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.

 

[GalileoG]

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.

 

[WpgKarl]

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?

 

[nutte]

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.

 

[WpgKarl]

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

 

[GalileoG]

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.

 

[nutte]

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.

 

[kslee1000]

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)

 

[kslee1000]

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

 

[youngstructural]

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...

 

[WpgKarl]

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!

 

[kslee1000]

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.

 

[BAretired]

"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:

P_cr = 16.94*(E*I_y*C)^1/2/ L²

where I_y = 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

 

[kslee1000]

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.

 

[BAretired]

kslee,

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

I_y = 0.08333 in⁴

C = 3.73e6 in²#

P_cr = 39,244#

M = 353,194"#

f_b = 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

 

[BAretired]

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 P_cr is reduced by a factor of approximately F:

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

Best regards,

BA

 

[BAretired]

I made an error in calculating f_b 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:

P_cr = 4,907#

f_b = 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

 

[WpgKarl]

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!

 

[BAretired]

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

 

[BAretired]

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

Best regards,

BA