A12 Design for Buckling-columns and Plates

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:

where c is a constant depending upon the end conditions:

clamped-free: c=0.25pinned-pinned: c=1clamped-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:


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

(units of length)

Equation becomes . So finally we write the

Euler critical buckling stress as:

The term 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 W4x0.15 cross-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:

Consequently, the

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


Example. Consider an aluminum column ( ) with the cross-section above that is pinned on each end (c=1) and L=100". The radius of gyration is " and the slenderness ratio is equal to . The buckling stress becomes:

For a typical aluminum, we note that the yield stress is around (or greater). Hence, buckling will occur well

before the yield stress is reached, and buckling for long, slender columns (large ) is thus geometrically dominated, not material yielding dominated.

For very short columns (small ), the column will not buckle but simply compress, and a simple model is sufficient. Failure will then be due to yielding of the material.


There is an intermediate range of 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

Note that Johnson's equation is Euler's solution inverted and offset by a constant (=yield stress). If one graphs equations and [For the case of c=1 (pinned-pinned) and aluminum with

and ], 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 for an slenderness ratio of zero. The tangent point can be found by setting the two solutions equal to each other:

Letting , the above can be written as the quadratic equation:

which has the solution . Now .

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


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

and , we have the following plot with the equal point at

.

Note that this plot, and the resultant slenderness ratio

where the Euler and Johnson models are equal, is a function of column end conditions (c) and the material being used ( and

). 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 , Euler's solution will over estimate the critical

stress. For , 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 ( ) and aluminum material as before ( and ). The slenderness ratio for the column is equal to:

. The transition point on the two curves

is given by . Hence, this

indicates that one should use the Johnson solution since 27.52<71.64. Johnson's solution gives the critical buckling stress

as:


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 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.

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:


The constant is compressive buckling coefficient and is a function of the edge boundary conditions and (a/b).


The value of 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., 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


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.

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").

The cross-sectional area of the plate is . The area of the tubes within the 20" length is (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


And the load carried by the tubes is

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

dashed curve. For , we find that . Gerard's

equation for plates with in-plane compression is

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

hence the allowable stress is .

Substituting into Gerard's equation gives:


Solving for t gives:

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: (using nominal dimensions

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

Radius of gyration:

Slenderness ratio:

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


The transition point between the equations is at

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

The load carried by the column is and the compressive load is

Note: and .


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: Tubes (2 of them at 20" spacing, each 2" square):

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