Width/Thickness Ratio 1

[WpgKarl]

I have a steel plate in flexure, length is 36", plate dimensions are 1/2" wide x 8" deep (in direction of bending).

How would I check the class of the rectangular plate in bending, using the Canadian steel code? (Want to see if section is Class 1, 2, 3 or 4)for calc. of Mr.

 

[kslee1000]

BA:

I am confused here:

Pcr = 33100 is acting in direction of span (L=36"), and P = 26070 is acting in direction of beam depth (h=8"), what is the relevance of the two?

 

[kslee1000]

Based on my limit knowledge on stability, and the lack of handy literatures, the most closely matched discussion/teaching on this subject I can find is from "Steel Structures - Design and Behavior", 2nd Ed. Ch. 6.15, by C. G. Salmon & J. E. Johnson.

For a plate subjected to uniform compression along the span length (a), the critical buckling stress is a function of E, poisson's ratio, b (cross section depth), t (plate thickness) & k (a constant depending on type of stress, edge support condition, and most interestingly, length to depth, a/b, aspect ratio). Fcr (ksi) = k*(Elastic Buckling Stress EQ).

From the figure provided, as a/b increases, k is approaching a constant (flat), its value is depending on the edge support condition. There is no case for both top and bottom edges free, but the worst case being top edge simply supported (clamped), and the bottom edge free. For such case, under axial compression, k = o.425. For your case, the corresponding Fcr = 43.5 ksi, which is lower than the stated Fy = 44 ksi. Not to mention it will get worse if the top edge is left free with concentrate loads. Watch out. You need to do more reasearch on this.

By the way, the article referred to reasearch/literature by Timoshenko & others.

 

[BAretired]

kslee,

We have been talking about a beam with vertical load P acting downward at midspan. As the load is increased, one of two things can happen, the beam can reach its factored moment capacity or it can buckle laterally at Pcr. If Pcr is greater than P(factored) then the beam is not subject to lateral buckling, i.e. flexure governs design.

Both 33100# and 26070# mentioned above are vertical loads acting at right angles to the longitudinal axis of the beam. The beam in this case reaches maximum fiber stress without lateral buckling, so buckling does not govern the design.

If the beam is reduced to 8" x 1/4" the opposite is true, i.e. it buckles before it can develop maximum bending capacity.

Best regards,

BA

 

[kslee1000]

BA:

I see your point. You could be 100% correct. Unfortunately, I don't have the referenced literature to verify the validity of the method used. My doubt on stability was triggered by the design of web plate for deep girders, which with the help of flanges, but still requiure stiffereners to provide lateral stability. I am kind wary on this type of application.

 

[BAretired]

WpgKarl,

You asked

How would I check the class of the rectangular plate in bending, using the Canadian steel code?

.

Article 13.6 on page 1-34 for Class 1 and 2 sections appears to be conservative:

M_u = (omega₂*pi/L)*(EIyGJ)^1/2

The second term under the square root sign is zero because Cw = 0.

Omega₂ = 1.0 (conservative)

P_cr = 4*M_u/L



Best regards,

BA

 

[BAretired]

kslee,

For a simple beam of narrow rectangular section, the warping rigidity may be taken as zero. With a moment applied at each end, the moment is constant throughout the span. The smallest root of the differential equation, according to Timoshenko & Gere gives:

M_cr = (pi/L)*(EIyGJ)^1/2

Constant moment throughout the span represents the most critical moment diagram so far as lateral torsional buckling is concerned. It seems to me this would be a safe value to use for other less critical moment envelopes.

Best regards,

BA

 

[kslee1000]

BA:

Thanks. Looks like I should collect the reference and read closely, if I can find time and put my hands on. Appreciate your effors to clear this matter, though I wouldn't put this application to task.