A specimen of aluminum having a rectangular cross section 9.9 mm × 12.4 mm (0.3898 in. × 0.4882 in.) is pulled in tension with 35900 N (8071

Question

A specimen of aluminum having a rectangular cross section 9.9 mm × 12.4 mm (0.3898 in. × 0.4882 in.) is pulled in tension with 35900 N (8071 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.

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Yến Oanh 6 months 2021-08-02T15:48:41+00:00 1 Answers 81 views 0

Answers ( )

    0
    2021-08-02T15:50:27+00:00

    Answer:

    Explanation:

    Given:

    Dimensions:

    Length, l = 9.9 mm

    = 0.0099 m

    Width, w = 12.4 mm

    = 0.0124 m

    Force, f = 35900 N

    Young modulus, E = 69 GPa

    = 6.9 × 10^10 Pa

    Young modulus, E = stress/strain

    Stress = force/area

    Area of a rectangle = length, l × width, w

    = 0.0099 × 0.0124

    = 1.2276 × 10^-4 m^2

    Therefore,

    Strain = stress/young modulus

    = (35900/1.2276 × 10^-4)/6.9 × 10^10

    = 4.24 × 10^-3 m/m

    = 0.00424 m/m

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