Azizi Applied Analysis in Geotechnics Solution Manual

APPLIED ANALYSES IN GEOTECHNICS: SOLUTIONS MANUAL

1

APPLIED ANALYSES IN GEOTECHNICS

Fethi AziziSOLUTIONS MANUAL

[email protected]

Non fissured rock

h = 14m

2m2 m

2 m2m

2m2m

2 m12m 10m 8m 6 4 2

datumD

AB

C

Chapter 1

Problem 1.1

a) w = eGs

= 0.752.7 = 28%

b) w = e−A (1+e)Gs

= 0.75−0.04×1.752.7 = 25.2%

c) bulk density : ρ = Gs(1+w)1+e ρw = 2.7×1.252

1.75 × 1 = 1.93Mg/m3

bulk unit weight: γ = ρ.g = 9.81 × 1.93 = 18.9 kN/m3

saturated density : ρsat = Gs+e1+e ρw = 2.7+0.75

1.75 × 1 = 1.97Mg/m3

saturated unit weight : γsat = g.ρsat = 9.81 × 1.97 = 19.3kN/m3

Problem 1.2

- the bulk unit weight : γ = Gs 1+w1+e

γw

- the saturated unit weight : γsat = Gs+e1+e γw

Hence : γsatγ = 1

1+w1 + e

Gs = 1

1.25 × 1 + 1

2.7 ≈ 1.1

Problem 1.3

a) the water content : w = Mw

Ms= 1320−1075

1075 = 22.8%

b) the void ratio : e = Vv

Vs

- the volume of solids : Vs = Ms

Gs.ρw= 1075

2.7×1 = 398.1cm3

- the volume of voids : Vv = V − Vs = π ×102

4 × 10 − 398.1 = 387.3cm3

therefore : e = 387.3398.1 = 0.97


2

c) degree of saturation : S = wGse = 0.228×2.7

0.97 = 63.3%

d) air content : A = e−wGs

1+e = 0.97−0.228×2.71.97 = 18%

e) bulk density : ρ = MVs+Vv

= 1320785.4 = 1.68g/cm3

f) dry density : ρd = ρ1+e = 1.68

1.97 = 1.37g/cm3

Problem 1.4

Subscripts s and f refer to soil and filter respectively.

From the graph, it is seen that :d10s = 0.105mm, d15s = 0.12mm, d30s = 0.2mm, d50s = 0.28mmd60s = 0.32mm, d85s = 0.85mm, dmax s = 4mm.

a) coefficient of curvature : Cu = d60

d10= 0.32

0.105 ≈ 3

coefficient of gradation : Cg =d30

2

d10d60= 0.22

0.105 × 0.32 = 1.2

b) coefficient of permeability : k ≈ d102 = 0.1052 = 1.1 × 10−2cm/s.

c) refer to figure:


3

0.01 0.1 1 10 100

Perc

enta

ge p

assi

ng

Particle size (mm)

silt sand gravel cobbles

0

20

40

60

80

100

F: fine, M: medium, C: coarse

M C F M C F M C

Problem 1.5

a) the soil plasticity index : Ip = A × % clay = 0.72 × 0.23 = 16.56% the liquid limit : wL = Ip + wP = 16.56 + 16 = 32.56% Thus : clay of low plasticitywL = 32.5%, Ip = 16.5%

b) the liquidity index : IL = w−wP

IP w = ILIP + wP

= 0.75 × 0.165 + 0.16 = 28.4%

Problem 1.6

wop = 1GsGs(1 − A) ρw

ρd max − 1 = 12.7 ×

2.7 × [1 − 0.03] × 11.65 − 1 ≈ 22%

Problem 1.7

a) From the graph : ρd max = 1.71 Mg/m3, wop ≈ 16%

b) the bulk density :

ρ = RC.ρd max(1 + wop) = 0.95 × 1.71 × 1.16 = 1.88Mg/m3.

c) the air content :

A = 1 − ρdf

ρwGs(1 + Gswop) = 1 − 0.95×1.71

2.7 × (1 + 2.7 × 0.16) ≈ 13.8%


4

1.5

1.6

1.7

1.8

0.10 0.15 0.20 0.25 0.30

moisture content

ρd (Mg/m3)

The degree of saturation :

S = 1 − A1 + A

Gswop

= 1 − 0.1381 + 0.138

0.16×2.7

= 65.2%

The void ratio : e = wopGs

S = 0.16×2.70.652 = 0.66

Therefore the porosity : n = e1+e ≈ 0.4


5

Chapter 2

Problem 2.1

The maximum capillary rise : hc = Ce.d10

= 1.5×10−6

0.94×10−6 = 15.95m

Whence a suction pressure : u = −hc.γw = −159.5kN/m2

The capillary saturation rise : hs = hcd10

d60= 15.95 × 1

3.5 = 4.56m

Problem 2.2

Prior to lowering the water table, the effective stress at a depth of 5m is:

σ = 2 × 18.5 + 3 × 10 = 67kN/m2

The maximum increase in effective stress at that depth is such that :

σmax = 1.2 × 67 = 80.4kN/m2

Accordingly:

80.4 = 18.5x + 10(5 − x) x = 30.48.5 = 3.58m

meaning that the water table can only be lowered by : .3.58 − 2 = 1.58m

Problem 2.3

a) the porwater pressure : uA = 10 × 6 = 60kN/m2

the total pressure : σA = 3 × 10 + 3 × 20 = 90kN/m2

the effective pressure : σA = σA − uA = 90 − 60 = 30kN/m2

b) the porewater pressure : uA = 10 × 3 = 30kN/m2

the total pressure : σA = 3 × 20 = 60kN/m2



6

c) the porewater pressure : uA = 2 × 10 = 20kN/m2

the total pressure : σA = 1 × 17 + 2 × 20 = 57kN/m2


d)

Problem 2.4

The depth at which is in all probability larger than ∆σv = 0.05σo zw = 5m.under these conditions, the effective vertical stress due to self weight is :

σo = γ.zw + (γsat − γw)(z − zw) = 9zw + 10z = 45 + 10z.

so that : ∆σv = 0.05[45 + 10z] = 2.25 + 0.5z

But is due to the (three) line loads, and using the superposition∆σv

principle, it is seen that at a given depth z, beneath the centre of the∆σv

loaded surface is such that :

∆σv =2Q1

πz3

x1

2 + z2

2 + 2 ×2Q2

πz3

x2

2 + z2

2

where, with reference to the figure : .x1 = 0, x2 = 3.5m

Thus, equating the two expressions, and rearranging:


7

3m

40 80 40 80 40 80

4

2

4

2

4

2

kN/m2 kN/m2 kN/m2

u σ σ σ

σ

σσ uu

3.534.z + 0.785.z2 = 2000 + 2400

12.25z2 + 1

2 z ≈ 72.6m

Problem 2.5

a) stresses at point A due to the vertical component of the (line) load:

- vertical stress : σA = 2F. cos 12π

z3

R4

since R = z = 1m σA = 2 × 620 × cos 12π ≈ 386 kN/m2

- shear stress : τ xz = 0The stresses at A due to the horizontal component of the force are zero.Therefore, at A the vertical and shear stresses generated by the inclinedload are respectively :

σA = 386kN/m2, τ xz = 0

b) at B, the stresses due to the vertical component of the load are:- vertical stress :

σB = 2F cos 12π

z3

R4 = 2 × 620 × cos 12π × 1

(22 + 12)2 = 15.4kN/m2

- shear stress:

τ xz = 2F cos 12π

z2.xR4 = 2 × 620 × cos 12

π × 1 × 2(22 + 12)2 = 30.9k/m2

The stresses due to the horizontal component of the load are :

σv = 2F sin 12π

z2.xR4 = 2 × 620 × cos 12

π × 2 × 125 = 6.6kN/m2

τ xz = 2F sin 12π

x2.zR4 = 2 × 620 × sin 12

π × 22 × 125 = 13.1kN/m2

Whence, trough superposition :


8

- the total vertical and shear stresses at B are respectively : σB = 15.4 + 6.6 = 22kN/m2

τ xz = 30.9 + 13.1 = 44kN/m2

Problem 2.6

The vertical stress at B is evaluated using superposition principle. Theuniform load applied by 1 is : .q = γH = 21H kN/m2

a) vertical stress at B due to 1:

σv1 = qπ[α + sin α cos (α + 2β)]

From the opposite figure:

tan β = −1.252 α = −2β = 1.18rd.

