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SOIL MECHANICS
(version Fall 2008)

Presented by:
Jerry Vandevelde, P.E.
Chief Engineer

GEM Engineering, Inc.
1762 Watterson Trail
Louisville, Kentucky
(502) 493-7100
1
National Council of Examiners
for Engineering and Surveying
http://www.ncees.org/

2
STUDY REFERENCES
• Foundation Engineering; Peck Hanson &
Thornburn
•Introductory Soil Mechanics and Foundations;
Sowers
•NAVFAC Design Manuals DM-7.1 & 7.2
•Foundation Analysis and Design; Bowles
•Practical Foundation Engineering Handbook;
Brown
3
Soil Classification Systems
* Unified Soil Classification System
* AASHTO

Need: Particle Sizes and Atterberg Limits

4
Particle Sizes (Sieve Analysis)

(Well Graded)

(Poorly Graded)
0.1

5
Atterberg Limits
Liquid, Plastic & Shrinkage Limits
Plasticity Index (PI)
PI = Liquid Limit - Plastic Limit
(range of moisture content over which soil is plastic or malleable)

6
UNIFIED SOIL
CLASSIFICATION
SYSTEM
ASTM D-2487

7
8

Ref: Peck Hanson & Thornburn 2nd Ed.
Effective Size = D10
10 percent of the sample is finer than this size

D60 = 1.6mm

D30 = 0.2mm
D10 = 0.03mm
0.1
0.1

9
Uniformity Coefficient (Cu) = D60/D10
Coefficient of Curvature (Cz) = (D30)2/(D10xD60)

D60 = 1.6mm

D30 = 0.2mm
D10 = 0.03mm

0.1

10
Well Graded - Requirements
50% coarser than No. 200 sieve
Uniformity Coefficient (Cu) D60/D10
>4 for Gravel
> 6 for Sand
Coefficient of Curvature (Cz)
= (D30)2/(D10xD60) = 1 to 3

11
Is the better graded material a
gravel?

81% Passing No. 4

18% Finer No. 200

0.1

0.1

12
Gravel if > 50 Percent Coarse

Fraction retained on No. 4 sieve

81% Passing No. 4

18% Finer No. 200

% Retained on No. 200 = 82%
1/2 = 41%
19% (100-81) retained on No. 4
sieve (gravel)
19< 41 half of coarse fraction

∴ sand

(“S”)

0.1

13
Well Graded Sand?

D60 = 1.6mm

Uniformity Coefficient (Cu)
> 6
= D60/D10

D30 = 0.2mm
D10 = 0.03mm

0.1

Coefficient of Curvature (Cz)
= 1 to 3
= (D30)2/(D10xD60)
14
Well Graded Sand?
Uniformity Coefficient (Cu)
D60/D10 = 1.6/.03 = 53 > 6
D60 = 1.6mm
D30 = 0.2mm
D10 = 0.03mm

0.1

Coefficient of Curvature (Cz)
= (D30)2/(D10xD60)
= 0.22/(.03x1.6)
= 0.83 <1 to 3

∴Poorly graded
15
What classification?
Unified Classification of Coarse Soils with Fines
< 5% Passing No. 200 sieve:GW,GP, SW, SP
81% Passing No. 4

5% - 12% Passing No. 200 sieve:
Borderline- use dual symbols
18% Finer No. 200

> 12% Passing No. 200 sieve: GM, GC, SM, SC

0.1

>12% passing No. 200 sieve
Since = “S” ∴ SC or SM
16
What Unified Classification if
LL= 45 & PI = 25?

From sieve data

SC or SM
0.1

A) “SC” B) “SM” C) “CL” or D) “SC & SM”
17
Unified Classification

Answer is “A”
⇒ SC

18
AASHTO
(American Association of State Highway and
Transportation Officials)

19
What is the AASHTO Classification?

65% Passing No. 10

40% Passing No. 40
18% Finer No. 200

1) 18 % passing No. 200 sieve
2) 65% passing No. 10 sieve
3) 40% passing No. 40 sieve
4) assume LL = 45 & PI = 25

20
18 percent passing No. 200 sieve; 65 percent passing No. 10 sieve
40 percent passing No. 40 sieve; assume LL = 45 & PI = 25

21
AASHTO Classification
1
2

3
4

4

1) 18 % passing No. 200 sieve
2) 65% passing No. 10 sieve
3) 40% passing No. 40 sieve
4) assume LL = 45 & PI = 25

