A12 - Design for Column and Plate Buckling
1
Design for Column and Plate Buckling
The critical buckling load for a long slender column was
previously obtained (see A10 and A11) by solving the governing
differential equation of equilibrium and is given by:
 2 EI
(12.1)
Pcr  c 2
L
where c is a constant depending upon the end conditions:
clamped-free: c=0.25
pinned-pinned: c=1
clamped-pinned: c=2
clamped-clamped: c=4
Equation (12.1) can be written as a critical buckling stress, and can
also be put in terms of a non-dimensional ratio called slenderness
ratio as follows. The critical buckling stress is simply:
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A12 - Design for Column and Plate Buckling
 cr
Pcr
 2 EI
 2E

c 2 c 2
A
L A
L (A/ I)
(12.2)
The term (A/I) is related to the radius of gyration defined by
  I/A
Equation (12.2) becomes  cr
(units of length)
(12.3)
 2E
. So finally we write
c 2
2
L (1/  )
the Euler critical buckling stress as:
 2E
E  c
( L /  )2
(12.4)
The term L /  is non-dimensional and is known as the slenderness
ratio of the column.
A12 - Design for Column and Plate Buckling
3
When Euler's equation (12.4) is compared to experimental results,
it found that the slenderness ratio must be "large" in order to obtain
acceptable correlation. What is large will be considered shortly.
For columns that have a cross-section such that the moments of
inertia are different about the two axes, the minimum moment of
y
inertia must be used. For example,
0.23”
suppose we have an aluminum W4x0.15
cross-section. This is a cross-section that
is 4" deep and has a web that is 0.15" thick. 3.54”
x
The top and bottom caps are 0.23" thick
0.15”
and the shear web is 3.54" long. We have
the following section properties:
0.23”
2
4
4
A  1.965 in , I xx  5.62 in , I yy  1.04 in
Consequently, the column will buckle so that bending occurs about
the y-axis ( I min  1.04 in4 ).
4”
A12 - Design for Column and Plate Buckling
4
Example. Consider an aluminum column ( E  10.4 x106 psi ) with
the cross-section above that is pinned on each end (c=1) and
L=100". The radius of gyration is   I min / A  0.727 " and the
slenderness ratio is equal to L /   100"/ 0.727"  137.6 . The
buckling stress becomes:
 2E
 2 (10.4 x106 psi )
E  c
1
 5, 425 psi
2
2
(L /  )
(100"/ 0.727")
For a typical aluminum, we note that the yield stress is around
 y  40,000 psi (or greater). Hence, buckling will occur well
before the yield stress is reached, and buckling for long, slender
columns (large L /  ) is thus geometrically dominated, not material
yielding dominated.
For very short columns (small L /  ), the column will not buckle
but simply compress, and a simple   P / A model is sufficient.
Failure will then be due to yielding of the material.
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A12 - Design for Column and Plate Buckling
There is an intermediate range of L /  where neither Euler's model
nor a P/A model matches experimental results. Johnson's solution
is often used in the intermediate range and is given by
  y ( L /  )2 

 J   y 1 
2

4c E 

Note that Johnson's equation
is Euler's solution inverted
and offset by a constant
( y =yield stress). If one
graphs equations (12.4) and
(12.5) [For the case of c=1
(pinned-pinned) and
aluminum with E  10.4 Mpsi
and  y  40 ksi ], we find that
the equations are equal and
(12.5)
Euler
Johnson
Slenderness ratio
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A12 - Design for Column and Plate Buckling
tangent to each other at a specific slenderness ratio. Note that
Euler's method goes to infinity when the slenderness ratio goes to
zero, whereas Johnson's solution is equal to  y for an slenderness
ratio of zero. The tangent point can be found by setting the two
solutions equal to each other:
  y ( L /  )2 
 2E

c
  y 1 
2
2


(L /  )
4
c

E


Letting a  ( L /  )2 , the above can be written as the quadratic
equation:
( 2y )a 2  (4c 2 E y )a  4c 2 4 E 2  0
which has the solution a  2c 2 ( E /  y ) . Now  L /    a .
Hence, the slenderness ratio at the equal point is given by
L
  2cE /  y

