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Width/Thickness Ratio

Width/Thickness Ratio

Width/Thickness Ratio

(OP)
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.

 

RE: Width/Thickness Ratio

This is elementary stuff. I would suggest you open the code (S16) and check out the tables for class designations.

Also, you may want to read "Limit States Design in Structural Steel" by Kulak. It's a good complimentary book to the S16.

Clansman

"If a builder has built a house for a man and has not made his work sound, and the house which he has built has fallen down and so caused the death of the householder, that builder shall be put to death." Code of Hammurabi, c.2040 B.C.

RE: Width/Thickness Ratio

(OP)
Sounds elementary at first glance, but I'm not so sure.  The logical starting point is the use the h/w ratio for the plate & check limits for the web of an I-shape.  However, there are different boundary conditions for the restraint of the web of an I-beam in flexure and a rectangular plate in flexure.

In the I-Beam, if the web were to try and buckle laterally (out of plane), it is restrained by the flanges - in my case there is no such restraint - so I am wondering if the boundary conditions used to develop the code Eqn for webs of I-beams are applicable to my case, or should it be adjusted to suit a rect. plate?  

RE: Width/Thickness Ratio

The equations for a web would not apply to your situation, for the reasons you note.  Your plate is an unstiffened element.  I'm not familiar with the Canadian code, so I don't know if it addresses your situation.  The AISC code didn't address this until the latest (13th) edition.

RE: Width/Thickness Ratio

(OP)
What does the latest AISC(13th) edition say for a case like this?

RE: Width/Thickness Ratio

Woops. Misunderstood your post.

It has always been an office policy here to always use the elastic section modulus for plates, even if they possess plastic capacity.

That does not directly answer your question, I too am now curious what h/w limit to check against.

Clansman

"If a builder has built a house for a man and has not made his work sound, and the house which he has built has fallen down and so caused the death of the householder, that builder shall be put to death." Code of Hammurabi, c.2040 B.C.

RE: Width/Thickness Ratio

The latest AISC specification can be downloaded here:  http://www.aisc.org/content.aspx?id=2884.  Section F11 (on page 60 of the printed specification) has the provisions for bending on a plate.

RE: Width/Thickness Ratio

Besides the code, by examing the geometric ratios, L/d = 36/8 = 4.5, d/b = 8/.5 = 4.0, I think this is not a simple problem, the stresses are non-linear, instability and inelastic buckling are likely the dominant factors for design. (Only personal thinking, no reference)

RE: Width/Thickness Ratio

correction: L/d 36/8 = 4.5, d/b = 8/.5 = 16

RE: Width/Thickness Ratio

This is where prescriptive codes fail us:  We can return to fundamentals and make this work, but it's just not going to fit into any of the boxes the code has created for you.

Stability is going to be your problem.  Even if the plate has the strength, in theory, to take the load in bending, lateral torsional buckling will happen almost instantly without flanges.  Further, LTB will be raced to the finish by the fundamental falling over of the plate when any load is placed on the section.

IF you solve the stability issue, there is no reason that you cannot use this section as a beam.  But then you are genuinely looking at a beam.

You could even analyze this as a strut and tie model.  In any case you have to solve the stability issue first.

Cheers,

YS

B.Eng (Carleton)
Working in New Zealand, thinking of my snow covered home...

RE: Width/Thickness Ratio

(OP)
Thx to all for your help.  It's funny that something this simple (a rectangular plate in bending) isn't properly accounted for in our steel codes - probably b/c a rectangle isn't used very often as a beam - not very efficient in bending as there are no flanges.

Having a thickness of 1/2" and a depth of 8", I just don't see localized buckling happening in this plate before lateral torsional buckling occurs.  I checked the depth/thickness ratio of the plate and if I used the limit for the web of an I beam as a rough estimate, I am way below the limit for plastic (Class 1) capacity.

I am going to check the beam by limiting the factored stresses to yielding (My) and I will also check the beam using the eqn for unbraced beams, using 3 ft (the beam length) as the unbraced length.  The Cw (warping constant)  in the second half of the unbraced moment capacity eqn is equal to zero for a member w/o flanges, so that will likely limit the bending resistance to something less than My, for this unbraced length.

I will also make sure the ends of the beam are restrained from twisting.

PS - This is for a pre-stressing application - the force is applied as a point load (2 strands at 4" apart) near the middle of the span.

Cheers!

RE: Width/Thickness Ratio

Not the code does not address this, it is because the rareness of your application. Beware, thin plate is very weak in resisting in-plane forces, as it will always buckle under compression prior to the force could fanned out. Try to grab a piece of 8"x11" paper and rotate your hands inward, that's what your beam will looks like upon loading. Torsion? Not to me. The compression zone has failed and buckled. Please send me a note if you can find any meaningful papers/reasearches on this type of application.   

