1. Chapter 8 Civil Engineering Part 2: Geotechnical Engineering
4th Edition
Chapter 8
Civil Engineering
Part 2: Geotechnical Engineering

2. Part 2: Geotechnical Engineering
The stability of a structure is no better than its foundations.
Ask the people who built this 
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3. Trouble in the Basement …
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4. Properties of Soils
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5. sat = (Ws + Wp)/V = dry + n w
Wet Soil
The mechanical properties of soils depend how much the soil’s pores are water filled
sat = (Ws + Wp)/V = dry + n w
where Ws and Wp are the weight of dry soil and the weight of water in the pores. The corresponding weight densities are dry and w. V is the volume of the soil. The term, n, is the volume fraction of the pores in the soil.
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6. Dry Weight Densities of Different Soils
Wet Soils
Dry Weight Densities of Different Soils
Soil Type
Dry Density (lbf/ft3)
Porosity
Wet density
Lbf/ft3
Sand
94.8
0.375
118
Clay
74.9
0.550
109
Silt
79.9
0.425
106
Water
62.4 —
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7. Effective Stress Principle
Pressure Is What You Feel when Your Fingers Press with Force Fn on the Contact Patch of Area A
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8. Effective Stress Principle
Soil section in next slide being analyzed for stress on the horizontal plane (dashed line) at depth z. This force is actually distributed uniformly over the bottom face creating a constant pressure (or stress).

9. Effective Stress Principle
’ =  - pp = (sat - w)z is the stress at depth z
FBD
Effective stress on
layer of wet soil
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10. Example
Two top layers of a soil sample taken from a potential building site are shown in the next figure. If the thicknesses of the layers are HA = 2.50 ft and HB = 2.00 ft, determine:
(a) the effective stress at a depth of ZA = 2.00 ft in Layer A
(b) The effective stress at a depth of ZB = 3.00 ft in Layer B. Assume both layers are fully saturated with weight densities of γA = 78.6 lbf/ft3 and γB = 91.2 lbf/ft3. The weight density of water is γw = 62.4 lbf/ft3. 
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11. Example Continued
Need: Find the values of effective stress at depths of 2.00 ft and 3.00 ft.
Know: The depths of interest are ZA = 2.00 ft and ZB = 3.00 ft; the layer thicknesses are HA = 2.50 ft and HB = 2.00 ft; the weight densities of saturated soil and water are γA = 78.6 lbf/ft3, γB = 91.2 lbf/ft3, and γw = 62.4 lbf/ft3; both soil layers are fully saturated.
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12. Example Continued
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13. Example continued
How: Since all of the soils above z = ZA are of a uniform compositions and fully saturated, use effective stress principle. With two layers of soil above z = ZB generalize the effective stress equation for the weights z = ZB in the free-body diagram.
(1) draw the free-body diagram of the soil section,
(2) find Fn from the summation of forces in the vertical direction,
(3) substitute Fn into the definition of pressure to get total stress,
(4) subtract pore pressure from total stress to get effective stress.
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14. Example continued Solve:
a) ’ =  - pp = (sat - w)z = (78.6 – 62.4)(2.00) [lbf/ft3] [ft] = 32.4 lbf/ft2
Generalize the equation for the effective stress
’ = AHA + BHB - wz
= (78.6)(2.50) + (91.2)( ) – (62.4)(3.00) [lbf/ft2] = 54.9 lbf/ft2

15. Summary of Part 2 Geotechnical Engineering
Structures stand on soil whose properties are variable, particularly if wet
Starting with a FBD, stress in wet soil can be estimated from the weight of dry soil and wet soil at any layer
The effective stress equation allows the calculation of stress at any layer of wet soil knowing its porosity
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