22
AASHTO
Group Index

23
Mass-Volume (Phase Diagram)
• Unit volume of soil
contains:
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

– Air (gases)
– Water (fluid)
– Solid Particles

Ws

24
Moisture Content = ω
weight of water/ weight of dry soil
ω = Ww/Wd
water loss/(moist soil weight - water loss)
ω = Ww/(Wm-Ww)
and
ω =(Wm-Wd)/Wd

25
Mass - Volume Relationships
Density or Unit Weight =
Moist Unit Weight = γm

γ

γm = Wm/Vt = γd + ω γd
ω = (γ m - γ d )/ γd
ω γd + γd = γm
γm= (1+ ω) γd
γd = γm/(1+ ω) b
26
Total Volume =
∑ Volume (solid + water + air)
= Vs+Vw+Va

∴
Va = Vt - Vs- Vw

Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws
27
Relationship Between Mass & Volume
Volume = Mass/(Specific Gravity x Unit Weight of Water)
= Ws/(SGxWw)

Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws

28
Specific Gravity =
weight of material/ weight of same volume of
water
Soil Specific Gravity
Typical Range
2.65 to 2.70
Specific Gravity of Water = 1

29
Saturation = S expressed as percent
S = volume of water/ volume of voids x 100
Total
Volume

Vt

Va

Vv
Vw

S = Vw/Vv x 100

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Always ≤ 100

Ws

30
Porosity
n = volume of voids/ total volume
n = Vv/Vt

Void Ratio
e = volume of voids/ volume of solids
e = Vv/Vs
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws
31
What is the degree of saturation for a soil
with:
SG = 2.68, γm = 127.2 pcf & ω = 18.6 percent
A) 88.4
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

B) 100.0

Wt
Vs

Soil

Ws

C) 89.1
32
What are the porosity and degree of saturation for a soil with:
SG = 2.68, γm = 127.2 pcf &

ω = 18.6 percent

γd = γm/(1+ ω) = 127.2/(1.186)
= 107.3pcf
Total
Volume

Vt

Va

Vv
Vw
Vs

Air
Total

Water

Soil

Ww Weight
Wt
Ws

Ww = γm- γd = 19.9 pcf
Vw = Ww/62.4 = 0.319 cf
Vs = γd /(SGx62.4) = 0.642 cf
Va = Vt - Vw - Vs
= 1- 0.319 - 0.642 = 0.039 cf
Vv = Vw + Va = 0.358 cf
33
What are the porosity and degree of saturation for a soil with:
SG = 2.68, γm = 127.2 pcf &

ω = 18.6 percent

Vw = 0.319 cf, Vs = 0.642 cf,
Vv = 0.358 cf
Total
Volume

Vt

Va

Vv
Vw
Vs

Air

Degree of Saturation = Vw/Vv x 100
Total

Water

Soil

Ww Weight
Wt
Ws

= 0.319/0.358 x 100 = 89.1%

Answer is “C”
34
Ref:
NAVFAC DM-7

35
Borrow Fill Adjustments
Borrow Material Properties:

γm = 110 pcf & ω = 10%

Placed Fill Properties: γd = 105 pcf & ω = 20%
How much borrow is needed to produce 30,000 cy of fill?
How much water must be added or removed from each cf of
fill?
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws
36
Borrow Fill Adjustments
Borrow Material Properties:

γm = 110 pcf & ω = 10%

γd = γm /(1+ω) = 110/(1.10) =100 pcf; Ww = 110-100=10 lbs
Placed Fill Properties: γd = 105 pcf & ω = 20%
Ww = ωx γd = 0.2x 105 = 21 lbs
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws
37
Borrow Fill Adjustments
Borrow Properties: γm = 110 pcf, γd =100 & ω = 10%
Placed Fill Properties: γd = 105 pcf & ω = 20%
Since borrow γd =100pcf & fill γd =105pcf, 105/100 =1.05
It takes 1.05 cf of borrow to make 1.0 cf of fill
For 30,000 cy, 30,000 x 1.05 = 31,500 cy of borrow
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws
38
Borrow Fill Adjustments
Borrow Material Properties: Ww = 10 lbs
Placed Fill Properties: Ww = 21 lbs
Water supplied from borrow in each cf of fill
= 10 x 1.05 = 10.5 lbs; 21 lbs - 10.5 = 10.5 lbs short/1.05 cf
10.5lbs/1.05 cy = 10 lbs of water to be added per cf borrow
Total
Volume

