Calculations involving a single random variable (SRV)

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Mercy Fisher
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1 Calclations involving a single random variable (SRV) Example of Bearing Capacity q φ = 0 µ σ c c = 100kN/m = 50kN/m ndrained shear strength parameters What is the relationship between the Factor of Safety (FS) of a conventional bearing capacity calclation (based on the mean strength) and the probability of failre (p f )?

2 First perform a deterministic calclation Conventional bearing capacity calclations typically involve high factors of safety of at least 3. The bearing capacity of an ndrained clay is given by the Prandtl eqation: q = ( +π ) c = 5.14c where c is a design mean vale of the ndrained shear strength. If c = 100kN/m, and FS = 3 this implies an allowable bearing pressre of: q all kN/m = = 3

3 Now perform a probabilistic calclation If additional data comes in to indicate that the same ndrained clay has a mean shear strength of µ = 100kN/m and a standard deviation of σ = 50kN/m c c and is lognormally distribted, what is the probability of bearing failre? In other words, what is the probability of the actal bearing capacity being less than the factored deterministic vale Pq [ < 171]? q = 5.14 c, hence if c is a random variable we can write: E[ q ] = 5.14E[ c ] ths µ =5.14 µ =514 q c and Var[ q] = 5.14 Var[ c] ths q =5.14 c =57 σ σ (Note that since q c, V = V q c = 1 ) Failre occrs if q < q all The probability of this happening can be written as Pq [ < 171]

4 First find the properties of the nderlying normal distribtion of ln q { } 1 σ ln q = ln 1+ V ln q = + = 1 1 µ ln q = ln µ ln ln(514) (0.47) 6.13 q σ q = = ln Pq [ < 171] =Φ 0.47 =Φ.10 ( ) = 1 Φ.10 ( ) = = (1.8%)

5 Example of Slope Stability φ = 0 µ, σ γ c sat c Dimensionless strength parameter C = c γ H sat What is the relationship between the Factor of Safety (FS) of a ope (based on the mean strength) and the probability of failre (p f ) of an ndrained clay ope for different vales of the mean and standard deviation of strength?

6 First perform a deterministic calclation Using limit eqilibrim or charts, find the FS corresponding to a niform ope of strength µ c for a homogeneos ope FS α C Linear relationship between C and FS for a cohesive ope with a ope angle of β =6.57 and a depth ratio of D = o

7 Now perform a probabilistic calclation 1) Noting that FS C or FS = KC, compte K, hence µ = Kµ and V = V FS C FS C ) If C is lognormal, so is FS, so compte the nderlying normal properties σ and µ 3) Find the probability P[ FS < 1 ] from standard tables. ln1 µ ln FS µ ln FS If FS is lognormal, P[ FS < 1] =Φ =Φ σ σ ln FS ln FS ln FS 1 µ FS If FS is normal, P[ FS < 1] =Φ σ FS ln FS

8 Single Random Variable (SRV) approach Defining FS 5.88µ C Probability of Failre p f or p f = p( FS < 1) = p( C < 0.17) High! V C σ C = µ C The mean strength may be too optiminstic. Probability of failre p vs. Factor of Safety FS (based on the mean) f

9 The median helps to interpret lognormal behavior p f MedianC < 0.17 MedianC > 0.17 p vs. V for different vales of Median f C C

10 Mean strength factoring All crves assme FS=1.47 based on C des = 0.5 linear scaling C = µ f µ des C 1 C C sd scaling = µ f σ des C C

11 All the examples considered so far were for a single random variable (sally the ndrained strength) How can we accont 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 inflence of several random inpt variables on a fnction is the First Order Second Moment (FOSM) method. The method is called First Order becase it incldes only first order terms in a Taylor expansion. The method is called Second Moment becase it enables estimates to be made of the variance (second moment) of fnction nder consideration.

18 Earth Pressre Example sing FOSM with 4 Random Variables 0.30 Cohesionless soil H = c = 0, φ Active Earth Pressre Coefficient K a φ = tan (45 ) γ c γ bf P a Units in kn and m tanδ 3.66

19 H = 5.03 Cohesionless soil c = 0, φ Active Earth Pressre Coefficient K a φ = tan (45 ) 1 P = γ H K = γ K Rankine's Theory: a bf a 1.65 bf a W = W + W γ c γ bf P a tanδ 3.66 Factor of Safety against iding: c bf 1 = 0.46(3.66) γ + (4.57)( ) γ (4.57) γ = 3.4γ γ c c c bf bf FS = Units in kn and m ( c + bf ) 3.4γ 8.36γ tanδ 1.65γ bf K a

20 Inpt data assming all variables are random FS is a fnction of 4 random variables: FS = f ( γ c, γbf, Ka,tan δ) Property γ c (kn/m 3 ) µ 3.58 σ 0.31 γ bf (kn/m 3 ) K a tan δ

