412 PROFESSOR A. H. GIBSON ON 
(b) Circular-Are Girder built in at the Ends, with Uniform Loading and with 
two Symmetrical Intermediate Supports. 
Let the angle subtended by the girder be (180-2@)° or 2a, and let the supports 
(at C and D, fig. 11) be distant y from each end. Let the upward reaction at each 
support=P. Let M,’, T,”, R,” represent such end conditions at A as would be 
produced by these two reactions alone, and let M,’, T,’, R,’ represent such end 
conditions as would be produced by the load alone, with supports removed. 
Fig. 11. 
The downward deflection at C and D due to the loading would be, by 
equation (21) :— 
sal (Me — wr”)y sin y — (R,'7 — T,’)(sin y — y cos y) — 2wr?(cos y — 1) | 
Yo= (T,’ - Ry'r)(sin y — y cos y) + (M,’ - wr?) {y sin y + 2 cos y - 2} - (26) 
2 
2CJ + 2R,'7(y - sin y) - Zeon? (E + cos y— 1) 
where R,/ = wr ¢ _— 6) , and where M,’ and T,’ for the particular value of ¢ obtaining in 
the girder, are given by the curves of fig. 7. 
The upward deflection at C and, from symmetry, at D, due to the two upward 
forces P is obtained by substituting y for 6, in equation (12), which becomes :— 
—— 
aq Me'y sin y — (R,r - T,,”) (sin y -— y cos 7) | 
Yy= iM (27) 
+ aa —R,’r) (sin y — y cos y) + 2R,"7 (y — sin y) + M,"(y sin y + 2 cos y — 2) | 
The values of M,”, R,", and T,” for use in this expression are the sum of the corre- 
sponding values produced by each of the two forces P acting at points distant y from 
A and from B, and may evidently be obtained by adding the values of M, and M,, R, 
and R,, T, and T,, as obtained from the curves of figs. 5 and 6 for a girder having 
the correct value of ~, and having the force P at y from A. 
