OS-C501

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Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 206 Table 14-31 Elastic properties Property after 25 years in water after 25 years in water with matrix cracking E 1 fibre 21.3 GPa 21.3 GPa E 2 matrix 6.8 GPa 0.08 GPa ν 12 0.29 0.003 G 12 linear 2.9 GPa 0.03 GPa 7.6.5 The laminate modulus in the 15 degree direction is obtained by calculation the Young's modulus in the main x-direction for the original laminate rotated by - 15 degrees, i.e.: Table 14-32 Rotated laminate (-15°) Ply no: Fibre angle (°) 1 0 2 - 30 3 70 4 - 100 5 70 6 - 100 7 - 100 8 70 9 - 100 10 70 11 - 30 12 0 The laminate modulus is obtained from laminate theory calculations. Even though the laminate is not symmetric and the laminate modulus is not a meaningful value, it is sufficient for the calculation γ a . The results are: 10.31 GPa for non-degraded properties, and: 4.04 GPa for the laminate with matrix cracks. Therefore: γ A = E lin /E nonlin , = 2.55 for the 15 degree ply. 7.6.6 The laminate modulus in the - 15 degree direction is obtained in a similar way by rotating the original laminate by +15 degrees. The result is the same as in [7.6.5]. 7.6.7 The laminate modulus in the 85 degree direction is obtained by calculation the Young's modulus in the main x-direction for the original laminate rotated by - 85 degrees, i.e.: Table 14-33 Rotated laminate (-85°) Ply no: Fibre angle (°) 1 - 70 2 - 100 3 0 4 - 170 5 0 6 - 170 7 - 170 8 0 9 - 170 10 0 11 - 100 12 - 70 The laminate modulus is obtained from laminate theory calculations. Even though the laminate is not symmetric and the laminate modulus is not a meaningful value, it is sufficient for the calculation γ a . The results are: 16.02 GPa for un-degraded properties, and: 12.01 GPa for the laminate with matrix cracks. Therefore: γ A = E lin /E nonlin , = 1.33 for the 85 degree ply. 7.6.8 The laminate modulus in the - 85 degree direction is obtained in a similar way by rotating the original laminate by + 85 degrees. The result is the same as in [7.6.7]. 7.6.9 Analysing the design criterion above (see [7.6.2]) we find the maximum allowable strain in fibre direction ε nk after 25 years of service to be: DET NORSKE VERITAS AS
Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 207 maximum strain in 15 o plies: maximum strain in 85 o plies: The maximum ply strains in fibre direction calculated by this method are 0.17%. This shows that fibres are not predicted to fail. Guidance note: Note that these strains are artificially low to compensate for the simplified calculation method used. ---e-n-d---of---G-u-i-d-a-n-c-e---n-o-t-e--- 7.7 Fibre dominated ply failure due to static long-term loads (ref. Sec.6 [10]) 7.7.1 The analysis method is the same as for the water vessel in [6.8]. 7.8 Fibre dominated ply failure due to cyclic fatigue loads (ref. Sec.6 [11]) 7.8.1 The analysis method is the same as for the water vessel in [6.9]. 7.9 Unacceptably large displacement 7.9.1 Same as in [6.10], no requirements to be checked. 7.10 Impact resistance 7.10.1 Same as in [6.11]. 7.11 Explosive decompression 7.11.1 Same as in [6.12]. 7.12 Chemical decomposition 7.12.1 Same as in [6.13]. 7.13 Summary evaluation 7.13.1 Summarising as follows: ε nk ε = γ γ Sd ε ε nk = γ γ Sd ∧ fiber k FM FM . γ ∧ fiber k . γ Rd Rd 0.87 = = 0.275% 1.05× 1.18× 2.55 0.87 = = 0.528% 1.05× 1.18× 1.33 — The analysis and results are identical to the non-linear analysis. — This method is more complicated in this example, but may be simpler in larger structures, because the analysis has to be done only once. — The conclusions are also identical to the non-linear analysis in [6.14]. DET NORSKE VERITAS AS
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OS-C501

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