CE2305 – Foundation Engineering. Two Marks Questions With Answers 2014

Anna University, Chennai

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CE2305 – Foundation Engineering. PART-B

1. (a) (i) Compute the ultimate load that an eccentrically loaded square footing of width 2 m width an eccentricity of 0.315 m can take at a depth of

0.45 m in soil with γ =17.75 kN/m3, , C=9 kN/m2 and f = 35° , NC =

y

52, N q = 35 and N y = 42. (8)

q

q f =  D N′

q

q n f =  D (N′

+ 0.4  B N′

y

– 1) + 0.4  B N′

q s =

q n f

F + D and F = 3

Maximum safe load = B2 ´ q s .

(ii) Write the step by step procedure for IS code method for computing bearing capacity in shallow foundation. (8)

IS Code Method:

q f can be computed using only one reduction factor, by rewriting as under:

1

q f = c NC + s N q + 2  B N R W

where s

= effective unit weight of the soil situated above the base

é Zw ù

ê 2 ú

ê

ú

R W = Rw 2 = 0.5 ë 1 + B û

Above equation has been recommended by Indian Standard. When the water table is situated at a depth D 1 below the ground level (D 1 < D) or D 2 above the base of the footing (such that D 1 + D 2 = D), we have

s = ( D 1 + sat D 2 ) –  w × D 2 =  D 1 + ′ D 2 .

Knowing s and R W( = Rw 2 ), q f can be computed.

This alternative method is preferred to the method of Equation in which two reduction factors Rw 1 and Rw 2 are used.

(b) (i) A group of 9 piles arranged in a square pattern with diameter and length of each pile as 25 cm and 10 m respectively, is used as a

foundation in soft clay deposit. Taking the un confined compressive

strength of clay

as 120 kN/m2 and the pile spacing

of the group

ensure the bearing capacity factor NC = 9.1 and adhesion factor =

0.81. A factor of safely of 2.51 may be taken. (8)

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(a) Piles acting individually

q u c = 2

120

2

= 2 = 60 kN/m

Q up = A p × r p + A s × r f

where,A p =

p

4 (0.25)2 = 0.04909 m2

A s = p (0.25) ´ 10 = 7.854 m2

r p = C NC = 60 ´ 9.1 = 546 kN/m2 r f = m c = 0.81 ´ 60 = 48.6 kN/m2

Q up = 0.4909 ´ 546 + 7.854 ´ 48.6 = 408.50 kN ≃ 409 kN

\ Load capacity of 9 piles = 9 ´ 409 = 3681 kN

(b ) Pile acting as a group:

B = 2 x + d = 2 ´ 1 + 0.25 = 2.25 m

Q ug = A p × r p + A s × r f

where,A p =

B2 = (2.25)2 = 5.0625 m2

A s = 4 B ´ 10 = 4 ´ 2.225 ´ 10 = 90 m2 r p = c = 60 kN/m2

Q ug = 5.0625 ´ 540 + 90 ´ 60

= 2733.75 + 5400

= 8133.75 kN

Qu.min = 3681 kg and Q a =

(ii) Write short notes on :

Qu.min

F =

3681

3.5 = 1051.71 kN

1. Under reamed pile (5)

Ans: Refer Page No: 4.6

2. Forces on pile cap (3)

Piles are seldom used as single individual members, but are generally used in groups. A number of piles (generally a minimum of three), are installed at distance 2B to 4B (B being the width or dia of a single pie), and joined at the top by a slab, known as the pile cap (Refer Fig.)

