Harr Paper

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1Slope Stability Santiago Chile, November 2009

Quantif the reliability of RockSlopes

In the introduction to Guidelines for Open Pit Slope Design (Guidelines) (1), it is stated, “It is often no longer sufficient to present slopedesign in deterministic (factor of safety) terms….” It is also noted, “A measure is required so that mine executives have sufficientinformation and understanding to be able to establish acceptable levels of risk.” To these assertions, as is well documented in theGuidelines, are the considerable uncertainties in geological and hydrological parameters (properties) and spatial variations andcorrelations that must be accommodated to obtain meaningful quantitative assessments of the adequacy of rock slopes.

A method of analysis is offered to meet these needs. Its outputs are the probabilit y of failure (PoF) and reli ability (R = 1 - PoF),the necessary ingredients for quantifying risk. The procedure can account for the contributions of many uncertain (random)

parameters, whether correlated or not. What is more, its versatility can provide reliability me asures for s tate of the art equations, graphs and char ts or computer generated studies. Its flexibili ty permits the evaluation and comparison of many possible scenarios. Many exam ples are presented to provide b ackground and to illus trate the m ethodology at the level of the practitioner.It will be assumed that the reader is familiar with Appendix 2 of the Guidelines. Reference to equations and figures in the Guidelineswill use the chapter and number designation: for example, Figure 6.43 will refer to the 43rd figure in Chapter 6; likewise, eqn. 10.2will designate the second equation in Chapter 10. Figures and equations in the present paper will use integer notation; Figure 9and Table 9 will refer to the 9 th figure and table in the present pape r.

Abstract

M.E. Harr

Professor Emeritus of Civil

Engineering

Purdue University

SOME BASIC CONCEPTS

Shown in Figure 1 are N derived factors of safety (FoS, see Tables 9.1 to 9.7) in selected ranges of values and their frequency ofoccurrence (fN). Note that the system is similar to a system of vertical forces on a beam: except, in the probabilistic case, the frequenciesmust sum to unity.

Example 1Given the following 10 tabular realizations for the factors of safety, obtain their expected value, variance, standard deviation, and

coefficient of variation.

FoS Occurrence Frequency, f

1.3 3 0.3

1.5 4 0.4

1.8 3 0.3

Sum 10 1.0

Figure 1 – Example 1



2Santiago Chile, November 2009 Slope Stability

Parameter Coefcient of variation, %

Friction angle Ø

Gravel 7Sand 12

c, strength parameter(cohesion) 40

Porosity 10

Specic gravity 2

Unit weight 3

Coefcient of permeability (240 at 80% saturation

to 90 at 100% saturation)

Example 2

Given the probabilistic parameters of the FoS in the previous example, obtain its probability distribution consistent to the principle of

maximum entropy.

In Table A2.2 it was shown that, given only the expected value and standard deviation of a distribution, the principle of maximum entropy

prescribes the symmetrical Normal (Gaussian) Distribution. Some examples were shown in Figure A2.4. Note that the transformation is made

from the discrete frequencies of Figure 1 to the continuous normal distribution, with probabilities measured as areas under the curve.

With the availability of computers today in the workplace, many probabilistic functions are readily and easily evaluated. Microsoft’s office

EXCEL functions will be used in the present paper. The EXCEL function for the normal distribution is:

+NORMDIST(x, expected value, standard deviation, true) (eqn. 1)

yielding the value for a normal distribution less than the variate x for the specified parameters. The word ‘true’ and the + sign

are required.

Example 3

Assuming a normal distribution and using the results of Example 1, obtain the probability that the factor of safety will lie between 1.30

and the value 1.70.

Here, P[ 1.30 ≤ FoS ≤ 1.70] = +NORMDIST[ 1.70, 1.53, 0.2, true] –

+NORMDIST[1.30,1.53,0.2,true]

= 0.802 – 0.125

= 0.677, call 68%

Example 4

If a FoS = 1 represents failure, estimate the probability of failure and reliability for the system in Example 1.

Here, +NORMDIST(1.00, 1.53, 0.20, true) = 0.004 (negligible) and

PoF = 0.004

= 0.4% and the reliability R = 99.6%.

