1

CE2305 - FOUNDATION ENGINEERING .

1. SITE INVESTIGATION AND SELECTION OF FOUNDATION

Scope and objectives Methods of exploration-averaging and boring Water boring and rotatory drilling Depth of boring Spacing of bore hole - Sampling Representative and undisturbed sampling sampling techniques Split spoon sampler, Thin tube sampler, Stationary piston sampler Bore log report Penetration tests (SPT and SCPT) Data interpretation (Strength parameters and Liquefaction potential) Selection of foundation based on soil condition.

2. SHALLOW FOUNDATION

Introduction Location and depth of foundation codal provisions bearing capacity of shallow foundation on homogeneous deposits Terzaghis formula and BIS formula factors affecting bearing capacity problems - Bearing Capacity from insitu tests (SPT, SCPT and plate load) Allowable bearing pressure, Settlement Components of settlement Determination of settlement of foundations on granular and clay deposits Allowable settlements Codal provision Methods of minimising settlement, differential settlement.

3. FOOTINGS AND RAFTS

Types of foundation Contact pressure distribution below footings & raft - Isolated and combined footings types proportioning - mat foundation types use - proportioning floating foundation.

4. PILES

Types of piles and their function Factors influencing the selection of pile Carrying capacity of single pile in granular and cohesive soil - Static formula - dynamic formulae

(Engineering news and Hileys) Capacity from insitu tests (SPT and SCPT) Negative skin friction uplift capacity Group capacity by different methods (Felds rule, Converse Labarra formula and block failure criterion) Settlement of pile groups Interpretation of pile load test Forces on pile caps under reamed piles Capacity under compression and uplift.

5. RETAINING WALLS

Plastic equilibrium in soils active and passive states Rankines theory cohesionless and cohesive soil - Coloumbs wedge theory condition for critical failure plane - Earth pressure on retaining walls of simple configurations Graphical methods (Rebhann and Culmann) - pressure on the wall due to line load Stability of retaining walls.

2

UNIT II

SHALLOW FOUNDATION

Introduction Location and depth of foundation codal provisions bearing capacity of

shallow foundation on homogeneous deposits Terzaghis formula and BIS formula

factors affecting bearing capacity problems - Bearing Capacity from insitu tests (SPT,

SCPT and plate load) Allowable bearing pressure, Settlement Components of

settlement Determination of settlement of foundations on granular and clay deposits

Allowable settlements Codal provision Methods of minimising settlement,

differential settlement

CONTENTS

S.NO Part A

PAGE.NO.

1 What is shallow foundation? 8

2 What are the factors to be consider while designing the foundation. 8

3 Define Bearing capacity and Ultimate bearing capacity.

8

4 Define Net ultimate bearing capacity and Net safe bearing capacity. 8

5 Define Safe bearing capacity and Allowable bearing pressure. 9

6 6. Explain ultimate bearing capacity with the help of load settlement curve.

9

7 State the different modes of shear failure. 10

8 In what way the local shear failure differs from General shear failure. 10

9 How the effective dimensions can be calculated in an eccentrically loaded

footing?

10

10 What are the Assumptions made in Terzahis Analysis? 10

11 State the Limitations of Terzahis Analysis. 11

3

12 State the factors affecting Bearing capacity. 11

13 What is the correlation between C.P.T. and S.P.T. values? 11

14 Define Settlement 11

15 What are components of settlement 11

16 Define Co-efficient of settlement 12

17 Define Immediate Settlement (Si) 12

18 Define primary consolidation 12

19 Define Secondary compression settlement 12

20 Define seat of settlement. 12

21 State the corrections to be made for the Settlement due to Consolidation 13

22 State the corrections made for the observed SPT values. 13

23 State the factors affecting Bearing capacity. 13

24 State Permissible Settlement as per IS code. 13

S.NO Part B

PAGE.NO.

1

What are the IS code recommendations for the location and depth of

foundation?

14

2 Explain the different modes of failure of foundation soil.

15

3 Explain the corrections made for the observed SPT values.

17

4 Explain the method of carrying out Plate Load Test

17

5 Explain the Application SPT values

20

6 Explain the Application of CPT values

21

4

7 Explain the components of Settlement?

22

8 Explain the Emprical Equations to determine the Allowable bearing

pressure.

23

9 Calculate the ultimate bearing capacity per unit area of

(1) a strip footing 1 m wide (2) a square footing 3m x 3m, and (3) a circular footing of 3m diameter.

Given:

Unit weight of the soil 1.8 t/m3, cohesion = 2t/m

2

And = 20o

Nc = 17.5, Nq = 7.5 and N = 5.

25

10

A strip footing 2 m wide carries a load intensity of 400 KN/m2 at a depth

of 1.2 m in sand. The saturated unit weight of sand is 19.5Kn/m3

and unit

weight above water table is16.8KN/m3

. The shear strength parameters

are c = 0 and = 35o . Determine the factor of safety with respect to shear failure for the following cases of location of water table.

28

11

Determine the ultimate bearing capacity of the footing in Ex. 15.1 if the

ground water table is located (a) at a depth of 0.5 m below the ground

surface, (b) at a depth of 0.5m below the base of the footing. sat = 20 kN/m

3. Use Terzaghi theory.

30

12

An R.C. Column footing square in the shape is to rest 1.5 m below level.

