Where L = the length of clear span, in feet, that beam, etc., is unsupported sideways.

Where b = the least breadth in inches of top flange, or Least thickness of beam, lintel or arch.

Where y =a constant, as found in Table XVI.

(In place of w we can use r = the moment of resistance of beam supported sideways, and in place of w1 we use r1 = the moment of resistance of beam not supported sideways.)

The above practice, however, would seem to diminish the weight unnecessarily, particularly where the beam, girder, etc., is of uniform section throughout; for while the beam in that case, would, be equally strong at all points, it would be strained to the maximum compression only at the point of greatest bending-moment, the strain diminishing towards each support, where the compression would

Table XVI - Value Of Y In Formula (78)

Material of beam, girder, lintel, straight arch, etc.

Value of y for girders, beams, etc., of variable cross-sections.

Value of y for beams, girders, lintels, straight arches, etc., of uniform cross-sec-tion throughout.







Steel .......................












cease entirely. To consider, therefore, the whole as a long column carrying a weight equal to this maximum compressive strain, seems unreasonable. Box has shown, however, that the maximum tendency to deflect laterally is when we consider the top flange (or top half in rectangular beams, lintels and straight arches) as a column equal to two thirds of the span (unsupported sideways) loaded with a weight equal to one-third of the greatest compressive strain at any point. This greatest compressive strain is always at the point of greatest bending moment (usually the centre of span), and is equal to the area of top flange, multiplied by (c/f)1 In case of plate girders the angle-irons and part of web between angle-irons should be included in the area. Box's theory is given in Formula (5); if then we take one-third of this "maximum tendency to deflect" as safe, we should have the same Formula as (78) but with a smaller value for y. The writer would recommend using the larger value for y, where, as in plate girders, trusses, etc., the section of top flange or chord is diminished, varying according to the compressive strain at each point; and using the smaller value for y, where the section of beam, girder or top chord is uniform throughout. Thus the 10 1/2"-90 pounds beam at 20 feet span will safely carry (if supported sideways) a uniform load of 5,9 tons or 11800 pounds (see Table XV.) The width of flange being 4 1/2", and this width and its thickness, of course, being uniform throughout the entire length of beam, we use the smaller value for y (second column) and have for the actual safe uniform load, if the beam is not secured against lateral flexure:

1 This is not quite correct. The greatest compressive strain is really a little less, as will be explained in Vol. II.

w1 = 11800/1+0,0192. 202/4 1/2 = 11800/1+0,379

11800/1,379 = 8557 pounds, or 4,28 tons.

Had we used the larger value for y =z 0,0432 we should have had w1= 11800/1+ 0,854 = 11800/1,854 = 6365 pounds, or 3,18 tons, which closely resembles the value (3,29) given in the Iron Companies' hand-books, but is an excessive reduction under the circumstances.

Where two or more beams are used to carry the same load, as girders for instance, or as lintels in a wall, they should be firmly bolted together, with cast-iron separators between. In this case use for b in Formula (78) the total width, from outside to outside of all flanges, and including in b the spaces between. The separators are made to fit exactly between the inner sides of webs and top and bottom flanges. The separator is swelled out for the bolt to pass through. Sometimes there are two bolts to each separator, but it is better (weakening the beam less) to have but one at the centre of web. The size of separators and bolts vary, of course, to suit the different sizes of beams. They should be placed apart about as frequently as twenty times the width of flange of a single beam. Where beams are placed in a floor, the floor arches usually provide the side bracing. But in order to avoid unequal deflections, and possible cracks in the arches, (from unequal or moving loads or from vibrations) and also to take up the thrust on the end beams of each floor, it is necessary to place

lines of tie-rods across the entire line of beams. The size of these rods can be calculated as already explained in the Chapter on Arches (p. 166); they are usually made, however, from 5/8" to 7/8" diameter. Each rod extends from the outside web of one beam to the outside web of the next beam. The next rod is a little to one side of it, so that the rods do not really form one straight line, but every other rod falls in the same line. Care must be taken not to get the rods too long, or there will have to be several washers under the head and nut, making a very unsightly job, to say the least. Contractors will do this, however, for the sake of the convenience of ordering the rods all of one or two lengths. Where, therefore, the beams are not spaced evenly the contractor should be warned against this. One end of the rod has a "head " welded on, the other has a "screw-end," which need not be "up-set;" the nut is screwed along this end, thus forcing both nut and head to bear against the beams solidly. The distance between lines of tie-rods, would depend somewhat on our calculation, if made; the usual practice, however, is to place them apart a distance equal to about twenty times the width of flange of a single beam.

Sometimes where wooden girders have heavier loads to carry than they are capable of doing, and yet iron girders cannot be afforded, a sheet of plate-iron is bolted between two wooden girders. In this case care must be taken to so proportion the iron, that in taking its share of the load, it will deflect equally with the wooden girders, otherwise the bolts would surely shear off, or else crush and tear the wood.

We consider the two wooden girders as one girder and calculate (or read from Table XIII) their safe load, taking care not to exceed 0,03 inches of deflection per foot of span. We then, from Table VII or Formulas (37) to (41) obtain the exact amount of their deflection under this load. We now calculate the iron plate, for deflection only, inserting the above amount of deflection, and for the load the balance to be borne by the iron-work. An example will best illustrate this: