PE Exam Review - Geotechnical

Transcription

1 PE Exam Reiew - Geotechnical Demonstration Problem Solutions I. Soled Problems Soil Classification.... Phase Relationships Effectie Stress and Stress Distribution Consolidation Strength and Bearing Capacity Earth Retaining Structures Piles Seepage... 8

2 Demonstration Problem Solutions Geotechnical Engineering Demo Problem Solutions Module 1 - Soil Classification Situation Laboratory gradation analyses and Atterberg Limits were performed on two soil samples. Results are summarized below. No organic odor or materials were noted in either sample. Soil A: U.S. Siee Size Percent Passing No No No No liquid limit (w l ) 35% plastic limit (w p ) % Soil B: U.S. Siee Size Percent Passing No No No No liquid limit (w l ) 37% plastic limit (w p ) 0%

3 Demonstration Problem Solutions Geotechnical Engineering 3 Requirements AASHTO Soil Classification System a Siee analysis: % passing #10 #40 #00 Characteristics of fraction passing #40: W l : Granular materials (35% or less passing #00 siee) Silt-clay material (more than 35% passing #00 siee) A-1 A-3 A- A-4 A-5 A-6 A-7 A-8 A1-a A1-b A-4 A-5 A-6 A-7 A7-5 A7-6 < 50 < 30 < 15 < 50 < 5 > 51 < 10 < 35 < 35 < 35 < 35 > 36 > 36 > 36 > 36 < 40 > 41 < 40 > 41 < 40 > 41 < 40 > 41 I p : < 6 NP < 10 < 10 > 11 > 11 < 10 < 10 > 11 > 11 1A) What is the classification of Soil A according to the AASHTO System (include the group index)? 1. First, find the plasticity index: I p w l w p I p Use the chart to find the soil group % passing #00 > 35% w l < 40 I p > 11 Soil is in group A-6 3. Find the group index: I F 35)[ ( w 40)] 0.01( F 15)( I 10) g ( 00 l 00 p I ( 51 35)[ (35 40)] 0.01(51 15)(13 10) 4. 8 g Round 4.8 to 4; Soil A is A-6(4)

4 Demonstration Problem Solutions Geotechnical Engineering 4 1B) What is the classification of Soil B according to the AASHTO System (include the group index)? 1. First, find the plasticity index: I p w l w p I p Use the chart to find the soil group % passing #00 < 35% w l < 40 I p > 11 Soil is in group A Find the group index. For A--6 and A--7 soils: I 0.01( F00 15)( I 10) I 0.01(0 15)(17 10) g Round 0.35 to 0; Soil B is A--6(0) g p Unified Soil Classification System b Soil Classification Criteria for assigning group symbols and group names using laboratory tests A Group Symbol Group Name B Course-grained Clean graels: C u > 4 and 1 < C c < 3 E GW Well-graded grael F soils: Graels: > 50% < 5% fines c C u > 4 and/or 1 < C c < 3 E GP Poorly graded grael F More than 50% of course retained on fraction retained #00 siee on #4 siee Fine-grained soils: 50% or more pass the #00 siee Sands: > 50% of course fraction passes #4 siee Silts and Clays: w l < 50 Fines classify as ML or Graels w/ fines: MH GM F, G, H Silty grael > 1% fines c Fines classify as CL or CH GC F, G, H Clayey grael Clean sands: C u > 6 and 1 < C c < 3 E SW Well-graded sand I < 5% fines D C u > 4 and/or 1 < C c < 3 E SP Poorly graded sand I Fines classify as ML or Sands w/ fines: MH SM G, H, I Silty sand > 1% fines D Fines classify as CL or CH SC G, H, I Clayey sand Inorganic I p > 7 and plots on or aboe A line J CL K, L, M Lean clay I p < 4 or plots below A line J ML K, L, M Silt K, L, M, N wl oen dried Organic clay < 0.75 Organic wl not dried OL K, L, M, O Organic silt Silts and Clays: w l > 50 Inorganic Organic I p plots on or aboe A line CH K, L, M Fat clay I p plots below A line MH K, L, M Elastic silt K, L, M, P wl oen dried Organic clay < 0.75 wl not dried OH K, L, M, Q Organic silt Highly organic soils Primarily organic matter, dark in color, and organic odor PT Peat

5 A: based on the material passing the 3-in (75-mm) siee B: if field sample contained cobbles or boulders, add with cobbles, boulders, or both to group name C: Graels with 5-1% fines require dual symbols: GW-GM: well-graded grael with silt GW-GC: well-graded grael with clay GP-GM: poorly-graded grael with silt GP-GC: poorly-graded grael with clay D: sands with 5-1% fines require dual symbols: SW-SM: well-graded sand with silt SW-SC: well-graded sand with clay SP-SM: poorly-graded sand with silt SP-SC: poorly-graded sand with clay D60 D30 E: C u C c D10 D10D60 F: If soil contains > 15% sand, add with sand to group name G: If fines classify as CL-ML, use dual symbol GC-GM or SC- SM H: If fines are organic, add with organic fines to group name I: If soil contains > 15% grael, add with grael to group name J: If Atterberg Limits plot in hatched area, soil is a CL-ML, silty clay K: If soil contains 15-9% plus #00, add with sand or with grael, whicheer is predominant L: If soil contains > 30% plus #00, predominately sand, add sandy to group name. M: If soil contains > 30% plus #00, predominately grael, add graelly to group name. N: I p > 4 and plots on or aboe A line O: I p < 4 or plots below A line P: I p plots on or aboe A line Q: I p plots on or aboe A line 1C) What is the classification of Soil A according to the Unified Soil Classification System? 1. Use USCS chart % passing #00 > 50 Fine-grained soils w l < 50 top row no organic odor or color inorganic I p > 4 w l 35; I p 13 plots aboe A line on plasticity chart c Soil A is CL 1D) What is the classification of Soil B according to the Unified Soil Classification System? 1. Use USCS chart % retained on #00 > 50 Course-grained soils % passing #4 > 50 Sands % passing #00 (i.e., % fines) > 1% classify fines according to plot location on plasticity chart c w l 37 I p 0 Plots in region labeled CL or OL fines classify as CL or CH Soil B is SC

6 Demonstration Problem Solutions Geotechnical Engineering 6 Module - Phase Relationships Situation Soil from a borrow pit will be used to construct a 50,000 cubic yard highway embankment. Project specifications require that the fill in the embankment be compacted to at least 95% of the Modified Proctor maximum dry density. The material in the borrow pit has the following characteristics: Total Unit Weight: 10.0 pcf Moisture Content: 8.0% Specific Graity of Soil:.66 The Modified Proctor maximum dry density and optimum moisture content are 18 pcf and 13.0%, respectiely. Requirements A) What is the dry unit weight of the material in the borrow pit? 1. Dry unit weight: γ d γ t 1+ w 10. γ d pcf B) What is the oid ratio of the borrow material in the borrow pit? 1. Void ratio: V s V e V Ws G γ s w s W s γ d for 1 ft 3 soil V s V V V e t s

7 Demonstration Problem Solutions Geotechnical Engineering 7 C) In order to meet the specified compaction criteria, what is the minimum dry unit weight of the compacted fill? 1. Maximum dry density 18 pcf. Minimum dry density is 95% of max 3. Minimum dry density pcf D) The minimum number of cubic yards of borrow material required to construct the embankment is approximately: 1. For each cubic foot of compacted fill, you need 11.6 lb of dry soil, but each cubic foot of excaated borrow weighs only lb, so you will need to excaate a larger olume. γ d borrowvborrow γd fillvfill Vborrow , 000 V borrow 73,750 yd 3

8 Demonstration Problem Solutions Geotechnical Engineering 8 E) How many gallons of water must be added per cubic yard of borrow to achiee the Modified Proctor optimum moisture content? Assume no loss of water by eaporation during transport from the borrow pit. 1. Find difference between current and optimum weight of water in borrow material Ww w opt W W w cur W s Ww pcf W w opt w s Ww pcf W w cur. Added water pcf ft 3 3 lb 7 ft gal 18 gal/yd 3 3 yd 8.34 lb F) If the compacted fill in the embankment becomes saturated, what is the moisture content? Assume that the compacted fill does not swell when water is added. 1. Se wgs S 1.0 for saturated soil e 0.49 G s w w 18.4%

9 Demonstration Problem Solutions Geotechnical Engineering 9 G) The borrow material is transported from the borrow pit in dump trucks that can hold 10 cubic yards of material. The aerage total unit weight of the borrow when it is placed in the truck is pcf. The minimum number of truck loads of borrow required to construct the embankment using 50,000 yd 3 is approximately: 1. First find the olume of borrow material that must be transported by truck to fill the embankment. γ d trucksvtrucks γd fillvfill γ t trucks 115 γd trucks w Vtrucks , 000 V trucks 85,446. Diide by the truck capacity 85,446 Truckloads 8,

10 Demonstration Problem Solutions Geotechnical Engineering 10 Module 3 - Effectie Stress and Stress Distribution Situation 1 A subsurface profile consists of a 6 foot layer of sand oerlying 0 feet of clay. The clay is underlain by a grael layer that is at least 50 feet thick. The total unit weights of each of the materials encountered are gien below. Sand Clay Grael Total Unit Weight 10 pcf 115 pcf 140 pcf The ground water leel is at the top of the sand layer. Assume that the total head is the same in the sand and the grael layer. Requirements 3-1A) The total ertical stress at a depth of 8 feet below ground surface is approximately: 1. Total ertical stress is the sum of stresses for each soil layer: σ γ H Sand: psf Clay: psf Grael: psf. σ 3300 psf t 3-1B) The pore water pressure at a depth of 8 feet below the ground surface is approximately: 1. Pore water pressure without seepage: u γ h. u psf w 3-1C) What is the ertical effectie stress at a depth of 8 feet below ground surface?, 1. Vertical effectie stress: σ σ u,. σ psf

11 Demonstration Problem Solutions Geotechnical Engineering D) What is the total ertical stress at a depth of 8 feet below ground surface if the water leel rises to 4 feet aboe the ground surface? 1. Add stress of extra 4-ft of water to total stress from requirement A. σ γ w h psf σ psf 3-1E) What is the change in the ertical effectie stress at a depth of 8 feet below ground surface if the water leel rises to 4 feet aboe ground surface?, 1. Vertical effectie stress: σ σ u u γ h σ' psf 3. σ' 0 w 3-1F) A 0-foot diameter water storage tank is constructed on the ground surface. The weight of water in the tank exerts a uniform pressure of 1,00 psf on the ground surface. What is the total ertical stress at a depth of 0 feet below ground surface at the center of the tank? Assume the sand and the clay hae the same elastic modulus. 1. Total σ σ from soil + σ from tank σ soil (10 pcf 6 ft) + (115 pcf 14 ft) 330 To find σ tank use the Boussinesq stress contour chart for uniformly loaded circular footings d x z 0 Stress is acting at center of tank, so 0 ; r r 10 I 0.31 (from chart) σ tank Ip foundation psf Total σ psf

12 Demonstration Problem Solutions Geotechnical Engineering 1 Situation A two mile long embankment was constructed for a new highway. The embankment is 1 feet high and 0 feet in width. Due to right-of-way restrictions, retaining walls are used to maintain the width of the embankment at 0 feet for its full height. The total unit weight of the embankment fill is 15 pcf. The embankment is constructed on a sand layer that is more than 100 feet thick. The water table is at a depth of 5 feet below the original ground surface. The total unit weight of the sand is 10 pcf aboe the ground water table and 130 pcf below the water table. Requirements 3-A) The stress increase at a depth of 15 feet under the center of the embankment is approximately: 1. Stress increase: σ I (influence factor from Boussinesq chart) σ total σ γ psf backfill h embankment total From Boussinesq stress contour chart for square footings d : depth 0.75B (width) distance from center 0B so, I σ psf 3-B) How much would the stress due to the embankment increase/decrease if the water table was lowered to 10 feet? 1. As long as the embankment is aboe the water table, raising or lowering water table won t affect the applied stress due to the embankment. The best answer here is It would not change.

