Page 277 - Handbook of Civil Engineering Calculations, Second Edition
P. 277

2.62    REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

                            3. Alternatively, evaluate F i by assigning an arbitrary depth to
                            the member
                            Thus, set h   10 in. (254 mm); y b   S t h/(S b   S t )   4.815 in. (122.301 mm); f cai   f bp
                            ( f bp   f tp )y b /h   2973   (2973   807)0.4815   1153 lb/sq.in. ( 7949.9 kPa); F i
                            1153(500)   576,500 lb (2,564,272.0 N).


                            EFFECT OF INCREASE IN BEAM SPAN


                            Consider that the span of the beam in the previous calculation procedure increases by 10
                            percent, thereby causing the midspan moment due to superimposed load to increase by 21
                            percent. Show that the member will be adequate with respect to flexure if all cross-
                            sectional dimensions are increased by 7.2 percent. Compute the new eccentricity in the
                            center interval, and compare this with the original value.


                            Calculation Procedure:
                            1. Calculate the new section properties and bending moments
                                            2
                                                                 2
                                                                                  3
                            Thus  A   500(1.072)   575 sq.in. (3709.9 cm );  S b   4584(1.072)   5647 in 3
                                                               3
                            (92,554.3 cm );  S t   4257(1.072)   5244 in (85,949.2 cm );  M s   9000(1.21)
                                                                          3
                                     3
                                                     3
                                                                2
                            10,890 in.·kips (1230.4 kN·m); M w   3500(1.072) (1.21)   4867 in.·kips (549.9 kN·m).
                            2. Compute the required section moduli, prestresses, prestressing
                            force, and its eccentricity in the central interval, using the same
                            sequence as in the previous calculation procedure
                                                               3
                                                    3
                                         3
                                                                          3
                            Thus S b   5649 in (92,587.1 cm ); S t   5246 in (85,981.9 cm ). Both these values are
                            acceptable. Then f bp   3046 lb/sq.in. ( 21,002.2 kPa); f tp   886 lb/sq.in. ( 6108.9
                            kPa); F i   662,800 lb (2,948,134.4 N); e   16.13 in. (409.7 mm). The eccentricity has
                            increased by 11.5 percent.
                              In practice, it would be more efficient to increase the vertical dimensions more than
                            the horizontal dimensions. Nevertheless, as the span increases, the eccentricity increases
                            more rapidly than the depth.
                            EFFECT OF BEAM OVERLOAD
                            The beam in the second previous calculation procedure is subjected to a 10 percent over-
                            load. How does the final stress in the bottom fiber compare with that corresponding to the
                            design load?
                            Calculation Procedure:
                            1. Compute the value of f bs under design load
                            Thus, f bs   M s /S b   9,000,000/4584   1963 lb/sq.in. ( 13,534.8 kPa).
                            2. Compute the increment or f bs caused by overload and the
                            revised value of f bf
                            Thus,  f bs   0.10( 1963)   196 lb/sq.in. ( 1351.4 kPa); f bf   200   196   396
                            lb/sq.in. ( 2730.4 kPa). Therefore, a 10 percent overload virtually doubles the tensile
                            stress in the member.
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