CE0029

GEOTECHNICAL ENGINEERING 1

LESSON 9
CONSOLIDATION
 OBJECTIVES

• Discuss the fundamental principles for estimating the consolidation

settlements of soil layers under superimposed loadings
INTRODUCTION
A stress increase caused by the construction of foundations or other loads
compresses soil layers.

The compression is caused by:

a) deformation of soil particles
b) relocations of soil particles
c) expulsion of water or air from the void spaces.
INTRODUCTION
The soil settlement caused by loads may be divided into three broad categories:

1. Elastic settlement (or immediate settlement), which is caused by the elastic deformation of
dry soil and of moist and saturated soils without any change in moisture content.

2. Primary consolidation settlement, which is the result of a volume change in saturated

cohesive soils because of expulsion of water that occupies the void spaces.

3. Secondary consolidation settlement, which is the result of the plastic adjustment of soil
fabrics. It is an additional form of compression that occurs at constant effective stress.
INTRODUCTION
The total settlement of a foundation is given as

𝑆𝑇 = 𝑆𝑐 + 𝑆𝑠 + 𝑆𝑒

𝑤ℎ𝑒𝑟𝑒:
𝑆𝑇 = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡
𝑆𝑐 = 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡
𝑆𝑠 = 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡
𝑆𝑒 = 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡
CONSOLIDATION OF SOIL
Soil volume reduction due to expulsion of water upon application of external
load/stress.
CONSOLIDATION DAMAGES
Consolidation can cause cracks in walls, foundations, etc.
CONSOLIDATION vs COMPACTION

Compaction Consolidation
Applicable to unsaturated soils Applicable to saturated soils
Decrease in water voids (air voids do not
Decrease in air voids
exist)
Applicable for both fine-grained and
Only applicable for fine-grained soils
coarse-grained soils
Instantaneous process Time-dependent process
May be accomplished by rolling,
In general, caused by static loading
tamping, or vibration
CONSOLIDATION OF SOIL
Magnitude of consolidation settlement
- expressed in terms of compression index 𝐶𝑐

Rate of consolidation/settlement
- expressed in terms of coefficient of consolidation 𝐶𝑣
COMPRESSION INDEX
The compression index for the calculation of field
settlement can be determined by graphic
construction as shown in the figure after one
obtains the laboratory test results for void ratio
and pressure.

𝑒 − 𝑒′
𝐶𝑐 =
∆𝑃 + 𝑃𝑜
𝑙𝑜𝑔
𝑃𝑜
𝑤ℎ𝑒𝑟𝑒:
𝑒 = 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑎𝑡 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃𝑜
𝑒 ′ = 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑎𝑡 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ∆𝑃 + 𝑃𝑜
COMPRESSION INDEX
Skempton (1944) suggested the following empirical expression for the
compression index for undisturbed clays:

𝐶𝑐 = 0.009 (𝐿𝐿 − 10)

𝑤ℎ𝑒𝑟𝑒:
𝐿𝐿 = 𝑙𝑖𝑞𝑢𝑖𝑑 𝑙𝑖𝑚𝑖𝑡
SWELLING INDEX
The determination of the swelling index is important in the estimation of
consolidation settlement of overconsolidated clays. In most cases, the value of the
swelling index is 1/4 to 1/5 of the compression index.

1
𝐶𝑠 ≈ 𝐶𝑐
5
DEFINITIONS OF CLAYS
There are two basic definitions of clay based on stress history:

1. Normally Consolidated: The present effective overburden pressure (also known as

effective in-situ stress) is the maximum the soil has ever experienced.

