Answer:
see explanation
Explanation:
By load resistance and factor design:
calculate factored load using following load combination
[tex]P_u = 1.2D + 1.6L[/tex]
[tex]P_u[/tex] is factored load
D is dead load
L is live load
Substitute 10kips for D and 25kips for L
[tex]P_u = 1.2D + 1.6L\\\\P_u = 1.2(10) + 1.6(25)\\\\P_u = 52.0kips[/tex]
Calculate the required gross area 0f the member
[tex]A_g = \frac{P_u}{0.9F_y} \\\\[/tex]
[tex]A_g[/tex] is gross area of section
[tex]F_y[/tex] is yield strength
substitute 52kips for [tex]P_u[/tex] and 35kips for [tex]F_y[/tex]
[tex]A_g = \frac{P_u}{0.9F_y} \\\\\\A_g = \frac{52.0}{0.9(35)} \\\\\\A_g = 1.65in^2[/tex]
Calculate required effective area of the member
[tex]A_e = \frac{P_u}{0.75F_u} \\\\[/tex]
[tex]A_e[/tex] is the effective area
[tex]F_u[/tex] is the ultimate strength
substitute 52kips for [tex]P_u[/tex] and 60ksi for [tex]F_u[/tex]
[tex]A_e = \frac{P_u}{0.75F_u} \\\\A_e = \frac{52.0}{0.75 (60)} \\\\= 1.16in^2[/tex]
Calculate minimum value for radius of gyration [tex](r_{min})[/tex]
[tex](r_{min}) = \frac{L}{300} \\\\(r_{min}) = \frac{8 \times 12 }{300} \\\\= 0.32in[/tex]
Corresponding to the above calculated value, try pipe 3 standard
For pipe 3 standard
Check the above section requirement
[tex]A_g = 2.07in^2\\2.07in^2 > 1.65in^2(OK)\\\\r_{min} = 1.17in\\1.17in > 0.32in (OK)[/tex]
Calculate effective net area of the section
[tex]A_e = A_g\\\\= 2.07in^2\\2.07in^2 > 1.65in^2[/tex]
therefore the section is OK
Hence , use pipe 3 standard