The total ground level. The total load to be transmitted including the

frequent flooding, the friction of the foot along the sides is to be

neglected and a factor of safety 2.4 gm/c.c. angle of internal friction 33

and value of N = 33 and Nq = 32 find a suitable a size of the footing for the above condition.

30

13

In a plate bearing test on pure clayey soil failure occurred at a load of

12.2 tones. The size of the plate was 45 cm x 45 cm and the test was one

at a depth of 1.0 m below ground level. Find out the ultimate bearing

capacity for a 1.5 m wide continuous wall footing with its base at a depth

of 2m below ground level. The unit wt. of clay may be taken as 1.9 gm/

c.c. and Nc = 5.7, Nq = 1 and N = 0.

30

14

A square footing located at a depth of 1.5 m below the ground surface in

cohesionless soil carries a column load of 1280 kN. The soil is

submerged having an effective unit weight of 11.5 kN/m3 and an angle of

shearing resistane of 30o. Find the size of the following for Fs = 3 by

Terzaghis theory of general shear failure

31

Determine the net bearing pressure for a 2X2 m footing at a depth of 1.5 32

5

15

m in a medium dense sand so that the total settlement does not exceed 25

mm. The average SPT blows below the footing are 20 per 30 cm. The

average moist unit weight of soil is 17 KN/m3. The water table is at 5 m

below the ground level

16

If a deposit happens to be silty clay (saturared with a value of qc = 88

kg/cm2, determine the unconfined compressive strength of clay as per

Use p0=127kN/m2.

33

17

A footing foundation of 3m X 3m is to be constructed at a site at a depth

of 1.5 m below ground level. The water table is at the base level of

foundation. The average static cone penetration resistance obtained at the

site is 20 Kg/m2.

The soil is cohesive determine the safe bearing capacity for a settlement

of 40mm.

33

18

A 30 cm square bearing plate settles by 1.5 cm in a plate loading test on

a cohesion less soil when the intensity of loading is 2 kg/cm2. What will

be the settlement of a prototype footing 1m square under the same

intensity of loading.

34

19

Two plate load test s were conducted at the level of a prototype

foundation in cohesionless soil close to each other. The following data

are given.

Size of plate Load applied Settlement

recorded

0.3m X 0.3m 30 KN 25 mm

0.6m X0.6m 90 KN 25

mm

If the footing is to carry a load of 100KN, determine the size of the

footing for the same settlement of 25 mm.

34

20

Soil investigation at a site gave the following information. Top soil up to

a depth of 10.6 m is fine sand, and below this lies soft clay layer of 7.60

m thick. The water table is at 4.60 m below the ground surface. The

submerged unit weight of sand is 2/4.10 mKNb and unit weight

above water table is 17.6 KN/m3. The water content of the normally

consolidated clay wn = 40% its liquid limit, wl = 45%, and specific

gravity of the solid particle is 2.78. The proposed construction will

transmit a net stress of 120 KN/m2

Find the average settlement of clay

layer.

35

6

PART A

TWO MARK QUESTIONS AND ANSWERS

1. What is shallow foundation?

When the Depth of the foundation Df is less than the Breath of the foundation B,

then the foundation is called shallow foundation.

BD f

2. What are the factors to be consider while designing the foundation

(1) The load applied to the soil should be such that the induced stress in the

soil is lesser than its capacity, as otherwise, soil will fail by shear.

(2) The settlement in the soil should be such that it is within the tolerance limit

of the super- structure. In particular, differential settlements should not cause any

unacceptable damage nor interfere with the function of the structure

3. Define Bearing capacity and Ultimate bearing capacity

Bearing capacity The supporting power of a soil or rock is referred to as its bearing

capacity.

Ultimate bearing capacity (qf) The ultimate bearing capacity of the soil (qf) is thus defined as that

pressure or load intensity, which just causes the shear failure of the supporting soil

immediately below and adjacent to a foundation.

(or)

The Ultimate bearing capacity is defined as the minimum gross pressure

intensity at the base of the foundation at which the soil fails in shear.

4. Define Net ultimate bearing capacity and Net safe bearing capacity

Net ultimate bearing capacity (qnf) It is the minimum net pressure intensity causing shear failure of soil.

qnf = qf . Df

Net safe bearing capacity (qns) The Net safe bearing capacity is the net ultimate bearing capacity divided by the

factor of safety.

qns =F

D f.qf

7

5. Define Safe bearing capacity and Allowable bearing pressure

Safe bearing capacity (qs).

The maximum pressure which the soil can carry selfly without risk of shear

failure is called the safe bearing capacity.

Allowable bearing pressure It is the net loading intensity at which neither the soil fails in shear nor there

is excessive settlement.

7. Explain ultimate bearing capacity with the help of load settlement curve

Load settlement Curves:

When a load is applied to a continuous footing of depth Df and width B, the

footing settles. The relation between the average load per unit area and the settlement is

generally of the type shown in fig (2) A visual inspection of any one such curve indicates

that the settlement increases proportionately with the load up to a certain value; and on

further loading, the settlement increases suddenly. In dense and stiff soils, the failure

occurs at a very distinct point (curve I); whereas in loose or soft soils, the point of failure

is slightly less distinctly defined (curve II) and may be taken as the point, where the

settlement curve becomes fairly steep and approaches a straight line. This failure stress

(qf) is called the ultimate bearing capacity of the soil.

7. State the different modes of shear failure

1. General shear failure 2. Local shear failure 3. Punching shear failure

8. In what way the local shear failure differs from General shear failure

The bearing capacity is less in local shear failure than in the general shear failure.

For such a case, Terzaghi has suggested that the values of Nc, Nq, and N may be worked

out by assuming the shear parameters c and , each equal to 2/3 of their respective effective ultimate values ( to be compared as tan ). In other words,

.c for local shear failure = ccl3

2

for local shear failure =

tan

3

2tan 1l

It is difficult to precisely define the limiting conditions for which the general or

local shear failure should be assumed at a given site. However, the following points may

be used as a guide;

8

(i) Angle of shear resistance For > 36: general shear failure For < 28; local shear failure

(ii) SPT N Test results N > 30: General shear failure

N < 5: Local shear failure

9. How the effective dimensions can be calculated in an eccentrically loaded

footing?

If the eccentricities are eL and eB, then the usefull or effective widths of the footing

shall be

B = B 2eB L = L -- 2eL

10. What are the Assumptions made in Terzahis Analysis?

On the basis of the failure patterns, Terzaghi first devised an approximate

method for computing the ultimate bearing capacity of soils. The various assumptions

which he made in this analysis are summarized below:

i) The soil is homogenious and Isotropic and its shear strength is represented by Colombs equation.

ii) The base of the footing is roug. iii) The Elastic zone has straight boundaries inclined at = to the

horizontal, and the plastic zone fully develop.

iv) The resultant passive load PP components which can be calculated separately and added, although thre critical surface for these components

are not identical.

v) zones do not extend above the horizontal plane through the base of the footing, ie The shear resistance of the soil above the base is neglected and

the effect of soil around the footing is considered equivalent to a surcharge

. Df. vi) The footing is shallow, i.e. Df

9

(iii) Error due to assumption 5 increases with depth of foundation, and hence the theory is suitable for shallow foundation only.

12. State the factors affecting Bearing capacity.

(i) angle of internal friction (or) angle of shearing resistance. (ii) Width of the footing, (iii) Depth of the footing, (iv) Unit weight of the soil, (v) Position of ground water table.

13. What is the correlation between C.P.T. and S.P.T. values?

Mayerhof correlatesSPT and CPT for fine or silty medium loose to medium-

dense sand as

qc = 0.4N

14. Define Settlement.

The term settlement is used to describe the vertical displacement of the base of a

structure. The effects of settlement depend only on its magnitude but also on its degree of

uniformity and on the nature of the engineering works affected.

15. What are components of settlement.

There are three components of settlements. They are:

1. Immediate settlement, Si

2. Primary consolidation settlement, Sc

3. Secondary compression settlement, Ss

The total settlement is the sum of these three which may be written as

S = Si + Sc + Ss

16. Define Co-efficient of settlement.

Co-efficient of settlement is defined as the ratio of ultimate bearing capacity and

settlement.

Co-efficient of settlement S

q

17. Define Immediate Settlement (Si)

Immediate settlement is a combination of elastic compression and plastic deformation

without change in volume or water content. This type of settlement develops are

construction proceeds. Immediate settlements are significant in non-saturated clays, silts

and granular soils

18. Define primary consolidation

10

primary consolidation settlement (Sc) which is due to a gradual expulsion of pore

water from the voids of the soil, resulting in a dissipation of excess pore water pressure

and an increase in the effective stress. This is computed using the Terzahis theory of

consolidation.

Sc = S0

19. Define Secondary compression settlement

Secondary compression settlement,(Ss), which occurs at constant effective stress,

with volume change occurring due to rearrangement of soil particles.

20. Define seat of settlement.

The Seat of settlement can be defined as the stressed zone beneath a foundation

within which the stresses induced by the load are large enough to cause significant orders

of settlement. Out side this zone the stresses are so small that they do not contribute to

any significant settlement.

21. State the corrections to be made for the Settlement due to Consolidation

1. Correction for the effect of three dimensional consolidation 2. Correction for the rigidity of foundation. 3. Correction for the depth of embedment.

22. State the corrections made for the observed SPT values.

Two types of corrections are normally applied to the observed N values in cohesion

less soil. They are:

1. Correction due to overburden pressure. 2. Correction due to dilitancy

Over burden correction should be made before dilatancy correction.

23. State the factors affecting Bearing capacity.

(vi) Angle of internal friction (or) angle of shearing resistance. (vii) Width of the footing, (viii) Depth of the footing, (ix) Unit weight of the soil, (x) Position of ground water table.

24. State Permissible Settlement as per IS code.

11

Permissible settlement based on Type of Structure.

Type of Structure permissible Settlement

Commercial and 25 mm

Industrial Buildings 38 mm

Ware Houses 50 mm

PART B

SIXTEEN MARK QUESTIONS AND ANSWERS

1. What are the IS code recommendations for the location and depth of foundation?

Foundation should be located at a minimum depth of 50 cm below the natural

ground surface.

The foundation must be placed below the zone of volume change.

For the structures in river, the foundation depth should be sufficiently below the

deepest scour level.

For footings in sloping Ground

(a) When the ground surface slopes downward adjacent to a footing, the

sloping should not encroach upon a frustum of bearing material under the

footing having sides which make an angle of 60 with the horizontal for

rock and 30 for soil and the horizontal distance from the lower edge of the

footing to the sloping surface shall be at least 60 cm for rock and 90 cm for

soil.

(b) For footings in granular soils, the line joining the lower adjacent edges

of adjacent footings should not have a slope steeper than two horizontal to

one vertical [Fig. (a)].

(c) In clayey soils, the slope of the line joining the lower adjacent edge of

the upper footing and the upper adjacent edge of the lower should not be

steeper than two horizontal to one vertical .

Type

of soil

Permissible Settlement

Permissible Differential Settlement

Isolated footing Raft footing

Isolated footing

Raft footing

Sandy

Clays

4.0 cm

6.5 cm

4 to 6.5 cm

6.5 to 10 cm

2.5 cm

4.0 cm

2.5 cm

4.0 cm

12

To ovoid damage to an existing structure, the foundation for a new structure at an

adjacent site should be located , as illustrated in the Fig. (c). THe adjacent edge of the

new footing must be atleast at a distance S from the edge of the existing footing where S is the width of the larger footing. The line from the edge of the new footingto the edge of the existing footing should make an angle of 45 or less with the horizontal

plane.

When a new footing is placed lower than an old footing, the existing structure may

be endangered because of the lateral flow of soil beneath the existing footing. The

excavation must not, therefore, be too close the existing footing. Suitable bracing should

be provided.

Construction should not be allowed above the underground utilities such as water

supply lines, sewage pipes` etc.

The presence of defects such as cavities, old mine tunnels, soft fill mineral, etc.

must be carefully examined.

2. Explain the different modes of failure of foundation soil

General shear failure

In the case of general shear failure, continuous failure surfaces develop between

the edges of the footing and the ground surface as shown in fig. As the pressure is

increased towards the value qf, a state of plastic equilibrium reaches initially in the soil

around the edges of the footing; and then it gradually spreads downward and outward.

Ultimately, the entire soil mass above the failure surfaces reaches in a state of plastic

equilibrium. Immediately after that the soil starts slipping, and heaving of the ground

surface occurs on the both sides of the footing; although the final slip movement would

occur only on the one side, accompanied by the tilting of the footing. This type of failure

mostly occurs in stiff clays or dense sands of low compressibility.

Local shear failure In the case of local shear failure, there is a significant compression of the soil

under the footing, and the plastic equilibrium develops only in the soil below the footing.

The failure surfaces, therefore, do not reach the ground, and only slight heaving occurs.

There would thus be no tilting of the footing. This type of shear failure mostly occurs in

loose or soft soils of high compressibility.

Punching shear failure The punching shear failure occurs when there is a compression of the soil under the

footing, accompanied by shearing in the vertical direction around the edges of the

footing, and there is no question of any tilting of the footing, and ultimate bearing

13

capacity is not well defined. This type of failure, generally does not occur in shallow

footings, and occurs in soil of low compressibility when the footings are considerably

deep. The width to depth ratio fD

B is, thus, quite important for such a failure.

3. Explain the corrections made for the observed SPT values.

Two types of corrections are normally applied to the observed N values in cohesion

less soil. They are:

1 Correction due to overburden pressure.

2 Correction due to militancy

Correction for overburden pressure

The corrected N value is given by Peck(1974)

N = CN. N

N = corrected value of observed.

CN. = correction factor for overburden pressure.

N = observed N value

NC = 0.77 log10

2000 proposed by Peck(1974)

Over burden correction factor that is generally used is

04.01

4

NC if

2/75 mKN

01.025.3

4

NC if

2/75 mKN proposed by Bazaraa(1967)

Correction for Dilatancy

In saturated fine silty dense or very dense sand deposits, the N value observed may be

greater than the actual value because of the tendency of such materials to dilate during

shear under undrained conditions. Terzaghi and peck recommended that if the observed

N value is greater that 15, it should be corrected for dilation effect as

N=15+0.5(N-15)

Where N= corrected value of N. N = corrected N for over burden pressure.

4. Explain the method of carrying out Plate Load Test.

The plate load test is a semi-direct method to estimate the allowable bearing

14

pressure of soil to induce a given amount of settlement.

Plates, round or square, varying in sizes, from 30 to 60 cm and thickness of about 2.5 cm

are employed for the test.

The load on the plate is applied by making use of a hydraulic jack. The reaction

of the jack load is taken by a cross beam or a steel truss anchored suitably at both the

ends. The settlement of the plate is measured by a set of three dial gauges of sensitivity

0.02 mm placed at 1200 apart. The dial gauges are fixed to independent supports which

do not get disturbed during the test.

The method of performing the test is essentially as follows:

1. Excavate a pit of size not less than 5 times the size of the plate. The bottom of the

pit coincides with the level of the foundation.

2. If water table is above the level of foundation, pump out the water carefully and it

should be kept just at the level of foundation.

3. A suitable size of plate if selected for the test. Normally a plate of size 30 cm is

used in sandy soils and bigger size in clay soils. The ground should be leveled

and the plate is seated over the ground

Figure shows the arrangement for a plate load test.

