Page 273 - Handbook of Civil Engineering Calculations, Second Edition
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2.58    REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

                            2. Compute the stresses at midspan due to the beam weight
                                               2
                            Thus, M w   ( /8)(83)(20) (12)    49,800 in.·lb (5626.4 N·m);  f bw   49,800/133
                                      1
                             374 lb/sq.in. ( 2578.7 kPa); f tw   374 lb/sq.in. (2578.7 kPa).
                            3. Set the critical stresses equal to their allowable values to
                            secure the allowable unit superimposed load
                            Use Fig. 32 or 33 as a guide. At support: f bi   2400 lb/sq.in. ( 16,548 kPa); f ti   190
                            lb/sq.in. ( 1310.1 kPa); at midspan, f bf   0.85(2400)   374   f bs   425 lb/sq.in.
                            ( 2930.4 kPa); f tf   0.85( 190)   374   f ts   2250 lb/sq.in. ( 15,513.8 kPa). Also,
                            f bs   2091 lb/sq.in. ( 14,417.4 kPa); f ts   2038 lb/sq.in. ( 14,052 kPa).
                              Since the superimposed-load stresses at top and bottom will be numerically equal, the
                            latter value governs the beam capacity. Or w s   w w , f ts /f tw   83(2038/374)   452 lb/lin ft
                            (6596.4 N/m).
                            4. Find F i,max and its eccentricity
                            The value of w s was found by setting the critical value of f ti and of f tf equal to their re-
                            spective allowable values. However, since S b is excessive for the load w s , there is flexibil-
                            ity with respect to the stresses at the bottom. The designer may set the critical value of ei-
                            ther f bi or f bf equal to its allowable value or produce some intermediate condition. As
                            shown by the calculations in step 3, f bf may vary within a range of 2091   2038   53
                            lb/sq.in. (365.4 kPa). Refer to Fig. 34, where the lines represent the stresses indicated.
                              Points B and F are fixed, but points A and E may be placed anywhere within the 53-
                            lb/sq.in. (365.4-kPa) range. To maximize F i , place A at its limiting position to the right;
                            that is, set the critical value of f bi rather than that of f bf equal to the allowable value. Then
                            f cai   F i,max /A   /2(2400   190)    1105 lb/sq.in. ( 7619.0 kPa); F i,max   1105(80)
                                        1
                            88,400 lb (393,203.2 N); f bp   1105   88,400e/133    2400; e   1.95 in. (49.53 mm).
                            5. Find F i,min and its eccentricity
                            For this purpose, place A at its limiting position to the left. Then f bp   2,400   (53/0.85)
                             2338 lb/sq.in. ( 16,120.5 kPa); f cai   1074 lb/sq.in. ( 7405.2 kPa); F i,min   85,920
                            lb (382,172.2 N); e   1.96 in. (49.78 mm).
                            6. Verify the value of F i,max by checking the critical stresses
                            At support: f bi   2400 lb/sq.in. ( 16,548.0 kPa); f ti   190 lb/sq.in. ( 1310.1 kPa).
                            At midspan: f bf   2040   374   2038   372 lb/sq.in. ( 2564.9 kPa); f tf   162
                            374   2038   2250 lb/sq.in. ( 15,513.8 kPa).






















                            FIGURE 34. Stresses at midspan under maximum prestressing force.
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