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(ii) A retaining wall, 4 m high support a back fill (c = 20 kN/m2 f =30°, γ

= 20 kN/m3) with horizontal top, flush with the top of the wall. The backfill carries a surcharge of 20 kN/m2. If the wall is pushed towards the backfill compute the total passive pressure on the wall, and it’s point of application. (8)

Solution:

C = 2 kN/m2

f = 30°; K p = N f = tan2 (45° + f / 2) = tan2 60° = 3

The passive pressure intensity due to surcharge is given by

p 1 = K p × q = N f q = 3 ´ q = 3 ´ 20 = 60 kN/m2

The passive pressure intensity due to backfill is given by equation a :

 

p

=

2 C

N f +  H N f = p 2 + p 3

where

p 2

=

2 C

N f = 2 ´ 20 3 = 69.28 kN/m2

clip_image003clip_image004and p 3 =  H N f = 20 ´ 4 ´ 3 = 240 kN/m2

The pressure distribution diagram consists of two rectangles and one triangle corresponding to p 1, p 2 and p 3, as shown in figure.

clip_image005clip_image006

Figure

\P 1 = H (q × N f) = 4 ´ 60 = 240 kN/m acting at 2 m above base

clip_image007P 2 = H (2C N f ) = 4 ´ 69.28 = 277.1 kN/m, acting at 2 m above base.

1 1

P 3 =

2 H ( H N f) = 2 ´ 4 ´ 240 = 480 kN/m,

4

acting at 3 m above base.

\ P = P 1 + P 2 + P 3 = 240 + 277.1 + 480 = 997.1 kN/m

4

(240 ´ 2) + (277.1 ´ 2) + (480 ´ )

=

3

z 997.1 = 1.68 m.

(b) A retaining wall 6 m high retains sand with f = 30° and unit weight

24 kN/m3 upto a depth of 3 m from top. From 3 m to 6 m the material

is a cohesive soil with c = 20 kN/m2 and f = 20°. Unit weight of cohesive soil is 18 kN/m3. A uniform surcharge of 100 kN/m2 acts on the top of soil determine the total lateral pressure acting on the wall and its points of applications.

(16)

Solution:

(a) Lateral pressure due to top soil

1 – sin 30° 1

Ka 1 =

1 + sin30° = 3 . Hence at any depth z 1 below top,

1 1

pa 1 = Ka 1 q 1 + Ka 1  1 z 1 = 3 ´ 100 + 3 ´ 24 z 1 = 33.33 + 8 z 1

At z 1 = 0 pa 1 = 33.33 kN/m2

At z 1 = 3 m pa 1 = 33.33 + 8 ´ 3 = 33.33 + 24 = 57.33 kN/m2

The active pressure distributions diagram is shown in Fig. (a)

clip_image008

(b ) Lateral pressure due to bottom soil:

For the bottom soil, the weight of the top soil and the initial surcharge

(q 1 = 100 kN/m2) become the surcharge.

q 2 = q 1 +  1 H1 = 100 + 24 ´ 3 = 172 kN/m2 cot a 2 = cot (45 + j/2) = cot (45 + 20/2) = 0.7

The active earth pressure at any depth z2 below the junction of the two soils is given by equation.

pa 2 =  2 z 2 cot2 a 2 + q 2 cot2 a 2

or pa 2 = 18 z2 (0.7)2 – 2 ´ 20 (0.7) + 172 (0.7)2 or pa 2 = 8.82 z 2 – 28 + 84.28 = 8.82 z 2 + 56.28

At z 2 = 0, pa 2 = 56.28 kN/m2

At z 2 = 3 m, pa 2 = 8.82 ´ 3 + 56.28

= 26.46 + 56.28 = 82.74 kN/m2

The active pressure distribution diagram for bottom soil is given in

Fig. (c ).

(c ) Total lateral pressure:

The lateral pressure diagram for both the soils is shown in Fig. (d ) with component diagrams completely marked.

Force P 1 = 33.33 ´ 3 = 100 kN/m acting at z 1 = 3 + 3/2 = 4.5 m above base

1

P 2 =

2 ´ 24 ´ 3 = 36 kN/m

acting at z 2 = 3 + 3/3 = 4 m above base

P 3 = 56.28 ´ 3 = 168.84 kN/m acting at z 3 = 3/2 = 1.6 above base

1

P 4 =

2 ´ 26.46 ´ 3 = 39.69 kN/m

acting at z 4 = 3/3 = 1 m above base

\ Total P = 100 + 36 + 168.84 + 39.69 = 344.53 kN/m

z

Acting at –

above base.

