For a span of from 15 to 30 feet, we can use the combination shown in
Plate II, Fig. 3. The piece A F must have the same dimensions as a
simple string piece of a length A B--so that it may not yield between
B and either of the points A or D. The two braces D F and E F must be
stiff enough to support the load coming upon them. Suppose the weight
on a pair of drivers of a Locomotive to be 10 tons, then each side
must bear 5 tons, and each brace 2-1/2 tons = 2-1/2 x 2240 = 5600 lbs.
Now, to allow for sudden or extra strains, call 8000 lbs. the strain
to be supported by each brace, and, accordingly, 8 square inches of
sectional area would be sufficient for compression only; but, as the
brace is inclined, the strain is increased. Let the vertical distance
from A to D be 10 ft., and, calling the span 30 ft.--A B will be 15
ft.--from whence D F must be 18 ft., then we shall have the proportion

10 : 18 :: 8000 : 14400 lbs.

which would require an area of about 15 square inches of section to
resist compression, or a piece 3x5 inches. Now, as this stick is more
than 6 or 8 diameters in length, it will yield by bending--and
consequently its area must be increased. The load, which a piece of
wood acting as a post or strut will safely sustain, is found by the
formula already given.

2240 bd³
W = --------
L²

[TeX: $W = \frac{2240 bd^3}{L^2}$]

Now substituting 3 for b, and 5 for d, we have