From eqn. A2.6, the expected value (E[FoS]) is:E[FoS] = (1.3)0.3 + (1.5)0.4 + (1.8)0.3 = 1.53E[FoS2] = (1.3)20.3 + (1.5)20.4 + (1.8)20.3 = 2.37

From eqn. A2.8, the variance (V[FoS]) is:V[FoS] = 2.37 – (1.53)2 = 0.04

From eqn. A2.9, the standard deviation ( σ[FoS]) is:σ[FoS] = √ 0.04 = 0.20

From eqn. A2.10, the coefficient of variation (shown as percentages and placed within parenthesis, variances are designated withinbrackets) is:

V(FoS) =(0.20/1.53) 100 = 13%

Some representative values of coefficients of variation from Harr (2) are given in Table 1.

Table I – Some representative coefficients of variation




The inverse normal function:

+NORMINV(probability, expected value, standard deviation) (eqn. 2)

yields the variate for the specified probability for a normal distribution

Example 5

Find the value of the FoS for the previous example such that the probability it will be exceeded is 90%. Hence, we seek the value for

a probability of 10%:

+NORMINV(0.10, 1.53, 0.20)= 1.27 and FoS10%

= 1.3

The normal distribution ranges from minus infinity to plus infinity. In general, no properties, parameters, or factors of safety can display

negative values. The normal distribution will always assign a finite probability for a negative value of a variate. Hence, care must be taken

before blindly invoking the normal distribution. To avoid this, recourse can be made to the Beta distribution, wherein extreme values can

be specified. Often, one has a feel for the extremes of a design variable. Lacking this knowledge, as noted in Table A2.1, an assignment

of extremes at three standard deviations from the expected value is seen to be reasonable. The principle of maximum entropy (Table

A2.2) indicates that the versatile beta distribution is indicated, given the expected value, standard deviation and extreme (minimum and

maximum) values. Its measure can be found in EXCEL as (note the expression forα below differs from Equation A2.23a)

+BETADIST(x,α, β, a, b) (eqn. 3)

Where, x is the variable of interest,

α = (X2 /Y2 )(1-X) – X

β = ( α + 1)/X – ( α + 2)

a = minimum value of x

b = maximum value of x

X = (E[x]-a)/(b – a)

Y = σ[x]/(b – a)

In spite of its formidable appearance, the function is well suited to spreadsheet computations. Note, for symmetrical distributions α = β.

Example 6

Repeat Example 4 assuming a beta distribution.

With E[FoS] = 1.53, σ[FoS] = 0.20, assuming minimum and maximum values of 3 standard deviations above and below the expected

value (no other information being given), we have a = 0.93 and b = 2.13. Thus, α = 4 and β = 4 (symmetry requires α = β ) and

+BETADIST yields PoF =P[ FoS ≤ 1.0 ] = 0.00035 = 0.035% (negligible).

The inverse beta distribution is:

+BETAINV(probability,α, β, a, b) (eqn. 4)

Example 7

Repeat Example 5 for the beta distribution.

+BETAINV(0.10, 4,4, 0.93,2.13) = 1.26. The normal distribution gave 1.27!

The variance of a distribution can also be expressed as the product of expectations

V[x] = E[(x - E[x])(x – E[x])]

Suppose x and y are two variables associated with the same event. Their relative dependency is given by their covariance:

cov[x , y] = E[(x – E[x])(y –E[y])] (eqn. 5)

The measure of their relationship is the correlation coefficient, defined as:

ρ = cov[x , y]/ σ[x]σ[y] (eqn. 6a)




The correlation coefficient satisfies the condition:

+1 ≤ ρ ≤ +1 (eqn. 6b)

Where ρ = ± 1 denote perfect correlations, all corresponding values will lie on the same straight line: ρ = -1 designates when onevariable decreases the other increases, ρ =0 indicates no linear relationship between the two. Some examples are shown in Figure 2.

Figure 2 - Some examples of the correlation coefficient.

The EXCEL function +CORREL(x1:x

N,y

1:y

N ) gives the correlation coefficient between columns of N corresponding x, y variates. The

subscripts indicate the spreadsheet ranges.

Example 8

Given the following data set arranged in an EXCEL spreadsheet as shown, obtain the correlation coefficient.

Row x y 2 1 56 3 3 28 4 5 20

5 7 16 6 9 14

+CORREL(x2:x6,y2:y6 ) = -0.883

As expected, the minus sign indicates that as x increases y decreases and that there is a strong negative correlation.

