DEFLECTION OF BEAMS 119 
» BDC. Let R, and fo Bey the reactions at the supports due 
y load represented by A then 
R= pee e eye, at Re Si 
: tt A,F,B, is the bending moment diagram corresponding to the load 
diagram "ADB, then the ordinate of the curve A,F,B, at any point will 
epresent the deflection at that point, and the maximum deflection will 
=: t the section of the beam where the shearing force due to the load 
A \DE is zero. The complete shearing force diagram ST for the load 
ADB is shown, but it is not necessary to draw this to find the deflection 
of the beam, but it is necessary to find the point F, where the shearing 
ke 2 is zero. Let AF =c, then the resultant P of the load represented 
by the triangle AHF is equal to _ a ~ wee 
a and P acts in a vertical 
' ine through the centre of gravity of the triangle AHF. 
7 _ The shearing force at F=R, —P, and if this is zero R, =P, hence 
” Whe? Wab(a + 2b) { 
) a ies and therefore c = 
a(a + 2b)\* 
—— 
ie * Tho tending momenta F=Ryc- Poe = 2Pe =3, = 3L 
: _ Wd Kot 2M as are 
a the deflection at F = Pisbed 3 . If a=aL, then b=(1 —n)L, 
2 
a deflection at F is given by the expression el ~ “(oe = ). 
aT 
3 7 2 
= The bending moment at C = Rya— PS = te + 2h) — wer b_W ha 
i. 
-—, Wa? 6? WL 2 
- | the deftection at C=— I> SEI n(1 — n). 
(128. Beam of Uniform Section Fixed at the Ends and Loaded at 
_ the Middle.—The beam AB (Fig. 160) is held at the ends in such a 
y that the tangents to the bent beam at A and B are horizontal. The 
load W at the centre of the beam will obviously bend the middle part of 
the beam so that it sags, that is, it becomes concave on the top, and the 
P t to the bent beam at © will be horizontal. Hence the curve 
a B, into which the beam bends must have points of inflexion E, and 
fa bot ween the centre of the beam and its ends, and the positions of 
se points have to be determined. 
. If the beam AB were simply supported at the ends it would be con- 
ave on its upper surface for the whole of its length, and the bending 
oment diagram would be the triangle acl, the altitude of which would 
equal to the bending moment at the centre, namely, }WL.. Also the 
ending moment would be everywhere positive. In order that the beam 
m Ly be concave on its under surface at A and B there must be negative 
7 
