dr c.m. martin department of engineering science university of oxford
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11th International Conference of IACMAG, Torino21 Giugno 2005
Exact bearing capacity calculations using the method of characteristics
Dr C.M. MartinDepartment of Engineering Science
University of Oxford
Outline
• Introduction
• Bearing capacity calculations using the method of characteristics
• Exact solution for example problem
• Can we solve the ‘N problem’ this way?
• The fast (but apparently forgotten) way to find N
• Verification of exactness
• Conclusions
• Idealised problem (basis of design methods):
Bearing capacity
Central, purely vertical loading
Rigid strip footing
Semi-infinite soilc, , , =
B
D
qu = Qu/B
Bearing capacity
Rigid strip footing
B
q = Dq = D
Semi-infinite soilc, , , =
• Idealised problem (basis of design methods):
Central, purely vertical loading
qu = Qu/B
Classical plasticity theorems
• A unique collapse load exists, and it can be bracketed by lower and upper bounds (LB, UB)
• LB solution from a stress field that satisfies– equilibrium– stress boundary conditions– yield criterion
• UB solution from a velocity field that satisfies– flow rule for strain rates– velocity boundary conditions
• Theorems only valid for idealised material– perfect plasticity, associated flow ( = )
Statically admissible}
Plastically admissible
Kinematically admissible}
Method of characteristics
• Technique for solving systems of quasi-linear PDEs of hyperbolic type
• Applications in both fluid and solid mechanics• In soil mechanics, used for plasticity problems:
– bearing capacity of shallow foundations– earth pressure on retaining walls– trapdoors, penetrometers, slope stability, …
• Method can be used to calculate both stress and velocity fields (hence lower and upper bounds)
• In practice, often gives LB = UB exact result• 2D problems only: plane strain, axial symmetry
Method of characteristics
• Technique for solving systems of quasi-linear PDEs of hyperbolic type
• Applications in both fluid and solid mechanics• In soil mechanics, used for plasticity problems:
– bearing capacity of shallow foundations– earth pressure on retaining walls– trapdoors, penetrometers, slope stability, …
• Method can be used to calculate both stress and velocity fields (hence lower and upper bounds)
• In practice, often gives LB = UB exact result• 2D problems only: plane strain, axial symmetry
Outline
• Introduction
• Bearing capacity calculations using the method of characteristics
• Exact solution for example problem
• Can we solve the ‘N problem’ this way?
• The fast (but apparently forgotten) way to find N
• Verification of exactness
• Conclusions
c
n
Z
2
x
z
3 = – R
1 = + R
X
13
cos sinR c
Lower bound stress field
• To define a 2D stress field, e.g. in x-z plane– normally need 3 variables (xx, zz, xz)
– if assume soil is at yield, only need 2 variables (, )
( , )R f M-C
general[ ]
c
n
Z
2
x
z
3 = – R
1 = + R
X
13
cos sinR c
Lower bound stress field
• To define a 2D stress field, e.g. in x-z plane– normally need 3 variables (xx, zz, xz)
– if assume soil is at yield, only need 2 variables (, )
( , )R f M-C
general[ ]
= – /2 c
n
Z
2
x
z
3 = – R
2
1 = + R
X
13
cos sinR c
Lower bound stress field
• To define a 2D stress field, e.g. in x-z plane– normally need 3 variables (xx, zz, xz)
– if assume soil is at yield, only need 2 variables (, )
2
( , )R f M-C
general[ ]
