Handbook of Civil Engineering Calculations

FIGURE 48

lively, are ea = 1 in (25.4 mm); em, = 30 in (762.0 mm); eb = -3 in (-76.2 mm). Evaluate
the prestress shear and prestress moment at section C (a) by applying the properties of the
trajectory at C; (b) by considering the prestressing action of the steel on the concrete in
the interval A C.

Calculation Procedure:

1. Compute tfie eccentricity and slope of the trajectory at C
   Use Eqs. 66 and 67. Let m denote the slope of the trajectory. This is positive if the trajec-
   tory slopes downward to the right. Thus ea - 2em + eb = I - 60 - 3 = -62 in (1574.8 mm);
   2>ea - 4em + eb, = 3 - 120 - 3 = -120 in (-3048 mm); em = 2(-62)(20/100)^2 + 120(20/100)

• 1 = 20.04 in (509.016 mm); mc = 4(-62/12)(20/100^2 ) - (-120/12 x 100) = 0.0587.

2. Compute the prestress shear
   and moment at C
   Thus Vp 0 = -mcFt = -0.0587(860,000) =
   -50,480 Ib (-224,535.0 N); Mpc = -Fte
   = -860,000(20.04) = -17,230,000 in-Ib
   (-1,946,645.4 N-m). This concludes the solu-
   tion to part a.


3. Evaluate the vertical
   component w of the radial force
   on the concrete in a unit
   longitudinal distance
   An alternative approach to this problem is to
   analyze the forces that the tendons exert on
   the concrete in the interval AC, namely, the FIGURE 49. Free-body diagram of
   prestressing force transmitted at the end and concrete.
   the radial forces resulting from curvature of
   the tendons.
   Consider the component w to be positive
   if directed downward. In Fig. 49, Vpr-Vpq =
   -Ft(mr - mq); therefore, A^/Ajc = -F,Am/Ax. Apply Eq. 68: dVpldx = -Fjdm/dx =
   -(4FJL^2 )(ea ~ 2em + eb)\ but dVpldx = -w. Therefore,

w = Fi ^L = / ' \ fe _ 2em + Cb) (71)

OX \ L, /

-Trajectory

Trajectory