Calculations involving a single random variable (SRV)
Example of Bearing Capacity
q
φu = 0
µc = 100kN/m 2
u
σ c = 50kN/m 2
undrained shear
strength parameters
u
What is the relationship between the Factor of Safety (FS) of a conventional
bearing capacity calculation (based on the mean strength) and the
probability of failure (pf)?
First perform a deterministic calculation
Conventional bearing capacity calculations typically involve high factors of
safety of at least 3.
The bearing capacity of an undrained clay is given by the Prandtl equation:
qu = (2 + π )cu
= 5.14cu
where cu is a design “mean” value of the undrained shear strength.
If cu = 100kN/m 2 , and FS = 3 this implies an allowable bearing pressure of:
qall =
5.14 × 100
= 171kN/m 2
3
Now perform a probabilistic calculation
If additional data comes in to indicate that the same undrained clay has a
mean shear strength of µcu = 100kN/m 2 and a standard deviation of σ cu = 50kN/m 2
and is lognormally distributed, what is the probability of bearing failure?
In other words, what is the probability of the actual bearing capacity being
less than the factored deterministic value P[ qu < 171] ?
qu = 5.14cu , hence if cu is a random variable we can write:
E[qu ] = 5.14E[cu ]
thus µ qu =5.14µcu =514
and Var[qu ] = 5.142 Var[cu ] thus σ qu =5.14σ cu =257
1
(Note that since qu ∝ cu , Vqu = Vcu = )
2
Failure occurs if qu < qall
The probability of this happening can be written as P[qu < 171]
First find the properties of the underlying normal distribution of ln qu
σ ln q
u
µln q
u
2
⎧
⎪ ⎛ 1 ⎞ ⎫⎪
2
= ln 1 + Vqu = ln ⎨1 + ⎜ ⎟ ⎬ = 0.47
⎪⎩ ⎝ 2 ⎠ ⎪⎭
1 2
1
= ln µqu − σ ln qu = ln(514) − (0.47) 2 = 6.13
2
2
{
}
⎛ ln171 − 6.13 ⎞
P[qu < 171] = Φ ⎜
⎟
0.47
⎝
⎠
= Φ ( −2.10 )
= 1 − Φ ( 2.10 )
= 1 − 0.982
= 0.018 (1.8%)
Example of Slope Stability
Dimensionless
strength
parameter
C=
cu
γ sat H
φu = 0
µc , σ c
u
u
γ sat
What is the relationship between the Factor of Safety (FS) of a slope
(based on the mean strength) and the probability of failure (pf) of an
undrained clay slope for different values of the mean and standard
deviation of strength?
First perform a deterministic calculation
Using limit equilibrium or charts, find the FS corresponding to a uniform slope of strength µcu
for a homogeneous slope
FS α C
Linear relationship
between C and FS
for a cohesive slope
with a slope angle
of β = 26.57o
and a depth ratio
of D = 2
Now perform a probabilistic calculation
1) Noting that FS ∝ C or FS = KC , compute K , hence µ FS = K µC and VFS = VC
2) If C is lognormal, so is FS , so compute the underlying normal properties σ ln FS and µln FS
3) Find the probability P [ FS < 1] from standard tables.
⎡ ln1 − µln FS ⎤
⎡ µln FS ⎤
If FS is lognormal, P [ FS < 1] = Φ ⎢
=
Φ
⎥
⎢−
⎥
⎣ σ ln FS ⎦
⎣ σ ln FS ⎦
⎡1 − µ FS ⎤
If FS is normal, P [ FS < 1] = Φ ⎢
⎥
⎣ σ FS ⎦
Single Random Variable (SRV) approach
Defining FS ≈ 5.88µC
Probability of Failure
p f = p( FS < 1)
VC =
or
p f = p(C < 0.17)
σC
µC
High!
The mean strength
may be too
optiminstic….
Probability of failure p f vs. Factor of Safety FS (based on the mean)
The median helps to interpret lognormal behaviour
pf
Median C < 0.17
Median C > 0.17
p f vs. VC for different values of Median C
Mean strength factoring
All curves assume FS=1.47 based on Cdes = 0.25
linear scaling
Cdes = µC − f1µC
sd scaling
Cdes = µC − f 2σ C
All the examples considered so far were for a single random variable
(usually the undrained strength)
How can we account for more than one random variable?