Thus : σv1 = 0.642q.

b) vertical stress at B due to 2: from the opposite figure :

β = tan−1

1.252 = 0.558rd

α 1 = tan−1 1.25+2.5H

2 − 0.558 rd.

using equation 2.35a:

σv2 = q2π

(1.25+2.5H)1.25H α 1 − 0.899


9

31 1

2.5

2.5m

2mB

123

H

3H 2.5H

B

H

2.5m

2mβ

H

2m

B

1.25

2.5H

βα 1

b) vertical stress at B due to 3: from the figure :

α 2 = tan−1 1.25+3H

2 − 0.558 rd

whence, according to equation 2.35a:

σv3 = q2π

(1.25+3H)1.5H α 2 − 0.899

Adding the three stress contributions at B, then equating the outcome to and substituting for , it follows that :150kN/m2 q = 21H

[150 = 13.482H + 3.342H (1.25+2.5H)1.25H tan−1

1.25+2.5H

1.25H +

](1.25+3H)1.5H tan−1

1.25+3H

2 − 1.023

H − 4.034

So that through trial and error : H ≈ 7.3m

Problem 2.7

The tank radius being : . The vertical stress increase at A (at aa = 10mdepth ) is such that . On the other hand :z = 1.2m ∆σA = 5kN/m2

∆σA = q.Iσ Iσ = 5250 = 0.02.

Because . From the charts of figure 2.20, itz = 1.2m za = 1.2

10 = 0.12follows that :

za = 0.12, Iσ = 0.02 r

a ≈ 2.5

but : ra = x+a

a = 1 + xa x = 1.5a = 15m.


10

H

2m

B

1.25

3H

βα 2

Problem 2.8

Because of the nature of Fadum's method,the following vertical stresses areevaluated using the superpositionprinciple. The irregular shape offoundation makes it inevitable to addsome artificial areas; in which case, caremust be taken so as to subtract the stressgenerated by any of the added areas.

Prior to evaluate the stress increase at any of the points, the expression ofthese stresses are presented in terms of the areas contributing to the overallstress increase. A minus sign indicates that the contribution to the stress ofthe corresponding area is to be subtracted. Thus, referring to the figureopposite :

∆σA = A2H7 − A1G7 + A1FE − A2D3 + ABC3 ∆σB = B5EA + B2H8 − B2DC − B1G8 + B1F5

∆σC = CBA3 + C5E3 + CDH8 − C4G8 + C4F5∆σD = DHG4 + D6E3 − D6F4 + D2A3 − D2BC∆σE = E6D3 + E5BA − E5C3 + E6H7 − EFG7∆σF = FGH6 + F6D4 + F1AE − F1B5 + F4C5∆σG = GHD4 + G1A7 − GFE7 − G1B8 + G4C8∆σH = H2A7 − H6E7 + H6FG − H2B8 + HDC8

For a uniform load q, the above stresses evaluated at a depth usingz = 1mFadum's charts are :

∆σA ≈ 0.245q, ∆σB = 0.25q, ∆σC ≈ 0.724q, ∆σD ≈ 0.244q∆σE ≈ 0.244Q, ∆σF ≈ 0.728q, ∆σG ≈ 0.247q, ∆σH ≈ 0.249q

Problem 2.9

The expressions of stresses are presented in terms of the areas contributingto the overall stress increase at different points. Referring to the figureabove and considering the two uniform surface loads separately :- stress increase at K due to is calculated from the area :q1

∆σK (q1) = B5EA


11

A

B C

D

E

G

H

F1

2

3

4

5

6

7

8

- stress increase at K due to is calculated from the area :q2

∆σK (q2) = B8H2 − B8G1 − BCD2 + B5F1

- stress increase at L due to is calculated from the area :q1

∆σL (q1) = E5BA


∆σL (q2) = E3D6 − E5C3 + E6H7 − EFG7

- stress increase at M due to is calculated from the area :q1

∆σM (q1) = F1AE − F1B5


∆σM (q2) = F4C5 + F6D4 + F6HG

Thus using the dimensions of different areas in conjunction with Fadum'scharts, then applying the influence factors appropriately and adding thecontribution of both loads at every point, it follows that :

∆σK ≈ 0.245q1 + 0.005q2

∆σL ≈ 0.242q1 + 0.002q2

∆σM ≈ 0.008q1 + 0.72q2

Problem 2.10

Use the scaled foundation according to Newmark method, then place thepoint at which the stress is to be estimated at the centre of the concentriccircles. Whence point M (see the upper foundation in the following figure)and K (the lower foundation). Now, only the number of elements needs beestimated in either case, taking care of the difference in loading. Whenceform the figure :

∆σM ≈ 0.005 [15q1 + 24q2] ≈ 32.5 kN/m2

∆σK ≈ 0.005 [19q1 + 10q2] ≈ 19.5kN/m2


12


13

z = scale line Influence factor : I = 0.005

Chapter 3

Problem 3.1

Please note: the answers given on page 160 of the book, relating to seepageforces and effective stresses for this problem, are erroneous.

The hydraulic gradients at different points, and the corresponding angleswith respect to the vertical (refer to enlarged figure on the RHS) are suchthat:iA = 0.21, αA ≈ 1 iB = 0.23, αB ≈ 6 , iC = 0.3, α c ≈ 12iD = 0.4, αD ≈ 19 , iE = 0.45, αE ≈ 10 , iF = 0.33, αF ≈ 4.5iG = 0.3, αG ≈ 3

The seepage forces:- on the upstream side : (downward force)F = A zuγw i cos α- on the downstream side : (upward force)F = −A zeγw i cos α

For a unit area , it follows that:A = 1m2

FA = 0.21 × 0.9 × 10 × cos 1 = 1.89 kNFB = 0.23 × 2.4 × 10 × cos 6 = 5.49 kNFC = 0.3 × 3.7 × 10 × cos 12 = 10.86 kNFD = 0.4 × 4.7 × 10 × cos 19 = 17.77 kNFE = −0.45 × 2.6 × 10 × cos 10 = −11.52kNFF = −0.33 × 2.6 × 10 × cos 4.5 = −8.55 kNFG = −0.3 × 0.6 × 10 × cos 3 = −1.8kN.


14

I m p e r v i o u s

A

B

C

D EF

G

datumzu

ze

A

B

C

D EF

G

The effective stresses :- on the upstream side : σ = zu γ + γw i cos α- on the downstream side : σ = ze γ − γw i cos α

Whence: σA = 0.9(11 + 0.21 × 10 × cos 1) = 11.8kN/m2, σB = 2.4(11 + 0.23 × 10 × cos 6) = 31.9kN/m2

σC = 3.7(11 + 0.3 × 10 × cos 12) = 51.6kN/m2

σD = 4.7(11 + 0.4 × 10 × cos 19) = 69.5kN/m2

σE = 2.6(11 − 0.45 × 10 × cos 10) = 17.1kN/m2

σF = 1.7(11 − 0.33 × 10 × cos 4.5) = 13.1kN/m2

σG = 0.6(11 − 0.3 × 10 × cos 3) = 4.8kN/m2

The porewater pressure : and the pressure head are withu = hp.γw

reference to the figure (note the datum) :hpA = 3.6 − 0.18 − 1.3 = 2.12m uA = 21.2kN/m2

hpB = 3.6 − 0.36 − 0.18 + 0.2 = 3.26m uB = 32.6kN/m2

hpC = 3.6 − 2 × 0.36 − 0.18 + 1.6 = 4.3m uC = 43kN/m2

hpD = 3.6 − 3 × 0.36 − 0.18 + 2.5 = 4.84m uD = 48.4kN/m2

hpE = 0.72 + 0.18 + 2.6 = 3.5m uE = 35kN/m2

hpF = 0.36 + 0.18 + 1.7 = 2.24m uF = 22.4kN/m2

hpG = 0.6 + 0.18 = 0.78m uG = 7.8kN/m2

Problem 3.2

The equivalent horizontal permeability is such that :

Q = THL .kh = 1 × 0.88

1.6 .kh = 0.55kh

kh = Q0.55 = 8.33

0.55 × 10−6 = 1.51 × 10−5m/s.

