This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

A rule for this is the same as that for a carriage-beam carrying one set of tail-beams, if to it there be added the effect of the second set of tail-beams. Equation (51.) with the addition named becomes -

b=f [gn(mn+s2)+5/8 cl3], (54.)

Fd3r

in which n is the length of one set of tail-beams, and s the length of the other set; and m + n = l.

If / be taken at 90, and r at 0.03, these two reduce to 3000, and we have -

b=3000[gn(mn+s2)+5/8 cl3]/Fd3 (55.)

a rule for the breadth of a carriage-beam carrying two sets of headers, for dwellings and for ordinary stores. It may be stated in words as follows:

Rule XLIII. - Multiply the length of the longer set of tail-beams by the difference between this length and the length of the carriage-beam, and to the product add the square of the length of the shorter set of tail-beams; multiply the sum by the length of the longer set of tail-beams, and by the length of the header; to this product add 5/8 of the product of the cube of the length of the carriage-beam into the distance apart from centres of the common beams; multiply this sum by 3000; divide this product by the product of the cube of the depth in inches into the value of F for the material of the carriage-beam, in Table III., and the quotient will be the required breadth.

Example. - In a tier of 3 x 12 inch beams, placed 14 inches from centres, what should be the breadth of a spruce carriage-beam 20 feet long in the clear of the bearings, carrying two sets of tail-beams, one of them 9 feet long, the other 5 feet; the headers being 15 feet long? The difference between the longer set of tail-beams and the carriage-beam is (20 - 9 =) 11 feet. Therefore, by the rule, 9 x 11 + 52 = 124; then (124x9x15=) 16740 +(5/8 203x 14/12 =) 5833 1/8 = 22573 1/3; then 22573 1/3x3000 = 67720000. Now the value of F for spruce, Table III., is 3500; this by 123, the cube of the depth, equals 6048000; by this dividing the aforesaid 67720000, we obtain a quotient of 11.197, the required breadth of the carriage-beam. If, in equation (54.), / be taken at 275, and r at 0.04, these reduce to 6875, and we obtain -

b=6875[gn(mn+s2)+5/8 cl3]/Fd3. (56.)

a rule for the breadth of carriage beams carrying two sets of tail-beams, in the floors of first-class stores. This is like the rule for dwellings, except that the constant is 6875 instead of 3000.

160. - Breadth of. Carriage - Beam with Well-Hole at Middle. - When the framed opening between the two sets of tail-beams occurs at the middle, or when the lengths of the two sets of tail-beams are equal, then equation (54.) reduces to

b=fl(gn2+5/8 cl2)/Fd3r; (57.)

and if f be taken at 90, and r at 0.03, these reduce to 3000, and we have -

b=3000l(gn2+5/8 cl2)/Fd3, (58.)

a rule for the breadth of a carriage-beam carrying two sets of tail-beams of equal length, in the floor of a dwelling or of an ordinary store; and which in words is as follows:

Rule XLIV. - Multiply the length of the header by the square of the length of the tail-beams, and to the product add -| of the product of the square of the length of the carriage-beam by the distance apart from centres of the common beams; multiply the sum by 3000 times the length of the carriage-beam; divide the product by the product of the cube of the depth into the value of F for the material of the carriage-beam, in Table III., and the quotient will be the required breadth.

Example. - In a tier of 3x12 inch beams, placed 12 inches from centres, what must be the thickness of a hemlock carriage-beam 20 feet long, carrying two sets of tail-beams, each 8 feet long, with headers 10 feet long? By the rule, 10x82 + 5/8 1 x202= 890; 890 x 3000 x20= 53400000. Now, the value of F, in Table III., for hemlock is 2800; this by the cube of the depth, 1728, equals 4838400; by this dividing the former product, 53400000, and the quotient, 11.0367, is the required breadth of the carriage-beam.

If, in equation (57.),f be taken at 275, and r at 0.04, these will reduce to 6875, and we shall have -

b = 6875l (gn2 + 5/8 cl2) (59.)

Fd3

a result the same as in equation (58.), except that the constant is 6875 instead of 3000. Equation (59.) is a rule for the breadth of carriage-beams carrying two sets of tail-beams of equal length, in the floor of a first-class store. In words at length, it is the same as Rule XLIV., except that the constant 6875 is to be used in place of 3000.

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