Column Buckling - Inelastic
A long topic
Effects of geometric imperfection
EIx v  Pv  0
EIy u Pu  0
Leads to bifurcation buckling of
perfect doubly-symmetric columns
P
M x  P(v  v o )  0
 EIx v  P(v  v o )  0

 v  Fv2 (v  v o )  0
v o  o sin
z
L
vo
v

v
 v  Fv2v  Fv2v o
Mx
 v  Fv2v  Fv2 ( o sin
z
L
)
Solution  v c  v p
v c  A sin( Fv z)  B cos( Fv z)
P
v p  C sin
z
L
 Dcos
z
L
Effects of Geometric Imperfection
Solve for C and D first
 v p Fv2v p  Fv2o sin
z
L
 2 

z
z 
z
z 
z
  C sin Dcos  Fv2 C sin Dcos  Fv2o sin
0



L  
L
L
L
L
L
2
2


z   
z   
2
2
2
 sin C   Fv C  Fv o  cos   D  Fv D 0
L  L 
L  L 


  2

 2
2
2
2
C   Fv C  Fv o  0 and   D  Fv D 0
L 
 L 

Fv2o
C 
 2
2
   Fv
L 
 Solution becomes
and D  0
Fv2o
z
v  A sin( Fv z)  Bcos( Fv z) 
sin
L
 2
2
   Fv
L 
Geometric Imperfection
Solve for A and B
Boundary conditions v(0)  v(L)  0
v(0)  B  0
v(L)  A sin Fv L  0
A0
 Solution becomes
Fv2o
z
v
sin
2
L
 
2
   Fv
L 
Fv2
2 o
 
P

 
z PE o
z
L 
v 
sin

sin
Fv2
L 1 P
L
1
PE
 2
 
L 
P
z
PE
v 
o sin
P
L
1
PE
Total Deflection
P
z
z
PE
 v  vo 
o sin  o sin
P
L
L
1
PE
 P

 P

z
1
z
  E  1o sin 
o sin
L 1 P
L
1 P



PE
 PE

z
 AFo sin
L
AF = amplification factor
Geometric Imperfection
AF 
1
P
1
PE
 amplification factor
M x  P(v  v o )
 M x  AF (Po sin
z
)
L
i.e., M x  AF  (moment due to initial crooked)
12
10
Amplification Factor AF

Increases exponentially
Limit AF for design
Limit P/PE for design
8
6
Value used in the code is 0.877
This will give AF = 8.13
Have to live with it.
4
2
0
0
0.2
0.4
0.6
P/PE
0.8
1
Residual Stress Effects
Residual Stress Effects
History of column inelastic buckling

Euler developed column elastic buckling equations (buried
in the million other things he did).




Take a look at: http://en.wikipedia.org/wiki/EuleR
An amazing mathematician
In the 1750s, I could not find the exact year.
The elastica problem of column buckling indicates elastic
buckling occurs with no increase in load.

dP/dv=0
History of Column Inelastic Buckling



Engesser extended the elastic column buckling theory in
1889.
He assumed that inelastic
buckling occurs with no
increase in load, and the
relation between stress
and strain is defined by
tangent modulus Et
Engesser’s tangent modulus theory is easy to apply. It
compares reasonably with experimental results.

PT=ETI / (KL)2
History of Column Inelastic Buckling

In 1895, Jasinsky pointed out the problem with Engesser’s
theory.


If dP/dv=0, then the 2nd order moment (Pv) will produce
incremental strains that will vary linearly and have a zero
value at the centroid (neutral axis).
The linear strain variation will have compressive and tensile
values. The tangent modulus for the incremental
compressive strain is equal to Et and that for the tensile
strain is E.
History of Column Inelastic Buckling

In 1898, Engesser corrected his original theory by
accounting for the different tangent modulus of the tensile
increment.



