version 1.0 solved examples - shallow foundation analysis...

SoFA version 1.0

Konstantinos Nikolaou Dimitris Pitilakis

ShallOw Foundation Analysis Software

Aristotle University of ThessalonikiThessaloniki 2012

Solved Examples

SoFA version.1.1

Solved Examples Konstantinos Nikolaou

Dimitris Pitilakis

ShallOw Foundation Analysis software

Aristotle University of Thessaloniki

Thessaloniki 2013

http://www.sofasoftware.weebly.com

SoFA: Shallow Foundation Analysis Software Solved Examples

2

Contents

Contents .............................................................................................................................................. 2

Acknowledgements ............................................................................................................................. 3

Introduction ......................................................................................................................................... 3

Example # 1 – Rectangular foundation - Cohesive soil ...................................................................... 4

Example # 2 – Rectangular Foundation Cohesionless soil ................................................................. 8

Example # 3 – Foundation on Cohesionless soil .............................................................................. 13

Example # 4 – Earthquake Bearing Capacity ................................................................................... 18

Example # 5 – Settlement Calculation .............................................................................................. 23

Further Reading ................................................................................................................................ 27

Appendix - List of symbols ............................................................................................................... 28


3

Acknowledgements We thank Prof. Christos Anagnostopoulos from the Dept. of Civil Engineering of AUTh for the

scientific guidance. We thank Konstantinos Trevlopoulos (AUTh) for his very useful remarks. Also

we thank Ivan Kraus (University of Osijek) for his comments and for the contribution of the symbols

list and the third solved example.

Introduction This manual contains solved examples that were used to validate SoFA v.1.1.

For an advanced description of the algorithms and formulas used, consult the analytical user’s

manual.

If you discover what you think is a bug, report it here. Please try to include all SoFA reports.

!

All loads are considered to act at the base of the footing and NOT at the

theoretical point of column fixity (for more information check out the analytical

users' manual).

SoFA does NOT calculate the footing self weight.

http://sofasoftware.weebly.com/documentation.html



http://sofasoftware.weebly.com/bug-report.html


4

Example # 1 – Rectangular foundation - Cohesive soil Calculate the ultimate static bearing capacity of the shallow foundation depicted in fig. 1.

Figure 1

𝑒x =𝑀y

𝑉=

154.5

1545= 0.1 m & 𝑒y =

𝑀x

𝑉=

1931.25

1545= 1.25 m

𝐵x′ = 𝐿𝑥 − 2 ∗ 𝑒x = 2 − 2 ∗ 0.1 = 1.8 m

𝐵y′ = 𝐿y − 2 ∗ 𝑒y = 4 − 2 ∗ 1.25 = 1.5 m

𝐵 = min Bx′ , By

′ = 1.5 m & 𝐿 = 𝑚𝑎 𝑥 Lx′ , Ly

′ = 1.8 m

𝒅𝐰 < 𝒅𝐟

𝑝𝑜 = 𝛾 ∗ 𝑑𝑤 + 𝛾𝑠𝑎𝑡 ∗ 𝑑𝑓 − 𝑑𝑤 + 𝑞 = 19 ∗ 1 + 18 ∗ 2 − 1 + 0 = 37 kPa

For cohesive soils under undrained loading conditions according to Eurocode 7:

𝒒𝐮 = 𝟓. 𝟏𝟒 ∗ 𝒄𝐮 ∗ 𝒔𝐜 ∗ 𝒊𝐜 ∗ 𝒃𝐜 + 𝒑𝒐

𝑠c = 1 + 0.2 ∗𝐵

𝐿= 1 + 0.2 ∗

1.5

1.8= 1.17

𝐻 = 𝐻x2 + 𝐻y

2 = 202 + 1002 = 102 kN

𝑖c =1

2+

1

2∗ 1 −

𝐻

𝐵 ∗ 𝐿 ∗ 𝑐u=

1

2+

1

2∗ 1 −

102

1.5 ∗ 1.8 ∗ 100= 0.89

𝑞𝑢 = 5.14 ∗ 100 ∗ 1.17 ∗ 0.89 ∗ 1 + 37 = 572.2 kPa

𝑉𝑢 = 𝑞𝑢 ∗ 𝐵 ∗ 𝐿 = 572.2 ∗ 1.5 ∗ 1.8 = 1545 kN


5

Calculation performed using SoFA:

Figure 2

Click the Bearing Capacity button to access window showing the Safety Factory for Static Load Case:

Figure 3


6

Shallow Foundation Bearing Capacity

-------------------------------------------------------------

Kostis Nikolaou ([email protected])

Dimitris Pitilakis

Aristotle University of Thessaloniki - 2012

-------------------------------------------------------------

Geometry of the problem

* Dimentions(dx/dy) = 2.000 x 4.000 [m]

* Depth of foundation (df) = 2.000 [m]

* Depth of water level(dw) = 1.000 [m]

* Foundation base inclination(omega) = 0.000 [rad.]

* Soil inclination(beta) = 0.000 [rad.]

Design Loads - Static Load Case

* Vd = 1545.000 [kN]

* Hdx = 20.000 [kN]

* Hdy = 100.000 [kN]

* Mdx = 1931.250 [kNm]

* Mdy = 154.500 [kNm]

Soil Properties

* Type = C [C: cohessive CL: cohesionless]

* Loading= UN [D: drained UN: undrained]

* phik = 0.0 [deg.]

* ck = 0.0 [kPa] - drained shear strength

* cuk = 100.0 [kPa] - undrained shear strength

* soil Weight = 19.00 [kN/m^3]

=============================================================

Eccentricities (Static Load Case): ex = 0.10 ey = 1.25 [m]

Effective Dimentions: 1.500 x 1.800 [m]

Effective Area: 2.700 [m^2]

-------------------------------------------------------------

Bearing Capacity Check -- Undrained Conditions -- Static Load Case

-------------------------------------------------------------

* Eurocode 7 (2004)

sc=1.167 | ic=0.894 | bc=1.000 |

- qu_un = 573.359 [kPa]

- Vu = qu*B*L = 1548.069[kN]

- Vu/1.40 = qu*B*L/1.40 = 1105.764[kN] < 1545.000 [kN] - NOT ok -

- FS = qu/N*Aeff = 1.002

* EAK (2000)

sc=1.167 | ic=0.971 |

- qu_un = 618.989 [kPa]

- Vu = qu*B*L = 1671.272[kN]

- Vu/1.40= qu*B*L/1.40 = 1193.765[kN] < 1545.000 [kN] - NOT ok -

- FS = qu/N*Aeff = 1.082

* DIN4017 (2006)

sc=1.167 | ic=0.894 | bc=1.000 | gc=1.000 |

- qu_un = 573.359 [kPa]

- Vu = qu*B*L = 1548.069[kN]


7


- FS = qu/N*Aeff = 1.002

* Meyerhof (1953,1963)

sc=1.167 | ic=0.918 | dc=1.267 |

sq=1.000 | iq=0.918 | dq=1.000 |

- qu_un = 631.535 [kPa]

- Vu = qu*B*L = 1705.145[kN]


- FS = qu/N*Aeff = 1.104

* Hansen (1970)

sc=0.003 | ic=0.019 | bc=0.000 | gc=0.000 | dc=0.444 |

- qu_un = 734.360 [kPa]

- Vu = qu*B*L = 1982.771[kN]


- FS = qu/N*Aeff = 1.283


8

Example # 2 – Rectangular Foundation Cohesionless soil Calculate the ultimate static bearing capacity of the shallow foundation depicted in fig. 2.

Figure 4

𝑒x =𝑀y𝑦

𝑉=

900.8

4504= 0.2 m & 𝑒y =

𝑀x

𝑉=

900.8

4504= 0.2 m

𝐵x′ = 𝐿x − 2 ∗ 𝑒x = 2 − 2 ∗ 0.2 = 1.6 m

𝐵y′ = 𝐿y − 2 ∗ 𝑒y = 2 − 2 ∗ 0.2 = 1.6 m

𝐵 = min Bx′ , By

′ = 1.6m & 𝐿 = 𝑚𝑎 𝑥 Lx′ , Ly

′ = 1.6 m

𝒅𝐟 < 𝒅𝐰 < 𝒅𝐟 + 𝑩 = 𝟒 𝐦

𝑝𝑜 = 𝛾 ∗ 𝑑𝑓 + 𝑞.

𝑝𝑜 = 𝛾 ∗ 𝑑f + 𝑞 = 18 ∗ 1 + 19 ∗ 1 + 0 = 37 kPa

𝛾 ′ =𝛾 ∗ 𝑑w − 𝑑f + 𝛾sat − 𝛾w ∗ 𝑑f + 𝐵 − 𝑑w

𝐵

𝛾 ′ =19 ∗ 2.5 − 2 + 20 − 10 ∗ (2 + 1.6 − 2.5)

1.6= 12.81 kN/m3

For cohesionless soils under drained loading conditions, according to Eurocode 7:

