Solution Manual for Foundation Design: Principles and Practices, 3rd Edition

Chap. 3 3.1 Soil Mechanics Explain the difference between moisture content and degree of saturation. Solution Moisture content of a soil is the ratio of the weight of its water to weight of its solids, whereas the degree of saturation is the ratio of the volume of water to volume of voids. The degree of saturation can range from 0 to 100%, whereas the moisture content can be larger than 100%. Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-1 © 2016 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. Chap. 3 3.2 Soil Mechanics A certain saturated sand (S = 100%) has a moisture content of 25.1% and a specific gravity of solids of 2.68. It also has a maximum index void ratio of 0.84 and a minimum index void ratio of 0.33. Compute its relative density and classify its consistency. Solution First we must compute the void ratio e using the given specific gravity and moisture content. The void ratio calculated using specific gravity and moisture content is = e wGs 25.1% × 2.68 = = 0.67 100% S Using Equation 3.1, the relative density is = D r emax − e 0.84 − 0.67 × 100% = ×100% = 33% 0.84 − 0.33 emax − emin Look up the consistency in Table 3.3 based on the calculated relative density of 33%: the soil is classified as loose. Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-2 © 2016 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. Chap. 3 3.3 Soil Mechanics Consider a soil that is being placed as a fill and compacted using a sheepsfoot roller (a piece of construction equipment). Will the action of the roller change the void ratio of the soil? Explain. Solution Yes, whenever a soil is compacted, the volume of the soil decreases. Since the volume of the solids does not change with compaction, the volume of the voids decreases; and since the void ratio is the ratio of the volume of voids to the volume of solids, the void ratio decreases as a result. Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-3 © 2016 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. Chap. 3 3.4 Soil Mechanics A sample of soil has a volume of 0.45 ft3 and a weight of 53.3 lb. After being dried in an oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven. Solution First we compute the moisture content of the sample w= W − Ws 53.3 − 45.1 ×100% = ×100% = 18% 45.1 Ws Then we compute the degree of saturation S= w γw 1 − γ d Gs × 100%= 0.18 × 100%= 71% 62.4 1 − ( 45.1 0.45) 2.70 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-4 © 2016 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. Chap. 3 3.5 Soil Mechanics A site is underlain by a soil that has a unit weight of 18.7 kN/m3 above the groundwater table and 19.9 kN/m3 below. The groundwater table is located at a depth of 3.5 m below the ground surface. Compute the total vertical stress, pore water pressure, and effective vertical stress at the following depths below the ground surface: a. 2.2 m b. 4.0 m c. 6.0 m Solution Using Equation 3.3, the total vertical stresses computed at the given depths are kN/m3 )(2.2 m) 41.1 kN/m 2 = = σ z (z = 2.2 m) (18.7 (18.7 kN/m3 )(3.5 m) + (19.9 kN/m3 )(4.0= m − 3.5 m) 75.4 kN/m 2 σ= z (z = 4.0 m) (18.7 kN/m3 )(3.5 m) + (19.9 kN/m3 )(6.0= m − 3.5 m) 115 kN/m 2 σ= z (z = 6.0 m) The pore water pressures computed at the given depths are = = u (z = 2.2 m) (9.81 kN/m3 )(0 m) 0.0 kN/m 2 u= (z = 4.0 m) (9.81 kN/m3 )(4.0= m − 3.5 m) 