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C.1. INTRODUCTION

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C.2. VERIFICATION PROBLEM ONE – TERZAGHI ONE-

DIMENSIONAL CONSOLIDATION

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C.3. VERIFICATION PROBLEM TWO – CONSOLIDATION OF

CONTIGUOUS CLAY LAYERS WITH DIFFERENT

PERMEABILITY

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Verification Problem Two - consolidation of contiguous clay layerswith different permeability

327

The purpose of this Mathcad worksheet is to evaluate the expressions for the excess pore pressure in two contiguous clay of unlike compressibility developed by Gray, H. (1944), "Simultaneous consolidation of contiguous layers of unlike compressible soils", ASCE Transactions, No. 2258, pp. 1327-1356.

The general cases of two adjacent compressible strata:

IIa2k 2 e2c2

a1k 1 e1c1

I

Case a . Free drainage attop and bottom

Hs h 2

= ν

h 1h 1 z

Case b . Free drainageat top only

Terminology and defin itions stress-stra in relations:the soil is assumed to be linearly elastic

k = coefficient of permeabilityav = coefficient of compressibilitye = void ratioe0 = initial void ratioc = coefficient of consolidationp = surface traction applied at t = 0

µ = poisson's ratio

E = Young's modulus

D = constrained modulus

mv = coefficient of compressibility

H sH

1 ez

h

1 ek s

k

1 eD

E 1 µ( ).

1 µ( ) 1 2 µ.( ).G

E

2 1 µ( ).

ck

a v γ. 1 e( ).m v

1

Da v

m v

1 e0

Define four dimensionless numbers (µ, σ, ν, T):

Note : The time factor, T, is based only on the properties of layer I and the time, t.µ 2 c 1

c 2σ 1

µ

k s1

k s2

. νh 2

h 1T

c 1

h 12

t.

u1 = excess pore pressure in layer Iu2 = excess pore pressure in layer IIU1 = average degree of consolidation in layer IU2 = average degree of consolidation in layer II

Appendix C 328

Gray (1944) developed the following analytical solutions for the two cases:

Case a

)sin()sin(1 1

1

2

nn

nTA

n Ah

zAeCu n µν∑

∞

=

−=

nn

nTA

n Ah

zAeCu n µν )1sin()sin(

1 12

2∑∞

=

− −+=

( )( ) 2

cos1))(sin)(sin(

)sin)sin()(sin(21

12221

nTAn

n nnn

nnn eAAAA

AAAU −

∞

=−

++−= ∑ µνµνσ

µνσµν

( )( ) ( )( ) 2

cos1sinsin

sinsinsin21

12222

nTAn

n nnn

nnn eAAAA

AAAU −

∞

=−

++−= ∑ µνµνµνσ

µνσµν

in which An must be a root of

( ) AAAAAF µνµνσ cossinsincos +=

and

nnn

nnn AAA

AApC 22 sinsin

sinsin2

µνµνσµνµνσ

++=

Case b

)sin()cos(1 1

1

2

nn

nTA

n Ah

zAeCu n µν∑

∞

=

−=

nn

nTA

n Ah

zAeCu n µν )1sin()cos(

1 12

2∑∞

=

− −+=

( )2

12221

))(cos)(sin(

)sin))(cos(sin(21 nTA

n nnn

nnn eAAA

AAAU −

∞

=∑ +

−=µνµνσ

µν

( )( ) ( )

2

1222

2

2cossin

cos1cos21 nTA

n nnn

nn eAAA

AAU −

∞

=∑ +

−−=

µνµνσµν

µν

in which An must be a root of

( ) AAAAAG µνµνσ coscossinsin +=and

nnnn

nn AAAA

ApC 22 cossin

cos2

µνµνσ +=


329

Develop solution for Case b using Gray's solution. First, set up procedure for finding roots of G(A).

G A σ, µ, ν,( ) σ sin A( ). sin µ ν. A.( ). cos A( ) cos µ ν. A.( ).

