Foundation Engineering II Introduction to Building Foundations PDF

Preview

Summary

Foundation Engineering by Sontao ZUE...

Description

IISEE Lecture Note 2011

FOUNDA FOUNDATION TION ENGINEERING 2

By Songtao XUE

Department of Architecture, Faculty of Engineering Tohoku Institute of T Teechnology

Feb. 25 , 2011

International Institute of Seismology and Earthquake Engineering (IISEE) Building Research Institute

Syllabus Subjec Subject:

Foundation Engineering 2

Lecturer Lecturer:

Songtao XUE

Day Day:

1

Contents: A foundation is the part of an engineered structure that transmits the structure’s forces into the soil and rock that supports it. The shape, depth, and materials of the foundation design depend on the many factors including the structural loads, the existing ground conditions, and local material availability. Proper design of building foundation requires the knowledge of (a) external loads and loads transmitted by the building superstructure, (b) local code requirements, (c) nature and composition of different types of soil at the site, (d) behavior and stress-related deformability of soils supporting the foundation system, and (e) general geological conditions of the site. Together with knowledge of such scientific principles, rational engineering judgment acquired through observation and experience is indispensable in the foundation engineering practice. It is evident that this one-day lecture cannot attempt to cover all these aspects. The objective here is to cover some basic aspects of the design and construction of building foundations, including scientific principles as well as practical aspects. It is expected that the balance between theoretical and practical aspects in the content would provide the trainees with the clear overview of the essentials of building foundations. Sp Special ecial Mentioning The level of this lecture achieves the PE (Professional Engineer) level and the contents, the examples and practice problems are all in the same level with the PE (Professional Examination), but the units are different. We used SI unit system instead of the American system.

1

Contents: Part 1. Shallow Foundations General Bearing Capacity Bearing Capacity of Clay and Sand Effects of Water table on Footing Design Eccentric Loads on Rectangular Footings Rafts on Clay Examples and Problems

Part 2. Pile and Deep Foundations Piles Capacity from Driving Data Theoretical Point-bearing Capacity Theoretical Skin-friction Capacity Pile Groups Examples and Practice Problems

Part 3. Retaining W Wa all Earth pressure and Vertical Soil Pressure Active Earth Pressure Passive Earth Pressure Surcharge Loading Effective Stress Cantilever Retaining Walls Examples and Practical Problems

2

This lecture

Soil Mechanics a and nd Foundation E Engineering ngineering Lectures

Part 1. Shallow Shallow Foundations Prof. Dr Dr.. Songtao XUE Tohoku Institute of Technology

Part 1. Part 2. Part 3. • The level of this lecture achieves the level and the contents, th e examples and practice problems are all in the sam e level with the , but t he units are different. We used SI unit system instead of the American system.

Tongji University

Februar y 14, 2011

CONTENT

• • •

Introduction of Foundations • Function : to transfer the forces from the structure to

Introduction of Foundations

the soil or rock without excessive settlement.

Bearing Capacity of Clay and Sand Effects of Water table on Footing Design Eccentric Loads on Rectangular Footings Rafts On Clay Examples and Problems

• Foundation Types:

¾S ¾

Structure

• Shallow Foundation : Foundation

Df

• • • •

Soil

the depth of the foundation is shallow relative to its width,

B Rock

Februar y 14 , 2011

• Category of Shallow Foundation

Februar y 14, 2011

• Category of Shallow Foundation ¾ Continuous (or wall) footings

¾Spread footings

¾ Mats or rafts

Square

Spread footings

Rectangular

Circular Other Shapes Februar y 14 , 2011

Mats or rafts Februar y 14, 2011

1

Main considerations in designing shallow foundation

General Bearing Capacity

Bearing Capacity failure and excessive Settlements

Footing and its Function： ： Widened parts of the foundation Accepts load from structure Transmit Forces to the soil, not to exceed bearing capacity

So

il

Column

General Consideration for Footings

Footing

General Considerations for design of footings: Bearing Capacity Failure

Located below the frost line and the moisture content change level. Be safe against overturning, sliding and uplift. Satisfy the allowable soil pressure

