# Soil Mechanics by Jerry Vandevelde

Topics: Soil mechanics, Soil classification, Unified Soil Classification System Pages: 25 (2665 words) Published: October 21, 2012
SOIL MECHANICS
(version Fall 2008)

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

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

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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

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Particle Sizes (Sieve Analysis)

0.1

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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)

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UNIFIED SOIL CLASSIFICATION SYSTEM ASTM D-2487

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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

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

D60 = 1.6mm

D30 = 0.2mm D10 = 0.03mm

0.1

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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

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Is the better graded material a gravel?

81% Passing No. 4

18% Finer No. 200

0.1

0.1

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Gravel if > 50 Percent Coarse

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

81% Passing No. 4

18% Finer No. 200

∴ sand
0.1

(“S”)
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Uniformity Coefficient (Cu) > 6 = D60/D10 Coefficient of Curvature (Cz) = 1 to 3 = (D30)2/(D10xD60) 14

D60 = 1.6mm D30 = 0.2mm D10 = 0.03mm

0.1

Uniformity Coefficient (Cu) D60/D10 = 1.6/.03 = 53 > 6
D60 = 1.6mm D30 = 0.2mm D10 = 0.03mm

Coefficient of Curvature (Cz) = (D30)2/(D10xD60) = 0.22/(.03x1.6) = 0.83 12% Passing No. 200 sieve: GM, GC, SM, SC

0.1

>12% passing No. 200 sieve Since = “S” ∴ SC or SM
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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”
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Unified Classification

Answer is “A” ⇒ SC

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AASHTO
(American Association of State Highway and Transportation Officials)

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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

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

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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

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AASHTO Group Index

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

Va

Air
Total

Vt

Vv Vw Vs

Water

Ww Ws

Weight

Wt Soil

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

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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

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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

Va

Air
Total

Vt

Vv Vw Vs

Water

Ww Ws

Weight

Wt Soil

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

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