A12_Design_for_buckling-columns_and_plates

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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

2cr

EI P c

L

π =

where c is a constant depending upon the end conditions:

clamped-free: c=0.25

pinned-pinned: c=1

clamped-pinned: c=2clamped-clamped: c=4

Equation 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:




2 2

2 2 ( / )

cr cr

P EI E c c

A L A L A I

π π σ = = =

The term (A/I) is related to the radius of gyration defined by

/ I A ρ = (units of length)

Equation becomes2

2 2(1/ )cr

E c

L

π σ

ρ = . So finally we write the

Euler critical buckling stress as:

2

2( / ) E

E c L

π σ

ρ =

The term / L ρ is non-dimensional and is known as the slenderness ratio of the column.




When Euler's equation 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

inertia must be used. For example,

suppose we have an aluminum W4x0.15cross-section. This is a cross-section that

is 4" deep and has a web that is 0.15"

thick. The top and bottom caps are 0.23"

thick and the shear web is 3.54" long. We

have the following section properties:

2 4 41.965 , 5.62 , 1.04 xx yy A in I in I in= = = Consequently, the

column will buckle so that bending occurs about the y-axis (4

min 1.04 I in= ).

x 4”

0.23”

0.23”

3.54”

0.15”




Example. Consider an aluminum column ( 610.4 10 E x psi= ) with

the cross-section above that is pinned on each end (c=1) and

L=100". The radius of gyration is min / 0.727 I A ρ = = " and theslenderness ratio is equal to / 100"/ 0.727" 137.6 L ρ = = . The

buckling stress becomes:2 2 6

2 2

(10.4 10 )1 5, 425

( / ) (100"/ 0.727")

E

E x psic psi

L

π π σ

ρ

= = =

For a typical aluminum, we note that the yield stress is around40,000 y psiσ = (or greater). Hence, buckling will occur well

before the yield stress is reached, and buckling for long, slender

columns (large / L ρ ) is thus geometrically dominated, not materialyielding dominated.

For very short columns (small / L ρ ), the column will not buckle

but simply compress, and a simple / P Aσ = model is sufficient.

Failure will then be due to yielding of the material.




There is an intermediate range of / L ρ 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 by2

2

( / )1

4

y J y

L

c E

σ ρ σ σ

π

= −

Note that Johnson's equationis Euler's solution inverted

and offset by a constant ( yσ

=yield stress). If one graphs

equations and [For the case

of c=1 (pinned-pinned) andaluminum with

10.4 E Mpsi= and40 y ksiσ = ], we find that the

equations are equal and

tangent to each other at a

Euler

Johnson

Slenderness ratio




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 yσ

for an slenderness ratio of zero. Thetangent point can be found by setting the two solutions equal to

each other:22

2 2

( / )1

( / ) 4

y y

L E c

L c E

σ ρ π σ

ρ π

= −

Letting 2( / )a L ρ = , the above can be written as the quadratic

equation:2 2 2 2 4 2( ) (4 ) 4 0 y ya c E a c E σ π σ π − + =

which has the solution 22 ( / ) ya c E π σ = . Now ( )/ L a ρ = .

Hence, the slenderness ratio at the equal point is given by

2 / y

equal

LcE π σ

ρ

=




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 thisvalue. For the case of c=1 (pinned-pinned) and aluminum with10.4 E Mpsi= and 40 y ksiσ = , we have the following plot with

the equal point at( / ) 71.64equal L ρ = .

Note that this plot, and the

resultant slenderness ratio/ L ρ where the Euler and

Johnson models are equal, is

a function of column end

conditions (c) and the

material being used ( E and

yσ ). Hence, the

determination of which model to use (Euler or Johnson) must be

determined for each problem. For this material (typical aluminum)

Use Johnson Use Euler

Slenderness ratio




and end condition (c=1), we see the following: for values of ( / ) 71.64 L ρ < , Euler's solution will over estimate the critical

stress. For ( / ) 71.64 L ρ >

, Johnson's solution will under estimatethe critical stress.