On the other hand : kh = Σ di.ki

Σ di= 0.2 (k1+k2+...+k5)

1

kh = 0.2(k1 + 0.5k1 + 0.25k1 + 0.125k1 + 0.0625k1) = 0.3875k1

Hence : k1 = 1.510.3875 × 10−5 = 3.91 × 10−5m/s.


15

Problem 3.3

The equivalent vertical permeability is such that :

Q = A.HL .kv = (1.6 × 0.8) × 0.88

1.6 .kv = 1.126kv

kv = 2.781.126 × 10−6 = 2.47 × 10−6m/s.

But: kv = Σ di

Σ di/ki= 1

0.2

1kv1

+ 10.5kv1

+ 10.25kv1

+ 10.125kv1

+ 10.0625kv1

= 0.1613kv1

therefore : kv1 = 2.470.1613 × 10−6 = 1.53 × 10−5m/s.

Problem 3.4

The permeability of the isotropic layer being :

k = [15.1 × 2.47] 1/2 × 10−6m/s = 6.11 × 10−6m/s.

The new tank length :

L = L.kv

kh

1/2= 1.6 ×

2.4715.1

0.5= 0.647m

The hydraulic gradient in the vertical direction : iv = 0.881 = 0.88

Whence the quantity of vertical flow:

Q = L × 0.8k.iv = 0.647 × 0.8 × 6.11 × 0.88 × 10−6 = 2.78 × 10−6 m3/s


16

Problem 3.5

The approximate solution is as per the following figure

Problem 3.6

- The radii at point A:

, r1 = 1.5m, r7 = 43.5m r2 = r12 = (1.52 + 152)1/2 = 15.07m

r3 = r11 = (62 + 27.52)1/2 = 28.14m r4 = r10 = (212 + 27.52)

1/2 = 34.6m

r5 = r9 = (27.52 + 362)1/2 = 45.3m r6 = r8 = (152 + 43.52)

1/2 = 46.01m


17

55 m

45 m

12.5 15 15 12.5

15

15

7.5

7.5 r6 r7

r4

r2

r3r10

r12

r9

r11

r8r5

r1

I m p e r v i o u s

- the radii at point B:

r1 = (1.52 + 62)1/2 = 6.18m

r2 = (1.52 + 212)1/2 = 21.05m

r3 = (1.52 + 362)1/2 = 36.03m

r4 = (112 + 43.52)1/2 = 44.87m

r5 = (262 + 43.52)1/2 = 50.68m r6 = (43.52 + 412)

1/2 = 59.78m

r7 = (362 + 53.52)1/2 = 64.48m r8 = (53.52 + 212)

1/2 = 57.47m

r9 = (53.52 + 62)1/2 = 53.83m r10 = (1.52 + 412)

1/2 = 41.03m

r11 = (1.52 + 262)1/2 = 26.04m r12 = (1.52 + 112)

1/2 = 11.1m

- the radii at point C:

r2 = r12 = (1.52 + 152)1/2 = 15.07m r3 = r11 = (112 + 22.52)

1/2 = 25.04m

r4 = r10 = (262 + 22.52)1/2 = 34.38m r5 = r9 = (22.52 + 412)

1/2 = 46.77m

r6 = r8 = (152 + 53.52)1/2 = 55.56m r1 = 1.5m, r7 = 53.5m


18

55 m

45 m

12.5 15 15 12.5

15

15

7.5

7.5

55 m

45 m

12.5 15 15 12.5

15

15

7.5

7.5

r6

r7

r4

r2

r3

r10r12

r9

r11

r8

r5

r1

r6

r7

r4r2 r3

r10

r12r9

r11

r8

r5

r1

- the drawdown at A is thus:

H − hA = dA = q2πDk Σ

j=1

12ln

Rorj

[ dA = 4.6×10−2

2π×15×1.9×10−3 ln 9001.5 + 2 ln 900

15.07 + 2 ln 90028.14 + 2 ln 900

34.6 + 2 ln 90045.3 +

]2 ln 90046.01 + ln 900

43.5

= 4.6×10−2

30π×1.9×10−3 × 42.98 = 11.04m

- the drawdown at B :

[dB = 4.6×10−2

30π×1.9×10−3 ln 9006.18 + ln 900

21.05 + ln 90036.03 + ln 900

44.87 + ln 90050.68 +

]ln 90059.78 + ln 900

64.48 + ln 90057.47 + ln 900

53.83 + ln 90041.03 + ln 900

26.04 + ln 90011.1

= 4.6×10−2

30π×1.9×10−3 × 39.77 = 10.21m

- the drawdown at C :

[dC = 4.6×10−2

30π×1.9×10−3 ln 9001.5 + 2 ln 900

15.07 + 2 ln 90025.04 + 2 ln 900

34.38 + 2 ln 90046.77 +

]2 ln 90055.56 + ln 900

53.5

= 4.6×10−2

30π×1.9×10−3 × 42.57 = 10.93m


19

Problem 3.7

- the radii at point D:

r1 = r3 = r7 = r9 = 27.04mr2 = r8 = 22.5mr4 = r6 = r10 = r12 = 31.32mr5 = r11 = 27.5m

- the drawdown at D (unconfinedflow) is :

and dD = H − hD H2 − hD2 = q

πk Σj=1

nln Ro

rj

Thus : dD = H −

H2 − q

πk Σj=1

nln Ro

rj

with : Σj=1

12Rorj = 4 ln 700

27.04 + 2 ln 70022.5 + 4 ln 700

31.32 + 2 ln 70027.5

= 38.79

dD = 20 − 202 − 8.33×10−3

4π×10−4 × 38.79 = 8.05m

- point E is in a similar position than point A in the previous problem (i.e.the same radii apply) :

[Σj=1

12Rorj = ln 700

1.5 + 2 ln 70015.07 + 2 ln 700

28.14 + 2 ln 70034.6 + 2 ln 700

45.3 +

] 2 ln 70046.01 + ln 700

43.5 = 39.96

and : dE = 20 − 202 − 8.33×10−3

4π×10−4 × 39.96 ≈ 8.38m

- point F is in a similar position than point B in the previous problem:

[Σj=1

12Rorj = ln 700

6.18 + ln 70021.05 + ln 700

36.03 + ln 70044.87 + ln 700

50.68 + ln 70059.78 +

] ln 70064.48 + ln 700

57.47 + ln 70053.83 + ln 700

41.03 + ln 70026.04 + ln 700

11.1 = 36.75

Whence : dF = 20 − 202 − 8.33×10−3

4π×10−4 × 36.75 = 7.50m


20

55 m

45 m

12.5 15 15 12.5

15

15

7.5

7.5

r6r7

r4r2 r3

r10

r12

r9

r11

r8

r5

r1

Problem 3.8

According to the figure, there are 15 equipotential drops, so that theporewater pressure at A is :

uA = γwhA = 10 × 14 − 2.5 × 14

15 ≈ 117kN/m2

as opposed to : with a drainageuA = 10 × (14 − 2.75 × 147) = 85kN/m2

blanket as per the following figure.


