THE EQUILIBRIUM OF THE CIRCULAR-ARC BOW-GIRDER. All 
or by the curves of fig. 5, for the corresponding value of a or (90°—¢). The 
downward deflection at the centre due to the load is obtained by substituting a for 4, 
and by substituting the corresponding values of M, and of T, as given by the curves of 
fig..7, m (21). 
E.g., «=90°; ¢=0; (semicircular girder). 
; Blin. ed | oye eos ae 1 a (77 
The upward deflection at centre = salt — (500 - 182) | + aa ( 182 — -500) a —I+ ae - 2) | 
UY seid ee 
=o OCs 
owe): 
The downward deflection at the centre, due to the loading = wrt | jae a | 
DEL * 20) 
and on equating these :— 
Bee Ue 
“467403 + -0382EL |" 
The value of this is sensibly independent of the ratio of El to CJ. Taking this 
ratio as 1°24, as in the experimental girder, gives 
( °7928 
= wr 9 Say } = 1:54wr. 
Again, 
R,+R,+ P=zwr 
2 R.=R,=% { r—154 | 
“SOlwr. 
Also 
{| 
M,+M, = 2wr? — Pr 
= 46? 
M,=M, = ‘237. 
The value of TT, is the difference between the values produced by the load and by 
the upward reaction P. The first of these is -298wr* (fig. 7); the second is °182Pr 
(fig. 5). 
aba als 
U) 
1, = {(298 - (182 x 1-54)} 072 
018 wr, 
This value may be obtained alternatively by substituting the foregoing values of 
M, and of R, in equation (20) with 6, = a and by equating to zero. 
The values of M, and of T, at any point distant @ from the end then become, on 
substituting in equations (18) and (19) :— 
M,=wr?{1 — ‘77 cos 6 — 783 sin 6}, 
T,=wr?{‘801 — -783 cos 6 + “76 sin 6 - 6} . 
Where the girder subtends an angle less than 180°, the problem may be solved in an 
exactly similar manner by making use of the requisite relationships from curves 
figs. 5 and 7. 
