OS-C501

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Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 194 5 Analysis of gas vessel with liner 5.1 General 5.1.1 All relevant failure mechanisms shall be evaluated for all loads of all phases. 5.1.2 The following failure mechanisms were identified in [3.1.3] for the gas vessel: — fibre failure: — short-term static — long-term static — long-term fatigue — unacceptably large displacement (for vessel with liner) — impact resistance — explosive decompression — chemical decomposition. 5.1.3 Matrix cracking does not need to be checked, since the liner keeps the fluid inside the vessel even if cracks are present inside the laminate. 5.1.4 The filament wound laminate as described in [2.3.2] is modelled as 12 layers of 0.5 mm thickness and the following lay-up in Table 14-17 is applied, in accordance with Sec.4 [1.4.10]: Table 14-17 Lay-up of laminate Ply no: Fibre angle (°) 1 15 2 - 15 3 85 4 -85 5 85 6 - 85 7 - 85 8 85 9 - 85 10 85 11 - 15 12 15 5.1.5 Since the laminate thickness is much smaller than the diameter of the vessel (see the table in [2.3.1]), the vessel is analysed using classical thin-wall theory and laminate theory. 5.1.6 For the gas vessel with liner matrix cracking in the plies will not lead to leakage. Therefore, we can apply a linear failure analysis with degraded material properties (see Sec.9 [2.5]) for the gas vessel. 5.1.7 Using degraded properties throughout the laminate may put higher stresses and strains into the fibre direction than in reality. This is a conservative way to model this vessel and fulfils the requirement of Sec.9 [2.1.9]. 5.1.8 Based on the above assumptions the simplified analytical analysis may be summarised as shown in [5.2]. 5.2 Analysis procedure (ref. Sec.9) 5.2.1 The (mean/laminate) axial (σ x ) and hoop (σ y ) stresses are calculated. From thin-wall theory these stress components are given by: PD PD σx = and σy = (which means that σ y = 2σ x ) 4t 2t where, P = pressure = 46 bar = 4.6 MPa = 4.6 N/mm 2 D = inner diameter of vessel = 250 mm t = laminate thickness = 6 mm. Remark that the laminate shear stress (σ xy ) is zero in the present example. 5.2.2 The mid-plane strains (ε 0 ) are calculated from the relation: N=Aε 0 DET NORSKE VERITAS AS
Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 195 where, N A = cross section force vector (containing N x , N y and N xy ), which is obtained by integrating the laminate stresses through the thickness of the laminate. = extensional stiffness matrix for the laminate (we refer to classical laminate theory), which is obtained by summing the stiffness matrices for all the plies after multiplication by the ply thickness. This means that the material properties for each ply (E 1 fibre , E 2 matrix , G 12 and ν 12 ) contribute to the A matrix. 5.2.3 When using the degraded failure analysis some of these parameters are assumed to be very small (see Sec.9 [2.5]). The elastic properties with matrix cracking and no water exposure were calculated in [4.1.4]. The values are the same at the beginning and at the end of the life of the component, as shown in Table 14-18. Table 14-18 Elastic properties E 1 fibre 23.7 GPa E 2 matrix 0.08 GPa ν 12 0.003 G 12 linear 0.03 GPa 5.2.4 Due to the assumption that t
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