POINT ESTIMATE METHOD (PEM)

The expected value and the standard deviation supply information concerning the central tendency and scatter about that value of arandom (uncertain) variable. The beam analogy previously referred to in Figure 1, is shown again in Figure 3a. It is seen that the beam issupported by a unit force acting at E[x]. Rosenblueth (3), recommended the analogue beam, shown in Figure 3b, be supported by forcesof 1/2, acting at E[x] – σ[x] and E[x] + σ[x] ( the derivation can be found in Harr (2), p206).

Figure 3 - Continuously distributed vertical force system on a rigid beam (a), single support (b), two supports




Example 9

The well-known coefficient of active earth pressure is K A = tan2(45 - Ø/2).

If E[Ø] =45o and V(Ø) = 12%, obtain the coefficient of variation of K A.

The standard deviation is σ[Ø] = 0.12(45°) = 5.4o. Hence, Ø- = 45 - 5.4 = 39.6o and Ø+ = 45 + 5.4 = 50.4o.

Substituting into the expression for K A yields K

A+ = 0.13 and K

A- = 0.22. It follows that:

E[K A] = 1/2 (0.13 + 0.22) =0.18 and

E[K 2 A] = 1/2 ( 0.017 + 0.048) = 0.033

From eqn. A2.8, V[K A] = 0.033 – 0.182 = 0.001 and σ[K

A] = 0.024.

Hence, V(K A ) = (0.024/0.18) 100 = 13.6%

Figure 4 shows the procedure followed to obtain the solution.

Figure 4 - Example 9.

Example 10

Using the results of the previous example, estimate P[KA ≤ 0.16 ].

Having only estimates of the expected value and standard deviation, the normal distribution will be employed with:

+NORMDIST(0.16, 0.18, 0.024, true) = 0.202. Hence, P[KA ≤ 0.16] = 20%

The point estimate method for one random variable, y = y(x), can be generalized as:

E[y] = p- y

- +p

+ y

+

E[y2] = p- y2

- +p

+ y2

+ (eqns. 7)

V[y] = E[y2] – E[y]2

where p- = p

+ = ½.

For two random variables y = y(x1, x

2 ), the equivalent expressions are given by Harr (2, p209):

E[y] = p++

y++

+ p+-y

+- + p

-+ y

-+ + p

—y

--

E[y2] = p++

y2++

+ p+-

y2+-

+ p-+

y2-+

+ p—

y2-- (eqns. 8a)

V[y] = E[y2] – E[y]2

where pij = 1/4 and y

±± = y(E[x

1] ± σ[x

1], E[x

2] ± σ[x

2])




If the variables are correlated, ρ is the correlation coefficient,

p++

= p-- = (1+ρ )/4 (eqns. 8b)

p+-

= p-+

= (1 – ρ )/4

Example 11

Equation 10.2 (of the Guidelines) gives an equation for the factor of safety, assuming a planar critical surface at an angle θ (joint dip

angle) for a slope of height H, inclined at an angle β (effective bench face angle) with the strength parameters c and Ø as:

FoS = tan(Ø)/tan( θ ) + 2c sin( β )/ γH sin ( β – θ )If E[c] = 40kPa, V(c) = 30%, E[Ø] = 35o, V(Ø) =12%, H = 250m, unit weight of γ =2.4 Mg/m3, β = 60o, θ = 47.5o, obtain the PoF =

P[ FoS ≤ 1.00] and the reliability (1 –PoF) for values of correlation coefficients of ρ = -1, 0, 1.

The solution was developed on an EXCEL spreadsheet, Figure 5. In the figure, the computer results have been rounded off. The FoS

was first obtained for ρ = 0 and then multiplied by the other correlation coefficients. The PoFs were obtained using the +NORMDIST

function. The versatility of the procedure should be apparent as answers are immediately available for any set of input parameters. Of

special note, the minimum PoF was obtained for ρ = -1, c and Ø are negatively correlated. This is generally the case. Implied is, if one

strength parameter gets weaker the other gets stronger.