21

14m

A

Non fissured rock

Non fissured rock

14m

8m

2.513

1

drainage blanket

A

Problem 3.9

From the opposite figure, it is seenthat :- the exit hydraulic gradientestimated from element a :

ie ≈ 0.390.96 = 0.41

- the factor of safety against piping F = 1

0.41 = 2.4

- the flownet contains 10 flowchannels (both sides count) and 14equipotential drops. Thus, thequantity of flow :

q = H. Nf

Nd.k = 5.5 × 10

14 .k = 1.96k (m3/s/m)

The profiles of both porewater pressure and effective stresses are as per thefollowing figure


22

2 m

impermeable

CL

element a

CL

100 60 0 60 100u , σ (kN/m2)

uu

σσ

Problem 3.10

With reference to the opposite figure, it isseen that for the (circular) cofferdam of theprevious problem :

d1

T1= 5.5

11 = 0.5, d2

T2= 0.5, T2

b = 116 = 1.833

Hence the form factors from figure 3.45 :

.φ1 ≈ 1, φ2 ≈ 1.38

For a circular cofferdam with a flow pathlength , it follows that :l = 5.5m

h2 = 1.3H φ2

φ1+φ2= 1.3 × 5.5 × 1.38

2.38 = 4.145m

whence an average exit hydraulic gradient :

ieav = h2

l = 4.1455.5 = 0.75

The quantity of flow being :

per metre of perimeter)q = 0.8H 1φ1+φ2

k = 0.8 × 5.5 × k2.38 = 1.85k (m3/s

Compared to the results obtained in the previous problem, it is seen that,while the quantities of flow are comparable, the average hydraulic gradientestimated from Davidenkoff & Franke method is higher than thatcalculated from the planar flownet. This reflects the fact that Davidenkoff& Franke method takes into account the nature of flow (circular cofferdamin this case) whereas the flownet solution in the previous problem can beapplied indiscriminately to any type of flow (circular cofferdam, trench,square cofferdam). The results obtained from the flownet should thereforebe interpreted cautiously.


23

CL

b

d

T dT

H

1

1 2

2

impervious

Problem 3.11

Referring to the figure opposite, itis seen that the sheet pile in thecase of problem 3.11 ischaracterised by the following :

hu = 5.5m, hd = 0, H = 5.5mL = 5.5m, T = D = 11m

Hence the parameter :

ξ = H

ln

TL +

T2

L2 −10.5

+ln

DL +

D2

L2 −10.5

= 5.52 ln 2+ 3

≈ 2.09m

The velocity potential can now be calculated at points A, F, and B :

ΦA = ξ ln

TL +

T2

L2 − 10.5

= 2.09 ln

2 + 3 = 2.752m

ΦF = 0

ΦB = −ξ ln

DL +

D2

L2 − 10.5

= −2.09 ln

2 + 3 = −2.752m

The velocity potential, total head, and hydraulic gradient at a depth y onthe upstream side along AF are respectively :

Φy = ξ ln

yL +

y2

L2 − 10.5

h = Φy − ΦA + hu + T = Φy + 13.748 (m)

iy = ΦA−Φy

T−y = 2.752−Φy

11−y


24

Impervious

L

F

y

x

T

D

A

B

Hhu

hdzu

ze

The porewater pressure and effective vertical stress along AF are thereafterevaluated as follows :

u = γw(h − y)

σ = zu.[γsat − γw(1 − iy)]

Note that y in the above relationships is measured from the top of theimpermeable soil layer.

Thus along AF: y(m) Φy(m) h(m) iy u (kN/m2) σ (kN/m2)

point A 11 2.752 16.50 - 55 0 10 2.518 16.27 0.234 63 13.39 2.248 16 0.252 70 278 1.924 15.67 0.276 77 417 1.51 15.26 0.310 82.5 56.4

point F 5.5 0 13.75 0.50 82.5 88

The velocity potential, total head, and hydraulic gradient at a depth y onthe downstream side along BF are respectively :

Φy = −ξ ln

yL +

y2

L2 − 10.5

h = Φy − ΦB + hd + D = (Φy + 13.752) m

iy = Φy−ΦB

D−y = Φy+2.75211−y

and the porewater pressure and vertical effective stress :

u = γw(h − y)

σ = ze[γsat − γw(1 + iy)]

Whence the corresponding numerical values:


25

y(m) Φy(m) h(m) iy u(kN/m2) σ (kN/m2)point B 11 -2.752 11 - 0 0

10 -2.518 11.23 0.234 12.3 8.669 -2.248 11.5 0.252 25 178 -1.924 11.8 0.276 38 24.77 -1.51 12.24 0.31 52.4 31.6

point F 5.5 0 13.75 0.50 82.5 33


26

Chapter 4

Problem 4.1

Schmertmann method yields apreconsolidation pressure of :

,σp ≈ 900 kN/m2

whence :

OCR = 900200 = 4.5

Problem 4.2

Casagrande method leads to :

σP ≈ 1, 180 kN/m2

corresponding to : OCR = 1180200 = 5.9

Note that, when applicable, Schmertmann method is more reliable.


27

0.9

0.70

0.50

0.30

e

100 1000 10000 σvo

eo

0

0.2

∆e

σv (kN/m2)

σp ≈ 900 kN/m2

0.9

0.70

0.50

0.30

e

100 1000 10000 σv (kN/m2)

σp

Problem 4.3

From the graph : t90 ≈ 9 min

Therefore :

or cv = T90.d2

t90= 0.848×10−4

81×60 = 1.74 × 10−8m2/s cv ≈ 0.55 m2/year

For a degree of vertical consolidation , the corresponding timeU = 0.8factor is : . Accordingly, the actual time of consolidation in theTv ≈ 0.54laboratory is :

t80 = Tv80.d2

cv = 0.540.55 × 10−4 = 9.8 × 10−5 years

On site, the clay layer thickness being 8m and is drained on both sides;hence Moreover, the relationship between consolidation times H = 4m. tl

in the laboratory and in the field is such that :tf

tf = t l

Hd

2 tf80 = 9.8 × 10−5 ×

4

10−2

2≈ 15.7 years


28

0 10 20 30 40

0

0.5

1.0

1.5

2.0

∆h (mm)

t (min)

Problem 4.4

a- The load can be assumed to have been applied instantaneously for aperiod of 5 months, at the end of which was achieved in theU = 90%vertical direction. Accordingly :

t90 = Tv90H2

cv = 0.848×42

2 = 6.785 years

b- again, assuming the full load was applied for a period of 5 months, anoverall degree of consolidation of 90% implies that the vertical degree ofconsolidation after 5 months is such that :

Tv = cvtH2 = 2

42 × 512 = 0.052 Uv ≈ 28%

The radial and vertical degrees of consolidation are related to the overalldegree as follows :

1 − U = (1 − Uv)(1 − Ur)

Therefore : (1 − 0.9) = (1 − Ur)(1 − 0.28) Ur = 86.1%

The number of sand drains needed to achieve such degree of radialconsolidation in a time is estimated using the followingt = 5 monthsprocedure :the drain radius being , first select a curve corresponding to arw = 32.5mmratio on figure 4.24. Next read on the selected curve the time factor n1 Tr

corresponding to a degree of radial consolidation , thenUr = 0.881calculate

. n2 = 1rw

ch.tTr

0.5= 1

0.0325 × 3.1Tr

× 512

0.5

Compare thereafter to : if different, restart the same calculationn2 n1

using the curve corresponding to in figure 4.24 (pencil in ann2

intermediate curve if the need arises).The results are as follows :

n1 = 20 Tr = 2.3 n2 = 23.06n1 = 23 Tr = 2.4 n2 = 22.6


29

Hence the number of drains needed : n = 23

. n = rerw re = 23 × 0.0325 ≈ 0.75m

So that for a square arrangement, the spacing between drains is such that :

re = 0.564S S = 1.32m

Problem 4.5

a- the average immediate elastic settlement can be estimated in thefollowing way : Si = α1α 2

B.qEu

Figure 4.28 yields : H/B = 0.6, Df/B = 0.15 α 1 ≈ 0.2, α 2 ≈ 0.9Whence:

Si = 0.2 × 0.9 × 10×16012000 = 0.024m

b- the clay layer being only drained on the top side, therefore the timeneeded to achieve in situ is estimated as follows :Uv = 90%

where laboratory consolidation time is : tf = t l6

10−2

2

tl = Tv90.d2

cv = 0.8482.1 × 10−4 = 0.403 × 10−4years

Thus: tf = 0.403 × 10−4 × 3610−4 = 14.51 years

c- an overall degree of consolidation is to be achieved in a timeU = 0.9, and accordingly, the corresponding degree of verticalt = 1.5years

consolidation is such that :

Tv = cvt

H2 = 2.1 × 1.536 = 0.0875 Uv ≈ 50%

Now the degree of radial consolidation :

1 − U = (1 − Ur)(1 − Uv) (1 − 0.9) = (1 − Ur)(1 − 0.5)


30

Ur = 80%.