100 ´ 4.5 + 36 ´ 4 + 168.84 ´ 1.5 + 39.69 ´ 1

= 344.53 = 2.57 m

2 (a) A square footing for a column is 2.5 m × 2.5 m and carries a load of

2000 kN. Find the factor of safety against bearing capacity failure, if the soil has the following properties. (16)

C = 50 kN/m2

f = 15°

g = 17.6 kN/m3

N′ ′ ′

C = 12.5, Nq = 4.5 and Ny = 2.5. The foundation is taken to a depth of

1.5 m.

N′ ′ ′

C = 12.5 Nq = 4.5 Ny = 2.5

y

D = 1.5 m B = 2.5 m f = 15°

 = 17.6 kN/m3C = 0

q

q f =  D N′

q

q n f =  D (N′

+ 0.4  B N′

y

– 1) + 0.4  B N′

q s =

q n f

F + D and F = 3

Maximum safe load = B2 ´ q s .

3. (a) Explain with neat sketches about pile load test method of determination of load carrying capacity of piles. (16)

PILE LOAD TESTS

The pile load test can be performed either on a working pile which forms the foundation of the structure or on a test pile. The test load is applied with the help of calibrated jack placed over a rigid circular or square plate which in turn is placed on the head of the pile projecting above ground level. The reaction of the jack is borne by a truss or platform which may have gravity loading (in the form of sand bags etc.) or alternatively, the truss can be anchored to the ground with the help of anchor piles. In the later case, under-reamed piles or soil anchors may be used for anchoring the truss. Both the arrangements are shown.

The load is applied in equal increments of about one-fifth of the estimated allowable load. The settlements are recorded with the help of three dial gauges of sensitivity 0.02 mm, symmetrically arranged over the test plate,and fixed to an independent datum bar. A remote controlled pumping

unit may be used for the hydraulic jack. Each load increment is kept for sufficient time till the rate of settlement becomes less than 0.02 mm per hour. The test piles are loaded until ultimate load is reached. Ordinarily, the

1

rest load is increased to a value 2 2 times the estimated allowable load or to

a load which causes a settlement equal to one-tenth of the pile diameter, whichever occurs earlier. The results are plotted in the form of load- settlement curve. The ultimate load is clearly indicated by load settlement curve approaching vertical. If the ultimate cannot be obtained from the loads settlement curve, the allowable load is taken as follows:

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(b) What are the different methods of soil stabilization? Explain with neat sketches. (16) INTRODUCTION:

Stabilization incorporates the various methods employed for modifying the properties of a soil to improve its engineering performance. Stabilization is used for a variety of engineering works, the most common application being in the construction of road and air-field pavements, where the main objective is to increase the strength or stability of soil and to reduce the construction cost by making best use of the locally available materials.

MECHANICAL STABILISATION:

Mechanical stabilization involves two operations: (i) changing the composition of soil by addition or removal of certain constituents, and (ii) densification or compaction. The particle size distribution and composition

are the important factors governing the engineering behavior of a soil. Significant changes in the properties can be made by addition or removal of suitable soil fractions. For mechanical stabilisation where the primary purpose is to have a soil resistant to deformation and displacement under loads, soil materials can be divided into two fractions: the granular fraction retained on a 75 micron IS sieve and the fine soil fraction passing a 75- micron sieve. The granular fraction impart strength and hardness. The fine fraction provides cohesion or binding property, water-retention capacity and also acts as a filler for the voids of the coarse fraction.

