Page 293 - Handbook of Civil Engineering Calculations, Second Edition
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2.78    REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN














                                FIGURE 50
                              This result discloses that when the trajectory is parabolic,  w is uniform across the
                            span. The radial forces are always directed toward the center of curvature, since the ten-
                            sile forces applied at their ends tend to straighten the tendons. In the present instance, w
                                  2
                            (4F i /100 )( 62/12)    0.002067F i lb/lin ft ( 0.00678F i N/m).
                            4. Find the prestress shear at C
                            By Eq. 69a, m a    [ 120/(100   12)]   0.1; V pa   0.1F i ; V pc   V pa   20w
                            F i ( 0.1   20   0.002067)   0.0587F i   50,480 lb ( 224,535.0 N).
                            5. Find the prestress moment at C
                            Thus, M pc   M pa   V pa (240)   20w(120)   F i   0.1   240   20   0.002067   120)
                             20.04F i   17,230,000 in.·lb (1,946,645.4 N·m).



                            PRESTRESS MOMENTS IN A
                            CONTINUOUS BEAM

                            The continuous prismatic beam in Fig. 50 has a prestressing force of 96 kips (427.0 kN)
                            on a parabolic trajectory. The eccentricities are e a   0.40 in. ( 10.16 mm); e d   0.60
                            in. (15.24 mm); e b   1.20 in. ( 30.48 mm); e e   0.64 in. (16.256 mm); e c   0.60
                            in. ( 15.24 mm). Construct the prestress-moment diagram for this member, indicating all
                            significant values.


                            Calculation Procedure:

                                               2
                            1. Find the value of wL /4 for each span by applying Eq. 71
                            Refer to Fig. 52. Since members AB and BC are constrained to undergo an identical rota-
                                                          tion at B, there exists at this section a bending
                                                          moment M kb in addition to that resulting from
                                                          the eccentricity of F i . The moment M kb induces
                                                          reactions at the supports. Thus, at every section
                                                          of the beam there is a moment caused by conti-
                                                          nuity of the member as well as the moment
                                                           F i e. The moment M kb is termed the continu-
                                                          ity moment; its numerical value is directly pro-
                                                          portional to the distance from the given section
                                                          to the end support. The continuity moment
                            FIGURE 51. Free-body diagram  may be evaluated by adopting the second
                            of concrete.                  method of solution in the previous calculation
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