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6StripandSpreadFoundations3.ppt

Geotechnical Engineering

6

Strip and Spread Foundations

Contents:

Introduction

1.1 Types of foundations

1.2 Footing depth

1.3 Settlement

1.4 Bearing capacity

Vertical stress distribution

2.1 Simple spread assumption

2.2 Bousinesq’s ‘surface point load’

2.3 Fadum’s ‘uniformly distributed rectangular load’

2.4 Newmark’s chart(s)

2.5 Uniform strip load

2.6 Bulb of pressure

1. Introduction

  • Foundation provides a critical interface between a structure and the ground beneath it – structural integrity of both sides must be considered
  • Three main types: strip, pad & raft
  • Three main design criteria
  • footing (foundation) depth
  • settlement
  • bearing capacity

1.1 Types of foundations

1.2 Design criteria:

1.2.1 Footing depth

At a depth as to prevent failure due to changes in surface conditions:

  • Water flow – removal of fine soil
  • Frost heave – (uniform fine sands and silts, etc) 450mm frost => 600mm foundation depth
  • Shrinkage and swelling – action of tree roots

New foundation outside the zone of influence

1.2.2 Settlement (total and differential)

  • Total settlement
  • Amount and rate is important (small settlement rapidly = more damaging than larger movements over a longer time period)
  • Differential settlement
  • Relative settlement between different parts of a structural

Total settlement = no problem

Differential settlement = problem

1.2.3 Bearing capacity

  • Bearing capacity is the average contact pressure between the footing and the soil which will produce shear failure in the soil divided by FOS (>3)

plastic

elastic

Steel

Soil

2. Vertical stress distribution

Assume that the soil is:

  • Homogeneous
  • Elastic (i.e. stress proportional to strain)
  • Elastically isotropic (Ex=Ey=Ez)

Basic methods:

  • Simple spread assumption
  • Boussinesq’s ‘surface point load’ solution
  • Fadum’s ‘uniformly distribution rectangular load’ solution
  • Newmark’s chart(s)
  • Uniform strip load solution

2.1 Simple spread assumption

Large loaded areas on thin beds of soil e.g. raft foundations, embankments, etc

Example 1: Simple spread assumption

Estimate the average vertical stress at 5 m depth below a 2 m square foundation, load 200 kN/m2.

200 kN/m2 x 2 m x 2 m = 800 kN

Stress on foundation x Area of foundation = load on foundation

800 kN / (7 m x 7 m) = 16.3 kN/m2

Example 2: Simple spread assumption

Estimate the average vertical stress at 5 m depth below a strip foundation 1.5 m, load 90 kN/m2.

90 kN/m2 x 1.5 m = 135 kN/m

Stress on foundation x width of foundation = load on foundation per unit length

135 kN/m / 6.5 m = 21 kN/m2

2.2 Bousinesq ‘surface point load’ solution

  • Single point load / Semi-infinite / Elastic Medium
  • Polar co-ordinates

for vertical stress

where Iz (influence factor)

Iz can be plotted against r/z

Example 3a: 1m square foundation point load 50kN, Calculate
(a) stress 4m directly beneath the centre

  • Points below the load

r = 0

and at all depths

r/z = 0

from graph Iz vs r/z

σ = 0.4775 x 50 kN = 1.49 kN/m2

42 m2

Example 3: A 1m square foundation point load 50 kN, Calculate
(b) a point 3m below the foundation offset by a distance of 5m

b) Points below the load

r = 5 m

and at

r/z = 5/3 = 1.667

from graph Iz vs r/z

s = 0.017 x 50 kN = 0.094 kN/m2

32 m2

0.017

Bousinesq’s method is only for a point load, however it can be extended by the principle of superposition to cover the case of a foundation exerting a uniform pressure:

Example 4: Using the theory of superposition find the stress at a depth of 3 m beneath the corner of a 3 m square raft which carries a UDL of 200 kN/m2

Divide area into number of equal sections

Total vertical stress at that point = sum of all the stresses

Calculate the load due to stress acting on that section

200 kN/m2 x 1.5 m x 1.5 m (area of small section) = 450 kN

Assume the load acts at the centre of each section

Calculate the vertical stress at the point in question for each section

A

r at point ‘A’

B

r at point ‘B’

1.06 x 3 = 3.18 m

C

D

r at point ‘C & D’

