Geotechnical Engineering Soubra Macuh 1074581022509

8/10/2019 Geotechnical Engineering Soubra Macuh 1074581022509

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functions and by using a powerful spreadsheet optimisation

tool. The analysis is made in the general case of an inclined

wall and a sloping backfill, and considers a frictional and

cohesive (c, ) soil. A uniform surcharge is assumed to act

on the ground surface. Active and passive earth pressure

coefficients due to soil weight, cohesion and surcharge loading

are presented for various governing parameters and compared

with those given by other authors.

2. FAILURE MECHANISMS AND GOVERNING

EQUATIONS

The variational analysis details and the equivalence between

the variational limit equilibrium method and the upper-bound

method in limit analysis for a rotational log-spiral failure

mechanism are given elsewhere.2

However, for clarity, only the

upper-bound technique (that is, the kinematical approach) of

limit analysis is briefly described here.

Two rotational log-spiral failure mechanisms are consideredin

the present analysis, one for the activestate M1 (Fig.1(a)) and

the other for the passive state M2 (Fig. 1(b)). For both M1 and

M2 mechanisms, the region ABC rotates as a rigid body aboutthe as yet undefined centre of rotation O relative to the

material below the logarithmic failure surface BC. Thus the

surface BC is a surface of velocity discontinuity. These failure

mechanisms can be specified completely by two variables

0 and 1. It should be emphasised that the earth pressure

coefficient due to soil weight, K, is calculated with the

assumption of a cohesionless soil with no surcharge loading.

The computation of the coefficients Kq and Kc due to

surcharge loading and cohesion is based on the assumption of

a weightless soil with c 0 for Kq and q 0 for Kc. The

formulation for the coefficients of earth pressure due to soil

weight, surcharge and cohesion follows.

2.1. Rate of work of external forces

As shown in Fig.1, the external forces acting on the soil

mass in motion consist of the self-weight of the soil, W, the

active or passive earth force ( Pa or Pp), the adhesive force,

Pad( cl tan = tan ), and the surcharge, qL , acting on the

ground surface. The rate of work for the different external

forces can be calculated as follows.

2.1.1. Rate of work of the soil weight. A direct integration of

the rate of work of the soil weight in the region ABCis very

complicated. An easier alternative is first to find the rate ofwork _WWOBC, _WWOAB and _WWOAC due to soil weight in the regions

OBC, OAB and OAC respectively. The rate of work for the

0

A

C

Br0

r1

Pad

Pa

r

l

L

1

0

> 0

> 0

q

r= r0e(

0)tan

(a)

W

Fig. 1. Log-spiral failure mechanisms: (a) M1 for active; (b) M2 for passive earth pressure analyses

Geotechnical Engineering 155 Issue 2 Soubra Macuh0 Earth pressure coefficients


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region ABC is then found by simple algebraic summation,_WWOBC _WWOAB _WWOAC. The steps of computation of the rate of

work due to self-weight of the soil are essentiallythe same as

those of an inclined slope considered by Chen.3

It is found that

the rate of work due to the soil weight in the region ABC is

_WWsoil r30(f1 f2 f3)1

where f1, f2 and f3 are non-dimensional functions, which are

given in Appendix 1.

2.1.2. Rate of work of the active or passive force and the

adhesive force. The rate of work of the active or passive force

(Pa or Pp) and the adhesive force, Pad, can be expressed as

follows:

_WW( Pa or Pp), Pad Pa,p r0f4 cr20f52

where f4 and f5 are non-dimensional functions, which are

given in Appendix 1. It should be mentioned that the active or

passive force is assumed to act at the lower third of the wall

length for the calculation of the coefficients Ka and Kp.However, the computation of Kac, Kpc, Kaq and Kpq is based on

the assumption that the point of application of the active or

passive force is applied at the middle of the wall length. These

hypotheses are in conformity with the classical earth pressure

distributions, and allow direct comparison with existing

solutions.

