OS-C501

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Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 196 5.3.2 The following values are selected for the gas vessel with liner: Table 14-20 Short term values used for gas vessel with liner Partial factor Value Explanation Characteristic value of the local response of the structure (strain) in the fibre direction n Guidance note: The characteristic strain to failure of 0.87% is the worst case in this example for short-term loads at the beginning of the life of the component and after exposure to cyclic and permanent loads. ---e-n-d---of---G-u-i-d-a-n-c-e---n-o-t-e--- 5.3.3 Evaluating the design criterion in [5.3.1] we find the maximum allowable strain in fibre direction ε nk after 25 years of service to be: This is more than the largest actual strain (in the fibre directions) ε 1 =0.65%. (Note that short term loads are not critical for the design, but long term loads as described below.) 5.4 Fibre dominated ply failure due to static long-term loads (ref. Sec.6 [10]) 5.4.1 The characteristic stress rupture curve is given by: log[ σ ( t) ] = log[ 348] − 0.0423log( t) (from [4.6.2]). 5.4.2 The time to stress rupture shall be checked by the criterion given in Sec.6 [10.4.8]: 5.4.3 The following values are selected for the gas vessel with liner: ε nk 0.65% Largest strain in fibre direction, see the table in [5.2.8] Characteristic fibre strain to failure ˆ ε fibre k 0.87% See [4.1.4], [4.4.5] and [4.4.6] Partial load effect factor Partial resistance factor γ F x γ M 1.18 From Sec.8 [2.4]: Maximum load is known with 0 COV Strain to failure: COV < 5% Load-model factor γ Sd 1.05 Due to simplifications in the analytical model used, see [5.2.4] and [5.2.5]. Partial resistance-model factor γ Rd 1.0 Degraded properties are used in the analysis abs γ fat γ Rd t N ∑ ε nk y j = 1 t 0.87% < = 0.70% 1.18⋅1.05 ⋅1.0 t charact actual { σ applied } j Sd < j { } 1 γ σ applied Table 14-21 Long term values used for gas vessel with liner Factors Value Explanation Design life t y 25 years Design life of 25 years. The total number of load conditions N 1 Only one load condition. Actual time at one permanent static load condition per year t actual 1 year The vessel is basically loaded all year, except for the short unloading times that are ignored here. Local response of the structure to the σ applied Calculated below. permanent static load conditions (max. stress) Characteristic time to failure under the permanent static load condition t charact Calculated below. Load-model factor γ Sd 1.05 Same as before, due to simplifications in the analytical model used, see [5.2.4]-[5.2.5]. Resistance-model factor γ Rd 0.1 Only one load condition. Partial fatigue safety factor γ fat 50 See Sec.8 [5]. 5.4.4 The criterion in [5.4.2] can be evaluated to show that the characteristic time to failure should be: γ Sd t charact actual = t γ Rdγ fatt = 125years y DET NORSKE VERITAS AS
Offshore Standard DNV-OS-C501, November 2013 Sec.14 Calculation example: two pressure vessels – Page 197 5.4.5 From the stress rupture formula in [5.4.1] the stress level corresponding to a characteristic life of 125 years is: 163 MPa. Dividing this value by the load-model factor γ Sd gives: σ j applied =155 MPa , and ε j applied 155 MPa 23 . 7 GPa This is the same as the largest actual strain (in the fibre directions) ε 1 =0.65%. 5.4.6 The stress rupture behaviour is critical for the design. For a high safety class application the stress rupture data should be confirmed by testing. 5.4.7 Short-term failure due to maximum loads after 25 years was already considered in [5.3]. 5.5 Fibre dominated ply failure due to cyclic fatigue loads (ref. Sec.6 [11]) 5.5.1 The characteristic S-N curve for R=0.1 is given by: [ ε( N )] = 0 − 0.101log( N ) (from [4.6.2]). 5.5.2 The number of cycles to fatigue failure shall be checked by the criterion given in Sec.6 [11.3.5]: γ fat 5.5.3 The following values are selected for the gas vessel with liner: γ Rd Table 14-22 Fatigue values used for gas vessel with liner Factors Value Explanation Number of years for the fatigue evaluation t y 25 years Design life of 25 years (typically equal to the design life) The total number of strain conditions N 1 Only one load condition Number of cycles per year at a particular strain n actual 52 1 cycle per week condition Local response of the structure to the strain condition applied ε applied Calculated below Characteristic number of cycles to failure n charact Calculated below under a given strain condition Load-model factor γ Sd 1.05 Same as before, due to simplifications in analytical model, see [5.2.4]-[5.2.5] Partial resistance-model factor γ Rd 0.1 Only one load condition Partial fatigue safety factor γ fat 50 See Sec.8 [5] 5.5.4 The criterion in [5.5.2] can be evaluated to show that the characteristic number of cycles to failure should be: 5.5.5 The strain amplitude corresponding to a characteristic life of 6500 cycles with an R ratio of 0.1 is 0.476% according to [5.5.1]. The maximum strain is then 0.952%. Therefore: {γ sd ε j applied} = 0.952% and ε j applied = 0.907% This is more than the largest actual strain (in the fibre directions) ε 1 =0.65%. = 5.5.6 Fatigue data do not have to be confirmed by testing if the factor γ Rd can be multiplied by 20 (Sec.6 [11.3.8]). In this case the strain amplitude for a life of 6500 x 20 = 130000 cycles should be found. The strain amplitude with an R ratio of 0.1 is 0.352% according to [5.5.1]. The maximum strain is then 0.704%. Therefore: {γ sd ε j applied } = 0.704% and ε j applied = 0.67% This is more than the largest actual strain (in the fibre directions): ε 1 = 0.65%. t σ y E j applied N 1 ∑ j = 1 fibre Fatigue data do not have to be confirmed by testing, provided the other similarity requirements (from Sec.6 [11.3.8]) are fulfilled. 0 . 65 5.5.7 Short-term failure due to maximum loads after 25 years was already considered in [5.3]. n n γ = actual charact charact n = t γ y fat = log . 063 { γ ε } j applied Sd < { ε } 1 j applied Rd γ n Sd actual = 6500 % DET NORSKE VERITAS AS
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