1.5 m + 0.75 m = 2.25 m

0.75 m

At Point A r = 1.06m

At Point B r = 3.18m

At Point C & D r = 2.37m

s = 0.356 x450 kN = 17.8 kN/m2

32 m2

s = 0.073 x 450 kN = 3.65 kN/m2

32 m2

s = 0.142 x 450 kN = 7.1 kN/m2

32 m2

Total stress = 17.8+3.65+7.1+7.1

= 35.7 kN/m2

2.3 Fadum‘Uniformly distributed rectangular load’ solution

Fadum (1948) developed influence factors for the distribution of vertical stress under the corner of a uniformly loaded rectangular foundation length = L, Breath =B

Using above method and superposition it is possible to calculate the vertical stress beneath a wide range of foundation shapes

Example 5: A 4 m x 3 m rectangular foundation, UDL of 100 kN/m2.
Calculate the vertical stress
(a) 2 m beneath the corner of the foundation

0.22

=22 kN/m2

Example 5: A 4 m x 3 m rectangular foundation, UDL of 100 kN/m2.
Calculate the vertical stress
(b) 2 m beneath the centre of the foundation

0.152

=60.8 kN/m2

Example 6: UDL of 100 kN/m2, Calculate the vertical stress (a) Z and (b) Z1 at a depth of 4 m

=31.7 kN/m2

Area 1

Area 2

Area 3

Iz=0.06

Iz=0.132

Iz=0.125

1 m

3 m

3 m

Z1

1 m

2 m

Iz=0.178

Iz=0.075

Iz=0.072

Iz=0.06

subtract

subtract

plus

equals

=11.5 kN/m2

6(c) Two foundations ‘A’ and ‘B’ each 2.5 m square are situated 4 m apart, as shown below, the load on the centre of each foundation is 1000 kN.

(i) Using Boussinesq’s point load solution calculate the increase in vertical stress at a depth of 5 m under the centre of footing ‘A’ due to its load and the load carried by footing ‘B’.

(ii) Using Fadum’s graphical solution calculate the increase in vertical stress at a depth of 5 m under the corner marked X on footing ‘A’ due to its load and the load carried by footing ‘B’.

Boussinesq’s point load solution

or

= 0.4775

For point A r = 0

= 0.04

or

Iz=0.04

= 20.7 kN/m2

r

Equivalent UDL =

160 kN/m2

Fadum’s graphical solution

Stress at 5 m beneath ‘X’ from foundation ‘A’

Stress at 5 m beneath ‘X’ from

foundation ‘B’ & area ‘C’

Stress at 5 m beneath ‘X’ for area ‘C’

C

0.085

0.126

0.11

=16.2 kN/m2

2.4 Newmark’s Chart(s)

Newmark (1942) developed a method which can cater for all foundation shapes – ‘influence diagram’

Vertical stress at the origin of the chart

where N=number of blocks

P=load per unit area

  • The influence of each of the small areas given on the chart are the same
  • Loaded area drawn onto the chart at correct scale
  • Iz = 0.001 for each block within the foundation area
  • Scale the foundation should be drawn is such that the distance A – B on the chart is equivalent to the depth z at which the stress is to be determined

Example 7: UDL of 100 kN/m2, Calculate the vertical stress Z and Z1 at a depth of 4 m

=29.5 kN/m2

For Z1

Number of blocks approx = 120

=12 kN/m2

1 m

3 m

3 m

Z1

1 m

2 m

2.5 Uniform strip load solution

Vertical stress at any point beneath a strip foundation can be calculated from:

for simplicity the above expression can be re-written as:

Standard values of Is are given in the following table

Standard values of Is have been calculated for varying values of x and z

Example 8: A 1 m wide strip foundation is to support a UDL of 90 kN/m2. Estimate the vertical stress 4 m beneath
a) the centre of the strip

=14.2 kN/m²

Example 8: A 1m wide strip foundation is to support a UDL of 90 kN/m2. Estimate the vertical stress 4 m beneath
b) the edge of the strip

=13.8 kN/m²

Example 8: A 1 m wide strip foundation is to support a UDL of 90 kN/m2. Estimate the vertical stress 4 m beneath
c) 1 m outside the edge of the strip

=11.0 kN/m²

2.6 Bulb of Pressure

  • Use contours to view the distribution of stress ‘isobars’ or ‘pressure bulb’
  • The size of pressure bulb is proportional to the size of the loaded area – settlement of foundation

Total se)lement = no problem

Differen3al se)lement = problem

Iz = 3 2π

1

1+ r z[ ]2( ) 5 2

I

z

=

3

2p

1

1+rz

[]

2

()

5

2

017

.

