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

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