FoS(c,Ø) = FoS(±, ±)

Factors Units

E[c] 40 kPa

E[Ø] 35 degrees 0.61 radians

β 60 degrees 1.05 radians

θ 47.5 degrees 0.83 radians

γ 2.4 Mg/m3

H 250 m

ρ = -1 ρ = 0 ρ = 1

ρ++ = ρ-- 0 0.25 0.5

ρ+- = ρ-+ 0.5 0.25 0

σ [c] 12

σ [Ø] 4.2

E[c]+σ [c] 52

E[c]- σ [c] 28

E[Ø]+σ [Ø] 39.2 degrees 0.68 radians

E[Ø]-σ [Ø] 30.8 degrees 0.54 radians

2sinβ/sin(β-θ) 8.00

For E[FoS]

ρ = -1 ρ = 0 ρ = 1

FoS(+,+) 0.00 1.44 0.72

FoS(+,-) 0.62 1.24 0.00

FoS(-,+) 0.56 1.12 0.00

FoS(-,-) 0.00 0.92 0.46

E[FoS] = 1.18 1.18 1.18




For E[FoS2]

ρ = -1 ρ = 0 ρ = 1

0.00 2.08 1.04

0.77 1.54 0.00

0.63 1.26 0.000.00 0.85 0.42

E[F0S2]= 1.40 1.43 1.46

V[FoS] = 0.0035 0.0357 0.0679

σ [FoS] = 0.06 0.19 0.26

P[ FoS ≤ 1] 0.12 17.01 24.45 %

Reliability, R 99.90% 83% 76%

Figure 5 - Spreadsheet solution of Example 11

Figure 6 is one of a number of design charts developed by Hoek and Bray (4).

Figure 6 - Examples 12 & 13 (after Hoek & Bray(4)) .




Example 12

Given H = 14m, c= 15kPa, Ø = 30o , γ = 18kN/m3 , and FoS = 1.3, using Figure 6 find the effective bench face angle β. The solution

is indicated by the arrows on the figure to yield β = 40o.

Example 13

If the coefficients of variation of c and Ø are V(c) = 40% and V(Ø) = 10% and β = 40o , estimate the PoF and reliability R for the data

of Example 12 using the same Hoek and Bray chart in Figure 6, assume the variates are not correlated.

The spreadsheet solution is given in Figure 7, PoF = 5.4% and the reliability = 94.6%.

FoS(c,Ø) = FoS(±, ±)

Factors Units

E[c] 15 kPa

E[Ø] 30 degrees 0.52 radians

β 40 degrees 0.70 radians

V(c) = 40%, V(Ø) = 10%

γ 18 Mg/m3

H 14 m

σ [c] 6

σ [Ø] 3

E[c]+σ [c] 21

E[c]- σ [c] 9

E[Ø]+σ [Ø] 33 degrees 0.58 radians

E[Ø]-σ [Ø] 27 degrees 0.47 radians

Entering Figure 6 with the following values of c, Ø, β

c, Ø c/γHtanØ tanØ/FoS FoS FoS2

+ + 0.13 0.39 1.67 2.79

+ - 0.16 0.35 1.46 2.13

- + 0.05 0.52 1.25 1.56

- - 0.07 0.48 1.06 1.12

E[ ] = 1.36 1.90

V[FoS] = 0.050

σ [FoS] = 0.22

P[FoS ≤ 1] 5.4 %

Reliability, R 94.6 %

Figure 7 - Spreadsheet solution to Example 13

The point estimate method for more than two random variables is given in Chapter 4 of Harr (2, pp 218-220). For very many variables,

whether correlated or not, see Harr (5). Care should be taken in the latter publication (5), as a mathematical error was found that

telescopes through the paper.

A three point estimate solution can be used for functions of random variables that display minimum or maximum values within the range

of E[x] ± σ[x]. Instead of p- = 1/2 and p+ = 1/2, as was shown for two points, values are calculated at p- = 1/6, p

+= 1/6, acting at E[x]

- √ 3 σ[x] and E[x] + √ 3 σ[x], respectively, and po = 2/3, acting at E[x].

Example 14Obtain the three point solution for Example 9.




From Example 9, K A = tan2(45 - Ø/2), with E[Ø] =45o and V(Ø) = 12%, σ[Ø]= 5.4, it follows:

E[Ø] + √ 3 σ[Ø ] = 54.35o; K A+ = 0.103 1/6 K

A+ = 0.017

E[Ø] - √ 3 σ[Ø ] = 35.65o; K A- = 0.264 1/6 K

A- = 0.044

E[Ø] = 45o K Ao

= 0.172 2/3 K Ao

= 0.115

E[K A] = 0.176

1/6 K 2 A = 0.002

1/6 K 2 A- = 0.012

2/3 K 2 Ao

= 0.020

E[K 2 A

] = 0. 034

Within round off accuracy, as the KA function is monotonic, the result is the same as in Example 9.

