UNIVERSITY OF CEBU-MAIN CAMPUS
SANCIANGKO ST., CEBU CITY
COLLEGE OF ENGINEERING
PRIMARY, SECONDARY AND TIME RATE
OF CONSOLIDATION
ABANGAN, ROSIE JANE P.
ENGBINO, MAE CLAIRE P.
BSCE-5
ENGR. FAELNAR
INSTRUCTOR
PRIMARY CONSOLIDATION SETTLEMENT
A stratum of normally consolidated clay 9 m thick is located at a depth 15m below ground
level. The natural moisture content of the clay is 50% and its liquid limit is 45%. The specific
gravity of the solid particles is 3.10. The water table is located at a depth 4 m below ground
surface. The soil is sand above the clay stratum. The submerged unit weight of the sand is 12
kN/m3 and the same weighs 16 kN/m3 above the water table. The average increase in pressure
at the center of the clay stratum is 110 kN/m2 due to the weight of a building that will be
constructed on the sand above the clay stratum. Estimate the expected settlement of the
structure.
Solution:
e Ww 4m
h3 & Y3
1+e 3m
h2 & Y2
W
1 Ws 9m
h1 & Y1 CLAY
(a) (b)
Determination of e and Yb for the clay.
Ww = Wn (Ws) = VsGsYw = (1)(3.10)(1) = 3.10 g
Ws
Ww = 50 (3.10) = 1.55 g
100
Vw = Ww = 1.55 = 1.55 cm3
Yw 1
e0 = V w = 1.55 cm3 = 1.55
Vs 1
W = Ww + Ws = 1.55 + 3.10 = 4.65 g
Yt = W = 4.65 = 1.82 g/cm3
1+ e0 2.55
Yb = (1.82 -1) = 0.82 g/cm3
Determination of overburden pressure P0.
P0 = Y1h1 + Y2h2 + Y3h3
P0 = (0.82)(9.81)(4.5) + (12)(3) + (16)(4)
P0 = 136.20 kN/m2
Compression index
Cc = 0.009 (wt – 10)
Cc = 0.009 (45 – 10)
Cc = 0.32
Excess pressure
Δp = 110 kN/m2
Total Settlement
St = Cc H log P0+ Δp
1+e0 P0
St = 0.32 (900) log 136.20+110
2.55 136.20
St = 29.04 cm
Estimated settlement = 29.04 cm
TIME RATE OF CONSOLIDATION SETTLEMENT
An 8m depth of sand overlies a 6m layer of clay, below which is an impermeable
stratum; the water table is 2m below the surface of the sand. Over a period of 1 year, a
3m depth of fill (unit weight 20 kN/m3) is to be dumped on the surface over an extensive
area. The saturated unit weight of the sand is 19 kN/m3 and that of the clay is 20 kN/m3;
above the water table the unit weight of the sand is 17 kN/m3.
For the clay, the relationship between void ratio and effective stress (units kN/m2)
𝜎′
can be represented by the equation e = 0.88 – 32 log (100) and the coefficient of
consolidation is 1.26 m2/year.
(a)Calculate the final settlement of the area due to consolidation of the clay and
the settlement after a period of 3 years from the start of dumping.
(b) If a very thin layer of sand, freely draining, existed 1.5m above the bottom of
the clay layer, what would be the values of the final and 3-year settlements?
Solution:
2m
8m W.T.
Sand
6m Clay
d=6m d = 2.25 m
d = 1.5 m
Impermeable
Since the fill covers a wide area, the problem can be considered to be one-dimensional.
The consolidation settlement will be calculated in terms of Cc, considering the clay layer
as a whole, and therefore the initial and final values of effective vertical stress at the
centre of the clay layer are required.
σ’0 = (17)(2) + (9.2)(6) + (10.2)(3)
σ’0 = 119.8 kN/m2
e0 = 0.88 – 0.32 log 1.198
e0 = 0.855
σ’1 = 119.8 +(3)(20)
σ’1 = 179.8 kN/m2
179.8
log (119.8) = 0.176
Scf = (0.32)(0.176)(6000)
1.855
Scf = 182 mm
In the calculation of the degree of consolidation 3 years after the start of dumping, the
corrected value of time to allow for the 1-year dumping period is:
1
t=3- = 2.5 years
2
The layer is half-closed, and therefore d= 6 m. Then:
Tv = cv t = 1.26(2.5)
d2 62
Tv = 0.0875
When Tv = 0.0875 & U=0.335
Sc = (0.335)(182)
Sc = 61 mm
Settlement after 3 years is 61 mm
The final settlement will still be 182mm (ignoring the thickness of the drainage layer): only
the rate of settlement will be affected. From the point of view of drainage there is now
an open layer of thickness 4.5m (d = 2.25 m) above a half-closed layer of thickness 1.5m
(d = 1.5 m).
By proportion:
Tv1 = (0.0875)(62/2.252)
Tv1 = 0.622
U1 = 0.825
and
Tv2 = (0.0875)(62/1.52)
Tv2 = 1.40
U2 = 0.97
Now for each layer, Sc= UScf which is proportional to UH . Hence if Ū is the overall degree
of consolidation for the two layers combined:
4.5U1 + 1.5U2 = 6.0 Ū
4.5(0.825) + 1.5(0.97) = 6.0 Ū
Hence Ū= 0.86 and the 3 year settlement is:
Sc = (0.86)(182)
Sc = 157 mm
SECONDARY CONSOLIDATION SETTLEMENT
For a normally consolidated clay layer in the field, the following values are given:
Thickness of clay layer is 4.2 m.
Void Ratio (e0) is 0.7
Compression Index (Cc) is 0.31
∆ơ is 38 kN/m2
Average effective pressure on the clay layer is 130 kN/m2
Secondary compression index (C∞) 0.04
In how many years will the total consolidation settlement of the clay layer becomes 90 mm
after the completion of primary consolidation settlement? (Note: time of completion of
primary settlement=2 years)
C’∞= C∞
1+ep
ep= e0 - ∆eprimary
∆e=Cc log (ơo+∆ơ)
ơo
=0.28 log (130+38)
130
∆e= 0.031
Primary consolidation
Sc= ∆eH
1+e0
= 0.031(4.2 x 1000)
1+.07
Sc =76.59 mm
e0= 0.7 and thus,
ep=0.7-.31
ep=0.669
Hence,
C’∞= 0.04
1+0.669
90=Sc+Ss
90=76.59+Ss
Ss=13.41
13.41=0.024(4.2x1000) log x
2
X=2.72 years
