Calculating minimum I-beam section area under load

[a13x]

Homework Statement

This seems to a reasonably basic question to go through but yet I seem to have a bit of a mental block on it. so any help in the right direction would be great..

An I-beam is 10mm thick on the web and flange, is 6m high and has a Young's modulus of 200 GN/m2. The design must be optimized to carry 92KN down the neutral axis of the section. Given the the depth of the section, d, must be half the breadth, b, calculate the minimum section requirements when:

Homework Equations

1, The upper end is free
2, The upper end is pinned and constrained to move vertically

The Attempt at a Solution

To do this problem I used the critical pressure equation and derived it for I:

Pcr=2π2 EI/L^2

I= (P_cr.L^2)/(2π^2.E)

I= (92〖x10〗^3×〖6x10〗^3)/(19.74×〖200x10〗^9 )

I=0.838

Now this is where I get stuck as I don't know how to input I value to calculate the breadth and depth of the I-beam section. Putting it into the 2nd moment of area equation for an I-beam gets very messy. So what do I do from here?

I know the answer for depth, d=79mm, and when calculated for a rectangular cross section (I=bd^3/12) d=150mm, almost exactly twice. How do I correct this from a rectangle to an I-beam equation?

 

[PhanthomJay]

Be careful with units. If you are using Newtons and millimeters, then E must be expressed in N/mm^2.

You don't state the bottom end condition. I assume it is fixed.

The critical buckling formula depends on the end conditions at each end (k factor for effective length).

It also depends on the weak axis moment of inertia.

You might want to simplify the problem by ignoring the small contributions from various parts of the cross section. Are you sure the flange width is twice the depth? I don't know why that optimizes the section. I've never seen an I beam with a flange dimension twice the depth. And no safety factor?

 

[a13x]

I have included a diagram from the question regarding the section.
For the first part the column is fixed-free
Second part it is Fixed-Pinned

There is no safety factor given in the question. Apart from what I have done myself shown in first post I'm totally stuck with it. All information I have given is what I am given (including diagram and end conditions)

 

[PhanthomJay]

I have included a diagram from the question regarding the section.
For the first part the column is fixed-free
Second part it is Fixed-Pinned

There is no safety factor given in the question. Apart from what I have done myself shown in first post I'm totally stuck with it. All information I have given is what I am given (including diagram and end conditions)

Oh, OK, that makes much more sense now seeing the picture, the beam flange width, d, is half its depth, b. Or b=2d. That's confusing lettering. But anyway, when calculating the weak axis I about the centroid, the contribution of the beam's web (the piece between the 2 flanges) is negligible. Thus, in determining it's moment of inertia, you need only to look at the 2 flanges (What is the I of each rectangular flange, if its width is 10 and it's height is d? What's the total I of the cross section? Set the result equal to your calculated I , and solve for d). But first remember, in calculating the numerical value of I, that

P_cr =(pi)^2(EI)/(kL)^2.

Where k depends on the end conditions (you can look those values up for the free-fixed or pinned-fixed case). Watch your units.