| —Dimensions of Strut.— | ||||
| Length Strut. | 3×4-in. | 4×4-in. | 6×6-in. | 8×8-in. |
| Feet. | Lbs. | Lbs. | Lbs. | Lbs. |
| 14 | ..... | 700 | 900 | 1,100 |
| 12 | 600 | 800 | 1,000 | 1,200 |
| 10 | 700 | 900 | 1,100 | 1,200 |
| 8 | 850 | 1,050 | 1,200 | 1,200 |
| 6 | 1,000 | 1,200 | 1,200 | 1,200 |
In using this table it must be borne in mind that bracing both ways reduces the length of a long strut. For example, if a strut 24 ft. long be divided into three panels by bracing the length of strut so far as the table is concerned is 8 ft.
As stated above wall forms are rarely computed. Experience has shown that the maximum spans of various thicknesses of lagging between supports are: 1-in. boards, 24 ins.; 1½-in. plank, 4 ft., and 2-in. plank, 5 ft. Studding will vary in size from 2×4 to 4×6 ins., strutted and braced horizontally to meet conditions. Column forms, like wall forms, are rarely computed, yokes being spaced 2 ft. apart for 1¼-in. lagging up to 3 to 3½ ft. apart for 2-in. lagging.
Floor forms, including girder and slab forms, are computed on the basis of a maximum deflection and not on the basis of strength. Sagging forms are liable to rupture the beam or slab. The amount of deflection considered allowable varies from no deflection up to ⅜ to ½ in. Assuming the deflection, permissible thickness of the timber necessary to carry the load is determined by the formulas:
| d = 5 W l³ ÷ 384 E I | (1) |
and
| bh³ | |||
| I | = | —— | (2) |
| 12 |
Formula (1) is the familiar one for computing deflection for a beam supported (not fixed) at the ends. Mr. Sanford F. Thompson suggests using the constant {3/384}, which is an approximate mean between {1/384} that for beams with fixed ends and {5/384} that for beams with ends supported. Formula (1) then becomes
d = 3 W l³ ÷ 384 E I,
in which as above: