= +363 + 390 + 253 = + 1006 lb/in
2
(+6936.4 kPa). At section 4:fbi = +349 + 318 = +667
lb/in
2
(+4599.OkPa);/^ = +349 -307 + 270 = +312 lb/in
2
(+2151.2kPa);or/ft/ = +349 +
512+270 = +1131 lb/in
2
(7798.2 kPa);/rf = -161 +1031 = + 870 lb/in
2
(+5998.TkPa);/^=
-161 - 237 + 876 = +478 lb/in^2 (+3295.8 kPa), arftf= -161 + 142 + 876 = +857 lb/in^2
(+5909.0 kPa). At section B:fbi = +2180 - 2243 = -63 lb/in^2 (-434.4 kPa);/^ = +2180 +
1280 - 1906 = + 1554 lb/in^2 (+10,714.8 kPa); fti = -1008 + 2215 = +1207 lb/in^2
(+8322.3kPa); ftf= -1008 -592 + 1883 = +283 lb/in^2 (+1951.3 kPa).
In all instances, the stresses lie within the allowable range.
- Establish the true trajectory by means of a linear
transformation
The imposed limits are emax = yb - 9 = 41.6 in (1056.6 mm), emin = -(yt - 9) = -14.4 in
(-365.76mm).
Any trajectory that falls between these limits and that is obtained by linearly trans-
forming the concordant trajectory is satisfactory. Set eb = -14 in (-355.6 mm), and com-
pute the eccentricity at midspan and the maximum eccentricity.
Thus, em = + 19.53 +^1 / 2 (39.05 - 14) = +32.06 in (814.324 mm). By Eq. 706, es =
-(^1 / 8 )(-4 x 32.06 - 14)^2 /(-2 x 32.06 - 14) = +32.4 in (+823.0 mm) < 41.6 in (1056.6
mm). This is acceptable. This constitutes the solution to part a of the procedure. Steps 15
through 20 constitute the solution to part b. - Assign eccentricities to the true trajectory, and check
the maximum eccentricity
The preceding calculation shows that the maximum eccentricity is considerably below the
upper limit set by the beam dimensions. Refer to Fig. 57. If the restrictions imposed by
line c' are removed, e 2 may be increased to the value corresponding to a maximum eccen-
tricity of 41.6 in (1056.6 mm), and the value of F 1 is thereby reduced. This revised set of
values will cause an excessive initial tensile stress at B 9 but the condition can be remedied
by supplying nonprestressed reinforcement over the interior support. Since the excess ten-
sion induced by F 1 extends across a comparatively short distance, the savings accruing
from the reduction in prestressing force will more than offset the cost of the added rein-
forcement.
Assigning the following eccentricities to the true trajectory and checking the maxi-
mum eccentricity by applying Eq. 1Ob, we get ea = O; em = + 41 in (1041.4 mm); eb = -l4
in (-355.6 mm); es = -(
1
X 8 )M x 41 - 14)
2
/(-2 x 41 - 14) = +41.3 in (1049.02 mm). This
is acceptable. - To analyze the stresses, obtain a hypothetical concordant
trajectory by linearly transforming the true trajectory.
Let y denote the upward displacement at B. Apply the coefficients C 3 to find the eccen-
tricities of the hypothetical trajectory. Thus, emleb = (41 -
1
Ay)/(-14 - y) =
+0.0625/-0.125O; y = 34 in (863.6 mm); ea = 0;em = +24 in (609.6 mm); eb = -48 in
(-1219.2 mm); C 1 = - 48 (+0.0550)/-0.1250 = +21.12 in (536.448 mm); e 2 = +26.88 in
(682.752 mm); ez = + 17.28 in (438.912 mm); e 4 = -7.68 in (-195.072 mm). - Evaluate F 1 by substituting in relation (b') of step 7
Thus, Ft = 2509(14,860)/(10.32 + 26.88) = 1,000,000 Ib (4448 kN). Hence, the introduc-
tion of nonprestressed reinforcement served to reduce the prestressing force by 14 per-
cent. - Calculate the prestresses at every boundary section; then find
the stresses at transfer and under design load
Record the results in Table 5. (At sections 1 through 4, the final stresses were determined
by applying the values on lines 5 and 8 in Table 4. The slight discrepancy between the fi-