0

3

5

1

1

2

3

2

5

2

=

÷

÷

ø

ö

ç

ç

è

æ

ú

û

ù

ê

ë

é

+

=

p

z

I

06

.

1

75

.

0

75

.

0

2

2

=

+

37

.

2

75

.

0

25

.

2

2

2

=

+

Iz = 3 2π

1

1+ 1.06 3

⎣⎢ ⎤

⎦⎥

2⎛

⎝ ⎜⎜

⎠ ⎟⎟

5 2

= 0.356

I

z

=

3

2p

1

1+

1.06

3

é

ë

ê

ù

û

ú

2

æ

è

ç

ç

ö

ø

÷

÷

5

2

=0.356

073

.

0

3

18

.

3

1

1

2

3

2

5

2

=

÷

÷

ø

ö

ç

ç

è

æ

ú

û

ù

ê

ë

é

+

=

p

z

I

142

.

0

3

37

.

2

1

1

2

3

2

5

2

=

÷

÷

ø

ö

ç

ç

è

æ

ú

û

ù

ê

ë

é

+

=

p

z

I

Iz = 3 2π

1

1+ r z[ ]2( ) 5 2

I

z

=

3

2p

1

1+rz

[]

2

()

5

2

z

L

m

=

z

B

n

=

P

I

z

=

s

2

2

4

=

=

=

z

L

m

5

.

1

2

3

=

=

=

z

B

n

σ = IzP = 0.22×100 kN /m 2

s=I

z

P=0.22´100kN/m

2

1

2

2

=

=

=

z

L

m

75

.

0

2

5

.

1

=

=

=

z

B

n

σ = IzP = 4×0.152×100 kN /m 2

s=I

z

P=4´0.152´100kN/m

2

75

.

0

4

3

=

=

=

z

L

m

25

.

0

4

1

=

=

=

z

B

n

σ =∑ IzP = (0.06+ 0.132+ 0.125)×100 kN /m2

s=

å

I

z

P=(0.06+0.132+0.125)´100kN/m

2

1

4

4

=

=

=

z

L

m

5

.

0

4

2

=

=

=

z

B

n

75

.

0

4

3

=

=

=

z

L

m

5

.

0

4

2

=

=

=

z

B

n

75

.

1

4

7

=

=

=

z

L

m

75

.

0

4

3

=

=

=

z

B

n

75

.

1

4

7

=

=

=

z

L

m

25

.

0

4

1

=

=

=

z

B

n

5

.

1

4

6

=

=

=

z

L

m

25

.

0

4

1

=

=

=

z

B

n

75

.

0

4

3

=

=

=

z

L

m

25

.

0

4

1

=

=

=

z

B

n

σ =∑ IzP = (0.178−0.075+0.072−0.06)×100 kN /m2

s=

å

I

z

P

=(0.178-0.075+0.072-0.06)´100kN/m

2

5

22

31

2

1

z

I

r

z

p

éù

ëû

=

æö

éù

+

ç÷

ëû

èø

2

vz

P

I

z

s

=

5

22

31

2

0

1

5

z

I

p

éù

ëû

=

æö

éù

+

ç÷

ëû

èø

5

22

31

2

6.5

1

5

z

I

p

éù

ëû

=

æö

éù

+

ç÷

ëû

èø

0

5

r

z

=

6.5

5

r

z

=

1.3

=

0.4775

=

2

2

1000

0.4775

5

1000

0.04

5

v

s

æö

´+

ç÷

èø

´

=

æö

ç÷

èø

(

)

1000

2.52.5

=

´

L

mn

z

===

L

m

z

==

B

n

z

==

L

m

z

==

B

n

z

==

2.5

0.5

5

=

6.5

1.3

5

=

2.5

0.5

5

=

4

0.8

5

=

2.5

0.5

5

=

(0.12

(0.11160

(0.085160)

6

)

160)

z

IP

s

-

´

=

´

=

+

´

Np

z

001

.

0

=

s

100

295

001

.

0

´

´

=

z

s

100

120

001

.