Define the function zbrak which will bracket the first n roots of G(A,σ,µ,ν). The brackets for each root are returned as a row in an array. The values of the left and right brackets are in column zero and one, respectively. The variable inc controls the step size of A that zbrak uses when incrementing A.

zbrak n σ, µ, ν, inc,( ) W 0

A 0

B A

fa G A σ, µ, ν,( )

check 0

B B inc

fb G B σ, µ, ν,( )

check 1

Wi 1 0, A

Wi 1 1, B

A B

fa fb

fa fb. 0<if

fa fb

A B

fa fb. 0>if

check 0while

i 1 n..∈for

W

Define the function bren t, which uses Brent's method for finding the root of a function known to lie between a and b. The root is refined until its accuracy is tol or the maximum no. of iterations is reached.

First, define the functions test and test 1 which will perform logical tests for bren t.

test fb fc,( )

1

break

fc 0>if fb 0>if

1

break

fc 0< fb 0<ifif otherwise

1Machine ε:

test1 e toli, fa, fb,( ) 1 fa fb> e toliifif

1 otherwise ε mach 1 10 15.

Appendix C 330

Reference for Brent's method: Brent, R.P. 1973, Algorithms for Minimization without Derivatives (Englewood Cliffs, NJ: Prentice-Hall), Chapters 3, 4.

brent a b, σ, µ, ν, tol,( ) itmax 100

fa G a σ, µ, ν,( )

fb G b σ, µ, ν,( )

break fa fb. 0>if

c a

fc fa

d b a

e d

c a

fc fa

d b a

e d

test fb fc,( ) 0>if

a b

b c

c a

fa fb

fb fc

fc fa

fc fb<if

toli 2 ε mach. b. 0.5 tol.

xm 0.5 c b( ).

b

break xm toliif

break fb 0if

sfb

fa

p 2 xm. s.

q2 1.0 s

a cif

q2fa

fc

rfb

otherwise

test1 e toli, fa, fb,( ) 0>if

i 1 itmax..∈for


331

rfc

p s 2 xm. q2. r q2( ). b a( ) 1.0 r( ).( ).

q2 q2 1.0( ) r 1.0( ). s 1.0( ).

q2 q2 p 0>if

p p

e d

dp

q2

2 p. min3 xm. q2. toli q2.

e q2.<if

d xm

e d

otherwise

d xm

e d

otherwise

a b

fa fb

b b d d toli>if

b b tolixm

xm. otherwise

fb G b σ, µ, ν,( )

b

Appendix C 332

Define remaining functions necessary for solving an example problem for Case b.

INC 0.1 i 12 the variable i will determine how many terms are evaluated when approximating the infinite series.

u 1 p σ, µ, ν, z, h 1, T, sum 0

W zbrak i σ, µ, ν, INC,( )

An brent Wn 1 0, Wn 1 1,, σ, µ, ν, TOL,

Cn2 p. cos An( ).

σ An. sin µ ν. An.( )2. µ ν. An. cos An( )2.

term Cn exp T An2.( ). cos Anz

h 1

.. sin µ ν. An.( ).

sum sum term

n 1 i..∈for

sum

:excess pore pressure in layer I

u 2 p σ, µ, ν, z, h 1, T, sum 0



Cn2 p. cos An( ).

σ An. sin µ ν. An.( )2. µ ν. An. cos An( )2.

term Cn exp T An2.( ). cos An( ). sin µ An. 1 ν z

h 1

..

sum sum term

n 1 i..∈for

sum

:excess pore pressure in layer II

u p σ, µ, ν, z, h 1, T, u 1 p σ, µ, ν, z, h 1, T, z h 1if

u 2 p σ, µ, ν, z, h 1, T, otherwise

:general expression for excess pore pressure at elevation z.

Data p σ, µ, ν, h 1, T, inc, a0 0, 0

a0 1, u p σ, µ, ν, a0 0,, h 1, T,

steph 1 ν h 1

.

inc

aj 0, aj 1 0, step

aj 1, u p σ, µ, ν, aj 0,, h 1, T,

j 1 inc..∈for

a

Data is a function for generating an excess pore pressure isochrones.


333

Example Problem (after Gray, H. 1944)

p 100 σ 2 µ 5 ν 4 h 1 2 e 1 e2

H Data p σ, µ, ν, h 1, 100, 100, I Data p σ, µ, ν, h 1, 200, 100,

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

T = 100T = 200 Values of u/p

z/H

The solution for this example agrees with Gray's solution.

Example of how the first i roots of G(A) are found

A 0 .02, .7.. i 5

G A σ, µ, ν,( )

A0 0.5 1

2

0

2

First, zbrak is called to bracket the roots

I zbrak i σ, µ, ν, .05,( )

j 1 i..

