Page 276 - Handbook of Civil Engineering Calculations, Second Edition
P. 276

PRESTRESSED CONCRETE                  2.61


                              stresses at midspan. Thus, f bi   f bp   374   2400; f ti   f tp   374   190; f bp   2774
                              lb/sq.in. ( 19,126.7 kPa); f tp   564 lb/sq.in. ( 3888.8 kPa); f bf   0.85(2774)   374
                              f bs   425; f tf   0.85( 564)   374   f ts    2250; f bs   2409 lb/sq.in. ( 16,610.1
                              kPa); f ts   2356 lb/sq.in.( 16,244.6 kPa). The latter value controls.
                                Also, w s   83(2356/374)   523 lb/lin ft (7632.6 N/m); 523/452   1.16. Thus the ca-
                              pacity is increased 16 percent.
                                When the foregoing calculations are compared with those in the earlier calculation
                              procedure, the effect of using parabolic tendons is to permit an increase of 374 lb/sq.in.
                              (2578.7 kPa) in the absolute value of the prestress at top and bottom. The accompanying
                              increase in f ts is 0.85(374)   318 lb/sq.in. (2192.6 kPa).
                              2. Find the minimum prestressing force and its eccentricity
                              at midspan
                              As before, F i   85,920 lb (382,172.2 N); f tp   1074   85,920e/133   564; e   2.54 in.
                              (64.516 mm).





                              DETERMINATION OF SECTION MODULI

                                                                           2
                              A beam having a cross-sectional area of 500 sq.in. (3226 cm ) sustains a beam-weight
                              moment equal to 3500 in.·kips (395.4 kN·m) at midspan and a superimposed moment that
                              varies parabolically from 9000 in.·kips (1016.8 kN·m) at midspan to 0 at the supports.
                              The allowable stresses are: initial,  2400 and  190 lb/sq.in. ( 16,548 and  1310.1
                              kPa); final,   2250 and  200 lb/sq.in. ( 15,513.8 and  1379 kPa). The member will be
                              prestressed by tendons deflected at the quarter points. Determine the section moduli cor-
                              responding to balanced design, the magnitude of the prestressing force, and its eccentrici-
                              ty in the center interval. Assume that the calculated eccentricity is attainable (i.e., that the
                              centroid of the tendons will fall within the confines of the section while satisfying insula-
                              tion requirements).



                              Calculation Procedure:

                              1. Equate the critical initial stresses, and the critical final stresses,
                              to their allowable values
                              Let M w and M s denote the indicated moments at midspan; the corresponding moments at the
                              quarter point are three-fourths as large. The critical initial stresses occur at the quarter point,
                              while the critical final stresses occur at midspan. After equating the stresses to their allowable
                              values, solve the resulting simultaneous equations to find the section moduli and prestresses.
                              Thus: stresses in bottom fiber, f bi   f bp   0.75M w /S b   2400; f bf   0.85f bp   M w /S b
                                                                                          3
                                                                                3
                              M s /S b   200. Solving gives S b   (M s   0.3625M w )/2240   4584 in (75,131.7 cm ) and
                              f bp   2973 lb/sq.in. ( 20,498.8 kPa); stresses in top fiber, f ti   f tp   0.75(M w /S t )   190;
                              f tf   0.85f tp   M w /S t   M s /S t   2250. Solving yields S t   (M s   0.3625M w )/2412   4257
                                          3
                               3
                              in (69,772.2 cm ) and f tp   807 lb/sq.in. ( 5564.2 kPa).
                              2. Evaluate F i and e
                              In this instance, e denotes the eccentricity in the center interval. Thus f bp   F i /A   F i e/S b
                               2973; f tp   F i /A   F i e/S t   807; F i   (2973S b   807S t )A/(S b   S t )   576,500 lb
                              (2,564,272.0 N); e   2973S b /F i   S b /A   14.47 in. (367.538 mm).
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