Question 7.3: A rock armoured breakwater has been designed deterministical......
A rock armoured breakwater has been designed deterministically, using the Van der Meer equations (see Section 9.4.3) with the following design parameters:
H_{S} = 3.0 m
T_{m} = 6.0
P = 0.1
cotα = 2
Δ = (ρ_{r}/ρ)−1 = 1.59
N = 2000
S_{d} = 2
The resulting necessary rock size, using the above is
D_{n50}= 1.3 m.
Use the partial Safety Factors for Stability Failure of Rock Armour, Plunging Waves, Van der Meer Formula and Design Without Model Tests to determine the failure probability for this rock armoured breakwater, at ultimate limit state collapse for which
S_{d} = 8.
Design equation taken from the CEM (cf. Table VI-6-5):
G=\frac{1}{\Gamma_r} 6.2 S^{0.2} _d P \Delta D_{n 50} \varepsilon_m{ }^{-0.5} N^{-0.1}-\Gamma_S H_S.The partial safety factors given in Table VI-6-5 are reproduced below. It may be seen that they are a function of both the probability of failure and the coefficient of variation of the design wave height (\sigma ^{\prime }_{H} = σ/µ). The values of D_{n50} also given in this table are found using the above design equation.
Hence, it can be seen that the originally selected D_{n50} results in an ultimate limit state failure probability of about 7.5% for a coefficient of variation of 5% or about 15% for a coefficient of variation of 20%.
| \sigma ^{\prime }_{H} = .05 \sigma ^{\prime }_{H} = 0.2 | ||||||
| P_{f} | Γ_{s} | Γ_{r} | D_{n50} | Γ_{s} | Γ_{r} | D_{n50} |
| 0.01 | 1.6 | 1.04 | 1.59 | 1.9 | 1 | 1.82 |
| 0.05 | 1.4 | 1.02 | 1.36 | 1.5 | 1.06 | 1.52 |
| 0.1 | 1.3 | 1 | 1.24 | 1.3 | 1.1 | 1.37 |
| 0.2 | 1.2 | 1 | 1.15 | 1.2 | 1.06 | 1.22 |
| 0.4 | 1 | 1.08 | 1.03 | 1 | 1.1 | 1.05 |