Application of geo-engineering knowledge in designing

Shallow Foundations -3

ENB371: Geotechnical Engineering 2

Combined & Matt (Raft) FoundationsField tests used in shallow foundation design

Chaminda

Chapter Chapter 5 5 & & 66

BRAJA M. DASBRAJA M. DAS

Content……

• Combined Footings (supports a line of two or more columns)

• Mat or Raft Foundations (supports more than one line of columns)

Field tests in Foundation system• Plate load test• Standard Penetration Test (SPT)• Cone Penetration Test (CPT)

Combined Footings

Under normal conditions, spread footings (square,rectangular, circular) and strip footings are economical t osupport columns and walls.

However, under certain circumstances (restrained byproperty line) it is desirable to construct a footing that

• Rectangular Combined Footing• Trapezoidal Combined Footing• Cantilever Footing

property line) it is desirable to construct a footing thatsupport a line of two or more columns

Combined Footings (cont..)Rectangular combined footing

Location of the resultant of the column loads, X

21

32

QQ

LQX

+=

For a uniform distribution of soil pressure: theresultant load should pass through the centreof the foundation . Thus,of the foundation . Thus,

foundation theoflength where

)(2 2

=+=

L

XLL

Once the L is determined, L2

321 LLLL −−=Assume B and calculate net allowable bearing pressure, q net(all) , using Meyerhof BearingCapacity Theory and FS .

FS

qq unet

allnet)(

)( = 21)( )( QQLBq allnet +≥××

Combined Footings (cont..)Trapezoidal combined footing

Assume B 1 and B 2 , then the effective width B:

21

32

QQ

LQX

+=

From the property of a trapezoid

Location of the resultant of the column loads, X2

21 BBB

+=

3

2

21

212

L

BB

BBLX

++=+

LBB

A2

21 +=Calculate net allowable bearing pressure, q net(all) , using Meyerhof Bearing CapacityTheory, FS , B, and L

FS

qq unet

allnet)(

)( =21)( QQAq allnet +≥×

Once the L is determined, area of the foundation, A

Combined Footings (cont..)Cantilever Footing

• Cantilever footing uses a strap beam to connecteccentrically loaded column foundation to the foundationof an interior column.

• Cantilever footings may be used in place of rectangular ortrapezoidal combined footings when the allowable bearingcapacity is high and distances between columns are large

Mat (Raft) Foundations

• Supports more than one line of columns

• Use with soil that has low bearing capacity

• If spread footings covering more than the• If spread footings covering more than thehalf of building area, Mat foundation mightbe more economical

Mat (Raft) Foundations (cont..)Types of Mat Foundations

(a) Flat Plate (b) Flat Plate thickened under column

Mat (Raft) Foundations (cont..)Types of Mat Foundations

(c) Beams and Slab (d) Slab with Basement Walls

Mat (Raft) Foundations (cont..)Bearing Capacity of Mat Foundation

Comparison of isolated foundation and mat foundatio n ( B = width, D f = depth)

Same procedure as square or rectangular spread footing

Mat (Raft) Foundations (cont..)Net pressure on soil caused by a mat foundation

The net average applied pressure on soil by foundation, q

fDA

Qq γ−=

q should be less than q net(all) and q net(u)

Mat (Raft) Foundations (cont..)Net pressure on soil caused by a mat foundation

This indicates that the net pressure increase in soil under t he foundation canbe reduced by increasing the depth (D f) of mat foundation

fDA

Qq γ−=

This relation for Df is referred to as the depth of a fully compensated foundation

For no increase in the net pressure on the soil below the mat fo undation, qshould be zero. Thus,

γγ

A

QDD

A

Qff =⇒−=0

Example -1A mat foundation on a saturated clay soil has dimen sions of 20 m x 20 m. Given: dead and live load = 48 MN, c u = 30 kN/m2, and γγγγclay = 18.5 kN/m3.

(a) Find the depth , D f, of the mat to be fully compensated foundation.(b) What will be the depth of the mat (D f) for a factor of safety of 2

against bearing capacity failure?

Example -1(Solution)

2

3

/30

/5.18

48

20

20

mkNc

mkN

MNQ

mL

mB

u

clay

=

====

γ

The net average applied pressure on soil by foundation, q

QfD

A

Qq γ−=

To be fully compensated foundation, q = 0. Thus,

mA

QD f 49.6

5.18)2020(

100048 =××

×==γ

Example -1(Solution)

2

3

/30

/5.18

48

20

20

mkNc

mkN

MNQ

mL

mB

u

clay

=

====

γ

Meyerhof bearing capacity equation

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq γγγγγ5.0' ++=For undrained condition, φφφφ=0, thus N c=5.14, Nq=1, Nγγγγ=0

Shape factors,

19.114.5

1

20

2011 =

+=

+=c

qcs N

N

L

BF

10tan20

201'tan1 =

+=

+= φL

BFqs

Example -1(Solution)

No load inclination 1=== FFF

Depth factors, assume 1≤B

D f

+=

+=

204.014.01 ff

cd

D

B

DF

120

)0sin1(0tan21)sin1('tan21 22 =

−+=

−+= ff

qd

D

B

DF φφ

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq γγγγγ5.0' ++=No load inclination 1=== iqici FFF γ

idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq γγγγγ5.0' ++=

ff

fu

DD

qD

q

γ+

+×=

××××+

+×××=

204.015.183

1111)20

4.01(19.114.530

Example -1(Solution)

+×=−+

+×=

204.015.183

204.015.183)(

fff

funet

DDD

Dq γγ

fuunet Dqq γ−=)(

fffappnet DDDA

Qq 5.181205.18

2020

48000)( −=−

×=−= γ

A 2020×

)(

)(

appnet

unet

q

qFS =

mDD

D

ff

f

4.15.18120

204.015.183

2 =⇒−

+×

=

Meyerhof (1963) Bearing Capacity Theory Table 03: Bearing capacity factors (Eq. 11, Eq. 12, and Eq. 13)

Field tests in Foundation system• Plate load test• Standard Penetration Test (SPT)• Standard Penetration Test (SPT)• Cone Penetration Test (CPT)

Plate Load Test

Plate load test is used in the field to determine t he ultimate bearing capacity

Test details:• Size of steel plate:

- thickness is 25 mm

≧≧≧≧

- Circular( Diameter (B) = 150 ~ 750 mm)/ Square (30 0 X 300 mm)• Excavated width ≧≧≧≧ 4B• Excavated depth D f (Df – depth of proposed foundation)

Plate Load Test (cont….)

• Load is applied to the plate in steps by means of a jack• One hour is allowed to elapse between loading steps• Test is conducted:

– Until failureOR– Until the plate settlement of 25 mm

Plate Load Test (cont….)

Plate Load Test (cont…)

qu(P) is obtained from plate loading test (stress at the failure of the test plate OR stress at 25 mm settlement of bearing plat e

For test in clay:

plate test theofcapacity mearing Ultimate

foundation proposed theofcapacity bearing Ultimate

)(

)(

=

=

Pu

Fu

q

q

qu(F) , SFq S

Plate Footing

For test in clay:

)()( PuFu qq =

For test in sandy soils:

P

FPuFu B

Bqq )()( =

plate test theofwidth

foundation theofwidth

==

P

F

B

B

BP BF

qu(F) , SFqu(P), SP

~2.5BF

~2.5BP

~ 0.1q

q

Plate Load Test (cont…)

For clayey soil:

FPF B

BSS =

plate test theof settlement

foundation theof settlement

==

P

F

S

S

Plate Footing

The allowable bearing capacity of a foundation, bas ed on the settlement consideration and for the given intensit y of load, q 0 , is

)foundation theof stress bearing is ( If 00)()( qqqq PF ==

PPF B

SS =

For sandy soils:

22

+=

PF

FPF BB

BSS

BP BF

qu(F) , SFqu(P), SP

Plate Footing

~2.5BF

~2.5BP

~ 0.1q

q

mmSF 25≤

)foundation theof stress bearing is ( If 00)()( qqqq PF ==

Example -1The results of a plate loading test on a sandy soil are shown in the following figure. The size of the plate is 300 x 30 0 mm. Determine the size of a square column foundation that should carr y a load of 2500 kN with the maximum settlement of 25 mm.