13 Demonstration Problem Solutions Geotechnical Engineering 13 Module 4 - Consolidation Situation 1 A soil profile consists of dense sand, 40 feet in thickness, underlain by a 10-foot layer of normally consolidated clay. The clay, in turn, is underlain by relatiely incompressible and impermeable bedrock. The present water table is at the existing ground surface. Ten feet of fill will be placed oer the existing ground surface in a 1,000-foot wide by 1,500-foot long area. The sand has a oid ratio of 0.45 and a specific graity of.67. The sand is saturated below the water table and has a moisture content of 5 percent aboe the water table. The fill has a total unit weight of 15 pounds per cubic foot and a moisture content of 10 percent. The water table will be lowered 0 feet with the addition of the fill material. The clay is saturated, has a specific graity of solids of.70 and a moisture content of 35 percent. The compression index of the clay has been determined to be 0.. The coefficient of consolidation 0.1 square feet per day. Requirements 4-1A) What is the total unit weight of the saturated sand? 1. Saturated unit weight: γ sat ( w Gs + e) γ 1+ e ( )6.4. γ sat pcf B) What is the total unit weight of the moist sand aboe the ground water leel? 1. Total unit weight: γ t ( Gs + S e) γ 1+ e wgs S e 0.45 ( ) 6.4. γ t 10.7 pcf w

14 Demonstration Problem Solutions Geotechnical Engineering C) What is the total unit weight of the saturated clay? 1. Saturated unit weight: ( ) 6.4. γ sat pcf γ sat (1 + w) γ w + 1 G w s 4-1D) What is the oid ratio of the saturated clay? 1. Se wgs wgs e S 1 4-1E) What is the total settlement in inches of the clay layer due to the combined effect of the imposed fill and lowered water table? 1. Settlement for normally consolidated clay ( σ ' σ' m ): f Cc ρ 1+ e 0 σ' H log σ' Initial stresses at mid-depth in clay σ 1 0 γsandhsand + γclay h psf clay + u γ h w psf 0 σ σ u 3149 psf ' Final stresses at mid-depth in clay σ γ h + γ h f fill fill moistsand moistsand satsand satsand clay clay + γ h + γ h f 0 σ f psf u γ h w psf f σ ' σ u 5373 psf f f f ρ 10 log 0. 6 ft 3.1 in

15 Demonstration Problem Solutions Geotechnical Engineering F) What is the estimated time required for 50% consolidation of the clay layer? 1. Consolidation time: TH t c d Single drainage, so H d thickness of soil layer 10 ft Read T from Time factor s. Degree of Consolidation chart e. U 50%, so T t 00 days G) How long will it take for 50% consolidation of the clay layer to occur if the clay layer is underlain by sand instead of bedrock? 1. Consolidation time: TH t c d Single drainage, so H d ½ thickness of soil layer 5 ft Read T from Time factor s. Degree of Consolidation chart e. U 50%, so T t 50 days 0.1 Situation A soil profile consists of a10-foot layer of dense sand oerlying an 8-foot layer of normally consolidated clay. The clay is also underlain by dense sand. The groundwater leel is at the ground surface. Also, the total head measured in a piezometer installed in the middle of the clay layer is equal to the existing ground surface eleation. A proposed deelopment requires that 13 feet of fill be placed aboe the existing ground surface. The fill material has a total unit weight of 130 pcf and a moisture content of 1 percent. The saturated unit weight of the sand and clay is 138 pcf and 113 pcf, respectiely. The clay has a specific graity of.68 and a moisture content of 40 percent. The clay has a compression index of 0.0 and a coefficient of consolidation of 0.05 square feet per day.

16 Demonstration Problem Solutions Geotechnical Engineering 16 Requirements 4-A) What is the pore pressure in the center of the clay immediately after placement of the fill layer? Assume that fill placement occurs instantaneously. 1. Pore pressure static + initial excess pore pressures: u u i + u 0 u γ h w psf i u σ γ h psf 0 fill fill. u psf 4-B) Assuming that the total consolidation settlement due to the weight of the added fill is 4.1 inches, how long after the fill is placed can the building be constructed? Assume that the building has no weight 1. Total settlement 4.1 in; allowable settlement 1 in. So, fill must settle 3.1 in before building can be constructed.. Consolidation time: TH t c d Use the Time factor s. Degree of Consolidation chart e to find T. ρt 3.1 U ρ 4.1 For U 76%, T 0.50 There is double drainage, so H d ½ h clay 4 ft c t 160 days 0.05

17 Demonstration Problem Solutions Geotechnical Engineering 17 4-C) What is the pore pressure in the middle of the clay layer 100 days after placement of the fill? Assume fill placement is instantaneous. 1. Pore pressure: u u e + u 0 u psf u e σ ' 1 u u i i u i 1690 psf σ' Use the chart that relates to depth for arious T alues f u z 4 depth 1 (for double drainage) H 4 Use this equation to find T: σ' 0.41 u ue u i i T T 0.31 u e 0.59 u i psf. u psf i TH t c d

