*Review text lumps density and unit weight together and uses symbol ρ
Gas Water Solid V g V w V s V v V t O W w W s W t
17.
Common practice is to assume V s = 1, then express other volumes and weights accordingly. From definitions Gas Water Solid wG s 1 e wG s ρ w G s ρ w ρ sub = ρ sat - ρ w Buoyant unit weight ρ sat = ρ m for S=100% Saturated unit weight ρ d = W s /V t Dry unit weight (dry density) ρ m = W t /V t Moist unit weight
18.
Table 35.7 - Useful for rapid calculation of phase relationships
19.
Given : e = 0.62 w = 15% G s = 2.65 Calculate : a. ⍴ d b. ⍴ m c. w for S = 100% d. ⍴ sat for S = 100% Example Gas Water Solid wG s 1 e wG s ρ w= S e ρ w G s ρ w
Given by slope of rebound portion of curve (OC portion)
Coefficient of Consolidation, C v e i e f High C v (fast settlement) Low C v (slow settlement) time e C c C r Log p e Normally consolidated Over-consolidated
Shear Strength of soils friction cohesion c s For drained loading, c = 0 S Typical for sands For un-drained loading , S Typical for clays c S nc
31.
Shear Strength and Principal Stresses Ϭ 3 Ϭ 1 Ϭ Շ c Փ Failure surface is always oriented at 45 + Փ/2 angle to minor principal stress axis At failure Shear strength Shear stress failure 45+ /2
The increase in stress produced by an areal load is constant with depth Local Load Spread footing imposes uniform load of 1,000 psf over 10 ft x 10 ft area What is σ v ' different below edge of footing than below center. Different at depth than shallow Examples Z=20’ 5 ‘ ρ m = 120 10 ‘ z ρ m = 132 5 ‘ ρ m = 110 15 ‘ ρ sub = 66
Important to be able to calculate subsurface stresses caused by loads or loaded areas on the ground surface.
Usually interested for settlement calculation problems.
Generally accomplished by stress distribution methods based on theory of elasticity.
Can use principle of superposition very useful.
Boussinesq – stresses caused by point load on surface.
Boussinesq solution widely used
For point load, use Eq. 40.1
Example 1
For strip footing loads, use Appendix 40a (left)
Example 2
For square footing loads, use Appendix 40a (right)
Example 3
For circular loaded areas, use Appendix 40b
Example 4
For loaded areas of arbitrary shape, use (Newmark)
Influence chart method – see Fig 40 3
Influence Chart
Represents entire ground surface
Divided into number of “squares” – see Fig 40.3
Squares set up so that uniform load on each
would cause same stress on subsurface
point below center of chart
38.
Example 1 Calculate vertical stress 5 ft. below and 2 ft. to the side of a surface point load of 1,000 lbs. 1000 lbs 5 ' 2 ' 5 ' P v Example 2 Calculate the vertical stress at a depth of 15 feet below the edge of a 5-foot-wide strip footing which imposes a bearing pressure of 2,000 psf on the ground surface. 15 ' P v 0.2p Chart in Appendix 40A p. A-69 (left side) 2,000 psf
40.
Example 3 Calculate the vertical stress at a depth of 14 feet below the center of a 4 ft. square footing that applies 10,000 psf bearing pressure to the ground surface 4 ' 14 ' P v 0.04p 10,000 psf p. 40.A Right side
42.
Example 3 Calculate the vertical stress at a depth of 14 feet below the center of a 4 ft. square footing that applies 10,000 psf bearing pressure to the ground surface 4 ' p. 40.A Right side Example 4 A 16 ft diameter water tank contains 20 feet of water. Calculate the vertical stress caused by the tank at a point 8 feet below the ground surface and 10 feet from the center of the tank. 8 ' P v 16 ft 10 ' 14 ' P v 0.04p 10,000 psf I = 0.2 Appendix D p. 40.B
44.
Determination of appropriate soil properties Compute C c or C r from e-log p curve Consolidation test C c applies to normally consolidated range C r applies to over-consolidated range Initial Conditions Final Conditions C c or C r e e 1 e 2 p 1 p 2 Log p
3. Definition of vertical strain P ‘ 1 P ‘ 2 Log p e e 1 e 2 P ‘ p e p OC NC initial final
49.
First, calculate initial effective stress at center of soft clay layer before new fill placed Next, calculate final stress after placement of new fill Then, calculate ultimate settlement as Example 5 Calculate the ultimate settlement of the soft clay layer due to placement of the new fill 4 ' 3 ' 2 ' 5 ' New Fill = 125 pcf; w= 10% Old Fill Same properties as new fill Soft Clay C c =1.06 e o =2.53 sub = 30 pcf Dense Sand
50.
Now, what would happen if half of the new fill was removed ? Since effective stress is decreasing, use C r Assuming C r = 0.10 rebound
51.
Let’s now assume that 4 more feet of new fill is placed, bringing the total thickness Of new fill to 6 ft. Then
Rate controlled by coefficient of consolidation, C v
High C v rapid consolidation
Low C v slow consolidation
Degree of consolidation
Fraction of ultimate settlement which has occurred by time t
Fraction of ultimate settlement which has occurred by time t Time required to reach given degree of consolidation Dimensionless time factor Settlement at given time t Where T v (t) and U(t) are related by Eq 40.23 and Table 40.1 50 90 100 .2 .85 T v 0% U Length of longest drainage path
54.
Example 6 If C v for the soft clay of Example 5 was 10ft 2 /yr, how long would it take for 2 in of settlement to occur? What if the soft clay was underlain by impermeable bedrock? Then z= 5 ft Double drainage Single drainage