Example: Consider the case of a column 20" long (L=20") with

the same W4x0.15 cross-section ( 0.727" ρ = ) and aluminum

material as before (10.4 E Mpsi

= and40

yksiσ =

). Theslenderness ratio for the column is equal to:/ 20"/ 0.727" 27.52 L ρ = = . The transition point on the two curves

is given by 2 / 71.64 y

transition

LcE π σ

ρ

= =

. Hence, this

indicates that one should use the Johnson solution since27.52<71.64. Johnson's solution gives the critical buckling stress

as:

2

2

( / )1 37, 049

4

y J y

L psi

c E

σ ρ σ σ

π

= − =




Buckling of Flat Plates

In the notes by Prof. Pollock (see A11), the buckling of flat plateswas briefly discussed. This included flat plates subjected to in-

plane 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




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).




Flat Plates in Compression

Consider a flat plate of thickness 't", dimensions a and b, and

subjected to in-plane compression as shown below.

a

b

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: 22

212(1 )

ccr

k E t

b

π σ

ν

= −

The constant ck is compressive buckling coefficient and is a

function of the edge boundary conditions and (a/b).




The value of ck can be plotted as follows:




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 aπ where x is the coordinate axis in the direction

of load application (a direction).




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.

The design parameters are

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

90”

60”




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.

simply supported between stiffeners clamped between stiffeners

“small” stiffeners “large” stiffeners

Design the plate for minimum weight.




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").

b=20”

2”

t2”t

The cross-sectional area of the plate is (20") p A t = . The area of

the tubes within the 20" length is 2(1" 2" 1") 8"T A t 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.

Hence the load carried by the plate is




40 / (20 / 28) 28.57 / plate N lbf in lbf in= =And the load carried by the tubes is

40 / (8 / 28) 11.43 /tubes N lbf in 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 thedashed curve. For

90"4.5

20"

a

b= = , we find that 7.2ck ≈ . Gerard's

equation for plates with in-plane compression is22

212(1 )

c

cr

k E t

b

π

σ ν

= − . The plate must carry 28.57 lbf/in of load;

hence the allowable stress is / 28.57( / ) /cr p N t lbf in t σ = = .

Substituting into Gerard's equation gives:




22 6

2

28.57 / (7.2)(10.4 10 )

20"12(1 (0.3) )

lbf in x psi t

t

π = −

Solving for t gives: 0.055"t =

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: 28"(.055") 0.44t A in=; (using nominal dimensions

only)

Moment of inertia (about centroid and axis parallel to plate):

{ }3 3 2 4

2 .055"(2") /12 2"(.055") /12 (2" .055")(1") 0.293 I x in= + + =Radius of gyration: 4 2/ 0.293 / 0.44 0.816" I A in in ρ = = =

Slenderness ratio:90"

1100.816"

L

ρ = =

Now determine which column equation to use: Euler or Johnson.




The transition point between the equations is at

62 / 2(4)(10.4 10 ) /(40 ) 143 y

transition

LcE x psi ksiπ σ π

ρ

= = =

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

2 2

2 2 6

( / ) 40 (110)1 40 1 28, 200

4 4(4) (10.4 10 )

y J y

L ksiksi psi

c E x psi

σ ρ σ σ

π π

= − = − = The load carried by the column is 11.43 / (8") 91.4 P lbf in lbf = =

and the compressive load is 2/ 91.4 / 0.44 208 P A lbf in psiσ = = =

Note: 208 28, 212 J psi psiσ σ = < = and 208 40 y psi ksiσ σ = < = .




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: 3(60")(90")(0.055")(0.098 / ) 29.11lbf in lb=Tubes (2 of them at 20" spacing, each 2" square):

2 3

2[(90")(0.44 )(0.098 / )] 7.76in lbf in lb=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

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A12_Design_for_buckling-columns_and_plates