𝑞u = 𝑐 ′ ∗ 𝑁c ∗ 𝑠c ∗ 𝑖c ∗ 𝑏c + 𝑝𝑜′ ∗ 𝑁q ∗ 𝑠q ∗ 𝑖q ∗ 𝑏q +1

2∗ 𝐵′ ∗ 𝛾 ′ ∗ 𝑁γ ∗ 𝑠γ ∗ 𝑖γ ∗ 𝑏γ

𝑁𝑞 = tan2 𝜋

4+

𝜙′

2 ∗ 𝑒𝜋∗𝑡𝑎𝑛𝜙 ′ = 𝑡𝑎𝑛2

𝜋

4+

35

2 ∗ 𝑒𝜋∗𝑡𝑎𝑛 35 = 33.3

𝑁𝛾 = 2 ∗ 𝑁q − 1 ∗ 𝑡𝑎𝑛𝜙′ = 2 ∗ 33.3 − 1 ∗ 𝑡𝑎𝑛35 = 45.2

𝑠𝑞 = 1 + 𝐵

𝐿∗ 𝑠𝑖𝑛ϕ′ = 1 +

1.6

1.6∗ 𝑠𝑖𝑛35 = 1.57


9

𝑠γ = 1 − 0.3 ∗𝐵

𝐿= 1 − 0.3 ∗

1.6

1.6= 0.7

𝐻 = 𝐻x2 + 𝐻y

2 = 450.42 + 450.42 = 636.96 𝑘𝑁

𝑚 = 𝑚𝐿 ∗ cos2 𝜃 + 𝑚𝐵 ∗ sin2 𝜃 = 1.5

𝑚𝐿 =2+𝐿/𝐵

1+𝐿/𝐵= 1.5 & 𝑚𝐵 =

2+𝐵/𝐿

1+𝐵/𝐿= 1.5

𝑖q = 1 −𝐻

𝑉 + 𝐵 ∗ 𝐿 ∗𝑐 ′

𝑡𝑎𝑛𝜙′

𝑚

= 1 −636.96

4504 + 0

1.5

= 0.795

𝑖𝛾 = 1 −𝐻

𝑉 + 𝐵 ∗ 𝐿 ∗𝑐 ′

𝑡𝑎𝑛𝜙′

𝑚+1

= 1 −636.96

4504 + 0

2.5

= 0.683

bq = b𝛾 = 1 − 𝜔 ∗ 𝑡𝑎𝑛𝜙′ 2

= 1.0

𝑞u = 0 + 1 ∗ 18 + 1 ∗ 19 ∗ 33.3 ∗ 1.57 ∗ 0.795 +1

2∗ 1.6 ∗ 12.81 ∗ 45.2 ∗ 0.7 ∗ 0683

= 1759.3 kPa

𝑉u = 𝑞u ∗ 𝐵 ∗ 𝐿 = 1759.3 ∗ 1.6 ∗ 1.6 = 4504 kN


Figure 5

Click the Bearing Capacity button to access the windows showing the Safety Factory for Static Load

Case and the Ultimate Bearing Capacity:


10

Figure 6

Figure 7


11


-------------------------------------------------------------


Dimitris Pitilakis


-------------------------------------------------------------








* Vd = 4504.000 [kN]

* Hdx = 450.400 [kN]

* Hdy = 450.400 [kN]

* Mdx = 900.800 [kNm]

* Mdy = 900.800 [kNm]

Soil Properties

* Type = CL [C: cohessive CL: cohesionless]

* Loading= D [D: drained UN: undrained]

* phik = 35.0 [deg.]




=============================================================




-------------------------------------------------------------

Bearing Capacity Check -- Drained Conditions -- Static Load Case

-------------------------------------------------------------

* Eurocode 7 (2004)

Nc=1.000 | sc=1.000 | ic=0.504 | bc=1.000 |

Ng=45.228 | sg=0.700 | ig=0.683 | bg=1.000 |

Nq=33.296 | sq=1.574 | iq=0.796 | bq=1.000 |

- qu_dr = 1761.196 [kPa]

- Vu = qu*B*L = 4508.662[kN]


- FS = qu/N*Aeff = 1.001

* EAK (2000)

Nc=1.000 | sc=1.000 | ic=1.000 |

Ng=45.228 | sg=0.700 | ig=0.815 |

Nq=33.296 | sq=1.700 | iq=0.852 |

- qu_dr = 2046.046 [kPa]

- Vu = qu*B*L = 5237.879[kN]

- Vu/1.40 = qu*B*L/1.40 = 3741.342 < 4504.000 [kN] - NOT ok -

- FS = qu/N*Aeff = 1.163 [kPa]

* DIN 4017 (2006)

Nc=1.000 | sc=1.000 | ic=1.000 | bc=1.000 | gc=1.000 |

Ng=45.228 | sg=0.700 | ig=0.683 | bg=1.000 | gg=1.000 |

Nq=33.296 | sq=1.574 | iq=0.796 | bq=1.000 | gq=1.000 |


12

- qu_dr = 1761.196 [kPa]

- Vu = qu*B*L = 4508.662[kN]


- FS = qu/N*Aeff = 1.001

* Meyerhof (1953,1963)

Nc=1.000 | sc=1.000 | ic=1.000 | dc=1.000 |

Ng=37.152 | sg=1.369 | ig=0.593 | dg=1.240 |

Nq=33.296 | sq=1.369 | iq=0.829 | dq=1.240 |

- qu_dr = 2112.816 [kPa]

