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 bd cubed
W = --------
L squared

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

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

2240 x 3 x 125 840000
W = -------------- = ------ = 2592 lbs.
324 324

[TeX: $W=\frac{2240 \times 3 \times 125}{324}=\frac{840000}{324}=2592$]

which is not enough. Using 6 for b and 8 for d, we have

2240 x 6 x 512
W = -------------- = 21238 lbs.
324

[TeX: $W = \frac{2240 \times 6 \times 512}{324} = 21238$]

which is something larger than is actually required, but it is no
harm to have an excess of strength. Now in many cases this arrangement
would be objectionable, as not affording sufficient head room on
account of the braces--and we can as well use the form of structure
given in Pl. I. Fig. 3, since it is evidently immaterial whether the
point B be supported on F or suspended from it, provided we can
prevent motion in the feet of the braces, which is done by notching
them into the stringer at that point. This of course creates a
tensional strain along the stringer, which is found as
follows:--Representing the applied weight by F B, Pl. II, Fig. 2, draw
B D parallel to F C, also D H parallel to A C--D H is the tension.
This is the graphical construction, and is near enough for practice.
Geometrically we have the two similar triangles A F B and D F H,
whence

A F : D F :: A B : D H

D F x A B
and D H = ---------
A F

[TeX: $DH = \frac{DF \times AB}{AF}$]

This style of structure may be used up to 50 feet, but it is not
employed for spans exceeding 30 feet in length. It is very customary
to make the braces in pairs so as to use small