soilmech ch5 settlement calculation

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Soil Mechanics Settlement Calculation page 1

Contents of this chapter :

CHAPITRE 5. SETTLEMENT CALCULATION......................................................................1

5.1 S ETTLEMENT OF A S INGLE INFINITE LAYER .................................................................1 5.2 EXAMPLE ..................................................................................................................2 5.3 EXERCISE .................................................................................................................5 5.4 S ETTLEMENT UNDER UNIFORMLY LOADED AREAS ......................................................5

5.4.1 S TRESSES UNDER CIRCULAR UNIFORM LOADING ........................................................5 5.4.2 S TRESSES UNDER RECTANGULAR UNIFORM LOADING .................................................7 5.4.3 EXERCISES .............................................................................................................13 5.4.4 S TRESSES UNDER LOADS OF ARBITRARY SHAPE ........................................................13 5.4.5 EXERCISE ...............................................................................................................15

Chapitre 5. Settlement Calculation

5.1 Settlement of a Single Infinite Layer

The settlement ∆h of a single relatively thin layer, shown in Fig. 1, can be calculated once thechange in voids ratio is known.

h

∆∆∆∆ h

Fig. 1 Settlement of a single layer

We have seen in the previous chapter that :

i

i f

e

ee

hh

+

−=

∆1

where e i and e f are the initial and final voids ratio.

Thus1 f i

i

e eh he

−∆ =+

the settlement of a thicker layer can be calculated by dividing the layer into a number of sublayers as shown in Fig. 2. This is necessary because both the initial and final effective stressvary with depth as do the voids ratio.




….

sub-layer 1

sub-layer 2

sub-layer n

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 individualsub-layers and adding them. In doing this it is assumed that the voids ratio and the effectivestress are constant throughout the sub-layer and equal to their values at the centre of the sub-layer.

1 1

1

1

k k k

i

n n k k tot k

i

thus

e h For sub layer k he

so that

e hTotal Setttlement h h

e

∆− ∆ =+

∆∆ = ∆ =

+∑ ∑

5.2 Example

A soil deposit, shown in Fig. 3 consists of 5 m of gravel overlaying 8 m of clay. Initially thewater table is 2 m below the surface of the gravel. Calculate the settlement if the water tablerises to the surface of the gravel slowly over a period of time and surface loading induces anincrease of total stress of 100 kN/m² at the point A and 60 kN/m² at the point B. Thepreconsolidation stress at A is 120 kN/m², and the deposit is normally consolidated at B. Thegravel has a saturated bulk unit weight of 22 kN/m 3 and a dry unit weight of 18 kN/m 3 and isrelatively incompressible when compared to the clay. The voids ratio of the clay is 0.8 and theskeletal particles have a specific gravity of 2.7. The compression index of the clay is 0.2 andthe recompression index is 0.05.

In solving this problem it will be assumed that the gravel is far less compressible than the clayand thus that the settlement of the gravel can be neglected. The settlement of the clay layer

will be calculated by dividing it into two sub-layers




2m 5m

4m

4m

Gravel

Clay

A

B

Water table

Fig. 3 Layered soil deposit

In order to commence the calculations it is first necessary to calculate the unit weight of theclay, this is shown schematically in Fig. 4.

Voids

Skeletalmaterial

Vs=1 m 3

Vv= e*V s =0.8 m 3 W VkN

v w v==

γ γγ γ *.8

Distribution of Volume Distribution of Weight

W V GkN

s s w s==

* *.γ γγ γ

27

Fig. 4 Determination of Saturated Unit Weight

38 2719.44 /

0.8 1w s

sat v s

W W kN m

V V γ

+ += = =

+ +

• Initial State at A: A I

Total stress σzz = 2 ×18 + 3 × 22 + 2 × 19.44 = 140.89 kN/m²Pore water pressure u w = 5 × 10 kN/m² = 50 kN/m²Effective stress σ′ zz = σzz - u w = 140.89 - 50 = 90.89 kN/m²

Notice the initial effective stress is less than σ′ pc =120 kN/m² thus the clay is initially over-consolidated.