The method of performing the test is essentially as follows:

4. A seating load of about 70 g/cm2 is first applied and released after some time

higher load is next placed on the plate and settlements are recorded by means of

the dial gauges. Observations on every load increment shall be taken until the rate

of settlement is less than 0.25 mm per hour. Load increments shall be

approximately one-fifth of the estimated safe bearing capacity of the soil. The

average of the settlements recorded by 2 or 3 dial gauges taken as the settlement

of the plate for each of the load increment.

5. The test should continue until a total settlement of 2.5 cm or the settlement at

Which the soil fails, whichever is earlier, is obtained. After the lad is released, the

Elastic rebound of the soil should be recorded.

From the test results, a load-settlement curve should be plotted . The allowable pressure

on a prototype foundation for an assumed settlement may be found out by making use of

the following equations suggested by Terzaghi and Peck,

15

for granular soils,

2

( 0.3)

0.3)

for claysoils,

p

f p

p

f p

p

B bS S

b B

BS S x

b

where, Sf = permissible settlement of foundation in mm, Sp = settlement plat in mm, B =

size of foundation in metres, bp = size of plate in metres.

5. Explain the Application SPT values

Footings of granular soils are sometimes proportioned using empirical relationships.

Teng (1969) proposed an equation for a settlement of 25 mm based on the curves

developed by Terzaghi and Peck (1948). The modified form of the equation is,

dwsFR

B

BNq 2

2

2

3.0)3(35

Teng (1969)

dwsFR

B

BNq 2

2

2

3.0)3(53

Modified Teng formula

Where qs = net allowable bearing pressure for settlement of 25 mm in kN/m2

Rw2 = water table correction factor = 21 1

2

wD

B

Fd = depth factor = 21

B

D f

B = width of footing in metres.

Df = depth of foundation in metres.

Dw2 = depth of WT below the base of foundation.

16

For Raft foundation on sand:

qs = 21 X N KN/m2

Where N = corrected N value

6. Explain the Application of CPT values

SAFE BEARING PRESSURE FROM CPT VALUES FOR FOOTING ON

COHESIONLESS SOIL

The static cone penetration test in which a standard cone of 10 cm2 sectional area

us pushed into the soil without the necessity of boring provides a much more accurate

and detailed variation in the soil. Meyerhof (1956) suggested a set of empirical equations

based on the Terzaghi and Peck curves (1984)

2

2

2

2

2

2

2

3.6 1.2

12.1 1 1.2

2.7 /

s c w

s c w

s c w

q q R kNm for B m

q q R kNm for B mB

Anapproximateformulaforallwidths

q q R kN m

where, qc is the cone penetration resistance in kg/cm2 and

qs in kN/m2

The cone penetration resistance, qc and cu may be related as

qc = Nkcu + p0

where, Nk = cone factor,

p0 = total over burden pressure.

Lunne and Kelvin (1981) investigated the value of cone factor Nk for both the

normally consolidated and over-consolidated clays. The values of Nk are.

Type of clay Cone factor

Normally consolidated 11 to 19

Over consolidated

17

At shallow depths 15 to 20

At deep depths 12 to 18

The above equation have been developed for a settlement of 25mm. and if the allowable

settlement is other than 25mm, say then corresponding increased bearing capacity is

25

ss qq

7. Explain the components of Settlement?

Immediate Settlement (Si)

Immediate settlement is a combination of elastic compression and plastic

deformation without change in volume or water content. This type of settlement

develops are construction proceeds. Immediate settlements are significant in non-

saturated clays, silts and granular soils. The equation for immediate settlement of a

loaded footing is

21i f

c

S qB IE

where q intensity of contact pressure,

B = width of footing,

p = Poissons ratio,

Ec = modulus of elasticity,

If = influence factor.

Typical range of values for Poissons ratio p is given in Table 18.7. Terzaghi has

suggested a value of p = 0.3 for sand and 0.4 to 0.43 for clays based on elasticity~

consideration.

Typical range of values for Poissons ratio

Type of soil

Clay, saturated 0.40.5

Clay, unsaturated 0.10.3

Sandy clay 0.20.3

18

Silt 0.30.35

Sand (dense) 0.20.4

Coarse (void ratio = 0.4 to 0.7) 0.15

Fine grained (void ratio = 0.4 to 0.7) 0.25

Rock 0.10.4

The influence factor %depends upon the shape of the loaded area and the

distribution o contact pressure. The values of If are given in Table 18.8 for various

shapes of flexible and rigid footings (Bowles).

Influence factor If

Shape Flexible footing Rigid footing

Circle 0.85 0.88

Square 0.95 0.82

Rectangle

L/B = 1.5 1.20 1.06

L/B = 5 1.83 1.70

L/B = 10 2.25 2.10

L/B = 100 2.96 3.40

primary consolidation settlement

primary consolidation settlement (Sc) which is due to a gradual explusion of pore

water from the voids of the soil, resulting in a dissipation of excess pore water pressure

and an increase in the effective stress. This is computed using the Terzahis theory of

consolidation.

Sc = S0

Where

= the coefficient of depending on the geometry of the footing and the history of the clay.

