FIGURE 50This result discloses that when the trajectory is parabolic, w is uniform across the
span. The radial forces are always directed toward the center of curvature, since the ten-
sile forces applied at their ends tend to straighten the tendons. In the present instance, w =
(4F//100^2 )(-62/12) = - 0.002067F 1 Ib/lin ft (-0.00678F 1 N/m).
- Find the prestress shear at C
By Eq. 690, ma = - [-120/(1OO x 12)] = 0.1; Vpa = -0.1F 1 ; Vpc = Vpa - 2Ow = F,(-0.1 + 20
x 0.002067) = -0.0587F 1 = -50,480 Ib (-224,535.0 N).
5. Find the prestress moment at C
Thus, Mpc = Mpa + VpJiTM) - 20w(120) = F 1 - 0.1 x 240 + 2Ox 0.002067 x 120) -
-20.04F 1 - = -17,230,000 in-lb (1,946,645.4 N-m).
PRESTRESS MOMENTS IN A
CONTINUOUS BEAM
The continuous prismatic beam in Fig. 50 has a prestressing force of 96 kips (427.0 kN)
on a parabolic trajectory. The eccentricities are ea = —0.40 in (—10.16 mm); ed — +0.60 in
(15.24 mm); eb, = -1.20 in (-30.48 mm); ee = +0.64 in (16.256 mm); ec = -0.60 in
(-15.24 mm). Construct the prestress-moment diagram for this member, indicating all
significant values.
Calculation Procedure:
- Find the value of wL
2
/4 for each span by applying Eq. 71
Refer to Fig. 51. Since members AB and BC are constrained to undergo an identical rota-
tion at B, there exists at this section a bending mo-
ment Mkb in addition to that resulting from the ec-
centricity of F 1. The moment Mkb induces
reactions at the supports. Thus, at every section of
the beam there is a moment caused by continuity
of the member as well as the moment -Fte. The
moment Mkb is termed the continuity moment; its
numerical value is directly proportional to the dis-
tance from the given section to the end support.
The continuity moment may be evaluated by
FIGURE 51. Free-body diagram adopting the second method of solution in the pre-
of concrete. vious calculation procedure, since this renders the
Front Rear