Shallow Foundations

CHAPTER 1 SHALLOW FOUNDATIONS

LEARNING OUTCOMES

At the end of this lesson, students, will be able to:

1. Explain criteria of shallow foundation, (C2, PLO1). 2. Explain the concept of bearing capacity, (C2, PLO1). 3. Analyze safety factor of shallow footing, (C4, PLO1) 4. Sketch and solve problem on shallow footing in any given problem, (C4, PLO1)

CHAPTER CONTENT

1.0 Introduction o Foundation is part of structure that transfer load directly the soil stratum underneath it. o A foundation should be qualified to the two basic functions: i. Safety factor against shear failure for the soil bearing between 2.5 and 3 (normally stated) ii. Footing settlement shall not be essessive. o If soil near surface has sufficient bearing capacity to support load structure; the soil is competent. o Footing is the small separate portion of the structure that support the whole load.

1

1.1 Types of foundations o Foundations can be categorized into several types such as single footing, Figure 1a. o Single footing – for the purpose analysis, is strong enough as flat footing, square, exerted on by concentrated load or distributed load, Figure 1b.

Figure 1: Single foundation o Combination of footing – footing supports two columns, Figure 2a o Continuous or wall footing – where the size is extended in one direction to support wall–like load, Figure 2b. o Two or more footing connected by a strap is known as strip footing, Figure 2c. o Huge pad footing that support several footing that are not in a straight line known as raft footing, Figure 2d.

2

Figure 2: Classification of footing : (a) combined (b) wall (c) strap (d) mat or raft foundation o Foundation shall meet to satisfy the three criteria : i.

Shall be located at the right location vertically and horizontally to reduce the influence of external load.

ii.

Free from damage and bearing capacity failure.

iii.

Save from excessive settlement.

o Foundation design procedures are as follows : i.

Calculate all loads acting on a footing.

ii.

Obtain on site sub-soil information randomly from laboratory and test.

iii.

Determine the depth and location of foundation.

iv.

Calculate bearing capacity for supporting soil.

v.

Determine size of foundation.

vi.

Check the allowable bearing capacity, allowable against sliding and allowable against overturning.

vii.

Estimate total settlement

viii.

Design the foundation structure

3

1.2 Foundation load o All loads has to be calculated in designing foundation or footing such as :- live load, dead load, wind load, snow, soil lateral pressure, hidrostatic pressure and earthquake. o Snow load may not be considered in Malaysia, however there is an awareness on considering earthquake safety factor. a)

Dead load Referring to total weight of materials and permanent

-

acessories installed on a structure that to be desinged (aircondition systems and finishings) b)

Live load -

Referring to the temporary weight of bodies or structures on building to be designed (human, furnitures)

-

Normally live intensity used in the design of structure is determined in local building codes.

c)

Wind load -

Not all considered as live load and applied on any surface of the structure.

d)

Estimated based on the building codes. Lateral earth pressure

-

Result in lateral load which acted on the substructure part of the building, will be explained more in Chapter 10.

e)

Assumed as dead load Hydrostatic pressure

-

Produce lateral pressure (dead load)

-

Also result in hydrostatic uplift (buoyancy) to the bottom of the structure base.

-

Normally hydrostatic lateral pressure is stable but uplift pressure is not.

-

Buoyancy shall be reduced by increasing the dead load portion of the whole structure.

4

f)

Earthquake force -

Earthquake force may act laterally, horizontally or torsionally on a building structure from various directions.

-

A building code need to be consulted for the specification earthquake forces used in design.