• Final State at A: A F

Total stress σzz = 100 + 2 × 22 + 3 × 22 + 2 × 19.44 = 248.89 kN/m²Pore water pressure u w = 7 × 10 kN/m² = 70 kN/m²




Effective stress σ′ zz = σzz - u w = 248.89 - 70 = 178.89 kN/m²

e

σσσσ 'pc= 120 91

Slope : -C c

Slope : -C r

179 σσσσ ' (Log scale)

AI

AF

Fig. 5 Initial and final states A I and A F

Notice that the final effective stress exceeds the initial preconsolidation stress and thus theclay moves 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 consolidatedand so the calculation of the change in voids ratio must be made in two stages.

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

∆e 1 = - C r × log 10(σ′ pc (initial) / σ′ I)

Stage 2 Soil normally consolidated ( σ′ = σ′ pc)

∆e 2 = - C c × log 10(σ′ F / σ′ pc (initial) )now

1 0 1 0

1

4 1 2 0 .0 0 1 7 8 .8 9[ 0 .0 5 lo g ( ) 0 .2 lo g ( ) ]1 .8 9 0 .8 9 1 2 0 .0 00 . 0 9

k k

i

h eh

e

m

∆∆ =

+

= − × + ×

= −

∑

• Initial State at B

Total stress σzz = 2 ×18 + 3 × 22 + 6 ×19.44 = 218.67 kN/m²Pore water pressure u w = 9 × 10 kN/m² = 90 kN/m²Effective stress σ′ zz = σzz - u w = 218.67 - 90 = 128.67 kN/m²

• Final State at B

Total stress σzz = 60 + 2 × 22 + 3 × 22 + 6 × 19.44 = 286.67 kN/m²




Pore water pressure u w = 11 × 10 kN/m² = 110 kN/m²Effective stress σ′ zz = σzz - u w = 286.67 - 110 = 176.67 kN/m²

• Settlement of the second sub-layer

The soil in the second is normally consolidated and thus:

∆e 2 = - C c × log 10(σ′ F / σ′ I)

now

10

1

4 176.670.2 log ( ).

1.8 128.670.061

k k

i

h eS

e

m

∆∆ =

+

= − ×

=

∑

• Total Settlement

Total settlement = 0.09 + 0.061 m

= 0.151m

5.3 Exercise 1. The construction of a new embankment induces a uniform load of 18 kN/m² on the top of

a 12 m thick layer of saturated clay, whose characteristics are: bulk unit weight γ = 16kN/m³ ; voids ratio e o = 1,8 ; compression index Cc = 0,7.The water table is at the initial ground level (top of the clay layer).Assuming that the clay is normally consolidated, estimate the settlement of the clay afterrealization of the embankment. Compare results obtained by dividing the clay layer in 1, 2or 4 sub-layers.

5.4 Settlement under Uniformly Loaded Areas

The calculation of settlement depends upon knowledge of the initial and final effective stresswithin each sub layer of the deposit. The initial effective stress state can be determined, fromknowledge of the bulk unit weight and the position of the water table. The increase in totalstress can be estimated using the theory of elasticity. (Note the soil is in general not reallyelastic however in the working stress range this assumption provides reasonably accurateestimates of the stress increases due to the applied loads)

5.4.1 Stresses under Circular Uniform LoadingA circular area of diameter 5 m, subjected to an average applied stress of 100 kN/m² is shownin Fig.6.




5m

p=100 kN/m²

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 for points on the centre line under a circular load:

∆σ∆σ∆σ∆σ zz pa

z

==== −−−− ++++ −−−−( [ ] )/1 12

23 2

where

p = the surface stress = 100 kN/m²a = the radius of the loaded area = 2.5mz = the depth of interest = 2m

