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