huangyhg

width/thickness ratio
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.

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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.fff">
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?
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.
what does the latest aisc(13th) edition say for a case like this?
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.fff">
the latest aisc specification can be downloaded here:
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)
correction: l/d 36/8 = 4.5, d/b = 8/.5 = 16
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...
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

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