• Substitute stresses-at-yield (in terms of , ) into equilibrium equations
• Result is a pair of hyperbolic PDEs in ,
• Characteristic directions turn out to coincide with and ‘slip lines’ aligned at
• Use and directions as curvilinear coords obtain a pair of ODEs in , (easier to integrate)
• Solution can be marched out from known BCs
0xx xz
x z
xz zz
x z
Lower bound stress field
• Substitute stresses-at-yield (in terms of , ) into equilibrium equations
• Result is a pair of hyperbolic PDEs in ,
• Characteristic directions turn out to coincide with and ‘slip lines’ aligned at
• Use and directions as curvilinear coords obtain a pair of ODEs in , (easier to integrate)
• Solution can be marched out from known BCs
0xx xz
x z
xz zz
x z
Lower bound stress field
> 0
• Marching from two known points to a new point:
(xB, zB, B, B)B (xA, zA, A, A)Az
x
Lower bound stress field
• Marching from two known points to a new point:
(xB, zB, B, B)B (xA, zA, A, A)A
(xC, zC, C, C)C
z
x
Lower bound stress field
• Marching from two known points to a new point:
2d d tan d d
cos
Rx z
d
tand
x
z
(xB, zB, B, B)B (xA, zA, A, A)A
(xC, zC, C, C)C
z
x
Lower bound stress field
d
tand
x
z
2d d tan d d
cos
Rx z
• Marching from two known points to a new point:
• ‘One-legged’ variant for marching from a known point onto an interface of known roughness
2d d tan d d
cos
Rx z
d
tand
x
z
(xB, zB, B, B)B (xA, zA, A, A)A
(xC, zC, C, C)C
z
x
Lower bound stress field
d
tand
x
z
2d d tan d d
cos
Rx z
• Marching from two known points to a new point:
• ‘One-legged’ variant for marching from a known point onto an interface of known roughness
2d d tan d d
cos
Rx z
d
tand
x
z
(xB, zB, B, B)B (xA, zA, A, A)A
(xC, zC, C, C)C
z
x
Lower bound stress field
d
tand
x
z
2d d tan d d
cos
Rx z
FD formFD form
• Substitute velocities u, v into equations for– associated flow (strain rates normal to yield surface)– coaxiality (princ. strain dirns = princ. stress dirns)
• Result is a pair of hyperbolic PDEs in u, v
• Characteristic directions again coincide with the and slip lines aligned at
• Use and directions as curvilinear coords obtain a pair of ODEs in u, v (easier to integrate)
• Solution can be marched out from known BCs
Upper bound velocity field
• Marching from two known points to a new point:
(xB, zB, B, B, uB, vB)B (xA, zA, A, A, uA, vA)Az,v
x,u
Upper bound velocity field
• Marching from two known points to a new point:
(xB, zB, B, B, uB, vB)B (xA, zA, A, A, uA, vA)A
(xC, zC, C, C, uC, vC)C
z,v
x,u
Upper bound velocity field
• Marching from two known points to a new point:
(xB, zB, B, B, uB, vB)B (xA, zA, A, A, uA, vA)A
(xC, zC, C, C, uC, vC)C
z,v
x,u
Upper bound velocity field
d sin( ) d cos( ) 0u v d sin( ) d cos( ) 0u v
• Marching from two known points to a new point:
• ‘One-legged’ variant for marching from a known point onto an interface of known roughness
(xB, zB, B, B, uB, vB)B (xA, zA, A, A, uA, vA)A
(xC, zC, C, C, uC, vC)C
z,v
x,u
Upper bound velocity field
d sin( ) d cos( ) 0u v d sin( ) d cos( ) 0u v
• Marching from two known points to a new point:
• ‘One-legged’ variant for marching from a known point onto an interface of known roughness
(xB, zB, B, B, uB, vB)B (xA, zA, A, A, uA, vA)A
(xC, zC, C, C, uC, vC)C
z,v
x,u
Upper bound velocity field
d sin( ) d cos( ) 0u v d sin( ) d cos( ) 0u v
FD form FD form
Outline
• Introduction
• Bearing capacity calculations using the method of characteristics
• Exact solution for example problem
• Can we solve the ‘N problem’ this way?
