticity, σ = stress, e = strain

    - The ratio of the stress applied on a body to resulting strain within the elastic limit is called as modulus of elasticity.

    Solution:

    Substituting the given values we get,

    E = σ / e

    150 x 103 = 200 / e

    e = 1.33 x 10-3

    Therefore, strain = change in length / original length

    1.33 x 10-3 = 0.5 / original length

    Change in length = 375.93 mm

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