0

´

´

=

z

s

P

I

s

z

=

s

(

)

[

]

b

a

b

b

p

s

+

+

=

2

cos

sin

p

z

(

)

[

]

p

b

a

b

b

+

+

=

2

cos

sin

s

I

x/b z/b 0 0.4 0.8 1 1.25 1.5 2 3 5 10 0.0 1.000 1.000 1.000 0.500 0.000 0.000 0.000 0.000 0.000 0.000 0.2 0.997 0.992 0.909 0.500 0.059 0.011 0.002 0.000 0.000 0.000 0.4 0.977 0.955 0.773 0.498 0.178 0.059 0.011 0.001 0.000 0.000 0.6 0.937 0.896 0.691 0.495 0.258 0.120 0.030 0.004 0.000 0.000 0.8 0.881 0.829 0.638 0.489 0.305 0.173 0.056 0.010 0.001 0.000 1.0 0.818 0.766 0.598 0.480 0.332 0.214 0.084 0.017 0.002 0.000 1.2 0.755 0.707 0.564 0.468 0.347 0.243 0.111 0.026 0.004 0.000 1.4 0.696 0.653 0.534 0.455 0.354 0.263 0.135 0.037 0.005 0.000 1.6 0.642 0.605 0.566 0.440 0.356 0.276 0.155 0.048 0.008 0.000 1.8 0.593 0.563 0.497 0.425 0.353 0.284 0.172 0.060 0.010 0.001 2.0 0.550 0.524 0.455 0.409 0.348 0.288 0.185 0.071 0.013 0.000 2.5 0.462 0.445 0.400 0.370 0.328 0.285 0.205 0.095 0.022 0.002 3.0 0.396 0.385 0.355 0.334 0.305 0.274 0.211 0.114 0.032 0.003 3.5 0.345 0.338 0.317 0.302 0.281 0.258 0.210 0.127 0.042 0.004 4.0 0.306 0.301 0.285 0.275 0.259 0.242 0.205 0.134 0.051 0.006 5.0 0.248 0.245 0.237 0.231 0.222 0.212 0.188 0.139 0.065 0.010 6.0 0.208 0.207 0.202 0.198 0.192 0.186 0.171 0.136 0.075 0.015 8.0 0.158 0.157 0.155 0.153 0.150 0.147 0.140 0.122 0.083 0.025 10 0.126 0.126 0.125 0.124 0.123 0.121 0.117 0.107 0.082 0.032 15 0.085 0.085 0.084 0.084 0.083 0.083 0.087 0.078 0.069 0.041 20 0.064 0.064 0.063 0.063 0.063 0.063 0.062 0.061 0.056 0.041 50 0.025 100 0.013

x/b

z/b00.40.811.251.523510

0.01.0001.0001.0000.5000.0000.0000.0000.0000.0000.000

0.20.9970.9920.9090.5000.0590.0110.0020.0000.0000.000

0.40.9770.9550.7730.4980.1780.0590.0110.0010.0000.000

0.60.9370.8960.6910.4950.2580.1200.0300.0040.0000.000

0.80.8810.8290.6380.4890.3050.1730.0560.0100.0010.000

1.00.8180.7660.5980.4800.3320.2140.0840.0170.0020.000

1.20.7550.7070.5640.4680.3470.2430.1110.0260.0040.000

1.40.6960.6530.5340.4550.3540.2630.1350.0370.0050.000

1.60.6420.6050.5660.4400.3560.2760.1550.0480.0080.000

1.80.5930.5630.4970.4250.3530.2840.1720.0600.0100.001

2.00.5500.5240.4550.4090.3480.2880.1850.0710.0130.000

2.50.4620.4450.4000.3700.3280.2850.2050.0950.0220.002

3.00.3960.3850.3550.3340.3050.2740.2110.1140.0320.003

3.50.3450.3380.3170.3020.2810.2580.2100.1270.0420.004

4.00.3060.3010.2850.2750.2590.2420.2050.1340.0510.006

5.00.2480.2450.2370.2310.2220.2120.1880.1390.0650.010

6.00.2080.2070.2020.1980.1920.1860.1710.1360.0750.015

8.00.1580.1570.1550.1530.1500.1470.1400.1220.0830.025

100.1260.1260.1250.1240.1230.1210.1170.1070.0820.032

150.0850.0850.0840.0840.0830.0830.0870.0780.0690.041

200.0640.0640.0630.0630.0630.0630.0620.0610.0560.041

500.025

1000.013

0

5

.

0

0

=

=

b

x

8

5

.

0

4

=

=

b

z

90

158

.

0

´

=

=

P

I

s

z

s

1

5

.

0

5

.

0

=

=

b

x

90

153

.

0

´

=

=

P

I

s

z

s

3

5

.

0

5

.

1

=

=

b

x

90

122

.

0

´

=

=

P

I

s

z

s