Next, bren t is called to find the roots. Brent uses the brackets found by zbrak.

rj 1 brent Ij 1 0, Ij 1 1,, σ, µ, ν, 1 10 13.,

solj 1 G rj 1σ, µ, ν, check the roots

brackets roots Value of G at roots

I

0.05

0.2

0.35

0.5

0.65

0.1

0.25

0.4

0.55

0.7

= r

0.071

0.215

0.36

0.508

0.657

= sol

1.87410 13

0

2.33110 15

0

0

=

Appendix C 334

Define the expressions for the degrees of consolidation of layers I and II for case b.

i 12 INC .1

Ucon1 σ µ, ν, T,( ) sum 0



termsin An( ) cos An( ). sin µ ν. An.( ).

An2 σ sin µ ν. An.( )2. µ ν. cos An( )2..

exp T An2.( ).

sum sum term

n 1 i..∈for

1 2 sum.

Ucon2 σ µ, ν, T,( ) sum 0



termcos An( )2 1 cos µ ν. An.( )( ).

An2 σ sin µ ν. An.( )2. µ ν. cos An( )2..

exp T An2.( ).

sum sum term

n 1 i..∈for

12

µ ν.sum.

Define expression for "resultant" degree of consolidation for both layers.

U res σ µ, ν, T,( ) U1 Ucon1 σ µ, ν, T,( )

U2 Ucon2 σ µ, ν, T,( )

ν U2. U1

1 ν


335

Verification problem 2 for SAGE

p 100 σ 1

2µ 1

2ν 1 h 1 5 e 1 e2

First, plot G(A) to find out nature of function for the parameters of the given problem.

A 0 .1, 10..

G A σ, µ, ν,( )

A0 5 10

1

0

1

INC = 0.1 is a reasonable increment for the function zbrak to bracket the roots of G(A) for this problem. Note that INC must be small enough to "capture" each root of G(A).

H Data p σ, µ, ν, h 1, .1, 100, I Data p σ, µ, ν, h 1, .5, 100,

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

T = 0.10T = 0.50 Values of u/p

z/H

Appendix C 336

T trial beg end, n,( ) T val010beg

k 1

num 10j

T valknum

10

n. jj.

k k 1

jj 1 n..∈for

j beg end 1..∈for

T val

Define the function Ttrial to generate the time factors, T, for a plot of resultant U versus the logarithm of T.

beg is the beginning log cycle (i.e. 0 for 100)

end is the ending log cycle (i.e. 1 for 101)

n is the number of increments for T in each log cycle

Example:

T trial 0 2, 3,( )

1

3.333

6.667

10

33.333

66.667

100

=

l 0 1, 40.. :range variable used for plot

Resultant U versus log T for Verification Problem 2

1 10 3 0.01 0.1 1 10

0

0.2

0.4

0.6

0.8

1

T

U

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C.4. VERIFICATION PROBLEM THREE – SURFACE

SETTLEMENT OF CLAY LAYER CONSOLIDATING

UNDER A STRIP FOOTING LOAD

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Appendix C 338

The purpose of this MathCAD worksheet is to evaluate the expression for the surface settlement of a clay layer loaded with a strip footing. The original derivations were done by Gibson, R.E., Schiffman, R.L.,and Pu, S.L. "Plane Strain and Axially Symmetric Consolidation of a Clay Layer on a Smooth Impervious Base," Quartly Journal of Mechanics and Applied Mathematics, Vol. XXIII, Pt. 4, 1970, pp. 505-520. The expression that they developed is difficult to evaluate, because it involves an semi-infinite integral and an infinite series. Furthermore, it also involves finding the roots of a characteristic equation.

b b

f

zh

x,r

Gibson et al found the following expression for the settlement of the surface of the clay layer.

w o( ),x t ..η f.2 G

d

0

∞

λ..Γ ( ),x λ tanh ( )λ 2

λ2

..2 η M ( )λ.2 η 1

P( ),λ t

where:

P λ t,( )

1

∞

n

exp α n2 λ2 ct

h2.

F α n λ,=

L λ( ) λ2 η 1. 1 λ csch λ( ). sech λ( ).( )1. λ2

M λ( ) λ η 1. coth λ( ). 1 λ csch λ( ). sechλ( ).( )1.