Excessive settlement

Bearing Capacity: Settlements: Februar y 14 , 2011

Februar y 14, 2011

Soil deformation and bearing force

Allowable Bearing Capacity

P ④

④

③

① ②

Df

③

The allowable bearing capacity (Also: net allowable bearing pressure or safe bearing pressure) is the net pressure in excess of the overburden stress that will not cause shear failure or excessive settlements. Typical Allowable Soil Bearing Capacities Type of soil

② Sliding surface

Allowable pressure (kPa) 200 100

Massive Crystalline Bedrock Sedimentary and foliated rock Sandy gravel and/or gravel Sand, silty sand, clayey sand, silty gravel and clayey gravel

100

Clay, sandy clay, silty clay, and clayey silt February 14, 2011

General Bearing Capacity Equation • The ultimate (or gross) bearing capacity for a shallow wall footing is given as,

Introduction of the coefficients

P

④ ③

p q : additional surface surcharge Nγ , Nc, Nq : capacity factors. N γ , Nc, Nq should be multiplied by factors 50 for other shape and N c 40 depth of footings: 30 Nq Capacity factors 20

④

①

Bearing capacity factors

Nγ

Shape and Depth factors

0 0

10

20

φ

30

40

D

③

②

f

② Sliding surface

60

10

pq : Surcharge

qult =

γ

+

+(

+

Nq

φ: angle of internal friction Februar y 14 , 2011

Februar y 14, 2011

2

Bearing Capacity factors Meyerhof and Vesic φ

Terzaghi

φ

Nc

Nq

Nγ

0.0

5.7

1.0

0.0

5.0

7.3

1.6

0.5

10.0

9.6

2.7

1.2

Nc

Nq

Nγ

N bγ

0

5.14

1.00

0.00

0.00

5

6.50

1.60

0.07

0.50

10

8.30

2.50

0.37

1.20

15

11.00

3.90

1.10

2.60

20

14.80

6.40

2.90

5.40

Shape factors Nc and N γ multipliers for various values of B/L B/L Nc Nγ 1(square)

1.25

0.85

0.50

1.12

0.90 0.95

15.0

12.9

4.4

2.5

25

20.70

10.70

6.80

10.80

0.20

1.05

20.0

17.7

7.4

5.0

30

30.10

18.40

15.70

22.40

0.00

1.00

1.00

25.0

25.1

12.7

9.7

32

35.50

23.20

22.00

30.20

30.0

37.2

22.5

19.7

1(circular)

1.20

0.70

34

42.20

29.40

31.20

41.10

34.0

52.6

36.5

35.0

36

50.60

37.70

44.40

56.30

35.0

57.8

41.4

42.4

38

61.40

48.90

64.10

78.00

40.0

95.7

81.3

100.4

40

75.30

64.20

93.70

109.40

45.0

172.3

173.3

297.5

42

93.70

85.40

139.30

155.60

48.0

258.3

287.9

780.1

44

118.40

155.30

211.40

224.60

50 50.0 .0

347.5

415.1

1153.2

46

152.10

158.50

328.70

330.40

48

199.30

222.30

526.50

496.00

50

266.90

319.10

873.90

762.90

Depth factor suggested to be applied for the Nc is Depth factor for Nc :

⋅

K is a constant which is 0.2~0.4 11

Net and allowable Bearing Capacity Net Bearing Capacity is just the foundation weight taken away from the ultimate capacity.

−ρ

=

L

Februar y 14, 2011

Bearing Capacity of Clay and Sand • Bearing Capacity of Clay Undrained case or φ Undrained shear strength

Allowable Bearing Capacity

= 0o case:

Su = c

Pore pressure

φ = 0o

cohesion

short time

If no surface surcharge

pq = 0 F :safety factor between

ased on qnet ). Smaller value is sometimes used for transient load conditions such as from wind and seismic forces.

