Compression BADI Y1

Structures in compression In many structures, such as houses, the materials used to build them are in compression rather than tension. The ultimate compressive strength of the structural materials used to build them (brick, steel, concrete) is enormous. So why do buildings fail?

Types of compression failure There are three modes of failure. [a] Under axial compression concrete fails in shear. [b] the separation of the specimen into columnar pieces by what is known as splitting or columnar fracture. [c] Combination of shear and splitting failure.

Buckling When steady compressive forces are applied to short “fat” members, they tend to fail by compressive yielding or fracture; s > sf When steady compressive forces are applied to long “thin” members, they sometimes induce sudden major change in geometry such as bowing, wrinkling, twisting, bending or buckling. Even under s < sf, failure can occur. Onset of buckling failure does not depend at all upon the strength of the material, but only on the dimension of the structure and the modulus of elasticity of the material.

A structure may fail to support its load when a connection snaps, or it bends until it is useless, or a member in tension either pulls apart or a crack forms that divides it, or a member in compression crushes and crumbles, or, finally, if a member in compression buckles, that is, moves laterally and shortens under a load it can no longer support. Of all of these modes of failure, buckling is probably the most common and most catastrophic.

Leonhard Euler long ago showed that there was a critical load for buckling of a slender column. A column, of course, is simply a common case of a compression member. With any smaller load, the column would remain straight and support it. With any larger load, the least disturbance would cause the column to bend sideways with an indefinitely large displacement; that is, it would buckle.

Bending moment for a column The simplest case is that of a column pinned at both ends, that is, free to rotate, under a load P. Let x be the distance along the column, and y its displacement to the side. The bending moment at any point x is Py, which increases with the displacement, it should be noticed. This means that any buckling merely encourages further buckling, explaining why such failure is catastrophic.

Solution for critical value of P The bending moment is related to the curvature by Py = M = -YI d2y/dx2, since for small y the second derivative is the reciprocal of the radius of curvature. Y is Young's modulus, and I is the moment of inertia of area of the cross-section of the column.

Solution The differential equation is easy to solve, with the result that y = A sin ax + B cos ax, where a2 = P/YI. B = 0, since y(0) = 0, and the condition A sin aL = 0 must also be satisfied, where L is the length of the column. This means that aL = π, 2π, ... , or the smallest value of P is given by P' = π2YI/L2. The constant A itself remains indeterminate.

Critical load for buckling The interpretation of this result is that for P < P' the column remains straight and A = 0. For P > P', the column is unstable and buckles. P' is the critical load for buckling. It is found from practice that this theory gives good results for columns that are more than 30 times longer than wide. P' is proportional to 1/L2, so the supporting capacity of a column decreases rapidly with an increase in length.

Unbalanced loads We also notice that an eccentric load will cause a decrease in strength, since there will be a bending moment and a curvature from the start. The column will be stable under sideways impacts and will return to straightness when such load is removed. The moment of inertia I will be different for different directions of buckling if the cross-section is not axially symmetrical. The direction with the minimum I will be the critical one.

Ways to prevent buckling There is a good incentive to look for ways to increase the critical buckling load. If the ends of the column are clamped, that is, prevented from rotating, the buckling curve changes from a half-wavelength sine to a full wavelength cosine, effectively halving the length of the column, and increasing the critical load by a factor of 4. A brace in the middle of the column that prevents it from moving to one side also halves the buckling length, and increases the critical load by a factor of 4.

Maximum safe load for a column For a short column, P' will become large, and at some point the compressive strength of the material will be exceeded. Let P" = Aσ be the maximum load for failure by compression. Rankine combined these results empirically to give a maximum load P by the formula 1/P = 1/P' + 1/P", which is quite conservative, and seems to give good results for columns between five and forty diameters long. Note that the critical load using this formula will be less than either P' or P".

Rules of thumb If the column is made from a circular pipe, an effective unit stress for buckling can be expressed as Y/n2, where n = 2L / d, where d is the diameter of the pipe. For mild steel, a safe compressive stress is about 17 000 psi, and Y = 30 000 000 psi, so P' and P" would be about equal at n = 42. A 3" diameter pipe with walls 1/4" thick and 10' 6" long could be expected to support about 40 000 lb, using a unit stress of 8 500 psi. This is a quite conservative design, with a factor of safety of at least 2 against buckling.

Local buckling If the walls of the pipe are thin, another type of failure can occur, local buckling, in which the side of the pipe under compression when it deflects slightly buckles over a short distance, which is not surprising in view of the ratio of length to thickness. This was also found to occur in the upper flange of I-beams when there was too little support for a broad flange. Provision was made for this in the standard design procedure. Attempts to save material by making thin members of strong materials risks failure by local buckling.

Moments of Inertia The moment of inertia of a rectangular cross-section is I = bh3/12, and for buckling b is taken as the larger of the two dimensions. For a solid circular cylinder, I = πd4/32. The dimensions and moments of inertia for rolled sections are given in handbooks.