- Vu = qu*B*L = 5408.809[kN]


- FS = qu/N*Aeff = 1.201

* Hansen(1970)

Nc=1.000 |

Ng=33.921 | sg=0.600 | ig=0.696 | bg=1.000 | gg=1.000 | dg=1.000 |

Nq=33.296 | sq=1.444 | iq=0.774 | bq=1.000 | gq=1.000 | dq=1.318 |

- qu_dr = 1957.807 [kPa]

- Vu = qu*B*L = 5011.985[kN]


- FS = qu/N*Aeff = 1.113


13

Example # 3 – Foundation on Cohesionless soil Calculate both the ultimate static bearing capacity and the safety factor for the strip foundation

founded on the layer of sand as depicted below.

Figure 8

The geometry of the foundation, the vertical load and the soil properties given here match the same of

the Camus 4 specimen tested within the framework of the Training and Mobility of Researchers of the

5th Topic “Shear Wall Structures” of the European Program ICONS (“Innovative Seismic Design

Concepts for New and Existing Structures”). The data used here is provided at http://www-

tamaris.cea.fr/media/resultats/CAMUS4/Camus4.pdf. The safety factor used for calculation of the

bearing capacity for the sand below the specimen was aimed to be between values of 2,34 and 3,24.

The foundation is not embedded into the sand.

For cohesionless soils under drained loading conditions, according to Eurocode 7:

bearing factors:

30,332

3545tan

2

'45tan 35tan2'tan2

eeNq

13,4635tan/130,33'tan/1 qc NN

23,4535tan130,332'tan12 qNN

shape factors:

22,135sin1,2

8,01'sin1

L

Bsq

23,1130,33

130,331

1

1

q

qq

cN

Nss

89,01,2

8,03,013,01

L

Bs

28,18,0/1,21

8,0/1,22

/1

/2

BL

BLmL

72,11,2/8,01

1,2/8,02

/1

/2

LB

LBmB

28,10sin72,10cos28,1sincos 2222 BL mmm

http://www-tamaris.cea.fr/media/resultats/CAMUS4/Camus4.pdf



14

inclination factors:

1

35tan

01,28,02,196

01

'tan

'1

28,1

m

q cLBV

Hi

1130,33

130,331

1

1

q

qq

cN

Nii

1

35tan

01,28,02,196

01

'tan

'1

28,111

m

cLBV

Hi

base inclination factors:

135tan01'tan122 bbq

1130,33

130,331

1

1

q

qq

cN

Nbb

bisNBbisNpobisNcq qqqqccccu ''5,0''

kPa 834,289834,28900189,023,45188,05,01122,130,3301123,113,460 uq

48,2

1,28,0/2,196

834,289

/

LBV

qSF u


15

Using SoFA:

Figure 9

Figure 10


16

Figure 11


-------------------------------------------------------------


Dimitris Pitilakis


-------------------------------------------------------------




* Depth of water level(dw) = Inf [m]




* Vd = 196.200 [kN]

* Hdx = 0.000 [kN]

* Hdy = 0.000 [kN]

* Mdx = 0.000 [kNm]

* Mdy = 0.000 [kNm]

Soil Properties



* phik = 35.0 [deg.]




=============================================================




-------------------------------------------------------------


-------------------------------------------------------------


17

* Eurocode 7 (2004)

Nc=1.000 | sc=1.000 | ic=1.000 | bc=1.000 |

Ng=45.228 | sg=0.886 | ig=1.000 | bg=1.000 |

Nq=33.296 | sq=1.219 | iq=1.000 | bq=1.000 |

- qu_dr = 288.425 [kPa]

- Vu = qu*B*L = 484.554[kN]

- Vu/1.00 = qu*B*L/1.00 = 484.554[kN] > 196.200 [kN] - ok -

- FS = qu/N*Aeff = 2.470

* EAK (2000)

Nc=1.000 | sc=1.000 | ic=1.000 |

Ng=45.228 | sg=0.886 | ig=1.000 |

Nq=33.296 | sq=1.267 | iq=1.000 |

- qu_dr = 288.425 [kPa]

- Vu = qu*B*L = 484.554[kN]

- Vu/1.00 = qu*B*L/1.00 = 484.554 > 196.200 [kN] - ok -

- FS = qu/N*Aeff = 2.470 [kPa]

* DIN 4017 (2006)

Nc=1.000 | sc=1.000 | ic=1.000 | bc=1.000 | gc=1.000 |

Ng=45.228 | sg=0.886 | ig=1.000 | bg=1.000 | gg=1.000 |

Nq=33.296 | sq=1.219 | iq=1.000 | bq=1.000 | gq=1.000 |

- qu_dr = 288.425 [kPa]

- Vu = qu*B*L = 484.554[kN]

