Page 111 - Trenchless Technology Piping Installation and Inspection
P. 111

Existing Pipeline Renewal and Replacement Methods        77

                 a.  Calculate coefficient of elastic support
                          ⎡        1         ⎤
                                                 .
                      B ′ = ⎢          0 065 ×   16  ⎥ = 0 414
                                       .
                          ⎣1  + 4 (  × 2 178  ⎦ )
                                   .
                 b.  Calculate ovality reduction factor
                              ⎡  ⎛    ⎞  ⎤  3
                              ⎢  ⎜1 –  20 .  ⎟  ⎥
                              ⎢  ⎝  100 ⎠ ⎥
                          C =   ⎢      ⎥  = = 0 836
                                            .
                              ⎢ ⎛  20 .  ⎞  2 ⎥
                              ⎢ ⎜1 –  100 ⎟ ⎠ ⎦ ⎥
                               ⎝ ⎣
                 c. Calculate E
                             L
                    E  = E – (E × reduction factor to account for long-term effects)
                     L
                    E  = 250,000 – (250,000 × 0.5) = 125,000 psi
                     L
                 d.  Determine minimum CIPP thickness in buckling
                                                     1
                                 ⎡     ⎛      2     ⎤ 3
                                 ⎢     ⎜ N ×  q ⎞ ⎟  ⎥
                                            t
                                 ⎢     ⎜ ⎝  1  ⎟ ⎠  ⎥
                     =
                    t  0 721  ×  D × ⎢              ⎥
                       .
                                 ⎢ E ×  R ×  B′ ×  C ×  E′ s⎥
                                       w
                                   L
                                 ⎢ ⎢                ⎥
                                 ⎣ ⎢                ⎦ ⎥
                                                                  1
                                  ⎡                  ⎞  2        ⎤ 3
                                                ×
                                                   .
                                  ⎢         ⎛ 2 0.    14 90 ⎟    ⎥
                                            ⎜
                    t   0 721   48 00    ⎢ ⎢  ⎝  1   ⎠           ⎥ ⎥
                                 ×
                     =
                           ×
                              .
                       .
                                                      ×
                                                            ×
                                         ×
                                      ,
                                                        .
                                                  .
                                  ⎢ 125 000  ×  0.835  × 0 414 0 836 1000 ⎥
                                  ⎢ ⎣                            ⎥ ⎦
                    t = 1.01 in. (external pressure, deteriorated conduit)
                 e.  Check minimum stiffness requirement
                               ×
                           EI       E
                                          ≥
                             3  =       3   0 093.
                                   ×
                           D     12   ( DR)
                    where DR = Dimension Ratio, DR = D/t
                    Solve for t,
                                             1
                                  ⎡         ⎤ 3
                                  ⎢     3   ⎥
                                =
                              t   ⎢   48    ⎥
                                  ⎛ ⎢  250 000  ⎞⎥
                                       ,
                                  ⎜ ⎢       ⎟⎥
                                        .
                                  ⎝ ⎣ 12  × 0 093 ⎠⎦
                    t = 0.790 in. (min. stiff. limitation, deteriorated conduit)
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