chapter 7: settlement of shallow...

Civil Engineering Department: Foundation Engineering (ECIV 4052)

Engr. Yasser M. Almadhoun

Chapter 7: Settlement of Shallow Foundations

Introduction

The settlement of a shallow foundation can be divided into two major

categories:

(a) elastic, or immediate settlement and

(b) consolidation settlement.

Immediate, or elastic settlement

Immediate, or elastic settlement of a foundation takes place during or

immediately after the construction of the structure.

Consolidation settlement

Consolidation settlement occurs over time. Pore water is extruded from the

void spaces of saturated clayey soils submerged in water. The total

settlement of a foundation is the sum of the elastic settlement and the

consolidation settlement.

Consolidation settlement comprises two phases: primary and secondary.

The fundamentals of primary consolidation settlement were explained in

detail in Chapter 2.

Secondary consolidation settlement occurs after the completion of primary

consolidation caused by slippage and reorientation of soil particles under

a sustained load. Primary consolidation settlement is more significant than

secondary settlement in inorganic clays and silty soils. However, in organic

soils, secondary consolidation settlement is more significant.

Elastic Settlement of Shallow Foundation on Saturated Clay

(𝝁𝒔 = 0.5)

The average settlement of flexible foundations on saturated clay soils

(Poisson’s ratio, 𝜇𝑠 = 0.5).



Example 7.1

See example 7.1 in textbook, page 340.

Elastic Settlement in Granular Soil

Settlement Based on the Theory of Elasticity



To calculate settlement at the centre of the foundation, use:



To calculate settlement at a corner of the foundation, use:

The elastic settlement of a rigid foundation can be estimated as:

Due to the nonhomogeneous nature of soil deposits, the magnitude of Es

may vary with depth. For that reason, Bowles (1987) recommended using

a weighted average of Es:

Example 7.2




Settlement of Sandy Soil: Use of Strain Influence Factor

The recommended variation of the strain influence factor Iz for square

(L/B =1) or circular foundations and for foundations with L/B = 10.



Note that the maximum value of Iz [ that is, Iz(m) ] occurs at z = z1 and

then reduces to zero at z = z2. The maximum value of Iz can be calculated

as:

where qc = cone penetration resistance



Substitute in the equation of Se in order to get the elastic settlement of

sandy soil.

Settlement of Foundation on Sand Based on Standard Penetration

Resistance (Meyerhof Method)

Meyerhof proposed a correlation for the net bearing pressure for

foundations with the standard penetration resistance, N60. The net pressure

has been defined as:

According to Meyerhof’s theory, for 25 mm (1 in.) of estimated maximum

settlement:

In English units:



In SI units:

where

Se = settlement, in mm.

N60 = the standard penetration resistance between the bottom of the

foundation and 2B below the bottom.

Later, Meyerhof (1965) suggested that the net allowable bearing pressure

should be increased by about 50%. Bowles (1977) proposed that the

modified form of the bearing equations be expressed as:

In SI units:

where B is in meters and Se is in mm.



Example 7.7


Effect of the Rise of Water Table on Elastic Settlement

Terzaghi suggested that the submergence of soil mass reduces the soil

stiffness by about half, which in turn doubles the settlement.

In most cases of foundation design, it is considered that, if the ground water

table is located 1.5B to 2B below the bottom of the foundation, it will not

have any effect on the settlement.

The total elastic settlement (S’e) due to the rise of the ground water table

can be given as:



The following are some empirical relationships for Cw:

Example 7.9


Consolidation Settlement

Primary Consolidation Settlement

Consolidation settlement occurs over time in saturated clayey soils

subjected to an increased load caused by construction of the foundation.



On the basis of the one-dimensional consolidation settlement equations

are:

𝜎′𝑜 + ∆𝜎′𝑎𝑣 > 𝜎′𝑐



Correlations for Cc and Cs

𝐶𝑐 =∆𝑒

log(𝜎2/𝜎1)

𝜎1 = 𝜎′𝑜

𝜎2 = 𝜎′𝑜 + ∆𝜎′𝑎𝑣 = 𝜎′𝑜 +1

6(∆𝜎′

𝑡 + 4∆𝜎′𝑚 + ∆𝜎′𝑏)

𝐶𝑐 = 0.009(𝐿𝐿 − 10)

𝐶𝑠 = (1

5−

1

10)𝐶𝑐

Secondary Consolidation Settlement

At the end of primary consolidation (i.e., after the complete dissipation of

excess pore water pressure) some settlement is observed that is due to the

plastic adjustment of soil fabrics. This stage of consolidation is called

secondary consolidation.

