Page 112 - Trenchless Technology Piping Installation and Inspection
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78    Cha pte r  T w o

          II. Deflection (spangler formula)

                                   ×
                                       +
                      Y  =    K × ( L W W )     × 100
                                          s
                      D        E
                                      + .0 061 × E′ s
                           . 15 × ( DR – )1  3
          where  Y = vertical deflection of Liner pipe, in
                Y
                  = deflection ratio, expressed as a percentage
                D
                L = empirical lag factor, taken as 1.25
                K = bedding constant, taken as 0.083
               W = earth load, psi
               W  = live load, psi (W  and/or W )
                 s               sc        sd
               DR = liner pipe dimension ratio =  D
                                            t

               E′ =  modulus of soil reaction, psi (typical value
                 s
                   700–1500 psi)
            a. Determine loads
               1.  Earth load (modified Marston Formula)

                                    ×
                                        ×
                                   Cw B
                               W =        d
                                     144
               where W = earth load, psi
                     B  = trench width, ft
                      d
                     w = soil density, pcf
                             – ⎛  2ku′ H ⎞
                            ⎜ ⎜  s ⎟ ⎟
                         1 – e ⎝  B d  ⎠
                     C =
                            2ku′
                      e = 2.718
                     k = a ratio of horizontal to vertical pressures
                     u′ =  coefficient of sliding friction between the backfill
                        and trench walls (Table 2.11)
                    H  = soil depth to top of pipe, ft
                      s
                     B  = 6.5 (Table 2.10)
                      d
                    ku′ = 0.13 (Table 2.11)
                                         ×
                                       .
                                   ⎛ 2–  × 0 130 16 ⎞
                                   ⎜ ⎝  65  ⎟ ⎠
                                       .
                             –.
                        C =  1 2 718         = 182.
                                  ×
                                     .
                                 20 13
                                    ×
                             .
                                       .
                                           .
                        W =  182  × 120 65  = 985 psi
                                144
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