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

The load sustained by a header is equally distributed, and is equal to the superficial area of the floor supported by the header multiplied by the load on every superficial foot of the floor. This is equal to the length of the header multiplied by half the length of the tail-beams, and by the load per superficial foot. Putting g for the length of the header, n for the length of the tail-beams, and f for the load per superficial foot; U, the uniformly distributed load carried by the header, will equal 1/2 fng. By substituting for U, in equation (35.), this value of it, we obtain -

bd3=1/2 fng13/1.6 Fδ

The symbols g and l here both represent the same thing, the length of the header; combining these, and for δ putting its value gr, we obtain -

bd3=fng3/3.2 Fr

To allow for the weakening of the header by the mortices for the tail-beams (which should be cut as near the middle of the depth of the header as practicable), the depth should be taken at, say, one inch less than the actual depth. With this modification, we obtain -

b = fng3/3.2 Fr(d - 1)3. . (48.)

If f be taken at 90, and r at 0.03, we have, by reducing -

. b=937.5 ng3/F(d-1)3 (49.)

which is a rule for the breadth of headers for dwellings and for ordinary stores. This, in words, is as follows:

Rule XLI - Multiply 937.5 times the length of the tail-beams by the cube of the length of the header, both in feet. The product divided by the cube of one less than the depth multiplied by the value of F, Table III., will equal the breadth of the header in inches for dwellings or ordinary stores.

Example. - K header of white pine, for a dwelling, is 10 feet long, and sustains tail-beams 20 feet long; its depth is 12 inches: what must be its breadth? By the rule, 937.5 x 20 x 103= 18750000. This divided by (12-1)3x 2900=

3859900, equals 4.858, say 5 inches, the required breadth. For first-class stores, f should be taken at 275, and r at 0.04. With these values the constants in equation (48.) reduce to 2148.4375, or, say, 2150. This gives -

b = 2150 ng3/F (d-1)3 (50.)

a rule for the breadth of a header for first-class stores. It is the same as that for dwellings, except that the constant 2150 is to be used in place of 937.5. Taking the same example, and using the constant 2150 instead of 937.5, we obtain 11 . 14 as the required breadth of the header for a first-class store. Modifying the question by using Georgia pine instead of white pine, we obtain 5 .476 as the required thickness, say 5 1/2 inches.

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