Thus estimating the number of sand drains needed to achieve such a degreein the knowledge that : :ch = 3.2m2/year, t = 1.5years, rw = 0.075m

n1 = 20 Tr ≈ 1.7 n2 = 10.075 ×

3.2×1.5

1.7

0.5= 22.4

n1 = 22 Tr ≈ 1.75 n2 ≈ 22

Whence : so that for a squaren = 22 = rerw re = 22 × 0.075 = 1.65m

arrangement, the spacing between drains is : S = 10.564re = 2.92m

Problem 4.6

a- the stress induced by the loadedfoundation will affect the settlementcalculations up to a depth of .3B = 30mHence the entire clay layer is affected.Select sublayers of a maximum thickness

, and therefore 2 sublayers eachB/2 = 5mwith a thickness of 3m will be adequate.Next, estimate the effective vertical stressdue to self weight of soil beneath thefoundation, and the initial void ratio (fromthe curve provided) at the middle of eachsublayer (points A and B):

point depth (m) σvo(kN/m2) eo

A 1.5 59 0.845 B 4.5 149 0.83

The increase in effective vertical pressure at both points is estimated asfollows :

∆σA = q.

1 − 1

1+[ a

z ] 2

3/2

= 160 ×

1 − 1

1+

51.5

2

3/2

≈ 156kN/m2


31

A

B

1.5m

4.5m

sand

clay

impermeable rock

∆σB = 160 ×

1 − 1

1+

54.5

2

3/2

≈ 112 kN/m2

The preconsolidation pressure at both points is :

σpA = 325kN/m2 , σpB = 375kN/m2

Accordingly :point ∆σv(kN/m2) σv + ∆σv (kN/m2) σp (kN/m2) A 156 215.5 < 325 B 112 261.3 < 375

Apply equation 4.48 to estimate the consolidation settlement at the surfaceof each layer :

Sc1 = 0.12 × 3 × 11+0.86 log

215.559

= 0.108m

Sc2 = 0.12 × 3 × 11+0.838 log

261.3149

= 0.048m

so that the total consolidation settlement (without correction) is :Sc = 0.108 + 0.048 = 0.156m

This value has to be corrected for soil type and foundation size. Accordingto figure 4.35 : . Whence theH/B = 0.6, A = 0.25 µ = 0.62corrected consolidation settlement :

Sc = 0.62 × 0.156 = 0.097m.

The total settlement of the flexible foundation consists of adding the elasticsettlement calculated in the previous problem :

S = 0.097 + 0.024 = 0.121m.

c- the creep settlement is estimated at the top of each sublayer :

Ss1 = Cα1+eo

log t2t1

= 0.0151+0.86 log 3.5 = 4.4 × 10−3m


32

Ss2 = 0.0151+0.838 log 3.5 = 4.4 × 10−3m

leading to a total magnitude of creep settlement between andt1 = 2years of : t2 = 7yeras Ss = 8.8 mm


33

Chapter 5

Problem 5.1

The measured results yieldthe following :

q (kN/m2) p (kN/m2) 64 71.5 129 143 193 214 257 286

The corresponding graph is characterised by a slope : M ≈ 0.9

Accordingly : M = 6 sinφ

3−sinφ φ = sin−1

3M6+M

= sin−1 2.76.9 ≈ 23

Problem 5.2

The graph in the figure opposite yields a slope correspondingto an average maximum angle offriction : φmax ≈ 45

Whence the angle of friction atthe critical density :

φc = φmax − υ = 45 − 10 = 35


34

0 100 200 3000

100

200

300

q (kN/m2)

p (kN/m2)

0

30

60

90

120

0 30 60 90 120

σ(kN/m2)

τmax (kN/m2)

Problem 5.3

a- the time needed for a given degree U of porewater pressure dissipation isestimated as follows :

t = h2

(1−U)η.cv

In this case, (sample drained on both sides without radialU = 0.97, η = 3drainage), cv = 2.7m2/year = 5.14 × 10−6m2/ min, h = 38mm.Therefore :

t97 = 382×10−6

(1−0.97)×3×5.14×10−6 = 3121 min (≈ 52h)

The shear speed is such that :

v = ε1. lt97

= 0.003×763121 = 7.3 × 10−5mm/ min

consequently, the speed used ( is in this case inadequate6 × 10−4mm/ min)

b- were the sample to be drained radially as well as vertically, then and :η = 40.4

t97 = 382×10−6

(1−0.97)×40.4×5.14×10−6 = 231.8 min (≈ 3.9h)

and the shear speed beyond the specified axial deformation is :

, v = 0.003×76231.8 = 9.83 × 10−4mm/ min

meaning that the minimum shear speed of is very6 × 10−4mm/ minadequate


35

The figure oppositeindicates that werethe sample to drainedonly on both sides,the measurements arenot representative ofsoil behaviour up to ashear strain of 2%beyond which aminimum degree ofporewater pressuredissipation of 95% isachieved

Problem 5.4

a- first calculate the different quantities as follows :

σ3f = σ3 − uf

(kN/m2 σ1f =

qf + σ3f kN/m2 pf = 1

3σ1f + 2σ3f

kN/m2

104 224 144176.3 379.3 244

Because the clay is normally consolidated, and at failure, thec = 0increments of stresses are related through the slope M :

∆p , ∆q

M = ∆qf

∆pf

= 203−120244−144 = 0.83

Moreover, M is related to the angle of shearing resistance as follows :

φ = sin−1 3M

6+M = sin−1

3×0.836.83

= 21.4


36

100

80

60

40

20

00 2 4 6

drainage :both ends only

radial + bothends

U(%)

εs (%)

v = 6 × 10−4mm/ min

b- at failure, the stresses are such that :

,τ f = qf

2 cos φ σf = τ f

tan φ

Whence :

σ3(kN/m2) τ f (kN/m2) σf (kN/m2) 200 55.9 142.5 300 94.5 241.1

Problem 5.5

The stress paths of the two tests are depicted in the figure both in terms ofeffective stress path (ESP) and total stress paths (TSP).

The difference between TSP and ESP in each test corresponds to theporewater pressure.

At failure, the porewater pressure parameter A is estimated as follows :- sample A: : heavily overconsolidated clayAf = uf

qf = − 12145 = −0.08

- sample B : : lightly overconsolidated clay.Af = 105195 = 0.54


37

0

100

200

0 100 200 300

ESP

TSPA

Bq(kN/m2)

p (kN/m2)

Chapter 6

Problem 6.1

a) The stress-strain relationship in the elastic range is represented byequation 6.42 :

. Whence the bulk modulus :

δεv

δεs

=

1K 0

0 13G

δpδq

K = δpδεv

= 5010−2 = 5MN/m2.

