Probability theory in geotechnics—an introductionby G. N. SMITH*, MSc, CEng, MICE,

IntroductionWHEN USING limit state design one acknowledges that, forany proposed structure, there will always be a risk of failure and,provided that this probability of failure is acceptable, there is noneed to strengthen the structure further.

In the above context 'failure's defined as either the failureof the structure to satisfy some limit serviceability state or asthe ultimate failure of the structure.

The acceptable probability of failure is usually referred to asthe 'target probability'nd is given the symbol P«. By consider-ing social and economic demands that will be made by the societyin which the structure will exist, it is possible to determinevalues for P«corresponding to particular limit states.

Obviously, for an economical design, the actual probability offailure of the structure, P<, need not be less than P«.

The object of any design calculation is to predict the valuesof the design parameters, i.e. the material properties and the ap-plied loads, that will operate at failure. The operating values ofthese parameters are often referred to as the 'design values'ndgiven the collective symbol x,

By the use of statistics and probability theory it is possibleto design a structure so that its probability of failure, P<, isequal to P«and, in so doing, determine the design values, x,.*.

Partial factors of safety can then be evaluated for eachparameter, either in terms of mean values:—

X.

mI

or in terms of characteristic values:

X.tyx,t =—

L.iConversely, if the partial factors of safety corresponding to

the target probability of failure are known, then it becomes arelatively simple matter to determine the x,* values and to designthe structure so that its Pi value equals P«.

The previous paragraph gives an indication of the possibleformat of future structural design codes, once limit state prin-ciples have been fully adopted. The important part of thesecodes will be tabulated values of partial factors of safety. Thesevalues, which will be applied to obtain the design parameters,will cover a suitable range of target probabilities, related to dif-ferent serviceability limit states.

With such knowledge a design engineer could obtain all

the economic advantage of limit state design with a standard ofmathematical knowledge no higher than that used at present in

routine design calculations.The object of this article is to investigate the feasibility of

the use of first order second moment methods in the determina-tion of the probability of failure of geotechnical structures. In

order that the methods used in the worked examples are easyto follow, various simplifying assumptions have been made sothat there should be no attempt to draw any quantitative con-clusions from the results.

The first order second moment method was pioneered byCornell (1969) and has gradually been extended by other work-ers. In its present form the method is generally referred to as the'Level II approach'nd has been fully reported in the CIRIA Re-port No. 63, "Rationalisation of safety and serviceability factorsin structural codes", (1976).

Although, as the worked examples in the article indicate, theLevel II approach can be used as a direct design method, a slightchange in the algorithm can allow the method to determine

x,'aluescorresponding to a specified Pr value.It is seen therefore that the Level II approach is probably

best regarded as a stepping stone in the evaluation of partialfactors of safety for use in a code of practice, rather than as adirect design method.

*Senior Lecturer in Civil Engineering, Heriot-Watt University, Edinburgh

Let us consider two sets of random variables, one setmaking up the resistance or strength of a particular structure, R,and the other making up the random loading system, S, to whichthe structure will be subjected.

Obviously failure will occur when the value of the loadingexceeds the value of the strength.

If we define Z as being the function of strength minus load,i.e. Z = (R —S), we can write that failure will have occurredwhenS) RorwhenZ(0.

Failure will be imminent when S = R, or when Z = 0.Hence we can define the probability of failure, Pr, as being:

Pf ——P(R = S) = P(Z = 0)

By means of a Taylor's expansion it is generally possible toapproximate this failure surface to a straight line, or plane, atany specified point which lies on it. The point usually chosen isthe point at which failure is most likely to occur and is known asthe design point, x* = (x,*, x,, x,.'. x„*),where x,'s thevalue of the variable X,. at the design point.

The method is referred to as an advanced first order secondmoment method and assumes that the variables are normallydistributed. If it is necessary, it is possible to use transformedvalues of the mean and standard deviation of a variable that doesnot have a normal distribution although, if several variables ofroughly equal weight are grouped together, the resulting func-tion Z = g (X,,X,,....X„) will tend to have a normal dis-tribution even if the separate variables are not themselves normal.

The approach was pioneered by Cornell (1969) and hasbeen extended by several workers notably Hasofer & Lind (1974)and Rackwitz (1976). The method is fully reported in the CIRIAreport No. 63 (1976) and a basic introduction to it has beenprepared by Smith (1981).

