The diagonal struts set between floor-beams, as in Fig. 43, are known as cross-bridging, or herringbone bridging. By connecting the beams thus at intervals, say, of from 5 to 8 feet, the stiffness of the floor is greatly increased. The absolute strength of a tier of beams to resist a weight uniformly distributed over the whole tier is augmented but little by cross-bridging; but the power of any one beam in the tier to resist a concentrated load upon it, as a heavy article of furniture or an iron safe, is greatly increased by the cross-bridging; for this device, by connecting the loaded beam with the adjacent beams on each side, causes these beams to assist in carrying the load. To secure the full benefit of the diagonal struts, it is very important that the beams be well secured from separating laterally, by having strips, such as cross-furring, firmly nailed to the under edges of the beams. The tie thus made, together with that of the floor-plank on the top edges, will prevent the thrust of the struts from separating the beams.

162. - Bridging: Value to Resist Concentrated Loads. - A rule for determining the additional load which any one beam connected by bridging will be capable of sustaining, by the assistance derived from the other beams, through the bridging, may be found in Chapter XVIII., Transverse Strains. This rule may be stated thus:

161 Cross Bridging Or Herring Bone Bridging 64

Fig. 43.

R =5 c3fl/4d2 (1+ 22 + 32+ 42+ etc.); (60.)

in which R is the increased resistance, equal to the additional load which may be put upon the loaded beam; c is the distance from centres in feet at which the beams in the tier are placed: f is the load in pounds per superficial foot upon the floor; l is the length of the beams in feet; and d is the depth of the beams in inches. The squares within the bracket are to be extended to as many places as there are beams on each side which contribute assistance through the bridging. The rule given in the work referred to, for ascertaining the number of spaces between the beams, is -

n=d/c2; (61.)

or, the depth of the beam in inches divided by the square of the distance from centres, in feet, at which the beams are placed will give the number of spaces between the beams which contribute on each side in sustaining the concentrated load. The nearest whole number, minus unity, will equal the required number of beams.

The value of c for beams in floors of dwellings is given in equation (46.), and lor those in first-class stores in equation (47.). By a modification of equation (34.), putting c f l for U, we have -

cfl=1.6Fbd3δ/l3

and c=1.6Fbd3δ/fl4, (62.)

or c = 1.6Fbd3r/fl3. (63.)

These equations give general rules for the value of c

Now, the rule, in words at length, for the resistance offered by the adjoining beams to a weight concentrated upon one of the beams sustained by cross-bridging to the others, is -

Rule XLV. - Divide the depth of the beam in inches by the square of the distance apart from centres in feet at which the floor-beams are placed; from the quotient deduct unity, and call the whole number nearest to the remainder the First Result. Take the sum of the squares of the consecutive numbers from unity to as many places as shall equal the above first result; multiply this sum by 5 times the length in feet, by the load per foot superficial upon the floor, and by the fifth power of the distance apart from centres in feet at which the beams are placed; divide the product by 4 times the square of the depth in inches, and the quotient will be the weight in pounds required.

Example. - In a tier of 3 x 12 inch floor-beams 20 feet long, placed in a dwelling 16 inches from centres and well bridged, what load maybe uniformly distributed upon one of the beams, additional to the load which that beam is capable of sustaining safely when unassisted by bridging? Here, according to the rule, 12 divided by (1 1/2+ 1 1/3 = ) 1 7/8 equals 6 3/4; 6 3/4 - 1 - 5 3/4, the nearest whole number to which is 6, the first result. The sum of the square of the first 6 numbers equals (1 + 22 + 32 + 42 + 52 + 62 =) 1 +4 + 9+ 16 + 25 + 36 = 91. Therefore, 91 x 5 x 20 x 90 x (4/3)5 == 3451266.* The square of the depth (12 x 12 = ) 144 x 4 = 576; by this dividing the above 3451266, we have the quotient 5991.78, say 5992 pounds, the required weight. This is the additional load which may be placed upon the beam. At 90 pounds per superficial foot, the common load on each beam, we have (4/3)5 = 45/35.. For the numerator we have 4x4x4x4x4=1024, and for the denominator 3x3x3x3x3 = 243. The former divided by the latter gives as a quotient 4.214, the value of (4/5)5. The process of involving a number to a high power, or the reverse operation of extracting high roots, may be performed by logarithms with great facility. (See Art. 427.)

* The value of c, 16 inches, equals 4/3 feet. The fifth power of this, or (J)', is obtained by involving both numerator and denominator to the fifth power, and dividing the fifth power of the former by the fifth power of the latter; for

90 x 20 x 4/3 = 2400 as the common load. To this add 5992, the load sustained through the bridging by the other beams, and the sum, 8392 pounds, will be the total load which may be safely sustained, uniformly distributed, upon one beam - nearly 3 1/2 times the common load.