2.1.3. Rate of work of the surcharge loading. The rate of work

of the surcharge loading q can be expressed as follows:

_WWq qr20f63

where f6 is a non-dimensional function, which is given in

Appendix 1.

The total rate of work of the external forces is the summation

of these three contributionsthat is, equations (1), (2) and (3):

X[ _WW]ext _WWsoil _WW( Pa or Pp), Pad

_WWq4

2.2. Rate of energy dissipation

Since no general plastic deformation of the soil is permitted

to occur, the energy is dissipated solely at the velocitydiscontinuity surface BC between the material at rest and the

material in motion. The rate of energy dissipationper unit area

of a velocity discontinuity can be expressed as3

0

A

C

B

r0

r1

Pad

Pp

r

l W

L

1

0

> 0

> 0

q

r= r0e(

0)tan

(b)

Fig. 1. (continued)

Geotechnical Engineering 155 Issue 2 Soubra Macuh 121Earth pressure coefficients


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_DD cVcos 5

whereVis the velocity that makes an angle with the

velocity discontinuity. The total rate of energy dissipation

along BC can be expressed as follows:

_DDBCcr

20f76


Appendix 1.

2.3. Energy balance equation

By equating the total rate of work of external forces (equation

(4)) to the total rate of energy dissipation (equation(6)),we

have

r3

0(f1 f2 f3) Pa,p r0f4 cr2

0f5 qr2

0f6 cr2

0f77

The energy balance equationof the rotational log-spiral

mechanism (i.e. equation(7)) is identical to the moment

equilibrium equation about the centre of the log-spiral. It

should be emphasised that the log-spiral function has a

particular property, that the resultant of the forces ( dl ) and

(tan dl ) passes through the pole of the spiral. Hence the

moment equilibrium equation of the soil mass in motion about

the centre of the log-spiral is independent of the normal stress

distribution along the slip surface. Based on equation (7), the

active and passive forces can be expressed respectively as

follows:

Pa Kal2

2 Kaq ql Kac cl8

Pp Kpl2

2 Kpqql Kpc cl9

where Ka, Kp, Kaq, Kpq, Kac and Kpc are the earth pressure

coefficients. The coefficients K, Kq and Kc representrespectively the effect of soil weight, vertical surcharge loading

and cohesion, and the subscripts a and p represent the active

and passive cases respectively. These coefficients are given as

follows, using the lower sign for the passive case:

Ka Kp 2

l

r0

2 (f1 f2 f3)f410

Kaq Kpq 1l

r0

f6f411

Kac,pc 1

l

r0

f7 f5f412

For a surcharge loading q0 normal to the ground surface, the

active and passive earth pressure coefficients, Kaq0 and Kpq0,

are given as follows:

Kaq0,pq0 1

l

r0

f8

f413


Appendix 1.

3. NUMERICAL RESULTS

The most critical earth pressure coefficients can be obtained by

numerical maximisation of the coefficients Ka, Kaq and Kaq0and minimisation of the coefficients Kac, Kp, Kpq, Kpq0

and Kpc. These optimisations are made with regard to theparameters0 and 1. The procedure can be performed using

the optimisation tool available in most spreadsheet software

packages. In this paper the Solver optimisation tool of

Microsoft Excel has been used. Two computer programs have

been developed using Visual Basic for Applications (VBA) to

define the active and passive earth pressure coefficients as

functions of the two angular parameters 0 and1 defined in

Fig.1.

In the following sections, the passive and active earth pressure

coefficients obtained from the present analysis are presented

and compared with those given by other authors. Then a

demonstration of the implementation of earth pressure

coefficients as user-defined functions in Microsoft Excel Visual

Basic is presented. An illustrative example shows the easy use

of spreadsheets in optimisation problems. The paper ends with

the presentation of two design tables giving some values of the

active and passive earth pressure coefficients for practical use

in geotechnical engineering.