2. Overconsolidated: The present effective overburden pressure 𝑃𝑜 is less than that which
the soil experienced in the past. The maximum effective past pressure is called the
preconsolidation pressure 𝑃𝑐 .
PRIMARY CONSOLIDATION SETTLEMENT
NORMALLY CONSOLIDATED CLAYS
The primary settlement 𝑆 for normally consolidated clays is as follows:

∆𝑒
𝑆= 𝐻
1 + 𝑒𝑜

𝐶𝑐 𝐻 ∆𝑃 + 𝑃𝑜
𝑆= 𝑙𝑜𝑔
1 + 𝑒𝑜 𝑃𝑜

𝑤ℎ𝑒𝑟𝑒:
𝐶𝑐 = 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑛𝑑𝑒𝑥
𝑒𝑜 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜
∆𝑒 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 (𝑓𝑖𝑛𝑎𝑙 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙)
𝐻 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑙𝑎𝑦𝑒𝑟
𝑃𝑜 = 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑜𝑣𝑒𝑟𝑏𝑢𝑟𝑑𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
∆𝑃 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑜𝑣𝑒𝑟𝑏𝑢𝑟𝑑𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
PRIMARY CONSOLIDATION SETTLEMENT
OVERCONSOLIDATED CLAYS 𝑷𝒐 < 𝑷𝒄
The primary settlement 𝑆 for overconsolidated clays is as follows:

Case 1: 𝑃𝑜 + ∆𝑃 ≤ 𝑃𝑐
𝑤ℎ𝑒𝑟𝑒:
𝐶𝑠 𝐻 ∆𝑃 + 𝑃𝑜
𝑆= 𝑙𝑜𝑔 𝐶𝑐 = 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑛𝑑𝑒𝑥
1 + 𝑒𝑜 𝑃𝑜 𝐶𝑠 = 𝑠𝑤𝑒𝑙𝑙 𝑖𝑛𝑑𝑒𝑥
𝑒𝑜 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜

Case 2: 𝑃𝑜 + ∆𝑃 > 𝑃𝑐 𝐻 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑙𝑎𝑦𝑒𝑟

𝑃𝑜 = 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑜𝑣𝑒𝑟𝑏𝑢𝑟𝑑𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝐶𝑠 𝐻 𝑃𝑐 𝐶𝑐 𝐻 ∆𝑃 + 𝑃𝑜 ∆𝑃 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑜𝑣𝑒𝑟𝑏𝑢𝑟𝑑𝑒𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝑆= 𝑙𝑜𝑔 + 𝑙𝑜𝑔 𝑃𝑐 = 𝑝𝑟𝑒𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
1 + 𝑒𝑜 𝑃𝑜 1 + 𝑒𝑜 𝑃𝑐
𝑃𝑐
𝑂𝐶𝑅 =
𝑃𝑜
SAMPLE PROBLEM 1
The soil profile shown in the figure consist of 9 layer of sand underlain by 6 m layer
of clay. The location of water table is 3 m from the ground level. The building to be
constructed will cause an additional pressure of 50 kPa on the ground. What is the
primary settlement of the clay layer if.
a. Clay is normally consolidated?
b. Preconsolidation pressure Pc=190 kPa?
3m 𝛾𝑑 = 16 kN/m3
c. Preconsolidation pressure Pc=170 kPa

6m 𝛾𝑠𝑎𝑡 = 18.5 kN/m3

𝑆𝑎𝑛𝑑

𝛾𝑠𝑎𝑡 = 19 kN/m3
6m 𝐶𝑐 = 0.36 𝐶𝑙𝑎𝑦
𝑒 = 0.95
1. The soil profile shown in the figure consist of 9 layer of sand underlain by 6 m layer of clay. The location of water table is 3 m from
the ground level. The building to be constructed will cause an additional pressure of 50 kPa on the ground. What is the primary
settlement of the clay layer if.
a. Clay is normally consolidated? ′
𝑃0 = ∑𝛾 ℎ
b. Preconsolidation pressure Pc=190 kPa?
c. Preconsolidation pressure Pc=170 kPa
= 16 (3) +(18.5 − 9.81)(6)
Solution: +(19 − 9.81)(3)
a. 𝐹𝑜𝑟 𝑛𝑜𝑟𝑚𝑎𝑙𝑙𝑦 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑒𝑑 𝑐𝑙𝑎𝑦 𝑃0 = 127.71 kPa
3m 𝛾𝑑 = 16 kN/m3
𝐶𝑐𝐻 𝑃0 + ∆𝑃
𝑆𝑐 = log
1+𝑒 𝑃0
6m 𝛾𝑠𝑎𝑡 = 18.5 kN/m3
𝑆𝑎𝑛𝑑
0.36 (6000) 127.71+ 50
= log
1 + 0.95 127.71
𝛾𝑠𝑎𝑡 = 19 kN/m3
𝑆𝑐 = 158.939 mm 6m 𝐶𝑐 = 0.36 𝐶𝑙𝑎𝑦
𝑒 = 0.95
1. The soil profile shown in the figure consist of 9 layer of sand underlain by 6 m layer of clay. The location of water table is 3 m from
the ground level. The building to be constructed will cause an additional pressure of 50 kPa on the ground. What is the primary
settlement of the clay layer if.
a. Clay is normally consolidated? ′
𝑃0 = ∑𝛾 ℎ
b. Preconsolidation pressure Pc=190 kPa?
c. Preconsolidation pressure Pc=170 kPa
= 16 (3) +(18.5 − 9.81)(6)
Solution: +(19 − 9.81)(3)
b. 𝑂𝑣𝑒𝑟𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑒𝑑? 𝑃0 = 127.71 kPa
∆𝑃 + 𝑃0 < 𝑃𝑐 3m 𝛾𝑑 = 16 kN/m3