• The fast (but apparently forgotten) way to find N
• Verification of exactness
• Conclusions
Example problem
B = 4 m
q = 18 kPaq = 18 kPa
c = 16 kPa, = 30°, = 18 kN/m3
Rough base
after Salençon & Matar (1982)
qu
Example problem: stress field (partial)
known (passive failure); = /2
Example problem: stress field (partial)
Symmetry: = 0 on z axis (iterative construction req’d)
known (passive failure); = /2
• Shape of ‘false head’ region emerges naturally
• qu from integration of tractions
• Solution not strict LB until stress field extended:
Example problem: stress field (partial)
Symmetry: = 0 on z axis (iterative construction req’d)
known (passive failure); = /2
Example problem: stress field (complete)
Minor principal stress trajectory
• Extension strategy by Cox et al. (1961)
• Here generalised for > 0
• Utilisation factor at start of each ‘spoke’ must be 1
Example problem: stress field (complete)
Minor principal stress trajectory
Extension technique
1
3
z0 + q
1 + (z z0)
1
z + q
z0
z
q
1 0
1 02 cos sin
z q
c z q
Extension technique
1
3
z0 + q
1 + (z z0)
1
z + q
z0
z
Critical utilisation is here:
q
Rigid
Rigid
Rigid
Rigid
Rigid
Example problem: velocity field
• Discontinuities are easy to handle – treat as degenerate quadrilateral cells (zero area)
Rigid
Rigid
Rigid
Rigid
Rigid
Example problem: velocity field
Some cautionary remarks
• Velocity field from method of characteristics does not guarantee kinematic admissibility!– principal strain rates may become ‘mismatched’
with principal stresses 1, 3
– this is OK if = 0 (though expect UB LB)– but not OK if > 0: flow rule violated no UB at all
• If > 0, as here, must check each cell of mesh– condition is sufficient
• Only then are calculations for UB meaningful– internal dissipation, e.g. using– external work against gravity and surcharge
0xx zz
1 3,
max
cosD c
• qu from integration of internal and external work rates for each cell (4-node , 3-node )
• Discontinuities do not need special treatment
Rigid
Rigid
Rigid
Rigid
Rigid
Example problem: velocity field
Convergence of qu (kPa) in example
Mesh
Initial
2
4
8
16
32
64
etc.
Convergence of qu (kPa) in example
Mesh Stress calc.
Initial 1626.74
2 1625.96
4 1625.76
8 1625.71
16 1625.70
32 1625.70
64 1625.70
etc. 1625.70
LB
Convergence of qu (kPa) in example
Mesh Stress calc. Velocity calc.
Initial 1626.74 1626.94
2 1625.96 1626.01
4 1625.76 1625.77
8 1625.71 1625.72
16 1625.70 1625.70
32 1625.70 1625.70
64 1625.70 1625.70
etc. 1625.70 1625.70
UBLB
Outline
• Introduction
• Bearing capacity calculations using the method of characteristics
• Exact solution for example problem
• Can we solve the ‘N problem’ this way?
• The fast (but apparently forgotten) way to find N
• Verification of exactness
• Conclusions
The solutions obtained from [the method of characteristics] are generally not exact collapse loads, since it is not always possible to integrate the stress-strain rate relations to obtain a kinematically admissible velocity field, or to extend the stress field over the entire half-space of the soil domain.
Hjiaj M., Lyamin A.V. & Sloan S.W. (2005). Numerical limit analysis solutions for the bearing capacity factor N. Int. J. Sol. Struct. 42, 1681-1704.
Why not?
N problem as a limiting case
c = 0, > 0, > 0, =
B
qu
N problem as a limiting case
c = 0, > 0, > 0, =
B
u0limq B q
N q q
qu
N problem as a limiting case
c = 0, > 0, > 0, =
B
u0limq B q
N q q
qu
tan 2tan 4 2e
N problem as a limiting case
c = 0, > 0, > 0, =
B
u0limq B q
N q q
ulim 2B q
N q B
qu
tan 2tan 4 2e
Stress field as B/q
c = 0, = 30°, Rough ( = )
B/q 2qu/B
0
Stress field as B/q
B/q 2qu/B
0.1 397.0
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
0.2 211.9
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
0.5 99.43
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
1 60.69
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
2 40.28
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
5 26.84
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
10 21.70
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
10 21.70
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
20 18.74
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
50 16.65
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
100 15.83
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
200 15.35
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
500 15.03
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
1000 14.91
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
104 14.77
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
105 14.76
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
106 14.75
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
109 14.75
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
1012 14.75
c = 0, = 30°, Rough ( = )
Stress field as B/q
B/q 2qu/B
1012 14.75 Take as N
Fan (almost) degenerate
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
0
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
0.1 397.0
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
0.2 211.9
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
0.5 99.43
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
1 60.69
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
2 40.28
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
5 26.84
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
10 21.70
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
10 21.70
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
20 18.74
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
50 16.65
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
100 15.83
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
200 15.35
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
500 15.03
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
1000 14.91
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
104 14.77
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
105 14.76
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
106 14.75
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
109 14.75
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
1012 14.75
c = 0, = 30°, Rough ( = )
Velocity field as B/q
B/q 2qu/B
1012 14.75 Take as N
Fan (almost) degenerate
c = 0, = 30°, Rough ( = )
Convergence of 2qu/B when B/q = 109
Mesh
Initial
2
4
8
16
32
64
etc.
