Handbook of Civil Engineering Calculations

Thus, using the ACI Code, E 0 = (145)

15

(33)(4000)°-

5

= 3,644,000 IMn

2

(25,125.4

MPa).

2. Construct the prestress-moment diagrams associated
   with the three cases described
   See .Fig. 38. By symmetry, the elastic curve corresponding to F 1 is horizontal at midspan.
   Consequently, A^ equals the deviation of the elastic curve at the support from the tangent
   to this curve at midspan.


3. Using the literal values shown in Fig. 38, develop an equation
   for Ap by evaluating the tangential deviation; substitute
   numerical values
   Thus, case a:

V^

or A^ = 430,000(8.8)(36)^2 (144)/[8(3,644,000)(40,000)] = 0.61 in (15.494 mm). For case
b:

= M(2L^2La-a^

p 24EJ ^ }

or A^ = 0.52 in (13.208 mm). For case c:

A

5ML

A * ^m

'=^7

(60)

or Ap = 0.51 in (12.954 mm).

4. Compute Aw
   Thus, Aw - -5ww/,^4 /(384£c/) = -0.09 in (-2.286 mm).


5. Combine the foregoing results to obtain A 7 -
   Thus: case a, A, = 0.61 - 0.09 = 0.52 in (13.208 mm); case b, A, = 0.52 - 0.09 = 0.43 in
   (10.922 mm); case c, A,- = 0.51 - 0.09 - 0.42 in (10.688 mm).

(a) StraigM tendons (b) Deflected tendons (c) Parabolic tendons

FIGURE 38. Prestress-moment diagrams.