In(10 mm, S/1000),(23)Appl. Sci. 2021, 11,eight of2.3. Calculation from the New Hanger Installation Procedure The installation in the new hanger is basically the reverse course of action from the hanger removal. Nevertheless, the tension method for the duration of the installation on the new hanger is the exact same as that with the unloading process, because the pocket hanging hanger is carried out via the jack pine oil with out the should reduce it. two.three.1. Initial State The initial state is the state ahead of the new hanger is installed: (a) New hanger: elasticity modulus is En , cross-sectional region is definitely an , and cable length s is L0 . (b) Pocket hanging hanger: elasticity modulus is E , cross-sectional region can be a , cable s s s length is L 0 , shear force is Q0 , and cable tension is T0 . Because the new hanger is installed just after the old hanger is removed, then there is certainly: L0=Ls d N, s T0 = TN , g(24)According to the displacement coordination and force balance, it has:s L0 = s T0 L 0 s + L 0, EA s(25) (26)s s T0 = Q0 + G,two.3.two. The ith(i = 1, 2, . . . , Nn ) Times Tension with the New Hanger Just after the ith times tension with the new hanger, let the new hanger internal force be Fiz , the pocket hanging hanger internal force be Tiz , the unstressed lengths of your new hanger z z and pocket hanging hanger be Li , L i , respectively, and the displacement in the ith instances tension of the new hanger be xiz . There is absolutely no distinction between this method as well as the ith instances on the pocket hanging; hence, the derivation is just not repeated and you will discover:s Tiz = Tis-1 – E A Tis-1 – G – Qi-1 + Fiz z , z Li(27) , (28) (29)=En An Ls i -s s s G + Qi-1 – Fiz – Tis-1 z + En An Li-1 + Li-1 Fis-Fiz + En An xiz = Ls i -1 s Tis-1 – G – Qi-1 + Fiz z , s i -1 k +where z = 1/ LEA .2.three.3. The ith(i = 1, two, . . . , Nn ) Times Unloading from the Pocket Hanging Hanger Soon after the ith times unloading on the pocket hanging hanger, let the new hanger internal force be Fis , the pocket hanging hanger internal force be Tis , the unstressed lengths in the s s new hanger and pocket hanging hanger be Li , L i , respectively, and also the displacement on the ith instances tension of the new hanger be xis .z Fis = Fiz – En An ( Fiz – G – Qi + Tis ) s ,(30)zLi =sz z E A Li G + Qi – Tis – Fiz s + E A L i + L i Tiz , Tis + E A z z xis = Li ( Fiz – G – Qi + Tis ) s ,z(31) (32)z exactly where s = 1/ Li k + En An .= = – – + ,Appl. Sci. 2021, 11,,(31) (32)9 ofwhere = 1/ + .2.3.four. Displacement Handle 2.3.4. Via the above calculation, it may be observed that soon after the ith = 1, 2, … , times Displacement Control Via the above calculation, it could be observed that just after the the = 1, end . , Nn occasions tension of your new hanger, the 18-Oxocortisol Cancer accumulative displacement ofith (ilower two, . . from the)hanger tension from the new hanger, the accumulative displacement from the lower finish of your hanger is: is:- = – 1 i1 z +s + , Xiz = (i – 1) n=1 ( xn + xn ) + xiz ,(33) (33)Immediately after the ith = 1, 2, … , times unloading from the pocket hanging hanger, the acAfter the ith (i = 1, two, . . . , Nn ) times unloading from the pocket hanging hanger, the cumulative displacement in the reduce end in the hanger to become replaced is: accumulative displacement Xis of your reduce finish with the hanger to be replaced is: = + , (34) i s z s Xi = n = 1 ( x n + x n ) , (34) , , and handle displacement threshold [D] have to satisfy the following connection: iz , Xis , and control displacement threshold [D] ought to satisfy the following relationship: X , g , Xid [ D ], Xi [.