Answer:
slenderness ratio = 147.8
buckling load = 13.62 kips
Step-by-step explanation:
Given data:
outside diameter is 3.50 inc
wall thickness 0.30 inc
length of column is 14 ft
E = 10,000 ksi
moment of inertia
![= (\pi)/(64 (D_O^2 -D_i^2))](https://img.qammunity.org/2020/formulas/engineering/college/6fh249dvf9bj6k2twnpt7zuksv8gmtozoq.png)
![I = (\pi)/(64)(3.5^2 -2.9^2) = 3.894 in^4](https://img.qammunity.org/2020/formulas/engineering/college/3qq6w6rqs1npalfztenwegpz4xe0kbr41k.png)
Area
![= (\pi)/(4) (3.5^2 -2.9^2) = 3.015 in^2](https://img.qammunity.org/2020/formulas/engineering/college/a4gxnz47xk8uvs18hx1n81b6bcf9ijsuim.png)
![radius = \sqrt{(I)/(A)}](https://img.qammunity.org/2020/formulas/engineering/college/5nix7piqb1v9jmx6j15uel9xaeov1jnuy5.png)
![r = \sqrt{(3.894)/(3.015)](https://img.qammunity.org/2020/formulas/engineering/college/ztlgzwihvbxoq8nvhli1y3fgtugcghrp48.png)
r = 1.136 in
slenderness ratio
![= (L)/(r)](https://img.qammunity.org/2020/formulas/engineering/college/mexw4o3xo87dw1mrg0bsnzbwzm9sd6y15u.png)
![= (14 *12)/(1.136) = 147.8](https://img.qammunity.org/2020/formulas/engineering/college/xkrueojo1gto56of6a3j8epgsmgn4gl9nu.png)
buckling load
![= P_cr = \frac{\pi^2 EI}}{l^2}](https://img.qammunity.org/2020/formulas/engineering/college/kbl5e9ltpltx9tdr8nciqndgjw8487mr2a.png)
![P_(cr) = (\pi^2 *10,000*3.844)/(( 14* 12)^2)](https://img.qammunity.org/2020/formulas/engineering/college/5j9tqivwvjvxujype8fow052bsrjlezya4.png)
![P_(cr) = 13.62 kips](https://img.qammunity.org/2020/formulas/engineering/college/7zq1g7ntzhgzo5pa0qmohmosof6xnf4tk3.png)