(e.g. the cohesion and friction angle in a drained analysis)
A well-established approximate method for estimating the influence of
several random input variables on a function is the
First Order Second Moment (FOSM) method.
The method is called “First Order” because it includes only first order terms
in a Taylor expansion.
The method is called “Second Moment” because it enables estimates
to be made of the variance (second moment) of function under consideration.
Earth Pressure Example using FOSM with 4 Random Variables
0.30
Cohesionless soil
c′ = 0, φ ′
H = 5.03
Active Earth Pressure Coefficient
′
K a = tan 2 (45 − φ )
2
4.57
γc
1.37
0.46
γ bf
1.83
tan δ
3.66
Pa
Units in kN and m
0.30
Cohesionless soil
c′ = 0, φ ′
H = 5.03
Active Earth Pressure Coefficient
′
K a = tan 2 (45 − φ )
2
4.57
γc
1.37
0.46
γ bf
Pa
Units in kN and m
1.83
tan δ
3.66
1
Rankine's Theory: Pa = γ bf H 2 K a = 12.65γ bf K a
2
W = Wc + Wbf
1
= 0.46(3.66)γ c + (4.57)(0.30 + 0.46)γ c + 1.83(4.57)γ bf
2
= 3.42γ c + 8.36γ bf
Factor of Safety against sliding: FSsl
3.42γ
(
=
c
+ 8.36γ bf ) tan δ
12.65γ bf K a
Input data assuming all variables are random
FSsl is a function of 4 random variables:
FSsl = f (γ c , γ bf , K a , tan δ )
Property
µ
σ
γc (kN/m3)
23.58
0.31
γbf (kN/m3)
18.87
1.10
Ka
0.333
0.033
tan δ
0.5
0.05
Now use the FOSM Method to estimate the statistics of FSsl
E[FSsl
3.42 E[γ
(
]≈
c
] + 8.36 E[γ bf ]) E[ tan δ ]
12.65 E[γ bf ] E[K a ]
2
= µ FSsl
⎛ ∂FSsl ⎞
⎛ ∂FSsl ⎞
⎛ ∂FSsl ⎞
⎛ ∂FSsl ⎞
+
+
+
Var[FSsl ] ≈ ⎜
Var[
γ
]
Var[
γ
]
Var[
]
K
⎜
⎟
⎟
⎜
⎟
⎜
⎟ Var[ tan δ ]
c
bf
a
⎜
⎟
⎝ ∂ (tan δ ) ⎠
⎝ ∂γ c ⎠
⎝ ∂K a ⎠
⎝ ∂γ bf ⎠
2
2
2
2
= σ FS
sl
The mean value of FS sl can be estimated in the FOSM Method by substituting all the
mean values of the input variables into the governing function, thus:
Hence, µ FSsl
3.42(23.58) + 8.36(18.87) ) 0.5
(
≈
= 1.50
12.65(18.87)0.333
Since we have a functional relationship between FS sl and the input variables, the
variance of FS sl can be estimated in this case by evaluating the derivatives
analytically at the mean.
Factor of Safety against sliding: FSsl
3.42γ
(
=
c
+ 8.36γ bf ) tan δ
12.65γ bf K a
∂FSsl 0.27 tan δ
=
= 0.02
γ bf K a
∂γ c
∂FSsl −0.27γ c tan δ
=
= −0.03
∂γ bf
γ bf2 K a
∂FSsl −(0.66γ bf + 0.27γ c ) tan δ
=
= −4.50
2
∂K a
γ bf K a
(0.66γ bf + 0.27γ c )
∂FSsl
=
= 3.00
γ bf K a
∂ (tan δ )
All derivatives evaluated at the mean
Hence:
Var[FSsl ] ≈ 0.022 × 0.312 + 0.032 × 1.102 + 4.502 × 0.0332 + 32 × 0.052
= 0.046
∴ σ FSsl = 0.046=0.21
⎧⎪ µ FSsl = 1.50 ⎫⎪
σ FSsl
Summary: ⎨
= 0.14
⎬ VFSsl =
µ FSsl
⎪⎩σ FSsl = 0.21⎪⎭
What is the probability of sliding failure? P [ FS sl < 1] ??
