Presentation on theme: "Flow through Soils II (ch8). 2D flow Vel. vectors are confined to a single plane 1D flow vel vectors parallel 2D flow vel vectors not necessarily parallel."— Presentation transcript:
Flow through Soils II (ch8)
2D flow Vel. vectors are confined to a single plane 1D flow vel vectors parallel 2D flow vel vectors not necessarily parallel wall soil
Laplace’s Equation 1D flow - Darcy’s law 2D flow - Laplace’s equation x z dz dx L Rate of in vel in z-dir
Laplace’s Equation Laplace’s equation “represents energy loss through any resistive medium” Assumptions Darcy is valid S = 100% is of constant volume Isotropic k x = k z Homogeneous k = same throughout
Remember main point = get Q Examples Q beneath a damQ in excavations Solving Mathematics Graphical Flownets
Flowline wall impervious Equipotential line Flowpath – “channel” between two flowlines Equipotential line – along any eqpl, the total head is the same
Flownets 1 st eqpl: starts in inlet last eqpl: ends at outlet impervious 1 st eqpl Of soil last eqpl
Flownets vel Flowline node eqpl Requirements: Perpendicular crossings at nodes Maintain “squareosity”
Flownets X Mistake – redraw!
Flownets - Example 10m 14m 2m L = 100 m k =.1 cm/s 2m A B
Flownets - Example h = 14 – 2 = 12 m h at first eq = 24 m Q = h L k (N FP /N ED ) Q = 0.3 m 3 /sec N FP = 3 N ED = 12 h at last eq = 12 m
Flownets - Example h at first eq = 24 m Head loss per drop = h/N ED = 12 / 12 = 1 m h e at A = 8 m h at A = 24 – 8 = 16 m h p at A = h – h e = 16 – 8 = 8 m u A = (h p )( w ) = 78.5 kPa
Flownets - Example h at first eq = 24 m Head loss per drop = h/N ED = 12 / 12 = 1 m h e at B = 10 m h at B = 24 – 9 = 15 m h p at B = h – h e = 15 – 10 = 5 m u B = (h p )( w ) = 49 kPa
Uplift pressures Concrete dam u uplift Concrete dam = W/A If u uplift ~ , structure can float away!
Uplift Pressures Case 1 (no flow) u = (h p )( w ) = u hydrostatic Draw flownet Compute h at several points along the structure’s base Determine h p at these points Case 2 (with flow) Find u at these points by u = (h p )( w )
Filters Problem - water pressure - soil filter - Soil particles migrate out Sol. - drains Problem Sol.
Filters 2 filter purposes Allow adequate drainage Soil retained is called base soil Disallow particle migration Filter soil
Filters 2 filter failure types Clogging: base clogs filter pores – (k decreases) Piping: base migrates through filter
D d max D Simple Cubic Packing D/d max = 2.4 Tetrahedral Packing D/d max = 6.5 Interstice Sizes Filters - principles
Arching Bridging Filters - principles
To disallow particle migration To allow adequate drainage (maintain proper k) Terzaghi’s Filter criteria
See table 8.2 in book… Example: Sherard’s Filter criteria
Example (see notes (pad (lined sheet))… Example: Filter selection