Test x1 x2 x3 dx1 dx2 dx3

1 2 1 2 -5.25 -6.67 -4.92

2 4 3 1 -3.25 -4.67 -5.92

3 6 7 5 -1.25 -0.67 -1.92

4 7 6 3 -0.25 -1.67 -3.92

5 5 4 7 -2.25 -3.67 0.08

6 8 9 6 0.75 1.33 -0.92

7 9 8 7 1.75 0.33 0.08

8 7 10 9 -0.25 2.33 2.08

9 10 11 11 2.75 3.33 4.08

10 8 9 8 0.75 1.33 1.08

11 12 11 10 4.75 3.33 3.08

12 9 13 14 1.75 5.33 7.08

E[xi] 7.25 7.67 6.92

Figure 8 - Tabular values of test results for three variates

Suppose that 12 tests are performed and sample measurements of variables x1, x2, and x3 yield the values in Figure 8. Also shown

are the sample expected values 7.25, 7.67, and 6.92. The columns labeled dx1, dx2, and dx3 are mean corrected; that is, the respective

expected values have been subtracted from the raw data (for example, for test 1, 2 – 7.25 = -5.25)

Designate a matrix D (2, pp 236-237 and Appendix A) composed of the 12 rows and 3 columns of dx1, dx2, and dx3. Performing the

matrix multiplication (DT is the transpose of D), produces the symmetrical covariance matrix C:

7.48 8.73 7.84

C = (1/11)DTD = 8.73 12.97 12.24

7.84 12.24 14.63

The elements on the principal diagonal are the respective variances of x1, x2, and x3 (7.48, 12.97, and 14.63). Hence the respectivestandard deviations are σ[x1] = 2.73, σ[x2] = 3.60, and σ[x3] = 3.82. The off-diagonal elements are the respective covariances (for

example, 8.73 = σ[x1]σ[x2]ρ1,2). Hence, the correlation coefficient between x1 and x2 is:

ρ1,2 = 8.73/(2.73)(3.60) = 0.89

Thus, it is seen that data such as in Figure 8 can readily be reduced to produce the expected values, standard deviations, and

correlation coefficients of random variables, the very fodder of probabilistic analysis. Computer software abound that yield the above

results directly.

CLOSURE

It was the intent of this paper to demonstrate how quantitative measures of reliability and probability can be obtained by a simple

yet very versatile methodology. It is believed that the wide use of examples in the paper has enhanced the reader’s ability to apply thedeveloped methodology to the design and evaluation of rock slopes.




APPENDIX

E [x] = x ̄ (eqn. A2.6)

V [xi ] = E [xi2

] - ( E [xi ])2

(eqn. A2.8)

σ[xi] = (eqn. A2.9)

V(x) = x 100 (%) (eqn. A2.10)

α = (eqn. A2.23a)

Figure A2.4 - Examples of normal probability distribution funcions

Table IX.1 - Examples of acceptable FoS values (Pries & Brown 1983)




H Chebyshev’s

inequality

Gauss’

inequality

Exponential

distribution

Normal

distribution

Uniform

distribution

0,5 0 0 0,780 0,380 0,29

1 0 0,56 0,860 0,680 0,58

2 0,75 0,89 0,950 0,960 1,00

3 0,89 0,95 0,982 0,997 1,00

4 0,94 0,97 0,993 0,999 1,00

Table A2.1 - Probabilities for range of expected values ± h sigma units

Table A2.2 - Maximum entropy probability distribution

REFERENCES

1. J.R.L. Read and P.F. Stacey, Guidelines for Open Pit Slope Design, CSIRO Publishing, Melbourne, Australia, 2009.

2. M.E. Harr, Reliability-based Design in Civil Engineering, Dover Publications, New York, 1987.3. E. Rosenblueth, “Point Estimates for Probability Moments”, Proc. Nat. Acad. Sci. USA, V72(2), 1975.

4. E. Hoek and J. Bray, Rock Slope Engineering (3rd edn.), IMM, London 1981.

5. M.E. Harr, “Probabilistic Estimate For Multivariate Analyses”, Appl. Math. Modelling, V13, May 1989.

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Harr Paper