Γ x λ,( )2 h.

π λ.sin λ b

h.. cos λ x

h.. plane strain

F α n λ, 1

M λ( )

1

21 tan α n

2 α n1

tan α n..

Γ r λ,( ) b J1 λ b

h.. J0 λ r

h.. axisymmetric

αn are the roots of the characteristic equation: material parameters:

ν = poisson's ratioα2

L λ( ) α M λ( ). tan α( ).G = shear modulus

k = permeability of the clay

γw = unit weight of pore fluid (water)η

1 ν

1 2 ν.G

E

2 1 ν( ).c 2 G. η. k

γ w

.c = coefficient of consolidation

Verification Problem Three - surface settlement of clay layer consolidatingunder strip footing load

339

For the purposes of an example problem :h 4 ν 0.30 k 0.0028 f 1000 TOL 1 10 12.

b 4 E 33333 γ w 62.4 ε mach 1 10 14.

ε 1 10 12.η

1 ν

1 2 ν.( )G

E

2 1 ν( ).c 2 G. η. k

γ w

.

plane straincaseΓ x λ,( )

2 h.

π λ.sin λ b

h.. cos λ x

h..

L λ( )1

η 1 λ csch λ( ). sech λ( ).( ).1 λ2. M λ( )

λ coth λ( ).

η 1 λ csch λ( ). sech λ( ).( ).

F α λ,( )1

M λ( )

1

21 tan α( )

2 tan α( )

α.

Determine the roots ,αn, of the characteristic equation α2 - L(λ) - α∗M(λ)*tan(α)

q α λ,( ) α2L λ( ) α M λ( ). tan α( ).

Define the function bisect which returns a vector containing a and b, where a and b define an interval (a,b) for which f(a)*f(b)<0, using the Bisection Method. For this case f(x) is q(a,l)

bisect n λ, φ, itmax,( ) a n 1( ) π.

b n1

2π.

iter 0

xnew ab a

2

iter iter 1

a

b

breakb a

πφ<if

fa q a λ,( )

fnew q xnew λ,( )

a xnew( ) fa fnew. 0>( )if

b xnew( ) fa fnew. 0( )if

iter itmax<while

a

b

The initial estimate of (a,b) is (nπ,n+1/2π)

The function, bisect, will be used to determine a very small interval (a,b) which contains αn.

Note: α1 occurs in the interval (0,π/2), α2 occurs in the interval (π,3π/2) and so on.

Appendix C 340

Define the function Mulle r which uses Muller's method to refine the estimate of αn, the root of q(α,λ).

Mulle r x0 x 1, x 2, maxit, λ,

h 0 x 1 x 0

h 1 x 2 x 1

P0 q x0 λ,

P1 q x1 λ,

P2 q x2 λ,

δ 0P1 P0

h 0

δ 1

P2 P1

h 1

a0 P2

a2

δ 1 δ 0

h 1 h 0

a1 a2 h 1. δ 1

D a12 4 a2

. a0.

E a1 D a1 D a1 D>if

E a1 D otherwise

h 2a0

E.

xstar x2 h

x 0 x 1

x 1 x 2

x 2 xstar

iter 1 maxit..∈for

xstar

Reference: Asaithambi, N.S. Numerical Analysis Theory and Practice. Saunders College Publishing.


341

Define the root finding function alph a. alpha finds the nth root of the characteristic equation: q(α,λ).Alph a uses the bisection method to narrow the interval around the nth root and then calls Mulle r to "polish" the root. Note: Muller's method finds real and complex roots.

alpha n λ,( ) x bisect n λ, ε, 42,( )

Re Muller x0

x0 x1

2, x1

, 1, λ,

Example: an alpha 2 5,( ) an 4.314963= q an 5,( ) 1.77635710 14=

term n λ, t,( ) α n alpha n λ,( )

num exp α n2 λ2 c t.

h2.

denom F α n λ,

num

denom

Define the function, term, which evaluatesthe nth term of P(l,t).

Define the function, H(λ,t), which estimates P(λ,t) by summing P(λ,t) until the terms computed become insignificant.

H λ t,( ) H 0

i 350

s i 0

check 10 TOL.

i i 1

s i term i λ, t,( )

checks i

HH 0>if

H H s i

break i 1if

check TOL>while

H

Note that roundoff and truncation errors pose serious problems to the evaluation of P(λ,t) using H(λ,t). Next, Euler's transformation of alternating series and van Wijngaarden's implementation will be examined as an alternative way of approximating P(λ,t).

Appendix C 342

Define the function Euler, which evaluates Euler's transformation of alternating series with van Wijngaarden's implementation. One term of the original alternating series are incorporated into the estimates of the partial differences.

sub1 nter m wk1, wk2, sum,( )

newsum sum 0.5 wk2.

n nterm 1

wk2 wk1if

newsum sum wk2

n nterm

otherwise

n

newsum

Sub1 is a subroutine used by Euler. Sub1 determines whether a new partial difference should be evaluated (increase nterm by 1) or just revise the estimate of S.

Eule r ter m jter m, wksp,( )

n 1

sum 0.5 term.

wksp2 term

jter m 1if

n wksp0

tmp wksp2

wksp2 term

dum wkspj 2

wkspj 2 0.5 wkspj 1 tmp.

tmp dum

j 1 n 1..∈for n 2if

wkspn 2 0.5 wkspn 1 tmp.

a sub1 n wkspn 1, wkspn 2

, wksp1,

n a0

sum a1

otherwise

wksp0 n

wksp1 sum

wksp

Euler estimates the summation of a convergent infinite series whose terms alternate in sign.

Wijngaarden's implementation adapts Euler's transformation to positive or negative convergent series.

Reference: Press, W.H., Teukolosky, S. A., Vetterling, W.T., and Flannery, B.P. Numerical Recipes in FORTRAN. 2nd ed. Cambridge University Press.


343

Set up a trial series to evaluate the function Euler. Solve the problem:A

1

∞

n

k( )n

=

trial k( ) j 0

s 0

W 0

check 10 TOL.

W old 0

j j 1

s 0

s s 2t k 2t

j..

t 8 7, 0..∈for

s 1( )j 1 s.

W Euler s j, W,( )

checkW1 W old

W oldW old 0>if

W old W1

check TOL>while

W1

j

for k<1:A

k

1 k

The function trial uses the Euler function to estimate A.

Use van Wijngaarden's procedure for evaluating a positive series with Euler's technique (i.e. convert the series into an alternating series)

Asum k n,( ) s 0

s s kj

j 1 n..∈for

Asu m evaluates A through n terms.

Asum1

422, 0.333333333333314= trial

1

4

0.33333333333335

22= A

1

3for k = 1/4

This verifies that Euler works, but it does not offer significant savings in computations in this particular case.

Appendix C 344

Define J(λ,t), which will estimate P(λ,t) using the Euler's transformation. Van Wijngaarden's transformation is used to convert P(λ,t) into an alternating series (terms in the sum alternate in sign). Euler's transformation only works for alternating series.

J λ t,( ) W 0

check 10 TOL.

jterm 0

Σ 0

r 0

r r 1

jterm jterm 1

w r 0

w r w r 2k ter m 2k r. λ, t,.

k 12 11, 0..∈for

w r 1( )jterm 1w r.

W Eule r wr jter m, W,

checkW1

Σ

ΣΣ 0>if

Σ W1

check TOL>while

Σ

r

Examples of estimating P(λ,t) with J(λ,t) and H(λ,t):

J 10 0,( )2.284385

23= H 10 0,( ) 2.266755= J .5 0,( )

0.13361

24= H .5 0,( ) 0.133549=

J 10 1,( ) 2.3602810 7

7= H 10 1,( ) 2.3602810 7=

J .5 1,( )0.087142

6= H .5 1,( ) 0.087142=

It is apparent that the functions J and H give slightly different results at times early in the solution, but there is almost no difference in the values later in the solution. The source of this difference is probably round-off error. The round-off is probably larger for H since it is a simple summation.


345

Define the function Gauleg which will return a matrix whose 1st column contains the Gauss pointsand whose 2nd column contains the Gauss weights for the m-point quadrature rule for the interval (a,b)

Gauleg a b, m,( ) jm 1

2

eps 3 10 14.

xm 0.5 a b( ).

xl 0.5 b a( ).

z cos π i .25

m .5.

check 2 eps.

p1 1

p2 0

p3 p2

p2 p1

p12 k. 1( ) z. p2. k 1( ) p3.

k

k 1 m..∈for

pp mz p1. p2

z z. 1.

z1 z

z z1p1

pp

check z z1

check eps>while

Ai 1 0, xm xl z.

