124 APPLIED MECHANICS ~ 
131. Beam of Uniform Section, Fixed at one End, Supported at 
the Other, and Loaded Uniformly.—The treatment of this case is similar 
_to that of the case in the pre- 
ceding Article, and the steps will 
be here stated briefly, and the 
results given. The load is w per 
unit length. Removing the sup- 
port at A (Fig. 164), the down- 
ward deflection at that point due 
wt 
to the uniform load will be SEI 
Wa 
An upward force P at A, the Am 
uniform load being removed, will — A, 
produce an upward deflection at om P 
PL3 
that point equal tos. Hence ‘hoon: mn A 
PL’ wl ; jumy C 
SRI SET’ therefore P = 3L. AB, E, 
The bending moment at a dis- Fia. 164. 
tance # from A is 3wLz— 4we?, 
and when this is zero, 3wLa= wz, and «= $L. This gives the point 
of inflexion E. When e= L, the bending moment is 
BL? — dL? = — dL. 
Between A and E the bending moment is greatest at C, where x= 3L, 
and is then equal to 73,7L?. 
The beam AB may-fow be considered as a cantilever B,E, fixed 
at B,, and a beam E,A, supported at the ends. The load on the 
cantilever B,E, is P= Gul at E,, and a uniform load of w per unit 
length. The load on the beam E,A, is a uniform load of w per unit 
length. 
_ s¥LGL) , ywh(ZL) _ 5eLt 
Deflection at E, = 3ET ~ SEL 2048EI° 
B(3wL)(3L)? 135A 
Deflection of C, below E,A, = 381E.  ~ 39768EL 
Total deflection of C, below A,B, 
5wLs 135wL4 wht 
A ly: 
2x 2048EI * 39768EL i871 
4 
The greatest deflection is ee at a point whose distance from A is 
0°4215L. 
Observe that this beam is not strengthened by fixing it at one end, 
the maximum bending moment being 47L?, the same as when the beam 
is simply supported at the ends. 
