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PIPES CONVEYING FLUID: LINEAR DYNAMICS I                81

                The kinetic energy of the pipe and the fluid may be evaluated by making use of  (3.23)
              and (3.26),

                           L
                  Tp = im 1 (x2 + i2) ds,   q = &M .IL [(i + Ux’I2 + (i + UZ’)~] ds,   (3.60)
                           0
              in  which  m  and M  have  been  defined  in  Section 3.3.2;  again, the  subscripts p  and  f
              stand for the pipe and fluid, respectively. The integrands in $ and Tf  may be simplified
              by  noting that X - 6(e2), x’ 2:  1 - ;w’~, and x’~ + z’~ = 1 from inextensibility condition
              (3.14). Hence, recalling also that X = U and z  = w, the expressions for $ and Tf  become
                             L                    L
                    $= lml W2ds,        Tf  = iMl [U2+W2+2Uww’+2UU]ds.             (3.61)

              It is noted that (3.61) could have been obtained directly with the aid of Figure 3.8(b); the
              various terms are obtained from Cartesian components of (3.24), which may be expressed
              as  (W + U  sin x) and  (U cos x + U) with sin x 2 w’ and cos x 2:  1  - ;w’~, neglecting
              terms smaller than 6(e2).
                The potential energy is given by
                                                                I’
                      v = vp + v, = ;EI lL ds + &(m + M)g lL w’~ ds ds.            (3.62)
                                           w”’

              The component of  V  associated with gravity may be simplified via integration by parts,
              as follows:
                                                                      :

                                                                                ds}
                                                              wI2
                   i(m + M)g/‘  1’ w’~ dsds = ;(m + M)g { [s 1’ ds] 1 - 1‘ ~w’~
                              0  0
                                                         L
                                                                   ds.
                                           = T(m + M)g / (L - s)~’~                (3.63)
                                              I
                                                        0
                Finally,  substituting (3.61)-(3.63)  into  (3.59) and  making  use  of  the  standard vari-
              ational  techniques  and  of  the  boundary  conditions  for  a  cantilever,  after  considerable
              manipulation, this reduces to




                                                 + (M + m)h) Swds dt  = 0.         (3.64)

              Two  items  should be  remarked  upon  in  the  derivation of  (3.64). Firstly,  the  terms in
              the second integral of  (3.59) cancelled out with identical ones originating from the first
              integral after integration by parts. For instance,
                    6 1  MU2uLdt = MU26 6’ lL $d2dsdt = MU2 h”          w’(6w)’ ds dt



                                  =MU2[      w‘Sw1 L dt-MU2 1” .IL W” 6w ds dt,
                                                  0
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