Moreover, it is seen that : Now thatδq3G = δεs δεs = 60

3×104 = 2 × 10−3.are known, use both equations 6.36 & 6.37 :δεv and δεs

δεv = δε1 + 2δε3 = 10−2

δε1 = 5.3 × 10−3, δε3 = 2.35 × 10−3

δεs = 23 [δε1 − δε3] = 2 × 10−3

b) Under undrained conditions, and according to equation 6.37 :δεv = 0 Thus for the applied increment of deviator stress, it follows δεs = δε1. that : . δq = 3Gδεs = 3Gδε1 δε1 = 30

3×104 = 10−3

The increment of excess porewater pressure generated by is such that :δq δu = δq

3 = 10 kN/m2.

c) If the drainage were opened, then whenall excess porewater pressure has dissipated,the effective mean pressure would haveincreased by exactly the same amount :

. δp = δu = 10 kN/m2

Hence an increment of volumetric strain :

δεv = δpK = 10

5×103 = 2 × 10−3.

d) See figure opposite.


38

200 240 280

0

40

80

p (kN/m2)

q (kN/m2)

Problem 6.2

a) For the stress path to be within the elastic wall, it suffices to show that is smaller than the mean effective pressure at yield p2 = 350 kN/m2 py.

Assuming the drained loading was pursued until the yield curve was met, itis clear that yielding occurs at the point of intersection of the yield curveand the effective stress path. Thus equating the yield curve equation 6.20 tothat of the effective stress path 6.4, it follows that :

and py = poM2

M2+

qy

py

2 py = p1 + qy

3

whence: p1

po+ qy

3po= M2

M2+

qy

p1+qy3

2 qy − 1215

0.81+

qy

300+qy3

2 + 900 = 0.

The solution can be obtained by trial and error : qy ≈ 199 kN/m2 py = 366.3 kN/m2.

It is therefore seen that , meaning thatp2 = 350 kN/m2 < py = 366.3 kN/m2

the loading applied is well within the elastic wall.

b) The increment of mean effective pressure that generated δεv = 7 × 10−3

is in fact :; δp = p2 − p1 = 350 − 300 = 50 kN/m2

accordingly : K = δpδεv

= 507 × 103 = 7.14 MN/m2.

On the other hand, the corresponding increment of deviatoric stress is suchthat : ; δq = 3p2 − p1

= 150 kN/m2

and therefore : G = δq3δεs

= 504.2 × 103 = 11.9 MN/m2.

c) The specific volume corresponding to the point where:is calculated using the elastic wallp2 = 350 kN/m2, q2 = 150 kN/m2


39

equation : where is the specific volumev2 = vo − κ ln

p2

po

vo

corresponding to , calculated from the NCL equation :po

.vo = Γ + (λ − κ)ln 2 − λ ln po = 2.4 + 0.1 ln 2 − 0.15 ln 500 = 1.537

Whence : Accordingly, if, from thev2 = 1.537 − 0.05 ln

350500

= 1.555.

point with the co-ordinates , an undrained loading is applied,v2, p2, q2

then the mean effective stress remains constant up to the yielding point(remember that within the elastic wall, an undrained loading implies

constant). δεv = δpK = 0 p = p2 =

Moreover, the undrained loading means that the specific volume remainsconstant : . Consequently, substituting for and v into the yieldv = v2 pcurve equation 6.20, it follows that :

350500 = 0.81

0.81+η2 η = q350 = 0.5891 q = 206.2 kN/m2.

Therefore the point where the yield curve is met has the co-ordinates :v2 = 1.555, q2 = 206.2kN/m2, p2 = 350 kN/m2.

Note that the same value can be found by substituting for p' and v intoq2

the Roscoe surface equation 6.18a : q = Mp22 exp

Γ−v2−λ lnp2

λ−κ − 1

1/2.

The corresponding porewater pressure generated by the undrained loadingat yield is estimated from the difference between the total and the effectivemean pressures at yield. The total mean pressure is such that :

whence :p2 = 350 + 206.2−1503 = 368.7 kN/m2;

u = p2 − p2 = 368.7 − 350 = 18.7 kN/m2.

d) At failure, the specific volume is constant since thevf = v2 = 1.555loading is undrained. Thus, using the CSL equations 6.7 & 6.8, it followsthat at failure :

pf = expΓ−vf

λ = exp (2.4−1.555)0.15 = 279.6 kN/m2,

qf = Mpf = 0.9 × 279.6 = 251.6 kN/m2.


40

The total mean pressure at failure is calculated as follows : pf = 350 + 251.6−150

3 = 383.9 kN/m2,

whence the porewater pressure at failure :

uf = pf − pf = 383.9 − 279.6 = 104.3 kN/m2.

e) Refer to the figure.


41

1.4

1.5

1.6

1.7

0

100

200

300

0 100 200 300 400 500 600

OA

B

C

D

O

A B,C

D

CSL

CSL

NCL

q (kN/m2)

p (kN/m2)

v

Problem 6.3

a) Because the clay sample was not unloaded, the undrained stress path ison the Roscoe surface. Moreover, undrained conditions imply that thespecific volume remains constant.Its value can be found from the NCL equation :

, with the quantity vo = Γ + (λ − κ)ln 2 − λ ln po po = σ3 = 220 kN/m2.

Thus : vo = 1.8 + 0.05 ln 2 − 0.1 ln 220 = 1.295.

Now the Roscoe surface equation is used to find corresponding to p1

and : q1 = 74 kN/m2 v1 = 1.295

q1 = Mp1

2 exp

Γ−v1−λ ln p1

λ−κ − 1

1/2

. Hence 0.92 exp 10.1 − p 2 1/2 − 74 = 0 p1 ≈ 204.8 kN/m2.

b) The total mean pressure corresponding to is such that :q1 = 74 kN/m2

Accordingly, the pore water pressure is:p1 = 220 + 743 = 244.7 kN/m2.

u = 244.7 − 204.8 = 39.9 kN/m2.

c) Once all excess pore water pressure has dissipated, the mean effectivestress will have increased by the quantity u, while the deviator stressremains constant :

p2 = 204.8 + 39.9 = 244.7 kN/m2, q2 = 74 kN/m2.

The corresponding specific volume is then calculated from the Roscoesurface equation 6.18b :

v2 = Γ − λ ln p − (λ − κ)ln

q

Mp 2

2

+ 12


42

1.8 − 0.1 ln 244.7 − 0.05 ln

740.9×244.7× 2

2+ 1

2

v2 = 1.279.

d) δεv = δpK = 39.9

7 × 10−3 = 5.7 × 10−3.

e) The loading being drained, hence at failure :

pf = 3po

3−M = 3×2203−0.9 = 314 kN/m2,

also, at failure the CSL equations apply : q = Mpf = 0.9 × 314 = 283 kN/m2,

vf = Γ − λ ln pf = 1.8 − 0.1 ln 314 = 1.225.

f) Refer to figure.


43

1.11.2

1.3

1.4

300

0 100 200 300 400 0

100

200

CSL

CSL

NCL

O

A B

C

OA

B

C

p (kN/m2)

q (kN/m2)

v

Problem 6.4

a) the sample was unloaded isotropically until the mean effective stressreached a value such that :p

OCR = po

p= 1.5 p = 600

1.5 = 400 kN/m2.

Because the sample is overconsolidated, its initial behaviour will be elasticuntil the yield curve is met. Also, whilst the behaviour remains elastic, theensuing loading will be characterised by a constant specific volume,

implying that within the elastic wall : constant. δεv = δpK = 0 p =

Accordingly, the mean effective stress at the point where the yield curve ismet is The corresponding specific volume can bep1 = 400 kN/m2.

calculated from the elastic wall equation : , where thev1 = vo − κ ln

p1

po

quantity represents the specific volume under and is estimated fromvo po

the NCL equation :

vo = Γ + (λ − κ)ln 2 − λ ln po = 1.8 + 0.05 ln 2 − 0.1 ln 600 = 1.195.Whence :

v1 = 1.195 − 0.05 ln 46 = 1.215.