A plot of deformation against the logarithm of time during secondary

consolidation is practically linear as shown in the figure below.

The magnitude of the secondary consolidation can be calculated as:



where

C’α = Cα / (1+ ep) (varies between 0.0005 to 0.001)

ep = void ratio at the end of primary consolidation

Hc = thickness of clay layer

From the figure, the secondary compression index can be defined as:

where

Cα = secondary compression index

∆e = change of void ratio

t1, t2 = time

Secondary consolidation settlement is more important in the case of all

organic and highly compressible inorganic soils. In overconsolidated

inorganic clays, the secondary compression index is very small and of less

practical significance.

Example 7.11


Field Load Test (Plate Load Test)

The ultimate load-bearing capacity of a foundation, as well as the allowable

bearing capacity based on tolerable settlement considerations, can be

effectively determined from the field load test, generally referred to as the

plate load test.

The plates that are used for tests in the field are usually made of steel and

are 25 mm (1 in.) thick and occasionally, square plates that are 305 mm X

305 mm (12 in. X 12 in.) are used.



To conduct a plate load test, a hole is excavated with a minimum diameter

of 4B (B is the diameter of the test plate) to a depth of Df, the depth of the

proposed foundation.

The plate is placed at the centre of the hole, and a load that is about 1/4 to

1/5 of the estimated ultimate load is applied to the plate in steps by means

of a jack.

During each step of the application of the load, the settlement of the plate

is observed on dial gauges. At least one hour is allowed to elapse between

each application. The test should be conducted until failure, or at least until

the plate has gone through 25 mm (1 in.) of settlement.

For tests in clay:

where

qu(Fd) = ultimate bearing capacity of the proposed foundation

qu(Pd) = ultimate bearing capacity of the test plate

For tests in sandy soils:

where

BF = width of the foundation

BP = width of the test plate



The allowable bearing capacity of a foundation, based on settlement

considerations and for a given intensity of load, qo, is:

Problems

Problem (1)

Find the size of the square footing, using Bowles theory, that carry

allowable load of 1000 KN, given that:

N60 = 10

Se(all) = 25 mm

Solution:

The load is relatively large, so assume that B > 1.22 m

𝑞𝑛𝑒𝑡(𝑎𝑙𝑙) =𝑁60

0.08(𝐵 + 0.3

𝐵)𝐹𝑑 (

𝑆𝑒25

)

𝑞𝑛𝑒𝑡(𝑎𝑙𝑙) =𝑁60

0.08(𝐵 + 0.3

𝐵) (1 + 0.33×

1.5

𝐵) (

𝑆𝑒25

)

1000

𝐵2=

10

0.08(𝐵 + 0.3

𝐵) (1 + 0.33×

1.5

𝐵) (

25

25)

Solving the previous equation for B:

B = 2.3 m



Problem (2)

Refer to the following figure, determine the average stress increase in the

clay layer below the centre of the foundation due to the net foundation load

of 50 ton. After that, determine the primary consolidation settlement for

the clay layer.

Solution:

First of all, you have to determine the average stress in the clay layer below

the centre of the foundation, and then you can calculate the settlement using

the appropriate equation:

∆𝜎𝑎𝑣 = 4𝑞𝑜 [𝐻2𝐼𝑎(𝐻2) −𝐻1𝐼𝑎(𝐻1)

𝐻2 −𝐻1]

𝑞𝑜 =𝑄

𝐴=

50

5×5= 2𝑡𝑜𝑛/𝑓𝑡2 = 4409.2𝑝𝑠𝑓

𝑚2 =𝐵′

𝐻1=𝐵/2

𝐻1=5/2

3= 0.83

𝑛2 =𝐿′

𝐻1=𝐿/2

𝐻1=5/2

3= 0.83

𝐼𝑎(𝐻1) = 0.21 (Figure 6.11, page 280 in textbook)



𝑚2 =𝐵′

𝐻2=𝐵/2

𝐻2=5/2

11= 0.23

𝑛2 =𝐿′

𝐻2=𝐿/2

𝐻2=5/2

11= 0.23

𝐼𝑎(𝐻1) = 0.11 (Figure 6.11, page 280 in textbook)

