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Solution HW#2 - 1 Solution of Homework No.2 CEE 330 Fall...

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1 Solution of Homework No.2 – CEE 330 – Fall 2009-10 Generated by Hung Tran Date: September 30, 2009 Problem 1 : ( 10 points ) T T W V γ = ; s s w G γ γ = ; w s W w W = ; v s V e V = s s s W V γ = 1 . 1 s w w G e γ γ + = + 1 . 1 w s w T s s s s s T w s v s v T w s T s s s W W W W W W W W W V V V V V V V V V γ γ γ γ γ + + = = = = + + - Basic definition. Problem 2 : ( 15 points ) (2) (3) (5) 100% 100% 100% (3) (1) (4) w s M w M - = × = × = × - . e wGs = - fully saturated soil 1 . 1 s w w G e γ γ + = + G s = 2.7, γ w = 10 kN/m 3 Boring Tin ID Mass Dry Soil Mass Water Water Content Void Ratio Unit Weight Top Depth Bottom Depth Tin Mass Tin + Moist Soil Tin + Dry Soil M d M w w e γ M M g g g g g % kN/m 3 (1) (2) (3) (4) (5) G-01 0.61 0.91 2C 24.5 62.5 55 30.5 7.5 24.6% 0.66 20.2 G-01 1.22 1.83 12B 31.1 72.9 59.9 28.8 13 45.1% 1.22 17.7 G-01 1.83 2.44 C3 31.2 81.9 62.4 31.2 19.5 62.5% 1.69 16.3 G-01 2.44 2.51 6 31 63 57 26 6 23.1% 0.62 20.5 G-01 2.51 3.05 4 30.5 80 46.5 16 33.5 209.4% 5.65 12.6
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2 Problem 3 : ( 15 points ) Given data φ H V T M ws M d G s γ w mm mm m 3 g g Mg/m 3 63 158 4.93 x 10 -4 893 788 2.68 1 Phase diagram analysis Solid Water Air Vs = 2.94 Vw = 1.05 Va = 0.935 Vv = 1.985 V = 4.925 Ms = 0.788 kg Mw = 0.105 kg M = 0.893 kg M w = M ws – M d = 893 – 788 = 105 g. 893 788 100% 100% 13.3% 788 w s M w M - = × = × = 3 4 0.788 10 ( ) 2.94 10 2.68( / 3) s s
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