50 +127.71 < 190

6m 𝛾𝑠𝑎𝑡 = 18.5 kN/m3
𝑆𝑎𝑛𝑑
𝐶𝑠𝐻 𝑃0 + ∆𝑃
𝑆𝑐 = log
1+𝑒 𝑃0
𝛾𝑠𝑎𝑡 = 19 kN/m3
1 6m 𝐶𝑐 = 0.36 𝐶𝑙𝑎𝑦
(0.36)(6000)
5 127.71+ 50 𝑒 = 0.95
= log
1 + 0.95 127.71

𝑆𝑐 = 31.787 mm 1
𝐶𝑠 = 𝐶𝑐
5
1. The soil profile shown in the figure consist of 9 layer of sand underlain by 6 m layer of clay. The location of water table is 3 m from
the ground level. The building to be constructed will cause an additional pressure of 50 kPa on the ground. What is the primary
settlement of the clay layer if.
a. Clay is normally consolidated? ′
𝑃0 = ∑𝛾 ℎ
b. Preconsolidation pressure Pc=190 kPa?
c. Preconsolidation pressure Pc=170 kPa
= 16 (3) +(18.5 − 9.81)(6)
Solution: +(19 − 9.81)(3)
C. 𝑂𝑣𝑒𝑟𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑒𝑑? 𝑃0 = 127.71 kPa
∆𝑃 + 𝑃0 > 𝑃𝑐 3m 𝛾𝑑 = 16 kN/m3

50 +127.71 > 170

6m 𝛾𝑠𝑎𝑡 = 18.5 kN/m3
𝑆𝑎𝑛𝑑
𝐶𝑠𝐻 𝑃𝐶 𝐶𝑐𝐻 𝑃𝑜 + ∆𝑃
𝑆𝑐 = log + log
1+𝑒 𝑃0 1+𝑒 𝑃𝐶
𝛾𝑠𝑎𝑡 = 19 kN/m3
1 6m 𝐶𝑐 = 0.36 𝐶𝑙𝑎𝑦
(0.36)(6000)
5 170 (0.36)(6000) 127.71 +50 𝑒 = 0.95
= log + log
1 + 0.95 127.71 1 + 0.95 170

𝑆𝑐 = 48.857 mm 1
𝐶𝑠 = 𝐶𝑐
5
First 5 + 10
A soil profile is shown in the figure. If a uniformly distributed load, ∆𝑃,
is applied at the ground surface, what is the settlement of the clay
layer caused by primary consolidation if

