Where δ = the greatest deflection at centre, in inches, of a wrought-iron beam or plate girder of uniform cross-section throughout, and carrying its total safe uniform load, calculated for rupture only.

Where L = the length of span, in feet.

Where d = the total depth of beam or girder in inches.

If beam or plate girder is of steel, use 64 1/2 instead of 75.

If the load is not uniform, change the result, as provided in cases (1) to (8), Table VII.

For a centre load we should use 93 3/4 in place of 75 or

δ = L2/93 3/4.d (80)

Where values are the same, as for Formula (79) except that beam or girder carries its total safe centre load, calculated for rupture only.

If beam or girder is of steel use 80§ instead of 93|.

Therefore not to crack plastering and yet to carry their full safe loads, wrought-iron beams or plate girders should never exceed in length (measured in feet) twice and a quarter times the depth (measured in inches), if the load is uniform, or

2 1/4.d = L (81)

Where L = the ultimate length of span (not to crack plastering), in feet, of a wrought-iron beam or plate girder, of uniform cross-section throughout and uniformly loaded with its total safe load.

Where d = the total depth of beam or girder in inches.

If beam or girder is of steel, use 2 instead of 2 1/4.

If the load is central the length in feet should not exceed 2 4/5 times the depth in inches, or

2 4/5.d = L. (82)

Where i = the ultimate length of span in feet (not to crack plastering), of a wrought-iron beam or plate girder, of uniform cross-section throughout, and loaded at its centre with its total safe load.

Where d = the total depth of beam or girder in inches.

If beam or girder is of steel use 2 2/5 instead of 2 4/5.

One thing should always be remembered, when using iron beams, and that is, that the deepest beam is always not only the stiffest, but the most economical. For instance, if we find it necessary to use a l0 l/2" beam - 105 pounds per yard,

Safe length, uniform Cross-section and Load.

Safe length, uniform Cross-section, Centre Load.

Deepest beam most economical.

it will be cheaper to use instead the 12" beam - 96 pounds per yard. The latter beam not only weighs 9 pounds per yard less, but it will carry more, and deflect less, owing to its extra two inches of depth. This same, rule holds good for nearly all sections.

To obtain the deflections of trussed beams or girders by the rules already given would be very complicated. For these cases, however, Box gives an approximate rule, which answers every purpose. He calculates the amount of extension in the tension (usually the lower) chord, and the amount of contraction in the compression (usually the upper) chord, due to the strains in each, and from these, obtains the defections. Of course the average strain in each chord must he taken and not the greatest strain at any one point in either. In a truss, where each part is proportioned in size to resist exactly the compressive or tensional strain on the part, every part will, of course, be strained alike; the strain in the compressive member being= (c/f) per square inch, throughout the whole length, and in the tension member = (t/f) per square inch, throughout the whole length.

The same holds good practically for plate girders, where the top and bottom flanges are diminished towards the ends, in proportion to the bending moment. But where, as in wrought-iron beams (and in many trusses), the flanges are made, for the sake of convenience, of uniform cross-section throughout their entire length, the "average" strain will, of course, be much less, and consequently the beam or girder stiffer.

If we construct the graphical representation of the bending moments at each point of beam (as will be explained in the next Chapter) and divide the area of this figure in inch-pounds by the length of span in inches, we will obtain the average strain in either flange, provided the flange is of uniform cross-section throughout, or v = a/l (83)

Where v = the average strain, in pounds, on top or bottom flange or chord, where beam or girder is of uniform cross-section throughout.

Where l = the length of span, in inches.

Where a = the area in pounds-inch of the graphical figure giving the bending moment at all points of beam.

Deflection of Trusses.

Average Strain in Chords.

Uniform Cross-section.

To obtain the dimensions of this figure measure its base line (or horizontal measurement) in inches, and its height (or vertical meas urement) in pounds, assuming the greatest vertical measurement as

= (c/f) or = ( t/f) in pounds, according to which flange we are examining.

Thus, in the case of a uniform load, this figure would be a parabola, with a base of length equal to the span measured in inches, and a height equal to the greatest fibre strains in pounds; the average strain therefore in the compression member of a beam, girder or truss, of uniform cross-section throughout would be, - (remembering that the area of a parabola is equal to two-thirds of the product of its height into its base), v = 2/3.l. (c/f)/l or v = 2/3. (c/f) (84)

Where v = the average strain, in pounds, in compression flange or chord of a beam, girder or truss of uniform cross-section throughout and carrying its total safe uniform load.

Where (c/f) = the safe resistance to compression per square inch of the material.

It is supposed, of course, that at the point of greatest bending moment - or where the greatest compression strain exists - that the part is designed to resist or exert a stress = (c/f) per square inch. If the greatest compression stress is less, insert its value in place of ( c/f). Of course, it must never be greater than (c/f).

Similarly we should have v= 2/3 (t/f) (85)