Question 7.8: It is required to assess the stability of a breakwater rock ......
It is required to assess the stability of a breakwater rock armour layer using Hudson’s formula, which is written as
\frac{H_{s} }{D_{n50}\Delta }= A(K_{D} cot(α) )^{1/3}(see Section 9.4.3 for a fuller explanation), where it is given that K_{D} = 4, cot(α) = 2 and Δ = 1.6. These variables are taken to be exact, but uncertainties in them are accounted for by the single multiplicative factor, A, which has a N(1, 0.18) distribution. We are also told that H_{s} ∼ N(4.4, 0.7) and D_{n50} ∼ N(1.5, 0.1) and that the variables are independent.
The failure criterion is that the rock size is no longer able to maintain stability against the wave conditions.
a. Rewrite Hudson’s formula as a reliability function, and hence determine the equation for the failure surface.
b. Rewrite the reliability function found in (a) in terms of standardised variables.
a. Hudson’s formula is rearranged so that the ‘load’ and ‘strength’ variables are grouped together in separate terms:
G = A{D_{n50}\Delta}\left(K_{D}\cot \left(\alpha \right) \right) ^{1/3} -H_{s}The first term on the right-hand side of the equation corresponds to the ‘strength’ and the second term to the ‘load’. The failure surface for this reliability function is found by setting G = 0. For K_{D} = 4 we find
1.59{D_{n50}\Delta} cot \left(\alpha \right) ^{1/3}-H_{s}=0b. Transforming to standardised variables using Equation 7.42
z_{i} = \frac{x_{i}-\mu _{xi} }{\sigma _{xi} } i=1,2,…..,n (7.42)
(and taking x_{1} as A, x_{2} as D_{n50} and x_{3} as H_{s}), the failure surface in the normalised coordinate system is given by
\left(1+0.18z_{1} \right).1.6.\left(1.5+0.1z_{2}\right).2^{1/3} .1.59- \left(4.4+0.7z_{3}\right)=0or
0.864z_{1} +0.32 z_{2}+0.058z_{1} z_{2}-0.7 z_{3}+0.4=0.The probability of failure may then be determined using MVA or FDA methods.