a. The clay is normally consolidated

b. The preconsolidation pressure 𝑃𝑐 = 190 𝑘𝑁/𝑚2
c. 𝑃𝑐 = 170 𝑘𝑁/𝑚2 ∆𝑷 = 𝟏𝟎𝟎 𝒌𝑵/𝒎𝟐

1
Use 𝐶𝑠 ≈ 𝐶𝑐 .
6
Solution:

a. Solve for the present effective overburden pressure at the middle of the clay layer.
4
𝑃𝑜 = 2𝛾𝑑𝑟𝑦 + 4 𝛾𝑠𝑎𝑡(𝑠𝑎𝑛𝑑) − 𝛾𝑤 + 𝛾 − 𝛾𝑤
2 𝑠𝑎𝑡(𝑐𝑙𝑎𝑦)
4
= 2(14) + 4 18 − 9.81 + 19 − 9.81
2
= 79.14 𝑘𝑃𝑎
SAMPLE PROBLEM 1 ∆𝑷 = 𝟏𝟎𝟎 𝒌𝑵/𝒎𝟐

Solution:
Since the clay is normally consolidated, the following equation is used:

𝐶𝑐 𝐻 ∆𝑃 + 𝑃𝑜
𝑆= 𝑙𝑜𝑔
1 + 𝑒𝑜 𝑃𝑜

𝐶𝑐 = 0.009 𝐿𝐿 − 10 = 0.009 40 − 10 = 0.27

𝐶𝑐 𝐻 ∆𝑃 + 𝑃𝑜 (0.27)(4) 100 + 79.14

𝑆= 𝑙𝑜𝑔 = 𝑙𝑜𝑔 = 0.213 𝑚 = 𝟐𝟏𝟑 𝒎𝒎
1 + 𝑒𝑜 𝑃𝑜 1 + 0.8 79.14

b. Compare ∆𝑃 + 𝑃𝑜 with 𝑃𝑐 to determine which equation to use

∆𝑃 + 𝑃𝑜 = 100 + 79.14 = 179.14 𝑘𝑃𝑎
𝑃𝑐 = 190 𝑘𝑃𝑎 (𝑔𝑖𝑣𝑒𝑛)
𝐶𝑠 𝐻 ∆𝑃 + 𝑃𝑜
𝑆𝑖𝑛𝑐𝑒 ∆𝑃 + 𝑃𝑜 < 𝑃𝑐 , the following equation will be used: 𝑆= 𝑙𝑜𝑔
1 + 𝑒𝑜 𝑃𝑜
1 1
𝐶𝑠 = 𝐶𝑐 = 0.27 = 0.045
6 6
SAMPLE PROBLEM 1 ∆𝑷 = 𝟏𝟎𝟎 𝒌𝑵/𝒎𝟐

Solution:

𝐶𝑠 𝐻 ∆𝑃 + 𝑃𝑜 0.045(4) 100 + 79.14

𝑆= 𝑙𝑜𝑔 = 𝑙𝑜𝑔 = 0.035 𝑚 = 𝟎. 𝟎𝟑𝟔 𝒎𝒎
1 + 𝑒𝑜 𝑃𝑜 1 + 0.8 79.14

c. Compare ∆𝑃 + 𝑃𝑜 with 𝑃𝑐 to determine which equation to use

∆𝑃 + 𝑃𝑜 = 100𝑞 + 79.14 = 179.14 𝑘𝑃𝑎
𝑃𝑐 = 170 𝑘𝑃𝑎 (𝑔𝑖𝑣𝑒𝑛)
𝐶𝑠 𝐻 𝑃𝑐 𝐶𝑐 𝐻 ∆𝑃 + 𝑃𝑜
𝑆𝑖𝑛𝑐𝑒 ∆𝑃 + 𝑃𝑜 > 𝑃𝑐 , the following equation will be used: 𝑆 = 𝑙𝑜𝑔 + 𝑙𝑜𝑔
1 + 𝑒𝑜 𝑃𝑜 1 + 𝑒𝑜 𝑃𝑐

𝐶𝑠 𝐻 𝑃𝑐 𝐶𝑐 𝐻 ∆𝑃 + 𝑃𝑜 0.045(4) 170 0.27(4) 179.14

𝑆= 𝑙𝑜𝑔 + 𝑙𝑜𝑔 = 𝑙𝑜𝑔 + 𝑙𝑜𝑔 = 0.0469 𝑚 = 𝟒𝟔. 𝟗 𝒎𝒎
1 + 𝑒𝑜 𝑃𝑜 1 + 𝑒𝑜 𝑃𝑐 1 + 0.8 79.14 1 + 0.8 170
COEFFICIENT OF CONSOLIDATION
2
𝐻𝑑𝑟 𝑇𝑣
𝐶𝑣 =
𝑡