- Vu/1.00 = qu*B*L/1.00 = 484.554[kN] > 196.200 [kN] - ok -

- FS = qu/N*Aeff = 2.470

* Meyerhof (1953,1963)

Nc=1.000 | sc=1.000 | ic=1.000 | dc=1.000 |

Ng=37.152 | sg=1.141 | ig=1.000 | dg=1.000 |

Nq=33.296 | sq=1.141 | iq=1.000 | dq=1.000 |

- qu_dr = 305.102 [kPa]

- Vu = qu*B*L = 512.571[kN]

- Vu/1.00 = qu*B*L/1.00 = 512.571[kN] > 196.200 [kN] - ok -

- FS = qu/N*Aeff = 2.612

* Hansen(1970)

Nc=1.000 |

Ng=33.921 | sg=0.848 | ig=1.000 | bg=1.000 | gg=1.000 | dg=1.000 |

Nq=33.296 | sq=2.506 | iq=1.000 | bq=1.000 | gq=1.000 | dq=1.000 |

- qu_dr = 207.015 [kPa]

- Vu = qu*B*L = 347.785[kN]

- Vu/1.00 = qu*B*L/1.00 = 347.785[kN] > 196.200 [kN] - ok -

- FS = qu/N*Aeff = 1.773


18

Example # 4 – Earthquake Bearing Capacity Calculate the ultimate earthquake bearing capacity of a 4mXinf strip foundation

Figure 12

For Cohesionless soil according to Eurocode 8 – part 5 - Annex F:

𝑎𝑣 = 0.5 ∗ 𝑆 ∗ 𝑎𝑔 ∗ 𝛾Ι = 0.5 ∗ 1.15 ∗ 2.943 ∗ 1.0 = 1.6922m

s2

𝜑d′ = tan−1

𝑡𝑎𝑛𝜑′

𝛾𝜑 = 𝑡𝑎𝑛−1

𝑡𝑎𝑛36′

1.25 = 30.167𝑜

𝑁𝑞 = 𝑡𝑎𝑛2 𝜋

4+

𝜙′

2 ∗ 𝑒𝜋∗𝑡𝑎𝑛𝜙 ′ = 𝑡𝑎𝑛2

𝜋

4+

30.2

2 ∗ 𝑒𝜋∗𝑡𝑎𝑛 30.2 = 18.75

𝑁𝛾 = 2 ∗ 𝑁𝑞 − 1 ∗ 𝑡𝑎𝑛𝜙′ = 2 ∗ 18.82 − 1 ∗ 𝑡𝑎𝑛30.2 = 20.64

𝑁mαx =1

2∗ 𝜌 ∗ 𝑔 ∗ 1 ±

𝑎v

𝑔 ∗ 𝐵2 ∗ 𝑁γ

𝑁mαx =1

2∗ 1650 ∗ 𝑔 ∗ 1 −

1.6922

𝑔 ∗ 42 ∗ 20.64 = 2211.67 kN

𝐹 =𝑎𝑔 ∗ 𝑆

𝑔 ∗ 𝑡𝑎𝑛ϕd′

=0.3𝑔 ∗ 1.15

𝑔 ∗ 𝑡𝑎𝑛30.2o= 0.594


19

𝑁 =

𝛾Rd ∗ 𝑁Ed

𝑁mαx=

1 ∗ 4504

2211.67= 2.037

𝑉 =𝛾Rd ∗ 𝑉Ed

𝑁mαx=

1 ∗ 450.4

2211.67= 0.204

𝑀 =𝛾Rd ∗ 𝑀Ed

𝐵 ∗ 𝑁mαx=

1 ∗ 4493

4 ∗ 2211.67= 0.102

𝛷 𝑁 , 𝑉 , 𝛭 , 𝐹 = 1 − 𝑒 ∗ 𝐹 𝐶T ∗ 𝛽 ∗ 𝑉 𝐶T

𝑁 𝛼 ∗ 1 − 𝑚 ∗ 𝐹 𝑘 𝑘 ′− 𝑁

𝑏 + 1 − 𝑓 ∗ 𝐹 𝐶M

′

∗ 𝛾 ∗ 𝛭 𝐶Μ

𝑁 𝑐 ∗ 1 − 𝑚 ∗ 𝐹 𝑘 𝑘 ′− 𝑁

𝑑 − 1 > 0

Foundation is UNSAFE


Figure 13


20

Click the Calculate button to access the window Input for Eurocode 8 – Annex F criterion:

Figure 14

You can modify Nmax or F calculated value at will. Click the Bearing Capacity button to access the

window showing the Safety Factory for Earthquake Load Case:

Figure 15


21


-------------------------------------------------------------


Dimitris Pitilakis


-------------------------------------------------------------


* Dimentions(dx/dy) = 4.000 x Inf [m]


* Depth of water level(dw) = Inf [m]




* Vd = 1.000 [kN]

* Hdx = 0.000 [kN]

* Hdy = 0.000 [kN]

* Mdx = 0.000 [kNm]

* Mdy = 0.000 [kNm]

Design Loads - Earthquake Load Case

* Ved = 4504.000 [kN]

* Hex = 450.400 [kN]

* Hey = 0.000 [kN]

* Mex = 0.000 [kNm]

* Mey = 900.800 [kNm]

Soil Properties



* phik = 30.2 [deg.]