Vt

Va

Vv
Vw

Air
Total

Water

Ww

Weight

Wt
Vs

Soil

Ws
39
Proctor: Moisture Density Relationships
Establishes the unique relationship of moisture to
dry density for each specific soil at a specified
compaction energy
MOISTURE-DENSITY RELATIONSHIP
108.0
106.0

D ry D ensity (pcf)

104.0
102.0
100.0
98.0
96.0
94.0
92.0
90.0
88.0
8.0

10.0

12.0 14.0 16.0

18.0 20.0 22.0 24.0

26.0 28.0

Moisture Content (%)

40
Proctor: Moisture Density Relationships
• 4” mold 25 blows
• 6” mold 56 blows

• Standard
– 5.5 lb hammer
– dropped 12 in
– 3 layers

Standard: ASTM D-698
AASHTO T-99
Modified: ASTM D-1557
AASHTO T-150

• Modified
– 10 lb hammer
– dropped 18 in
– 5 layers
41
PROCTOR COMPACTION TEST

Maximum Dry Density - Highest density for
that degree of compactive effort
Optimum Moisture Content - Moisture content
at which maximum dry density is achieved for
42
that compactive effort
Proctor: Moisture Density Relationships
What density is required
for 95% Compaction?

MOISTURE-DENSITY RELATIONSHIP
108.0
106.0

Dry Density (pcf)

104.0
102.0
100.0
98.0
96.0
94.0
92.0
90.0
88.0
8.0

10.0

12.0

14.0

16.0

18.0

20.0

22.0

24.0

26.0

28.0

What range of moisture
would facilitate
achieving 95%
compaction?

Moisture Content (%)

43
Proctor: Moisture Density Relationships
MOISTURE-DENSITY RELATIONSHIP

104 x .95 = 98.8 pcf

108.0
106.0

Dry Density (pcf)

104.0

A

102.0
100.0

B

Range of moisture is within
the curve A to B
(14 to 24 %)

95%

98.0
96.0
94.0
92.0
90.0
88.0
8.0

10.0

12.0

14.0

16.0

18.0

20.0

22.0

24.0

26.0

28.0

Moisture Content (%)

44
Proctor: Zero Air Voids Line
Relationship of density to
moisture at saturation for
constant specific gravity
(SG)

MOISTURE-DENSITY RELATIONSHIP
108.0
106.0

Dry Density (pcf)

104.0

Can’t achieve fill in zone
right of zero air voids line

102.0
100.0
98.0

Z

96.0
94.0
92.0
90.0
88.0
8.0

10.0

12.0

14.0

16.0

18.0

20.0

22.0

24.0

26.0

28.0

Moisture Content (%)

45
Proctor: Moisture Density Relationships
If SG = 2.65 & moisture
content is 24%
What dry density achieves
100% saturation?

MOISTURE-DENSITY RELATIONSHIP
108.0
106.0

Dry Density (pcf)

104.0
102.0
100.0
98.0
96.0
94.0
92.0
90.0
88.0
8.0

10.0

12.0

14.0

16.0

18.0

20.0

22.0

24.0

26.0

28.0

A) 100.0 pcf

Moisture Content (%)

B) 101.1 pcf
46
Proctor: Moisture Density Relationships
MOISTURE-DENSITY RELATIONSHIP

γd=SG62.4/(1+ωSG/100)
γd=2.65x62.4/(1+24x2.65/100)
γd=101.1 pcf

108.0
106.0

Dry Density (pcf)

104.0
102.0

X

100.0
98.0
96.0
94.0
92.0

Answer is “B”

90.0
88.0
8.0

10.0

12.0

14.0

16.0

18.0

20.0

22.0

24.0

26.0

28.0

Moisture Content (%)

47
Ref: Peck Hanson & Thornburn

Static Head

48
Calculate effective stress at point x
Ref: Peck Hanson & Thornburn

Saturated Unit Weight δsat
5’
δsat = 125 pcf

Moist Unit Weight δM
Dry Unit Weight δDry

7’ Submerged (buoyant) Unit Weight
= δsat - 62.4

x

49
Calculate effective stress at point x
Ref: Peck Hanson & Thornburn

Total Stress at X
5’

= 5 x 62.4+ 7x 125= 1187psf
Pore Pressure at X

δsat = 125 pcf

7’

= 12 x 62.4 = 749 psf
Effective Stress at X
= 1187-749= 438 psf

x

or (125-62.4) x 7=438 psf

50
Ref: Peck Hanson & Thornburn
Downward Flow
Gradient

51
Downward Flow Gradient
3’

5’

Total Stress at X
= 5 x 62.4+ 7x 125= 1187psf
Pore Pressure at X

δsat = 125 pcf

7’