𝑤ℎ𝑒𝑟𝑒:
𝐻𝑑𝑟 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑑𝑟𝑎𝑖𝑛𝑎𝑔𝑒 𝑝𝑎𝑡ℎ
= 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑜𝑛𝑒 𝑙𝑎𝑦𝑒𝑟 𝑖𝑓 𝑑𝑟𝑎𝑖𝑛𝑒𝑑 1 𝑠𝑖𝑑𝑒
= ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑖𝑓 𝑑𝑟𝑎𝑖𝑛𝑒𝑑 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠
𝑇𝑣 = 𝑡𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 (𝑛𝑜𝑛𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑎𝑙)
𝑡 = 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛

For 𝑈 = 0 𝑡𝑜 60%,
2
𝜋 𝑈%
𝑇𝑣 =
4 100

For U > 60%,

𝑇𝑣 = 1.781 − 0.933 log(100 − 𝑈%)

𝑤ℎ𝑒𝑟𝑒:
𝑈 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛 (%)
COEFFICIENT OF CONSOLIDATION
The average degree of consolidation for the entire
depth of the clay layer at any time t can be written
as:

𝑆𝑐(𝑡)
𝑈=
𝑆𝑐
𝑤ℎ𝑒𝑟𝑒:
𝑆𝑐(𝑡) = 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑦𝑒𝑟 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡
𝑆𝑐 = 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑦𝑒𝑟 𝑓𝑟𝑜𝑚 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛
SAMPLE PROBLEM 2
Under a given surcharge, a 5 m thick clay layer has a consolidation settlement of 305 mm. Assume Cv =
0.004 cm2/sec. How long will it take (in days) for 50% consolidation to occur if the layer is drained at
the top only? Tv = 0.197
Solution:
2 2
𝐻𝑑𝑟 𝑇𝑣 𝐻𝑑𝑟 𝑇𝑣
𝐶𝑣 = ; 𝑡=
𝑡 𝐶𝑣 100 cm
2

5 m 2 (0.197) 1m
𝑡=
0.004 cm2 /s
1 day
𝑡 = 12,312,500 secs
86,400 secs
𝑡 = 142.506days
SAMPLE PROBLEM 3
The time required for 50% consolidation of a 25-mm-thick clay layer (drained at both top and bottom)
in the laboratory is 2 min. 20 sec. How long (in days) will it take for a 3-m-thick clay layer of the same
clay in the field under the same pressure increment tor each 50% consolidation? In the field, there is a
rock layer at the bottom of the clay.
Solution: 2
25 mm 𝜋 50 2
2
𝐻𝑑𝑟 𝑇𝑣 2 4 100
𝐶𝑣 = =
𝑡 2.33333333 min
𝐶𝑣 = 13.148 mm2 /min 50
2
𝜋
2
𝐻𝑑𝑟 𝑇𝑣 3000 mm 2
4 100
𝑡= =
𝐶𝑣 13.148 mm2 /min
1 day
𝑡 = 134,400 min
1440 min

