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Allowable bending stress of rectangular plate

I'm using the 9th edition of the AISC steel manual and am getting myself all confused. I have a situation where I'm trying to use a steel .25" x 2.5" steel plate to stiffen an aluminum "beam" in bending. I'm orienting the plate such that I'm loading it in bending along the strong axis, where my Mom of I = .325in^4. I can only laterally brace this piece at 12" intervals. I suspect that when this thing is bent about the strong axis that my compression edge(not sure I can call it a flange as I'm basically just analyzing a prismatic rectangle) is going to buckle out of plane. I'm confused as to what my allowable bending stress would be for 36ksi plate steel.
How do I look at this in regards to Table B5.1? Do I even need to? It seems to me that I need to look at the "Unstiffened elements simply supported along one edge......." which gives me a limiting widthtothickness ratio of 76/Fy^.5. Is that right? If so, then what do I do with that then in Chapter F? There is no axial load carried by this piece.....only bending.
Thanks for your help. 

nutte (Structural) 
31 Mar 10 10:02 
As you've noticed, the 9th edition doesn't do a good job of addressing strongaxis bending on a rectangular plate. Use the 13th edition, where specification F11 does deal with this case, including lateral torsionla buckling. 

Wish I had one! 

rb1957 (Aerospace) 
31 Mar 10 10:09 
remember too, dissimiliar materials ... "rule of mixtures" 

ash060 (Structural) 
31 Mar 10 10:18 
PostFrameSE
The 2005 specification is available from AISC's website for free.
The manual tables are not there, but all of the spec is and that is all you need. 

Dissimilar metals corrosion and expansion/contraction  plus the extreme difficulty of 100% joining of the reinforcing plate (in steel) to the (more flexible!) aluminum base structure?
I don't recommend this configuration at all. Better to use a thicker Al plate. Then weld the AlAl. 

DHKpeWI (Structural) 
6 Apr 10 10:36 
Be careful when designing welded aluminum sections.
Depending on the type of weld (longitudinal or transverse) the allowable stresses of a welded section can be significantly reduced. See the Aluminum Design Manual. 

JAE (Structural) 
6 Apr 10 12:57 
PostFrameSE  simply go to this link: http://www.aisc.org/content.aspx?id=2884...and download the FREE AISC specification (Specification for Structural Steel Buildings (ANSI/AISC 36005) ) which is inside the 13th Edition. 

I appreciate all of your responses. I'm dealing with a 2.5" x .25" thick member bent about the strong axis. Given 36ksi steel, a Cb=1.0 and Lb=12", could somebody confirm that my calcs are correct, in that my allowable bending stress would be 29,246psi and my moment capacity is 7,604inlbs? If that's right, I'm comfortable with this methodology.
Thanks


Lb*d/t^2 = 480 < 1.9 E/Fy Z= bt^2/4 = 0.3906 in^3 FyZx = 14.0625 kin
I am getting an Mn based on equation F112 19.0791 kin. 

Thanks slickdeals,
If I break out my formula more it would look like this: Mn = [1.52  .274(480)*36/29,000]*My
That becomes 1.3567*My
My I believe should be 36*.26 = 9.36ink since Sy=.25*2.5^2/6 = .26
Did you use Mp rather than My???
Therefore, I think Mn = 1.3567*9.36 = 12.69ink.
My allowable moment capacity = 12.69/1.67 = 7.6ink.
Assuming I'm wrong and you're right, where did I go wrong?
Thanks 

Ref. Theory of Elastic Stability by Timoshenko and Gere For pure bending of a narrow rectangular section about its major axis: M _{cr} = π√(E*Iy*G*J)/L E = 29,000,000 psi G = 12,000,000 psi Iy = b*h ^{3}/12 = 0.003255 in^4 J = b*h ^{3}/3 = 0.01302 in^4 L = 12" So M _{cr} = 31,800"# > My = 9375"# Plastic moment, Mp = 14060"# For an unbraced length of 12", lateral buckling does not appear to govern, so you can use the allowable stress specified by your code. I would not expect it to be as high as 29,246 psi, however. BA 

PostFrameSE: Yes, you are right. It appears My should be used and not Mp because you are not reaching Mp. 

