FW4 E61.pdf

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Microsoft PowerPoint - Lecture_4
Calculations involving a single random variable (SRV)
Example of Bearing Capacity
q
φ =
0
µ
=
100kN/m
2
undrained shear
c
u
σ
=
50kN/m
2
strength parameters
c
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 ( p f )?
300862800.002.png
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:
q
u
= +
(2
π
)
c
u
=
5.14
c
u
where c
is a design “mean” value of the undrained shear strength.
If
c
=
100kN/m , and
2
FS
=
3 this implies an allowable bearing pressure of:
q
=
5.14 100
×
=
171kN/m
2
all
3
300862800.003.png
Now perform a probabilistic calculation
If additional data comes in to indicate that the same undrained clay has a
mean shear strength of
µ
=
100kN/m and a standard deviation of
2
σ
=
50kN/m
2
c
u
c
and is lognormally distributed, what is the pro
bability of bearing failure?
In other words, what is the probability of the actual bearing capacity being
less than the factored deterministic value [
Pq <
u
171] ?
q
u
=
5.14 , hence if is a random variable we can write:
u
c
u
E[ ] 5.14E[ ] thus =5.14 =514
q
u
=
c
u
µ µ
u
c
u
and Var[ ] 5.14 Var[ ] thus =5.14 =257
q
=
2
c
σ σ
u
u
q
u
c
u
(Note that since
qcVV
∝ =
u
,
q
u
c
u
=
1 )
2
Failure occurs if
qq
<
all
The probability of this happening can be written as [
Pq
u
<
171]
u
c
q
u
u
300862800.004.png
First find the properties of the underlying normal distribution of ln
q
⎧
⎛⎞
1
2
⎫
σ
= + = + =
{ }
V
2
ln 1
⎨
⎜ ⎝⎠
⎬
0.47
ln
q
u
q
u
⎪
2
⎪
⎩
⎭
µ µ σ
= − =
1
2
ln(514)
−
1
(0.47) 6.13
2
=
ln
q
u
q
u
2
ln
q
u
2
Pq
<= ⎜
]
⎛
ln171 6.13
−
⎞
u
⎝
0.47
⎠
=Φ −
( )
2.10
=−Φ
1
( )
2.10
=−
10.982
=
0.018 (1.8%)
⎪
⎪
ln 1
ln
[
⎟
300862800.005.png
Example of Slope Stability
Dimensionless
strength
parameter
C
=
c
u
γ
H
sat
φ
=
0
µ σ
u
,
c
u
γ
What is the relationship between the Factor of Safety ( FS ) of a slope
(based on the mean strength) and the probability of failure ( p f ) of an
undrained clay slope for different values of the mean and standard
deviation of strength?
u
c
sat
300862800.001.png

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