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26 August, 10:48

1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used in a medical device. From these data estimate the following for a 1 x 2 cm cross-section bar of the alloy having length 12 cm: a. The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.

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  1. 26 August, 11:56
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    F_x = 100,200 N

    x' = 21.321 cm ... Length

    y' = 0.7825 cm

    z' = 1.565 cm

    A' = (0.783 x 1.565) cm

    Explanation:

    Given:

    - The Modulus of Elasticity E = 645 MPa

    - The poisson ratio v = 0.28

    - The Yield Strength Y = 501 MPa

    - The Length along x-direction x = 12 cm

    - The length along y-direction y = 1 cm

    - The length along z--direction z = 2 cm

    Find:

    The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.

    Solution:

    - The Tensile forces within the limit of proportionality is given as:

    F_i = б_i*A_jk

    - A maximum tensile Force F_x along x direction can be given as:

    F_x = Y*A_yz

    F_x = 501 * (0.01*0.02) * 10^6

    F_x = 100,200 N

    - The corresponding strains in x, y and z direction due to F_x are:

    ξ_x = Y / E

    ξ_x = 501 / 645 = 0.7767

    ξ_y = ξ_z = - v*Y / E

    ξ_y = ξ_z = - 0.28*501 / 645 = - 0.2175

    - The corresponding change in lengths at tensile elastic stress are:

    Δx = x*ξ_x = 12*0.7767 = 9.321 cm

    Δy = y*ξ_y = - 1*0.2175 = - 0.2175 cm

    Δz = z*ξ_z = - 2*0.2175 = - 0.435 cm

    - The final lengths are:

    x' = x + Δx = 12 + 9.321 = 21.321 cm

    y' = y + Δy = 1 - 0.2175 = 0.7825 cm

    z' = z + Δz = 2 - 0.435 = 1.565 cm
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