Flow Channel-Soil Mechanics and Foundations-Lecture 16 Slides-Civil and Environmental Engineering, Slides of Soil Mechanics and Foundations

Download Flow Channel-Soil Mechanics and Foundations-Lecture 16 Slides-Civil and Environmental Engineering and more Slides Soil Mechanics and Foundations in PDF only on Docsity! CE 240 Soil Mechanics & Foundations Lecture 6.2 Seepage II (Das, Ch. 7) Class Outline • Seepage calculation from a flow net • Flow nets in anisotropic soil Here, I would like to call your attention for the definition of the Darcy’s law. If q=kiA, as the originally defined, then q should be in the dimension of volume/time. But here the equation (7.18) is written in a 2D fashion with q=kiL, so it is in volume/time/length. It can be interpreted as the hydraulic conductivity per lateral length in the direction perpendicular to the face of the paper (or the slide screen). Then, what is Nd? Id=1 Id=2 Id=3 Id=4 Id=5 Id=6 If =1 If =2 If =3 If =4 Although drawing square elements for a flow net is convenient, it is not always necessary. Alternatively, one can draw a rectangular mesh for a flow channel, as shown in Figure 7.7, provided that the width-to-length ratios for all the rectangular elements in the flow net are the same. In this case, Eq. (7.18) for rate of flow through the channel can be modified to Dean lo h,—h Aq = (2 )o, = (2 Jo = (A—**)o, oi (7.225 If b,/l, = boll, = ball, = +--+ =n (i.e., the elements are not square), Eqs. (7.20) and (7.21) can be modified to n = kH(| — a Aq ku( (72 Figure 7.7 Example 7.2, Figure 7.9 Sheet pile aceon | i 15 ft yy : | 5 ft ; Ground surface _Y_ ae : > scoot 30 ft | pee Impermeable layer — Example 7.2 A flow net for flow around a single row of sheet piles in a permeable soil layer is shown in Figure 7.9. Given that k, = k, = k = 5 x 10 ° emésec, determine a. How high (above the ground surface) the water will rise if piezometers are placed at points a, b, c, and d b. The rate of seepage through flow channel II per unit length (perpendicu- lar to the section shown) c. The total rate of seepage through the permeable layer per unit length Solution Parta From Figure 7.9, N; = 3 and N, = 6. The difference of head between the upstream and downstream sides = 10 ft. So, the loss of head for each drop = 10/6 = 1.667 ft. Point a is located on equipotential line 1, which means that the potential drop at ais 1 X 1.667 ft. So the water in the piezometer at a will rise to an elevation of (15 — 1.667) = 13.333 ft above the ground surface. Similarly, the piezometric level for b = (15 — 2 X 1.667) = 11.67 ft above the ground surface c = (15 — 5 X 1.667) = 6.67 ft above the ground surface d = (15 — 5 X 1.667) = 6.67 ft above the ground surface Example 7.2 (cont.) Flow nets for anisotropic soil 2 2 2 2 2 0 H V k h h k x z H V k k + 2 2 2 2 0 h h x z (7.26) Step 1: Adopt a vertical scale (that is, z axis) for drawing the cross section. Step 2: Adopt a horizontal scale (that is, x axis) such that horizontal scale = Vk,/k, X vertical scale. Step 3: With scales adopted as in Steps 1 and 2, plot the vertical section through the permeable layer parallel to the direction of flow. Step 4: Draw the flow net for the permeable layer on the section obtained from Step 3, with flow lines intersecting equipotential lines at right angles and the elements as approximate squares. The rate of seepage per unit length can be calculated by modifying Eq. (7.21) to (7.27) Example: Flow net for anisotropic soil Example: Flow net for anisotropic soil Fig. 4 Shows the dam drawn at its natural scale x z Impermeable bedrock L H1 H2 Impermeable dam Soil layer Z Example: Flow net for anisotropic soil Fig. 5 Shows the dam drawn to its transformed scale z Impermeable bedrock L/2 H1 H2 x Soil layer Z Example: Seepage under a dam h1 = 13.0 m h2 = 2.5 m kV = 10-6 m/s kH = 4 x10-6 m/s Nd = 14 Nf = 6 eq H Vk k k (13 2.5) 0.75 14 h m-6 -6 -6 eqk = (4×10 )×(10 ) = 2×10 m/s q =( ) ( . )2 10 0 756 6. / /1 5 10 3m s m thus / /9 10 3m s m. / /6 1 5 3 6s mq m