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CE-319 (3 Credit Hours)

Geotechnical and Foundation Engineering

Foundation Settlements-immediate

settlement

Instructor:

Dr Irshad Ahmad

Lecture on 21-3-15

Department of Civil Engineering

University of Engineering and Technology, Peshawar

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Foundation Settlement- Introduction

Some Typical Loads on Building Foundations

Number and Depth of boring

Immediate settlement calculations using Theory of Elasticity

SPT Procedure

SPT corrections

SPT correlations

Immediate settlement using SPT

Contents

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Foundation Settlement

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Foundation Settlement

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Some Typical Loads on Building Foundations

Settlements are cause by

change in stress in the soil mass

Change in stress occurs because

Of building loads

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Foundation Settlement

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Number of borings

Depth of borings

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Depth of boring

For Square footing:

10% qo at 2B

For Strip footing

10% qo at 6.5B

For circular footing

10% qo at 2 Dia

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Immediate Settlement

We will find immediate settlement through

THEORY OF ELASTICITY

USING STANDARD PENETRATION TEST RESULTS

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Immediate Settlement- Theory Of Elasticity

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Immediate Settlement

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Shape and Rigidity factor “I”

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Example

Method-1: Taking weighted average of Es

Method-2: Method of superposition

Both methods are discussed in class

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Example-Using Weighted Average of Es

Es (average) = (42.53 + 60 8)/11=55 Mpa,

H=11m, B=B/2=33.5/2=16.75,

H/ B=11/16.75=0.66 (use 0.7)

L/B = (39.5/2)/(33.5/2)= 1.18

Assume, =0.33, I=0.12 (after interpolation)

H=qoB(1-2)/Es m I

= 134 16.75 (1-0.332)/(55 1000) 4 0.12

= 0.0175m = 17.5 mm

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Example- Method of Superposition

Methodology: Calculate Hi

(1) Consider the 3m Sand layer with Es overlying

sand stone.

(2) Consider whole 11m layer as Clay with Es=60

MPa overlying Sand stone.

(3) Consider upper 3m layer as Clay with Es=60

Mpa overlaying Sandstone.

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Es = 42.5Mpa,

H=3m,B=B/2=16.75,

H/ B=3/16.75=0.18, L/B= 1.18

I=0.0316

H=qoB(1-2)/Es m I

134 16.75 (1-0.332)/(42.5 4

0.0316

= 5.96 mm

Es = 60 Mpa,

H=11m,B=B/2=16.75,

H/B=3/16.75=0.18, L/B= 1.18

I=0.12

H=qoB(1-2)/Es m I

134 16.75 (1-0.332)/(60 4

0.0316

= 16 mm

Es = 60 Mpa,

H=3m, B=B/2=16.75,

H/B=3/16.75=0.18, L/B= 1.18

I=0.0316

H=qoB(1-2)/Es m I

134 16.75 (1-0.332)/(60 4

0.0316

= 4.2 mm

H= 5.96 + 16 - 4.2 = 17.7 mm

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Standard Penetration Test (ASTM D-1586)

Practicing Engineers use the SPT widely in estimating the bearing

capacity of soils and to assess the in-situ relative density of a sand

deposit.

The test is performed using a split spoon barrel sampler 50mm

external diameter, 35mm internal diameter and about 650mm in length

and connected to the end of boring rods.

After boring has been advanced to the desired sampling elevation and

excessive cutting has been removed, attach the split spoon sampler to

the sampling rods and lover into borehole.

Drive the sample with blows from 140lb (63.5 kg) hammer falling

freely through a height of 760mm (30 inch).

The sampler is advanced under the impact of the hammer into three

successive (6 inch) increments. (i.e. total 18 inch).

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Standard Penetration Test (ASTM D-1586)

Count the number of blows in each 6 inch increment until one of the

following occurs

A total of 50 blows have been applied during any of the three 6

inch increments.

A total of 100 blows have been applied

There is no observed advance of the sampler during the

application of 10 successive blows of the hammer.

The sampler is advanced the complete 18 inch without the

limiting blow counts occurring as describe in 1,2 and 3.

The 1st 6 inch is considered to be a seating drive. The sum of the

number of blows required for the 2nd and 3rd 6-inch penetration is

termed the standard penetration resistance or the N-value.

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Disturbance caused by Split spoon sampler

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Standard Penetration Test (ASTM D-1586)

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Thin wall tube sampler (Shelby tubes)

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Hammers

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Standard Penetration Test (ASTM D-1586)

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Corrected N- value

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N-value corrections

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SPT Correlations

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SPT Correlations

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SPT Correlations

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SPT Correlations

Undrained shear strength cu(kN/m2) = 29 (N60)0.72 [Hara, et al. 1971]

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Immediate Settlement: SPT (Bowels P263)

N60

SI Fps

F1 0.05 2.5

F2 0.08 4

F3 0.3 1

F4 1.2 4

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Design N Values

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Problem (Bowel P-266)

Fill up the below table using equation for the widths given and prepare

a graph

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PROBLEM

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PROBLEM

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Example (Bowels P-543)

Find allowable BC qa to ensure

FOS=3 against Shear failure

Si <= 50 m

Solution:

(1) Shear Criteria:

qu = c Nc sc + D Nq + ½ B N s

BC factors: For ⱷ = 0, Nc = 5.14, Nq = 1 and

N = 0

Shape factors: sc=1.3, and s=0.8

qu (net)= qu(gross) - D =

(1505.14 1.3 + 18.7 1.5 1) - 18.7 1.5

= 1000 kPa

qsafe = qu/FOS = 334 kN/m2

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Example (Bowels P-543)

Es1= 1000 cu = 1000(300/2)=150 000 kPa

Es2= 500(N55+15)= 500[18(70/55)+15]=18950 kPa

Es2= 500(N55+15)= 500[22(70/55)+15]=22 000 kPa

Es2= 500(N55+15)= 500[40(70/55)+15]=32 900 kPa

(2) Settlement Criteria

Es(average)= 3.4(150000)+3.3(18950)+7(22000)

+13.8(32900)]/27.5 = 42 930 kPa

H from base of the mat to rock = (4.9-

1.5)+3.3+7+13.8 = 27.5m,

Estimate the mat will be on the order of 14m,

giving

B=B/2=14/2=7, H/ B=27.5/74, L/B = 1

For =0.33, I=0.42 (after interpolation)

H=qoB(1-2)/Es m I

0.05 = qo 7 (1-0.332)/(42930) 4 0.42

qo= 204 kPa

Allowable Bearing Capacity:

qa=200 kPa, settlement controls:

As the design proceeds and B is found to be

substantially different from 14m, it may be

necessary to revise qa.