of the stress in the web bars cut
by the section must be equal to S.

[Illustration: FIG. 38.]

21. _Method of Sections. A. Ritter's Method._--In the case of braced
structures the following method is convenient: When a section of a girder
can be taken cutting only three bars, the stresses in the bars can be found
by taking moments. In fig. 38 m n cuts three bars, and the forces in the
three bars cut by the section are C, S and T. There are to the left of the
section the external forces, R, W_1, W_2. Let s be the perpendicular from
O, the join of C and T on the direction of S; t the perpendicular from A,
the join of C and S on the direction of T; and c the perpendicular from B,
the join of S and T on the direction of C. Taking moments about O,

R_x-W_1(x+a)-W_2(x+2a) = Ss;

taking moments about A,

R3a-W_12a-W_2a = Tt;

and taking moments about B,

R2a-W_1a = Cc

Or generally, if M_1 M_2 M_3 are the moments of the external forces to the
left of O, A, and B respectively, and s, t and c the perpendiculars from O,
A and B on the directions of the forces cut by the section, then

Ss = M_1; Tt = M_2 and Cc = M_3.

Still more generally if H is the stress on any bar, h the perpendicular
distance from the join of the other two bars cut by the section, and M is
the moment of the forces on one side of that join,

Hh = M.

[Illustration: Fig. 39.]

[Illustration: Fig. 40.]

22. _Distribution of Bending Moment and Shearing Force._--Let a girder of
span l, fig. 39, supported at the ends, carry a fixed load W at m from the
right abutment. The reactions at the abutments are R_1 = Wm/l and R_2 =
W(l-m)/l. The shears on vertical sections to the left and right of the load
are R_1 and -R_2, and the distribution of shearing force is given by two
rectangles. Bending moment increases uniformly from either abutment to the
load, at which the bending moment is M = R_2m = R_1(l-m). The distribution
of bending moment is given by the ordinates of a triangle. Next let the
girder carry a uniform load w per ft. run (fig. 40). The total load [v.04
p.0551] is wl; the reactions at abutments, R_1 = R_2 = 1/2wl. The
distribution of shear on vertical sections is given by the ordinates of a
sloping line. The greatest bending moment is at the centre and = M_c =
1/8wl^2. At any point x from the abutment, the bending moment is M =
1/2wx(l-x), an equation to a parabola.

[Illustration: Fig. 41.]

[Illustration: Fig. 42.]

23. _S