= (12-3) x 62.4 = 562 psf
Effective Stress at X
= 1187-562 = 625 psf

x

or 438 + 3 x 62.4 = 625psf
see previous problem
52
Upward Flow Gradient

Ref: Peck Hanson & Thornburn

53
One Dimensional Consolidation

Δe/pn

54
Primary Phase Settlement (e log p)
ΔH = (H x Δe)/(1+eo)
eo

ΔH

H

55
Consolidation Test
Pre-consolidation Pressure

Cc = slope of e log p virgin curve
est. Cc = 0.009(LL-10%) Skempton
Rebound or recompression curves
56
56
Calculate Compression
Index; Cc

e- l o g p

1.50

1.40

Void Ratio (e)

1.30

1.20

1.10

ksf
0.1
1
4
8
16
32

(e)
1.404
1.404
1.375
1.227
1.08
0.932

1.00

0.90

0.80
0.1

1

10

Pr essur e ( ksf )

A) 0.21
B) 0.49

100

57
Cc is the slope of the virgin e-log p
e- l o g p

Cc = -(e1-e2)/log (p1/p2)
1.50

1.40

Void Ratio (e)

Cc

Cc=-(1.375-1.227)/log(4/8)
Cc = 0.49

1.30

Answer is “B”

1.20

1.10

1.00

0.90

0.80
0.1

1

10

Pr essur e ( ksf )

100

ksf
0.1
1
4
8
16
32

(e)
1.404
1.404
1.375
1.227
1.08
0.932

58
Permeability
Constant Head Conditions
• Q=kiAt
• Q= k (h/L)At
• k=QL/(Ath)

59
If Q =15cc & t = 30 sec
what is the permeability
k=QL/(Ath)
10cm
5cm

A) 0.01 cm/sec
B) 0.01x10-2 cm/sec

25cm2

C) 0.1 cm/sec

60
Constant Head Permeability

10cm
5cm

Calculate k
Q =15cc & t = 30 sec
• k=QL/(Ath)
• k= 15(5)/(25(30)10)
• k= 0.01 cm/sec
Answer is “A”

25cm2

61
Falling Head Permeability
• k=QL/(Ath)
(but h varies)
• k=2.3aL/(At) log (h1/h2)
• where a = pipette area
• h1 = initial head
• h2 = final head

62
If t = 30 sec; h1= 30 cm; h2 = 15 cm
L= 5 cm; a= 0.2 cm2; A= 30 cm2; calculate k
A) 2.3x10-3 cm/sec
B) 8.1x10-6 cm/sec
C) 7.7x10-4 cm/sec

63
Falling Head Permeability
k=2.3aL/(At) log (h1/h2)
k= 2.3 (0.2) 5 /(30x30) log (30/15)
k= 7.7x10-4 cm/sec
Answer is “C”

64
•Flow lines & head drop lines must
intersect at right angles
•All areas must be square
•Draw minimum number of lines
•Results depend on ratio of Nf/Nd

Flow Nets

6ft

2ft

65
Q=kia=kHNf /Nd wt (units = volume/time)
w= unit width of section
t=time

Flow Nets

6ft

66
Flow Nets

What flow/day?
assume k= 1x10-5 cm/sec
=0.0283 ft/day
Q= kH (Nf /Nd) wt
Q= 0.0283x8x(4.4/8)x1x1
Q= 0.12 cf/day

6ft

2ft

67
Flow Nets

Check for “quick conditions”
pc =2(120)= 240 psf (total stress)

μ= 2(62.4) = 124.8 (static pressure)
Δμ=

1/8(8)(62.4)= 62.4 (flow gradient)

p’c = pc -(μ+ Δμ)

= 240-(124.8+62.4)

p’c = 52.8 psf >0, soil is not quick

6ft

Below water level
use saturated unit
weight for total
stress

2ft
2ft

γsat=120 pcf

68
Stress Change Influence (1H:2V)

For square
footing

Δσz=Q/(B+z)2

69
If Q= 20 kips, Calculate the vertical stress
increase at 7 feet below the footing bottom

5’

8’

7’

70
If Q= 20 kips, Calculate the vertical stress
increase at 7 feet below the footing bottom

5’

8’

Δσz =

20000
(8+7)(5+7)

7’

Δσz = 111 psf

71
Westergaard (layered elastic & inelastic material)
If B= 6.3’ in a square
footing with 20 kips
load, what is the vertical
stress increase at 7’
below the footing
bottom?