3.1. Passive earth pressure coefficients

There are a great many solutions for the passive earth pressure

problem in the literature based on

(a) the limit equilibrium method415

(b) the slip line method 1620

(c) limit analysis theory.2, 2128

The tendency todayin practice is to use the values given by

Kerisel and Absi.29

3.1.1. Comparison with Rankine solution. For the general case

of an inclined wall and a sloping backfill (= 6 0,= 6 0),

the Rankine passive earth pressure is inclined at an angle

with the normal to the wall irrespective of the angle of friction

at the soilwall interface,30

where

tan sin( 2)sin

1 sin cos( 2)14



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and

sin sin

sin 15

The inclination of the slip surface with the horizontal direction

is given as follows:

2

4

216

and the coefficient Kp is given by

Kp cos( )sin cos sin( )

[1 sin cos( 2)]17

In order to validate the results of the present analysis, one

considers a soilwall friction angle equal to the value

given by equation(14).The numerical solutions obtained by

the computer program have shown that, in these cases, the

present results are similar to theexact solutions given by

Rankine (that is, equations(16) and(17)); the log-spiral slip

surface degenerates to a planar surface with radii approaching

infinity.

It should be emphasised that the results obtained from the

computer program indicate that the coefficient Kpc is related to

the coefficient Kpq0 by the following relationship (cf. Caquots

theorem of corresponding states31

):

Kpc Kpq0

1

cos tan

18

Also, it should be mentioned that the critical angular

parameters0 and1 obtained from the minimisation of both

Kpq0 and Kpc give exactly the same critical geometry.

3.1.2. Comparison with Kerisel and Absi. Figures2 and3

show the comparison of the present solutions of Kp and Kpq

with those of Kerisel and Absi29

for the case of a vertical wall

and an inclined backfill, and for the case of an inclined wall

and a horizontal backfill respectively, when 408.

For the Kp coefficient, the present results are greater than

those of Kerisel and Absi. However, the maximum difference

does not exceed 13%. For the Kpq coefficient, the present

solutions continue to be greater than those of Kerisel and Absi;

a maximum difference of 22% is obtained for the extreme case

when 408,= 1,= 0 and 308.

To conclude, the present solutions of Kp and Kpq are greater

than the ones given by Kerisel and Absi. However, for practical

configurations ( < 408, 1=3 < = < 2=3,= < 1=3 and

08), the maximum difference does not exceed 5% for Kp

and 7% for Kpq.

3.1.3. Comparison with the existing upper-bound

solutions. Rigorous upper-bound solutions of the passive earth

pressureproblem are proposed in the literature by Chen and

Rosenfarb23

and Soubra.27

Chen and Rosenfarb considered six

translational failure mechanisms and showed that the log-

sandwich mechanism gives the least (that is, the best) upper-

bound solutions. Recently, Soubra27

considered a translational

multiblock failure mechanism and improved significantly the

existing upper-bound solutions given by the log-sandwich

mechanism for the Kp coefficient, since he obtained smaller

upper bounds. The improvement (that is, the reduction relative

to Chen and Rosenfarbs upper-bound solution) attains 21%when 458,= 1,= 1 and 158.

The results of Kp and Kpq given by the present rotational

failure mechanism and those given by Soubra27

using a

translational failure mechanism are presented in Fig. 4 for the

general case of an inclined wall and a sloping backfill when

458 and= 1.

For the Kp coefficient, the present upper-bound solutions

are smaller (that is, better) than those of Soubra.27

The

improvement (that is, the reduction relative to Soubras upper-

bound solution) is 27% when 458,= 1,= 1 and

158. For the Kpq coefficient,it should be mentioned that

the values obtainedby Soubra27

are identical to those given by

Kerisel and Absi29

and correspond to the exact solutions for a

100

90

80

70

60

50

40

30

20

10

0

Kp

50

40

30

20

10

5

15

25

35

45

0

Kpq

Present solution

Krisel and Absi

40 30 20 10 0 10 20 30 40

/= 1

/= 1/2

/= 0

Present solution

Krisel and Absi/= 1

/= 1/2

/= 0

Fig. 2. Comparison of present Kp and Kpq coefficientswith those of Kerisel and Absi

29(for vertical wall and

inclined backfill)



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weightless soil. The present upper-bound solutions of the Kpqcoefficient are greater than those of Soubra

27and thus

overestimate the exact solutions. However, for practical


08) the maximum difference does not exceed 75%.

