BEAMS AND BENDING 89 
forces above the base line, and negative bending moments and negative 
shearing forces below the base line. 
102. Examples of Bending Moment and Shearing Force Diagrams. 
—Two examples of bending moment and shearing force diagrams 
have already been given in Art. 75, pp. 59-61, where the graphic 
method of constructing these diagrams was explained. In_ this 
chapter the bending moments and shearing forces will be found by 
calculation. 
In what follows M will denote the bending moment and F the 
shearing force at a section which is at a distance x from some fixed 
point, generally. the free end of a cantilever, and either the centre or 
one end of a beam. M,, will denote the maximum bending moment, 
and F,, the maximum shearing force. XX will denote the base line 
upon which the bending moment diagram or shearing force diagram is 
plotted. 
Exampre I.—Cantilever (Fig. Bed Boh loads W, and W,. 
Between A and B, M= -— Wt which is the equation to a straight 
line. M,,= -—-W,a at B. F=W 
Between Band C. M= - {W, 2+ W,(a- Mh which is the equation to 
_astraight line. M,,= — 'W, (a+b) -+W, b}. F=W,+ Wo. 
Ke OC He OC 
erate 
Fig. 117. Fig. 118. 
Exampte II.—Beam (Fig. 118) supported at’ the ends and carrying 
a load w per unit of length uniformly distributed. 
Reactions at supports = wl. 
Between A and B, M=wl(l—2)-w(l- ai) = (22), which 
is the equation to a parabola whose axis is the vertical through B, 
M=0 at A where x=/. M becomes M,, where x=0. Hence M,=“ 
at B. 
~ F=wl -w(l— x) =wzx, which is the equation to a straight line. F=0 
at B where #=0. F becomes F,, where a=. Hence F,,=wi at A. 
e 