1.3 Theory of bearing capacity o Conventional method of designing foundation often based on the theory of bearing capacity. o Bearing capacity, q – referring to the soil ability to support the whole foundation and superstructure. o Ultimate bearing capacity, qult or qu – referring to loading per area that will just cause shear failure in soil. o Allowable bearing capacity, qall or qa – referring to loading per area that the soil is able to support without unsafe movement which is ultimate bearing capacity divide bysafety factor. Also frequently referred to as design bearing capacity, qdes.

qa 

qult FS

(1.1)

o The safety factor, FS used is between 2.5 to 3. o Footing design should ensure the following :(i)

Collapse of footing structure

(ii)

Excessive settlement

o Basic principles of (Terzaghi and Peck, 1967) is shown in Figure 3.

5

Figure 3 : Plastic analysis of bearing capacity o Least load Q in Figure 3, applied that cause failure is the most critical; hence the ultimate bearing capacity, qult is the least load Q divide by footing’s area – the area that in contact with the soil beneath. o A wedge of soil under the footing moves downward with the footing. The downward movement is resist by the shear resistance along slip surface cde and cfg and by the sliding wedge of acfg and bcde. o For each asssumed slip surface, the corresponding load Q that would cause failure can be determine. o The set of slip surface giving the last applied load Q (that would cause failure) is the most critical, hance the soils ultimate bearing capacity (qult) is equal to the least load divided by the footing’s area. o The following equations for calculating ultimate bearing capacity, developed by :-

(i)

Continuous footing : qult  cN c   1D f N q  0.5 2 BN 

(ii)

(1.2)

Circular footing : qult  1.2cN c   1D f N q  0.6 2 RN  or qult  1.2cN c   1D f N q  0.3 2 BN 

(iii)

Square footing: qult  1.2cN c   1D f N q  0.4 2 BN 

6

(1.3)

(1.4)

Where : qult

– ultimate bearing capacity

c

– soil cohesion

Nc, Nq, Nγ

– Terzaghi’s bearing capacity factors

γ1

– effective unit weight for above base soil (kN/m3)

γ2

– effective unit weight for below base soil (kN/m3)

Df

– footing depth, or distance between soil surface and base of footing.

B

– breadth of footing, R – radius of circular footing

o Terzaghi’s bearing capacity factors are functions of soil’s angle of internal friction , Ø. o Every bearing capacity factor has influence on the value of ultimate bearing capacity : 

Value of Nq : influence of surcharge



Value of Nc : influence of cohesion,



Value of Nγ : influence of soil weight and breadth or radius of footing

o Values of Nc, Nq, Nγ for all bearing capacity shall be determined from Figure 4

7

Figure 4: Chart of Terzaghi’s bearing capacity factors and angle of internal friction , Ø. o Curve on Figure 4 shows values of Nc, Nq, and Nγ based on the following equations :

8

  N q  e  tan  tan 2  45   2 

(1.5)

N c  cot  N q  1

(1.6)

N  N q  1tan 1.4 

(1.7)

o Equations 1.2 to 1.4 are used for cohesive and cohesionless soil o Figure 5 shows dense sand and hard clay that produce general shear while loose sand and soft clay produce local shear.

Figure 5 : Relation between load and settlement of a footing for cases of local shear and general shear. o In local shear equations 1.2 to 1.4 cohesion, c is replaced by c` where :

c'

2 c 3

(1.8)

o Terms of Nc, Nq and Nγ are replaced by N’c, N’qdan N’γ where the latter is determined from Figure 4 by using value of Ø modified, Ø’ by the following equation :2 3

 

 '  tan 1  tan  

(1.9)

o For loose sand and soft clay, terms N’c, N’q, N’γ and c’ are used in equations 1.2 to 1.4.

9

o For cohesive soils, shear strength at its critical state only after the completion of construction process, at which where the shear strength only has cohesion component (c) and in this case Ø = 0 (internal friction angle is taken as zero). o For cohesionless soil, c terms in equations 1.2 to 1.4 is zero, c = 0. o Other correlation for N value for (SPT), Nq, Nγ and Ø are as shown in Figure 6.