18 Demonstration Problem Solutions Geotechnical Engineering 18 Module 5 - Strength and Bearing Capacity Situation A continuous footing is constructed at a depth of 5 feet in a uniform sand deposit. The width of the footing is 10 feet. The sand has total unit weight of 130 pcf and a friction angle (φ) of 37. The depth to groundwater is greater than 100 feet. Assume concentric ertical loading. Use Vesic s bearing capacity factors for all calculations. Requirements 5A) What is the approximate ultimate bearing capacity of the footing? 1. For sand with c 0, bearing capacity: qult 0.5B' γeff Nγ Sγ ssγi + σdnqsqssqi B B e B 0 10 ft gwt > B below footing, so γ eff γ t 130 pcf From table of Bearing Capacity Factors g, for φ 37 : N γ 66 N q 43 Strip footing with concentric ertical load, so: S S S S 1. 0 Sand deposit, so σ ' at bottom of footing D σ u 0, so σ ' σ γ D t psf t. q ult ,850 psf 70,850 lb 1 ton tsf ft 000 lb γ s qs γi qi

19 Demonstration Problem Solutions Geotechnical Engineering 19 5B) What is the approximate ultimate bearing capacity if the bottom of the footing is 10 feet deep? 1. For sand with c 0, bearing capacity: qult 0.5B' γeff Nγ Sγ ssγi + σdnqsqssqi B B e B 0 10 ft gwt > B below footing, so γ eff γ t 130 pcf From table of Bearing Capacity Factors g, for φ 37 : N γ 66 N q 43 Strip footing with concentric ertical load, so: S S S S 1. 0 Sand deposit, so σ ' at bottom of footing D σ u 0, so σ ' σ γ D psf t t. q ult ,800 psf 98,800 lb 1 ton tsf ft 000 lb γ s qs γi qi

20 Demonstration Problem Solutions Geotechnical Engineering 0 5C) What is the bearing capacity if the water table is at ground surface? Assume that the saturated unit weight of sand is equal to 138 pcf. 1. For sand with c 0, bearing capacity: qult 0.5B' γeff Nγ Sγ ssγi + σdnqsqssqi B B e B 0 10 ft γ eff γ sat - γ w From table of Bearing Capacity Factors g, for φ 37 : N γ 66 N q 43 Strip footing with concentric ertical load, so: S S S S 1. 0 Sand deposit, so σ ' at bottom of footing σ σ u D t D σ γ s qs γi qi σ D t γsat u γ w D psf psf σ psf D. q ult ,0 psf 41,0 lb 1 ton tsf ft 000 lb

21 Demonstration Problem Solutions Geotechnical Engineering 1 Module 6 - Earth Retaining Structures Situation A reinforced concrete retaining wall is proposed to support 14 feet of cohesionless sand backfill as shown on the figure below. The sand has a total unit weight of 10 pcf and an angle of internal friction of The unit weight of the concrete is 150 pcf. Assume that the friction angle for shear along the bottom of the wall is Use Rankine s theory to complete the following requirements. Requirements 6A) What is the actie earth pressure coefficient? 1. Actie lateral earth pressure coefficient: φ 30 1 sin 30. K a sin 30 K a 1 sin φ 1+ sin φ 6B) What is the total actie force per foot of wall? 1 1. Actie force: Pa γ th Ka 1. Pa lb 6.4 k 6C) What is the oerturning moment per foot of wall about point A? 1. Oerturning moment: 18. M o k/ft 3 M o P a H 3

22 Demonstration Problem Solutions Geotechnical Engineering 6D) What is the resisting moment per foot of wall about point A, neglecting passie pressure at the toe of the wall? A Diide the wall into 4 rectangular areas, as shown in the figure aboe.. Sum the moments for each of the four areas: M r Σ (F i arm i ) F Aγ Arm ½ d Section H W A γ 18 F (kips) Arm (ft) M (ft-k) Total E) What is the approximate factor of safety against oerturning, neglecting passie pressure at the toe of the wall? 1. Factor of safety: F.S Mr F.S. M 0