CEMENT STABILISATION:

1. Soil cements and its influencing factors

The soil stabilized with cement (Portland) is known as soil cement. The cementing action is believed to be the result of chemical reaction of cement with the siliceous soil during hydration. The binding action of individual particles through cement may be possible only in coarse-grained soils. In fine grained, cohesive soils, only some of the particles can be expected to have cement bonds, and the rest will be bonded through natural pollution. The important factors affecting soil cement are: nature of soil, cement content, conditions of mixing, compaction and curing, and admixtures.

2. Construction methods

The normal construction sequence for soil-cement bases is as follows: (i) Shaping the sub-grade and scarifying the soil, (ii) Pulverising the soil, (iii) Adding and mixing cements, (iv) Adding and mixing water, (v) Compacting,(vi) Finishing, (vii) Curing and (viii) Adding wearing surfacing. There are three methods of carrying out these operations : (i) mix- in place method, (ii) travelling plant method and (iii) stationary plant method.

LIME STABILISATION:

Hydrated (or slaked) lime is very effective in treating heavy, plastic clayey soils. Lime may be used alone, or in combination with cement, bitumen or fly ash. Sandy soils can also be stabilized with these combinations. Lime has been mainly used for stabilizing the road bases and sub-grades on addition of lime to soil, two main types of chemical reactions occur:

(i) alteration in the nature of the absorbed layer through base exchange phenomenon, and

(ii) cementing or puzzolanic action. Lime reduces the plasticity index of highly plastic soils making them more friable and easy to be handled and Pulverised. The plasticity index of soils of low plasticity generally increases. There is generally an increase in the optimum water content and a decrease in the maximum compacted density, but the strength and durability increase.

BITUMEN STABILAISATION:

Asphalts and tars are the bituminous materials which are used for stabilisation of soil, generally for pavement construction. These materials are normally too viscous to be incorporated directly with soil. The fluidity of

asphalts is increased by either heating, emulsifying or by cut-back process. Tars are heated or cut back. The bituminous materials when added to a soil impart cohesion or binding action and reduced water absorption. Thus either the binding action or the water proofing action or both the actions, may be utilized for stabilisation. Depending upon these actions and the nature of soils, bitumen stabilisation is classified under the following four types: (i) sand-bitumen, (ii) soil-bitumen (iii) water-proofed mechanical stabilisation and (iv) oiled earth.

CHEMICAL STABILISATION:

There are a great many chemicals which are used for stabilisation. Only the chemicals which are commonly used for stabilizing moisture in the soil and for cementation of particles will be described here.

1. Calcium chloride

2. Sodium chloride

3. Sodium silicate

STABILISATION BY HEATING:

Heating a fine grained soil to temperatures of the order of 400 to 600° C causes irreversible changes in clay minerals. The soil becomes non-plastic, less water sensitive and non-expansive. Also the clay clods get converted into aggregates. Soil can be baked in kilns, or in-situ downwards draft slow moving furnaces. The artificial aggregates so produced can be used for mechanical stabilisation.

ELECTRICAL STABILISATION:

The stability or shear strength of fine-grained soils can be increased by draining them with the passage of direct current through them. The process is also known as electro-osmosis, as discussed in Chapter 11. Electrical drainage is accompanied by electro-chemical composition of the electrodes and the deposition of the metal salts in the soil pores. There may also be some changes in the structure of soil. The resulting cementing of soil due to all these reactions, is also known as electro-chemical hardening and for this purpose the use of aluminium anodes is recommended.

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(i) One-half to one-third of the final load which causes settlement equal to

10% of pile diameter,

(ii) Two-thirds of the final load which cause a total settlement of 12 mm, or

(iii)Two-thirds of final load which causes a net settlement (residual settlement after the removal of load) of 6 mm.

4. Write a detailed note on various samplers. a) Standard split spoon sampler

This is a most commonly used sampler for obtaining disturbed soil samples. It is

mostly used in SPT test.

The sampler essentially consists of three parts:

1. Driving shoe of 75mm length with internal dia 38mm and external dia 50mm.

2. Split steel tube of 450mm length which can be split into two halves.

3. Coupling at top which is 150mm length. The coupling head is provided with a ball check

and 4 venting part of 10mm dia to make easy of sample recovery.

