Selecting Lightest Wide Flange Steel Column for 450kN Load

[Emzielou83]

Hi all,

I need some help.

Select the lightest wide flange section that can be used as a steel column 7m long tosupport an axial load of 450kN with a factor of safety of 3. Use 200MPa as the limitof elasticity, 200GPa as the modulus of elasticity and assume that the column issimply supported.

I have a printout with my assignment of universal columns dimensions and properties. I don't reall yknow what I'm supposed to be working out here.

Can anyone help?

Thanks

Emma

 

[nvn]

Hi, Emzielou83. There is a template you must use when you post a question. You must list relevant equations yourself, and you must show your work. Then someone might check your math. Is this a test question? Also, engineering coursework questions go https://www.physicsforums.com/forumdisplay.php?f=158") states that there should always be a space between a numeric value and its following unit symbol. E.g., 200 MPa, not 200MPa. You can see this by looking in any textbook.

 

[Mene]

Hi Emma,

Based on the information provided, you will need to calculate the required section modulus and moment of inertia for the column to withstand the given axial load of 450kN with a factor of safety of 3. The section modulus and moment of inertia are important factors in determining the strength and stiffness of a column.

To calculate the required section modulus, you can use the formula S = M/F, where S is the section modulus, M is the bending moment, and F is the axial load. The bending moment can be calculated using the formula M = PL/4, where P is the axial load and L is the length of the column. In this case, M = (450kN)(7m)/4 = 787.5kNm. Therefore, the required section modulus is S = (787.5kNm)/(450kN) = 1.75m^3.

To calculate the required moment of inertia, you can use the formula I = (PL^2)/8, where P is the axial load and L is the length of the column. In this case, I = (450kN)(7m)^2/8 = 3937.5m^4.

Once you have calculated the required section modulus and moment of inertia, you can refer to your printout of universal columns dimensions and properties to find a suitable section that meets these requirements. It is important to keep in mind that the section modulus and moment of inertia should be equal to or greater than the required values to ensure the column can safely support the given load.

I hope this helps. Good luck with your assignment!