90 APPLIED MECHANICS 
Between B and C. M=5(0 —2x*), the same as between A and B. 
F=w(l-—2) —wl= — wz, 
which is numerically the 
same as between A and B, 
but of the opposite sign. 
Exampte III.—Beam 
(Fig. 119) supported at 
two points equally dis- 
tant from the ends, and 
carrying a load w per 
unit of length uniformly 
distributed. 
Reactions at sup- 
ports = w(/, +/,). 
Between A and D. 
M= = which is the 
equation to a parabola, 
vertex at A and axis ver- 
tical. M=0 at A where 
2 
Gud Me ae at De: 
F=—wa, which is 
the equation toa straight 
lines F=0at A where -Fia. 119. 
c=0. F,= —wl, at D. 
Between D and B. 
M=w(l, +2,)(lp -2) -S(h +1,-2) = la ~ 1? — 22), 
which is the equation to a ie whose axis is es vertical through B. 
At D where «=/1,, M= — un as before. 
At B where x=0, M=<(ls —1). M=0 where x=b= tJ/T wis 
F=w(/,+1,) —w(l, +1, -2)=wa, which is the equation to a straight 
line. F=0 at B where x=0, F,,=wl, at D where x=/,. 
Between B and C the bending moment diagram is the same as 
between B and A, and the only difference in the shearing force diagrams 
is in sign. 
The maximum bending moment on the beam will be least when the 
bending moment at B is equal to the bending moment at D, that is, when 
5a a) = M1 oF y= J2. If l=1,4+4,, then 1,=1( /2—1)=0-4141. 
This shows be va the supports should be placed when the beam is 
uniformly loaded. An example of this is found in the floats of a paddle- 
wheel. 
ExampiLE IV.—Two equal cantilevers (Fig. 120) carrying a beam 
with a load W at its centre. 