23 Demonstration Problem Solutions Geotechnical Engineering 3 6F) What is the approximate eccentricity of the resultant of the ertical and horizontal forces? Wbase 1. Eccentricity: e x Mr M x ΣF o Σ F k x 3.8 ft from point A e ft 6G) What is the factor of safety against sliding without passie earth pressure considered? 1. Factor of safety: s ΣF tan φ s F.S. ΣF 18.1 k φ s 30 s 18.1 tan k F.S s P a

24 Demonstration Problem Solutions Geotechnical Engineering 4 6H) What is the Approximate factor of safety against bearing capacity failure? Use Meyerhof s bearing capacity and correction factors. 1. Factor of safety: F.S. Q ΣF ult Q ult q ult B For sand with c 0, bearing capacity: q 0.5B' γ N S S + σ N S S ult eff γ γs γi D q qs qi B B e ft γ eff 10 pcf From graph of Bearing Capacity Factors g, for φ 30 : N γ 15.7 N q 18.4 For strip footing: S S 1. 0 γ s qs For inclined load: S γ i 1 P θ arctan 19. ΣF a o 5 θ φ 19.5 Sγ i θ For inclined loading: Sqi Sand deposit, so σ ' at bottom of footing σ D D σ psf q ult ,96 psf Q ult ,850 psf k F.S

25 Demonstration Problem Solutions Geotechnical Engineering 5 Module 7 - Piles Situation A bridge is proposed to span a rier. The bridge piers will be supported on 1 diameter, steel, closed-end pipe piles (concrete-filled) drien into the medium-dense sand rier bottom. The water in the rier is 15 feet deep. Laboratory tests indicate that the sand has the following properties: specific graity G s.67; oid ratio, e 0.60 The results of a pile load test on a 1 steel pipe pile drien 15 feet into the sand indicated that the unit skin friction on the test pile at a depth of 10 feet into the sand was 350 psf at failure. Also, the ultimate end bearing capacity of the test pile was estimated to be 3.0 kips. Neglect the structural capacity of the pile and settlement in answering the following questions. Requirements 7A) What is the saturated unit weight of the sand? γ 1. Saturated unit weight: ( w Gs + e) γ sat 1+ e G s.67 e ( ). γ sat 17.5 pcf

26 Demonstration Problem Solutions Geotechnical Engineering 6 7B) The ultimate skin friction capacity of a pile embedded 5 ft into the sand rier bottom is approximately: 1. You need the unit skin friction to find the ultimate friction capacity. You know the skin friction for the pile test at 10 feet is 350 psf. f Kσ ' tan δ, so s 350 K tan δ ' σ effectie stress at 10 ft: σ ' σ u σ γ w h w + γ sat h sat σ psf u γ h w w 1560 psf σ ' psf 350 K tan δ f 0.538σ ' s. Now you need the effectie stress at 15 ft: σ ' σ u σ γ wh w + γsathsat σ psf u γ h psf w w σ ' psf 3. So, unit skin friction at 15 ft: f 0.538σ ' psf s π 1 4. Ultimate friction capacity: Q s π 1 9 k

27 Demonstration Problem Solutions Geotechnical Engineering 7 7C) What is the ultimate end bearing capacity of a pile embedded 5 feet into the sand rier bottom? 1. Ultimate end-bearing capacity: Qt σ' NqAtip Q t from test at 15 feet 3 k N q is the same at 15 and 5 feet 5 feet is deeper than the critical depth of 15 feet, σ ' 5 ' σ ' 15'. So, Q t at 5 feet 3 k 7D) Using a factor of safety of 3.0, what is the allowable capacity of a pile embedded 5 feet into the sand if the water leel rises 10 feet? 1. Allowable capacity: FS 3 Q all Qult FS Q ult Q s + Q t Change in water leel does not change Q s and Q t from preious requirements. Q ult k 61. Q all 0.3 k 3

