414 PROFESSOR A. H. GIBSON ON 
E.g., Senicircular Girder with uniform loading and with two piers at 45° from 
ends of span. 
In this special case, the end constants and pier reactions are :— 
P=1'460wr : M,= —:031wr? ; 
RR Selina T, =-010wr?. 
As before, between A and C— 
M,=M., cos 6 — (R,r — T,) sin 0, 
T,=(T, — R,”) cos 0+ R,r — M, sin 6, 
while between C and the centre— 
M,e=M, cos 6 ~(R,r—T,,) sin 6 + Px sin (6 - 45°), 
Ts=(T, — R,v) cos 6 -— R,v — M, sin 6 — Pr{1 — cos (6 - 45°)} 
(c) Semicircular Girder, built in at the Ends, carrying Uniform Load w 
per foot rise, and having Three Intermediate Supports. 
Let the supports be arranged symmetrically, P,; and P, being the reactions at 
the outer supports and Q that at the central support. These reactions may be 
obtained by expressing the facts (1) that the downward deflection at the centre 
due to the loading is equal to the sum of the upward deflections at the centre due 
to the forces P,, P,, and Q, in their respective positions; and (2) that the downward 
deflection at P, due to the loading is equal to the upward deflection at this poimt 
due to forces P,, P,, and $; e.g., if, for example, P, and P, are each at 45° from 
the ends, we have— 
Downward deflection at @ due to loading 
See 053 | 
om * oT i 
Downward deflection at P, or P, due to loading 
3928 0213 
ema ae 
these values being obtained from (21’) by substituting the values of 0, viz., 90° and 
45°, and of M, and T,, from fig. 7. 
Again, the upward deflection at Q due to force Q 
9 af 4674 , 0382 
Fl oat 
and the upward deflection at Q due to the two forces P, and P, = P 
=aprs| 2110 ‘0594 
z SEI | 2CJ 
| from (14) and fig. 5, 
| from (13) and fig. 5. 
Also the upward deflection at P, due to force P, 
= prof 1860 0055 
ORI ar | from (14) and fig. 5, 
