49_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

# 49_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - e v...

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p33 Consolidation Rate Example Pc = 7 cm H = 10m (clay) = 2 k = 1 x1Q·7 cm/s a, = 3.06x1Q" m'/kN eo = 1.50 u constant with depth (Case 1) Yw = 9.81 kN/m' a) How much settlement after 1 yr? b) How long for 5 cm of settlement? a) Cv; then calculate T and find U = Cv T k(1+e o ) 1x10- 7 em/s 2 (1+1.50) = avrw 3.06x10- 5 em 2 /kN (9.81 kN/m 3 ) ),n 'l( 10"j ""l)J' ~.~j"".<'l),11 (.4 \'/J<--7 cvt = (H/N)' 8.33x10-4 em 2 /s (6051 min)(60min/hr)(24hr 1 d)(365dl yr)(1yr) = (1 Om/2)2 (1 00em/1m)2 = O,loS' ('''''' ,... ~t(rI1IM,d' ill d".,+: r~r T: O.fUSj 0-= 1,. >;t. 50 i" I ye.-"", Pc.:. v.lLs (1 ".~~ .ls.J""e-, b) calculate U, find T, then calculate t U = 5em /7em = 71.4% Go to find T= 0, 'I ';tIl T(H/N)2 t= --'----''-
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Unformatted text preview: e v 0.424(1000/2)2 _' 1'1 n d~, S ";. L (0.000833)(60 x 60 x 24) -7 I yeo'S • Superposition works for the Time Factor Table (because the 1-D equation is linear) To determine the net consolidation, weight the % consolidation calculated for each shape by the area of the initial pore water pressure for that shape. Example: under center of a footing, single drainage at layer top I y////////////./. ///////7//////. //. z . a = b Case1 c ase where Uj = % consolidation at time t A, = area of initial pressure distribution...
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