S0 = Settlement calculated by Terzahis theory of consolidation.

19

S000

log1 P

PPH

e

C oc

Where

Compression Index Cc = 0.009(wl --10)

H = The thickness of the clay strata.

P0 =The effective vertical stress at the mid height of the layer.

e0 = initial voids ratio.

Values of

Types of clay

Very sensitive clays (soft alluvial and marine clays) 1 to 1.2

Normally consolidated clays 0.7 to 1.0

Over consolidated clays 0.5 to 0.7

Heavily over consolidated clays 0.2 to 0.5

Secondary compression settlement

Secondary compression settlement,(Ss), which occurs at constant effective stress,

with volume change occurring due to rearrangement of soil particles.

8. Explain the Emprical Equations to determine the Allowable bearing pressure.

(a) Peck, Hanson and Thornburn procedure.

qa = 0.44CwNSa KN/m2

where

qa = Allowable Bearing pressure in KN/m2 .

Cw = Correction factor for water table position.

BD

DC

f

ww 5.05.0

Sa = Permissible Settlement in mm

20

N = Corrected N value from S.P.T. Test

(b) Tengs Empirical Equation.

aDwaSCR

B

BNq

2

2

3.0)3(4.1

KN/m2

where,

Rw = Water table correction factor.

Sa = Permissible Settlement in mm

N = Corrected N value from S.P.T. Test

CD = Depth correction factor = 21 B

D

Mayerhof (1974)

qa = 0.49 N RDSa

where,

qa = Allowable Bearing pressure in KN/m2

RD = Depth correction factor.

Sa = Permissible Settlement in mm

N = Corrected N value from S.P.T. Test

9. Calculate the ultimate bearing capacity per unit area of

(4) a strip footing 1 m wide (5) a square footing 3m x 3m, and (6) a circular footing of 3m diameter.

Given:

Unit weight of the soil 1.8 t/m3, cohesion = 2t/m

2

And = 20o

Nc = 17.5, Nq = 7.5 and N = 5.

21

Solution.

In question nothing is mentioned about the depth of the foundation. Hence

in all the three cases the foundation is assumed to be on surface of the ground.

(i) Strip footing:

Ultimate bearing capacity from equation

qu = (cNc + dDq + b N ) Since d = 0, 2

nd term will be zero.

qu = cNc + b Nr

= 2 x 17.5 + x 1.8 x 1 x 5 = 35 + 4.5 = 39.5 tonnes / m

2.

(ii) Square footing :

From equation (8.12)

qu = 1.3 cNc + dNq + 0.4 bN (since d = 0, 2

nd term will be zero)

qu = 1.3 cNc + 0.4b x N = 1.3 x 2 x 17.5 x 0.4 x 1.8 x 3 x 5

2.6 x 17.5 x 0.72 x 15

= 45.5 + 10.8 = 56.3 tonnes/m2.

(iii) Circular footing :

From equation,

qu = 1.3 cNc + dNa + 0.4 rN Here dia = 3m, r = 1.5m

qa = 1.3 cNa + 0.6 rN = 1.3 x 2 x 17.5 + 0.6 x 1.8 x 1.5 x 5

= 45.5 + 8.12 = 53.62 tonnes/m2.

10. A strip footing 2 m wide carries a load intensity of 400 KN/m2

at a depth of 1.2

m in sand. The saturated unit weight of sand is 19.5Kn/m3

and unit weight above

water table is16.8KN/m3 . The shear strength parameters are c = 0 and = 35o .

Determine the factor of safety with respect to shear failure for the following cases of

location of water table.

(i) Water table is 4m below G.L. (ii) Water table is 1.2 m below G.L. (iii) Water table is 2.5 m below G.L. (iv) Water table is 0.5m below G.L. (v) Water table is at G.L. itself.

Use Terzahis Equations.

22

SOLUTION

Given Data;

The ultimate Bearing capacity for a strip footing for general shear is given by,

qf = CNc + DNq + 0.5BN

Taking in to the account the water reduction factor, we have ,from EQ.24.34.

qf = CNc + DNq RW1+ 0.5BN RW2

For the present case c=0

qf = DNq RW1+ 0.5BN RW2 For =350 assuming general shear failure,

Nq = 41.4 and Nq = 42.4

qf = 41.4x1.2x DNq RW1 =0..5x2x42.2 DNq RW2

Case a) Water table is 4m below G.L.

Zw2 = 4-1.2 = 2.8m

Rw1 = 1

Since Zw2>B, RW2 = 1

hence there will be no effect of water table also = 16.8

qf= 49.69X16.8X1=42.4X16.8X1

= 1546.9 KN/m2

now actual footing load

qa = 400 KN / m2

F.S. = a

f

q

q

400

9.1546= 3.87

Case b) Water table is just at base of the footing

RW1 = .5 (1+ ZW1/D)

= .5 (1+1) = 1

RW2 =. 5 (1+ Zw2/B)

= .5 (1+0)

= .5

For the surcharge term, use r + 16.8 KN/m2

, because the surcharge soil is situated above water table . for the wedge term use r= 19.5

KN/ m2, since the wedge soil is situated below the water table .

qf = 49.68 RW1+ 42.4satRW

23

=

49.68x16.8x1+42.4 x19.5x.5

= 1248 KN/m2

F.S. = qf / qa

= 1248 /400

= 3.12

Case (C) Water table at 2.5 m below ground the G.L.

