Solution Manual Vol 7

# 19 sat gs e 325 w 100 210 mgm3 1e 155 from equation

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Unformatted text preview: s e 3:25 w 1:00 2:10 Mg=m3 1e 1:55 From Equation 1.14 w e 0:55 0:204 Gs 2:70 20:4% 1.3 Equations similar to 1.171.20 apply in the case of unit weights; thus, d sat Gs 2:72 9:8 15:7 kN=m3 w 1:70 1e Gs e 3:42 w 9:8 19:7 kN=m3 1e 1:70 Using Equation 1.21 0 Gs 1 1:72 w 9:8 9:9 kN=m3 1e 1:70 Using Equation 1.18a with Sr 0.75 Gs Sr e 3:245 w 9:8 18:7 kN=m3 1e 1:70 Basic characteristics of soils 3 Using Equation 1.13 w Sr e 0:75 0:70 0:193 Gs 2:72 19:3% The reader should not attempt to memorize the above equations. Figure 1.10(b) should be drawn and, from a knowledge of the definitions, relevant expressions can be written by inspection. 1.4 Volume of specimen 382 76 86 200 mm3 4 Mass 168:0 Bulk density 1:95 Mg=m3 Volume 86 200 103 Water content w From Equation 1.17 1 e Gs 1 w w 1:00 1:80 2:73 1:287 1:95 ; e 0:80 168:0 130:5 0:287 130:5 28:7% Using Equation 1.13 Sr wGs 0:287 2:73 0:98 0:80 e 98% 1.5 Using Equation 1.24 d 2:15 1:92 Mg=m3 1 w 1:12 From Equation 1.17 1 e Gs 1 w w 1:00 1:38 2:65 1:12 2:15 ; e 0:38 Using Equation 1.13 Sr wGs 0:12 2:65 0:837 0:38 e 83:7% 4 Basic characteristics of soils Using Equation 1.15 A e wGs 0:38 0:318 0:045 1:38 1e 4:5% The zero air voids dry density is given by Equation 1.25 d Gs 2:65 1:00 1:95 Mg=m3 w 1 0:135 2:65 1 wGs i.e. a dry density of 2.00 Mg/m3 would not be possible. 1.6 Mass (g) 2010 2092 2114 2100 2055 (Mg/m3) 2.010 2.092 2.114 2.100 2.055 w 0.128 0.145 0.156 0.168 0.192 d (Mg/m3) 1.782 1.827 1.829 1.798 1.724 d0 (Mg/m3) 1.990 1.925 1.884 1.843 1.765 d5 (Mg/m3) 1.890 1.829 1.790 1.751 1.676 d10 (Mg/m3) 1.791 1.733 1.696 1.658 1.588 In each case the bulk density () is equal to the mass of compacted soil divided by the volume of the mould. The corresponding value of dry density (d ) is obtained from Equation 1.24. The dry densitywater content curve is plotted, from which wopt 15% and dmax 1:83 Mg=m3 Figure Q1.6 Basic characteristics of soils 5 Equation 1.26, with A equal, in turn, to 0, 0.05 and 0.10, is used to calculate values of dry density (d0, d5, d10 respectively) for use in plotting the air content curves. The experimental values of w have been used in these calculations; however, any series of w values within the relevant range could be used. By inspection, the value of air content at maximum dry density is 3.5%. 1.7 From Equation 1.20 e G s w 1 d The maximum and minimum values of void ratio are given by emax emin Gs w 1 dmin Gs w 1 dmax From Equation 1.23 ID Gs w 1=dmin 1=d Gs w 1=dmin 1=dmax 1 dmin =d 1=dmin 1 dmin =dmax 1=dmin d dmin dmax dmax dmin d 1:72 1:54 1:81 1:81 1:54 1:72 0:70 70% Chapter 2 Seepage 2.1 The coefficient of permeability is determined from the equation k 2:3 where 0:0052 m2 ; l 0:2 m 4 A 0:12 m2 ; t1 3 602 s 4 h0 1:00 0:456 log log 0:35 h1 2:3 0:0052 0:2 0:456 ;k 4:9 108 m=s 0:12 3 602 a 2.2 The flow net is drawn in Figure Q2.2. In the flow net there are 3.7 flow channels and 11 equipotential drops, i.e. Nf 3.7 and Nd 11. The overall loss in total head is 4.00 m. The quantity of seepage...
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## This note was uploaded on 09/21/2009 for the course CVEN 3718 taught by Professor Dobroslavznidarcic during the Spring '08 term at Colorado.

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