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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 +=
Verification Problem Two - consolidation of contiguous clay layerswith different permeability
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
Verification Problem Two - consolidation of contiguous clay layerswith different permeability
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
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 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.
Verification Problem Two - consolidation of contiguous clay layerswith different permeability
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
W zbrak i σ, µ, ν, INC,( )
An brent Wn 1 0, Wn 1 1,, σ, µ, ν, TOL,
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
W zbrak i σ, µ, ν, INC,( )
An brent Wn 1 0, Wn 1 1,, σ, µ, ν, TOL,
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
Verification Problem Two - consolidation of contiguous clay layerswith different permeability
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.
Verification Problem Three - surface settlement of clay layer consolidatingunder strip footing load
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.
Verification Problem Three - surface settlement of clay layer consolidatingunder strip footing load
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.
Verification Problem Three - surface settlement of clay layer consolidatingunder strip footing load
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=
Verification Problem Three - surface settlement of clay layer consolidatingunder strip footing load
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.