Page 140 - Water Engineering Hydraulics, Distribution and Treatment
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118
                                          Chapter 5
                                                   Water Hydraulics, Transmission, and Appurtenances
                                         y = [I /y A] + y ,where A = b × h,and I = bh ∕12.
                                         cp
                                              cg
                                                cg
                                                      cg
                                                               3
                                         y = [I /y A] + y = [(4 × 8 ft ∕12)/(20 ft × 32 ft )] + 20 ft
                                              cg
                                         cp
                                                      cg
                                                cg
                                           = 20.27 ft from point O.
                                      Solution 2 (SI System):
                                                            3
                                         P = (  h )A = (9.8kN∕m )(4.88 m + 2.44 m × 0.5)(1.22 m × 2.44 m) = 177.95 kN.
                                               cg
                                          f
                                      The above resultant force acts at the center of pressure, which is at a distance y from water surface O.
                                          At vertical position, h = y .
                                                         cg
                                                             cg
                                         y = 4.88 m + 2.44 m × 0.5 = 6.1 m.
                                         cg
                                                               2
                                         A = 1.22 m × 2.44 m = 2.98 m .  4  cg  3  2           cp
                                                                              3
                                         y = [I /y A] + y ,where A = b × h,and I = bh ∕12.
                                                      cg
                                                                         cg
                                              cg
                                                cg
                                         cp
                                                                      4
                                                                                      2
                                                                   3
                                         y = [I /y A] + y = [(1.22 × 2.44 m ∕12)/(6.1 m × 2.98 m )] + 6.1 m
                                                cg
                                         cp
                                                      cg
                                              cg
                                           = 6.18 m from point O.
                                      EXAMPLE 5.13 DETERMINATION OF RESULTANT FORCE DUE TO WATER ACTING ON AN INCLINED
                                      RECTANGULAR GATE
                                      Determine the total resultant force and its location acting on the gate shown in Fig. 5.7. The gate is 4.5 ft × 6.5 ft (1.37 m ×
                                      1.98 m) in dimension. The apex of the triangle is at C. The gate top is 5.5 ft (1.68 m) below the water surface. The related dimensions
                                      in the figure are a = 5.5 ft = 1.68 m; h = 4.5 ft = 1.37 m; b = 6.5 ft = 1.98 m.
                                                                Water surface                           O
                                                                                                      60°
                                                                     a
                                                                                h cp  h cg
                                                                                       C
                                                                                              y cg
                                                                             D      h         y cp
                                                                     b
                                      Figure 5.7 Inclined rectangular gate in a water tank.
                                      Solution 1 (US Customary System):
                                                             ◦
                                         h = 5.5 ft + 0.5(4.5 Cos 60 ) = 6.625 ft.
                                         cg
                                                           3
                                         P = (  h )A = (62.4 lb/ft )(6.625 ft)(4.5 ft × 6.5 ft) = 12,092 lb.
                                          f
                                               cg
                                                    ◦
                                         y = h /Sin 30 = 6.625 ft∕0.5 = 13.25 ft, or
                                              cg
                                         cg
                                                       ◦
                                         y = (5.5 ft/Cos 60 ) + (4.5 ft∕2) = 13.25 ft.
                                         cg
                                              3
                                                           3
                                                                         4
                                                             4
                                         I = bh ∕12 = (6.5 × 4.5 ft )∕12 = 49.35 ft .
                                         cg
                                                                              3
                                         y = [I /y A] + y ,where A = b × h,and I = bh ∕12.
                                                      cg
                                              cg
                                                                         cg
                                                cg
                                         cp
                                                                4
                                         y = [I /y A] + y = [49.35 ft /(13.25 ft × 4.5 ft × 6.5 ft)] + 13.25 ft = 13.38 ft.
                                                      cg
                                                cg
                                              cg
                                         cp
                                                    ◦
                                         h = y Sin 30 = 13.38 ft × 0.5 = 6.69 ft.
                                              cp
                                          cp
                                      Solution 2 (SI System):
                                                                 ◦
                                         h = 1.68 m + 0.5(1.37 m Cos 60 ) = 2.02 m.
                                         cg
                                                            3
                                         P = (  h )A = (9.79 kN/m )(2.02 m)(1.37 m × 1.98 m) = 53.64 kN.
                                          f    cg
                                                    ◦
                                         y = h /Sin 30 = 2.02 m∕0.5 = 4.04 m, or
                                         cg   cg
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