Figure 6 : Correlation for values of N (SPT), Nq,Nγ and Ø

10

Example 1.1

Given ; a. A strip of wall footing 1.2 m wide supported by uniform deposit of clay, Figure 7. b. Unconfined compressive strength of the soil, qu = 122kN/m2 c. Soil unit weight, γ = 19.68 kN/m3 d. Groundwater was no encountered during soil exploration. e. Footing depth, Df = 0.4 m

Find : a. Ultimate bearing capacity of the footing b. Allowable load for the wall with safety factor, FS = 3. Solution:-

For continuous wall footing : qult  cN c   1D f N q  0.5 2 BN 

c

qu 122kN / m 2   61kN / m 2 2 2

Using c>0, Ø =0 analysis for cohesive soil, when Ø =0. Figure 4 gives Nc=5.14, Nq=1.0, dan Nγ=0

Clay : 0.4 m

γ1 = 19.68 kN/m3 γ2 = 20.52 kN/m3 qu = 122 kN/m2

1.2 m

Figure 7 : A strip footing in clay soil

11

qult  cN c   1 D f N q  0.5 2 BN  

wa  qa  B 

Example 1.2

Given : a. Square pad footing with sides of 1.6 m located 1.4 m below the ground surface b. The effect of groundwater is negligible c. The subsoil consist of stiff cohesive soil with unconfined compressive strength of qu = 145 kN/m2. d. The unit weight is 19.68 kN/m2. Find :

Allowable bearing capacity, qall using a safety factor of 3.5. Solution;

1.4 m

1.6 m

Figure 8: Shallow footing in cohesive soil

For square footing : qult  1.2cN c   1D f N q  0.4 2 BN 

c=qu/2=145/2 = 72.5 kN/m2

12

Cohesive soil : Assuming γ1= γ2 = 19.68 kN/m3 qu = 145 kN/m2

Using c>0, Ø =0 analysis for cohesive soil, when Ø =0. Figure 4 gives Nc=5.14, Nq=1.0, dan Nγ=0 qult 

qall =

Qall=

Example 1.3 Given :

1.25 m

Dense soil : Ø = 30° Assuming : γ1= γ2 = 21.25 kN/m3 c = 46.8 kN/m2

1.45 m dia

Figure 9 : Footing in dense soil

a. Circular footing with diameter of 1.45 m is to be constructed below the ground surface at 1.25 m. b. Subsoil is consist of dense soil having the following strength parameters:

Internal friction angle,

Ø = 30o

Cohesion,

c = 46.8 kN/m2

c. The effect of groundwater is negligible

13

Find: The total allowable load (including column load, weight of footing, and weight of soil surcharge) that the footing can carry if FS=3. Solution : (for dense soil – hence general shear condition)

14

1.4

Effect of ground water table on bearing capacity o It has been assumed that the water was well below the footing and that did not affect the bearing capacity. However, this condition does not always the case, terms γ2BNγ and γ1DfNq require modification. o If water table is at or above the footing’s base, soil submerged unit weight shall be used, γ’ which equals (saturated unit weight, γsat – unit weight of water, γw) this term should be used in equations (1.2) hingga (1.4). o If the water table is at the distance of B (footing width) or more from the footing base, then the effect of the water is negligible – hence the full unit weight is to be used, Figure 10

Earth’s surface

Df

B B

Figure 10: Location of the depth B (equal footing width) below footing base o If the water table below the footing but at less than B distance – the value of unit weight shall be linearly interpolate value for effetive unit weight γ’.

15

o If the water table on the earth surface, therefore the value of effective unit, γ’ shall be used entirely.

Example 1.4

Given : a.

A square footing with dimensions of 2.2 m X 2.2 m constructed as shown in Figure 11.

b.

Water table is on the surface.

c.