Briefly the method consists of solving the set nf simultan-eous equations:

xr = m,.—x; Po.;

where x,.*

mtOixt

= the design value of X,. (its value at thepoint x")

= the mean value of variable X,.= the standard deviation of the variable X,= the sensitivity coefficient of the variable X,= the reliability index of the system,

xl =g,.'x") rr,.

n

(m, —x,.*) g,.'x')i=1

I g, 'x ) ~,. I

-'ote

g,.'x*) = the first derivative of g (X) at the design pointwith respect to X,.

October, 1981 29

Note that the term 'failure's meant to be used in its mostgeneral sense and implies the failure of the structure to satisfysome particular limit state criterion which may or may not beactual structural failure.

The failure boundary of a system, a line for a two-dimen-sional system, a surface for a multi-dimensional system, will in-variably be curved.

If we have a set of n variables, X = (X,, X,,....X„)then the equation of the failure boundary, or specified limit stateboundary, Z, can be expressed as

Z = g(X,,X, ....X)

Note. A check may be made onagainst the tabulated values:

(degrees) 10(radians) .1745

N, 9.48N., 1.22

the efficacy of these equations

15 20 25.2618 .3491 .436313.05 17.98 24.752.45 4.92 9.87

Fig. 1

The solut'on involves an iterative procedure and CIRIA sug-gest the following approach.1. Define the limit state function for the specified design and

specified failure mode in terms of the relevant number ofbasic variables, X, and design constants, C,

Zg(X~X2X2X)C~C)2. Guess a value for P3. Set x,.* = mr for all i4. Compute ag/ax,. for alii, at x = x"5. Compute x,. for all i6. Compute new x,.* values7. Repeat steps 4 to 6 until stable values of x,.'re achieved8. Evaluate Z = g (x,', x,', x,',....x„')9. Modify P to achieve Z = 010. Compute the notional failure probability y (- P)Note q (-P) is the general symbol for the area under the normaldensity function, shown in Fig. 1.

EXAMPLE ITests on undisturbed samples of soil taken from a thick soil

deposit showed that the soil had the following properties:Cohesion: m, = 10kN/m"-; o-,. = 1.5kN/m2

Friction: m~ = 25'; o.f, =3'nit

weight: m> ——20k N/m2; o-> ——2kN/m'he

soil is to support a 4m wide surface footing which willbe uniformly loaded with a vertical pressure having a mean valueof 200kN/m'nd a coefficient of variation of 20%. This loadingcan be assumed to include the weight of the foundation.

Ignoring any effects due to ground water or time, determine:(i) The factor of safety against bearing capacity failure, with

respect to mean values.(ii) The probability of bearing capacity failure.

Solution(i) Factor of safety

Terzaghi's formula for the ultimate bearing capacity of a sur-face strip footing is:

q„= cN„+ 0.5y BN~

where B = width of foundation

N, and N~ are bearing capacity coefficients dependent upon yFor y = 25': N,, = 25.1; N> ——9.7Hence q„= 10 x 25.1 + 0,5 x 20 x 4 x 9.7 = 639kN/m2

639Factor of safety = —= 3.2

200(ii) Probability of failure

Values for N, and Nt are given in most soils textbooks, e.g.Smith Ik Pole (1981).

(degrees) 10 15 20 25(radians) .1745 .2618 .3491 .4363

N, 9.6 12.9 17.7 25.1N 1.2 2.5 5.0 9.7

In order that the various differential operations may be car-ried out the relationships between N,, and y and between N, and

y must be established in the form of equations,If a microcomputer is available, or a programmable calculator

with a statistics module, such as supplied by Texas Instruments,it is relatively simple to obtain the equations that best fit thetabulated values. In this case both N,, and N~ follow exponentialcurves such that

N,. = 5.0e'""'P; N, = 0.3037e""""where y is in radians.

