180. - Floor - Arches; Tie-Rods: Dwellings. - From Transverse Strains, Art. 507, we have -

(79).

(79).

which is a rule for the diameter in inches of a tie-rod for an arch in the floor of a bank, office building, or assembly-room; in which d is the diameter in inches of the rod, s is the span of the arch, and c is the distance apart between the rods (s and c both in feet). This rule requires that the arch rise 1 1/2es per foot of the span, and that the brick-work and the superimposed load each weigh 70 pounds, or together 140 pounds. This rule, in words, is as follows:

Rule LVIII. - Multiply the span of the arch by the distance apart at which the rods are placed, and by the decimal 0.198; the square root of the product will be the diameter of the required rod.

Example. - What should be the diameter of the wrought-iron ties of brick arches of 5 feet span, in a bank or hall of assembly, where the ties are 8 feet apart? For this we

Deflection Of Iron Beams 71

or, the diameter of the required rods should be, say, 7/8 an inch.

181. - Floor-Arches; Tie-Rods: First-Class Stores. - From the same source as in last article, we have -

(80.)

(80.)

which is a rule for the size of tie-rods for the brick arches of the floors of first-class stores, where the arches have a rise of 1inches for each foot of the span, and where the weight of the brick arch and concrete is not over 70 pounds per superficial foot of the floor, and the loading does not exceed 250 pounds per superficial foot. As the rule is the same as the one in the preceding article, except the decimal, a recital of the rule, in words, is not here needed. To obtain the required diameter, proceed as directed in Rule LVIII., using the decimal 0.4527 instead of the one there given.

Tubular Iron Girders

182. - Tubular Iron Girders: Description. - The use of wooden beams for floors is limited to spans of about 25 feet. When greater spans than this are to be covered, some expedient must be resorted to by which intermediate bearings for the floor-beams may be provided. Wooden girders may be used, but these need to be supported by posts at intervals of from 10 to 15 feet, unless the girders are trussed, or made up of top and bottom chords, struts, and ties. And even this is objectionable, owing to the height such a piece of framing requires, and which encumbers the otherwise free space of the hall. A substitute for the framed girder has been found in the tubular iron girder, as in Fig, 46, made of rolled plate iron and angle irons, riveted. They require to be stiffened by an occasional upright T iron along eachside, and a cross-head at least at each bearing.

183. - Tubular Iron Girders: Area of Flanges; Load at Mddle. - In wrought-iron tubular girders it is usual to make the top and bottom flanges of equal thickness. From Transverse Strains, Art. 551, we have a rule for the area of the bottom flange; in which a' equals the area of the flange in inches, W the weight in pounds at the middle, l the length and d the depth of the girder, both in feet, and k the safe load in pounds per inch with which the metal may be loaded, and which is usually taken at 9000. The rule may be stated thus:

Tubular Iron Girders 73

Fig. 46.

a' = wl/4dk' (81)

Rule LIX. - Multiply the weight by the length; divide the product by 4 times the depth into the value of k, and the quotient will be the required area of the bottom flange.

Example. - In a girder 40 feet long and 3 feet high, to carry 75,000 pounds at the middle, what area of metal is required in the bottom flange, putting k at 9000? For this we have, by the rule -

, WI 75000 x 40

a = Wl/4 dk=75000x40/4x3x9000=27.77;

or, the area required is 27 7/8 inches. This is the amount of uncut metal. An allowance is required for that which will be cut by rivet-holes. This is usually an addition of one sixth.

184. - Tubular Iron Girders: Area of Flanges; Load at any Point. - The equation suitable for this (Transverse Strains, Art. 553) is -

a = W mn/dkl; (82.)

in which m and n are the distances respectively from the location of the load to the two ends of the girder. The other symbols are the same as in the last article. This rule may be thus stated:

Rule LX. - Multiply the weight by the values of m and of n; divide the product by the product of the depth into the length and into the value of k, and the quotient will be the required area of the bottom flange.

Example. - -In a girder 50 feet long between bearings and

3 1/2 feet high, what area of metal is required in the bottom flange to sustain 50,000 pounds at 20 feet from one end, when k equals 9000? By the rule, we have -

a' = W mnxdkl = 50000 x 20 x 30/3 1/2x9000x50=19.05;

or, each flange requires 19 inches of solid metal uncut for rivets.

185. - Tubular Iron Girders: Area of Flanges; Load Uniformly Distributed. - The equation appropriate here is (Transverse Strains, Art. 555) -

a' = U mn/2dkl (83.)

This is a rule by which to obtain the area of cross-section of the bottom flange at any point in the length of the girder, the load uniformly distributed; m and n being the respective distances from the point measured to the two ends of the girder, and U representing the uniformly distributed load in pounds. This, in words, is described as follows:

Rule LXI. - Divide the weight by the product of twice the depth into the length and into the value of k; then the quotient multiplied by the values of m and of n will be the required area of the bottom flange at the point measured, the distance of which from the ends equals to and n.

Example. - ln a girder 50 feet long and 3 1/2 feet high, to carry a uniformly distributed load of 120,000 pounds, what area of cross-section is required in the bottom flange, at the middle and at intervals of 5 feet thence, to each support; k being taken at 9000? Here we have, first -