After the bore hole of sufficient depth is made, the sampler is lowered into the bottom of the

hole. It is driven into the soil by hammering action. Suppose if there is any presents of fine

sand below water table, the sampling become difficult and use of spring catcher is necessary.

This type of sampling gives highly disturbed soil and it is possible to determine only the

index properties of the soil sample.

b) Shelby and thin walled tube sampler

It is made up of steel. The outer dia of the tube may be between 40mm and

125mm.\

The commonly used sampler has the outer dia of either 50.8mm or 76.2mm. The bottom of

the tube acts as a cutting edge. The area ratio will be less than 15% and the inside clearance

varies between 0.5 and 3%.

The length of the thin walled tube sampler is 5 to 10 times the dia for sandy soil and 10 to 15

times the dia for clayey soil. The dia generally varies between 40 to 125mm. the thickness

varies between 1.25 to 3.15mm. The sampler tube is attached to the drilling rod and lowered

to the bottom of tube. It is then pushed into the soil.

Care shall be taken to push the tube into the soil by a continuous rapid motion without impact

or twisting. At least 5minutes after pushing the tube into the final position, it turned two

revolutions and to shear the soil sample at the bottom of the hole, before the sampler is

withdrawn.

c) Piston sampler

It consists of a thin walled tube with a piston inside. The piston keeps the lower end of the

sampling tube closed when the sampler is lowered to the bottom of the hole. After the

sampler has been lowered to the desire depth the piston is prevented from moving downwards

by suitable arrangements. The thin tube sampler is pushed downwards to obtain the sample.

The piston remains in close contact with the top of the sample. The presence of piston

prevents the rapid squeezing of the soft soil into the tube and reduced the disturbance of the

sample. A vacuum is created on the top of the sample which helps in retaining the sample.

They are used for obtaining undisturbed soil sample.

5. Write a detailed on various types of boring. a) Augur boring

b) Wash (or) water boring c) Percussion boring

a) Augur boring :

An augur is a type tool which is used for understanding the characteristics of the sub surface

soil.

Generally there are two types of Augurs,

i) Manually operated Augurs (upto 6 m)

ii) ii) Mechanically operated Augurs (upto 12m)

Augur boring is highly useful to understand the subsurface soil profile for constructing

highways, railways, Airport etc., where the depth of exploration is small.

It is best suited for partially saturated soil. It is highly unsuitable if Boulders are present

below the ground.

The soil sample obtained will be highly disturbed.

b) Wash (or) Water Boring :

Soil exploration by wash boring is done by the following ways:

1. A casing pipe of 2 to 3 m length is inserted into soil.

2. A hollow drill rod is inserted inside the casting.

3. Water is allowed inside the drilling rod from water pipe and the water emerges out at the

bottom of drilling rod through chopping bit with pressure.

4. The water pressure lubricates the soil below and the soil is transformed into soil slurry

paste and it rises through the annular space between the casing and the drilling rod.

5. The disintegrated soil with water will be collected separately in a tub.

6. The continuous chopping and jetting action will be done till sufficient depth of boring is

reached.

c) Percussion boring :

This type of boring is highly suitable at the places where Boulders, rocks are present.

A vertical hole is made is heavy chisel is allowed to drop inside the vertical hole continuously

and boring of sufficient depth will be reached.

6. Explain SPT test in detail.

SPT is inside test especially for cohesion less soil which cannot be easily sampled. They

are used to find the relative density & angle of shearing resistance of cohesion less soil. The SPT

is conducted in a box hole using Standard Split Spoon samplers. When the bore hole has been

drilled to the desired depth, the drilling tools are removed & the sample is lowered to the bottom

of the hole. The sample is driven into the soil by drop hammer of weight

63.75kg mass falling

from height of 750mm at the rate of 30 blows per minute. The no of hammer blows necessary to

driven 150mm of the sample is counted, the sample is further driven by 150mm

& the no of

blow is recorded likewise the sampler is once again driven by 150mm & the no of blows is

recorded. The number of blows recorded for the first 150mm will not be taken for the

calculations.

