PSI - Issue 6

Irina Stareva et al. / Procedia Structural Integrity 6 (2017) 48–55 Author name / Structural Integrity Procedia 00 (2017) 000 – 000

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3

3. Solution of the problem

3.1. Algorithm

Suppose that during the whole process of dissolution the maximum in absolute value principal stress is the longitudinal stress, r R      , changing along the tube axis and growing with time. Let the z axis coincide with the tube axis. In the case of a standing tube, let the origin be located in the plane of the lower cross-section of the tube and the z axis be directed upwards. In this case, for z l  , the longitudinal stress, ( , ) z t  , is negative at any time. For a hanging tube, the origin is located in the plane of the upper cross-section, the z axis is directed downwards, and ( , ) 0  z t  . Since the corrosion kinetics constants r m and R m in (1) and (2) have same sign as the corresponding stresses, the same algorithm can be used for the both cases with the assumption that r m , R m and ( , ) z t  indicate the absolute values of the constants r m , R m and the longitudinal stress ( , ) ( , ) z t z t R r    , correspondingly. Then, the longitudinal stress (in absolute value) at the initial moment of time is determined by the equation

z ( ,0)

l z g 

(3)

( )



zz  0   

where ρ is the density of steel, g is the gravitational acceleration. Further, since the cross-section area of the tube decreases non-uniformly, the stress at any time is determined by the formula

l

( , ) S z t g S t d z    ( , )



( , ) z t

,





(4)

2

2

where ( , )] S z t R z t r z t    is the cross-section area decreasing with time in accordance with (1) and (2). Thus, we have to solve the system of integral and differential equations (1), (2), and (4) satisfying initial conditions (3). For this purpose, explicit integration procedure with a constant time step t  is used. For all discrete time points i t , all the quantities , , r R and  are calculated at nodal points j z (with equal spatial step z  ): ( , )], [ ( , ) ( , ) 1 j i r r j i j i t a m z t r z t r z t       , 0,... )], ( , [ ( , ) ( , ) 1 j N t a m z t R z t R z t j i R R j i j i        where 0 0  z and l z N  . Initial conditions are given by the equations 0 ( ,0) r r z j  , 0 ( ,0) R z R j  and (3). The integral in (4) for every point ( 0,... 1)   z j N j is calculated by the method of medium rectangles using the value of this integral at the point 1  j z , while at the point N z it equals zero. This step-by-step procedure continues until either the minimum thickness (0, ) (0, ) (0, ) i i i h t R t r t   becomes equal to any given limit value * h (e.g., zero) or the maximum stress (0, ) i t  reaches a given limit *  or t reaches a given service life. Here, *  is either a strength limit (taking into account safety factors), or any other critical stress. We do not pay any attention to the type of fracture, which strongly depends on the operating conditions (see Evstifeev et al. (2013)). If tubes are subjected to complex irregular loading programs, the models of the fatigue damage accumulation (e.g., proposed by Melnikov and Semenov (2014)) can be used. The presence of surface or near-surface defects causes stress concentration (Grekov (2004), Grekov and Kostyrko (2016), Savelyeva and Pronina (2015)). This fact should also be taken into account for the assessment of durability (see Pavlov and Melnikov (1992), Pronina (2017), Pronina and Khryashchev (2017)). For the problem analysis the described algorithm has been realized in MatLab. ( , ) [ ( , )