Solution HW#2

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 = ; w s W w W = ; v s V e V = s s s W V = 1 . 1 s w w G e + = + 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 + + = = = = + + - 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 + = + 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
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This note was uploaded on 03/03/2011 for the course CEE 330 taught by Professor Schneider during the Fall '10 term at Wisconsin.

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

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