version 1.0 Solved Examples -...
Transcript of version 1.0 Solved Examples -...
SoFA version 1.0
Konstantinos Nikolaou Dimitris Pitilakis
Ivan Kraus
ShallOw Foundation Analysis Software
Aristotle University of ThessalonikiThessaloniki 2012
Solved Examples
SoFA version.1.1
Solved Examples Konstantinos Nikolaou
Dimitris Pitilakis
Ivan Kraus
ShallOw Foundation Analysis Software
Aristotle University of Thessaloniki
Thessaloniki 2013
SoFA Solved Examples Shallow Foundation Analysis Software
2
Contents
Contents .................................................................................................................................................... 2
Acknowledgements ................................................................................................................................... 3
Introduction ............................................................................................................................................... 3
Example # 1 – Rectangular foundation - Cohesive soil ............................................................................ 4
Example # 2 – Rectangular Foundation Cohesionless soil ....................................................................... 8
Example # 3 – Strip Foundation Cohesionless soil ................................................................................. 13
Example # 4 – Earthquake Bearing Capacity ......................................................................................... 18
Example # 5 – Settlement Calculation .................................................................................................... 24
Further Reading ...................................................................................................................................... 28
Appendix - List of symbols ..................................................................................................................... 29
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Acknowledgements We gratefully acknowledge the contribution of Ivan Kraus (University of Osijek) to this manual.
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 considered 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 .
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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
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Calculation performed using SoFA:
Click the Bearing Capacity button to access window showing the Safety Factory for Static Load Case:
SoFA Solved Examples Shallow Foundation Analysis Software
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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
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* 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]
- Vu/1.40= qu*B*L/1.40 = 1105.764[kN] < 1545.000 [kN] - NOT ok -
- 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]
- Vu/1.40= qu*B*L/1.40 = 1217.961[kN] < 1545.000 [kN] - NOT ok -
- 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]
- Vu/1.40= qu*B*L/1.40 = 1416.265[kN] < 1545.000 [kN] - NOT ok -
- FS = qu/N*Aeff = 1.283
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Example # 2 – Rectangular Foundation Cohesionless soil Calculate the ultimate static bearing capacity of the shallow foundation depicted in fig. 2.
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
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𝑠𝑞 = 1 + 𝐵
𝐿∗ 𝑠𝑖𝑛ϕ′ = 1 +
1.6
1.6∗ 𝑠𝑖𝑛35 = 1.57
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
Calculation performed using SoFA:
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Click the Bearing Capacity button to access the windows showing the Safety Factory for Static Load Case
and the Ultimate Bearing Capacity:
SoFA Solved Examples Shallow Foundation Analysis Software
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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 2.000 [m]
* Depth of foundation (df) = 2.000 [m]
* Depth of water level(dw) = 2.500 [m]
* Foundation base inclination(omega) = 0.000 [rad.]
* Soil inclination(beta) = 0.000 [rad.]
Design Loads - Static Load Case
* 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.]
* ck = 0.0 [kPa] - drained shear strength
* cuk = 0.0 [kPa] - undrained shear strength
* soil Weight = 18.50 [kN/m^3]
=============================================================
Eccentricities (Static Load Case): ex = 0.20 ey = 0.20 [m]
Effective Dimentions: 1.600 x 1.600 [m]
Effective Area: 2.560 [m^2]
-------------------------------------------------------------
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]
- Vu/1.40 = qu*B*L/1.40 = 3220.473[kN] < 4504.000 [kN] - NOT ok -
- 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]
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* 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 |
- qu_dr = 1761.196 [kPa]
- Vu = qu*B*L = 4508.662[kN]
- Vu/1.40 = qu*B*L/1.40 = 3220.473[kN] < 4504.000 [kN] - NOT ok -
- 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]
- Vu/1.40 = qu*B*L/1.40 = 3863.435[kN] < 4504.000 [kN] - NOT ok -
- 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]
- Vu/1.40 = qu*B*L/1.40 = 3579.989[kN] < 4504.000 [kN] - NOT ok -
- FS = qu/N*Aeff = 1.113
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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.
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
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
SoFA Solved Examples Shallow Foundation Analysis Software
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28,10sin72,10cos28,1sincos 2222 BL mmm
inclination factors:
1
35tan
01,28,02,196
01
'tan
'1
28,1
m
q cLBV
Hi
1130,33
130,331
1
1
q
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
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
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Using SoFA:
SoFA Solved Examples Shallow Foundation Analysis Software
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Shallow Foundation Bearing Capacity
-------------------------------------------------------------
Kostis Nikolaou ([email protected])
Dimitris Pitilakis
Aristotle University of Thessaloniki - 2012
-------------------------------------------------------------
Geometry of the problem
* Dimentions(dx/dy) = 0.800 x 2.100 [m]
* Depth of foundation (df) = 0.000 [m]
* Depth of water level(dw) = Inf [m]
* Foundation base inclination(omega) = 0.000 [rad.]
