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Statistics Help!

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Assignment No.4. Harbour Breakwater Evaluation.

    Consider a harbour breakwater constructed with massive concrete tanks filled with sand. It is necessary to evaluate the risk that the breakwater will slide under the pressure of a large wave during a major storm. Stability against sliding exists when the ratio of the resultant horizontal force Rh to the resultant of the vertical force Rv does not exceed the coefficient of friction, X1.

    The resultant of the vertical forces Rv is given by the algebraic sum of the weight of the tank reduced for buoyancy, X2, and the vertical component of dynamic uplift pressure due to the breaking wave, Fv.

    Rv = X2 - Fv

    In turn Fv is proportional to the height of the wave, Hb, when the slope of the sea bottom is known. A simplification of the shoaling effect makes the wave height proportional to the deepwater value, X4.

    Fv = a1X3X4

    where a1 is the constant of proportionality and X3 is an additional variate to represent the uncertainties caused by the above simplification.

    The resultant Rh of the horizontal forces depends on the balance between the static and dynamic pressure components and under a simplifying hypothesis on the depth of the breakwater can be taken to be a quadratic function of the wave height Hb which in turn is proportional to X4

    Rh = X3(a3X4+a2X24)

    The constants a1 to a3 depend on the detailed geometry of the breakwater system.
    The following data refers to conditions in La Spezia harbour Italy. From the sea bottom profile and the geometry of the breakwater wall at La Spezia, engineers have estimated that

    a1 = 70 , a2 = 17 m/kN and a3 = 145.

    The variables X1 to X4 are independent variates (because for example the friction coefficient along a unit width of the vertical breakwater wall at La Spezia is not known precisely) with the following means and standard deviations

    mu1 = 0.64, mu2 = 3400 kN/m, mu3 = 1 and mu4 = 5.461 m .

    sigma1 = 0.096, sigma2 = 170 kN/m, sigma3 = 0.2, sigma4 = 1.081 m .

    Assume that Rh and RvX1 follow weibull distributions.

    -- Find the mean value for Rv

    -- Find the standard deviation for Rv

    and more questions which i havent got to yet

    2. Relevant equations



    3. The attempt at a solution

    Right well i think i've done the first bit, just using the mean values for the x's and the actual values for a1 to get 1931.347

    by first doing

    Fv = a1*X3*X4 and then Rv = X2 - Fv

    Do i just do the same for the standard deviation, by using the standard deviations in the equations???? the standard deviation question is worth a lot more marks.... :S
     
  2. jcsd
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