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                    178  Chapter 5  Water Hydraulics, Transmission, and Appurtenances
                                             3. Water hammer or increased internal pressure caused by sudden reduction in the veloc-
                                                ity of the water—by the rapid closing of a gate or shutdown of a pump, for example
                                             4. External loads in the form of backfill and traffic
                                             5. Their own weight between external supports (piers or hangers)
                                             6. Temperature-induced expansion and contraction.
                                             Internal pressure, including water hammer, creates transverse stress or hoop tension.
                                         Bends and closures at dead ends or gates produce unbalanced pressures and longitudinal
                                         stress. When conduits are not permitted to change length, variations in temperature like-
                                         wise create longitudinal stress. External loads and foundation reactions (manner of sup-
                                         port), including the weight of the full conduit, and atmospheric pressure (when the conduit
                                         is under a vacuum) produce flexural stress.
                                             In jointed pipes, such as bell-and-spigot cast-iron pipes, the longitudinal stresses must
                                         either be resisted by the joints or relieved by motion. Mechanical joints offer such resist-
                                         ance. The resistance of joints in bell-and-spigot cast-iron pipe to being pulled apart can be
                                         estimated from Prior’s (1935) observational equations:

                                                            3,800
                                                          p =     - 40           (U.S. Customary Units)     (5.39a)
                                                            d + 6
                                                              3,000       2
                                                         P = a     - 31bd        (U.S. Customary Units)     (5.40a)
                                                         f
                                                             d + 6
                                         where d is the diameter, in.; p is the intensity of pressure, psig; and P is the total force, lb.
                                                                                                 f
                                             The equivalent equations using the SI units are the following:
                                                                 670,000
                                                              p =       - 278          (SI Units)           (5.39b)
                                                                d + 152
                                                                    525          2
                                                              P = a       - 0.2bd      (SI Units)           (5.40b)
                                                             f
                                                                  d + 152
                                          where d is the diameter, mm; p is the intensity of pressure, kPa gauge; and P is the total force, N.
                                                                                                  f
                                             Tables of standard dimensions and laying lengths are found in professional manuals,
                                         specifications of the American Water Works Association (AWWA), and publications of
                                         manufacturers and trade associations.


                     EXAMPLE 5.11  PRESSURE INTENSITY AND FORCE OF PIPE’S MECHANICAL JOINTS

                                         In jointed cast-iron pipes, the longitudinal stresses can be resisted by the mechanical joints.
                                         Determine the pressure intensity and total force resisted by the mechanical joints of a 24-in.
                                         (609.6 mm) bell-and-spigot cast-iron pipe.

                                         Solution 1 (U.S. Customary System):
                                                                  p = [3,800>(d + 6)] - 40
                                                                   = [3,800>(24 + 6)] - 40
                                                                   = 86.66 psig.
                                                                 P = [3,000>(d + 6) - 31] d 2
                                                                f
                                                                   = [3,000>(24 + 6) - 31](24) 2
                                                                   = 39,744 lb.
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