1 d Settlement Calculation

7/27/2019 1 d Settlement Calculation

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9. CALCULATION OF SETTLEMENT

9.1 Settlement of a Single Layer

The settlements of a single relatively thin layer, shown in Fig. 1, can be calculated once the changein void ratio is known.

H

Fig. 1 Settlement of a single layer

Sz

x

For confined compression the horizontal strains are negligible i.e. xx = 0, yy= 0 and thus:

zz vS

H

e

e

thus

Se H

e

= = = +

=

+

1

1

(1)

the settlement of a thicker layer can be calculated by dividing the layer into a number of sub layers asshown in Fig. 2. This is necessary because both the initial and final effective stress vary with depthas do the voids ratio and the OCR.

sub-layer 1

sub-layer 2

sub-layer n

Notation

e voids ratio at the centre of layer i

e increase in voids ratio at the centre of layer i

H thickness of layer i

i

i

i

=

=

Fig. 2 Soil profile divided into a number of sub-layers

The settlement of the soil layer is calculated by calculating the settlement of the individual sub-layersand adding them, in doing this it is assumed that the voids ratio and the effective stress are constantthroughout the sub-layer and equal to their values at the centre of the sub-layer.

1

Fig. 1 Settlement of a soil layer

Fig. 2 Division of soil layers into sub-layers


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Fig. 4 Determination of Saturated Unit Weight

sat

w s

v s

W W

V VkN m=

+

+=

+

+=

7 84 26

0 8 119 06 3

. .46

.. /

or

sats wG e

ekN m=

+

+=

( ). /

119 06 3

Initial State at A

Total stress zz = 2 18 + 3 22 + 2 19.06 = 140.12 kPa

Pore water pressure uw = 5 9.8 kPa = 49 kPa

(3a)

Effective stress zz = zz - uw = 140.12 - 49 = 91.12 kPa

Notice the initial effective stress is less than pc =120 kPa thus the clay is initially over-

consolidated.

Final State at A

Total stress zz = 100 + 2 22 + 3 22 + 2 19.06 = 248.12 kPa

Pore water pressure uw = 7 9.8 kPa = 68.6 kPa

(3b)

Effective stress zz = zz - uw = 248.12 - 68.6 = 179.52 kPa

Notice that the final effective stress exceeds the initial preconsolidation stress and thus the claymoves from being initially over-consolidated to finally normally consolidated.

Settlement of the first sub-layer

The soil in the first sub layer moves from being over-consolidated to normally consolidated and sothe calculation of the change in voids ratio must be made in two stages.

Stage 1 Soil over-consolidated ( < pc (initial))

e1 = - Cr log10(pc (initial)/I)

Stage 2 Soil normally consolidated ( = pc)

Voids

Skeletalmaterial

Vs=1 m3

Vv= e*Vs =0.8 m3 W V

kN

v v=

=

*.7 84

Distribution of Volume Distribution of Weight

W V G

kN

s=

=

* *

.

2646

sats

s

sats

W W

V V

kN m

kN m

orG e

ekN m

=+

+

=+

+

=

=+

+

=

7 8 4 26 4 6

0 8 119 06

119 06

3

3

3

. .

.

/

. /

( ). /

3


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(3c)

e2 = - Cc log10(F/pc (initial))

now

S H ee

m

= +

=

=

+

14

1 80 05

120 00

9112

0 0911

0 2179 52

120 0010 10

.[ . log (

.

.)

.

. log (.

.)] (3d)

Initial State at B

Total stress zz = 218 + 3 22 + 6 19.06 = 216.36 kPa


(4a)Effective stress zz = zz - uw = 216.36 - 88.20 = 128.16 kPa

Final State at B

Total stress zz = 60 + 2 22 + 3 22 + 6 19.06 = 284.36 kPa


(4b)

Effective stress zz = zz - uw = 284.36 - 107.80 = 176.56 kPa

Settlement of the second sub-layer

The soil in the second is normally consolidated and thus:

e2 = - Cc log10(F/I) (4c)

now

SH e

e

m

= +

=

=

14

1 80 2

176 56

12816

0 0620

10.

. log (.

.)

.

. (4d)

Total Settlement

Total settlement = 0.0911 + 0.0620 m(5)

= 0.1531m

4


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9.2 Calculation of Stress Changes

The calculation of settlement depends upon knowledge of the initial and final effective stress withineach sub layer of the deposit. The initial effective stress state can be determined, from knowledge of

the bulk unit and the position of the water table. The increase in total stress can be estimated usingthe theory of elasticity. (Note the soil is in general not really elastic however in the working stressrange this assumption provides reasonably accurate estimates of the stress increases due to theapplied loads)

A fundamental solution of the equations of elasticity is Boussinesq's solution. This relates to a pointload applied to the surface of a half-space (very deep layer) and is shown schematically in Fig. 5.