72
Westergaard

Q = 20 kips
B = 6.3’
Z = 7’
Δσz = ?

73
Westergaard
7’/6.3’ = 1.1B
Δσz = 0.18 x 20000/6.32
= 90.7 psf

74
Boussinesq
(homogeneous elastic)

Q = 20 kips
B = 6.3’
Z = 7’
Δσz = ?

75
Boussinesq
Z/B = 1.1
Δσz = 0.3 x 20000/6.32
= 151 psf

76
Thanks for participating in the PE review course on
Soil Mechanics!

More questions or comments?

You can email me at:
gtv@gemeng.com

77

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Soil mechanics

  • 1. SOIL MECHANICS (version Fall 2008) Presented by: Jerry Vandevelde, P.E. Chief Engineer GEM Engineering, Inc. 1762 Watterson Trail Louisville, Kentucky (502) 493-7100 1
  • 2. National Council of Examiners for Engineering and Surveying http://www.ncees.org/ 2
  • 3. STUDY REFERENCES • Foundation Engineering; Peck Hanson & Thornburn •Introductory Soil Mechanics and Foundations; Sowers •NAVFAC Design Manuals DM-7.1 & 7.2 •Foundation Analysis and Design; Bowles •Practical Foundation Engineering Handbook; Brown 3
  • 4. Soil Classification Systems * Unified Soil Classification System * AASHTO Need: Particle Sizes and Atterberg Limits 4
  • 5. Particle Sizes (Sieve Analysis) (Well Graded) (Poorly Graded) 0.1 5
  • 6. Atterberg Limits Liquid, Plastic & Shrinkage Limits Plasticity Index (PI) PI = Liquid Limit - Plastic Limit (range of moisture content over which soil is plastic or malleable) 6
  • 8. 8 Ref: Peck Hanson & Thornburn 2nd Ed.
  • 9. Effective Size = D10 10 percent of the sample is finer than this size D60 = 1.6mm D30 = 0.2mm D10 = 0.03mm 0.1 0.1 9
  • 10. Uniformity Coefficient (Cu) = D60/D10 Coefficient of Curvature (Cz) = (D30)2/(D10xD60) D60 = 1.6mm D30 = 0.2mm D10 = 0.03mm 0.1 10
  • 11. Well Graded - Requirements 50% coarser than No. 200 sieve Uniformity Coefficient (Cu) D60/D10 >4 for Gravel > 6 for Sand Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 1 to 3 11
  • 12. Is the better graded material a gravel? 81% Passing No. 4 18% Finer No. 200 0.1 0.1 12
  • 13. Gravel if > 50 Percent Coarse Fraction retained on No. 4 sieve 81% Passing No. 4 18% Finer No. 200 % Retained on No. 200 = 82% 1/2 = 41% 19% (100-81) retained on No. 4 sieve (gravel) 19< 41 half of coarse fraction ∴ sand (“S”) 0.1 13
  • 14. Well Graded Sand? D60 = 1.6mm Uniformity Coefficient (Cu) > 6 = D60/D10 D30 = 0.2mm D10 = 0.03mm 0.1 Coefficient of Curvature (Cz) = 1 to 3 = (D30)2/(D10xD60) 14
  • 15. Well Graded Sand? Uniformity Coefficient (Cu) D60/D10 = 1.6/.03 = 53 > 6 D60 = 1.6mm D30 = 0.2mm D10 = 0.03mm 0.1 Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 0.22/(.03x1.6) = 0.83 <1 to 3 ∴Poorly graded 15
  • 16. What classification? Unified Classification of Coarse Soils with Fines < 5% Passing No. 200 sieve:GW,GP, SW, SP 81% Passing No. 4 5% - 12% Passing No. 200 sieve: Borderline- use dual symbols 18% Finer No. 200 > 12% Passing No. 200 sieve: GM, GC, SM, SC 0.1 >12% passing No. 200 sieve Since = “S” ∴ SC or SM 16
  • 17. What Unified Classification if LL= 45 & PI = 25? From sieve data SC or SM 0.1 A) “SC” B) “SM” C) “CL” or D) “SC & SM” 17
  • 18. Unified Classification Answer is “A” ⇒ SC 18
  • 19. AASHTO (American Association of State Highway and Transportation Officials) 19
  • 20. What is the AASHTO Classification? 65% Passing No. 10 40% Passing No. 40 18% Finer No. 200 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 20
  • 21. 18 percent passing No. 200 sieve; 65 percent passing No. 10 sieve 40 percent passing No. 40 sieve; assume LL = 45 & PI = 25 21
