DEFLECTION OF BEAMS 127 
the rectangle and G, is the centre of gravity of the triangle, it follows 
(Sots = deta + LgMy - $y + $144(Me — My)» $14 
= dyty + 4L3M, + 4L3M,. 
: [Note that in the foregoing L rile expression, as applied to Fig. 167, 
if a, is positive, M, and M, would be negative. } 
Ba ~ Henee BT = wgylo + SLM, +}15M,).. 
me Ex like manner DT’ = — pyls% + 4M, oh 4LIM.). 
- Bat if the three supports are at the same level as in Fig. 167, 
BT -DT’. 4 
L, 1 Lae 
Therefore — + i. +3L.M, +4(L, + L;)M, + $L5M, =0, ° 
2 
which is the form of the theorem of three moments for the case where the 
supports are at the same level. 
/ 
____ If the intermediate support at C is at a distance 6 below or above the _ 
_ supports at B and D, as shown in Fig. 169, then 
BrFs Dts UE ee #3(2 r) 
itt 
= ends = — 
L, Ls L, I, 
where the stl sign applies to the case where C is below BD, and the 
“minus sign applies to the case where C is above BD. It then follows 
that 2", “9% 4 31M, + 4(Ly+14)Mo+4L,M, = 23(7 + +e | bwin 
pis the most general form of the theorem of three moments. 
at er the common and simple case in which the three supports are 
at the same level, and the load over the span BC is uniformly distributed 
and equal to w, per unit of length, and the load over the span CD is 
also uniformly distributed and equal to w, per unit of length. Here 
ay =3- fe LeLy = yywyLs, %)= $Lg. 