This is known as the reduced modulus or double modulus
The assumptions are the same as before. That is, there is
no increase in load as buckling occurs.
The corrected theory is shown in the following slide
History of Column Inelastic Buckling





The buckling load PR produces
critical stress R=Pr/A
During buckling, a small curvature
d is introduced
The strain distribution is shown.
The loaded side has dL and dL
The unloaded side has dU and dU
dL  ( y  y1  y) d
dU  ( y  y  y1 ) d
 d L  E t ( y  y1  y) d
 dU  E( y  y  y1 ) d
History of Column Inelastic Buckling
d  v 
d L  E t ( y  y1  y) v 
dU  E( y  y  y1 ) v 
But, the assumption is dP  0
y
y y1
y y1
( d y )
  dU dA   d L dA  0
y
y y1
y y1
( d y )
  E( y  y  y1 ) dA   E t ( y  y1  y) dA  0
 ES1  E t S2  0
y
where, S1   ( y  y  y1 ) dA
y y1
and S 2 
y y1
 ( y  y1  y) dA
( d y )
History of Column Inelastic Buckling

S1 and S2 are the statical moments of the areas to the left
and right of the neutral axis.


Note that the neutral axis does not coincide with the centroid
any more.
The location of the neutral axis is calculated using the
equation derived ES1 - EtS2 = 0
M  Pv
y
y y1
y y1
( d y )
 M   dU ( y  y  y1) dA   d L ( y  y1  y) dA
 M  Pv  v ( EI1  E t I2 )
y
where, I1   ( y  y  y1 ) 2 dA
y y1
and I 2
y y1
 ( y  y1  y) 2 dA
( d y )
History of Column Inelastic Buckling
M  Pv  v ( EI1  E t I2 )
 Pv  ( EI1  E t I2 )v  0
P


v 
v 0
EI1  E t I2
 v  Fv2v  0
P
P

EI1  E t I2 EIx
I1
I2
and E  E  E t
Ix
Ix
where, Fv2 
PR 
 2 EI x
E is the reduced or double modulus
(KL) 2
PR is the reduced modulus buckling load
History of Column Inelastic Buckling


For 50 years, engineers were faced with the dilemma that
the reduced modulus theory is correct, but the
experimental data was closer to the tangent modulus
theory. How to resolve?
Shanley eventually resolved this dilemma in 1947. He
conducted very careful experiments on small aluminum
columns.



He found that lateral deflection started very near the
theoretical tangent modulus load and the load capacity
increased with increasing lateral deflections.
The column axial load capacity never reached the calculated
reduced or double modulus load.
Shanley developed a column model to explain the
observed phenomenon
History of Column Inelastic Buckling
History of Column Inelastic Buckling
History of Column Inelastic Buckling
History of Column Inelastic Buckling
Column Inelastic Buckling

Three different theories
 Tangent modulus
 Reduced modulus
 Shanley model
P
dP/dv=0

Tangent modulus theory
assumes
 Perfectly straight column
 Ends are pinned
 Small deformations
 No strain reversal during
buckling
Slope is zero at buckling
P=0 with increasing v
v
Elastic buckling analysis
PT
Tangent modulus theory

Assumes that the column buckles at the tangent modulus load such
that there is an increase in P (axial force) and M (moment).
 The axial strain increases everywhere and there is no strain
reversal.
Strain and stress state just before buckling
PT
T
Mx - Pv = 0
T=PT/A
Strain and stress state just after buckling
v
v
Mx
T
T
T
T=ETT
Curvature =  = slope of strain diagram
 
PT
T
h
h

T    y 
where y  dis tance from centroid
2

h

 T    y  ET
2

Tangent modulus theory

Deriving the equation of equilibrium
M x     ydA
A
   T   T
   T   ( y  h / 2)  E T
 M x    T   ( y  h / 2)E T   ydA
A
 M x   T  y dA  E T   y 2 dA h / 2)E T   y dA
A
A
A
 M x  0  E T  Ix  0
 M x  E T Ix v 


The equation Mx- PTv=0 becomes -ETIxv” - PTv=0
2
2
 Solution is PT=  ETIx/L
Example - Aluminum columns

Consider an aluminum column with Ramberg-Osgood
E 10100 ksi
stress-strain curve

40.15 ksi
0 .2
n
  
   0.002 
E
 0.2 
 1 0.002 n1

  n n
 E  0.2





1
0.002
n
 0.2
E
  n1
1
nE  
 0.2
 0.2 

E
E

n1  E T


0.002

1
nE  
 0.2
 0.2 
0.002






nE n1
n

0.000E+00
1.980E-04
3.960E-04
5.941E-04
7.921E-04
9.901E-04
1.188E-03
1.386E-03
1.584E-03
1.782E-03
1.980E-03
2.178E-03
2.376E-03
2.575E-03
2.775E-03
2.979E-03
3.198E-03
3.458E-03
3.829E-03
4.483E-03
5.826E-03
8.771E-03
1.529E-02
2.949E-02
5.967E-02
1.221E-01
18.55

