Smith's Elements of Soil Mechanics. Ian Smith
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2.26 Transformed and natural ‘squares’. (a) Transformed. (b) Natural.
and, in Fig. 2.26b:
(2.33)
Example 2.8 Seepage loss through dam (i)
The cross‐section of an earth dam is shown in Fig. 2.27a. Assuming that the water level remains constant at 35 m, determine the seepage loss through the dam. The width of the dam is 300 m, and the soil is isotropic with k = 5.8 × 10−7 m/s.
Solution:
The flow net is shown in Fig. 2.27b. From it, we have Nf = 4.0 and Nd = 14.
Fig. 2.27 Example 2.8. (a) The problem. (b) Flow net.
Example 2.9 Seepage loss through dam (ii)
A dam has the same details as in Example 2.8 except that the soil is anisotropic with kx = 5.8 × 10−7 m/s and kz = 2.3 × 10−7 m/s.
Determine the seepage loss through the dam.
Solution:
Transformed scale for x direction,
This means that, if the vertical scale is 1 : 500, then the horizontal scale is 0.63 : 500 or 1 : 794.
The flow net is shown in Fig. 2.28. From the flow net, Nf = 5.0 and Nd = 14.
Example 2.10 Seepage loss through dam (iii)
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:
The flow net is shown in Fig. 2.29 and from it we see Nf = 4.0 and Nd = 18 (average). From the flow net, it is also seen that a + Δa = 22.4 m. Now α = 45°, and hence:
Hence Δa = 7.6 m
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:
where A = total area and i = hydraulic gradient.
This total flow must equal the sum of the flow through each layer, therefore:
Considering unit width of soil:
hence
(2.34)
Considering flow perpendicular to the bedding planes:
Considering unit area:
Now
where h1, h2, h3, etc., are the respective head losses across each layer.
Now
hence
(2.35)
Example 2.11 Quantity of flow
A three‐layered soil system consisting of fine sand, coarse silt, and fine silt in horizontal layers is shown in Fig. 2.30.
Beneath the fine silt layer, there is a stratum of water‐bearing gravel