Subsoil is soft and loose uniform deposit. The result of laboratory test is as shown : Qallowable = ? Groundwater table

1.8 m

2.2 m x 2.2 m

Loose soil : Ø = 25° Assuming : γ1= γ2 = 16.32 kN/m3 c = 16.8 kN/m2

Figure 11: Shallow footing submerged under water

Internal friction angle, Ø

= 25°

Cohesion,c

= 16.8 kN/m2

Unit weight, γ1= γ2

= 16.32 kN/m3

Find

Allowable load,Qall that the footing can sustain with SF of 3.

16

Calculation: Assume local shear condition because of soft and loose soil

qult  1.2c' N 'c  1D f N 'q 0.4 2 BN '

2  tan 25 o   17.3; 3 

 '  tan 1 

c' 





2 2 c  16.8kN / m 2  11.2kN / m 2 3 3

From Figure 4 : N’c=10.2, N’q=5, dan N’γ=1.8 γ’= (16.32 – 9.81 ) = 6.51 kN/m3

qult  Qa  qa  A 

1.5 Inclined load o A footing subjected to by an inclined load can be resolved into vertical and horizontal load, Figure 12. o Bearing capacity analysis is conducted in the same manner as previously explined. The bearing capacity calculated then must be corrected by an Ri factor which obtained in Figure 13. o The footing’s stability shall be checked by calculating the FS against sliding, in Chapter on Retaining Wall.

17

Figure 12: Footing subjected to an inclined load

Figure 13: Inclined load factor, Ri for (a) horizontal footing (b) inclined footing

Example 1.5

Given

A square footing (1.5 m x 1.5 m) exerted by an inclined load as shown in Figure 14.

18

Figure 14: Shallow footing with inclined load Required : The safety factor against bearing capacity failure

Solution; For square : qult  1.2cN c   1D f N q  0.4 2 BN 

1.6 Eccentric load o

Foundation design is more complicated if subjected to eccentric load.

o

Two methods of analizing bearing capacity for eccentric load :  Useful width concept  Reduction factors

19

1.6.1 Useful width concept : o

Figure 15, shows the point of concentrated load with eccentricity in b direction. Shaded region is assumed to be the main contact area of the footing baes due the load. The shaded area equal to :

A  L  B  2eb 

(1.10)

Figure 15: Useful width method for bearing capacity determintaion on cohesive soil o

This method based on linear bearing capacity reduction and strictly for cohesive soil.

1.6.2 Reduction factors: o

Assuming that the load is applied at the centroid of the footing.

o

Corrected bearing capacity is estimated by multiplying with reduction factor (Re) obtained from Figure 16.

20

Figure 16: Eccentric load reduction factor. Cohesive – linear correlation Cohesionless – parabolic correlation

Example 1.6

Given : 1. A foundation with dimension of 1.5 m X 1.5 m located 1.4 m below ground surface. 2. The footing is subjected to an eccentric load of 350 kN, Figure 17. 3. Subsoil is the deposit of cohesive soil with the following parameters : qu = 200 kN/m2, γ1 = 20.40 kN/m3 and γ2 = 19.80 kN/m3 4. The effect of groundwater is negligible. Find : Factor of safety against bearing capacity failureby using both methods:

21

Figure 17: Footing with eccentric load Solution : 1.

Useful width concept :

qult  1.2cN c   1D f N q  0.4 2 BN 

c

qu 200kN / m2   100kN / m2 ; 2 2

γ1 = 20.40 kN/m3 and γ2 = 19.80 kN/m3 Df = 1.2 m;

B’ = B - 2(eb) = 1.5 – 2(0.2) = 1.1 m

Figure 18: Footing with eccentric load

22

Using c>0, Ø =0 analysis for cohesive soil, when Ø =0. Figure 4 gives Nc=5.14, Nq=1.0, dan Nγ=0 qult  1.2cN c   1D f N q  0.4 2 BN 

















 1.2 100kN / m 2 5.14  20.4kN / m3 1.4m 1.0  0.4 19.8kN / m3 1.1m 0  616.8  28.6  0  645.4kN / m 2

