- Determine whether the section moduli are excessive
Do this by setting fbf and/, equal to their allowable values and computing the correspon-
ding values of/w and/^. Thus,^ 7 = 0.85/^ - 2334 - -425; therefore,/^ - +2246 lb/in
2
(+15,486.2 kPa);/ri = ftp = -190 lb/in
2
(-1310.1 kPa);/w =4, ='+2246 < 2400 lb/in
2
(+16,548.0 kPa). This is acceptable. Also, ^= 0.85(-19O) + 1067 = +905 < 2250 lb/in
2
(+15,513.8 kPa); this is acceptable. The section moduli are therefore excessive.
- Find the minimum
prestressing force and its
eccentricity
Refer to Fig. 40. Thus,/^ = +2246 lb/in^2
(+15,486.2 kPa); ftp = -190 lb/in^2
(-1310.1 kPa); slope of AB = 2246 -
(-190)716 = 152.3 Ib/(in^2 -in) (41.33
MPa/m); F 1 IA = CD = 2246 - 10.98
(152.3) = 574 lb/in
2
(3957.7 kPa); F 1 =
574(316) = 181,400 Ib (806,867.2 N);
slope of AB = Fie/I = 152.3; e =
152.3(7240)/!81,400 = 6.07 in (154.178
mm^ FIGURE 40. Prestress diagram. - Determine the number of
strands required, and establish
their disposition
In accordance with the ACI Code, allowable initial force per strand = 0.1089
(0.70)(248,000) = 18,900 Ib (84,067.2 N); number required - 181,400/18,900 = 9.6.
Therefore, use 10 strands (5 in each web) stressed to 18,140 Ib (80,686.7 N) each.
Referring to the ACI Code for the minimum clear distance between the strands, we
find the allowable center-to-center spacing = 4(Yi 6 ) = I^3 A in (44.45 mm). Use a 2-in (50.8-
mm) spacing. In Fig. 41, locate the centroid of the steel, or y = (2^2+1 x 4)/5 = 1.60 in
(40.64 mm); v = 10.98 - 6.07 - 1.60 = 3.31 in (84.074 mm); set v = 3Vi 6 in (84.138 mm). - Calculate the allowable ultimate moment of the member
in accordance with the ACI Code
Thus, As = 10(0.1089) = 1.089 in^2 (7.0262 cm^2 ); d = yt + e = 5.02 + 6.07 - 11.09 in
(281.686 mm);;? = AJ(bd) = 1.089/[72(11.09)] = 0.00137.
FIGURE 41. Location of tendons.