0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
ET
ET
differences
equation
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10100.0
10099.9
10099.8
10099.5
10098.8
10097.6
10094.2
10088.7
10075.1
10054.2
10005.7
9934.0
9779.8
9563.7
9142.0
8602.6
7697.4
6713.6
5394.2
4251.9
3056.9
2218.6
1488.8
1037.0
679.2
468.1
306.9
212.4
140.8
98.5
66.3
46.9
32.1
23.0
Tangent Modulus Buckling
Ramberg-Osgood Stress-Strain
Stress-tangent modulus relationship
60
12000
Tangent Modulus (ksi)
50
Stress (ksi)
40
30
20
10000
8000
6000
4000
2000
10
0
0
0.000
0
0.010
0.020
0.030
Strain (in./in.)
0.040
0.050
10
20
30
40
Stress (ksi)
ET differences
ET equation
50
Tangent Modulus Buckling
(KL/r) c r
223.2521046
157.8630771
128.8946627
111.6260523
99.84137641
91.1422898
84.3813604
78.93150275
74.41710153
70.59690679
67.3048795
64.4113691
61.77857434
59.17430952
56.09208286
51.5097656
44.14566415
34.1419685
24.00464013
15.9961201
10.48827475
6.902516144
4.596633406
3.105440361
2.129145204
Column Inelastic Buckling Curve
60
Tangent Modulus Buckling Stress

0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
PT 
50
 2 E T Ix
L2
PT
 2 E T Ix
 2 ET

 T 

2
A
AL2
KL / r
40
 KL / rcr
30
 2 ET

T
20
10

0
0
30
60
90
KL/r
120
150
Residual Stress Effects




Consider a rectangular section
with a simple residual stress
distribution
Assume that the steel material
has elastic-plastic stress-strain
 curve.
Assume simply supported end
conditions
Assume triangular distribution
for residual stresses
b
x
d
rc
y
rt
y
y/b

rc
y
y
y
E

Residual Stress Effects

One major constrain on residual
stresses is that they must be
such that   r dA  0

2
  0.5 y  y
b
b / 2 
b / 2 

2 y 
x d  dx   0.5 y 
x d  dx
b 
0 


2d y b 2  2d y b 2 
 0.5 y d b 2  0.5 y d b 2 
 
 
b  8 
b  8 
0
0

Residual stresses are produced
by uneven cooling but no load
is present
Residual Stress Effects
b

Response will be such that elastic behavior when
x
d
  0.5 y
Px 
 2 EIx
2
and Py 
 2 EIy
L
Yielding occurs when
L2
b
b
x
  0.5 y i.e., P  0.5PY
Inelastic buckling will occur after   0.5 y
y
Y


2
 Y  Y b  Y (1 2 )


b

y
Y
Y/b
Residual Stress Effects
Total axial force corresponding to the yielded sec tion
   Y (1 2 ) 
 Y b  2bd   Y
bd  2


2
  Y 1 2 bd   Y (2  2 )bd
  Y bd  2bd Y  2 Y bd  2 2bd Y
  Y bd(1 2 2 )  PY (1 2 2 )
 If inelastic buckling were to occur at this load
Pcr  PY (1 2 2 )
 

1  Pcr 
1 
2  PY 
If inelastic buckling occurs about x  axis
 2E
d3
Pcr  PTx  2 (2b)
L
12
 2 EIx
 PTx 
2
L2
1  Pcr 
 PTx  Px  2 
1 
2  PY 
 PTx  Px  2 
1  PTx 
1

2  PY 
PTx Px
1  PTx 


2
1

PY PY
2  PY 
PTx 1
1  PTx 

 2
1

PY 2x
2  PY 
 2x 
 PTx 
21

 PY 
PTx
PY
b
b
x
y
Pcr  PTx
2


P
1
E
r
Let, x  2   2  x 
PY x
 Y K x Lx 
If inelastic buckling occurs about y  axis
Pcr  PTy 
 2E
(2b) 3
d
12
L2
 2 EIy
3
 PTy 
2



