The usual formulae for rupture and deflection assume the beam, girder or truss to be supported against possible lateral flexure (bending sideways). Now, if the top chord of a truss or beam is comparatively narrow and not supported sideways, the heavy, compressive strains caused in same may bend it sideways. To calculate this lateral flexure, use the formula given for long columns in compression, but in place of l use only two-thirds of the span of the beam, girder or truss, that is 2/3 l, and for w use one-third of the greatest compressive strain in top chord, which is usually at the centre.

Inserting this in Formula (3) we have: w/3 = a(c/f) transposing, we have, w = 3a (c/f) (5)

1 +(4l2n)/9p2 1+(4l2n)/9p2 where a the area of the cross-section of the top chord in inches,

Q2 is the square of the radius of gyration of the top chord around its vertical axis; we must therefore reverse the usual positions of b and d, that is the breadth of top chord, becomes the depth or d, and the depth of top chord becomes the thickness, or b (both in formulae given in last column of Table I.) - is the greatest allowable compressive strain in pounds at any point to resist lateral flexure safely at that point.

(c/f ) is the safe resistance of the material to compression per square inch in pounds.

l is the total length of span in inches.

n is given in Table II.


A trussed girder is 60' long between bearings, and not supported sideways: the top chord consists of two plates each 22" deep and 1" thick; the plates are 2" apart, as per Figure 5. The greatest compressive strain on top chord has previously been ascertained to be on the central panel, and to be 525000 pounds. Is there danger of the girder bending sideways?

The girder is safe against lateral flexure so long as the strain at centre does not exceed - in Formula (5). Now, the area a= 2.1.22 =44.

Using 48000 pounds per square inch for ultimate resistance to compression of wrought-iron, and a factor-of-safety of 4, we have

(c/f) = 48000/4 =12000

The length is 60', or 720", therefore l2 = 518400. From Table II we have n = 0,000025. And from Table I, section Number 16, we have for the above cross-section, Lateral Flexure In Top Flanges Of Beams Girders Or 10063 =( d3-d13) / 12 (d-d1)

As we are considering the section for bending sideways, we must, of course, take the neutral axis x - y vertically, therefore d becomes 4" and d, becomes 2". This supposes the plates to be stiffly latticed or bolted together, with separators between. We have then

Lateral Flexure In Top Flanges Of Beams Girders Or 10064

Fig. 5.

Lateral Flexure In Top Flanges Of Beams Girders Or 10065

= (43-23)/12(4-2) = 2 1/3

Then for w we have,

3.44.12000 w= 1+4.518400 0.000025

9. 2 1/3 = 1584000 = 1584000 = 456 484 lbs. 1+2,47 3,47

Or, we find that there is danger of the girder bending sideways long before the actual compressive strain of 525000 pounds has been reached. It will, therefore, be necessary to re-design the top chord, so that it will be stiffer sideways. This subject will be more fully treated when considering trusses.