Convergence of 2qu/B when B/q = 109
Mesh Stress calc.
Initial 14.7166
2 14.7446
4 14.7518
8 14.7537
16 14.7541
32 14.7542
64 14.7543
etc. 14.7543
LB
Convergence of 2qu/B when B/q = 109
Mesh Stress calc. Velocity calc.
Initial 14.7166 14.8239
2 14.7446 14.7713
4 14.7518 14.7585
8 14.7537 14.7553
16 14.7541 14.7545
32 14.7542 14.7543
64 14.7543 14.7543
etc. 14.7543 14.7543
UBLB
Completion of stress field (coarse)
c = 0 = 30°B/q = 109
Rough ( = )
N = 14.7543
Completion of stress field (fine)
c = 0 = 30°B/q = 109
Rough ( = )
N = 14.7543
Completion of stress field (fine)
EXACT
c = 0 = 30°B/q = 109
Rough ( = )
N = 14.7543
It also works for smooth footings…
c = 0 = 30°B/q = 109
Smooth ( = 0)
N = 7.65300
… and other friction angles
c = 0 = 20°B/q = 109
Rough ( = )
N = 2.83894
Outline
• Introduction
• Bearing capacity calculations using the method of characteristics
• Exact solution for example problem
• Can we solve the ‘N problem’ this way?
• The fast (but apparently forgotten) way to find N
• Verification of exactness
• Conclusions
Notice anything?
• Tractions distance from singular point• Characteristics self-similar w.r.t. singular point
c = 0 = 30°B/q = 109
Smooth ( = 0)
N = 7.65300
Recall N problem definition
q = 0
Semi-infinite soil c = 0, > 0, > 0
Recall N problem definition
Semi-infinite soil c = 0, > 0, > 0
q = 0
r
• No fundamental length can solve in terms of polar angle and radius r
• Along a radius, stress state varies only in scale:– mean stress r– major principal stress orientation = const
• Combine with yield criterion and equilibrium equations to get a pair of ODEs:
Governing equations
( ) ( )r s
2
sin 2 2 sin 2
cos 2 2 sin
cos cos 2 sin cos
2 sin cos 2 2 sin
sds
d
sd
d s
von Kármán (1926)
Direct solution of ODEs
r Edge of passive zone:
1
11
1
4 2
cos
1 sin
2
s
Underside of footing ( = 0):
0
0
0
2
?
0
s
( ), ( )r s
solve
(iteratively)
• Use any standard adaptive Runge-Kutta solver– ode45 in MATLAB, NDSolve in Mathematica
• Easy to get N factors to any desired precision
• Much faster than method of characteristics
• Definitive tables of N have been compiled for– = 1°, 2°, … , 60°– / = 0, 1/3, 1/2, 2/3, 1
• Values are identical to those obtained from the method of characteristics, letting B/q
Direct solution of ODEs
< 10 s to generate}
Selected values of N
• Exactness checked by method of characteristics: LB = UB, stress field extensible, match
[°] Smooth / = 1/3 / = 1/2 / = 2/3 Rough
5 0.08446 0.09506 0.1001 0.1048 0.1134
10 0.2809 0.3404 0.3678 0.3929 0.4332
15 0.6991 0.9038 0.9940 1.072 1.181
20 1.579 2.167 2.411 2.606 2.839
25 3.461 5.030 5.626 6.060 6.491
30 7.653 11.75 13.14 14.03 14.75
35 17.58 28.46 31.60 33.34 34.48
40 43.19 73.55 80.62 83.89 85.57
45 117.6 209.7 225.9 231.9 234.2
1 3, 1 3,
Selected values of N
[°] Smooth / = 1/3 / = 1/2 / = 2/3 Rough
5 0.08446 0.09506 0.1001 0.1048 0.1134
10 0.2809 0.3404 0.3678 0.3929 0.4332
15 0.6991 0.9038 0.9940 1.072 1.181
20 1.579 2.167 2.411 2.606 2.839
25 3.461 5.030 5.626 6.060 6.491
30 7.653 11.75 13.14 14.03 14.75
35 17.58 28.46 31.60 33.34 34.48
40 43.19 73.55 80.62 83.89 85.57
45 117.6 209.7 225.9 231.9 234.2
1 3, 1 3, • Exactness checked by method of characteristics: LB =
UB, stress field extensible, match
Influence of roughness on N
0.504719 0.500722 0.500043
Smooth
/ = 1/3
/ = 2/3
/ = 1/2
Outline
• Introduction
• Bearing capacity calculations using the method of characteristics
• Exact solution for example problem
• Can we solve the ‘N problem’ this way?