L2
3




1
P
 PTy  Py  2 1 cr 

 2  PY 


 PTy  Py  


Pcr  PTy
3
 PTy 
21

P

Y 

PTy 1 


 
PY 2y 

 PTy 
21

P

Y 

3
3
 PTy 
21

 PY 

PTy
PY
x
y
3
 PTy 
21

 PY 

PTy Py 


 
PY PY 




 2y  
b
b
 r 2
Py
1
E
y
Let,  2   2 


PY y
 Y 
K
L
 y y 
Residual Stress Effects
x
2.236
2.000
1.826
1.690
1.581
1.491
1.414
1.313
1.221
1.135
1.052
0.971
0.889
0.803
0.705
0.577
0.317
y
2.236
2.000
1.826
1.690
1.581
1.491
1.414
1.246
1.092
0.949
0.815
0.687
0.562
0.440
0.315
0.182
0.032
Column Inelastic Buckling
Normalized column capacity
P/PY
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
0.650
0.700
0.750
0.800
0.850
0.900
0.950
0.995
1.200
1.200
1.000
1.000
0.800
0.800
0.600
0.600
0.400
0.400
0.200
0.200
0.000
0.000
0.0
0.5
1.0
Lambda
1.5
2.0
Tangent modulus buckling - Numerical
1
Discretize the cross-section into fibers
Think about the discretization. Do you need the flange
To be discretized along the length and width?
2
For each fiber, save the area of fiber (Afib), the
distances from the centroid yfib and xfib,
Ix-fib and Iy-fib the fiber number in the matrix.
Afib
yfib
Centroidal axis
3
Discretize residual stress distribution
4
Calculate residual stress (r-fib)
each fiber
5
Check that sum(r-fib Afib)for
Section = zero
Tangent Modulus Buckling - Numerical
14
6
Calculate effective residual
strain (r) for each fiber
r=r/E
13
7
Assume centroidal strain

Calculate the critical (KL)X and (KL)Y for the 
(KL)X-cr =  sqrt [(EI)Tx/P]
(KL)y-cr =  sqrt [(EI)Ty/P]
Calculate the tangent (EI)TX and (EI)TY for the 
(EI)TX = sum(ET-fib{yfib2 Afib+Ix-fib})
(EI)Ty = sum(ET-fib{xfib2 Afib+ Iy-fib})
Calculate average stress =  = P/A
8
Calculate total strain for each fiber
tot=+r
Calculate Axial Force = P
Sum (fibAfib)
9
12
Assume a material stress-strain
curve for each fiber
Calculate stress in each fiber fib
11
10
Tangent modulus buckling - numerical
Section Dimension
b
d
y
12
4
50
No. of fibers
20
A
Ix
Iy
48
64
576.00
fiber no.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Afib
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
2.4
xfib
-5.7
-5.1
-4.5
-3.9
-3.3
-2.7
-2.1
-1.5
-0.9
-0.3
0.3
0.9
1.5
2.1
2.7
3.3
3.9
4.5
5.1
5.7
yfib
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
r-fib
-22.5
-17.5
-12.5
-7.5
-2.5
2.5
7.5
12.5
17.5
22.5
22.5
17.5
12.5
7.5
2.5
-2.5
-7.5
-12.5
-17.5
-22.5
r-fib
-7.759E-04
-6.034E-04
-4.310E-04
-2.586E-04
-8.621E-05
8.621E-05
2.586E-04
4.310E-04
6.034E-04
7.759E-04
7.759E-04
6.034E-04
4.310E-04
2.586E-04
8.621E-05
-8.621E-05
-2.586E-04
-4.310E-04
-6.034E-04
-7.759E-04
Ix fib
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
3.2
Iy fib
78.05
62.50
48.67
36.58
26.21
17.57
10.66
5.47
2.02
0.29
0.29
2.02
5.47
10.66
17.57
26.21
36.58
48.67
62.50
78.05
Tangent Modulus Buckling - numerical
Strain Increment