Hopefully this is my last question. If we can agree that the yield moment is around 7.6ink (including the 1.67 reduction factor), doesn't it follow that the allowable fiber bending stress is = to the M / S? If S is = .26in^3, then the fiber bending stress has to 29.2ksi right? Am I missing something BA??
Thanks all. 

nutte (Structural) 
8 Apr 10 17:39 
Your nominal moment, Mn, is higher than the yield moment, My, but not all the way to the plastic moment, Mp.
So, in theory and in line with the factor of safety, your stress diagram will not be linear. The extreme fiber stress will be 21.6 ksi (Fy/1.67). As you come down the member from the top fiber, the stress will remain 21.6 ksi, until you reach a point where it does start to decrease, until you have zero stress at the neutral axis.
In actuality, the extreme fiber stress is 29.2 ksi. Since this is below yield, the stress diagram will remain linear. 

PostFrameSE, The moment, 'My' at first yield is: My = Fy*S = 36,000*0.2604 = 9.375"k I don't use the AISC code and I don't use ASD, so I can't remember for certain, but as I recall, the maximum allowable stress you can use is Fy/1.67 = 21.65"k which corresponds to an allowable moment of 5.6"k. This would suggest that one of us is missing something. BA 

am kinda lazy to look for it.. where can i find how they come up with 0.75 for base plate
baseplate ASD Fb = 0.75 Fy


westy, that is in the compact shape direction, he is bending in the other direction, the noncompact direction. Michael. Timing has a lot to do with the outcome of a rain dance. 

nutte (Structural) 
9 Apr 10 10:19 
BA, the AISC equation is checking for lateral torsional buckling, which does control.
If it did not control, you'd be allowed to go to Mp, not just My, in which case the allowable moment would be 50% higher than the values you listed. 

nutte, The original question in this thread was in respect to allowable stresses. Under no circumstances would you be able to go to Mp if you are using ASD. The maximum allowable stress using ASD is Fy/1.67 as I understand it. As for the AISC equation, I can't comment because I don't have it, but the equation I provided earlier was for lateral torsional buckling of a narrow rectangular section according to Timoshenko and Gere. Are you saying it is wrong? BA 

nutte (Structural) 
9 Apr 10 11:01 
Under ASD, you are allowed to go to Mp. In the old ASD, for a wide flange, this was done using 0.66 Fy for the allowable, being a 10% increase over the normal 0.6 Fy, correlating to a general ratio of Zx/Sx for wide flange shapes equal to 1.10.
Under the current AISC 13th edition ASD, you're allowed to go to Fy Zx / 1.67 for your moment, if LTB doesn't control.
In this case, using AISC's LTB equations, you get a nominal moment higher than My, but less than Mp.
I'm not as familiar with the Timoshenko/Gere equations, so I can't comment on them. Regardless, the AISC equations do say that LTB controls. 

nutte, I wasn't aware of the change to ASD, but I admit it makes sense. I cannot believe there is such a huge difference in results between AISC and Timoshenko for lateral torsional buckling. Could you post the AISC equation for comparison? BA 

nutte (Structural) 
9 Apr 10 13:10 
Here you go. A previous section says phi=0.90 and omega=1.67. 

nutte (Structural) 
12 Apr 10 10:30 
BA:
The Timoshenko equation you posted is the same as Equation F114 in the AISC specification. Equation F113, the one that applies to our specific case, must be an for an inelastic buckling region.
Timoshenko: Mcr = pi*√(E*Iy*G*J)/L
F114: Fcr = 1.9 E Cb / (Lb*d/t^2)
We can manipulate Timoshenko's equation as such:
Iy = d*t^3/12 G = E/2.6 J = d*t^3/3
Mcr = pi*E^0.5*(E/2.6)^0.5*(d*t^3/12)^0.5*(d*t^3/3)^0.5/L
Mcr = 1.948*E*d*t^3/6/L
Multiply both top and bottom by d, resulting in:
Mcr = 1.948*E*d^2*t^3/6/L/d
Replace d*t^2/6 on top with Sx:
Mcr = 1.948*E*t^2*Sx/(L*d)
Remove Sx, to get Fcr, and pull t^2 to the bottom.
Fcr = 1.948*E/(L*d/t^2)
For Cb=1, this is AISC's equation, with 1.948 rounded to 1.9.


nutte, Yes, thanks for that. I guess we are in the inelastic buckling range with this problem. Tricky little devil, isn't it? BA 