  • 22. AASHTO Classification 1 2 3 4 4 1) 18 % passing No. 200 sieve 2) 65% passing No. 10 sieve 3) 40% passing No. 40 sieve 4) assume LL = 45 & PI = 25 22
  • 24. Mass-Volume (Phase Diagram) • Unit volume of soil contains: Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil – Air (gases) – Water (fluid) – Solid Particles Ws 24
  • 25. Moisture Content = ω weight of water/ weight of dry soil ω = Ww/Wd water loss/(moist soil weight - water loss) ω = Ww/(Wm-Ww) and ω =(Wm-Wd)/Wd 25
  • 26. Mass - Volume Relationships Density or Unit Weight = Moist Unit Weight = γm γ γm = Wm/Vt = γd + ω γd ω = (γ m - γ d )/ γd ω γd + γd = γm γm= (1+ ω) γd γd = γm/(1+ ω) b 26
  • 27. Total Volume = ∑ Volume (solid + water + air) = Vs+Vw+Va ∴ Va = Vt - Vs- Vw Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 27
  • 28. Relationship Between Mass & Volume Volume = Mass/(Specific Gravity x Unit Weight of Water) = Ws/(SGxWw) Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 28
  • 29. Specific Gravity = weight of material/ weight of same volume of water Soil Specific Gravity Typical Range 2.65 to 2.70 Specific Gravity of Water = 1 29
  • 30. Saturation = S expressed as percent S = volume of water/ volume of voids x 100 Total Volume Vt Va Vv Vw S = Vw/Vv x 100 Air Total Water Ww Weight Wt Vs Soil Always ≤ 100 Ws 30
  • 31. Porosity n = volume of voids/ total volume n = Vv/Vt Void Ratio e = volume of voids/ volume of solids e = Vv/Vs Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 31
  • 32. What is the degree of saturation for a soil with: SG = 2.68, γm = 127.2 pcf & ω = 18.6 percent A) 88.4 Total Volume Vt Va Vv Vw Air Total Water Ww Weight B) 100.0 Wt Vs Soil Ws C) 89.1 32
  • 33. What are the porosity and degree of saturation for a soil with: SG = 2.68, γm = 127.2 pcf & ω = 18.6 percent γd = γm/(1+ ω) = 127.2/(1.186) = 107.3pcf Total Volume Vt Va Vv Vw Vs Air Total Water Soil Ww Weight Wt Ws Ww = γm- γd = 19.9 pcf Vw = Ww/62.4 = 0.319 cf Vs = γd /(SGx62.4) = 0.642 cf Va = Vt - Vw - Vs = 1- 0.319 - 0.642 = 0.039 cf Vv = Vw + Va = 0.358 cf 33
  • 34. What are the porosity and degree of saturation for a soil with: SG = 2.68, γm = 127.2 pcf & ω = 18.6 percent Vw = 0.319 cf, Vs = 0.642 cf, Vv = 0.358 cf Total Volume Vt Va Vv Vw Vs Air Degree of Saturation = Vw/Vv x 100 Total Water Soil Ww Weight Wt Ws = 0.319/0.358 x 100 = 89.1% Answer is “C” 34
  • 36. Borrow Fill Adjustments Borrow Material Properties: γm = 110 pcf & ω = 10% Placed Fill Properties: γd = 105 pcf & ω = 20% How much borrow is needed to produce 30,000 cy of fill? How much water must be added or removed from each cf of fill? Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 36
  • 37. Borrow Fill Adjustments Borrow Material Properties: γm = 110 pcf & ω = 10% γd = γm /(1+ω) = 110/(1.10) =100 pcf; Ww = 110-100=10 lbs Placed Fill Properties: γd = 105 pcf & ω = 20% Ww = ωx γd = 0.2x 105 = 21 lbs Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 37
  • 38. Borrow Fill Adjustments Borrow Properties: γm = 110 pcf, γd =100 & ω = 10% Placed Fill Properties: γd = 105 pcf & ω = 20% Since borrow γd =100pcf & fill γd =105pcf, 105/100 =1.05 It takes 1.05 cf of borrow to make 1.0 cf of fill For 30,000 cy, 30,000 x 1.05 = 31,500 cy of borrow Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 38