Now that and are known, use the yield curve equation to determinev1 p1

the corresponding value of : q1

.p1

po= M2

M2 +

q1

p1

2 q1 ≈ 254.5 kN/m2

Note that an identical value could have been found using the Roscoeq1

surface equation 618a since the point of intersection belongs to both yieldand Roscoe surfaces.

b) The undrained loading is pursued until the mean effective stress reaches. Obviously the corresponding specific volume remainsp2 = 360 kN/m2

constant (undrained load) and thus The deviator stress canv2 = v1 = 1.215.now easily be found from the Roscoe surface equation 6.18a :


44

q2 = Mp2

2 exp

Γ−v2−λ ln p2

λ−κ − 1

1/2

q2 = 0.9 × 3602 exp (1.8−1.215−0.1 ln 360)0.05 − 1

0.5≈ 300 kN/m2.

The total mean pressure corresponding to is such that :q2

p2 = po + q2

3 = 400 + 3003 = 500 kN/m2.

Thus, the excess porewater pressure generated under is as follows :p2

u = p2 − p2 = 500 − 360 = 140 kN/m2.

c) Once all excess pore water pressure has dissipated, the total meanpressure become an effective pressure, while the deviator stress remainsconstant : and p3 = p2 = 500 kN/m2 q3 = q2 = 300 kN/m2.The corresponding specific volume decreases because of the waterexpelled from within the clay sample, and its magnitude can be calculatedfrom the Roscoe surface equation 618b :

v3 = Γ − λ ln p3 − (λ − κ)ln

q3

Mp3 2

2

+ 12

v3 = 1.8 − 0.1 ln 500 − 0.05 ln

3000.9×500 2

2+ 0.5

≈ 1.195.

d) Now the sample is subjected to drained shear so that at failure :

pf = 3po

3−M = 3×4003−0.9 = 571.4 kN/m2.

Using the CSL equations, it is seen that :qf = Mpf = 0.9 × 571.4 = 514.3 kN/m2,

vf = Γ − λ ln pf = 1.8 − 0.1 ln 571.4 = 1.165.


45

e) Seefigure

Problem 6.5

a) The specific volume corresponding to is found from the elastic wallp1

equation : with as calculated previously inv1 = vo − κ ln p1

povo = 1.195

6.4. Whence :v1 = 1.195 − 0.05 ln 4

6 = 1.215


46

1.11.16

1.18

1.20

400

0 200 400 600 800 0

200

CSL

1.22

600

OA

BC

D

E

O

A,BC

D

E

p (kN/m2)

q (kN/m2)

v

b) the point where yielding occurs corresponds to the intersection betweenthe yield curve and the effective stress path. Thus, equating the twofollowing equations :

p2 = p1 + q2

3

p1

po+ q2

3po= M2

M2 +

q2

p1+q23

2

p2 = poM2

M2+

q2

p2

2

Substituting for p1, po and M :

q2 − 1458

0.81 +

q2

400 + q2

3

2 + 1200 = 0

so that by trial and error :

. q2 ≈ 220.1 kN/m2, p2 = 473.4 kN/m2

Inserting these two quantities into the yield curve equation 6.19 yields thecorresponding specific volume :

v2 = Γ + (λ − κ)ln 2po

− κ ln p2 = 1.8 + 0.05 ln 2600 − 0.05 ln 473.4 = 1.207.

Note that an identical value could have been found using the Roscoesurface equation 6.18b.

c) Because of the drained nature of loading, it is seen that : q3 = 3p3 − p1

q3 = 3(500 − 400) = 300 kN/m2.

Since the point with the co-ordinates is on the Roscoe surface, usep3, q3

equation 6.18b to calculate the corresponding value of the specific volume:


47

v3 = Γ − λ ln p3 − (λ − κ)ln

q3

Mp3 2

2

+ 12

.v3 = 1.8 − 0.1 ln 473.4 − 0.05 ln

220.10.9×473.4 2

2+ 0.5

≈ 1.195

d) Undrained shear occurs under a constant specific volume, thence atfailure : But failure occurs on the CSL, and thereforevf = v3 = 1.195.using the two corresponding equations :

pf = exp

Γ−vf

λ = exp

1.8−1.195

0.1 = 425 kN/m2,

qf = Mpf = 0.9 × 425 = 382.5 kN/m2.

The total mean stress at failure is thence calculated as follows :

pf = p1 + qf

3 = 400 + 382.53 = 527.5 kN/m2.

Therefore the porewater pressure at failure is :

uf = pf − pf = 572.5 − 425 = 102.5 kN/m2.

e) As per the figure.


48


49

1.11.16

1.18

1.20

400

0 200 400 600 800 0

200

CSL

1.22

600

OA

O

B

C

D

A

B

CD

NCL

CSL

p (kN/m2)

q (kN/m2)

v

Problem 6.6

a) Given that anv2 = 1.207, p2 = 473.4 kN/m2, q2 = 220.1 kN/m2,undrained loading implies a constant specific volume so that at failure :

Also, from the CSL equations :vf = v2 = 1.207.

pf = exp

Γ−vf

λ = exp

1.8−1.207

0.1 = 376.2 kN/m2,

qf = Mpf = 0.9 × 376.2 = 338.5 kN/m2.

Moreover, the total mean stress at failure is : hence a porewater pressure atpf = p1 + qf

3 = 400 + 338.53 = 512.8 kN/m2;

failure :uf = pf − pf = 512.8 − 376.2 = 136.6 kN/m2.

The stress path isas per the figure.


50

1.11.16

1.18

1.20

400

0 200 400 600 800 0

200

CSL

1.22

600

OA

O

B

A

NCL

CSL

C

BC

p (kN/m2)

q (kN/m2)

v

Problem 6.7

a) The specific volume corresponding to the mean effective stress isp1

calculated from the elastic wall equation : . The quantityv1 = vo − κ ln p1

po

is obviously determined from the NCL equation :vo

vo = Γ + (λ − κ)ln 2 − λ ln po = 2 + 0.06 ln 2 − 0.1 ln 550 = 1.411.

Whence :. v1 = 1.411 − 0.04 ln 100

550 = 1.479

Since an undrained load up to was applied, the specificq = 120 kN/m2

volume remains constant, and more importantly, the mean effective stressremains equal to until yielding occurs. If failure occurs under undrainedp1

conditions, then the mean effective stress at failure can be calculated fromthe CSL equation :

. Since , yielding occursph = exp Γ−v1λ = exp 2−1.479

0.1 = 183.1 kN/m2 ph > p1

on the left hand side of the CSL, i.e. on the Hvorslev surface.

The point at which yielding occurs is such that :

(undrained loading), v2 = v1 = 1.479

and (mean effective stress constant up to the yieldp2 = p1 = 100 kN/m2

curve).

Accordingly, inserting these two quantities into the Hvorslev equation6.23a, it follows that :

q2 = (M − h)exp Γ−v2

λ + h.p2

.= (0.95 − 0.75)exp

2−1.4790.1

+ 0.75 × 100 = 111.6 kN/m2

Note that the same value is obtained from the yield curve equation 6.25,q2

since the yielding point is common to both Hvorslev and yield surfaces.


51

b) If the undrained load is pursued until then the specificq3 = 120 kN/m2,volume remains constant : , and the mean effective stressv3 = v2 = 1.479can be found from the Hvorslev surface equation 6.23a :

p3 = 1h q3 − (M − h)exp Γ−v3

λ p3 = 111.2 kN/m2.

The corresponding total mean pressure being :

, p3 = p1 + q3

3 = 10 + 1203 = 140 kN/m2

the porewater pressure generated is thence :

u = p3 − p3 = 140 − 111.2 = 28.8 kN/m2.

c) If the drainage was opened, then the excess pore water pressure starts todissipate. Accordingly, while the deviator stress remains constant at

, both the mean effective pressure and the specific volumeq = 120 kN/m2

vary according to the Hvorslev equation :

v = Γ − λ ln

q−hpM−h

= 2 − 0.1 ln

120−0.75p0.2

.

Since failure occurs on the CSL, it is seen that :

.qf = Mpf pf = 1200.95 = 126.3 kN/m2

and vf = Γ − λ ln pf = 2 − 0.1 ln 126.3 = 1.516.

An identical value of the specific volume at failure could be found usingthe above Hvorslev equation, which is used to calculate the co-ordinates ofthe stress path in the space (v,p') [refer to the figure] between the yieldpoint and that at failure.

d) See figure.