FS 

qult 645.4kN / m 2   3.04 qa  350kN     1.1m  1.5m 

(1) Reduction factors method Eccentricity ratio = ex/B = 0.2m/1.5m = 0.13 For cohesive soil; Figure 16, Re=0.76

qult  1.2cN c   1D f N q  0.4 2 BN 









 1.2 100kN / m 2 5.14  20.4kN / m3 1.4m 1.0  0.4 19.8kN / m3 1.1m 0  645.4kN / m

2





qultcorrected   qult  Re  645.4kN / m2 0.76  490.5kN / m2 FS 

1.7

490.5kN / m2  3.15 350kN / 1.5m  1.5m

Footing on slopes o

Ultimate bearing capacity for continuous foundation on slopes can be determined by using the following equation :

1 qult  cN cq  BN q (on slope) 2

(1.11)

Where Ncq and Nγq are bearing capacity factors for footing on slope o

Bearing capacity factors used in equation 1.11 are obtained from Figure 19 and Figure 20.

23

o

For square and circular footing on slopes, the assumption to made is the ratios of their bearing capacities on the slope to their bearing capacities on level ground are in the same proportions as the ratio of bearing capacities of continuous footings on slopes to the bearing capacities of the continuous footings on level ground.

 qult c..or.. s..on.. ground..level     q ult  continuous.. on.. ground..level  

qult c..or.s. footing..on..slope  qult continuous.. footing..on..slope 

(1.12)

Figure 19 : Bearing capacity factors for continuous footing on face of slope (a) cohesive soil (b) cohesionless soil

24

Figure 20: Bearing capacity factors for continuous footing on top of slope (a) cohesive soil (b) cohesionless soil

25

Example 1.7

A bearing wall is to be located close to the slope as shown in Figure 1.21. Ground water located at a great depth. Calculate the allowable bearing capacity if factor safety of 3 is used.

Cohesionless soil : γ=19.5kN/m3 Ǿ=30° c=0

Figure 21: Footing for wall on slope Solution :

1 qult  cN cq  BN q ; 2 c = 0; γ = 19.5kN/m3; B = 1.0m

from Figure 8.20; with Ǿ = 30°; β=30°;

b/B =1.5/1.0 =1.5;

Df/B = 1.0/1.0 = 1;



Nγq=40



1 1 qult  cN cq  BN q  0  19.5kN / m 2 1m 40   390kN / m 2 2 2 2 390kN / m qa   130kN / m 2 3

Example 1.8

26

Same conditions as in Example 1.7 except the footing is constructed on slope surface with safety factor of 3.

Solution : From Example 8.7 : (qult)continuous footing on slope = 390kN/m2 Figure 4, with Ǿ=30°; values of Nc = 30, Nq = 18, dan Nγ = 16

qult square.. footing..on.. ground..level  1.2cN c  D f N q  0.4BN   0  19.50kN / m 3 1m18  0.419.50kN / m 3 1m16  475.8kN / m 2 qult continuous.. footing..on.. ground..level  cN c  D f N q  0.5BN   0  19.50kN / m 3 1m18  0.519.50kN / m 3 1m16  507kN / m 2

Substitute into equation :

 qult c..or.. s..on.. ground..level    qult continuous..on.. ground..level 

qult square. footing..on.. slope  qult continuous.. footing..on.. slope 

 475.8kN / m 2   390kN / m   366kN / m 2 2   507kN / m  366kN / m 2 thus.....qa square.. footing..on.. slope..   122kN / m 2 3



2



Re-calculate for circular footing : (qa=114 kN/m2; Qa=89.5kN)

27

1.8

Footing size o

After the allowable footing size is determined, the sizing of the footing shall be determined.

Area of footing 

column load kN  bearing capacity kN / m 2



 (1.13)

Example 1.9

Given : Footing in Figure 1.22 was constructed below stif clay uniform deposit. Footing is subjected to load of 168 kN/m. Calculate the size of the footing with safety factor of 3.5.