Z = 2.5 1.2 =1.3 m ,B. RW1 = 1

RW2 =. 5 (1+ Zw2/B)

= . 5 (1+ 1.3/2)

= .825

For surcage term ,

= 16.8 KN/m3

for the wedge term, will be taken as average unit weight of soil situated below the footing level, since the soil upto the depth B below the footing is partly above water table

and partly below water table.

)7.03.1(

)7.05.19()3.18.16(

av

= 17.75KN/m3

qf = 49.68 RW1+ 42.4satRW

= 49.68x16.8x1+42.4 x17.75x0.825

=1455.5 KN/m 3

64.3400

5.1455.

a

f

q

qSF

11. Determine the ultimate bearing capacity of the footing in Ex. 15.1 if the ground

water table is located (a) at a depth of 0.5 m below the ground surface, (b) at a

depth of 0.5m below the base of the footing. sat = 20 kN/m3. Use Terzaghi theory.

Solution:

qu = qNq + 0.5 BN

24

where q is the effective surcharge and is the effective unit weight of the soil beneath the footing.

Case A

q is calculated by assuming the unit weight of the top 0.5m soil to be

unchanged, that is, 17 kn/m3 while the remaining 0.5 m is submerged, with a submerged

unit weight equal to 10 kN/m3 ( = sat-w).

Thus q = 0.5 x 17+0.5 x 10 = 13.5 kN/m2

In the second part, = = 10 kN/m2

Qu = 13.5 x 60 + 10 x 1.5 x 75

= 1372.5 kN/m2

compared with qu = 1976 kN/m2 when there was no effect of water table.

(b) For this case, q = 17kN/m2

In the term 0.5 BN, is given by Eq.15.30

= + (Dw/B) (t - )

Dw 0.5 m, t = 17 kN/m3 and = 10 kN/m3

= 10 + 0.5/0.5 (17 10) = 12.3 kN/m3

Hence qu = 17 x 60 + 0.5 x 12.3x 1.5 x 75

= 1711.9 kN/m2

12. An R.C. Column footing square in the shape is to rest 1.5 m below level. The

total ground level. The total load to be transmitted including the frequent flooding,

the friction of the foot along the sides is to be neglected and a factor of safety 2.4

gm/c.c. angle of internal friction 33 and value of N = 33 and Nq = 32 find a suitable a size of the footing for the above condition.

Solution. Assuming size of the footing as 2 m x 2 m

B = 2 and d = 1.5b > d

Hence the foundation is shallow.

qu = 1.3cNc + ydNq +0.4 ybN (8.19) = 0 + 2.4 x 1.5 x 32 x 0.4 x2.4 x 2 x 33 ( .: for and c = 0 )

= 115.2 + 63.4 = 178.6

25

Allowing factor of safety 2.5,

Safe bearing capacity

=178.6/2.5 = 71.4 t/m2.

Area of footing = 2 x 2 = 42

Total load it can carry

= -4x71.4 = 285.6 tones

And total is only 200 tones

Hence the section is safe.

To be more economical side of square footing less than 2 metres may be

taken trial be made for most suitable section.

13. In a plate bearing test on pure clayey soil failure occurred at a load of 12.2

tones. The size of the plate was 45 cm x 45 cm and the test was one at a depth of 1.0

m below ground level. Find out the ultimate bearing capacity for a 1.5 m wide

continuous wall footing with its base at a depth of 2m below ground level. The unit

wt. of clay may be taken as 1.9 gm/ c.c. and Nc = 5.7, Nq = 1 and N = 0.

Solution: For a square footing equation for ultimate bearing capacity is given as

qu = 1.3 cNc + . dNq + 0.4 bN .(1)

In case of plate bearing test failure stress

= 12.2/0.45 x 0.45

= 12.2/0.2025 = 60.2 t/m2

Now putting numerical values in Eq. (1)

60.2 = 1.3 x c x 5.7 + 1.9 x 1 x 1+0 = 7.4c + 1.9

c = 7.88 t/m2. Now taking the case of wall footing equation, for ultimate bearing capacity is

qu = cNc + .d.Nq + b.N

Since N = 0, 3rd

term will not be there

qu = 7.88 x 5.7 + 1.9 x 2 x 1

= 44.9 + 3.8

= 48.7 t/m2.

14. A square footing located at a depth of 1.5 m below the ground surface in

cohesionless soil carries a column load of 1280 kN. The soil is submerged having an

effective unit weight of 11.5 kN/m3 and an angle of shearing resistane of 30

o. Find

the size of the following for Fs = 3 by Terzaghis theory of general shear failure.

26

Solution

Use Eq. (18.16) for c= 0

0.4d f qq D N BN

Since Wt is close to GL

11.5 /kN m

For 30 , 22.5 and N = 19.7.o

qN Substituting the known values, we have

qd = 11.5 x 15 x 22.57 + 0.4 x 115 x 19.7B

= 388.13 + 90.62B (a)

Column load, 21280

1280 , /a aQ kN or q kN mBxB

Since 2 2

1280 3 38403,s d

xF q

B B (b)

Now equating Eq. (a) with (b)

2

3840388.13 90.62B

B

or B3 + 4.28B

2 42.37 = 0

Solving for B, we have B = 2.5m

Size of square footing = 2.5 x 2.5 m.

