Page 303 - Handbook of Civil Engineering Calculations, Second Edition

2.88 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

363 298 661 lb/sq.in. ( 4558.0 kPa); 363 390 253 1006 lb/sq.in.
( 6936.4 kPa). At section 4: f bi 349 318 667 lb/sq.in. ( 4599.0 kPa); f bf
349 307 270 312 lb/sq.in. ( 2151.2 kPa);or f bf 349 512 270 1131
lb/sq.in. (7798.2 kPa); f ti 161 1031 870 lb/sq.in. ( 5998.7 kPa); f tf 161
237 876 478 lb/sq.in. ( 3295.8 kPa), or f tf 161 142 876 857 lb/sq.in.
( 5909.0 kPa). At section B: f bi 2180 2243 63 lb/sq.in. ( 434.4 kPa); f bf
2180 1280 1906 1554 lb/sq.in. ( 10,714.8 kPa); f ti 1008 2215 1207
lb/sq.in. ( 8322.3 kPa); f tf 1008 592 1883 283 lb/sq.in. ( 1951.3 kPa).
In all instances, the stresses lie within the allowable range.
14. Establish the true trajectory by means of a linear
transformation
The imposed limits are e max y b 9 41.6 in. (1056.6 mm), e min (y t 9)
14.4 in. ( 365.76 mm).
Any trajectory that falls between these limits and that is obtained by linearly trans-
forming the concordant trajectory is satisfactory. Set e b 14 in. ( 355.6 mm), and
compute the eccentricity at midspan and the maximum eccentricity.
Thus, e m 19.53 /2(39.05 14) 32.06 in. (814.324 mm). By Eq. 70b, e s
1
2
( /8)( 4 32.06 14) /( 2 32.06 14) 32.4 in. ( 823.0 mm) < 41.6 in.
1
(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.
15. 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 i is thereby reduced. This revised set of
values will cause an excessive initial tensile stress at B, but the condition can be remedied
by supplying nonprestressed reinforcement over the interior support. Since the excess ten-
sion induced by F i 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. 70b, we get e a 0; e m 41 in. (1041.4 mm); e b
2
1
14 in. ( 355.6 mm); e s ( /8)( 4 41 14) /( 2 41 14) 41.3 in.
(1049.02 mm). This is acceptable.
16. 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 eccentrici-
ties of the hypothetical trajectory. Thus, e m /e b (41 1 /2y)/( 14 y)
0.0625/ 0.1250; y 34 in. (863.6 mm); e a 0; e m 24 in. (609.6 mm); e b 48 in.
( 1219.2 mm); e 1 48 ( 0.0550)/ 0.1250 21.12 in. (536.448 mm); e 2 26.88
in. (682.752 mm); e 3 17.28 in. (438.912 mm); e 4 7.68 in. ( 195.072 mm).
17. Evaluate F i by substituting in relation (b
) of step 7
Thus, F i 2509(14,860)/(10.32 26.88) 1,000,000 lb (4448 kN). Hence, the intro-
duction of nonprestressed reinforcement served to reduce the prestressing force by
14 percent.
18. 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