 equal
(12.6)
A12 - Design for Column and Plate Buckling
7
From experimental observation, one finds that the Euler solution is
good for slenderness ratios greater then this value, while the
Johnson solution is good for slenderness ratios smaller than this
value. For the case of c=1 (pinned-pinned) and aluminum with
E  10.4 Mpsi and  y  40 ksi , we have the following plot with
the equal point at
( L /  )equal  71.64 .
Note that this plot, and the
Use Johnson Use Euler
resultant slenderness ratio
L /  where the Euler and
Johnson models are equal, is
a function of column end
conditions (c) and the
material being used ( E and
Slenderness ratio
 y ). Hence, the
determination of which model to use (Euler or Johnson) must be
determined for each problem. For this material (typical aluminum)
A12 - Design for Column and Plate Buckling
8
and end condition (c=1), we see the following: for values of
( L /  )  71.64, Euler's solution will over estimate the critical
stress. For ( L /  )  71.64 , Johnson's solution will under estimate
the critical stress.
Example: Consider the case of a column 20" long (L=20") with
the same W4x0.15 cross-section (   0.727") and aluminum
material as before ( E  10.4 Mpsi and  y  40 ksi ). The
slenderness ratio for the column is equal to:
L /   20"/ 0.727"  27.52 . The transition point on the two curves
L
is given by 
  2cE /  y  71.64 . Hence, this
 transition
indicates that one should use the Johnson solution since
27.52<71.64. Johnson's solution gives the critical buckling stress
  y ( L /  )2 
  37,049 psi
 J   y 1 
as:
2


4
c

E


A12 - Design for Column and Plate Buckling
9
Buckling of Flat Plates
In the notes by Prof. Pollock (see A11), the buckling of flat plates
was briefly discussed. This included flat plates subjected to inplane compression or shear. Also, due to bending loads, but note
that the bending moment was about an axis perpendicular to the
plate; not the usual plate bending discussed in A05 where the
bending moment is about an x or y axis which lies in the plane of
the plate.
The buckling load for a flat plate is obtained by starting with the
governing differential equation for displacements for a plate as was
derived in A05 but modified so as to include the coupling between
in-plane and out-of-plane displacements (as was done for the
column in A10). For a particular set of edge boundary conditions,
a series solution of sine and cosine functions is assumed that
satisfy the governing differential equation. As for the column, the
result is an eigenvalue problem that must be solved to determine
A12 - Design for Column and Plate Buckling
10
the critical load under consideration (compression, shear or
bending moment). Much of the early work on the subject was
done by Gerard and Becker and is reported in Handbook of
Structural Stability, NACA TN 3781, 1957, and also in
Introduction to Structural Stability Theory, Gerard, McGraw-Hill,
1962.
The result of their work is still utilized today. Although the finite
element method may be used to predict bucking and collapse of
plates and more complex structures, it is quite computer intensive
and not done often in practice (because it requires the solution of
nonlinear equations of equilibrium).
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A12 - Design for Column and Plate Buckling
Flat Plates in Compression
Consider a flat plate of thickness 't", dimensions a and b, and
subjected to in-plane compression as shown below.
b
a
Note that "b" is width of the plate (edge where the load is applied),
and "a" is the length of the plate. Gerard's solution for a flat plate
in compression with various edge boundaries can be summarized
with the following equation:
2
 2 kc E  t 
(12.7)
 cr 
2 b
12(1  )  
The constant kc is compressive buckling coefficient and is a
function of the edge boundary conditions and (a/b).
A12 - Design for Column and Plate Buckling
The value of kc can be plotted as follows:
12
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A12 - Design for Column and Plate Buckling
Note that there are 5 edge condition cases presented for the
unloaded edges (length of "a"); and for each of these cases a curve
for the loaded edges (width of "b") being either clamped or simply
supported. Notation is: c=clamped, ss=simply supported, f=free.
Each one of the "scalloped" portions of a curve in Fig. C5.2 is the
solution for a particular buckling mode: n=1 (half sine wave), n=2
(full sine wave), etc. For clamped, would be cosine waves.
n=1
n=2
n=3
For the top curve (Case A, loaded edges clamped), you can
identify up to n=7. Thus for (a/b)=2, the plate will buckle with
n=3, i.e., sin(3 x / a) where x is the coordinate axis in the direction
of load application (a direction).
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A12 - Design for Column and Plate Buckling
Example: Consider the problem outlined in Pollock's notes (A11).
A 90"x60" flat plate with square tube stiffeners as shown below is
to withstand an in-plane load of 40 lbf/in. All plate edges are
assumed to be fully clamped. The material for both the plate and
tubes is aluminum with a
Young's modulus of 10.4
Mpsi, Poisson's ratio of
0.3, yield stress of 40 ksi
and specific weight of
0.098 lbf/in^3.
90”
The design parameters are
60”
the plate thickness (t), the
number of added stiffeners
running parallel to the 90" edge and size of the stiffeners. The
stiffeners are square tubes that have a wall thickness equal to that
chosen for the plate. As many stiffeners as desired may be used,
so long as the total cross-sectional area of the stiffeners does not
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A12 - Design for Column and Plate Buckling
exceed 30% of the area of plate (area over which the load is
applied - on one end).
The added stiffeners will relieve some of the load from the plate.
The amount of load carried by the square tubes depends on the
cross-sectional area of each tube and that of the plate. You may
reduce the amount of the edge loading on the plate, accordingly.
Similarly, the addition of stiffeners breaks the plate into two or
more smaller plates that are constrained along all four edges.
Small tubes (less than 1.5" x 1.5") can be taken act as simply
supported constraints for the plate (on edges parallel to tubes).
Tubes larger than this size act as clamped constraints for the plate.
“large” stiffeners
“small” stiffeners
simply supported between stiffeners
clamped between stiffeners
Design the plate for minimum weight.
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A12 - Design for Column and Plate Buckling
We could start the design in one of two ways: 1) Assume the
stiffener spacing and solve for plate thickness t, or 2) assume the
plate thickness t and solve for the stiffener spacing.
Suppose we start the design with a 2" x 2" stiffener every 20" (total
of 2 stiffeners). This will mean that the plate size between
stiffeners is b=20" (and the length is a=90").
2”
t
2”
t
b=20”
The cross-sectional area of the plate is Ap  (20")t . The area of
the tubes within the 20" length is AT  2(1" 2" 1")t  8"t (same
as area of one tube). The total area is 28t. We assume that the
load carried of the plate and tubes will be in the ratio of their areas.
A12 - Design for Column and Plate Buckling
17
Hence the load carried by the plate is
N plate  40 lbf / in(20 / 28)  28.57 lbf / in
And the load carried by the tubes is
Ntubes  40 lbf / in(8/ 28)  11.43 lbf / in
The problem stated that the edges are clamped where the loads are
applied. Since we have chosen 2"x2" tubes, we assume these are
large enough so that they provide a clamped edge for the plate
along their length. Hence the 90" x 20" is assumed to be clamped
on all edges. From Gerard's plot, we choose Case A and the
a 90"
 4.5 , we find that kc  7.2 . Gerard's
dashed curve. For 
b 20"
equation for plates with in-plane compression is
2
2
 kc E  t 
. The plate must carry 28.57 lbf/in of load;
 cr 
2 b
12(1  )  
hence the allowable stress is  cr  N p / t  28.57(lbf / in) / t .
Substituting into Gerard's equation gives:
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A12 - Design for Column and Plate Buckling
28.57 lbf / in  2 (7.2)(10.4 x106 psi)  t 



