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                    156  Chapter 5  Water Hydraulics, Transmission, and Appurtenances
                                             The moment of inertia of the most common rectangular shape (width b and height h)
                                          can be calculated by the following equation:
                                                                       I   (1>12)Ah 2
                                                                        cg
                                                                            (1>12)(bh) h 2
                                                                            (1>12)bh 3                        (5.9)





                     EXAMPLE 5.1   WATER, SPECIFIC WEIGHT, FORCE AND MOMENT OF INERTIA
                                         Water rises to level E in the pipe attached to water storage tank ABCD in Fig. 5.1. Neglecting the
                                         weight of the tank and riser pipe, determine (a) the resultant force acting on area AB, which is 8 ft
                                         wide (2.4 m); (b) the location of the resultant force acting on area AB; (c) the pressure on the bot-
                                         tom BC; and (d) the total weight of the water.



                     Figure 5.1 Water Tank for                                             E
                     Example 5.1.
                                                                                A   1 ft 2           12 ft (3.7 m)
                                                                                      2
                                                                                (0.093 m )






                                             6 ft (1.8 m)       A                                   D



                                                                B                                   C




                                                                                  20 ft (6.1 m)



                                         Solution 1 (U.S. Customary System):
                                             a. The resultant force, P f , acting on area AB:
                                                              P f   ( h cg ) A
                                                                         3
                                                                  (62.4 lb/ft )(12 ft   3 ft)(6 ft   8 ft)
                                                                  45,000 lb
                                             b. The location of the resultant force acting on area AB:
                                                              y cp   [I cg >y cg A]   y cg
                                                              I cg   (1>12)bh 3
                                                                      3
                                                              y cp   [8(6) >12]>[15(6   8)]   15
                                                                  0.20   15
                                                                  15.2 ft
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