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156            SECTION 2                                        STRENGTHENING AND REPAIR WORK








                                                                         Max. sheer stress
                                                                           3 w
                                                                         3
                                                                           4 bd





                        Figure 4.11  Parabolic shear stress distribution across a rectangular beam section.

                                              Actual shear stress    3 F A y˘/I   b              (4.19e)
                                                                xy
                                            Approximate shear stress 3    3 F/A
                                                                     xy     w
                            It can be shown that:








                            Total deflection: Consider a 100-foot girder subjected to total distributed load W.

                                                           Total u.d.l. 3 w
                                             ________________________________________
                                                         L

                                                              3

                             Max. deflection due to bending 3 5 WL /384 EI
                                                      I 3 bd /12;
                                                           3
                                                         3WL     6WL                E
                             Max. deflection due to shear 3   3       ,  assuming G 3
                                                        20bdG   2 20bdE             2
                                             WL   5  L 2  3 
                             Total deflection 3        +                                        (4.20)
                                                 
                                            2bdE 16  d 2  5 


                        4.9.8 Estimating Additional Deflection due to Shear
                            There is higher shear deflection on cantilever spans than on continuous spans (Figure

                        4.13).
                        1. Consider a 5-foot Overhang: W 3 1.0 KIP, I 3 12 8 9 /12 3 729 inch units, w 3 0. 33
                                                                           3
                            kips/ft
                            Deflection due to bending 3  WL /3EI 4 wL /8EI 3
                                                       3

                                                                 4
                                                           3
                                                    1.0 8 5  8 1728/3 EI 4 0.33 8 25 8 25 8 1728/8EI 3
                                                    125 8 1728/(3 8 3000 8 729 4 5/3/8 8 3000 8 729)
                                                    3 0.033 4 0.02
                                                                           2
                            Total deflection due to shear 3 6 WL/5bd G 4 (3/5) wL /bdG

                                                      6
                            G 3 E/2(1 4  ); E 3 3.0 8 10 psi;   3 0.15; G 3 3000/2.3 ksi
                            Defl ection 3  6 8 1.0 8 5 8 2.3 8 12/5 8 12 8 9 8 3000 3 0.0592
                                                 ×
                                                    2
                                          3  0.33 5 × 23 .
                                        +  ×              ×12  3 0.00042 4 0.0005 = 0.00092
                                          5  12 9
                                                ×× 3000
                                                                                       3 0.009/0.059 3 15%
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