Example -1 (solution)This problem has to be solved by trial and error

Q0 [kN] Assume widthBF [m]

[kPa] SP [mm]From given chart

SF [mm]Settlement of foundation

2500 4.0 156.25 4.0 13.85

2500 3.0 277.80 8.0 26.45

2500 3.2 244.10 6.8 22.74

FA

Qq 0

0 =

2500 3.2 244.10 6.8 22.74

2500 3.1

22

+=

PF

FPF BB

BSS

BF = width of foundationBP = width of plate = 300 mm

Standard Penetration Test (SPT)

• 63.5 kg mass is dropped through 760 mm to make one blow on the SPT spoon.

• Number of blows to penetrate three 150 mm distances are measured given as x/y/z

• The blows for first 150 mm are discarded.• The number of blows for next 300 mm is

taken as the “raw” blow count or standard taken as the “raw” blow count or standard penetration number, i.e., N=y+z

• N is measured at every 0.75 m or 1.5 m depth

The “raw” blow count or standard number (N) should be corrected for:

(1) Water table(2) Effective overburden pressure

Refusal100mm) 450 allfor ( N if

50 mm 300(for N if OR

>>

Standard Penetration Test (SPT) – Cont…

SPT Drilling rig required; Test within boreholes; Recovers a sample; Slow testing over soil profile;

Split with sampling tip (silty and clayey sand)

soil catchers

Standard Penetration Test (SPT) – Cont…

Split with sampling tip (silty and clayey sand)

Assembled with solid tip for sand and gravel

Standard Penetration Test (SPT) – Cont…Corrections to “raw” N

Correction 1 - Water table correction:When SPT test is carried out in fine sand or silty sand below the water

table, the measured N, if greater than 15, should b e corrected for increased resistance due to negative excess pore wa ter pressure during driving and unable to dissipate immediately

Conditions:Conditions:

1. Be below the water table

2. N> 15

3. Indication of reduced permeability (i.e., fine or silty/clayey sand)

Ncorr=15+0.5(N-15)

Standard Penetration Test (SPT) – Cont…Corrections to “raw” N

Correction 2 - For effective overburden pressureN depends on the effective stress at the depth of m easurement; effective stress can be represented by effective ov erburden pressure

Ncorr=Ncorr(1) × CN

Overburden correction factor C Ncan be obtained from the chart

Standard Penetration Test (SPT) – Cont…Corrected blow count (standard penetration resistan ce) Ncorr can be used with the effective overburden pressure at the depth where N is measured to obtain the effective friction angle φφφφ’ of soil.

Example -2

Say SPT was performed on silty sand medium (unit weight=18kN/m3) at 5 m depth. Water table was 2 m below the surface. SPTblow counts were 10/16/19. Calculate the corrected N value.

Nraw = 16 + 19 =35

Correction 1 – applies (silty sand, below water table)

Ncorr 1 = 15+0.5 (N-15) = 15+0.5(35-15)=25

Correction 2 - Vertical eff. Stress = 18 x 5- 3x 9.8

=60.6 kPa

From graph, C.F. (CN) = 1.25

Ncorr = Ncorr(1)×CN

Ncorr = 25 x 1.25 = 31

Standard Penetration Test (SPT) – Cont…Procedure to obtain the allowable bearing capacity of

sand using SPT results

Step 1: Conduct the SPT test to thedepth B below the foundationlevel (B is foundation width)

Step 2: Correct N (raw) asappropriated

Step 3: Calculate the average NStep 3: Calculate the average Ncorrvalue to the depth B below thefoundation

Step 4: Use the average N corr and Bto obtain the provisionalallowable bearing pressurefrom the chart (maximumsettlement of the foundation islimited to 25 mm)

Standard Penetration Test (SPT) – Cont…Procedure to obtain the allowable bearing capacity of

sand using SPT results

Step 5: To correct the provisional value of allowable bearin g pressure, itshould be multiplied by a factor C w ,

D

If )(0 BDD fw +≤≤

D fDwD

If )( BDD fw +>

BD

DC

f

ww +

+= 5.05.0

0.1=wC

Thenwprovallall Cqq ×= )(

Example -3

A footing 3 x 3 m is to be located at a depth of 1.5 m in a sanddeposit, the water table being 3.5 m below the surface. The values ofstandard penetration resistance were determined as given below.Determine the allowable bearing capacity of the foundation. Theunit weights of sand above and below water table are 17 kN/m2 and19.8 kN/m2, respectively.Depth (m) N