=============================================================


Effective Dimentions: 4.000 x 1[m] - Strip


-------------------------------------------------------------


-------------------------------------------------------------

* Eurocode 7 (2004)

Nc=1.000 | sc=1.000 | ic=1.000 | bc=1.000 |

Ng=20.637 | sg=1.000 | ig=1.000 | bg=1.000 |

Nq=18.753 | sq=1.000 | iq=1.000 | bq=1.000 |

- qu_dr = 1275.432 [kPa]

- Vu = qu*B*L = 5101.727[kN]

- Vu/1.40 = qu*B*L/1.40 = 3644.091[kN] > 1.000 [kN] - ok -

- FS = qu/N*Aeff = 5101.727

* EAK (2000)

Nc=1.000 | sc=1.000 | ic=1.000 |

Ng=20.637 | sg=1.000 | ig=1.000 |

Nq=18.753 | sq=1.000 | iq=1.000 |


22

- qu_dr = 1275.432 [kPa]

- Vu = qu*B*L = 5101.727[kN]

- Vu/1.40 = qu*B*L/1.40 = 3644.091 > 1.000 [kN] - ok -

- FS = qu/N*Aeff = 5101.727 [kPa]

* DIN 4017 (2006)

Nc=1.000 | sc=1.000 | ic=1.000 | bc=1.000 | gc=1.000 |

Ng=20.637 | sg=1.000 | ig=1.000 | bg=1.000 | gg=1.000 |

Nq=18.753 | sq=1.000 | iq=1.000 | bq=1.000 | gq=1.000 |

- qu_dr = 1275.432 [kPa]

- Vu = qu*B*L = 5101.727[kN]

- Vu/1.40 = qu*B*L/1.40 = 3644.091[kN] > 1.000 [kN] - ok -

- FS = qu/N*Aeff = 5101.727

* Meyerhof (1953,1963)

Nc=1.000 | sc=1.000 | ic=1.000 | dc=1.000 |

Ng=16.116 | sg=1.000 | ig=1.000 | dg=1.087 |

Nq=18.753 | sq=1.000 | iq=1.000 | dq=1.087 |

- qu_dr = 1227.150 [kPa]

- Vu = qu*B*L = 4908.599[kN]

- Vu/1.40 = qu*B*L/1.40 = 3506.142[kN] > 1.000 [kN] - ok -

- FS = qu/N*Aeff = 4908.599

* Hansen(1970)

Nc=1.000 |

Ng=15.478 | sg=1.000 | ig=1.000 | bg=1.000 | gg=1.000 | dg=1.000 |

Nq=18.753 | sq= Inf | iq=1.000 | bq=1.000 | gq=1.000 | dq=1.000 |

- qu_dr = 1195.723 [kPa]

- Vu = qu*B*L = 4782.894[kN]

- Vu/1.40 = qu*B*L/1.40 = 3416.353[kN] > 1.000 [kN] - ok -

- FS = qu/N*Aeff = 4782.894

-------------------------------------------------------------

Bearing Capacity Check -- Earthquake Load Case

-------------------------------------------------------------

* EAK (2000)

Nc=0.000 | sc=1.000 | ic=1.000 |

Ng=601.400 | sg=20.637 | ig=1.000 |

Nq=607.210 | sq=18.753 | iq=1.000 |

- qu_dr = 1208.610 [kPa]

- Vu = qu*B*L = Inf[kN]

- Vu/1.3 = qu*B*L/1.3 = Inf > 4504.000 [kN] - ok -

- FS = qu/N*Aeff = Inf [kPa]

* Eurocode 8_5 (2004)- Annex F(Informative)

shape factor sc = 1.00

Ntot = Nmax*L*sc = 2211.34

X direction

- n = grd*Ned/Ntot = 2.0368

- v = grd*Ved/Ntot = 0.2037

- m = grd*Med/(B*Ntot) = 0.1018

- F = 0.5936

- c = 99.0000 > 0.00 - NOT ok -

- SF = 0.1934

Y direction

- n = grd*Ned/Ntot = 2.0368

- v = grd*Ved/Ntot = 0.0000

- m = grd*Med/(B*Ntot) = 0.0000

- F = 0.5936

- c = 99.0000 > 0.00 - NOT ok -

- SF = 0.3533


23

Example # 5 – Settlement Calculation Calculate the settlements for example 1.