* Soil inclination(beta) = 0.000 [rad.]
Design Loads - Static Load Case
* Vd = 196.200 [kN]
* Hdx = 0.000 [kN]
* Hdy = 0.000 [kN]
* Mdx = 0.000 [kNm]
* Mdy = 0.000 [kNm]
Soil Properties
* Type = CL [C: cohessive CL: cohesionless]
* Loading= D [D: drained UN: undrained]
* phik = 35.0 [deg.]
* ck = 0.0 [kPa] - drained shear strength
* cuk = 0.0 [kPa] - undrained shear strength
* soil Weight = 18.00 [kN/m^3]
=============================================================
Eccentricities (Static Load Case): ex = 0.00 ey = 0.00 [m]
Effective Dimentions: 0.800 x 2.100 [m]
Effective Area: 1.680 [m^2]
SoFA Solved Examples Shallow Foundation Analysis Software
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-------------------------------------------------------------
Bearing Capacity Check -- Drained Conditions -- Static Load Case
-------------------------------------------------------------
* 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
SoFA Solved Examples Shallow Foundation Analysis Software
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Example # 4 – Earthquake Bearing Capacity Calculate the ultimate earthquake bearing capacity of a 4mXinf strip foundation
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
SoFA Solved Examples Shallow Foundation Analysis Software
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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
Calculation performed using SoFA:
SoFA Solved Examples Shallow Foundation Analysis Software
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Click the Calculate button to access the window Input for Eurocode 8 – Annex F criterion:
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:
SoFA Solved Examples Shallow Foundation Analysis Software
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Shallow Foundation Bearing Capacity
-------------------------------------------------------------
Kostis Nikolaou ([email protected])
Dimitris Pitilakis
Aristotle University of Thessaloniki - 2012
-------------------------------------------------------------
Geometry of the problem
* Dimentions(dx/dy) = 4.000 x Inf [m]
* Depth of foundation (df) = 2.000 [m]
* Depth of water level(dw) = Inf [m]
* Foundation base inclination(omega) = 0.000 [rad.]
* Soil inclination(beta) = 0.000 [rad.]
Design Loads - Static Load Case
* 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
* Type = CL [C: cohessive CL: cohesionless]
* Loading= D [D: drained UN: undrained]
* phik = 30.2 [deg.]
* ck = 0.0 [kPa] - drained shear strength
* cuk = 0.0 [kPa] - undrained shear strength
* soil Weight = 16.19 [kN/m^3]
=============================================================
Eccentricities (Static Load Case): ex = 0.00 ey = 0.00 [m]
Effective Dimentions: 4.000 x 1[m] - Strip
Effective Area: 4.000 [m^2]
-------------------------------------------------------------
Bearing Capacity Check -- Drained Conditions -- Static Load Case
-------------------------------------------------------------
* 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
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* 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 |
- 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]
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* 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
SoFA Solved Examples Shallow Foundation Analysis Software
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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
SoFA Solved Examples Shallow Foundation Analysis Software
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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.
Settlements due to consolidation are calculated:
If 𝜍tot′ > ζ’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
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Calculation performed using SoFA:
All other data is the same as in the Example #1
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
SoFA Solved Examples Shallow Foundation Analysis Software
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Shallow Foundation Settlements
--------------------------------------------
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]
* 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]
------------------------------
Settlements due to Consolidation
------------------------------
Initial Stresses
Depth: 2.50|Initial Stress: 31.00[kPa]|Is= 0.646 |Load Stress: 124.73 [kPa]|Total Stress: 155.73 | DHi: 21.76 [mm]
Depth: 3.50|Initial Stress: 39.00[kPa]|Is= 0.385 |Load Stress: 74.35 [kPa]|Total Stress: 113.35 | DHi: 14.38 [mm]
Depth: 5.00|Initial Stress: 51.00[kPa]|Is= 0.208 |Load Stress: 40.21 [kPa]|Total Stress: 91.21 | DHi: 15.67 [mm]
Depth: 8.00|Initial Stress: 75.00[kPa]|Is= 0.082 |Load Stress: 15.89 [kPa]|Total Stress: 90.89 | DHi: 10.36 [mm]
Settl.consol:
- DHcons = 62.1657 [mm]
TOTAL Settl.: 77.9781 [mm]
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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).
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[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
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
cohesion , q for surcharge and γ for weight density
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
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gi the design values of the factors for the soil inclination, with subscripts c for
cohesion , q for surcharge and γ for weight density
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
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
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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
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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