Fiber no.
-0.0003
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
tot
fib
-1.076E-03
-9.034E-04
-7.310E-04
-5.586E-04
-3.862E-04
-2.138E-04
-4.138E-05
1.310E-04
3.034E-04
4.759E-04
4.759E-04
3.034E-04
1.310E-04
-4.138E-05
-2.138E-04
-3.862E-04
-5.586E-04
-7.310E-04
-9.034E-04
-1.076E-03
-31.2
-26.2
-21.2
-16.2
-11.2
-6.2
-1.2
3.8
8.8
13.8
13.8
8.8
3.8
-1.2
-6.2
-11.2
-16.2
-21.2
-26.2
-31.2
Efib
Tx-fib
Ty-fib
Pfib
29000
92800 2.26E+06
-74.88
29000
92800 1.81E+06
-62.88
29000
92800 1.41E+06
-50.88
29000
92800 1.06E+06
-38.88
29000
92800 7.60E+05
-26.88
29000
92800 5.09E+05
-14.88
29000
92800 3.09E+05
-2.88
29000
92800 1.59E+05
9.12
29000
92800 5.85E+04
21.12
29000
92800 8.35E+03
33.12
29000
92800 8.35E+03
33.12
29000
92800 5.85E+04
21.12
29000
92800 1.59E+05
9.12
29000
92800 3.09E+05
-2.88
29000
92800 5.09E+05
-14.88
29000
92800 7.60E+05
-26.88
29000
92800 1.06E+06
-38.88
29000
92800 1.41E+06
-50.88
29000
92800 1.81E+06
-62.88
29000
92800 2.26E+06
-74.88
Tangent Modulus Buckling - Numerical

P


-0.0005
-0.0006
-0.0007
-0.0008
-0.0009
-0.001
-0.0011
-0.0012
-0.0013
-0.0014
-0.0015
-0.0016
-0.0017
-0.0018
-0.0019
-0.002
-0.0021
-0.0022
-0.0023
-0.0024
-0.00249
Tx
-417.6
-556.8
-696
-835.2
-974.4
-1113.6
-1252.8
-1384.8
-1510.08
-1624.32
-1734.72
-1832.16
-1924.8
-2008.32
-2083.2
-2152.8
-2209.92
-2263.2
-2304.96
-2340.48
-2368.32
-2386.08
-2398.608
Ty


1856000
1856000
1856000
1856000
1856000
1670400
1670400
1484800
1299200
1299200
1113600
1113600
928000
928000
742400
556800
556800
371200
371200
185600
185600


16704000
16704000
16704000
16704000
16704000
12177216
12177216
8552448
5729472
5729472
3608064
3608064
2088000
2088000
1069056
451008
451008
133632
133632
16704
16704
KLx-cr
KLy-cr
T/Y
209.4395102
628.3185307
0.174
181.3799364
544.1398093
0.232
162.231147 486.6934411
0.29
148.0960979 444.2882938
0.348
137.1103442 411.3310325
0.406
128.254983
384.764949
0.464
120.9199576 362.7598728
0.522
109.11051 294.5983771
0.577
104.4864889 282.1135199
0.6292
94.98347542
227.960341
0.6768
85.97519823 180.5479163
0.7228
83.65775001
175.681275
0.7634
75.56517263 136.0173107
0.802
73.97722346 133.1590022
0.8368
66.30684706 99.46027059
0.868
65.22619108 97.83928663
0.897
57.58118233
69.0974188
0.9208
49.27629185 44.34866267
0.943
48.8278711 43.94508399
0.9604
39.56410897 23.73846538
0.9752
39.33088015 23.59852809
0.9868
27.70743725 8.312231176
0.9942
27.63498414 8.290495243
0.99942
(KL/r) x
(KL/r) y
181.3799364 181.3799364
157.0796327 157.0796327
140.4962946 140.4962946
128.254983
128.254983
118.7410412 118.7410412
111.0720735 111.0720735
104.7197551 104.7197551
94.49247352 85.04322617
90.48795371 81.43915834
82.25810265 65.80648212
74.45670576 52.11969403
72.44973673 50.71481571
65.44135914 39.26481548
64.06615482 38.43969289
57.423414
28.711707
56.48753847 28.24376924
49.86676668 19.94670667
42.67452055 12.80235616
42.28617679 12.68585304
34.26352344 6.852704688
34.06154136 6.812308273
23.99534453 2.399534453
23.9325983
2.39325983
Tangent Modulus Buckling - Numerical
Inelastic Column Buckling
1
( T/ Y)
Normalized critical stress
1.2
0.8
0.6
0.4
0.2
0
0
20
40
60
80
100
120
KL/r ratio
(KL/r)x
(KL/r)y
140
160
180
200
Normalized column capacity
Column Inelastic Buckling
1.2
1.2
1
1.0
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0.0
0.0
0.5
Num-x
Elastic
1.0
Num-y
Lambda
AISC-Design
1.5
Analytical-x
Analytical-y
2.0