Question

Use LRFD and choose a pipe to be used as a tension member to resist a service dead load of 10 kips and service live load of 25 kips. The ends will be connected by welding completely around the circumference of the pipe. The length is 8 feet.

Answer 1

Answer:

see explanation

Explanation:

By load resistance and factor design:

calculate factored load using following load combination

$P_u = 1.2D + 1.6L$

$P_u$ is factored load

D is dead load

L is live load

Substitute 10kips for D and 25kips for L

$P_u = 1.2D + 1.6L\\\\P_u = 1.2(10) + 1.6(25)\\\\P_u = 52.0kips$

Calculate the required gross area 0f the member

$A_g = \frac{P_u}{0.9F_y}$

$A_g$ is gross area of section

$F_y$ is yield strength

substitute 52kips for $P_u$ and 35kips for $F_y$

$A_g = \frac{P_u}{0.9F_y} \\\\\\A_g = \frac{52.0}{0.9(35)} \\\\\\A_g = 1.65in^2$

Calculate required effective area of the member

$A_e = \frac{P_u}{0.75F_u}$

$A_e$ is the effective area

$F_u$ is the ultimate strength

substitute 52kips for $P_u$ and 60ksi for $F_u$

$A_e = \frac{P_u}{0.75F_u} \\\\A_e = \frac{52.0}{0.75 (60)} \\\\= 1.16in^2$

Calculate minimum value for radius of gyration $(r_{min})$

$(r_{min}) = \frac{L}{300} \\\\(r_{min}) = \frac{8 \times 12 }{300} \\\\= 0.32in$

Corresponding to the above calculated value, try pipe 3 standard

For pipe 3 standard

Check the above section requirement

$A_g = 2.07in^2\\2.07in^2 > 1.65in^2(OK)\\\\r_{min} = 1.17in\\1.17in > 0.32in (OK)$

Calculate effective net area of the section

$A_e = A_g\\\\= 2.07in^2\\2.07in^2 > 1.65in^2$

therefore the section is OK

Hence , use pipe 3 standard