4.91 kN/m 2 u= (z = 6.0 m) (9.81 kN/m3 )(6.0= m − 3.5 m) 24.5 kN/m 2 Using Equation 3.5, the effective vertical stresses computed at the given depths are σ z ‘ (z = 2.2 m) = 41.1 kN/m 2 − 0.0 kN/m 2 = 41.1 kN/m 2 σ z ‘ (z = 4.0 m) = 75.4 kN/m 2 − 4.9 kN/m 2 = 70.5 kN/m 2 σ z ‘ (z = 6.0 m) =115 kN/m 2 − 24.5 kN/m 2 =90.5 kN/m 2 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-5 © 2016 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. Chap. 3 3.6 Soil Mechanics The subsurface profile at a certain site is shown in Figure 3.20. Compute u, σx, σz, σʹx, and σʹz at Point A. Solution u = γ w zw = (62.4 lb/ft 3 )(11 ft) = 686 lb/ft 2 σ z = ∑γ H = (120 lb/ft 3 )(12 ft) + (117 lb/ft 3 )(10 ft) + (121 lb/ft 3 )(11 ft) = 3940 lb/ft 2 σ= σz −u z ‘ = 3940 lb/ft 2 − 686 lb/ft 2 = 3254 lb/ft 2 σ x ‘ = Kσ z ‘ = (0.70)(3254 lb/ft 2 ) = 2278 lb/ft 2 σ= σ x ‘+ u x = 2278 lb/ft 2 + 686 lb/ft 2 = 2964 lb/ft 2 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-6 © 2016 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. Chap. 3 3.7 Soil Mechanics The vertical load of 300 kN is applied to a 1.5 m × 1.5 m area at the ground surface that is level. a. Compute the induced vertical stress, Δσz, at a point 2.0 m below the corner of this square loaded area. b. Compute the induced vertical stress, Δσz, at a point 2.0 m below the center of this square loaded area. Solution a. This solution uses the Newman solution to the Boussinesq’s Method to compute the induced stress; there are many other methods to solve this problem. q= P = A 300 kN = 133 kPa 2 (1.5 m ) B 2 + L2 + z 2f = 1.52 + 1.52 + 2.02 = 8.5 B 2 L2 (1.52 )(1.52 ) = = 1.26 z 2f 2.02 B 2 + L2 + z 2f > B 2 L2 therefore, use Equation 3.12: z 2f 2 2 2 2 2 2  1  2 BLz f B + L + z f   B + L + 2 z f  Is =   4π  z 2f ( B 2 + L2 + z 2f ) + B 2 L2   B 2 + L2 + z 2f    + sin Is = −1   z 2f ( B 2 + L2 + z 2f ) + B 2 L2   2 BLz f B 2 + L2 + z 2f 1  2(1.5)(1.5)(2.0) 8.5   (1.5) 2 + (1.5) 2 + 2(2.0) 2     8.5 4π  (2.0) 2 ( 8.5 ) + (1.5) 2 (1.5) 2    + sin −1 2(1.5)(1.5)(2.0) 8.5   (2.0) 2 ( 8.5 ) + (1.5) 2 (1.5) 2  = 0.137 ∆s z = Is q = (0.137)(133 kPa) = 18.2 kPa Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-7 © 2016 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. Chap. 3 Soil Mechanics b. This solution uses the chart method to compute the induced stress; this problem can also be solved by other methods. q = 133 kPa, from part a. xf 0m = = 0 B 1.5 m z f 2.0 m = = 1.33 B 1.5 m Iσ = 0.23, from Figure 3.6 Iσ q ∆σ z = = (0.23)(133 kPa) = 30.6 kPa Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-8 © 2016 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. Chap. 3 3.8 Soil Mechanics A vertical load of 20 k is applied to a 6.0 ft × 4.0 ft area at the ground surface that is level. a. Compute the induced vertical stress, Δσz, at a point 6.0 ft below the corner of this rectangular loaded area. b. Compute the induced vertical stress, Δσz, at a point 6.0 ft below the center of this rectangular loaded area. Solution a. This solution uses Figure 3.8 for a chart solution; there are many other methods to solve this problem. P = A q= 20 k = 833 lb/ft 2 ( 6.0 ft )( 4.0 ft ) B 6.0 ft = = 1.0 z 6.0 ft L 4.0 ft n= = = 0.67 z 6.0 ft Iσ = 0.15, from Figure 3.8 m= Iσ q ∆σ z = = (0.15)(833 lb/ft 2 ) = 125 lb/ft 2 b. This solution uses Figure 3.8 for a chart solution; there are many other methods to solve this problem. Separate the loaded area into four equal