For a Cohesive soil both immediate and settlements due to consolidation have to be calculated

A. Immediate Settlements

𝛥𝐻im = 𝐼 ∗ 𝑝 ∗ 𝐵 ∗1 − 𝑣2

𝐸

I = 0.770 (tables from NAVFAC [14])

𝑝 =1545

2 ∗ 4= 193.125 kPa

B = 2 m

v = 0.45 Poisson’s ratio

E = 15 MPa

𝛥𝐻im = 15.812 mm

B. Due to Consolidation

We define four points and calculate the stresses

Layer

Depth

z

Initial

stress Is

Load

stress

Total

stress

OCR

stress Thickness Settlement

[m] [kPa] [kPa] [kPa] [m] [mm]

1 2.5 31 0.646 124.73 155.73 465 1 21.76

2 3.5 39 0.385 74.35 113.35 585 1 14.38

3 5 51 0.208 40.21 91.21 765 2 15.67

4 8 75 0.082 15.89 90.89 1125 4 10.36

62.17

Load stresses and Is factor are calculated using the formulas described in Bowles [5], chapter 5.

Integrating Boussinesq equation over a rectangle of dimentions B x L (Newmark 1935 [15]) gives:

𝑞v = 𝑞0 ∗ 𝐼s

𝐼s =1

4𝜋 2 ∗ 𝑀 ∗ 𝑁 ∗ 𝑉

𝑉 + 𝑉1

𝑉 + 1

𝑉+ 𝑡𝑎𝑛−1 2 ∗ 𝑀 ∗ 𝑁 ∗

𝑉

𝑉 − 𝑉1

𝑀 =𝐵

𝑧 & 𝑁 =

𝐿

𝑧 (𝐼s = 1 𝑓𝑜𝑟 𝑧 = 0 m)

𝑉 = 𝑀2 + 𝑁2 + 1

𝑉1 = 𝑀 ∗ 𝑁 2

Is is calcutated for point A.


24

Settlements due to consolidation are calculated:

If 𝜍tot′ > 𝜍’OCR

𝛥𝛨i = 𝐻i ∗0.1 ∗ 𝐶C

1 + 𝑒0∗ log

𝜍OCR′

𝜍init′

+ 𝐻i ∗𝐶C

1 + 𝑒0∗ log

𝜍tot′

𝜍OCR′

If 𝜍tot′ < ζ’OCR

𝛥𝛨i = 𝐻i ∗0.1 ∗ 𝐶C

1 + 𝑒0∗ log

𝜍tot′

𝜍init′

Total Settlements: 15.8 + 62.2 = 78.0 mm

Figure 16


All other data is the same as in the Example #1


25

Figure 17

Shallow Foundation Settlements

--------------------------------------------


Dimitris Pitilakis


--------------------------------------------





* Maximum depth of interest zmax = 10.000 [m]

* Stratum thickness H = Inf [m]

Loads

* Vd/Area = 193.125 [kPa]

Soil Properties

* modulus of elasticity E = 15000.000 [kPa]

* Poison ratio n = 0.450 [-]

------------------------------

Immediate Settlements

------------------------------

I = 0.770 [-]

- DHimed = I*p*B*(1-v^2)/E = 15.812 [mm]

10 15 20 25 30 35 40-10

-9

-8

-7

-6

-5

-4

-3

-2

w(z) [mm]

z [

m]

Settlements

Immediate

Due to consolidation

Total


27

Further Reading Greek

[1] Αναγνωζηόποςλορ Χ., Χαηδεγώγορ Θ., Αναζηαζιάδερ Α., Πιηιλάκερ Δ.(2012).” Θεμελιώζειρ,

ανηιζηεπίξειρ και γεωηεσνικά έπγα”. Εκδόζειρ Αϊβάδε, Θεζζαλονίκε

[2] Γεωπγιάδερ Κ., Γεωπγιάδερ Μ. (2009), "Σηοισεία Εδαθομεσανικήρ", εκδόζειρ ΖΗΤΗ,

Θεζζαλονίκε

[3] EAK 2000 (2000), “Ελλενικόρ Ανηιζειζμικόρ Κανονιζμόρ”

[4] Πιηιλάκερ Κ., Γεωπγιάδερ Μ., Μπανηήρ Σ., Χαηδεγώγορ Θ., Αναγνωζηόποςλορ Χ., Τίκα Θ.

(1999), "Ανηιζειζμικόρ Σσεδιαζμόρ Θεμελιώζεων, Ανηιζηεπίξεων και Γεωκαηαζκεςών", Α.Π.Θ.

