Solution Manual For Introduction To Geotechnical Engineering, An, 2nd Edition

Clay Minerals and Rock Classification Chapter 4 CHAPTER 4 CLAY MINERALS, SOIL AND ROCK STRUCTURES, AND ROCK CLASSIFICATION 4-1. Calculate the specific surface of a cube (a) 10 mm, (b) 1 mm, (c) and (d) 1 nm on a side. Calculate the specific surface in terms of both areas and m2/kg. Assume for the latter case that ρs = 2.65 Mg/m3. SOLUTION: Solve using Eq. 4.2: specific surface = surface area unit volume 6(100 mm2 ) = 60 mm 10 mm3 (60 mm )(1000 mm m ) 2 2 and specific surface = = 22,641m Mg = 22.6 m kg Mg 2.65 3 (a) specific surface = m 2 6(1mm ) = 6 mm 1mm3 (6 mm )(1000 mm m ) 2 2 and specific surface = = 2264 m Mg = 2.26 m kg Mg 2.65 3 (b) specific surface = m 6(1 μm ) = 6 μm = 6000 mm 1 μm3 (6000 mm )(1000 mm m ) 2 2 and specific surface = = 2,264,151m Mg = 2264 m kg Mg 2.65 3 2 (c) specific surface = m 2 6(1nm ) = 6 nm = 6,000,000 mm 1nm3 (6,000,000 mm )(1000 mm m ) 2 2 and specific surface = = 2,264,151,943 m Mg = 2,264,151m kg 2.65 Mg 3 (d) specific surface = m © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Clay Minerals and Rock Classification Chapter 4 4-2. Calculate the specific surface of (a) tennis balls, (b) ping pong balls, (c) ball bearings 1.5 mm in diameter, and (d) fly ash with approximately spherical particles 60 υm in diameter. SOLUTION: Solve using Eq. 4.2: specific surface = surface area sphere surface area = 4πr 2 ; sphere volume = unit volume 4 3 πr 3 (3)(4)π(67 mm)2 2 (a) tennis ball (Dia. = 67 mm): specific surface = = 0.089 mm 3 67 mm (4)π 2 (b) ping pong ball (Dia. = 40 mm): specific surface = (c) ball bearings (Dia. = 1.5 mm): specific surface = (d) fly ash (Dia. = 60 μm): specific surface = 3 60 ( 2) ( ) 3 40 ( 2) = 0.15 mm 3 1.5 = 4.0 mm ( 2) = 0.10 μm = 1 × 105 mm = 100,000 mm © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Clay Minerals and Rock Classification Chapter 4 4-5. Verify that the maximum and minimum void ratios for perfect spheres given in Table 4.5 are reasonable. SOLUTION: Three-dimensional particle arrangement of equal spheres has been studied in depth by mathematicians, statisticians, and materials scientists since the 1600s. A quick internet search on packing of equal spheres will reveal numerous mathematical theories and approaches for estimating the densest and loosest possible packing. In general, the loosest arrangement of equal spheres yields a void fraction of about 0.48, regardless of sphere size. The densest possible packing of equal-size spheres yields a solids volume of about: Vs = π = 0.7405. 18 (These values are approximate – there is not a unified consensus in the literature.) Loosest packing For Vt = 1.0, Vv = 0.48, and Vs = 1 − 0.48 = 0.52 thus; nmax = Vv × 100 = Vt emax = Vv 0.48 = = 0.92 Vs 0.52 0.48 × 100 = 48 % 1.0 Densest packing Vs = 0.7405, Vv = 1.0 − 0.7405 = 0.2595 thus; 0.2595 × 100 = 26% 1.0 0.2595 emin = = 0.35 0.7405 nmin = These agree with the values of emax and emin in Table 4.5 for equal spheres. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Clay Minerals and Rock Classification Chapter 4 4-7. A specially processed clay has particles that are 500 nm thick and 10,000 nm x 10,000 nm wide. The specific gravity of solids is 2.80. The particles lie perfectly parallel with an edge-to-edge spacing of 400 nm (i.e., they look like thin bricks stacked perfectly parallel). (a) Initially, the cation valence in the double layer is +1, resulting in a face-to-face spacing of 1500 nm. How many particles per cm3 will there be at this spacing? What are the void ratio and water content, assuming that the soil is at 100% saturation? (b) Another sample of the clay is mixed such that the cation valence is +2. What are the new void ratio and water content under these conditions? SOLUTION: Assume symmetrical