30 Ground Engineering

Hence q„= 5.0cea.""'»f' 0.5 x y x 4 x 0.3037e'"'8'P

5.0ce2.62r2 P + 06074 ye'.2 > ( —R)

Applied pressure = p (= S)

Z = (R —S) = q„—p

Z = 5.0ce' 'P + 0.6074ye"'""> -p= g(X,, X2, X, X,) where Xy c X2=y,'Xa = p; X~ = p

a2g 'x~) — —5 0e2.G003$

ac

a2g2'x*) = —= 0.6074e'""'P

ay

/3 = 3.130 <r,. = 1.5kN/m-"; o ~——2kN/m';

o.> = 3'.0524 radians); o- = 40kN/m2

Iteration Variables g (x*) Xi

START X, (c)X2 (y)X, (y)X,'p)

1 X, .1666X,, .1771X,, .9532X4 -0.1795

2 X, .2504X2 .1357X, .8308X~ -0.4782

3 X, .2447X2 .1446X., .8548X~ -0.4342

4 X, .2463X„ .1431X, .8498X4 O.4434

24.758419.7369

4057.07-1.0013.96305.6755

1327.29-1.0015.02886.6607

1503.92-1.0014.81376.4551

1464.11-1.00

x

10kN/m'0k

N/m'5"

200kN/m2

9.2218.8916.05

222.478.82

19.1517.20

259.868.85

19.0916.97

254.368.84

19.1017.02

255.52

(Z = -0.0004)Probability of failure = P<

——y (-3.130) = 0.00087There is a chance of failure of some 0.087%

Note. It is important in the determination of o, where units aremultiplied together, squared and then summated, that all unitsare consistent. It is for this reason that o<, must be expressed inradians.

Target probabilityA structure can suffer failure from several causes, accidents,

e.g. a gas explosion, irresponsible behaviour of the occupants, etc.The total probability of failure of a particular structure is thesummation of the probabilities of failure from all causes. It isestimated that the probability of failure due to design faults,

azg,'x') = —= 18.332ce'"""'"'P + 4.8466ye'"'l'Q

azg4 'x') = —= —1,0

ap

It now becomes possible to commence the iteration pro-cedure, starting with the mean values and adjusting P until avalue for Z is obtained that is close enough to zero to be ac-ceptable. Generally several iterations are necessary but a pro-grammable calculator removes most of the tedium.

For this example a value of 3.130 for P gives an acceptablevalue for Z. The full iteration is set out below.

Pl (design) say, is approximately one-tenth of the total proba-bility of failure, Pl (total).

The foregoing example has illustrated how it is possibles todetermine a value for Pl (design), the probability of failure of astructure due to a design or a construction fault. This value, al-though useful for comparative purposes, is fairly meaninglessunless it can be compared with some predetermined value forPr that is considered suitable for both the structure and the soc-ial environment in which it will exist. This predetermined valueof P< is known as the target probability and is given the symbol

PitObviously the task of determining sets of logical values for

P« that will embrace structural, regional and social aspects willbe the major concern of a committee dealing with the prepara-tion of a new structural design code which is to include probabil-ity theory. The task may prove to be impossible until a vastamount ol'tatistical data, collected over several years to come,is available.

A rough guide to P«values has been suggested by Cole(1980) who states that the maximum allowable target probabilitycan be related to the degree of damage that can be consideredas acceptable.

P«(total)10

'0'o10310'o

10'0

'egree of damage Ptt (design)Inconvenient (cracked paving slabs, etc.) ) 10 '-

Minor repairs necessary 10 'o 10 4

Major repairs necessary 10'o 100Major damage and/or casualties 10 o

Wt

~.25m w

.I

I

EXAMPLE IIThe cross-section of a cantilever retaining wall is shown in

Fig. 2. The retained soil has the following properties:

Unit weight, y, m~ = 18kN/m'; tr., ——1.0kN/m"

Angle of friction, tj, m> ——30'; o~, =2'he

applied loadings are:

Horizontal surcharge, w„,, m„, = 10kN/m-'; tr„= 2kN/m'li

Vertical line load, W~; this load can have any value ranging from0 to 1 000kN with a coefficient of variation of 20%.

The unit weight of concrete, y,, = 24kN/m'nd can beassumed to be of constant value.

The properties of the foundation soil may be taken as thesame as those of the retained soil and the coefficient of friction,p,, between the foundation and the foundation soil may be as-sumed equal to tan y.(i) Determine the value of W~ that corresponds to the maxi-

mum factor of safety against failure (using mean values).(ii) Determine the value of W~ that corresponds to the least

probability of failure. (All variables may be assumed tohave normal distributions).

