FIGURE 47. Parabolic arc.- Present the results
The equations are
y = 2(yl-2y 2 +y 3 )(~-}^2 -(3yl-4y 2 +y 3 )^+yl (66)
\L I Ldy ( x \ 3V 1 -4y 2 +Vs
»-;£-4(y.-2fc+*)(jr)- * (67)^
=
S
=
^^-^
+
^
(68)-(3yi-4y 2 +y 3 )
W 1 = — (69a)yl-4y 2 + 3y 3
m 3 = (696)(LU)(Zy 1 -4y 2 +y 3 )
xs = : (7Oa)
yi-2y 2 +y 3-(1/S)Qy 1 -^ 2 +y 3?
^ y,-2y2+y 3 +* <™ALTERNATIVE METHODS OFANALYZING A
BEAM WITH PARABOLIC TRAJECTORY
The beam in Fig. 48 is subjected to an initial prestressing force of 860 kips (3825.3 kN)
on a parabolic trajectory. The eccentricities at the left end, midspan, and right end, respec-