RE: Width/Thickness Ratio

"Lateral Buckling of Simply Supported Beam of Narrow Rectangular Cross Section" is treated in Theory of Elastic Stability by Timishenko and Gere(6.5).  The solution involves Bessel functions which are Greek to me, but for a load applied at the centroid of the middle cross section:

Pcr = 16.94*(E*Iy*C)1/2/ L2

where Iy = hb3/12
and   C = GJ = hb3G/3
L = span, h = height of section and b = thickness

The formula applies only within the elastic range and there is a correction for the load being applied higher or lower than the centroid.

There is also an expression for a load P located distance c from one support.

 

Best regards,

BA

RE: Width/Thickness Ratio

BA:

Excellent reference. How it defines "Narrow Rectangular Cross Section"? Does it has limits on h/b & b? Sorry for asking you to do the extra works, just not available to me at this time. Thanks.

RE: Width/Thickness Ratio

kslee,

No mention of a limit.  Just for fun, I worked it out for the parameters given.  I get:

Iy = 0.08333 in4

C  = 3.73e6 in2#

Pcr =  39,244#

M = 353,194"#

fb = 8278 psi

This would have to be reduced if the load is applied above the centroid.  It represents the buckling load, not an allowable load, so you would need to include a safety factor.

Best regards,

BA

RE: Width/Thickness Ratio

If the load P is applied a distance 'a' above the centroid and if a is small (the book doesn't say how small) the expression for Pcr is reduced by a factor of approximately F:

    where F = {1 - (1.74a/L)(E*Iy/C)1/2}

Best regards,

BA

RE: Width/Thickness Ratio

I made an error in calculating fb earlier.  It should be M/S = 353,194/5.33 = 66,200 psi which is above the yield strength for most structural steels.  So it appears that lateral buckling would not occur for this plate.  It would fail at its plastic moment.

I did a quick check for a plate 8 x 1/4 and found:

Pcr = 4,907#

fb = 44169/2.67 = 16,500 psi

So an 8" x 1/4" plate would be governed by lateral buckling and would fail under a load of about 4900# at midspan (less if load application was above centroid).

Best regards,

BA

RE: Width/Thickness Ratio

(OP)
Wow, great reference BAretired!  

In my case, I limited the extreme fiber stress in bending to Fy (300 MPa / 44 ksi).  Assuming a linear stress distribution, then the resultant compression force in the comp. zone of the rectangular beam (1/2" wide x 8" deep) would be 1/3 of 0.5*(8") = 4/3" = 1.33" from the top of the beam, while the bottom half is in tension.

So to check buckling using that eqn, would I only look at the top 1/2 of the beam (1/2" wide x 4" high) and use "a" = (2"-1.33" = 0.66"?)

Cheers!

RE: Width/Thickness Ratio

That book and a couple of others co-authored by Stephen P. Timoshenko are excellent references for structural engineers.

If you are using elastic design and stressing the plate to a maximum of 44,000 psi at ultimate, then the factored moment will be 234,600"#, corresponding to a factored point load of 26,070#.  The allowable load will be the factored load divided by your load factor, which is code dependent.

If, as is probable, the load is applied at the top of the beam, the term 'a' is 4" because it is 4" above the centroid.  If, on the other hand, the load is hanging from the bottom of the beam,'a' will be -4" and the critical load will be greater than I calculated (except for the fact that the stress exceeds the yield strength and Timoshenko's theory is valid only within elastic limits).

In lateral buckling problems, the point of application of load is an important variable.

Best regards,

BA

RE: Width/Thickness Ratio

WpgKarl,

Taking it a step further, if the load is applied at the top of the beam, then F = 0.844 and Pcr = 33,100#.  This value is not strictly valid because the beam would be stressed beyond the elastic range.  

Since Pcr is greater than 26,070#, the design is governed by maximum fiber stress and not by lateral buckling.

Best regards,

BA

RE: Width/Thickness Ratio

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?

RE: Width/Thickness Ratio

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.   

RE: Width/Thickness Ratio

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

RE: Width/Thickness Ratio

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. 3eyes

RE: Width/Thickness Ratio

WpgKarl,

You asked

Quote:

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:  

Mu = (omega2*pi/L)*(EIyGJ)1/2

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

Omega2 = 1.0 (conservative)

Pcr = 4*Mu/L

 

Best regards,

BA

RE: Width/Thickness Ratio

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:

Mcr = (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

RE: Width/Thickness Ratio

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.

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