09.10.2015 Views

OS-C501

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

Offshore Standard DNV-<strong>OS</strong>-<strong>C501</strong>, November 2013<br />

Sec.14 Calculation example: two pressure vessels – Page 198<br />

5.6 Matrix cracking (ref. Sec.6 [4])<br />

5.6.1 Matrix cracking does not have to be considered for the gas vessel with liner.<br />

5.7 Unacceptably large displacement (ref. Sec.6 [9])<br />

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<br />

value should not be exceeded to ensure that the liner will not yield.<br />

5.7.2 The liner in this example is thin and does not contribute to the load bearing capabilities of the vessel. The<br />

liner will follow the deformations of the laminate body of the vessel.<br />

5.7.3 For simplicity we assume that the liner has the same strain as the laminate. In reality the liner will have<br />

slightly larger strain since it is located on the inside of the cylindrical vessel, see [5.2.4]-[5.2.5]. This is taken<br />

into account by introducing a load-model factor, γ Sd = 1.05.<br />

5.7.4 The criterion for unacceptably large displacements from Sec.6 [9.1.1] shall be used.<br />

5.7.5 The following values are selected for the gas vessel with liner:<br />

Table 14-23 Displacement values used for gas vessel with liner<br />

Factors Value Explanation<br />

Specified requirement on maximum d spec 5% See [5.3.2]<br />

displacement<br />

Characteristic value of the local response of d n<br />

Calculated below<br />

the structure (here strain)<br />

Partial load effect factor γ F 1.15 From Sec.8 [2.4]:<br />

Maximum load is known with 0 COV<br />

Strain to failure COV 5%<br />

Target reliability level C<br />

Load-model factor γ Sd 1.05 Same as before, due to simplifications in analytical<br />

model, see [5.2.4]-[5.2.5]<br />

5.7.6 The maximum principle strain in the laminate should be less than 5/(1.15 × 1.1) = 4.14%.<br />

5.7.7 The highest elastic strain in fibre direction is only 0.65%. However, in this case we have to look at the<br />

elastic strain and the plastic strain due to creep. A method to calculate elastic and plastic strain is given in Sec.4<br />

[3.2.11].<br />

ε = ε + elastic ε plastic<br />

or<br />

σ σ<br />

ε = +<br />

E elastic E plastic<br />

5.7.8 Creep in the ±15 o plies: The elastic strain is 0.65%. The plastic strain can be calculated according to the<br />

representative data of App.F for creep:<br />

σ 0.2<br />

ε with time in hours and strain in %.<br />

1520 t<br />

plastic<br />

=<br />

The total strain for the maximum stress of 154 MPa (see [5.2.8]) and 219000 hours is: 0.65%+1.18%=1.83%.<br />

5.7.9 Creep in the ±85 o plies can be calculated the same way. Since the ply stresses are slightly lower, the creep<br />

strain is also slightly less.<br />

5.7.10 The principle strains of the laminate should be calculated from the ply strains and applied to the design<br />

criterion. Since the ply strains are so much below the acceptable levels this calculation is not done here.<br />

5.8 Impact resistance (ref. Sec.6 [12])<br />

γ<br />

d<br />

F<br />

.γ<br />

Sd<br />

. <<br />

n<br />

5.8.1 Impact may be caused by dropped tools etc. The possible impact scenarios, if any, should be defined.<br />

5.8.2 There is no good theoretical criterion to evaluate the resistance to impact. According to Sec.6 [12], the<br />

resistance of a structure to impact shall be tested experimentally.<br />

5.8.3 The critical failure mechanisms in this example is fibre failure. It would have to be shown that the defined<br />

impact scenarios do not cause any fibre failure. This could be shown on full scale specimens or on<br />

representative laminates.<br />

d<br />

spec<br />

DET NORSKE VERITAS AS

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!