2 3/ 2100 (1 [1 (1.25) ] ) 75.6 / ² zz kN mσ −∆ = × − + =

(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 Figure 7. Note that this chart can also be used for points on thecentre line for which r = 0.Now z/a = 2/2.5 = 0.8

r/a = 5/2.5 = 2using the chart ∆σ zz /p = 0.03 and so ∆σ zz = 3.0 kN/m²




10 -3 10 -2 10 -1 10

2

4

6

8

10

10

9

8

76

5

4

3

2.52.0

1.51.25

1.00

0.0

z/a

Values on curvesare values of r/a

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

Ipzz

σσσσσσσσ

====

5.4.2 Stresses under Rectangular Uniform Loading

Plan(top view)

Elevation(side view)

L

B

z

Point immediatelybeneath one of therectangle’s corners

Uniformly distributedsurface stress p

Fig. 8 Rectangular uniform loading on a deep elastic layer

Many loads which occur in practice are applied to foundations that may be considered toconsist of a number of rectangular regions. It is thus of interest to be able to calculate thevertical stress increases due to a uniformly distributed load acting on a rectangular loadedarea. This is shown schematically in Fig. 8.The vertical stress change at a distance z below one of the corners of the rectangular loadmay be determined from a chart which is given in the data sheets and is reproduced in Fig. 9

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




m=B/z=0.0

0.2

0.4

0.6

0.81.0

2.03.0 8

1010.10.01

(n=L/z)

0.00

0.05

0.10

0.15

0.20

0.25

Iqzz

σσσσσσσσ

====

Note m & n areinterchangeable

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 elasticlayer, the method for doing this is illustrated below.

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 zbelow point O.

The first step in using the influence charts is to break the rectangular loading up into a numberof components 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 ∆σ zz(ABCD) is given by:

∆σ ∆σ ∆σ ∆σ ∆σzz zz zz zz zzABCD OXAY OYBZ ZCT OTDX( ) ( ) ( ) ( ) ( )= + + +0

Fig. 9 Influence factors for uniformly loaded rectangular areas




O Point of interest

O

A B

D C

X

Y

Z

T

z

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

Example

Suppose we wish to evaluate the increase in stress at a depth of 2m below the point O due tothe rectangular loading shown in shown in Fig. 11, when the applied stress over ABCD is 100kN/m².

O

A B

D C

X

Y

Z

T

2m

3m

3m 2m

Fig. 11 Dimensions of rectangular loaded area

For rectangular loading OZCT

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

thusIσ = 0.175

and so∆σ zz = p I σ = 100 × 0.175 = 17.5 kN/m² (9a)

For rectangular loading OTDX

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

thusIσ = 0.194

Plan(top view)

Elevation(side view)

O

A B

D C

X

Y

Z

T

2m

3m

3m 2m

O

A B

D C

X

Y

Z

T

2m

3m

3m 2m




and so∆σ zz = p I σ = 100 × 0.195 = 19.4 kN/m²

For rectangular loading OXAY

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

Iσ = 0.216and so

∆σ zz = p I σ = 100 × 0.216 = 21.6 kN/m²

For rectangular loading OYBZ

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

thusIσ = 0.194

and so∆σ zz = p I σ = 100 × 0.194 = 19.4 kN/m²

Thus the increase in stress ∆σ zz = 17.5 + 19.4 + 21.6 + 19.4 = 78.9 kN/m²

This must of course be added to the existing stress state prior to loading to obtain the actualstress σ zz.

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 area canalso be found using the influence charts shown in Fig. 9.

A B

D C

X

Y

Z

T

O

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

The method is illustrated in Fig. 13.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

O

A B

D C

X

Y

Z

T

2m

3m

3m 2m

O

A B

D C

X

Y

Z

T

2m

3m

3m 2m




4. A stress -q acting over OXDT

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 2 Stage 1

(0)

(0) (-q)

(-q)

(0)

(0) (0)

(q)

(0)

A B

D C

X

Y

Z

T

O

(q)

(q) (q)

(q)

(0)

(-q)(-q)

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

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

∆σ ∆σ ∆σ ∆σ ∆σzz zz zz zz zzABCD OXAY OYBZ OZCT OTDX( ) ( ) ( ) ( ) ( )= − + − and thus

σ σ σ σ σzz ABCD q I OXAY I OYBZ I ZCT I OTDX( ) [ ( ) ( ) ( ) ( )]= − + −0




Example

A B

D C

Y

T

2m

10m

1m

X Z O

1m

Suppose the rectangular area ABCD, shown in Fig. 14 is subjected to a surface stress of 100kN/m² AND it is required to calculate the vertical stress increase at a point 1.5m below thepoint O.