Note. The investigation of the stability of a retaining wall in-volves the examination of four aspects:

(i) Sliding (2.0)(ii) Bearing capacity (2.0)

(iii) Overturning (2.0)(iv) Rotational slip in the surrounding soil (1.5)The figures in the brackets are the values quoted in CP2

"Earth retaining structures" (1951) as minimum factors of safetythat are acceptable for each mode of potential failure. Thesevalues apply in calculations that use design parameters that arethe mean values of laboratory test results.

In order to keep the example to a reasonable length it willbe assumed that the risk of failure by rotational slip is negligibleand may be ignored.

SolutionAs the retained soil is supported on a foundation slab there

will be no movement of soil relative to the back of the wallso that no friction can develop and the coefficient of activeearth pressure, K,, can be obtained from Rankine's formula:

K„ = tan'5'—It should be noted, for the probability analysis, that:

I)K„—= -tan 45' — 1 + tan'5'—i)Q 2 2

Active pressure at top of wall = K, w,

Active pressure at base of wall = K, w, + 5.5 K, y

5.5't K y.'. Total horizontal thrust on back of wall = + 5.5K, w,2

= (15.125y + 5.5 w,) K~(i) Sliding

Total resistance = R„= R, tan y, where R,. = total verticalreaction

R,, = 24 (.5 x 4 + 5 x .375) + 5 x 2.5 x y + 2.5w, + W~i.e. Total resistance (= R) = (93 + 12.5y + 2.5w, + W~) tan yTotal loading (= S) = Total horizontal thrust

(15.125y + 5.5w,,) K„

(ii) Bearing capacityTotal resistance = q„. Terzaghi's formula for q„ for a surface

footing on sand is:

q„= 0.5yBN> (For y = 30', N~ = 19.7)i.e. Total resistance (= R) = q„= 0.5./BN~

(Note: For simplification the reduction in bearing capacity due toinclined load effects has been ignored)

Total loading can be regarded as the maximum value ofbearing pressure that the foundation will experience, p „.

Taking moments about point A, the heel of the foundation:

2.5M due to weight of retained soil = 2.5 x 5 x,— y ——15.625y

22.5'-'

due to w„= —w,. = 3.125w,,2

Sm

fI

+90'

0.5m

li-. ~ C '.-''1m

25m

+0.5mFig. 2

A

M due to weight of concrete =24 (.375 x 5 x 2.75 + .5 x 4 x 2)= 219.75kNm

M due to horizontal thrust = (15.125y + 5.5w,) K, x heightabove foundat on level.

If we consider mean values then w, is equivalent to a height10m

of soil of —.Hence we will assume that the line of action18

1 10of total horizontal thrust acts at ——+ 5.5 = 2.0m above

3 18foundation level.

M due to WL, = 2.625W,Hence total applied M = 219,75 + 15.625y + 3.125w,. + 2.625 W~

+ 2 (15.125y + 5.5w,,) K„

October, 1981 31

For equilibrium the moment of R, about A must equal the totalapplied moment. If R, acts at a distance x from A then the ec-centricity of R,, about the centre line of the foundation, e, is

Bgiven by the equation e = x ——= x —2

2

2.1975 + 15.625y + 3.125ws + 2.625W~ +2 (15.125y + 5.5,) K

Hence e93 + 12'5y + 2'5w + Wc

R 6eand p,„= — 1 + — (=S)

B B(iii ) 0v crt urni ng

Taking moments about point C, the toe of the wall:

M due to concrete = 24 (.5 x 4 x 2 + .375 x 5 x 1.25)= 152.5kNm

M due to retained soil = 5 x 2.5 x 2.75 x ./——34.375 y

M due to surcharge and W~ = 2.5 x 2.75 x ws + 1.375W~= 6,875w, + 1,375WL

Total resistance = total resistive moment152.5 + 34.375y + 6.875ws + 1 375Wr,

(= R)

Total loading = M due to horizontal thrust on back of wall

= 2 (15.125y + 5.5ws) K„(=S)

I, Determination of factors of safetyW~ can have any value from 0 to 1 000kN, A suitable range

for W, values would be:0, 25, 50, 75, 100, 200, 300, 400, 500, 750, 1 000 k N.

