used, and is near enough. Beams of this form are much used. to sustain brick walls of buildings; the brickwork resting upon the bottom flange, and laid, not arching, but horizontal. In the cross section, the bottom flange is made to contain in area four times as much as the top flange. The strength will be in proportion to the area of the bottom flange, and to the height or depth. Hence, to obtain the greatest strength from a given amount of material, it is requisite to make the upright part, or the blade, rather thin; yet, in order to prevent injurious strains in the casting while it is cooling, the parts should be nearly equal in thickness. The thickness of the three parts—blade, top flange and bottom flange, may be made in proportion as 5, 6 and 8. For a beam of this form, the weight equally dif fused over it equals

where w equals the weight in pounds equally diffused over the length; d, the depth, or height in inches of the cross section at middle; a, the area of the bottom flange in inches; 7, the length of the beam in feet, in the clear between the bearings; and t, a decimal in proportion to unity as the safe weight is to the breaking weight. This is usually from 0.2 to 0.3, or from one-fifth to one-third, at discretion.

430.-Beams of this form, laid in series, are much used in sustaining brick arches turned over vaults and other fire-proof rooms, forming a roof to the vault or room, and a floor above; the arches springing from the flanges, one on either side of the beam, as in Fig. 278.

Fig. 2.8.

For this use the depth of cross section at middle equals

where the symbols signify as before, and c equals the distance apart from centres in feet at which the beams are placed, and f the weight per superficial foot, in pounds, including the weight of the material of which the floor is constructed.

Practical Rules and Examples.

431.-For a single girder the dimensions may be found by the following rule, ((200) and (201):)

Rule LXVIII.-Divide the weight in pounds equally dif fused over the length of the girder by a decimal in proportion to unity as the safe weight is to the breaking weight, multiply the quotient by the length in feet, and divide the product by 9000. Then this quotient, divided by the depth of the beam at middle, will give the area of the bottom flange; or, if divided by the area of the bottom flange, will give the depththe area and depth both in inches.

Example.-Let the weight equally diffused over a girder equal 60000 pounds; the decimal that is in proportion to unity as the safe weight is to be to the breaking weight, equal 0-3

the length in the clear of the bearings equal 20 feet. Then 60000 divided by 0-3 equals 200000, and this by 20 equals 4000000; this divided by 9000 equals 4444. Now if the depth is fixed, say at 20 inches, then 444, divided by 20, equals 22%, equals the area of the bottom flange in inches. But if the area is given, say 24 inches, then to find the depth, divide 4444 by 24, and the quotient, 18.5, equals the depth in inches; and such a girder may be made with a bottom flange of 2 by 12 inches, top flange, (equal to of bottom flange,) 11⁄2 by 4 inches, and the blade 14 inches thick.

432.-For a series of girders or iron beams, the dimensions may be found by the following rule: (202) and (203).

Rule LXIX.-Divide the weight per superficial foot, in pounds, by a decimal in proportion to unity as the safe weight is to the breaking weight, and inultiply the quotient by the square of the length of the beams and by the distance apart at which the beams are placed from centres, both in feet, and divide the product by 9000. Then this quotient, divided by the depth of the beams at middle, will give the area of the bottom flange; or, if divided by the area of the bottom flange, will give the depth of the beam-the depth and area both in inches. Example.-Let the weight per superficial foot resting upon an arched floor be 200 pounds, and the weight of the arches, concrete, &c., equal 100 pounds, total 300 pounds per superficial foot. Let the proportion of the breaking weight to be trusted on the beams equal 0:3, the length of the beams in the clear of the bearings equal 12 feet, and the distance apart from centres at which they are placed equal 4 feet. Then 300 divided by 0-3 equals 1000; this multiplied by 144 (the square of 12), equals 144000, and this by 4, equals 576000; this divided by 9000, equals 64. Now if the depth is fixed, and at 8 inches, then 64 divided by 8 equals 8, equals the area of the bottom flange. But if the area of the bottom flange is fixed, and at 6 inches, then 64, divided by 6, equals 10%, the depth

required. Such a beam may be made with the bottom flange 1 by 6 inches, the top flange, (equal to one-quarter of the bottom flange,) by 2 inches, and the blade & inch thick.

433. The kind of girder shown at Fig. 280, (a cast iron arch with a wrought iron tie rod,) is extensively used as a support upon which to build brick walls where the space below is required to be free. The objections to its use are, the disproportion between the material and the strains, and the enhanced cost over the cast iron girder formed as in Figs. 276 and 277. The material in the cast arch, (Fig. 280,) is greatly in excess over the amount needed to resist effectually the compressive strains induced by the load through the axis of the arch, while the wrought metal in the tie is usually much less than is required to resist the horizontal thrust of the arch; absolute failure being prevented, partly by the weight of the walls resting on the haunches, and partly by the presence of adjoining buildings, their walls acting as buttresses to the arch. Some instances have occurred where the tie has parted.

Where this arched girder is used it is customary to lay the first courses of brick in the form of an arch. This brick arch of itself is quite sufficient to sustain the compressive strain, and, were there proper resistance to the horizontal thrust provided, the brick arch would entirely supersede the necessity for the girder. Indeed, the instances are not rare where constructions of this nature have proved quite satisfactory, the horizontal thrust of the arch being sustained by a tie rod secured to a pair of cast iron heel plates, as in Fig. 279. The

Fig. 279.

brick arch, in this case, being built upon a wooden centre, which was afterwards removed.

The diameter of the rod required for an arch of this kind is equal to

where w equals the weight in pounds equally diffused over the arch; s, the length of the rod, clear of the heel plates, in feet; and h, the height at the middle, or rise, of the arc, in inches; D, the diameter, being also in inches.

When the diameter found by formula (204) is impracticably large, this difficulty may be overcome by dividing the metal into two rods. In the bow-string girder, (Fig. 280,) two rods cannot be used with advantage, because of the difficulty in adjusting their lengths so as to ensure to each an equal amount of the strain. But in the case of the brick arch, the two heel plates being disconnected, any discrepancy of length in the rods is adjusted simply by the pressure of the arch acting on the plates. When there are to be two rods, the diameter of each rod equals

434. To obtain the diameter of wrought iron tie-rods for heel plates, as in Fig. 279, proceed by this rule.

Rule LXX.-Multiply the weight in pounds equally distri buted over the arch by the length of the tie-rod in feet, clear of the heel plates, and divide the product by the height of the arc in inches, (that is, the height at the middle, from the axis of the tie-rod to the centre of the depth of the brick arch,) then, if there is to be but one tie-rod, divide the quotient by 3000;