3.2. Active earth pressure coefficients

As in the case of passive earth pressures, the numerical solutions

obtained by the computer program have shown that the presentmodel gives the exact solutions proposed by Rankine (when

they exist). In these cases, the log-spiral slip surface degenerates

to a planar surface with radii approaching infinity. Also, it

should be mentioned that the following relationship between

Kacand Kaq0is valid in the present analysis:

Kac

1

cos Kaq0

tan 19

and that the calculation of Kaq0 and Kac gives exactly the same

critical geometry.

3.2.1. Comparison with Kerisel and Absi. Figures5 and6

show the comparison of the present solutions of Ka and Kaq

with those of Kerisel and Absi

29

in the case of a vertical walland an inclined backfill, and in the case of an inclined wall

and a horizontal backfill respectively, when 458 and

= 1.

70

60

50

40

30

20

10

0

Kp

40

30

20

10

5

15

25

35

45

0

Kpq

Present solution

Krisel and Absi

40 30 20 10 0 10 20 30 40

/= 1

/= 1/2

/= 0

Present solution

Krisel and Absi/= 1

/= 1/2

/= 0

Fig. 3. Comparison of present Kp and Kpq coefficientswith those of Kerisel and Absi

29(for inclined wall and

horizontal backfill)

100

200

300

500

400

600

700

800

0

Kp

200

150

50

100

0

Kpq

45 30 15 0 15 30 45

Present solution

Soubra=15

= 15

= 0

Present solution

Soubra

= 15

= 15

= 0

Fig. 4. Comparison of present Kp

and Kpq

coefficientswith those of Soubra 27

Ka

Ka

0

02

0

01

02

03

04

05

06

07

08

04

06

08

10

12

Kaq

Kaq

45 35 25 15 5 5 15 25 35 45

Present solution

Krisel and Absi

Fig. 5. Comparison of present Ka and Kaq coefficientswith those of Kerisel and Absi

29(for vertical wall and

inclined backfill)



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The present results are smaller than those of Kerisel and Absi.

For the Ka coefficient, the maximum difference does not

exceed 3% when < 158; however, for 208 a significant

difference is observed. After careful examination of the values

proposed by Kerisel and Absi for similar configurations (see for

instance their values for= 0:66 or 0), it seems that their

Ka value for 458,= 1, 08 and 208 is not

correct. For the Kaq coefficient the underestimation does not

exceed 8%.

The preceding comparisons allow one to conclude that, for

practical configurations ( < 408,= < 1,= > 1=3 and 08), there is good agreement with the currently used results

of Kerisel and Absi for both Ka and Kaq. The maximum

difference does not exceed 3%.

4. IMPLEMENTATION

OF USER-DEFINED

FUNCTIONS IN VISUAL

BASIC FOR

APPLICATIONS, AND

THE USE OF SOLVER

To implement the functions

defining the passive earthpressure coefficients and to

run the Solver optimisation

tool of Microsoft Excel, one

has to follow the following

steps (for the active case, use

the appropriate equations

given in Appendix 1):

(a) Create the user-defined

functions shown in

Appendix 2. This is done

in Microsoft Excel 97 byfirst clicking Tools/

Macro/Visual Basic Editor

and then clicking Insert/

Module and, in the module sheet, typing Option explicit

. . ., etc. The functions are simple and self-explanatory.

(b) As shown in Fig.7, cells C5, C6, C7 and C8 are input data

that define the mechanical and geometrical parameters,

, and . Cells C12 and C13 contain values of angular

parameters of the log-spiral slip surface 0 and 1. Finally,

cell C19 contains the formula to compute the passive earth

pressure coefficient Kp. Arbitrary values were initially

entered in cells C12 and C13 for0 and 1, say 05 for0and 15 for1.