N γ = 0, Nq = 0 Cohesion dominate bearing capacity

+ρ

=

Februar y 14 , 2011

Bearing Capacity of Sand For ideal sand , c = 0 , then: qult = 12 ρgBNγ + (pq + ρgD f )N q

Februar y 14, 2011

Bearing Capacity of Sand (continued) (continued)

If no surface surcharge:

qnet = qult − ρ gD d f = ρ gBNγ + ρ gDf ( Nq −1 )

Cn

0

2.00

24

1.45

φ

1 2

e N dominate bearing capacity Depth term ρgD f q

qa =

q net = F

p e ⎛ B⎜ n d F ⎜⎝

⎛D 1 ρ gNγ + ρ g( N q − 1)⎜⎜ f 2 ⎝ B

⎞⎞ ⎟⎟ ⎟⎟ ⎠⎠

depend on oφ governed by Df /B n Assuming: F = 2, ρ g = 15.7 kN/m3

[in kPa]

Februar y 14 , 2011

48

1.21

96

1.00

144

0.87

192

0.77

240

0.70

287

0.63

335 383

0.58 0.54

431

0.50

479

0.46

Overburden load γ ⋅ D f in Eq.

qa = 10.53Cn N is assumed approximately 95.76kPa Corresponding N-values is from 3~4.5m below the surface. Correction factor needed for shallower foundations . In practice, Df lowest average N-value from B depth of Df +B 3~4.5m

Overburden (kPa)

N: standard penetration test value

Februar y 14, 2011

3

Effects of Water table on Footing Design •General principle 1: For cohesive (φ = 0 o) soils

Effects of Water table on Footing Design •General principle 2: For sand, use the submerged density ρ b = ρ d − 1000kg/m 3 in the equation for bearing capacity: 1 ρ ρ qult =

2

gBNγ + cN c + ( pq + gD f ) N q

The bearing capacity of a footing with the water table at the ground surface is half of the dry bearing capacity. For accurate estimate: (a) Water table is at the base of the footing: Df =Dw For ideal sand , c = 0:

q ult = 21 ρb gBN γ + ρd gD f N q Februar y 14 , 2011

Df

Dw B Water Table Februar y 14, 2011

Effects of Water table on Footing Design

General principle 2 (continue) (b) When water table is at the surface: Dw=0

•General principle 3: (c) When water table is between the surface and the footing: 0 0.4 ]

0.64Lq

With submerged construction or when drains become plugged, a water table can exist behind the wall.

water table The saturated soil density:

ρ sat

h

[m ≤ 0.4 ]

)

2

s = 100% m=

x y , n= H H

n : the porosity; e: void ratio.

Februar y 14 , 2011

Effective Stress (continued)

Cantilever Retaining Walls: Analysis

The effective pressure (generally referred to as the effective stress) is the difference between the total pressure and the pore pressure.

(

Februar y 14, 2011

)

The total horizontal pressure form the submerged sand is

p h = gρ w h + k a g ( ρ sat H − ρ w h) = g( k aρ sat H + (1− ka )ρ w h)

Targets of analysis: 9Have sufficient resistance against overturning and sliding 9Have adequate structural strength against bending outward 9The maximum soil pressure under the base must be less than the allowable soil pressure. Nine steps to analyze a retaining wall: ¾Step 1:

The increase in horizontal pressure above the saturated condition is the equivalent hydrostatic pressure (equivalent fluid pressure ) caused by the equivalent fluid weight: Februar y 14 , 2011

Cantilever Retaining Walls: Analysis step ¾Step Step 2: 2: disregarding restraint from passive distribution. (If it is to be considered, determine the passive earth pressure.) ¾ Step 3: Find all of the vertical forces acting at the base. Including: weights of the retaining wall, the 1 soil directly above the hell and the toe: Dividing the concrete and soil into areas with simple geometric shapes. 3 2 Find the centroid of each 54 shape and its moment arm: y x6 6 toe

x

Februar y 14, 2011

Cantilever Retaining Walls: Analysis step ¾Step 4: Find the moment about the toe from the forces found in steps 1 though 3. h: horizontal, M toe = Wi xi − Ra ,h ya + Ra ,v xa v: vertical ¾Step 5: Determine the location, x R , and eccentricityε of the vertical force component:

∑

¾Step 6: Check the safety factor, F, against overturning: heel r

Februar y 14 , 2011

FOT =

M resisting M overturning

granular soils : F ≥ 1.5

=

∑W x

i i

+ Ra ,v xa ,v

R a, h y a, h cohesive soils : F ≥ 2 Februar y 14, 2011

4

tan δ

Cantilever Retaining Walls: Analysis step

Cantilever Retaining Walls: Analysis step ¾ Step 8: (to be continued) For a keyless base and for tensioned soil behind the key,

