frame supported at the two end joints, and loaded at each
top joint. The loads and the supporting forces are indicated by arrows.
Fig. 67a shows the reciprocal figure or polygon for the external forces on
the assumption that the reactions are slightly inclined. The lines in fig.
67 a, lettered in the usual manner, correspond to the forces indicated by
arrows in fig. 66, and lettered according to Bow's method. When all the
forces are vertical, as will be the case in girders, the polygon of
external forces will be reduced to two straight lines, fig. 67 b,
superimposed and divided so that the length AX represents the load AX, the
length AB the load AB, the length YX the reaction YX, and so forth. The
line XZ consists of a series of lengths, as XA, AB ... DZ, representing the
loads taken in their order. In subsequent diagrams the two reaction lines
will, for the sake of clearness, be drawn as if slightly inclined to the
vertical.

[Illustration: FIG. 67.]

If there are no redundant members in the frame there will be only two
members abutting at the point of support, for these two members will be
sufficient to balance the reaction, whatever its direction may be; we can
therefore draw two triangles, each having as one side the reaction YX, and
having the two other sides parallel to these two members; each of these
triangles will represent a polygon of forces in equilibrium at the point of
support. Of these two triangles, shown in fig. 67 c, select that in which
the letters X and Y are so placed that (naming the apex of the triangle E)
the lines XE and YE are the lines parallel to the two members of the same
name in the frame (fig. 66). Then the triangle YXE is the reciprocal figure
of the three lines YX, XE, EY in the frame, and represents the three forces
in equilibrium at the point YXE of the frame. The direction of YX, being a
thrust upwards, shows the direction in which we must go round the triangle
YXE to find the direction of the two other forces; doing this we find that
the force XE must act down towards the point YXE, and the force EY away
from the same point. Putting arrows on the frame diagram to indicate the
direction of the forces, we see that the member EY must pull and therefore
act as a tie, and that the member XE must push and act as a strut. Passing
to the point XEFA we find two known forces, the load XA acting downwards,
and a push from the strut XE, which, being in compression, must push at
both ends, as in