15. Determine the net bearing pressure for a 2X2 m footing at a depth of 1.5 m in a

medium dense sand so that the total settlement does not exceed 25 mm. The average

SPT blows below the footing are 20 per 30 cm. The average moist unit weight of soil

is 17 KN/m3. The water table is at 5 m below the ground level.

Solution:

Correction for overburden pressure.

The corrected N value is given by Peck(1974)

N = CN. N

N = corrected value of observed.

27

CN. = correction factor for overburden pressure.

N = observed N value

NC = 0.77 log10

2000 proposed by Peck(1974)

5.255.15.17 D

Corected N = 1.5 X 20 = 30

dwsFR

B

BNq 2

2

2

3.0)3(35

Teng (1969)

Rw2 = 1

Fd = depth factor = 21

B

D f

depth factor =

2

5.11

Fd = 1.75

75.11

22

3.02)330(35

2

sq

qs = 546.75 KN/m2

16. If a deposit happens to be silty clay (saturared with a value of qc = 88 kg/cm2,

determine the unconfined compressive strength of clay as per Use p0=127kN/m2.

Solution

As per Eq. (17.11)

0 02( )c cu u

k

q p q pc or q

N N

28

Assume Nk = 20

Substituting for qc, p0 and Nk we have,

22(88 100 127) 867 /20

u

xq kN m

17. A footing foundation of 3m X 3m is to be constructed at a site at a depth of 1.5

m below ground level. The water table is at the base level of foundation. The

average static cone penetration resistance obtained at the site is 20 Kg/m2.

The soil is cohesive determine the safe bearing capacity for a settlement of 40mm

Given data:

B = 3 m

D = 1.5 m

qc = 20 Kg/cm2

Rw2 = 0.5

21

11.2 wcs RB

qq

5.03

11201.2

sq

qs = 37.3 KN/m2 ;

For 40 mm settlement the value of qf is

qs = 37.3 2/60

25

40mKN

;

18. A 30 cm square bearing plate settles by 1.5 cm in a plate loading test on a

cohesion less soil when the intensity of loading is 2 kg/cm2. What will be the

settlement of a prototype footing 1m square under the same intensity of

loading.(C.E.S. 1981)

Solution : bp = size of the plate = 0.3 m

Sp = settlement of the plate = 15 mm

B = width of the footing = 1.0 m

Sf = settlement of the footing required

29

Sf = Sp[ B(bp + 0.3)/bp (B+0.3) ]2

= 15[1x (0.3 + 0.3)/0.391+0.3)]2

= 35.5 mm.

19. Two plate load test s were conducted at the level of a prototype foundation in

cohesionless soil close to each other. The following data are given.

Size of plate Load applied Settlement recorded

0.3m X 0.3m 30 KN 25 mm

0.6m X0.6m 90 KN 25 mm

If the footing is to carry a load of 100KN, determine the size of the footing for

the same settlement of 25 mm.

Solution

q = Apm+Ppn

For Plate 1,

Ap = 0.3 X 0.3 = 0.09m2

; Pp = 0.3 X 4 = 1.2m ; q = 30 KN

For Plate2,

Ap = 0.6 X 0.6 = 0.16m2

; Pp = 0.6 X 4 = 2.4m ; q = 90 KN

30 = 0.09 m + 1.2 n -------(1)

90 = 0.16 m + 2.4 n ------ (2)

On solving (1) & (2),

m = 166.67 and n = 12.5

For prototype foundation,

q = 166.67 Af + 12.5 Pf

where

Af = B

2 Pf = 4 B

30

Substituting we have,

1000 = 166.67 B2

+ 12.5 ( 4B )

1000 = 166.67 B2

+ 50 B

B2 + 0.3 B 6 = 0

Solving,

B = 2.3 m

The size of the footing is : 2.3 m X 2.3 m

20. Soil investigation at a site gave the following information. Top soil up to a depth

of 10.6 m is fine sand, and below this lies soft clay layer of 7.60 m thick. The water

table is at 4.60 m below the ground surface. The submerged unit weight of sand is

2/4.10 mKNb and unit weight above water table is 17.6 KN/m3. The water

content of the normally consolidated clay wn = 40% its liquid limit, wl = 45%, and

specific gravity of the solid particle is 2.78. The proposed construction will transmit

a net stress of 120 KN/m2

Find the average settlement of clay layer.

Given data 2/4.10 mKNsub

2/6.17 mKN

P = 120 KN/m2

w = 40%

G = 2.78

We know that

S000

log1 P

PPH

e

C oc

Cc = 0.009(wl --10) ,

Cc = 0.009(45-10) = 0.32

rs

wGe 0

1

78.240.00

e

31

= 1.11

The effective vertical stress at the mid height of the layer.

2

0 /9.17429.82

7604.10660.1760.4 mKNP

S0 = cmm 2626.09.174

1209.174log

11.11

60.732.010

The consolidation settlement

Sc = S0

Sc = 1.1 S0

Sc = 1.1 X 26 cm

=28.6 cm