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Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 198 5.6 Matrix cracking (ref. Sec.6 [4]) 5.6.1 Matrix cracking does not have to be considered for the gas vessel with liner. 5.7 Unacceptably large displacement (ref. Sec.6 [9]) 5.7.1 It is assumed for the purpose of this example that the long-term yield strain of the liner is 5%, and this value should not be exceeded to ensure that the liner will not yield. 5.7.2 The liner in this example is thin and does not contribute to the load bearing capabilities of the vessel. The liner will follow the deformations of the laminate body of the vessel. 5.7.3 For simplicity we assume that the liner has the same strain as the laminate. In reality the liner will have slightly larger strain since it is located on the inside of the cylindrical vessel, see [5.2.4]-[5.2.5]. This is taken into account by introducing a load-model factor, γ Sd = 1.05. 5.7.4 The criterion for unacceptably large displacements from Sec.6 [9.1.1] shall be used. 5.7.5 The following values are selected for the gas vessel with liner: Table 14-23 Displacement values used for gas vessel with liner Factors Value Explanation Specified requirement on maximum d spec 5% See [5.3.2] displacement Characteristic value of the local response of d n Calculated below the structure (here strain) Partial load effect factor γ F 1.15 From Sec.8 [2.4]: Maximum load is known with 0 COV Strain to failure COV 5% Target reliability level C Load-model factor γ Sd 1.05 Same as before, due to simplifications in analytical model, see [5.2.4]-[5.2.5] 5.7.6 The maximum principle strain in the laminate should be less than 5/(1.15 × 1.1) = 4.14%. 5.7.7 The highest elastic strain in fibre direction is only 0.65%. However, in this case we have to look at the elastic strain and the plastic strain due to creep. A method to calculate elastic and plastic strain is given in Sec.4 [3.2.11]. ε = ε + elastic ε plastic or σ σ ε = + E elastic E plastic 5.7.8 Creep in the ±15 o plies: The elastic strain is 0.65%. The plastic strain can be calculated according to the representative data of App.F for creep: σ 0.2 ε with time in hours and strain in %. 1520 t plastic = The total strain for the maximum stress of 154 MPa (see [5.2.8]) and 219000 hours is: 0.65%+1.18%=1.83%. 5.7.9 Creep in the ±85 o plies can be calculated the same way. Since the ply stresses are slightly lower, the creep strain is also slightly less. 5.7.10 The principle strains of the laminate should be calculated from the ply strains and applied to the design criterion. Since the ply strains are so much below the acceptable levels this calculation is not done here. 5.8 Impact resistance (ref. Sec.6 [12]) γ d F .γ Sd . < n 5.8.1 Impact may be caused by dropped tools etc. The possible impact scenarios, if any, should be defined. 5.8.2 There is no good theoretical criterion to evaluate the resistance to impact. According to Sec.6 [12], the resistance of a structure to impact shall be tested experimentally. 5.8.3 The critical failure mechanisms in this example is fibre failure. It would have to be shown that the defined impact scenarios do not cause any fibre failure. This could be shown on full scale specimens or on representative laminates. d spec DET NORSKE VERITAS AS
Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 199 5.8.4 Alternatively, the vessel could be protected against impact by covers or other protection devices. 5.9 Explosive decompression (ref. Sec.6 [15]) 5.9.1 If the air diffuses through the liner more rapidly than it can diffuse out of the laminate, a layer of pressurised air may build up in the interface between liner and laminate. In such a case the interface should be vented or experiments should be made to show that the liner will not collapse when the internal pressure is reduced (Sec.6 [15.2]). 5.9.2 The rate of air flow through the liner is most likely much less than through the laminate and explosive decompression should be no problem, as long as the vessel is not exposed to external pressures as well, like under water usage. 5.10 Chemical decomposition (ref. Sec.6 [17]) 5.10.1 It is proven by many applications that composite laminates do not chemically decompose in air within 25 years. 5.11 Summary evaluation 5.11.1 Summary is as follows: — Impact resistance should be evaluated experimentally if the vessel may be exposed to impact loads. Experiments should show that possible impact loads will not cause fibre damage. — The gas vessel passed all other requirements for service at 46 bar. — The reduction of fibre dominated ply strength due to permanent loads is the design limiting factor for this vessel. — Obtaining material data from tubular specimens instead of flat plates (as used in this example) would allow better utilisation of the cylinder. 6 Non-linear analysis of vessel for water without liner 6.1 General 6.1.1 For the water vessel (without liner) it is assumed that leakage will occur if matrix cracking is present in one of the plies. We apply the 2-D in-plane progressive failure analysis (Sec.9 [2.2]) to evaluate fibre failure. An alternative method is given under [7]. 6.1.2 All relevant failure mechanisms shall be evaluated for all loads of all phases. 6.1.3 The following failure mechanisms were identified in [3.1.3] for the water vessel: — fibre failure: — short-term static — long-term static — long-term fatigue — matrix cracking (for vessel without liner): — short-term static — long-term static — long-term fatigue — impact resistance — explosive decompression — chemical decomposition 6.1.4 The filament wound laminate is the same as for the gas vessel in [5.1.4]. 6.2 Analysis procedure (ref. Sec.9 [2]) 6.2.1 The thin shell method for calculations of laminate stresses is the same as for the gas vessel. The elastic properties used in the analysis are different. 6.2.2 The elastic properties without and with matrix cracking were calculated in [4.1.4]. The values are different at the beginning and the end of the life of the component. DET NORSKE VERITAS AS
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OS-C501

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