2
t
20"


12(1  (0.3) )
2
Solving for t gives: t  0.055"
Check to see if the stiffener (a column) will buckle under the
compressive load that it must carry (neglecting that it is attached to
the plate).
Area of tube: At 8"(.055")  0.44in2 (using nominal dimensions
only)
Moment of inertia (about centroid and axis parallel to plate):
I  2 .055"(2")3 /12  2"(.055")3 /12  (2" x.055")(1") 2  0.293in 4


Radius of gyration:   I / A  0.293in 4 / 0.44in 2  0.816"
L
90"
Slenderness ratio:

 110
 0.816"
Now determine which column equation to use: Euler or Johnson.
A12 - Design for Column and Plate Buckling
19
The transition point between the equations is at
L
6


2
cE
/



2(4)(10.4
x
10
psi ) /(40ksi )  143
y

 transition
Since the slenderness ratio for the tube is 110, which is less than
143, then the Johnson equation should be used. Johnson's equation
gives the buckling stress as
  y ( L /  )2 


40ksi (110)2
  40ksi 1 
 J   y 1 
  28, 200 psi
2
2
6



4
c

E
 4(4) (10.4 x10 psi) 


The load carried by the column is P  11.43lbf / in(8")  91.4lbf
and the compressive load is   P / A  91.4lbf / 0.44in2  208 psi
Note:   208 psi   J  28, 212 psi and   208 psi   y  40ksi .
A12 - Design for Column and Plate Buckling
20
Hence, the tube stiffener is not even close to buckling or yielding.
With this design, when the plate buckles, the stiffened plate will
still carry significantly more load (via the tube stiffeners).
Weight of the stiffened plate as designed:
Plate only: (60")(90")(0.055")(0.098lbf / in3 )  29.11lb
Tubes (2 of them at 20" spacing, each 2" square):
2[(90")(0.44in2 )(0.098lbf / in3 )]  7.76lb
Hence, stiffened plate weighs 36.87 lb.
Short Design Project:
1. Review my work for accuracy.
2. Determine a better design (less total weight for stiffened plate)
following T. Pollock's requirements on tube size and associated
boundary condition on plate due to tube size (see further A11).
Due: Wednesday, April 18