areas, and then use superposition to compute the induced stress at below the center of the area. q = 833 lb/ft 2 B 3.0 ft m= = = 0.50 z 6.0 ft L 2.0 ft n= = = 0.33 z 6.0 ft Iσ = 0.062, from Figure 3.8 Iσ q ∆σ z = = (4)(0.062)(833 lb/ft 2 ) = 207 lb/ft 2 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-9 © 2016 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. Chap. 3 3.9 Soil Mechanics A 3 m × 3 m footing is to be built on the surface of a 15 m thick layer of unsaturated sand. The sand is underlain by a very dense gravel layer. The water table is at a great depth. The sand is relative uniform and in situ testing indicates it has a constrained modulus of 10 MPa. The footing load is 200 kN. Compute the settlement under the center of the footing. Solution The sand was separated into 8 layers each 2 m thick with the exception of layer no. 8 which was 1 m thick. Using Equation 3.14, change in the stress at the center of each layer was computed. Using Equation 3.20, the strain of each layer was computed. The results are shown in the following table. Layer No. Depth to layer midpoint (m) Layer Thickness (cm) ∆σ ʹ (kPa) M (kPa) εv Layer compression (cm) 1 1 200 19.43 10,000 0.19% 0.38 2 3 200 7.22 10,000 0.07% 0.14 3 5 200 3.13 10,000 0.03% 0.06 4 7 200 1.69 10,000 0.02% 0.04 5 9 200 1.05 10,000 0.01% 0.02 6 11 200 0.71 10,000 0.01% 0.02 7 13 200 0.51 10,000 0.01% 0.02 8 14.5 100 0.41 10,000 0.00% 0.00 δ = 0.68 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-10 © 2016 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. Chap. 3 Soil Mechanics 3.10 A 3-foot square footing carries a sustained load of 10 k. It is placed on the surface of a 30 foot thick saturated overconsolidated clay underlain by dense sand. Based on laboratory tests, the clay can be adequately modeled using the e-log-p method. The laboratory tests provide the following compressibility information for the clay: γ = 123 lb/ft3 Cc = 0.06 1 + e0 Cr = 0.002 1 + e0 σʹm = 900 lb/ft2 The groundwater table is located at the ground surface. Compute the settlement of the footing. Solution From Example 3.5 we know it should be adequate to compute the compressibility to only a depth of 6 feet, even though the compressible layer is much thicker. The clay later will be separated into 3 layers, each 2 feet thick. The compression of the first layer is computed in the following fashion. Compute the initial vertical stress considering the groundwater table at the ground surface = σ z′0 1 ft (123 lb/ft 2 ) − 1 ft ( 62.4 lb/ft 2 ) = 60.6 lb/ft 2 Using Equation 3.14 to compute the induced stress: 2.60 1   2   Iσ = 1 −   3    1 +  2(1)       = 0.953 10 k = q = 1,111 lb/ft 2 (3 ft)(3 ft) ∆σ z = Iσ q = (0.953)(1,111 lb/ft 2 ) = 1, 059 lb/ft 2 2 ′ σ z′0 + ∆σ= ′ 60.6 lb/ft 2 + 1,059 lb/ft= σ= 1,120 lb/ft 2 zf z σ c′ = σ z′0 + σ m′ = 60.6 lb/ft 2 + 900 lb/ft 2 = 960 lb/ft 2 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-11 © 2016 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. Chap. 3 Soil Mechanics Since σʹzf > σʹc , this is case OC-II and the layer compression is computed using Equation 3.31:   960   1,120   = = δ 24 in ( 0.002 ) log   + ( 0.06 ) log    0.15 in  60.6   960    This process is repeated for the remaining layers. The following table shows the results of the calculations. Computed at layer midpoint Layer No. 1 Depth to layer midpoint (ft) 1 Layer Thickness (ft) 2 σz0ʹ (lb/ft2) 60.6 ∆σz (lb/ft2) 1,059 σzfʹ (lb/ft2) 1,120 σcʹ (lb/ft2) 961 Case OC-II 2 3 2 182 489 671 1,082 3 5 2 303 223 526 1,203 Cc 1 + e0 Cr 1 + e0 0.06 0.002 δ (in) 0.15 OC-I 0.06 0.002 0.03 OC-I 0.06 0.002 0.01 δ =0.19 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-12 © 2016 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. Chap. 3 Soil Mechanics 3.11 A 2m thick fill is to be placed on the soil shown in Figure 3.21. Once it is compacted this fill will have a unit weight of 19.5 kN/m3. Compute the ultimate consolidation settlement. Solution Note that the sand layer is dense, its consolidation is negligible compared to that of the stiff clay layer, and that overconsolidation was assumed for the stiff clay. Also note that since the fill is areal, the influence factor is 1.0. Using methods similar to those used to solve Problem 3.10 and Example 3.5, the following table shows the results of the calculations. Computed at layer midpoint σz0ʹ ∆σz σzfʹ Layer Thickness (cm) (kPa) (kPa) (kPa) 100 35.9 39 75 0.046 δ (cm) 1.5 83 0.046 1.3 39 92 0.046 1.1 61.0 39 100 0.046 1.0 69.4 39 108 0.046 0.9 100 77.8 39 117 0.046 0.8 8.4 100 86.2 39 125 0.046 0.7 9.4 100 94.6 39 134 0.046 0.7 Layer No. 1 Depth to layer midpoint (m) 2.4 2 3.4 100 44.2 39 3 4.4 100 52.6 4 5.4 100 5 6.4 100 6 7.4 7 8 Cr 1 + e0 δ =8.0 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-13 © 2016 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. Chap. 3 Soil Mechanics 3.12 Estimate the effective friction angle of the following soils: a. Silty sand with dry unit weight of 100 lb/ft 3 . b. Poorly-graded gravel with relative density of 70%. c. Very dense well-graded sand. Solution Using Figure 3.14, estimate the effective friction angle of the given soils Soil Silty sand with dry unit weight of 100 lb/ft 3 Poorly-graded gravel with relative density of 70% Very dense well-graded sand Estimated effective friction angle (degrees) 32-37 35-37 35-40 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-14 © 2016 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. Chap. 3 Soil Mechanics 3.13 Explain the difference between the drained condition and the undrained condition. Solution • Drained condition is a condition where excess pore water does not exist under change in loading. In other words, the pore water pressure is equal to the hydrostatic pore water pressure. Drained conditions can develop in one of two ways. First, if the hydraulic conductivity of a saturated soil is sufficiently high and that loading rate is sufficiently low, then the excess pore water pressure can dissipate quickly. Secondly, drained conditions can also develop given sufficient time has elapsed since the end of loading, any excess pore water pressure will have been dissipated. • Undrained condition is where excess pore water pressure is generated in the saturated soil, or when the pore water pressure is not equal to the hydrostatic pore water pressure. It develops when water cannot flow quickly from the loaded soil. Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-15 © 2016 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. Chap. 3 Soil Mechanics 3.14 A soil has cʹ = 5 kPa and ϕʹ = 32°. The effective stress at a point in the soil is 125 kPa. Compute the shear strength normal to this stress at this point. Solution Using Equation 3.32, compute the shear strength = s 5 kPa + (125 kPa)( tan 32 = °) 83 kPa Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-16 © 2016 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. Chap. 3 Soil Mechanics 3.15 A footing with an embedment of 2 m is embedded in a sand with a unit weight of 125 lb/ft3 and a ϕʹ of 36°. If the footing is subjected to a horizontal load that causes it to move horizontally, compute the total active and passive resultant forces acting on the footing. Solution Using Equation 3.38, compute the active resultant force (125 lb/ft 3 ) (2m × 3.28 ft/m) 2 ( tan 2 (45° − 36° / 2)) 2 = 698 lb/ft Pa = Using Equation 3.40, compute the passive resultant force (125 lb/ft 3 ) (2m × 3.28 ft/m) 2 ( tan 2 (45° + 36° / 2)) Pp = 2 = 10,360 lb/ft Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-17 © 2016 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. Chap. 3 Soil Mechanics 3.16 The soil profile at a certain site is as follows: cʹ (lb/ft2) 0-12 γ (lb/ft3) 12-20 126 200 20 20-32 129 0 32 Depth (ft) 119 ϕʹ (degree) Su(lb/ft2) 1000 The groundwater table is at a depth of 15 ft. Develop plots of pore water pressure, total vertical stress, effective total stress, and shear strength on a horizontal plane vs. depth. All four of these plots should be superimposed on the same diagram with the parameters on the horizontal axis (increasing to the right) and depth on the vertical axis (increasing downward). Hint: Because the cohesion and friction angle suddenly change at the strata boundaries, the shear strength also may change suddenly at these depths. Solution 4500 4000 3500 3000 2500 2000 1500 1000 500 0 (lb/ft2) 0.0 2.0 4.0 Pore Water Pressure 6.0 8.0 10.0 Total Vertical Stress Depth (ft bgs) 12.0 14.0 Effective Vertical Stress 16.0 18.0 Shear Strength 20.0 22.0 24.0 26.0 28.0 30.0 32.0 34.0 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-18 © 2016 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. Chap. 3 Soil Mechanics 3.17 Repeat problem 3.16 using the following data: cʹ (kPa) 0-5 γ (kN/m3) 5-12 20.0 8.4 21 12-20 20.5 0 35 Depth (m) 18.5 ϕʹ (degree) Su(kPa) 50 The groundwater table is at a depth of 7 m. Solution 450 400 350 300 250 200 150 100 50 0 kPa 0.0 2.0 Pore Water Pressure 4.0 Depth ( m bgs) 6.0 Total Vertical Stress 8.0 Effective Vertical Stress 10.0 Shear Strength 12.0 14.0 16.0 18.0 20.0 22.0 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-19 © 2016 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. Chap. 3 Soil Mechanics 3.18 A 9 ft thick fill is to be placed on the soil shown in Figure 3.22. Once it is compacted this fill will have a unit weight of 122 lb/ft3. Compute the ultimate consolidation settlement caused by consolidation of the underlying clay. Solution Using methods similar to those used to solve Problem 3.10 and Example 3.5, the following table shows the results of the calculations. Computed at center of layer Layer No. 1 Depth to layer midpoint (ft) 18 Layer Thickness (ft) 6 σz0ʹ (lb/ft2) 1,658.6 ∆σz (lb/ft2) 1,098 σzfʹ (lb/ft2) 2,757 σcʹ (lb/ft2) 2,757 Case NC 2 24 6 1,830.2 1,098 2,928 2,928 3 30 6 2,001.8 1,098 3,100 3,100 4 36 6 2,173.4 1,098 3,271 5 50 6 2,792.8 1,098 6 56 6 3,048.4 7 62 6 3,304 Cc 1 + e0 Cr 1 + e0 0.14 0.06 δ (in) 2.22 NC 0.14 0.06 2.06 NC 0.14 0.06 1.91 3,271 NC 0.14 0.06 1.79 3,891 5,774 OC-I 0.12 0.05 0.52 1,098 4,146 6,030 OC-I 0.12 0.05 0.48 1,098 4,402 6,285 OC-I 0.12 0.05 0.45 δ =9.43 Solutions Manual Foundation Engineering: Principles and Practices, 3rd Ed 3-20 © 2016 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.