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.
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
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
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
RE: Width/Thickness Ratio
RE: Width/Thickness Ratio
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
RE: Width/Thickness Ratio
RE: Width/Thickness Ratio
RE: Width/Thickness Ratio
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
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
RE: Width/Thickness Ratio
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
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
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
where F = {1 - (1.74a/L)(E*Iy/C)1/2}
Best regards,
BA
RE: Width/Thickness Ratio
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
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
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
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
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
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
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
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.
RE: Width/Thickness Ratio
You asked .
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
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
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.