Axial loads and columns selection

[Emzielou83]

Hi,

I have a question from an engineering science assignment that I'm stuck on.

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

I have a printout with my assignment of universal columns dimensions and properties. As part of this printout I have the final answer on a answer sheet = Second moment of area, I = 3351 cm^4.

Im not sure if I need to work out bending moments and bending stresses, or if there is a formula to work this out using the figures I have been given.

I have looked in my textbook for this, however as I'm slowly finding out my textbook is not brilliant.

If anyone can help or knows of any good tutorial websites for this it would be a big help.

Thanks

Emma

 

[Mapes]

Why would you need to calculate bending stresses for an axial load?

 

[Emzielou83]

Hi,

Thank you for your reply.

I don't really know what an axial load is!

I can't find it in my textbook. Is it known as something else?

Am I right in thinking the symbol for an axial load is P?

I have looked on the internet but can't seem to find any formulas to work this out.

Thanks

Emma

 

[Mapes]

An axial load is one whose line of action is through the axis. The beam will stretch or compress lengthwise rather than bend.

[STRIKE]So you probably have a transverse load (one perpendicular to the axis), which makes the "simply supported" condition make more sense. Yes, you do need to calculate the maximum bending moment and then calculate the bending stresses as a function of beam shape. Any good mechanics book (e.g., Beer and Johnston) will describe how to do this.[/STRIKE]

EDIT: I missed the description "column" in the original question, which makes it more likely that the load is axial after all. Take nvn's advice below.

 

[nvn]

Emzielou83: Don't compute the bending stresses. Look for the Euler buckling formula. Ensure FS*P does not exceed the Euler buckling formula load, where FS = factor of safety, and P = axial load = 450 kN. Perhaps (?) also ensure FS times the column axial stress does not exceed the elastic limit listed in post 1.

 

[Emzielou83]

Thanks both for your help.

I have found the eulers formula:

P = pi^EI/ L^2

After transposing the formula to find I :

putting in my numbers of:

P= 450 kN
E = 200 GPa
L = 7 m

I am not sure what these units are (m2/cm2 etc) because E is in GPa and I don't know if I was supposed to put this as MPa first.

Even if I got the units wrong though, the answer still does not match. (I = 3351 cm^4.)

Thanks once again for your help

Emma

 

[nvn]

Emzielou83: Before you work the problem, convert all numbers to either N, mm, MPa, or N, m, Pa. Try it again. Also, didn't you forget to multiply P by FS in your above formula? See post 5. Also, there is a minor typo in your above formula.

 

[Emzielou83]

nvn:

I have done the FS*P calculation and it works out as 1350kN, (3*450kN). So in MPa would be 1350/7^2 = 27.55 kPa, (kN/m^2). Is this correct?

I'm still confused over the rest of it though, mainly because I keep coming out with the wrong answer.

Using the formula: P = pi^2EI/L^2

450= pi^2*200*I/7^2

450 = 9.869*0.2*I/49

49*450 = 1973.92*I

22050 = 1973.92*I

22050/1973.92 = I

11.1706 = I

I think that I should be in cm^4, but my answer is wrong (correct answer 3351 cm^4). Can you see what I am doing that is incorrect?

Much appreciated,

Emma

 

[nvn]

Emzielou83: First convert all quantities to N, mm, and MPa. Therefore, go ahead and convert (and list) all values, so we can see where the trouble is coming in. I.e., convert P in kN to N, convert E in GPa to MPa, etc. Try it again. Regarding your formula, go ahead and change P in your equation to FS*P.