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28 September, 18:28

The ultimate normal stress in members AB and BC is 350 MPa. Find the maximum load P if the factor of safety is 4.5. AB has an outside diameter of 250mm and BC has an outside diameter of 150mm. Both pipes have a wall thickness of 8mm

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  1. 28 September, 18:33
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    P_max = 278 KN

    Explanation:

    Given:

    - The ultimate normal stress S = 350 MPa

    - Thickness of both pipes t = 8 mm

    - Pipe AB: D_o = 250 mm

    - Pipe BC: D_o = 150 mm

    - Factor of safety FS = 4.5

    Find:

    Find the maximum load P_max

    Solution:

    - Compute cross sectional areas A_ab and A_bc:

    A_ab = pi * (D_o^2 - (D_o - 2t) ^2) / 4

    A_ab = pi * (0.25^2 - 0.234^2) / 4

    A_ab = 6.08212337 * 10^-3 m^2

    A_bc = pi * (D_o^2 - (D_o - 2t) ^2) / 4

    A_bc = pi * (0.15^2 - 0.134^2) / 4

    A_bc = 3.568212337 * 10^-3 m^2

    - Compute the Allowable Stress for each pipe:

    sigma_all = S / FS

    sigma_all = 350 / 4.5

    sigma_all = 77.77778 MPa

    - Compute the net for each member P_net, ab and P_net, bc:

    P_net, ab = sigma_all * A_ab

    P_net, ab = 77.77778 MPa*6.08212337 * 10^-3

    P_net, ab = 473054.0399 N

    P_net, bc = sigma_all * A_bc

    P_net, bc = 77.77778 MPa*3.568212337 * 10^-3

    P_net, bc = 277577.1721 N

    - Compute the force P for each case:

    P_net, ab = P + 50,000

    P = 473054.0399 - 50,000

    P = 423 KN

    P_net, bc = P = 278 KN

    - P_max allowed is the minimum of the two load P:

    P_max = min (423, 278) = 278 KN
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