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Beam buckling help.

Beam buckling help.

Beam buckling help.

(OP)
So i have a 'c' channel beam as shown here: The beam i used is marked with red


I assumed the whole beam to be of thickness C (.487") in this case.. found my moment of inertia to be Ixx=1.80621in^4

Here is my eulers formula

Fcr  = (pi^2 x E x I)/L^2    I used english standard units which i think are correct but see if i am correct here:

E for astm a36= 200gpa = 29007547.54psi
I is my moment of inertia in inches^4 = 1.80621  (using the 8" beam with thickness of .487" and spars of 2.527")

Fcr = (3.14^2 x 2907547.54psi x 1.80621in^4 )/216"^2 = 11072.126 lbs  before one beam buckles..

does this sound right or did i loose something in the units/math error?


 

As always thanks in advance and if you need ANY automotive engineering or performance advice feel free to email me as I am not on here that often.   

RE: Beam buckling help.

AISC gives Iyy=1.98, assume your L = 216 in

dont forget to add in KL is the effective length of the column
 
K = column effective length factor, whose value depends on the conditions of end support of the column, as follows.
For both ends pinned (hinged, free to rotate), K = 1.0.
For both ends fixed, K = 0.50.
For one end fixed and the other end pinned, K = 0.699....
For one end fixed and the other end free to move laterally, K = 2.0.

 

RE: Beam buckling help.

(OP)
do you have the formula with KL added?

As always thanks in advance and if you need ANY automotive engineering or performance advice feel free to email me as I am not on here that often.   

RE: Beam buckling help.

You called it a beam, but are apparently using the channel as a column.  Is this correct?  Channel shapes are very inefficient when used for compression members.

RE: Beam buckling help.

Quote:

I assumed the whole beam to be of thickness C (.487") in this case..

I hope you know that the thickness of the web is different than the thickness of the flanges.

 

RE: Beam buckling help.

The 11 kips you came up with is a nominal critical buckling load.  This needs to be reduced by a factor of safety to compare to an actual load.

This is a grossly inefficient column, though.  I don't have a steel manual in front of me, but using 490pci for steel with 18.75 plf, I come up with an area of 5.5 sq in.  That equates to a radius of gyration of 0.6.  This means your kl/r is 216/.06 = 360.  This is incredibly high.  

Not to mention, the nominal buckling stress is 11kips/5.5 sq in = 2 ksi.  2000 psi is an incredibly low buckling stress and indicates an inefficient use of material.

RE: Beam buckling help.

Limit States Per AISC (360-05 Specs Section F):

If you're using the angle for flexure (as a beam):

     -channel bent about major axis (rotated 90 degrees from your picture):

             1. Yielding: Mn=Mp=Fy*Zx
             2. Lateral Torsional Buckling:
                  a) L<Lp, the limit state doesn't apply
                  b) Lp<L<Lr, Mn=Cb(Mp-0.7Fy*Sx((L-Lp)/(Lr-Lp)))<Mp
                  c) L>Lr, Mn=Fcr*Sx<Mp

              where:
              Fcr=Cb*pi^2*E/(L/rts)^2(1+0.078J*c/(Sx*ho)(L/rts)^2)^0.5
              Lp=1.76ry*sqrt(E/Fy)
              Lr=1.95rts*E/(0.7Fy)*(J*c/(Sx*ho))^0.5(1+(1+6.76(0.7Fy*Sx*ho/(E*J*c))^2)^0.5)^0.5
              rts^2=(Iy*Cw)^0.5/Sx
              c=ho/2(Iy/Cw)^0.5 -> for channels
              Cw= Warping Constant
              ho=dist. between flange centroids

     -channel bent about minor axis (same as picture):

            1. Yielding: Mn=Mp=Fy*Zx<1.6Fy*Sy
            2. Flange Local Buckling:
                 a)Compact flanges (lambda=bf/(2tf)<lambda_p):
                  FLB doesn't apply
                 b)Non-Compact flanges (lambda_p<lambda<lambda_r):
                  Mn=(Mp-(Mp-0.7Fy*Sy(lambda-lambda_p)/(lambda_r-lambda_p)
                 c)Slender flanges (lambda>lambda_r):
                  Mn=Fcr*Sy

             where Frc=0.69*E/lambda^2
             lambda_p and r come from table B4.1 of the AISC 360-05 Specifications:

             lambda_p (flanges)=0.38(E/Fy)^0.5
             lambda_r (flanges)=1.0(E/Fy)^0.5

If you're using in unifrom compression (as a column), just say and I'll write the equations.

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