Πανεπιζηεμιακέρ Σεμειώζειρ ΑΣΤΕ, Θεζζαλονίκε

English

[5] Bowles J.E. (1997), "Foundation Analysis and Design", 5th edition, McGrow-Hill, New York

[6] Bond Α. (2008), "Decoding Eurocode 7", Taylor & Francis

[7] CE de Normalisation (1998), “Eurocode 7 – Eurocode 7: Geotechnical design – Part 1: General

rules”

[8] CE de Normalisation (1998), “Eurocode 8 – Design of Structures for earthquake resistance–Part 1:

General rules, seismic actions and rules for buildings”

[9] Combescure, D. and. Chaudat, Th. (2000) “ICONS european program seismic tests on r/c walls

with uplift; camus iv specimen”, Individual Study, ICONS project Rapport, SEMT/EMSI/RT/00-27/4,

CEA, Direction des Réacteurs Nucléaires, Département de Mécanique et de Technologie, 2000, Paris,

France. Available at: http://www-tamaris.cea.fr/media/resultats/CAMUS4/Camus4.pdf

[10] Deutsche Norm (2006), “DIN 4017 Soil: Calculation of design bearing capacity of soil beneath

shallow foundations”

[11] Frank R., Bauduin C., Driscoll R., Kavvadas M., Krebs Ovesen N., Orr T. and Schuppener B.

(2004) ,“ Designers' Guide to EN 1997-1 Eurocode 7: Geotechnical Design - General Rules”, Thomas

Telford

[12] Hansen J. B. (1970), “A Revised and Extended Formula for Bearing Capacity” Danish

Geotechnical Institute, Copenhagen, bulletin No. 28.

[13] Meyerhof G. G. (1963), “Some recent research on the bearing capacity of foundations” Canadian

Geo-technical Journal,No. 1, 16±26.

[14] Meyerhof G. G. (1953), “The bearing capacity of foundations under eccentric or inclined loads”

3rd

Int. Conf. on Soil Mech. and Found., Zurich,,Vol.1,pp.440-445

[15] Naval Facilities Eng. Command – NAVFAC (1986), ”Foundations and earth structures Design

manual 7.2 ”

[16] Newmark, N.M., (1935) “Simplified computation of vertical stress below foundations.”,Univ. of

Illinois Engineering Experiment Station, Circular 24, Urbana, Illinois, 19 p. (as referenced by Holtz

and Kovacs, 1981).

[17] Smith Θ. (2006), "Smith's Elements of Soil Mechanics", Wiley

[18] Terzaghi K., Peck P.B. (1967), “Soil Mechanics in Engineering practice”, John Wiley and Sons,

New York



28

Appendix - List of symbols

Latin letters

av design ground acceleration in the vertical direction

ag design ground acceleration on type A ground

Aeff effective area of the footing

bi the design values of the factors for the inclination of the base, with subscripts c for

cohesion , q for surcharge and γ for weight density

B the foundation width

B' the effective foundation width

c soil cohesion

c' cohesion of soil in terms of effective stress

cu undrained shear strength of soil

c'k cohesion (for drained loading)

Cc compression index

di the design values of the factors for the depth of foundation, with subscripts c for


df depth of foundation

dw depth of water level

eo gap percentage

e the eccentricity of the resultant action

E Young’s modulus of elasticity

F seismic inertia force (dimensionless)

FS factor of safety

gi the design values of the factors for the soil inclination, with subscripts c for


H stratum thickness

Hx horizontal force in x direction

Hy horizontal force in y direction

ii the inclination factor of the load, with subscripts c for cohesion , q for surcharge and

γ for weight density

Is influence factor

L the foundation length


29

L’ the effective foundation length

m exponent in formulas for the inclination factor i

Mx moment about xx axis

My moment about yy axis

N the bearing capacity factors, with subscripts cohesion c, surcharge q and weight

density γ

Nmax ultimate bearing capacity of the foundation under a vertical load

q overburden or surcharge pressure at the level of the foundation base

q’ the design effective overburden pressure at the level of the foundation base

qu bearing capacity of the shallow foundation

si the shape factors of the foundation base, with subscripts c for cohesion , q for

surcharge and γ for weight density

S soil factor as defined in EN 1998-1

V the vertical load

Vu ultimate vertical load

p actual soil pressure

po loading on foundation depth

po’ effective overburden pressure

z vertical distance (depth)

Greek letters

α the inclination of the foundation base to the horizontal

ß soil inclination according to DIN 4017

γ weight of the soil under the foundation

γ’ the design effective weight of the soil below the foundation level

γI importance factor

γs dry soil unit weight

γsat saturated soil unit weight

ΔH the settlement of the corner of a rectangular base

ν Poisson’s ratio

Φ angle of internal friction

Φ'k effective angle of internal friction

Φuk friction angle for undrained cohesive soils


30

σ'init initial stress

σ'OCR OCR stress

σ'tot total stress

Abbreviations

OCR over-consolidation ratio

For calculations, the following units or their multiples are recommended:

force [kN]

mass [kg]

mass density [kg/m3]

moment [kNm]

pressure [kPa]

stiffness [kPa]

strength [kPa]

stress [kPa]

weight density [kN/m3]

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