Smith's Elements of Soil Mechanics. Ian Smith

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Smith's Elements of Soil Mechanics - Ian  Smith


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2.26 Transformed and natural ‘squares’. (a) Transformed. (b) Natural.

      (2.33)equation

       Solution:

      The flow net is shown in Fig. 2.27b. From it, we have Nf = 4.0 and Nd = 14.

equation Schematic illustration of seepage loss through dam (a) The problem. (b) Flow net.

      Determine the seepage loss through the dam.

       Solution:

      Transformed scale for x direction, images

      This means that, if the vertical scale is 1 : 500, then the horizontal scale is 0.63 : 500 or 1 : 794.

equation Schematic illustration of seepage loss through dam.

      A dam has the same details as in Example 2.8, except that there is no filter drain at the toe. Determine the seepage loss through the dam.

       Solution:

equation

      Hence Δa = 7.6 m

equation Schematic illustration of seepage loss through dam.

      2.15.3 Permeability of sedimentary deposits

      A sedimentary deposit may consist of several different soils and it is often necessary to determine the average values of permeability in two directions, one parallel to the bedding planes and the other at right angles to them.

       Let there be n layers of thicknesses H1, H2, H3, … Hn.

       Let the total thickness of the layers be H.

       Let k1, k2, k3, … kn be the respective coefficients of permeability for each individual layer.

       Let the average permeability for the whole deposit be kx for flow parallel to the bedding planes, and kz for flow perpendicular to this direction.

      Consider flow parallel to the bedding planes:

equation

      where A = total area and i = hydraulic gradient.

      This total flow must equal the sum of the flow through each layer, therefore:

equation equation

      hence

      (2.34)equation

      Considering flow perpendicular to the bedding planes:

equation

      Considering unit area:

equation

      Now

equation

      where h1, h2, h3, etc., are the respective head losses across each layer.

      Now

equation

      hence

      (2.35)equation

      Beneath the fine silt layer, there is a stratum of water‐bearing gravel


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