The number of blows recorded for the last 150 intervals all added to give the

Standard

Penetration Number. The Standard Penetration Number is equal to the number of blows

necessary for 300mm penetration beyond sealing drive of 150mm.Its notation is

"N". If no of

blows for 150mm driven exceeds so the test is discontinued. The SPT is corrected for:

1) Dilatancy correction

2) Overburden pressure correction

1) Dilatancy Correction

Silty, fine sand & fine sand present below water table develops pore pressure which

cannot be easily dissipated. The pore pressure increases the resistance of soil &

hence SSN.

Terzaghi & Peck recommended the following correction in the case of silty sand

Where, NC=corrected value NR=observed value

If NR less than or equal to 15 then NC = NR

2) Overburden pressure Correction

In granular soil the Overburden Pressure affects the penetration resistance. If two soils

having the same relative density but with different confining pressure are tested, the with high

confining pressure gives a higher penetration number for soil .as a confining pressure in

cohesion less soil increases with depth, the penetration number for soil at shallow depth is under

estimate & that of deeper depth is over estimated for uniformly the N value obtained from field test under different effective Overburden Pressure are corrected to a standard effective

Overburden Pressure.

Gibbs & Holts recommended the use of the following equation for dry or moist clean

sand,

NC =NR * 350 / σ+70

Where,

NC = corrected value

NR = recorded values

σ = effective O.P. in KN/mm2

The above equation is applicable if σ less than or = 280 KN/mm2

The ratio of NC/NR should lie between 0.45 and 2. If ratio NC/NR is greater than 2,

NC should be divided by 2 to obtain the design value used in finding the bearing capacity of soil.

The corrected given by Bazaara & also by Peck & Bazaara is one of the commonly

used correction.

1) N = NR

If σ = 71.8kN/m2

2) If σ < 71.8KN/m2

Then,

N = 4NR / (1+0.0418 σ)

3) If σ > 71.8 KN/m2

Then,

N = 4NR / (3.25+0.0104 σ)

7. Explain the various parameters which affect the sampling in detail. a) Area ratio (Ar)

It is defined as the ratio of maximum cross section area of cutting edge to the area of the soil

sample.Ar = Maximum cross sectional area of the cutting edge x 100

Area of soil sample

Ar = (D2

2

-D1

2

) x 100 / D2

2

For good quality undisturbed sample the area ratio should be 10 % or less than

10 %.

b) Inside clearance (Ci)

Ci = (D3-D1) X 100 / D1

The inside clearance allow the elastic expansions of soil when enter the tube.

For obtaining good quality of undisturbed soil sample, the inside clearance should be between

0.5 – 3%.

This is also reduces the frictional drag on the soil sample.

c) Outside clearance (Co)

C0 = (D2-D4) x 100 / D4

The outside clearance reduces the force of application on the sample while driving. For

obtaining good quality undisturbed soil sample it’s the value be between 0 – 2%.

d) Inside wall friction:

Greater the inside wall friction, the lesser will be the amount of undisturbed soil sample

obtained. The inside wall friction can be reduced by applying oil inside the sampling tube

before driving.

e) Position of non return wall

The non return wall present in the sample should contain a larger orifice which permits the

escape of air, water, slurry to escape the sampling tube during driving and orifice should be

close as soon as the sampler is withdrawn.

f) Methods of applying force

Good quantity of undisturbed soil sample depends upon the methods of applying force. The

sampler should be pushed and not driven.

g) Recovery ratio (Lr)

It is the ratio of recovered length of sample to the penetration length of sample. Lr = recovered length of sample

Penetration length of sample

When, Lr = 1 indicates good recovery

Lr < 1 indicates soil is compressed

Lr > 1 indicates soil is swelled.