(i) Sliding

(93 + 12.5y + 2.5ws + Wr,) tan gF = —=

S (15.125/ + 5.5ws) K„

Putting in mean values for y and ws and substituting

30'ana45' —for K„2

198.03 + 0.5774WF =

109.08and, substituting for Wc gives:

WI ( k N ) 0 25 50 75 100 200 300 400 500 750 1 000

F 1.82 1.95 2.08 2.21 2.34 2.87 3.40 3 93 4.46 5.78 7.11

(ii) Bearing capacityq„= 0.5y /3 Nv = 0.5 x 18 x 4 x 19.7 = 709.2kN/m-

(= R)219.75 + 15.625y + 3.125w + 2 625WI

+ 2 (15.125y + 5.5ws) K,Now e

93 + 12.5y + 2.5ws + Wc

andp„,= —1+

When mean values are inserted, the expression for e becomes

750.42 2 ~625WI—2

343 + W,,

Substituting for W, gives values for e and, consequently,values for p„„s and F, as tabulated below.

Wc ( k N ) 0 25 50 75 100 200 300 400 500 750 1 000

e( m) .1878 2175 2434 .2663 .2865 .3489 .3918 .4232 .4471 .4878 5133

p „„(k N /ma) 109.9 122.0 134.1 146.2 158.4 236.8 255.2 303.7 352.1 473.2 594.3

F = R/S 645 581 529 485 448 343 278 234 201 149 119

(iii) Overturning

R 152.5 + 34.375y + 6.875ws + 1.375W<F = —=

S 2 (15.125y + 5.5ws) Ka

By substituting mean values the values for F tabulated below

may be obtained.

Wr ( k N ) 0 25 50 75 100 200 300 400 500 750 1 000

F 3 85 4.01 4,17 4.32 4.48 5.11 5.74 6.37 7.00 8,58 10.15

II. Determination of probabilities of failure(i) Sliding

As R and S have already been established the expression forZ = R —S can be written down immediately:

Z = (93 + 12.5y + 2.5ws + W~) tan y—(15.125y + 5.5w,) Ka

rP

and, substituting tan'5' — for Ka gives:2

Z = (93 + 12.5y + 2.5ws + Wc) tan <5 —(15.125y + 5.5ws)

tans 45'—

g(C, X,, X,, Xs) where X, = y, X,. = w„;

azg,'x*) = —= 12.5 tan y —15.125 tana 45'—

By 2

azg,'x*) = —= 2.5 tan y —5.5tana 45'—

8w 2

Iteration Variables g/ (x*)

START X,( /)X„(w„)X.,(y)Xa(Wr )

X,X,,XsX,X,XsX.,X4

X,X.,XsX,

2.1752-0.3900775.920.5774-2.0071-1.5421761.350.3971

-2.0302-1.5486789.900.3962

.0783—.0281

.9746

.2078-0.0740-0.11370.97990.1464

-0.722-0.11020.98120.1410

18kN/ms10k

N/m'0'0I<N

17.6710.2421.6641.1118.3210.9721.6143.7318.3110.9421.6143.97

(Z = -0.0113)Pr ——<t. (-4.28) = 0.00000935Log„, P/ ———5.03

(ii) Bearing capacityThe ultimate bearing capacity of the foundation can be

found in a similar manner to the previous example, but for thefollowing range of values:

(degrees) 15 20 25 30 35(radians) .2618 .3491 .4363 .5237 .6109

N~ 2.5 5.0 9.7 19.7 42.4which gives the relationship Nr = 0.2971e"""449, where y is in

radians.

Hence qa = 0.5 x 4 X y x 0.2971es 044"f'

5942 es assails ( —R)

Now p „=— 1 + — = — 1 +

azg;,'x*) = —= (93 + 12.5y + 2.5w„+ Wz) sec"- y +

il gp

(15.125y + 5.5w„) tan 45' — 1 + tans 45'—2 2

azg,'x*) = —= tan f

aw,

The determination of P, and hence the probabilities of slidingfailure that correspond to the chosen W, values can now beundertaken.

As an example the final set of iterations for W~ = 50kN

(P = 4.280) are set out below.