  • 39. Borrow Fill Adjustments Borrow Material Properties: Ww = 10 lbs Placed Fill Properties: Ww = 21 lbs Water supplied from borrow in each cf of fill = 10 x 1.05 = 10.5 lbs; 21 lbs - 10.5 = 10.5 lbs short/1.05 cf 10.5lbs/1.05 cy = 10 lbs of water to be added per cf borrow Total Volume Vt Va Vv Vw Air Total Water Ww Weight Wt Vs Soil Ws 39
  • 40. Proctor: Moisture Density Relationships Establishes the unique relationship of moisture to dry density for each specific soil at a specified compaction energy MOISTURE-DENSITY RELATIONSHIP 108.0 106.0 D ry D ensity (pcf) 104.0 102.0 100.0 98.0 96.0 94.0 92.0 90.0 88.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 Moisture Content (%) 40
  • 41. Proctor: Moisture Density Relationships • 4” mold 25 blows • 6” mold 56 blows • Standard – 5.5 lb hammer – dropped 12 in – 3 layers Standard: ASTM D-698 AASHTO T-99 Modified: ASTM D-1557 AASHTO T-150 • Modified – 10 lb hammer – dropped 18 in – 5 layers 41
  • 42. PROCTOR COMPACTION TEST Maximum Dry Density - Highest density for that degree of compactive effort Optimum Moisture Content - Moisture content at which maximum dry density is achieved for 42 that compactive effort
  • 43. Proctor: Moisture Density Relationships What density is required for 95% Compaction? MOISTURE-DENSITY RELATIONSHIP 108.0 106.0 Dry Density (pcf) 104.0 102.0 100.0 98.0 96.0 94.0 92.0 90.0 88.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 What range of moisture would facilitate achieving 95% compaction? Moisture Content (%) 43
  • 44. Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP 104 x .95 = 98.8 pcf 108.0 106.0 Dry Density (pcf) 104.0 A 102.0 100.0 B Range of moisture is within the curve A to B (14 to 24 %) 95% 98.0 96.0 94.0 92.0 90.0 88.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 Moisture Content (%) 44
  • 45. Proctor: Zero Air Voids Line Relationship of density to moisture at saturation for constant specific gravity (SG) MOISTURE-DENSITY RELATIONSHIP 108.0 106.0 Dry Density (pcf) 104.0 Can’t achieve fill in zone right of zero air voids line 102.0 100.0 98.0 Z 96.0 94.0 92.0 90.0 88.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 Moisture Content (%) 45
  • 46. Proctor: Moisture Density Relationships If SG = 2.65 & moisture content is 24% What dry density achieves 100% saturation? MOISTURE-DENSITY RELATIONSHIP 108.0 106.0 Dry Density (pcf) 104.0 102.0 100.0 98.0 96.0 94.0 92.0 90.0 88.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 A) 100.0 pcf Moisture Content (%) B) 101.1 pcf 46
  • 47. Proctor: Moisture Density Relationships MOISTURE-DENSITY RELATIONSHIP γd=SG62.4/(1+ωSG/100) γd=2.65x62.4/(1+24x2.65/100) γd=101.1 pcf 108.0 106.0 Dry Density (pcf) 104.0 102.0 X 100.0 98.0 96.0 94.0 92.0 Answer is “B” 90.0 88.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 22.0 24.0 26.0 28.0 Moisture Content (%) 47
  • 48. Ref: Peck Hanson & Thornburn Static Head 48
  • 49. Calculate effective stress at point x Ref: Peck Hanson & Thornburn Saturated Unit Weight δsat 5’ δsat = 125 pcf Moist Unit Weight δM Dry Unit Weight δDry 7’ Submerged (buoyant) Unit Weight = δsat - 62.4 x 49
  • 50. Calculate effective stress at point x Ref: Peck Hanson & Thornburn Total Stress at X 5’ = 5 x 62.4+ 7x 125= 1187psf Pore Pressure at X δsat = 125 pcf 7’ = 12 x 62.4 = 749 psf Effective Stress at X = 1187-749= 438 psf x or (125-62.4) x 7=438 psf 50
  • 51. Ref: Peck Hanson & Thornburn Downward Flow Gradient 51