¾ Step 7: The maximum pressure (at the toe) should not exceed the allowable pressure:

c A :the adhesion, is zero for granular soil tanδ :coefficient of friction is approximately 0.45 for

¾Step 8: Calculate the resistant against sliding. The active pressure is resisted by: friction and adhesion between the base and the soil, and in the case of a keyed base, also by the shear strength of the soil. When the base has a key, and has the compressed soil in front of the key

and with silt, 0.35 for silt, and 0.3 for clay ¾Step 9: Calculate the factor of safety against sliding: A lower factor of safety 1.5 is permitted when the passive resultant is disregarded. If the passive resultant is included , the factor of safety should be higher. If the factor is too low, the base length (B) can be increased, or a vertical key can be used.

Februar y 14 , 2011

Februar y 14, 2011

Examples and Problems: (solution,

Examples and Problems

Appendix 37.A necessary)

SOLUTION

⎛1 ⎞ ⎛ 0.56 m ⎞ ⎟ = arctan ⎜ ⎟ = 18.4o ⎝3 ⎠ ⎝ 1. 68m ⎠

Step 1: β = arctan ⎜

From the appendix 37.A kv = 1.57kPa, k h = 6.28kPa The active earth pressure resultants:

)(

)

5.59

)

R a, h located 5.59/3=1.86m above the bottom of the base.

0.46

4.57

EXAMPLE: The unkeyed retaining wall shown has been designed for a backfill of coarse-grained sand with silt having a density of 2000kg/m3. The angle of internal friction, φ = 30o, and the angle of external friction δ = 17o. 0.3 The backfill is sloped as β shown. The maximum allowable soil pressure 0.61 is 143.5 kPa. The adhesion is 45.5 kPa. Check the factor of safety against sliding. 0.91 0.46 1.68

Step 2: disregard the passive earth pressure.

Februar y 14 , 2011

Februar y 14, 2011

Examples and Problems: (solution) Ra,v

Step 3: calculate the weights of soil and concrete

β 0.61

5.59

4.57

Per meter of wall i

1

0.46

Ra,h 0.91 0.46

ε

1.68

3

xR

2

5 4 y toe

heel

6 x

r Februar y 14 , 2011

area

ρ

Wi

xi

Mi

m2

kg/m3

kN

m

kN·m

1

(0.5)(1.68)(0.56)=0.47

2002

9.22

2.49

22.96

2 3

(1.68)(4.57)=7.67 (0.3)(4.57)=1.37

2002 2402

150.49 32.26

2.21 1.22

332.57 39.35

4 5 6

(0.5)(0.15)(4.57)=0.34 (0.61)(0.91)=0.56 (0.46)(3.05)=1.4

2402 2002 2402

8.00 10.99 32.96

1.01 0.46 1.52

8.09 5.05 50.10

totals

243.92

458.13 Februar y 14, 2011

5

Examples and Problems: (solution) Step 4: find the moment about the toe:

Examples and Problems: (solution) Step 7: the the max max pressure pressure at at the the toe: toe:

∑

= 458.13kN ⋅ m − (98.12 kN )(1.86 m ) + (24.53kN )(3.05m ) = 350.44kN ⋅ m ⎛ 243.92kN + 24.53kN ⎞⎛ ⎛ 0.22m ⎞ ⎞⎟ =⎜ ⎟⎜⎜1 + 6 ⎜ ⎟⎟ 3.05m ⎝ 3.05m ⎠ ⎠ ⎠⎝ ⎝ = 126.1kPa ≤ 143.5kPa, [OK]

Step 5: Location and eccentricity of the vertical force:

=

350.44 kN ⋅ m = 1.3m 243. 92kN + 24.53kN

= 3.05m 2 − 1.3m = 0.22m Step 6: this step skipped

Solution Done

Februar y 14 , 2011

Examples and Problems: (solution)

Examples and Problems: (Problems)

RSL = (∑ Wi + Ra ,v ) tan δ + cA B

( )

= (243.92 kN + 24.53kN )tan 17 + (45.5kPa )(3.05m ) 　　 = 220.85kN o

FSL ...