P = 4.280; u~ = 1kN/m; <ra,s = 2kN/m'; <r,I, ——2'.0349 radians);<ru t = 10 kN

32 Ground Engineering

Mand substituting e = ——2 gives

R,

3 R„,

8 4

= 35.906—0.391~i—0,078w,. + 0.484WJ +(11.34y + 4.125w,,) K„(=S)'. Z = 0.5942y e" " I I< —35.906 + 0.391y + 0.078w3 —0.484W~

—(11.344' 4,125w,,) tana 45'—2

where <p is expre sed in radians in the expression e"'""'z

g,'x*) = —= 0.5942e" '""af'0.391—11.344tana 45'—ay 2

azg.,'x') = —= 0.078 —4.125 tana 45" ——

aw„, 2

It should be noted that the values for p„„, were evaluatedB

on the assumption that e < —(0.667m). This is the normal6

state of affairs in gravity and cantilever walls. However, it is asimple matter to evaluate e for each set of design values obtainedfor each value of W~ . When this is done it is found that e variesbetween 0.517 and 0.582m over the entire range of Wr .

(iii) Overturning

Z = 152.5 + 34.375y + 6.875w,. + 1'375Wr

2 (15.125y + 5.5w,.) tana 45'—2

aZg,'x') = —= 34.375—30.25 tan'5'—

ay 2

az(x') = —= 6.875 —11.0tan'5'—

aw 2

azga (X') = —=

a I/I

azg., (x)= —=

aw,

4.7919ye'"""'!' (11 344y + 4 125w )

tan 45' — 1 + tana 45'—2 2

—0.484

azga (x') = —= 2 (15.125y + 5.5w„) tan 45'—

aI/I 2

I/I

1 + tana 45'—2

azg,'x*) = —= 1.375

aw,

Iteration Variables g,.'x')START X,

X.,XaX,X,XaX3X4

X,X,XaX,X,XaX3X~

36.7458—1.29706071.86-0.4840

4.4214—1.94151624.64-0.4840

4.3222—1.94611658.89-0.4840

0.1708-0.0121

0.9850-0.0225

0.0773-0.0679

0.9911-0.0846

0.0740—0.0o66

0.9916-0.0829

(Z = 0.0276)

P! ——0.0000002144Log„, P< —— —6.67

An example of the ensuing iterations, for W,(P = 5.056) is set out below.

P = 5.056

50k N

X.!1810305017.1410.1220.0451.1417.6110.6919.9854.2817.6310.6719.9754.19

P =70Wi =0Iteration Variables g,.'x*) X(

X,X,X,X,

24.29173.2083503.831.3750

0.79210.20920.57350

12.467.07

21.970

(Z = 422.24)Hence P, ( y (—7.0) = 1.29 x 10-'3log„, Pr ( —11.89

h!oteThere are occasions when R is so much greater than S

that the value of Z is massive and can only be reduced tozero if at least some of the variables are reduced to ridiculous,or even negative values.

In these cases the simplest approach appears to be to deter-mine the value of P that gives the smallest value of Z withoutreducing any variable to less than a credible value, approximatelym; —4e-; .

Then we can say that the probability of failure is consider-ably less than y (-P).

This state of affairs is seen to occur in this example foroverturning. The least risk of overturning failure is when W~ = 0and the appropriate value for P in these circumstances was foundto be equal to 7.0, the iteration being shown below.

0- 0— 0-

I3.

o —60

—10—

—4 Log I3Pr (min)

I3,= —5.58

o —6-0

—10-

1- Max. F = 2.9

0I0o 4-CO

LL

—12

Fig. 3

I I

200 400 600 800 1 000Wc (kN)

(a)

I

200 400 600 800 1 000WL (kN)

(b)

6 I I I I

200 400 600 800 1 000Wc (kN)

(c)

October, 1981 33

The various probabilities of failure, best expressed aslog „, P,, are set out below.

SlidingWi (kN) 0 25 50 75 100 200 300 400 500 750 1 000LogtpP/ -3.74 -4,41 -5.03 -5.61 -6.15 -7.88 -8.95 -9.37 (-9,37

Bearing capacityWi (kN) 0 25 50 75 100 203 300 400 500 750 1 000Log„,P/ -7.75 -7.20 -6.67 -6.18 —5.72 -4.23 -3.18 —2.66 -1.87 -1.01 -0.57

OverturningLog tp Pi is less than -1 1 .89 for all values of WL

Fig. 3(a) shows the plots of logtp Pi against Wi for slidingand bearing capacity.