• The fast (but apparently forgotten) way to find N
• Verification of exactness
• Conclusions
N by various methods
0
5
10
15
20
25
Terz
aghi
(194
3)
Mey
erho
f (19
51)
Kumbh
ojkar
(199
3)
Zhu e
t al. (
2001
)
Silves
tri (2
003)
Caquo
t & K
erise
l (19
53)
Booke
r (19
70)
Graha
m &
Stu
art (
1971
)
Salenc
on &
Mat
ar (1
982)
Bolton
& L
au (1
993)
Kumar
(200
3)
Mar
tin (2
004)
Lund
gren
& M
orte
nsen
(195
3)
Hanse
n & C
hrist
ense
n (1
969)
Mar
tin (2
005)
Chen
(197
5)
Mich
alowsk
i (19
97)
Soubr
a (1
999)
Zhu (2
000)
Wan
g et
al. (
2001
)
Griffith
s (19
82)
Man
ohar
an &
Das
gupt
a (1
995)
Frydm
an &
Bur
d (1
997)
Yin et
al. (
2001
)
Sloan
& Yu
(199
6)
Ukritc
hon
et a
l. (20
03)
Hjiaj e
t al. (
2005
)
Mak
rodim
opou
los &
Mar
tin (2
005)
Mey
erho
f (19
63)
Brinch
Han
sen
(197
0)
Vesic
(197
5)
Euroc
ode
7 (1
996)
Poulos
et a
l. (20
01)
FELALimit Eqm Characteristics Upper Bd FE/FDODEs Formulae
= 30°, =
N by various methods
0
5
10
15
20
25
Terz
aghi
(194
3)
Mey
erho
f (19
51)
Kumbh
ojkar
(199
3)
Zhu e
t al. (
2001
)
Silves
tri (2
003)
Caquo
t & K
erise
l (19
53)
Booke
r (19
70)
Graha
m &
Stu
art (
1971
)
Salenc
on &
Mat
ar (1
982)
Bolton
& L
au (1
993)
Kumar
(200
3)
Mar
tin (2
004)
Lund
gren
& M
orte
nsen
(195
3)
Hanse
n & C
hrist
ense
n (1
969)
Mar
tin (2
005)
Chen
(197
5)
Mich
alowsk
i (19
97)
Soubr
a (1
999)
Zhu (2
000)
Wan
g et
al. (
2001
)
Griffith
s (19
82)
Man
ohar
an &
Das
gupt
a (1
995)
Frydm
an &
Bur
d (1
997)
Yin et
al. (
2001
)
Sloan
& Yu
(199
6)
Ukritc
hon
et a
l. (20
03)
Hjiaj e
t al. (
2005
)
Mak
rodim
opou
los &
Mar
tin (2
005)
Mey
erho
f (19
63)
Brinch
Han
sen
(197
0)
Vesic
(197
5)
Euroc
ode
7 (1
996)
Poulos
et a
l. (20
01)
FELALimit Eqm Characteristics Upper Bd FE/FDODEs Formulae
= 30°, =
N by FE limit analysis
Ukritchon et al. (2003)
SmoothRough
Rough
SmoothLOWER BOUND
UPPER BOUND
N by FE limit analysis
Hjiaj et al. (2005)
Smooth
Rough
Rough
SmoothLOWER BOUND
UPPER BOUND
N by FE limit analysis
Hjiaj et al. (2005)
LOWER BOUND
UPPER BOUND
Smooth
Rough
Rough
Smooth
N by FE limit analysis
Hjiaj et al. (2005)
LOWER BOUND
UPPER BOUND
Smooth
Rough
Rough
Smooth
N by FE limit analysis
Hjiaj et al. (2005)
• Structured meshes (different for each )
LOWER BOUND
UPPER BOUND
Smooth
Rough
Rough
Smooth
N by FE limit analysis
Makrodimopoulos & Martin (2005)
Smooth
Rough
RoughSmooth
LOWER BOUND
UPPER BOUND
N by FE limit analysis
Makrodimopoulos & Martin (2005)
Smooth
Rough
RoughSmooth
LOWER BOUND
UPPER BOUND
• Single unstructured mesh (same for each )
N by various methods
0
5
10
15
20
25
Terz
aghi
(194
3)
Mey
erho
f (19
51)
Kumbh
ojkar
