A carriage-beam or trimmer, in addition to its load as a common beam, carries one half of the load on the header, which, as has been seen in the last article, is equal to one half of the superficial area of the floor supported by the tail-beams multiplied by the weight per superficial foot of the load upon the floor; therefore, when the length of the header in feet is represented by g, and the length of the tail-beams by n, w equals

g/2xn/2xf, equals 1/4 f g n*

For a load not at middle, we have (25.) -

bd2=4 Wamn/Bl.

*The load from the header, instead of being 1/4 f g n, is, more accurately, 1/4 f n(g-c): because the surface of floor carried by the header is only that which occurs between the surfaces carried by the carriage-beams, each of which carries so much of the floor as extends half way to the first tail-beam from it, or the distance -; therefore, the width of the surface carried equals the length of the header less ( 2 x c/c = ) c, or g - c. When, however, it is considered that the carriage-beam is liable to receive some weight from a stairs or other article in the well-hole, the small additional load above referred to is not only not objectionable, but is really quite necessary to be included in the calculation.

This is a rule based upon resistance to rupture. By substituting for a, the factor of safety, Bl/Fdr, its value in terms of resistance to flexure (Transverse Strains, (154.)), we have -

bd2=4 WBlmn/B1Fdr=4 Wmn/Fdr;

or- bd3=4 Wmn/Fr

In this expression, W is a concentrated weight at the distances m and n from the two ends of the beam. Taking the load upon a carriage-beam due to the load from the header, as above found, and substituting it for W, we obtain -

bd3=4x1/4 fgnmn=fgmn2.

Fr Fr

This is the expression required for the concentrated load. To this is to be added the uniformly distributed load upon the carriage-beam; this is given in equation (35.). Substituting for U of this equation its value, fc l, gives -

bd3=fcl4/1.6 Fδ=5/8 fcl3/Fr.

Combining these two equations, we have for the total load -

bd3=f(gmn2+5/8cl3)/Fr (51.)

If, in this equation, fbe taken at 90, and r at 0.03, these reduce to 3000; therefore, with this value of f/r, we have -

b=3000(gmn2+5/8cl3)/Fd3. (52.)

This rule for the breadth of carriage-beams with one header, for dwellings and for ordinary stores, is put in words as follows:

Rule XLII. - Multiply the length of the framed opening by its breadth, and by the square of the length of the tail-beams; to this product add 5/8 of the cube of the length into the distance of the common beams from centres - all in feet; divide 3000 times the sum by the cube of the depth in inches multiplied by the value of F for the material of the beam, in Table III., and the quotient will be the breadth in inches.

Example. - In a tier of 3 x 10 inch beams, placed 14 inches from centres, what should be the breadth of a Georgia-pine carriage-beam 20 feet long, carrying a header 12 feet long, having tail-beams 15 feet long? Here the framed opening is 5 x 12 feet. Therefore, according to the rule, 12 x 5 x 152 = 13500; to which add(5/8x 203 x 14/12 =) 5833 1/3; the sum is 19333 1/3, and this by 3000= 58000000. The value of F for Georgia pine, in Table III., is 5900; the cube of the depth is 1000; the product of these two is 5900000; therefore, dividing the above 58000000 by 5900000 gives a quotient of 9.83, the required breadth in inches. If, in equation (51.), f be taken at 275, and r at 0.04, then - becomes 6875, and the equation becomes -

b=6875(gmn2+5/8 cl3)/Fd3. (53.)

a rule for the breadth of carriage-beams for first-class stores; the same as that for dwellings, except that the constant is 6875 instead of 3000.