(c) Invoke the Solver tool by clicking Tools/Solver. Fig.8

shows the Solver dialog box. To calculate the Kp

coefficient, set the C19 cell equal to minimum, by

changing cells C12 and C13, namely0 and 1, subject to

the constraints that C13 > C12 0:0001 (1 . 0),

C12 < 3:14 (0 < ), C13 < 3:14 (1 < ),

C13 > 0 (1 > 0) and C19 > 0 (Kp > 0).

If Solver reports a converged solution, one should accept the

solution and re-invoke Solver, until it reports it has found

a solution. It should be mentioned that the initial values of0and1 may influence the ability of the Solver to find a

solution. This need for judicious choice of starting values of0and1 is not a major inconvenience because different starting

values can be tried with ease using the proposed spreadsheet

approach. The appealing feature of the spreadsheet approach is

that once the spreadsheet is set up as shown in Fig. 7, running

other problems with different geometry and soil properties

merely requires changing the input data (that is,,, and ).

4.1. Illustrative example

Consider the following characteristics: 408,= 2=3,

= 0 and 08. Initial values of0 and1 are arbitrarilychosen, say 05 for0 and 15 for1. Fig.7 shows the critical

coefficient Kp 12:59 and the corresponding critical slip

surface.

Ka

Ka

0

02

03

01

0

01

02

03

04

0504

Kaq

Kaq

20 10 0 2010

Present solution

Krisel and Absi

Fig. 6. Comparison of present Ka and Kaq coefficientswith those of Kerisel and Absi

29(for inclined wall and

horizontal backfill)

Fig. 7. Calculation of passive earth pressure coefficient Kp using spreadsheet



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5. DESIGN TABLES

Tables 1 and 2 present the coefficients Kp, Kpq, Kpc, Ka, Kaq

and Kac obtained from the computer programs for practical use

in geotechnical engineering. These values are given for

ranging from 208 to 408, for five values of=, for 08 and

for four values of=. For practical configurations, the passive

(or active) earth pressure coefficients are given for negative (orpositive) values.

6. CONCLUSIONS

A simple method using spreadsheet software has been proposed

for computing the active and passive earth pressure

coefficients. The method is based on the upper-bound theorem

of limit analysis. The failure mechanism is of the rotational

type. It is bounded by a log-spiral slip surface. The energy

balance equation is shown to be equivalent to the moment

equilibrium equation about the centre of the log-spiral. The

present approach gives rigorous solutions for the active and

passive earth pressures in the framework of the kinematical

approach of limit analysis.

Numerical optimisation is performed automatically by a

spreadsheet optimisation tool. The implementation of the

proposed method has been illustrated using an example. Once

the spreadsheet has been set up, the same template can be used

for analysing other problems merely by changing the input

data. The advantage of this method is its simplicity in use.

Comparison with the currently used solutions of Kerisel and

Absi29

leads to the following conclusions:

(a) For the passive case, the present solutions of Kp and Kpq

are greater than those given by Kerisel and Absi.29

However, for practical configurations ( < 408, 1=3 < =

< 2=3,= < 1=3 and 08), the maximum difference

does not exceed 5% for Kp and 7% for Kpq.

(b) For the active case, the present solutions of Ka and Kaqallow one to conclude that for practical configurations

( < 408,= < 1,= > 1=3 and 08), there is good

agreement with the currently used results of Kerisel and

Absi. The maximum difference does not exceed 3%.

On the other hand, the present analysis improves the best

upper-bound solutions given in the literature by Soubra

27

forthe Kp coefficient. The improvement (that is, the reduction

relative to Soubras upper-bound solution) is 27% when

458,= 1,= 1 and 158. For the Kpq

coefficient, the present analysis overestimates the upper-bound

solutions given by Soubra. However, for practical


08) the maximum difference does not exceed 75%.

Numerical results of the active and passive earth pressure

coefficients are given in a tabular form for practical use. The

proposed method, being simple and rigorous, may be an

attractive alternative to other existing solutions, and can be

easily extended to other stability problems in geotechnical

engineering.