Am i 0, xm xl z.

Ai 1 1, 2xl

1 z z.( ) pp. pp.( ).

Am i 1, Ai 1 1,

i 1 j..∈for

A

Examples:

Gauleg 1 1, 2,( )0.5773502692

0.5773502692

1

1= Gauleg 0 π, 3,( )

0.354062724

1.5707963268

2.7875299296

0.872664626

1.3962634016

0.872664626

=

Appendix C 346

Now, define a MathCAD function, w, to evaluate the settlement, w, at any (x,t). The variable quad control how the integration is performed. Gauss-Legendre quadrature with a quad-point rule is used. The integral is broken up into subintegrals each of length π. This is done because the period of the function, w, i s π.

w x t, quad,( ) w 0

i 0

GAUS Gauleg 0 π, quad,( )

check 1

i i 1

a i 1( ) π.

b i π.

sum 0

weight GAUSj 1 1,

λ GAUSj 1 0, a

sum sum Γ x λ,( )tan hλ( )

2

λ2. 2 η. M λ( ).

2 η. 1J λ t,( )

0. weight.

j 1 quad..∈for

w w sum

checksum

w

check 1 10 3.>while

wη f.

2 G.w.

w 0 0, 5,( ) 0.079367=


347

Define two functions:wfinal - evaluates the final settlement at xwimmed - evaluates the immediate settlement at x

w final x n, quad,( ) w 0

check 1

a i 1( )π2

.

b iπ2

.

GAUS Gauleg a b, quad,( )

sum 0

weight GAUSj 1 1,

λ GAUSj 1 0,

sum sumΓ x λ,( ) tanh λ( ).

λ 1 λ csch λ( ). sech λ( ).( ).weight.

j 1 quad..∈for

w w sum

checksum

w

i 1 2 n...∈for

wη f.

2 η. 1( ) G.w.

w immed x n, quad,( ) w final x n, quad,( )2 η. 1

2 η..

For our example problem:

w final 0 20, 5,( ) 0.111079= G w final 0 20, 5,( ).

b f.0.356018=

w immed 0 5, 5,( ) 0.07976= G w immed 0 20, 5,( ).

b f.0.254298=

These results agree with Figure 5 from Gibson et al, 1970.

The following pages contain plots that investigate the nature of the functions involved in Gibson et al's solution.

Appendix C 348

jj 1 30..ll 1

ter m j j l l, 0.02,( )

ter m j j l l, 0,( )

jj0 10 20 30

0.2

0.1

0

Plot of the first 30 terms of the infinite series P(λ,t). for λ = 1and t = 0 and 0.02

q alph a j j l l,( ) l l,( )

jj0 10 20 30

1 10 8

5 10 9

0 The effects of noise and roundoff error appear in the higher values of αn (~n>10). λ = 1

ll 20 Now set λ = 20 and replot the data.

ter m j j l l, 0.02,( )

ter m j j l l, 0,( )

jj0 10 20 30

0.4

0.2

0

J l l 0,( )4.566113

25= J l l .02,( )

0.428654

29=

H l l 0,( ) 4.4956= H l l .02,( ) 0.428654=

Note the "pulse" type shape of the plot of the infinite series P(λ,t). This "pulse" shape appears when λ is greater than π.

q alph a j j l l,( ) l l,( )

jj 10 10 20 30

5 10 10

0

5 10 10

ter m j jπ, 0.02,( )

ter m j jπ, 0,( )

ter m j jπ .2, 0,( )

jj 10 1 2

0.3

0.2

0.1

0

Verification Problem Three - surface settlement of clay layer consolidatingunder a strip footing load

349

ll 0 .2, 20..

0 5 10 15 200.02

0

0.02

0.04

0.06

0.08Plot of integrand of wfinal vs. lambda

This plot clearly shows the damped oscillating nature of the integrand of Gibson et al's expression for settlement.

ll 20 20.2, 40..

20 25 30 35 404 10 4

2 10 4

0

2 10 4

4 10 4 Plot of integrand of wfinal vs. lambda

SAGE results are compared with this solution in Chapter 3.

c.1. introduction c.2. verification problem one – …verification problem two - consolidation of...

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