28 Demonstration Problem Solutions Geotechnical Engineering 8 Module 8 - Seepage Situation 1 A 10-foot thick layer of Soil A oerlies Soil B, as shown in the figure below. The ground surface is at eleation 100 feet. Soil A has a coefficient of permeability of 0 ft/day and a total unit weight of 15 pcf. Soil B has a coefficient of permeability of 10 ft/day and a total unit weight of 10 pcf. Groundwater is obsered at the ground surface, eleation 100. A piezometer is installed in Soil B at eleation 75. The total head measured in the piezometer is eleation 85. Eleation (ft) Soil A k 0 ft/day γ t 15 pcf Soil B k 10 ft/day γ t 10 pcf Piezometer Requirements 8-1A) What is the total ertical stress at eleation 90? 1. Total ertical stress: σ γh. σ psf 8-1B) What is the total ertical stress at eleation 75? 1. Total ertical stress: σ γ Ah A + γ Bh B. σ psf

29 Demonstration Problem Solutions Geotechnical Engineering 9 8-1C) What is the ertical effectie stress at eleation 75? 1. Vertical effectie stress: σ ' σ u u h p γ w h h h p t e u psf. σ ' psf

30 Demonstration Problem Solutions Geotechnical Engineering D) What is the ertical effectie stress at eleation 90? 1. Vertical effectie stress: σ ' σ u σ 150 psf u h p γ w h p h t h e h e 90 ft h t h. To find head loss through layer A, start with Darcy s equation: Q kia For A 1 ft, Q A Q B k A i A k B i B i i A B h H A A h H k A 0; H A 10 k B 10; H B 15 k A h h H A A A B B k h 3 B B A h H h B We know that h A + h B 15, so + h B 15 3 h A 3.7 ft B B h B 11.3 ft 3. Returning to the equations for head: h t h A ft h t 90 h p ft 4. So, pore pressure, u And σ ' psf

31 Demonstration Problem Solutions Geotechnical Engineering E) What is the rate of ertical seepage for a 1 square foot area? 1. With one-dimensional flow, use Darcy s equation to find seepage rate: Q kia k A 0 ft/day 3.7 i A A 1 ft. Q ft 3 /day Situation A concrete dam is constructed on a fine sand layer, as shown on the figure below. The sand is underlain by imperious bedrock. The sand is 40 feet thick and has a coefficient of permeability of 10-4 ft/minute and a saturated unit weight of 1 pounds per cubic foot. The bottom of the dam is 5 feet below the top of the sand layer. The tail water is at the ground surface, eleation 100, and the leel of the head water is eleation 10. Use the flow net shown below to respond to the arious requirements. 60 ft E L 10 DAM A H 0 f t EL ft 35 ft Imperious B

32 Demonstration Problem Solutions Geotechnical Engineering 3 Requirements 8-A) What is the rate of seepage (per foot of length) below the dam? N f 1. Seepage: Q k h N d k 10 4 ft/min N f 4 (number of flow channels, from flow net) N d 1 (equipotential drops, from flow net) h 0 ft Q ft / min 1 8-B) What is the total head at point A? 1. h h h T A T each equipotential drop represents an equal amount of head loss. There are 6 drops from the headwater side of the dam to point A, so: 6 h 0 10 ft ft h A T 8-C) What is the uplift pressure at the bottom of the dam at point A? 1. Uplift pressure: u h p γ w h h h p T e h T 110 ft h e 95 ft h p ft. u psf

33 Demonstration Problem Solutions Geotechnical Engineering 33 8-D) What is the pore pressure at point B? 1. Pore pressure at B: u h p γ w h p h T h e Points A and B are on the same equipotential line, so they hae the same total head, so h T 110 ft h e 60 ft h p ft. u psf

34 Demonstration Problem Solutions Geotechnical Engineering 34 References: a. Adapted from: Standard Specifications for Transportation Materials and Methods of Sampling and Testing, Part I, Specifications, 13 th ed., AASHTO, 198. b. Adapted from: 1989 Annual Book of ASTM Standards, ASTM, Philadelphia, c. See: 1989 Annual Book of ASTM Standards, ASTM, Philadelphia, d. See, for example: Michael R. Lindeburg, Ciil Engineering Reference Manual, 7 th ed., Professional Publications, Inc., Belmont, CA, e. See, for example: Wayne C. Teng, Foundation Design, Prentice-Hall, Inc., Englewood Cliffs, NJ, 196. f. See Module 4 Visual Aids g. From: Karl Terzaghi, Ralph B. Peck, and Gholamreza Mesri, Soil Mechanics in Engineering Practice, 3 rd ed., John Wiley & Sons, Inc., New York, 1996.