is in equilibrium, the left portion also induces compressive stresses on the concrete at C,
these stresses having a resultant that is numerically equal to and collinear with F 1.
- Express the prestress shear and moment in terms of F 1 -
Using the sign convention described, express the prestress shear and moment in terms
of Ff and Q. (The latter is positive if the slope of the trajectory is positive.) Thus Vp =
-F 1 sin 8; Mp = -Ffe cos 0. - Compute the prestress shear and moment
Since 9 is minuscule, apply these approximations: sin 9 = tan 0, and cos 0=1. Then
K, =-F 1 -tan 0 (53)
Or, Vp = -300,000(0.014) = -4200 Ib (-18,681.6 N).
Also,
Mp = -Fte (54)
Or, Mp = -300,000(8) = -2,400,000 in-lb (-271,152 Nm).
STRESSES IN A BEAM WITH
STRAIGHT TENDONS
A 12 x 18 in (304.8 x 457.2 mm) rectangular beam is subjected to an initial prestressing
force of 230 kips (1023.0 kN) applied 3.3 in (83.82 mm) below the center. The beam is on
a simple span of 30 ft (9.1 m) and carries a superimposed load of 840 Ib/lin ft (12,258.9
N/m). Determine the initial and final stresses at the supports and at midspan. Construct di-
agrams to represent the initial and final stresses along the span.
FIGURE 31
(b) Free-body diagram of CB
Force on -
steel at C
Force on
concrete at C-
(a) Beam
Trajectory of steel