APPENDIX 1

The non-dimensional functions f1, f2, . . . , f8 are given as

follows, using the lower sign for the passive case:

f1

e3(10)tan(3tan sin 1 cos 1) 3 tan sin 0 cos 0

3(9tan 2 1)

264

375

20

f2 1

6

L

r02sin 0 2

l

r0sin

L

r0cos

cos(1 ) e(10) tan

21

f3 1

6

l

r0sin 0 2sin 0

l

r0sin

22

f4

cos( ) cos 0 1

3

l

r0cos

sin( ) sin 0 1

3

l

r0sin

for K

cos( ) cos 0 1

2

l

r0cos

sin( ) sin 0 1

2

l

r0sin

for Kq, Kq0 and Kc

8>>>>>>>>>>>>>>>>>>>>>>>>>>>:

23

f5 l

r0

tan

tan sin( 0)24

f6 L

r0sin 0

l

r0sin

1

2

L

r0cos

25

f7 1

2tan (e2(10)tan 1)26

Fig. 8. Solver dialog box



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Kp Kpq

= = = 0 1=3 1=2 2=3 1 0 1=3 1=2 2=3 1

20 0 204 239 257 275 313 204 238 254 270 300 21=3 171 194 207 221 249 172 196 209 221 246 21=2 155 173 183 194 216 158 175 185 196 218 22=3 139 151 158 166 183 143 155 162 170 187 2

25 0 246 307 341 376 454 246 304 334 365 424 31=3 194 232 254 277 330 196 234 256 279 326 21=2 171 197 213 231 272 175 202 218 237 275 2

2=3 147 165 175 188 217 154 172 183 196 226 230 0 300 403 465 534 693 300 398 453 510 628 31=3 220 277 314 356 454 224 282 318 357 444 21=2 187 225 250 280 352 193 233 259 289 358 22=3 155 179 194 213 262 165 190 207 227 276 1

35 0 369 544 659 795 1130 369 535 634 742 982 31=3 250 335 395 467 653 255 343 402 470 631 21=2 203 257 296 344 470 213 270 310 359 479 22=3 162 193 215 243 322 176 210 234 265 347 1

40 0 460 762 981 1260 2001 460 742 927 1140 1643 41=3 283 410 509 635 993 291 422 520 639 942 21=2 220 295 353 430 655 234 314 375 455 669 22=3 167 208 238 279 406 187 232 266 312 447 1

Table 1. Passive earth pressure coefficients Kp, Kpq and Kpc

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Ka Kaq

= = = 0 1=3 1=2 2=3 1 0 1=3 1=2 2=3 1

20 0 0490 0459 0449 0442 0436 0490 0460 0450 0444 0441 141=3 0537 0507 0497 0490 0485 0541 0511 0501 0494 0491 151=2 0569 0541 0531 0525 0520 0578 0549 0539 0533 0530 152=3 0611 0586 0577 0572 0570 0628 0602 0593 0588 0586 16

25 0 0406 0378 0369 0364 0363 0406 0378 0370 0366 0368 121=3 0451 0423 0415 0410 0410 0456 0427 0419 0415 0417 11=2 0482 0455 0447 0443 0444 0493 0466 0458 0454 0456 142=3 0523 0499 0492 0489 0493 0546 0521 0514 0511 0515 14

30 0 0333 0309 0303 0300 0304 0333 0310 0304 0302 0309 1

1=3 0374 0350 0343 0341 0347 0379 0355 0349 0347 0354 121=2 0402 0379 0373 0371 0379 0416 0392 0386 0384 0393 12=3 0441 0420 0415 0414 0425 0469 0446 0442 0441 0453 1

35 0 0271 0251 0247 0247 0256 0271 0252 0248 0248 0260 101=3 0306 0286 0282 0282 0293 0312 0292 0288 0288 0302 11=2 0330 0311 0308 0308 0322 0346 0326 0323 0323 0339 12=3 0365 0347 0345 0347 0364 0398 0378 0376 0378 0397 12