For rectangular loading OZCT

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

thusIσ = 0.121

and so∆σ zz = p I σ = + 100 × 0.121 = +12.1 kN/m²

For rectangular loading OXDT

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

thusIσ = 0.167

and so ∆σ zz = p I σ = -100 × 0.167 = -16.7 kN/m²

For rectangular loading OXAY

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

thusIσ = 0.240

and so∆σ zz = p I σ = + 100 × 0.240 = + 24.0kN/m²

Fig. 14 Dimensions of rectangular loaded area




For rectangular loading OZBY

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

thus Iσ = 0.164and so

∆σ zz = p I σ = -100 × 0.164 = -16.4 kN/m²

Thus the increase in stress ∆σ zz = 12.1 - 16.7 + 24.0 + -16.4=3.0 kN/m²

5.4.3 Exercises

1. Using the graphed relation between e and σ′ from Chapter 4 - Exercise 1, calculate themaximum differential settlement of a flexible rectangular foundation with dimensions 3 m x6 m located on the surface of a 4.5 m layer of the clay. The stress on the foundation is 500kPa. The clay overlies an incompressible stratum. For the purpose of calculation, dividethe clay into three layers, each 1.5 m thick. (Use influence curves for stresses beneathcorners of rectangles). The water table is 1 m below the surface and the saturated unitweight of the soil is 17 kN/m 3. It may be assumed that the clay remains fully saturatedabove the water table.

2. A soil deposit consists of 2 m of gravel overlaying an 8 m thick deposit of an over-consolidated clay overlaying rigid permeable sandstone. The gravel has a unit weight of 22kN/m3 when saturated and a unit weight of 18 kN/m3 when dry. The properties of the clayare uniform throughout the layer and indicated below:

Void Ratio e 1.2Specific Gravity G s 2.7Pre-consolidation pressure σ′ pc 70 kPaCompression Index C c 0.1Recompression Index C r 0.02Consolidation coefficient c v 0.5 m 2 /year

Calculate the settlement under the centre of a square tank of side 5 m exerting an averagepressure of 175 kPa constructed on the ground surface after a long period of time.Assume that initially the water table is at the surface of the gravel but that after constructionit is lowered to the surface of the clay. The clay should be divided into two equal sub-layers.

5.4.4 Stresses under loads of arbitrary shape

Newmark’s chart provides a graphical method for calculating the stress increase due to auniformly loaded region, of arbitrary shape resting on a deep homogeneous isotropic elasticregion.

Newmark’s chart is given as separate .doc file on the Moodle website and is reproduced inpart in Fig 15. The procedure for its use is outlined below1. 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 verticalprojection on the surface) is at the center of the concentric circles, the orientation of thedrawing does not matter

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

count the square, otherwise neglect it.

4. The vertical stress increase ∆σ 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 kN/m². It isrequired to calculate the vertical stress at a depth of 4 m below the edge of the circle.

The loaded area is drawn on Newmark’s chart to the appropriate scale (i.e. the length OQ isset to represent 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

∆σ zz = 194 × 0.001 × 100 = 19.4 kN/m²

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

z/a = 2, r/a = 1. ∆σ zz /p = 0.2 and so ∆σ zz = 20 kN/m²




Loaded

Area

O Q

INFLUENCEVALUE = 0.001

4m

Fig 15 Newmark’s Chart

5.4.5 ExerciseUse the Newmark's influence chart given as separate .doc file on the Moodle website to findthe vertical stresses caused by a flexible hexagonal shaped foundation with all sides 3 m longcarrying a uniform load of 500 kN/m², at depths of 3 and 4.5 m beneath a corner of thefoundation.Note : To be able to draw the complete loading area, it may be necessary to move the centerof the chart to the left.

soilmech ch5 settlement calculation

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