0.75 80.75 8

1.55 7

2.30 9

3.00 13

3.70 12

4.45 16

5.20 20

Example -3

Water table correction is not necessary - sandDepth (m) N σ’v [kN/m2] CN Ncorr=N*CN

0.75 8 13

1.55 7 26 2.0 14

2.30 9 39 1.6 14

3.00 13 51 1.4 18

3.70 12 62 1.25 153.70 12 62 1.25 15

4.45 16 69 1.2 19

5.20 20 77

(av) = 16

B = 3 m, and the corrected average N = 16, therefore

The provisional allowable bearing capacity = 165 kN/m2 (from chart)

89.0)5.13(

5.35.05.05.05.0 =

+×+=

++=

f

ww DB

DC

qall = Cw×qall(prov)=0.89×165 =150 kN/m2

Cone Penetration Test (CPT)

A cone having an apex angle 600 and the diameter of 35.7 mm2

(projected area of cone is 1000 mm2) is pushed directly into the ground at a rate of 20 mm/sec by means of static thrust.

Cone Penetration Test (CPT) (cont..)

Cone Penetration Test (CPT) (cont…)

– No drilling

– No soil sample

– Much information

– penetration resistance, qc & sleeve friction, fs,

Cone Penetration Test (CPT) (cont…)

– penetration resistance, qc & sleeve friction, fs,

Pore water pressure, ..

– FR = fs/qc used to distinguish soil types

– Direct application to piling

– E = fn (qc)

Cone Penetration Test (CPT) (cont…)Typical test results

resistance tip=cq pressurewater -Pore=uFriction Sleeve=sf 100(%) ×=c

s

q

fFR

CPT & soil type

10

100

Con

e R

esis

tanc

e, q

c M

Pa

sands

silty

san

ds

sand

y si

lts,

silts

clay

ey s

ilts,

si

lty c

lays

0.1

1

0 1 2 3 4 5 6

Friction Ratio, F r %

Con

e R

esis

tanc

e, q

clays

Robertson & Campanella 1982

CPT and soil properties

Undrained shear strength

of clays,

• Nk = 12 to 15voc

u N

σqc

−=• Nk = 12 to 15

• Correlation improved if

OCR taken into account

• Direct application to piling

ku N

pressure overburden Total

resistance Tip

strengthshear Undrained

vo ===

σu

u

q

c

CPT and soil properties

Shear strength of clean sands

′+=′

vo

cq

σφ log38.01.0arctanmax

Shear strength parameters

obtained from CPT can be

used for bearing capacity

calculation of shallow

foundation

Settlement of shallow foundation -1

Width of the foundation (B) is decided

from ultimate bearing capacity analysis

to satisfy the given FS ( Q all is

foundation load from super-structure)

allQ

B

Sand

x

Clay

Calculate the total settlement of the foundation (elastic & consolidation)

considering soil properties and foundation stress distrib ution to the

depth of 1.5 B (square, rectangular, circular footing) OR 3B for strip

foundations

Ultimate Limit State

Settlement of shallow foundation -2

(Stresses in soil & elastic deformation of soil) – lecture 2 a nd 3

ionconsolidatelasticTotal SSS +=foundation under the SandClay / of settlement )(immediate Elastic=elasticS

foundation under theclay of settlemention Consolidat=ionconsolidatS

(Stresses in soil & consolidation) – lecture 4 and 5(Stresses in soil & consolidation) – lecture 4 and 5

IF)max(allowableTotal SS ≤

)foundation shallow isolatedfor mm 45 ~ 25 ( standarddesign in given is )max(allowableS

Satisfied Serviceability Limit State

IF NOT, increase the foundation size, calculate S Total and checkserviceability limit state

Flowchart for foundation design

Assume foundation

dimensions ( B & L)

FS

QQ ultnet

allnet)(

)(

Calculate

=

structuresuper from

load Foundation - Q

heoriescapacity t bearing From

Calculate )(ultnetQ

NO

QQ allnet ≥)(

B & L

From Ultimate Limit State

TotalS Calculate

Design Safe

)max(

Check

allowableTotal SS ≤

NO

OK

OK