40 0 0217 0202 0200 0202 0215 0217 0203 0201 0203 0219 091=3 0246 0231 0229 0231 0248 0253 0237 0235 0238 0256 101=2 0267 0252 0250 0253 0272 0284 0268 0266 0269 0291 102=3 0296 0282 0282 0286 0310 0332 0316 0316 0320 0347 10

Table 2. Active earth pressure coefficients Ka, Kaq and Kac

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f8 L

r0sin( 0)

l

r0sin( )

1

2

L

r0

27

where

L

r0

e3(10)tan(sin 1 cos 1 tan) sin 0 cos 0 tan

sin tan cos

28

l

r0

e(10)tan cos(1 ) cos(0 )

cos( )29

APPENDIX 2

The user-defined functions for passive earth pressure

coefficients coded in Microsoft Excel Visual Basic are asfollows:

' Program for evaluation of Kpgama, Kpc and Kpq for rotational

' mechanism using log-spiral slip surface

Option Explicit ' All variables must be declared

' Definition of the global constants

Public Const Pi = 3.141592654

' Variables

Public Phi As Double ' internal friction angle of the soil

Public Delta As Double ' angle of friction between soil and wall

Public Lambda As Double ' inclination of the wall

Public Beta As Double ' inclination of the backfill

Public Sl_r0 As Double ' l/r0

Public CL_r0 As Double ' L/r0' Unknown variables

Public Theta0 As Double ' first unknown angle (in radians)

Public Theta1 As Double ' second unknown angle (in radians)

' Defining the initial values

Sub Define()

Phi = Cells(5, 3).Value * Pi / 180# ' Values from the cells

Delta = Cells(6, 3).Value * Pi / 180#

Lambda = Cells(7, 3).Value * Pi / 180#

Beta = Cells(8, 3).Value * Pi / 180#

Theta0 = Cells(12, 3).Value

Theta1 = Cells(13, 3).Value

Sl_r0 = (-Exp((Theta1 - Theta0) * Tan(Phi)) * Cos(Theta1 - Beta) _

+ Cos(Theta0 - Beta)) / Cos(Beta - Lambda)

CL_r0 = (Exp((Theta1 - Theta0) * Tan(Phi)) _* (Sin(Theta1) - Cos(Theta1) * Tan(Lambda)) - _

Sin(Theta0) + Cos(Theta0) * Tan(Lambda)) / _

(Sin(Beta) * Tan(Lambda) + Cos(Beta))

End Sub

'***********************************************

Function f_1() As Double

Dim C1#, C2#

C1 = Exp(3# * (Theta1 - Theta0) * Tan(Phi))

C2 = 3# * (9# * (Tan(Phi)) ^ 2 + 1)

f_1 = -(C1 * (3# * Tan(Phi) * Sin(Theta1) - Cos(Theta1)) - _

3# * Tan(Phi) * Sin(Theta0) + Cos(Theta0)) / C2

End Function

'***********************************************

Function f_2() As DoubleDim C1#, C2#

C1 = 2# * Sin(Theta0) - 2# * Sl_r0 * Sin(Lambda) + CL_r0 * Cos(Beta)

C2 = CL_r0 * Cos(Theta1 - Beta) * Exp((Theta1 - Theta0) * Tan(Phi))

f_2 = -(1# / 6#) * C1 * C2

End Function

'***********************************************

Function f_3() As Double

f_3 = -(1# / 6#) * Sl_r0 * Sin(Theta0 - Lambda) * _

(2# * Sin(Theta0) - Sl_r0 * Sin(Lambda))

End Function

'***********************************************

Function f_4_Kpg() As Double

Dim C1#, C2#

C1 = Cos(Delta - Lambda) * (Cos(Theta0) - Sl_r0 / 3# * Cos(Lambda))C2 = Sin(Delta - Lambda) * (Sin(Theta0) - Sl_r0 / 3# * Sin(Lambda))

f_4_Kpg = C1 - C2

End Function



12/13


13/13

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