In order to compute probabilities we need to assume a suitable PDF
Let’s assume that FSsl is lognormally distributed
First find the mean and standard deviation of the underlying normal distribution
{
σ ln FS = ln 1 + V 2
sl
µln FS = ln µ FS
sl
FS sl
sl
}
= ln {1 + 0.142 } = 0.14
1 2
1
− σ ln FSsl = ln(1.5) − (0.14) 2 = 0.40
2
2
⎛ ln1 − 0.40 ⎞
P [ FS sl < 1] = Φ ⎜
⎟
0.14
⎝
⎠
= Φ(−2.86)
= 1 − Φ( 2.86)
= 1 − 0.999781
= 0.00022 or 0.02%
If we instead assume that FSsl is normally distributed
⎛ 1 − 1.5 ⎞
P [ FS sl < 1] = Φ ⎜
⎟
⎝ 0.21 ⎠
= Φ(−2.38)
= 1 − Φ( 2.38)
= 1 − 0.991343
= 0.0087 or 0.9 %
Numerical Differentiation
In this case we were able to differentiate analytically, but in some problems
we do not have a function to work with (e.g. slope stability). In these case
we can use numerical differentiation
Reworking of the earth pressure problem using numerical differentiation.
FSsl = f (γ c , γ bf , K a , tan δ )
2
⎛ ∂FSsl ⎞
⎛ ∂FSsl ⎞
⎛ ∂FSsl ⎞
⎛ ∂FSsl ⎞
+
+
+
Var[FSsl ] ≈ ⎜
Var[
γ
]
Var[
γ
]
Var[
]
K
⎜
⎟
⎟
⎜
⎟
c
bf
a
⎜
⎟ Var[ tan δ ]
⎜
⎟
⎝ ∂ (tan δ ) ⎠
⎝ ∂γ c ⎠
⎝ ∂K a ⎠
⎝ ∂γ bf ⎠
2
2
2
We can evaluate derivatives using a central difference formula in which we sample
the function at one standard deviation below and one standard deviation above the mean.
∂FS sl f ( µγ c , µγ bf , µ Ka + σ Ka , µ tan δ ) − f ( µγ c , µγ bf , µ Ka − σ Ka , µ tan δ ) ∆f Ka
e.g.
≈
=
∂K a
2σ Ka
2σ Ka
Note all variables are fixed at their mean except for the derivative variable
After squaring the derivative, the standard deviations cancel
and we are left with:
2
⎛ ∆fγ c ⎞ ⎛ ∆fγ bf ⎞ ⎛ ∆f Ka ⎞ ⎛ ∆f tan δ ⎞
Var[FSsl ] ≈ ⎜
⎟ +⎜
⎟ + ⎜⎜
⎟ +⎜
⎟
⎟
2
2
2
2
⎝
⎠
⎝
⎠ ⎝
⎠
⎠ ⎝
2
2
2
Compute the values using a tabular approach
Property
γc
µ
σ
(kN/m3)
23.58
0.31
γbf (kN/m3)
18.87
1.10
Ka
0.333
0.033
tan δ
0.5
0.05
µ± σ
23.89
23.27
19.97
17.77
0.367
0.300
0.55
0.45
f
1.50
1.49
1.47
1.53
1.36
1.66
1.65
1.35
∆f
0.01
-0.06
-0.30
0.30
Now substitute these values into the variance equations
⎛ 0.01 ⎞ ⎛ −0.06 ⎞ ⎛ −0.3 ⎞ ⎛ 0.3 ⎞
Var[FSsl ] ≈ ⎜
⎟ +⎜
⎟ +⎜
⎟ +⎜
⎟
2
2
2
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝ 2 ⎠
= 0.046
2
2
2
2
Hence, σ FSsl = 0.21 which is the same value (to 2 decimal places)
as the one we obtained analytically.
The fact that the analytical and numerical results are very similar in this case indicates
that there is almost no nonlinearity in this problem.
The Reliability Index
The Reliability Index is a measure of the margin of safety in
“standard deviation units”.
For example, if dealing with a Factor of Safety, the reliability Index is
given by:
µ FS − 1
β=
σ FS
In a normal variate, the “reliability index” (β) is uniquely related to the
“probability of failure” (pf) through the expression:
p f = 1 − Φ( β )
Consider a normal distribution of the Factor of Safety (FS)
f FS
µ FS = 1.5
σ FS = 0.21
pf is this area
FS
β is this distance ÷ the standard deviation
β=
(1.5 −1)
= 2.38
0.21
p f = 1 − Φ( 2.38) = 1 − 0.991343 = 0.0087
Probability of Failure: pf
0.5
Reliability Index: β
Probability of Failure vs. Reliability Index for a Normal Distribution