As the probabilities of failure for sliding and for rotationalslip are negligible, the total probability of failure can be effective-ly determined by simply summating the two curves of Fig. 3(a).This total probability curve is shown in Fig. 3(b) and from itwe see that the log of the least probability of failure is -5.58,when Wi ——100kN.

Fig. 3(c) shows the plot of the minimum F values. Fromthis figure we see that the maximum factor of safety possibleequals 2.9, when Wr ——250kN.

Obviously Figs. 3(b) and 3(c) should give the same valueof W, for least probability of failure and for maximum factor ofsafety. This is seen not to be so and indicates the need for someform of calibration.

Another approach would be to think in terms of target pro-bability. If P«(design) was 10 4 then, from Fig. 3(b) it is seenthat WL would have to be within the range 0 to 200kN.

If the factors of safety suggested by CP2 are used, it isfound that Wi wou'Id have to be within the range 50 to 500kN.

Acknowledge mentsThe author would like to acknowledge the interesting dis-

cussions and advice that he received-from Mr. R. A. Nichollsand Dr. E. Murray, of Nicholls Colton 8t Partners, testing engin-eers, Leicester, during the preparation of the material for thisPaper.

The Paper represents part of the work carried out by theauthor on behalf of the Transport and Road Research Laboratory.

In order to keep the presentation simple each example hasassumed stochastic independence of the variables involved. Itmay be that, for soil mechanics, this assumption will not alwaysbe acceptable and the author would welcome comments on thisor any other points from interested readers.

ReferencesBritish Standards Institution (1972): CP 110 (Part I) "The structural use ofconcrete".Coie, K. W. (1980): "Factors of safety and limit state design in geotechnicalengineering". Lecture given to the Scott sh Geotechnical Grouo, Glasgow.Construction Industry Research and Information Association (CIRIA) (1976):Report 63 "Rationalisation of safety and serviceability factors in structuralcodes".Cornell, C A. (1969) ''Structural safety specifications based on secondmoment reliability analysis", Final report of IABSE symposium on conceptsof safety of structures and reliability of design, London,Civil Engineering Code of Practice (1951): CP 2 "Earth retaining structures",Inst. of Structural Engineers, landon.Hasofer, A. M, & Lind N C. (1974): "An exact and invariant first-orderreliability format", Proc. Am Soc. Civ, Eng. —Jour. Eng. Mech. Div., Jan.Rackwirz, R. (1976): "Principles and methods for a practical probabilisticapproach to structural safety". Comite European du Baton.Smith, G. N. (1981) "Statistics and probability theory applied to earth

!retaining structures", Transport and Road Research Laboratory —Supple-mentary Report (to be published).Smith, G. N & Pole, E, L, (1981): "Elements of foundation design", GrenadaPublishing, St. Albans.

ONE PAPER at the Fifth International Sym-posium on Jet Cutting Technology, or-ganised jointly by BHRA and the Institutfur Werkstoffkunde (B) at Hanover, des-cribed methods of assisting the drivingand extraction of steel piles using high-speed water jets. In the instance of pile-driving reviewed, the water jet was usedto remove material and break up hard

rock ahead of the tip of sheet piling soas to reduce vibration which could havedamaged a nearby effluent disposal pipe.The actual driving was carried out with avibro-hammer, the piles penetrating med-ium sand and sandy mudstone strata intomudstone. The fan jet had an angle of30'n one plane (parallel to the sheetp'.le) and was supplied with water at a

(Left). Diagramma-tic illustration of thesheet pile drivingtechnique with highvelocity water jetto assist penetra-ti on

Hydraulicunit

Pile driving and extractionassisted by water jet

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pressure of 60MPa, but a flow of only0.05m'/min. It was found that the rate ofinstallation was approximately doubledand the vibration halved using the waterjet, compared with using the vibro-ham-mer on its own.

At another site, H-piles had been drivenin to a depth of 3.5m into mudstone afterpre-boring to a depth of Bm. When it wasnecessary to remove these piles a yearlater, the ground had consolidated somuch that no normal equipment wouldremove them. It was decided to free the

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Engineering

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Medium sand layer

G.L. —6.0m

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