∆𝜎𝑎𝑣 = 4(4409.2) [(11)(0.21) − (3)(0.11)

11 − 3] = 1278.67𝑝𝑠𝑓

𝜎𝑐 = 2000𝑝𝑠𝑓

𝜎𝑜 = 100×4.5 + (122 − 62.4)×3 + 4(120 − 62.4) = 859.2𝑝𝑠𝑓

𝜎𝑜 + ∆𝜎𝑎𝑣 = 859.2 + 1278.67𝑝𝑠𝑓

Note that:

𝜎𝑜 < 𝜎𝑐 < 𝜎𝑜 + ∆𝜎𝑎𝑣

So that:

𝑆𝑐 =𝐶𝑠𝐻𝑐

1 + 𝑒𝑜log (

𝜎𝑐𝜎𝑜) +

𝐶𝑐𝐻𝑐

1 + 𝑒𝑜log (

𝜎𝑜 + ∆𝜎𝑎𝑣𝜎𝑐

)

𝑆𝑐 =0.06×8

1 + 0.7log (

2000

859.2) +

0.25×8

1 + 0.7log (

2137.87

2000)

𝑆𝑐 = 0.137𝑓𝑡 = 1.65𝑖𝑛



Problem (3)

Refer to the following figure, the net load per unit area at the level

foundation is 3200 Ib/ft2. Assume that the foundation is rigid, determine

the elastic settlement that the foundation will undergo based on the theory

of elasticity.

Solution:

Solution:

𝑚′ =𝐿

𝐵=

10

6.5= 1.54

𝑛′ =𝐻

𝐵′=

𝐻

𝐵/2=

32

6.5/2= 9.85

𝐹1 = 0.5874 (Table 7.2, page 305 in textbook)

𝐹2 = 0.0247 (Table 7.3, page 307 in textbook)

𝐼𝑠 = 𝐹1 +1 − 2𝜇𝑠1 − 𝜇𝑠

𝐼𝑠 = 0.5874 +1 − 2(0.3)

1 − (0.3)= 0.016

𝐼𝑓 = 𝑓 (𝐷𝑓

𝐵, 𝜇𝑠,

𝐿

𝐵) = 0.828 (Table 7.4, page 309 in textbook)

𝑆𝑒(𝑓𝑙𝑒𝑥𝑖𝑏𝑙𝑒,𝑐𝑒𝑛𝑡𝑒𝑟) =𝑞𝑜𝐵

′𝛼

𝐸𝑠(1 − 𝜇𝑠

2)𝐼𝑠𝐼𝑓

𝑆𝑒(𝑓𝑙𝑒𝑥𝑖𝑏𝑙𝑒,𝑐𝑒𝑛𝑡𝑒𝑟) =𝑞𝑜𝐵

′𝛼

𝐸𝑠(1 − 𝜇𝑠

2)𝐼𝑠𝐼𝑓



𝑆𝑒(𝑓𝑙𝑒𝑥𝑖𝑏𝑙𝑒,𝑐𝑒𝑛𝑡𝑒𝑟) =(32006.5×10

) (6.52 ) (4)

(3200)(1 − 0.32)(0.016)(0.828)

= 0.4914𝑖𝑛

𝑆𝑒(𝑟𝑖𝑔𝑖𝑑) = 0.93×𝑆𝑒(𝑓𝑙𝑒𝑥𝑖𝑏𝑙𝑒,𝑐𝑒𝑛𝑡𝑒𝑟)

𝑆𝑒(𝑟𝑖𝑔𝑖𝑑) = 0.93×(0.4914) = 0.457𝑖𝑛

Problem (4)

For the following figure, determine the settlement of the foundation shown.

Solution:

Solution:

𝐻

𝐵=

3

1.5= 2

𝐿

𝐵=

3

1.5= 2

𝐴1 = 0.66 (Figure 7.1, page 300 in textbook)

𝐷𝑓𝐵

=1.2

1.5= 0.8

𝐴2 = 0.93 (Figure 7.1, page 300 in textbook)

𝑆𝑒 = 𝐴1𝐴2

𝑞𝑜𝐵

𝐸𝑠

𝑆𝑒 = (0.66)(0.93)(150)(1.5)

(600)= 0.230175𝑐𝑚

chapter 7: settlement of shallow...

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