Point load of magnitude P

Hz

x

Fig 5 Point load on an elastic half-space

Boussinesq found that :

zz

xx yy zz

z

Pz

RPz

R

uP

ER

z

Rwhere

R x y z

and

E Young s ulus

Poisson s ratio

=

+ + =+

=+

+

= + +

=

=

3

21

1

2

2 1

3

5

3

2

2

2 2 2

( )

( )[ ( ) ]

' mod

'

(6)

uz = vertical displacement due to load

The symbol is used to indicate that each of the quantities in equation (6) represents the increase in

the particular quantity, due to the applied load.

The solution for a point load is important because it can be used to develop solutions for distributedloads by integration. Some of these solutions are presented in the Soil Mechanics Data Sheets.

9.3Calculation of Stress Changes

5

Fig. 5 Point load acting on a half space


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9.3.1Stresses due to Circularfoundation loads applied at the ground surface

A circular foundation of diameter 5 m, subjected to an average applied stress of 100 kPa is shown in

Fig. 6.

5m

p=100 kPa

z

r

A

2m

B5m

Fig 6 Circular loaded area on a deep elastic layer

(a) Calculate the increase in vertical stress at point A

There is a simple analytic expression (given in the Data Sheets) for points on the centre line under acircular load:

zz p az

= + ( [ ] )/1 1 22

3 2 (7a)

where

p = the surface stress = 100 kPaa = the radius of the loaded area = 2.5mz = the depth of interest = 2m

zz kPa= + =100 1 1 125 75 6

2 3 2( [ ( . ) ] ) .

/ (7b)

(b) calculate the increase in vertical stress at point B

In this case there is no simple analytic expression and the solution must be found by using theinfluence charts given in the data sheets, reproduced in part in Figure 7. Note that this chart can alsobe used for points on the centre line for which r = 0.Now z/a = 2/2.5 = 0.8

r/a = 5/2.5 = 2(8)

6

Fig. 6 Circular loaded area on a deep elastic layer


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using the data sheets zz/p = 0.03 and so zz = 3.0 kPa

10-3 10-2 10-1 1

0

2

4

6

8

10

10

9

8

7

6

5

4

3

2.5

2.01.5

1.251.00

0.0

z/a

Values on curvesare values of r/a

Fig.7 Influence Factors for a Uniformly Loaded Circular Area of radius a

Ip

zz=

9.3.1Stresses due toRectangularfoundation loads applied at the ground surface

Plan

Elevation

L

B

zPoint immediately

beneath one of the

rectangles corners

Uniformly distributed

surface stress p

Fig. 8 Rectangular surface loading on a deep elastic layer

Many loads which occur in practice are applied to foundations that may be considered to consist of anumber of rectangular regions. It is thus of interest to be able to calculate the vertical stress increasesdue to a uniformly distributed load acting on a rectangular loaded area. This is shown schematicallyin Fig. 8.The vertical stress change at a distance z below one of the corners of the rectangular load may bedetermined from a chart which is given in the data sheets and is reproduced in Fig. 9

7

Fig. 7 Influence factors for a uniformly loaded circular area of radius a

Fig. 8 Rectangular uniform loading on a deep elastic layer


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m=B/z=0.0

0.2

0.4

0.6

0.8

1.0

2.03.0

8

1010.10.01

(n=L/z)

0.00

0.05

0.10

0.15

0.20

0.25

Iq

zz=

Note m & n are

interchangeable

Fig. 9 Influence factors for a uniformly loaded rectangular area

This chart can be used to determine the value of stress increase at any point in an elastic layer, themethod for doing this is illustrated below.

9.3.2.1 Calculation of Stress below an interior point of the loaded area

This situation is shown schematically in Fig.10. The stress change is required at a depth z below

point O.

The first step in using the influence charts is to break the rectangular loading up into a number ofcomponents each having a corner at O, this is relatively simple as can be seen in Fig.(10)

It thus follows that at the point of interest, the stress increase zz(ABCD) is given by:

zz zz zz zz zzABCD OXAY OYBZ ZCT OTDX( ) ( ) ( ) ( ) ( )= + + +0 (9)

8

Fig. 9 Influence factors for uniformly loaded rectangular areas


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O Point of interest

O

A B

D C

X

Y

Z

T

Fig. 10 stress increase at a point below a loaded rectangular region

z

Example

Suppose we wish to evaluate the increase in stress at a depth of 2m below the point O due to therectangular loading shown in shown in Fig. 11, when the applied stress over ABCD is 100 kPa.