8. Explain the Geophysical methods.

Geophysical method is one of the methods for exploration of soil. In which way, electricity are

used for exploration. Based on above, it classified into two types, they are, a) Seismic refraction

b) Electrical resistivity

a) Seismic refraction method :

This method is devised based on fact that seismic wave have different velocity in different type

of soil & rock stratums. Further, the waves are refracted when they are cross the boundary

between different types of soil. The method enables the determination of general soil type & the

approximate depth of boundaries of strata.

The method consists of inducing impact or generating shock by exploding a small charge at or

near the ground surface. The radiating shock waves are recorded by device called geophone

which records the time of travel of the waves. The geophones are installed at suitable known

distance on the ground in line from the source of shock or the same is moved away from the

geophone to produce shock waves at given intervals.

As the distance between the shock source & the geophone increases the refracted waves reach

the geophone earlier than the direct waves. The arrival time is plotted against the distance

between the source & the geophone .the depth of boundary between the two strata can be

estimated from the equation:

D = (d/2). ((v2-v1)/ (v2+v1))^ (1/2)

b) Electrical resistivity method :

The electrical resistivity method is based on the measurement & recording of changes in the

mean resistivity or apparent specific resistance of various soils. The resistivity is usually defined

as the resistant between the opposite forces of a unit cube of the material. Significant variations

in resistivity can be detected between different type of soil strata, above & below the water table between unissued rocks & soils between voids & soil or rocks.

The test is carried out by driving 4 metal spikes to serve as electrodes into the ground along a

straight line into equal distances current from a battery , flows though a soil between the outer

electrodes , producing an electrical field within the soil. The potential difference

(E) between the

two inner electrodes is then measured .the apparent resistivity is given by the equation:

Resistivity, p = 6.28DE / I Where,

D in cm, E in volts, I in amperes & p in ohm-cm.

9.Types of shear failure:

Vesic (1973) classified bearing capacity failures into three categories: General shear failure:

 Fig (a) shows a strip footing resting on the surface of a dense sand or a stiff

clay.

 The figure also shows the load settlement curve for the footing, where “q” is the load per unit area and “s” is the settlement. At a certain load intensity

equal to qu(ultimate bearing capacity),the settlement increases suddenly.

 A shear failure occurs in the soil at the load and the failure surfaces extent to the ground surfaces.

 This type of failure is known as general shear failure.

 A heave on the sides is always observed in general shear failure.

Local shear failure:

 Fig(b) shows strip footing resting on a medium dense sand or on a clay of medium consistency.

 The figure also shows the load settlement curve.

 When the load is equal to certain value qu, the foundation movement is

accompanied by sudden jerks.

 The failure surfaces gradually extend outwards from the foundation

 However a considerable movement of the foundation is required for the

failure surfaces to extend to the ground surface.

 The load at which this happens is equal qu. Beyond this point, An increase of

load is accompanied by a large increase in settlement.  This type of failure is

known as local shear failure.

 A heave is observed only when there is substantial vertical settlement.

Punching Shear failure:

 Fig(c) shows a strip footing resting on a loose sand or a soft clay.

 In this case, the failure surfaces do not extend up to the ground surface.

 There are jerks in foundation at a load of qu the footing fails at load of qu at

which stage load-settlement curve becomes steep and practically linear.

 This type failure is known as punching shear failure.

 No heave is observed. There is only vertical movement of footing

 Vesic proposed a relationship for the mode of failure based on the relative density Dr and Df/ B*, where B* = 2B (L/(B+L)), in which B is the width of the footing L is the length of footing.It is worth nothing that even for the same relative density (Dr), the mode of failure may change with a change in Df/B* ratio.

 It is generally observed that for shallow foundation, the ultimate load occurs

at a foundation settlement of 4 to 10% of B in the case of general shear failure, and at a settlement of 15 of 25% of B in local or punching shear failure.