21 Now se the FOSM Method to estimate the statistics of FS ( c + bf ) 3.4 E[ γ ] 8.36 E[ γ ] E[ tan δ] E[ FS] = µ 1.65 E[ γ ] E[ K ] bf a FS FS FS FS FS Var[ FS] Var[ γ c] + Var[ γbf] + Var[ Ka] + Var[ tan δ] γ c γ bf Ka (tanδ) = σ FS The mean vale of can be estimated in the FOSM Method by sbstitting all the FS mean vales of the inpt variables into the governing fnction, ths: ( + ) 3.4(3.58) 8.36(18.87) 0.5 Hence, µ FS = (18.87)0.333 Since we have a fnctional relationship between FS and the inpt variables, the variance of FS can be estimated in this case by evalating the derivatives analytically at the mean.

22 Factor of Safety against iding: FS = ( c + bf ) 3.4γ 8.36γ tanδ 1.65γ bf K a FS γ 0.7 tanδ = = γ K c bf a 0.0 FS γ 0.7γ tanδ = = 0.03 c bf γbf Ka FS K (0.66γ γc ) tanδ = = 4.50 bf a γ bfka FS (0.66γ 0.7 ) bf + γ c = = 3.00 (tanδ) γ K bf a All derivatives evalated at the mean

23 Hence: Var[ FS ] = σ FS = 0.046=0.1 µ FS = 1.50 σ FS Smmary: VFS = = 0.14 σ FS = 0.1 µ FS What is the probability of iding failre? P[ FS < ] 1??

24 In order to compte probabilities we need to assme a sitable PDF Let s assme that FS is lognormally distribted First find the mean and standard deviation of the nderlying normal distribtion σ ln FS { } { V } = ln 1+ = ln = 0.14 FS 1 1 µ ln FS = ln µ ln ln(1.5) (0.14) 0.40 FS σ FS = = [ 1] P FS ln < =Φ 0.14 =Φ(.86) = 1 Φ(.86) = = or 0.0%

25 If we instead assme that FS is normally distribted [ 1] P FS < =Φ 0.1 =Φ(.38) = 1 Φ(.38) = = or 0.9 %

26 Nmerical Differentiation In this case we were able to differentiate analytically, bt in some problems we do not have a fnction to work with (e.g. ope stability). In these case we can se nmerical differentiation Reworking of the earth pressre problem sing nmerical differentiation. FS = f ( γ, γ, K,tan δ) c bf a FS FS FS FS Var[ FS] Var[ γ c] + Var[ γbf] + Var[ Ka] + Var[ tan δ] γ c γ bf Ka (tanδ) We can evalate derivatives sing a central difference formla in which we sample the fnction at one standard deviation below and one standard deviation above the mean. e.g. FS K f( µ γ, µ γ, µ K + σ K, µ tanδ) f( µ γ, µ γ, µ K σ K, µ tanδ) fk = σ σ c bf a a c bf a a a a K K Note all variables are fixed at their mean except for the derivative variable a a

27 After sqaring the derivative, the standard deviations cancel and we are left with: fγ bf Ka tanδ fγ f f c Var[ FS] Compte the vales sing a tablar approach Property γ c (kn/m 3 ) µ 3.58 σ 0.31 µ± σ f f 0.01 γ bf (kn/m 3 ) K a tan δ

28 Now sbstitte these vales into the variance eqations Var[ FS] = Hence, σ FS = 0.1 which is the same vale (to decimal places) as the one we obtained analytically. The fact that the analytical and nmerical rets are very similar in this case indicates that there is almost no nonlinearity in this problem.

29 The Reliability Index The Reliability Index is a measre of the margin of safety in standard deviation nits. For example, if dealing with a Factor of Safety, the reliability Index is given by: β = µ FS σ 1 FS In a normal variate, the reliability index (β) is niqely related to the probability of failre (p f ) throgh the expression: p f = 1 Φ( β )

30 Consider a normal distribtion of the Factor of Safety (FS) f FS µ σ FS FS = 1.5 = 0.1 p f is this area FS β is this distance the standard deviation (1.5 1) β = =.38 p f = 1 Φ(.38) = =

31 0.5 Probability of Failre: p f Reliability Index: β Probability of Failre vs. Reliability Index for a Normal Distribtion

Prandl established a universal velocity profile for flow parallel to the bed given by

Prandl established a universal velocity profile for flow parallel to the bed given by EM 0--00 (Part VI) (g) The nderlayers shold be at least three thicknesses of the W 50 stone, bt never less than 0.3 m (Ahrens 98b). The thickness can be calclated sing Eqation VI-5-9 with a coefficient