(199
3)
Zhu e
t al. (
2001
)
Silves
tri (2
003)
Caquo
t & K
erise
l (19
53)
Booke
r (19
70)
Graha
m &
Stu
art (
1971
)
Salenc
on &
Mat
ar (1
982)
Bolton
& L
au (1
993)
Kumar
(200
3)
Mar
tin (2
004)
Lund
gren
& M
orte
nsen
(195
3)
Hanse
n & C
hrist
ense
n (1
969)
Mar
tin (2
005)
Chen
(197
5)
Mich
alowsk
i (19
97)
Soubr
a (1
999)
Zhu (2
000)
Wan
g et
al. (
2001
)
Griffith
s (19
82)
Man
ohar
an &
Das
gupt
a (1
995)
Frydm
an &
Bur
d (1
997)
Yin et
al. (
2001
)
Sloan
& Yu
(199
6)
Ukritc
hon
et a
l. (20
03)
Hjiaj e
t al. (
2005
)
Mak
rodim
opou
los &
Mar
tin (2
005)
Mey
erho
f (19
63)
Brinch
Han
sen
(197
0)
Vesic
(197
5)
Euroc
ode
7 (1
996)
Poulos
et a
l. (20
01)
FELALimit Eqm Characteristics Upper Bd FE/FDODEs Formulae
= 30°, =
N ( = ) by common formulae: error [%]
[°] Meyerhof (1963)
Hansen (1970)
Vesić (1975)
Eurocode (1996)
Poulos et al. (2001)
5 -38.5 -34.3 296.3 -12.4 114.9
10 -15.3 -10.2 182.6 19.8 30.0
15 -4.4 0.1 124.1 33.4 10.1
20 1.1 3.8 89.7 38.4 5.9
25 4.2 4.1 67.6 38.8 7.1
30 6.2 2.1 51.8 36.2 8.9
35 7.8 -1.6 39.3 31.2 7.7
40 9.5 -7.0 27.9 23.9 0.3
45 12.2 -14.3 16.0 14.3 -15.3
Bearing capacity factors for design
• If we use Nc and Nq that are exact for = …
… then we should, if we want to be consistent, also use N factors that are exact for =
• Then start worrying about corrections for– non-association ( < )– stochastic variation of properties– intermediate principal stress– progressive failure, etc.
tan 2tan 4 2
1 cot
q
c q
N e
N N
Bearing capacity factors for design
• If we use Nc and Nq that are exact for = …
… then we should, if we want to be consistent, also use N factors that are exact for = .
• Then start worrying about corrections for– non-association ( < )– stochastic variation of properties– intermediate principal stress– progressive failure, etc.
less capacity!
tan 2tan 4 2
1 cot
q
c q
N e
N N
Conclusions
• Shallow foundation bearing capacity is a long-standing problem in theoretical soil mechanics
• The method of characteristics, carefully applied, can be used to solve it c, , (with = )
• In all cases, find strict lower and upper bounds that coincide, so the solutions are formally exact
• If just values of N are required (and not proof of exactness) it is much quicker to integrate the governing ODEs using a Runge-Kutta solver
• Exact solutions provide a useful benchmark for validating other numerical methods (e.g. FE)
Downloads
• Program ABC – Analysis of Bearing Capacity
• Tabulated exact values of b.c. factor N
• Copy of these slides
www-civil.eng.ox.ac.uk
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