168 kN/m

1.2 m

B=? Figure 1.22 : Calculation of footing size

28

Stiff clay : γ1 = 17.8 kN/m3 γ2 = 19.6 kN/m3 qu = 142.4 kN/m2

Solution :

Example 1.10

Given :

a) A square footing located on deposit of stiff clay with unconfined compression strength (qu) of 115 kN/m2. b) Footing is located 1.2 m below the surface and sustain a load of 1250 kN. Unit weight of clay is 19.60 kN/m3. Ground water has negligible effect on the bearing capacity of the footing. c) Determine the dimension of square footing with FS = 3. d) Re-calculate the diameter of a circular footing used if using the same safety factor. Footing is located 1.5 m below and sustain 1500 kN of load has a value of qu= 124 kN/m2

Solution

29

Contact area needed :

; B2=9.92m2; B=3.15m

The possible size : 3.2m x 3.2m

Re-calculate if circular footing is used (B=3.55)

Example 1.11

Given

a)

A deposit of uniform soil with soil parameter criteria as shown in Figure 1.23. (Note : most examples are at extremme cases of cohesive or non-cohesive but the intermediate cases often referred to as c-Ǿ soil.

b)

The proposed footing is located at 1.5m deep and to sustain load of 2650 kN.

c)

Ground water table is very deep an its effect is negligible.

30

Figure 1.23: Shallow footing in c-Ǿ soil

Determine The dimensions of the square footing to sustain the load with factor of safety = 3.

Solution : With Ǿ = 30°;

Nc = 30;

Nq = 18;

Nγ = 16

First trial :

31

Second trial :

1.9 Contact pressure Contact pressure is referred to as the pressure between a footing’s base and the underlying soil below.

Contact pressure can be calculated by using the flexural formula: q

Q Mxy M yx   A Ix Iy

(1.14)

Where :

q

=

contact pressure

Q

=

total vertical axial load

A

=

footing area

Mx,My =

total moment in x and y direction

x,y

distance from centroid to the outer most point where where

=

the contact pressure is computed along respective x and y axes lx,ly

32

=

moment of about x and y axes (right hand rule)

Example 1.12

Given

A footing of 1.5 m X 1.5 m in dimension with centric axial loading of 225 kN. Unit weight of soil is 18.84 kN/m3 and the unit weight of concrete is 23.55 kN/m 3. The unconfined compression for the cohesive soil is 144 kN/m2.

Determine

1. Sketch the problem and label the necessary data. 2. Contact pressure and safety factor against bearing capacity failure.

Solution

γsoil = 18.84 kN/m3 γconcrete = 23.55 kN/m3

Figure 1.24a : Sketch of problem example 1.12

a.

Soil contact pressure ; q 

Q Mxy M yx   A Ix Iy

Column load is on the centroid therefore Mx = O and My = O (no eccentricity, e) Q = total axial vertical load on footing’s base

33

Q = column load + weight of footing base pad + weight of stum (pedestal) + weight of backfill

Column load

=

Weight of footing’s base

= = =

Weight of stum

= = =

Weight of backfill

= = Q=225 kN

Figure 1.24b : example 1.12

Contact pressure

=

c. Factor of safety against bearing capacity failure qult  1.2cN c  D f N q  0.4BN 

So :

34

Using c>0, Ø =0 analysis for cohesive soil, when Ø =0. Figure 4 gives Nc=5.14, Nq=1.0, dan Nγ=0

Example 8.13

Given :

1. A footing with the size of 2 m x 2 m, Figure 1.25 2. Base of the column is pinned 3. Load on base P = 300 kN 4. Weight of concrete footing including pedestal and pad (W 1) = 42 kN 5. Weight of backfill (W 2) = 50 kN 6. Horizontal load = 18 kN 7. Bearing capacity for the base soil taken as 145 kN/m 2.