Fig. 11 Dimensions of loaded area

O

A B

D C

X

Y

Z

T

2m

3m

3m 2m

For rectangular loading OZCT

m = L/z =1n = B/z =1

thus

I = 0.175

and so

zz = p I = 100 0.175 = 17.5 kPa (9a)

For rectangular loading OTDX

m = L/z = 1.5n = B/z = 1

thus

I = 0.194

and so

9

Fig. 11 Dimensions of rectangular loaded area

Plan

Elevation


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zz = p I = 100 0.195 = 19.4 kPa (9b)

For rectangular loading OXAY

m = L/z = 1.5n = B/z = 1.5

thus

I = 0.216

and so

zz = p I = 100 0.216 = 21.6 kPa (9c)

For rectangular loading OYBZ

m = L/z = 1.5n = B/z = 1

thus

I = 0.194

and so

zz = p I = 100 0.194 = 19.4 kPa (9d)

Thus the increase in stress zz = 17.5 + 19.4 + 21.6 + 19.4 = 78.9 kPa (9e)

This must of course be added to the existing stress state prior to loading to obtain the actual stress

zz.

9.3.2.2 Calculation of stress below a point outside the loaded area

The stress increase at a point vertically below a point O which is outside the loaded are can also befound using the influence charts shown in Fig. 9.

A B

D C Fig. 12 Rectangular loaded area

X

Y

Z

T

O

This is achieved by considering the stress q acting on ABCD to consist of the following:

1. A stress +q acting over OXAY

2. A stress +q acting over OZCT

3. A stress -q acting over OZBY

4. A stress -q acting over OXDTThis is illustrated in Fig. 13.

10

Fig. 12 Rectangular loaded area ABCD and point of interest O


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It thus follows that at the point O, the stress increase zz(ABCD) is given by:

zz zz zz zz zzABCD OXAY OYBZ OZCT OTDX( ) ( ) ( ) ( ) ( )= +

and thus (10)

zz ABCD q I OXAY I OYBZ I ZCT I OTDX( ) [ ( ) ( ) ( ) ( )]= + 0

A B

D C

X

Y

Z

T

O

A B

D C

X

Y

Z

T

O

Stage 4Stage 3

A B

D C

X

Y

Z

T

O

Stage 2Stage 1

(q)

(q) (0)

(q)

(q)

(q) (0)

(0)

(q)

Fig. 13 Decomposition of Loading over a rectangular region (exterior point)

A B

D C

Y

Z

T

O

(q)

(q) (q)

(q)

(0)

(0)(0)

11

Fig. 13 Decomposition of loading over a rectangular area (for stress at external point)


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Example

AB

D C


X

Y

Z

T

O

2m

10m

1m

1m

Suppose the rectangular area ABCD, shown in Fig. 14 is subjected to a surface stress of 100 kPaAND it is required to calculate the vertical stress increase at a point 1.5m below the point O.

For rectangular loading OZCT

m = L/z = 0.67n = B/z = 0.67

thus

I = 0.121

and so

zz = p I = + 100 0.121 = +12.1 kPa (11a)

For rectangular loading OTDX

m = L/z = 7.67n = B/z = 0.67

thus

I = 0.167and so

zz = p I = -100 0.167 = -16.7 kPa (11b)

For rectangular loading OXAY

m = L/z = 7.67n = B/z = 2.00

thus

I = 0.240

and so

zz = p I = + 100 0.240 = + 24.0kPa (11c)

For rectangular loading OYBZ

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m = L/z = 2n = B/z = 0.67

thus

I = 0.164and so

zz = p I = -100 0.164 = -16.4 kPa (11d)

Thus the increase in stress zz = 12.1 - 16.7 + 24.0 + -16.4=3.0 kPa (11e)

9.3.3 Stresses due to foundation loads ofarbitrary shape applied at the ground surface

Newmarks chart provides a graphical method for calculating the stress increase due to a uniformlyloaded region, of arbitrary shape resting on a deep homogeneous isotropic elastic region.

Newmarks chart is given in the data sheets and is reproduced in part in Fig 15. The procedure for itsuse is outlined below

1. The scale for this procedure is determined by the depth z at which the stress is to be evaluated,thus z is equal to the distance OQ shown on the chart.

2. Draw the loaded area to scale so that the point of interest (more correctly its vertical projectionon the surface) is at the origin of the chart, the orientation of the drawing does not matter

3. Count the number of squares (N) within the loaded area, if more than half the square is in countthe square otherwise neglect it.

4. The vertical stress increase zz = N [scale factor(0.001)] [surface stress (p)]

The procedure is most easily illustrated by an example.

Example

Suppose a uniformly loaded circle of radius 2 m carries a uniform stress of 100 kPa. It is required tocalculate the vertical stress at a depth of 4 m below the edge of the circle.

The loaded area is drawn on Newmarks chart to the appropriate scale (i.e. the length OQ is set torepresent 4 m) as shown in Fig. 15.It is found that the number of squares, N = 194 and so the stress increase is found to be

zz = 194 0.001 100 = 19.4 kPa (12)

This result can also be checked using the influence charts for circular loading and it is then foundthat:

z/a = 2, r/a = 1. zz /p = 0.2 and so zz = 20 kPa (13)

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O Q4m

Loaded

Area

Fig 15 Newmarks Chart

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1 d Settlement Calculation