𝑡 = 93.333 days
 SAMPLE PROBLEM 4 PLUS 10
The estimated primary consolidation settlement of a clay layer having a thickness of 4.5 m. is 300mm. The clay layer is drained
on top only.
a. Compute the average degree of consolidation for the clay layer when the settlement is 75 mm.
b. How long will it take for 50% consolidation to occur if the coefficient of consolidation is 0.004 cm 2/sec
c. If the 4.5 m clay is drained on both sides how long will it take for 50% consolidation 2
Solution: 2
𝜋 𝑈%
𝐻𝑑𝑟 𝑇𝑣 𝑇𝑣 =
a. 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑙𝑎𝑦 𝑎𝑡 b. 𝐶𝑣 = 4 100
𝑡
𝑈=
𝑎𝑛𝑦𝑡𝑖𝑚𝑒(𝑆𝑐𝑡 ) 𝑇𝑣 = 0.196
2
𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑙𝑎𝑦 𝑙𝑎𝑦𝑒𝑟
𝑑𝑢𝑒 𝑡𝑜 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑜𝑙𝑖𝑑𝑎𝑡𝑖𝑜𝑛 (𝑆𝑐 )
(4.5 × 100) (0.196) 2
0.004 = [ (4.5/2) × 100]
𝑡 c.
1 𝑑𝑎𝑦 𝑡
75 𝑡 = 9922500 s ×
= (100%) 86400 𝑠
300 𝑡 = 2480625 s
𝑈 = 25% 𝑡 = 115.049 𝑑𝑎𝑦𝑠
𝑡 = 28.762 days
2
𝜋 𝑈%
𝐻𝑑𝑟 = 0.5 thickness of sample if drained at both side 𝑓𝑜𝑟 𝑈 = 0 𝑡𝑜 60%, 𝑇𝑣 =
4 100
𝐻𝑑𝑟 = thikness of sample if drained at one side 𝐹𝑜𝑟 𝑈 > 60 %, 𝑇𝑣 = 1.781 − 0.933log(100 − 𝑈%)
SAMPLE PROBLEM 2
Under a given surcharge, a 5 m thick clay layer has a consolidation settlement of 305 mm. Assume 𝐶𝑣 = 0.004 𝑐𝑚2 /
𝑠𝑒𝑐. How long will it take (in days) for 50% consolidation to occur if the layer is drained at the top only? 𝑇𝑣 = 0.197

Solution:
2
𝐻𝑑𝑟 𝑇𝑣
𝐶𝑣 =
𝑡
2
2 100 𝑐𝑚
2
𝐻𝑑𝑟 𝑇𝑣 (5 𝑚) (0.197) 1 𝑚
𝑡= = = 12,312,500 s
𝐶𝑣 0.004 𝑐𝑚2 /𝑠

1 𝑑𝑎𝑦
12,312, 500 𝑠 = 𝟏𝟒𝟐. 𝟓𝟏 𝐝𝐚𝐲𝐬
86,400 𝑠
SAMPLE PROBLEM 3
The time required for 50% consolidation of a 25-mm-thick clay layer (drained at both top and bottom) in the
laboratory is 2 min. 20 sec. How long (in days) will it take for a 3-m-thick clay layer of the same clay in the field under
the same pressure increment tor each 50% consolidation? In the field, there is a rock layer at the bottom of the clay.

Solution:

• Solve for the coefficient of consolidation based on the laboratory results

25 𝑚𝑚 2 𝜋 50 2
2
𝐻𝑑𝑟 𝑇𝑣 2 4 100
𝐶𝑣 = = = 13.148 𝑚𝑚2 /𝑚𝑖𝑛
𝑡 2.33 𝑚𝑖𝑛

• Use the 𝐶𝑣 to obtain the time of consolidation in the field

2
𝜋 50
2
2
𝐻𝑑𝑟 𝑇𝑣 3000 𝑚𝑚
4 100
𝑡= = = 134,400 𝑚𝑖𝑛𝑠
𝐶𝑣 13.148 𝑚𝑚2 /𝑚𝑖𝑛

1 ℎ𝑟 1 𝑑𝑎𝑦
134,400 𝑚𝑖𝑛𝑠 = 𝟗𝟑. 𝟑𝟑 𝒅𝒂𝒚𝒔
60 𝑚𝑖𝑛𝑠 24 ℎ𝑟𝑠
REFERENCES
Das, Braja M., and Khaled Sobhan. (2014). Principles of Geotechnical Engineering. Cengage Learning.

Cairo University Soil Compressibility & Settlement Module

https://www.slideshare.net/1mirfan/geotechnical-engineeringi-lec-17-consolidation?qid=c7a0d72e-6aec-
406b-af88-53d903e56653&v=&b=&from_search=3

Likos, W. J. Consolidation [PowerPoint slides]. Retrieved from https://slideplayer.com/slide/10533550/