The typical W Shape steel beam is selected such as to provide the needed strength and stiffness for any particular application and in the safest and most efficient way possible. The best (and safest) way for the beam to provide needed strength is for the beam to develop a plastic hinge at the location of maximum moment if subject to overload. The plastic hinge can be counted on (`developed’) only if the beam is stable. Alternately stated, for the steel in the beam to yield (deform) into a plastic hinge, it must otherwise be kept from `buckling’. The potential buckling that must be considered is `local’ buckling of the relatively thin flanges and web of the beam, as well as the lateral torsional buckling of the *whole* beam. Local buckling is checked using `compactness’ criteria. Lateral torsional buckling of the whole beam is checked by assuring that the distance between brace points does not exceed the critical buckling length. Bracing, or bracing points, are typically provided by deliberate attachment of joists or other framing members to the beam flanges.

To illustrate the stability checks we will examine the W 16 x 31 Shape A992, 50 ksi beam from an earlier article. That example demanded that a plastic hinge be developed in the beam.

The compactness checks for steel sections in general are as follows:

For flange stability the `width to thickness’ ratio, ‘b/t’, for each half of each flange, must not exceed 65 / sqrt Fy. Cast in terms of the beam section geometry,

( bf / 2 ) / tf ≤ 65 sqrt Fy,

where

bf is the width of the flange,

tf is the thickness of the flange,

and

Fy is the Specified yield strength for the beam (50 ksi).

The values of bf and tf for the W 16 x 31 are 5.525 in. and 0.44 in., respectively, thus, the flange compactness check becomes,

… (is) (5.525 in. / 2) / 0.44 in. = 6.3 ≤ 65 sqrt 50 = 9.2 (?) …

The answer is `Yes’.

For web stability (compactness) the `b/t’ ratio is dc / tw and the limit is 640 / sqrt Fy, where dc is the effective depth of the web, and tw the thickness of the web.

For our example dc is estimated conservatively using the whole beam depth, 15.88 in., and the thickness is 0.275 in. So,

… (is) dc / tw = 15.88 in. / 0.275 in. = 58 ≤ 640 sqrt 50 = 91 (?) …

The answer is `Yes’.

Our beam is `compact’!

Now let’s look at lateral torsional buckling and the bracing length.

The critical bracing length, Lp, is given by the equation,

Lp = 1.76 ry sqrt (E / Fy)

where

ry is the radius of gyration of the beam section with respect to the y (weak) axis,

and

E is the Modulus of the Elasticity of the steel.

Since all structural steel used for W shapes has a value of 29,000 ksi for E, the above equation simplifies (a little) to,

Lp = 300 ry / sqrt Fy.

Looking up ry in the *Manual of* *Steel Construction* for the W 16 x 31 we find 1.17 in.

Thus,

Lp = 300 (1.17 in.) / sqrt 50 = 49.6 in. = 4.1 ft.

The floor joists/trusses are known in this example to frame into the beam at 2 ft o.c. intervals, thus the actual bracing length, Lb, is 2 ft.

Since the actual bracing length does not exceed the critical length, the beam is deemed stable with respect to lateral torsional buckling and the plastic hinge may be developed.

Since the beam has been found to be locally stable and stable with respect to the whole thing buckling, the plastic hinge may be developed in the beam.

References

Specification of a Steel W-Shape Beam Using LRFD, Jeff Filler, Yahoo! Contributor Network.

*Manual of Steel Construction – Load and Resistance Factor Design*, American Institute of Steel Construction, Chicago, IL.