edge-to-edge spacing of 400 nm and only include whole particles. (a) no. of particles in horizontal plane along one edge = (1× 107 ) − (2)(10,000) = 959.6 10,000 + 400 truncate and add 2 for edges = 959 + 2 = 961 no. of particles in plan = 961× 961 = 923,521 no. of particles in vertical plane along one edge = (1× 107 ) − (2)(500) = 4999.5 500 + 1500 truncate and add 2 for edges = 4999 + 2 = 5001 total number of particles = 923,521× 5001 = 4,618,528,521 = 4618.53 × 106 volume of one particle = 10,000 × 10,000 × 500 = 5.0 × 1010 nm3 = 5 × 10 −11 cm3 Vs = 4,618,528,521× (5.0 × 1010 ) = 2.30926 × 1020 nm3 = 0.23093 cm3 For S = 100%: Vw = Vv = Vt − Vs = 1 − 0.23093 = 0.7691 cm3 ; Ms = Gs × Vs × ρs = (2.80)(0.23093)(1 g cm ) = 0.6466 g; 3 w= e= Vv 0.7691 = = 3.33 Vs 0.23093 Mw = Vw × ρ w = (0.7691)(1 g cm ) = 0.7691 g 3 Mw 0.7691 × 100% = × 100% = 118.9 = 119% Ms 0.6466 (b) assume the-face-to-face spacing doubles for a cation valence of +2 no. of particles in horizontal plane along one edge = (1× 107 ) − (2)(10,000) = 959.6 10,000 + 400 truncate and add 2 for edges = 959 + 2 = 961 no. of particles in plan = 961× 961 = 923,521 (same as part a) no. of particles in vertical plane along one edge = (1× 107 ) − (2)(500) = 2856.9 500 + 3000 truncate and add 2 for edges = 2856 + 2 = 2858 total number of particles = 923,521× 2858 = 2,639,423,018 = 2639.423 × 106 volume of one particle = 10,000 × 10,000 × 500 = 5.0 × 1010 nm3 = 5 × 10 −11 cm3 Vs = 2,639,423,018 × (5.0 × 1010 ) = 1.31971× 1020 nm3 = 0.13197 cm3 For S = 100%: Vw = Vv = Vt − Vs = 1 − 0.13197 = 0.868 cm3 ; Ms = Gs × Vs × ρs = (2.80)(0.13197)(1 g cm ) = 0.3695 g; 3 w= e= Vv 0.868 = = 6.58 Vs 0.13197 Mw = Vw × ρ w = (0.868)(1 g cm ) = 0.868 g 3 Mw 0.868 × 100% = × 100% = 234.9 = 235% Ms 0.3695 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Clay Minerals and Rock Classification Chapter 4 4-15. Three sections of rock core are shown in Fig. 4.32. The rock comes from near Cumberland, RI, and is called Corbormite (Capt. James T. Kirk, personal communication, 2007). The length of the first (top) run is 56 in. and the computed RQD is 82%. For the second run (middle), a length of 60 in. was recovered and the RQD is 100%. Finally, the third run (bottom) is also 5 ft long and the RQD is 95%.Verify that the calculated RQD values for the top and bottom runs are correct. SOLUTION: RQD = ∑ Length of sound pieces > 4 in Total core run length Determine numerator values by closely examing photographs in Fig. 4.32 7 + 9 + 4 + 8.5 + 8 + 9 45.5 × 100 = × 100 = 81.3% 56 56 Thus, the given value of 82% appears reasonable. (a) Top run RQD = 7.5 + 8.5 + 7 + 11 34 × 100 = × 100 = 57% 60 60 The RQD value of 57% assumes all joints are natural (i.e., no mechanical breaks in this core run). Even if all the breaks and joints shown in Fig. 4.32 are mechanical, the reported value of 95% is still too high based on the percent recovery visible in the photo. (b) Bottom run RQD = © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Clay Minerals and Rock Classification Chapter 4 4-16. In one core run of 1500 mm selected from cores obtained during drilling for a bridge foundation in hard limestone, the following core recovery information was obtained: Determine (a) the percent core recovery, and (b) the RQD. Based on this RQD, what is the rock quality? SOLUTION: Total length of rock recovered × 100 Total core run length 250 + 50 + 50 + 75 + 100 + 125 + 75 + 100 + 150 + 100 + 50 + 125 1250 CR = = × 100 = 83.3% 1500 1500 (a) Core Recovery, CR = (b) RQD = ∑ Length of sound pieces > 100 mm × 100 Total core run length 150 + 100 + 125 + 100 + 150 + 100 + 125 850 RQD = × 100 = × 100 = 56.7% 1500 1500 Based on the Rock Mass Classifications shown in Table 4.13, the limestone rock quality would be deemed fair. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.