  • 52. Downward Flow Gradient 3’ 5’ Total Stress at X = 5 x 62.4+ 7x 125= 1187psf Pore Pressure at X δsat = 125 pcf 7’ = (12-3) x 62.4 = 562 psf Effective Stress at X = 1187-562 = 625 psf x or 438 + 3 x 62.4 = 625psf see previous problem 52
  • 53. Upward Flow Gradient Ref: Peck Hanson & Thornburn 53
  • 55. Primary Phase Settlement (e log p) ΔH = (H x Δe)/(1+eo) eo ΔH H 55
  • 56. Consolidation Test Pre-consolidation Pressure Cc = slope of e log p virgin curve est. Cc = 0.009(LL-10%) Skempton Rebound or recompression curves 56 56
  • 57. Calculate Compression Index; Cc e- l o g p 1.50 1.40 Void Ratio (e) 1.30 1.20 1.10 ksf 0.1 1 4 8 16 32 (e) 1.404 1.404 1.375 1.227 1.08 0.932 1.00 0.90 0.80 0.1 1 10 Pr essur e ( ksf ) A) 0.21 B) 0.49 100 57
  • 58. Cc is the slope of the virgin e-log p e- l o g p Cc = -(e1-e2)/log (p1/p2) 1.50 1.40 Void Ratio (e) Cc Cc=-(1.375-1.227)/log(4/8) Cc = 0.49 1.30 Answer is “B” 1.20 1.10 1.00 0.90 0.80 0.1 1 10 Pr essur e ( ksf ) 100 ksf 0.1 1 4 8 16 32 (e) 1.404 1.404 1.375 1.227 1.08 0.932 58
  • 59. Permeability Constant Head Conditions • Q=kiAt • Q= k (h/L)At • k=QL/(Ath) 59
  • 60. If Q =15cc & t = 30 sec what is the permeability k=QL/(Ath) 10cm 5cm A) 0.01 cm/sec B) 0.01x10-2 cm/sec 25cm2 C) 0.1 cm/sec 60
  • 61. Constant Head Permeability 10cm 5cm Calculate k Q =15cc & t = 30 sec • k=QL/(Ath) • k= 15(5)/(25(30)10) • k= 0.01 cm/sec Answer is “A” 25cm2 61
  • 62. Falling Head Permeability • k=QL/(Ath) (but h varies) • k=2.3aL/(At) log (h1/h2) • where a = pipette area • h1 = initial head • h2 = final head 62
  • 63. If t = 30 sec; h1= 30 cm; h2 = 15 cm L= 5 cm; a= 0.2 cm2; A= 30 cm2; calculate k A) 2.3x10-3 cm/sec B) 8.1x10-6 cm/sec C) 7.7x10-4 cm/sec 63
  • 64. Falling Head Permeability k=2.3aL/(At) log (h1/h2) k= 2.3 (0.2) 5 /(30x30) log (30/15) k= 7.7x10-4 cm/sec Answer is “C” 64
  • 65. •Flow lines & head drop lines must intersect at right angles •All areas must be square •Draw minimum number of lines •Results depend on ratio of Nf/Nd Flow Nets 6ft 2ft 65
  • 66. Q=kia=kHNf /Nd wt (units = volume/time) w= unit width of section t=time Flow Nets 6ft 66
  • 67. Flow Nets What flow/day? assume k= 1x10-5 cm/sec =0.0283 ft/day Q= kH (Nf /Nd) wt Q= 0.0283x8x(4.4/8)x1x1 Q= 0.12 cf/day 6ft 2ft 67
  • 68. Flow Nets Check for “quick conditions” pc =2(120)= 240 psf (total stress) μ= 2(62.4) = 124.8 (static pressure) Δμ= 1/8(8)(62.4)= 62.4 (flow gradient) p’c = pc -(μ+ Δμ) = 240-(124.8+62.4) p’c = 52.8 psf >0, soil is not quick 6ft Below water level use saturated unit weight for total stress 2ft 2ft γsat=120 pcf 68
  • 69. Stress Change Influence (1H:2V) For square footing Δσz=Q/(B+z)2 69
  • 70. If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5’ 8’ 7’ 70
  • 71. If Q= 20 kips, Calculate the vertical stress increase at 7 feet below the footing bottom 5’ 8’ Δσz = 20000 (8+7)(5+7) 7’ Δσz = 111 psf 71
  • 72. Westergaard (layered elastic & inelastic material) If B= 6.3’ in a square footing with 20 kips load, what is the vertical stress increase at 7’ below the footing bottom? 72
  • 73. Westergaard Q = 20 kips B = 6.3’ Z = 7’ Δσz = ? 73
  • 74. Westergaard 7’/6.3’ = 1.1B Δσz = 0.18 x 20000/6.32 = 90.7 psf 74
  • 75. Boussinesq (homogeneous elastic) Q = 20 kips B = 6.3’ Z = 7’ Δσz = ? 75
  • 76. Boussinesq Z/B = 1.1 Δσz = 0.3 x 20000/6.32 = 151 psf 76
  • 77. Thanks for participating in the PE review course on Soil Mechanics! More questions or comments? You can email me at: gtv@gemeng.com 77