Respuesta :
Answer:
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
