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10 January, 00:07

If the specimen is loaded until it is stressed to 65 ksi, determine the approximate amount of elastic recovery after it is unloaded. Express your answer as a length. Express your answer to three significant figures and include the appropriate units.

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  1. 10 January, 00:22
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    ER = 0.008273 in

    Explanation:

    Given:

    - Length of the specimen L = 2 in

    - The diameter of specimen D = 0.5 in

    - Specimen is loaded until it is stressed = 65 ksi

    Find:

    - Determine the approximate amount of elastic recovery after it is unloaded.

    Solution:

    - From diagram we can see the linear part of the curve we can determine the Elastic Modulus E as follows:

    E = stress / strain

    E = 44 / 0.0028

    E = 15714.28 ksi

    - Compute the Elastic strain for the loading condition:

    strain = loaded stress / E

    strain = 65 / 15714.28

    strain = 0.0041364

    - Compute elastic recovery:

    ER = strain*L

    ER = 0.0041364*2

    ER = 0.008273 in
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