35

Find :

1. Contact pressure and the soil pressure diagram 2. Safety factor against bearing capacity 3. Shear and moment at cross-section A-A, Figure 1.25. 4. Safety factor against sliding if the coefficient of friction = 0.40 5. Safety factor gainst overturning

Figure 1.25: Section A-A of a footing

36

Solution

1.

q

Contact pressure and soil pressure diagram:

Q Mxy M yx   A Ix Iy

Q = P + W 1 + W 2 = 300 + 42 + 50 = 392 kN A = 2m X 2m = 4m2 My = 18kN x 1.5m = 27kN.m (take moment at point C). x = 2m/2 = 1m Iy 

2m2m3  1.33m4

Mx = 0;

q

12

Mxy 0 Ix

27kN.m1m  98  20kN / m2 Q M x y M y x 392kN    0 2 A Ix Iy 4m 1.33m4

qright  98  20  118kN / m2  145kN / m2  OK qleft  98  20  78kN / m2  145kN / m2  OK

(If ultimate bearing capacity is given then; q ult /q right > 3.0 then OK)

The pressure is as shown

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8. Shear and moment at section A-A: From the principal of similar triangle :

DE EH  DF FG

DF  118  78  40kN / m 2

EH 

2m 0.5m   0.75m 2 2

FG = 2m

DE EH 0.75m   DE   40kN / m 2  15kN / m 2 DF FG 2m









Shear A  A  0.75m  103kN / m 2 2m   12 0.75m  15kN / m 2 2m   154.5kN  11.25kN  165.75kN

 0.75m  Moment A  A  154.5kN    11.25kN  23  0.75m 2    57.94kN.m  5.63kN.m  63.57kN.m

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TUTORIAL

QUESTION 1

A footing with size of 3 m x 3 m constructed 1.5 m below the ground surface. The construction of the footing was on the highly cohesive clay soil with the unconfined compression strength of 100 kN/m2. The unit weight of the soil is 20kN/m3. Calculate the ultimate bearing capacity of the soil.

QUESTION 2

A square footing is to be constructed on a uniform thick deposit of clay with the unconfined compressive strength of 150 kN/m2. The footing will be located 1.5 m below the ground surface and to carry a total load of 1500 kN. The unit weight of the supporting soil is 20.1 kN/m3. Effect of ground water is negligible. Considering general shear condition and safety facor of 3. Determine the dimension of square footing.

QUESTION 3

A footing with the size of 2.5 m x 2.5 m was buried 2 m below the ground surface in a dense cohesionless soil. The results of laboratory and field tests on the supporting soil shows that the soil’s unit weight is 20.4 kN/m3. Average corrected SPT N-value beneath the footing is 37. Compute the allowable (design) load with SF = 3.

QUESTION 4

40

A construction company proposed a footing that will sustain a load of 2500 kN. The footing was constructed on a high density cohesionless. Unit weight of the soil was 21.20 kN/m 3 and the internal friction was 38o. The footing was constructed at 1.5 m depth.

Calculate the size of the footing if the using the safety facor of 3.

QUESTION 5

A footing was constructed as shown below. Axial vertical loading on the footing (including column load, backfill and footing weight) was 600 kN. Horizontal load was 50 kN and a moment of 67.5 kN-m were also applied on the footing.

Calculate the following :

a. Shear on a-a cross-section. b. Moment on a-a cross-section. c. Safety factor against overturning. d. Safety factor against sliding, if the coefficient of friction 0.6. e. Safety factor against bearing capacity of soil, if the ultimate bearing capacity is 540 kN/m2.

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TUTORIAL 1: SHALLOW FOUNDATIONS

SUBJECT

:

GEOTECHNICAL ENGINEERING

CODE SUBJECT

:

DFC 3043 / DAC 21103

DURATIONS

:

1 WEEKS

INSTRUCTIONS

:

ANSWER ALL

NAME

:

NO. MATRIC

:

(SEPARATE THIS PAGE)

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