52


53

0 50 100 150 2000

50

100

1.5

1.45

1.55

A

BC

D

A,BC

D

NCL

CSL

CSL

Hvorslev surface

elastic line

NCL

p (kN/m2)

q (kN/m2)

v

Problem 6.8

a) Obviously, because of the undrained loading from the yield point, thetotal elastoplastic volumetric strain at failure is zero. The total shear strainin the elastoplastic range is calculated in the same way used in conjunctionwith example 6-. The yield point is such that :

p3 = 350 kN/m2, q3 = 206.2 kN/m2, v3 = 1.555.

Up to failure, the stress path is on the Roscoe surface, and since the volumeremains constant, the Roscoe surface equation reduces to a relationship

between p' and q : .q = Mp

2 exp

Γ−v3−λ lnp

λ−κ − 1

0.5

Whence the following calculations :

Averagep (kN/m2) q (kN/m2) v 350 206.2 1.555 p q v δq δp η = q/p 340 214.4 1.555 -20 16 0.63 330 222.2 1.555

320 228.9 1.555 -20 13.4 0.715 310 235.6 1.555

300 241.1 1.555 -20 11.1 0.804 290 246.7 1.555

248.8 249.1 1.555 -10.4 4.9 0.875 279.6 251.6 1.555

Next, the average values are used to calculate the increments of elasticvolumetric strains :

, and plastic shear strains : ; thence :δεve = κ δp

v.pδεs

p = − 2ηM2−η2 δεv

e

p q v δεve (10−3) δεs

p (10−3)340 214.4 1.555 -1.89 5.78320 228.9 1.555 -2 9.57300 241.1 1.555 -2.14 21284.8 249.1 1.555 -1.17 46.1


54

Finally, the cumulative plastic shear strains are such that :

q (kN/m2) εsp

206.2 0222.2 5.78 × 10−3

235.6 1.53 × 10−2

246.7 3.63 × 10−2

251.6 8.24 × 10−2

b) Refer to the figure.

Problem 6.9

The stress path being on the Roscoe surface from the onset of undrainedloading. Thus, for the undrained part of the stress path, the volume remainsconstant (refer to the solution of problem 6.3), and the Roscoe surfaceequation is used to calculate the value of q for each given value of p'. Forthe drained stress path, all three quantities v, p' and q are variable, andtherefore both the Roscoe surface equation and the effective stress pathequation are used in a way that, for a selected value p', the corresponding qvalue is calculated from the effective stress path equation, then the Roscoe


55

200

220

240

260

0 2 4 6 8

q (kN/m2)

εsp (10−2)

surface equation is used to determine the value of v. Whence the followingvalues :

Averagep (kN/m2) q(kN/m2) v220 0 1.295 p q v δp δq η = q/p

215 30.5 1.295 -10 -60.9 0.142210 60.9 1.295

207.4 67.5 1.295 -5.2 13.1 0.325204.8 74 1.295

224.8 74 1.287 39.9 0 0.329244.7 74 1.279

252.4 97 1.273 15.3 46 0.384 260 120 1.267

270 150 1.258 20 60 0.556280 180 1.250

290 210 1.242 20 60 0.724300 240 1.235

307 261.5 1.230 14 43 0.852314 283 1.225

The average quantities are thence used to estimate the increments of

elastoplastic volumetric strains: , δεve = κδp

p

, .δεv

p = λ−κv.p (M2+η 2)

(M

2 − η2)δp + 2ηδq δεsp = 2η

M2−η 2 δεvp

p q v δεve(10−3) δεv

p(10−3) δεsp(10−2)

215 30.5 1.295 -1.79 1.79 0.06207.4 67.5 1.295 -0.97 0.97 0.09224.8 74 1.287 6.89 5.27 0.49252.4 97 1.273 2.38 7.39 0.86270 150 1.258 2.94 10 2.22290 210 1.242 2.78 9.63 4.88307 261.5 1.230 1.85 6.42 13.0


56

The cumulative volumetric and shear elastoplastic strains are thence asfollows :

p (kN/m2) q(kN/m2) εv(10−2) εs(10−2 ) 220 0 0 0

210 60.9 0 0.06204.8 74 0 0.15244.7 74 1.22 0.64 260 120 2.19 1.5290 180 3.49 3.72300 240 4.73 8.6314 283 5.55 21.6

The corresponding graphs are as shown.


57

200

250

300

350

0 2 4 6

p (kN/m2)

εv (10−2)

0

120

240

0 5 10 15 20

q (kN/m2)

εs (10−2)

Problem 6.10

Referring to the solution of problem 6-4, then using a similar approach tothe calculations related to the solution of problem 6-9, the followingquantities can easily be found (note that up to the yield point, the behaviouris undrained and hence the elastic strains are zero) :

Average values p (kN/m2) q(kN/m2) v

400 245.5 1.215 p q v δp δq η = q/p 390 267.2 1.215 -20 25.4 0.685

380 279.9 1.215 370 289.9 1.215 -20 20.1 0.783

360 300 1.215 430 300 1.205 140 0 0.698

500 300 1.195 512.5 337.5 1.189 25 75 0.658

525 375 1.184 537.5 412.5 1.179 25 75 0.767

550 450 1.174 560.7 482.2 1.169 21.4 64.3 0.860

571.4 514.3 1.165

The strains increments corresponding to the average values are thus :


p(10−3) δεsp(10−2 )

390 267.2 1.215 -2.1 2.1 0.84370 289.9 1.215 -2.2 2.2 1.75430 300 1.205 13.5 3.36 1.45512.5 337.5 1.189 2.05 7.14 2.49537.5 412.5 1.179 1.97 6.8 4.7560.7 482.2 1.169 1.63 5.52 13.49


58

Whence the following cumulative elastoplastic strains :

p (kN/m2) q(kN/m2) εv(10−2) εs(10−2) 400 254.5 0 0

380 279.9 0 0.84360 300 0 2.59500 300 1.69 4.04 525 375 2.60 6.53550 450 3.48 11.23571.4 514.3 4.20 24.72

It is seen from the latter table that the total cumulative strains at failure are:

εv = 4.2 × 10−2, εs = 24.72 × 10−2.

Problem 6.11

Just prior to the yield point, the drained behaviour of the clay is elastic;therefore the ensuing elastic volumetric strain is calculated as follows :

δεv = δpK = 473.4−400

8200 = 8.95 × 10−3.

Once yielding occurs, the elastoplastic strains must be calculated in asimilar way to that used previously in conjunction with problem 6-10 :

Average valuesp (kN/m2) q(kN/m2) v

473.4 220.1 1.207 p q v δp δq η = q/p 486.7 260 1.201 26.6 79.9 0.534500 300 1.195

487.5 314.8 1.195 -25 29.6 0.646475 329.6 1.195

462.5 343.2 1.195 -25 27.3 0.742450 356.9 1.195

437.5 369.7 1.195 -25 25.6 0.845425 382.5 1.195


59

and the strain increments corresponding to the average values :


p(10−3) δεsp(10−2)

486.7 260 1.201 2.27 7.75 1.58487.5 314.8 1.195 -2.14 1.99 0.65462.5 343.2 1.195 -2.26 2.26 1.29437.5 369.7 1.195 -2.39 2.39 4.21

Whence the following cumulative elastoplastic strains :

p (kN/m2) q(kN/m2) εv(10−3) εs(10−2)

473.4 220.1 0 0500 300 10.02 1.58475 329.6 9.87 2.24 450 356.9 9.87 3.52 425 382.5 9.87 7.73

The corresponding graphs are depicted in the figures.


60

400

450

500

0 4 8 12

200

300

400

0 2 4 6 8

p (kN/m2)

εv (10−3)

q (kN/m2)

εs (10−2)

Chapter 7

Problem 7.1

The factor of safety in this case is :

F = 2cz.γsat . sin2β + tanφ

tanβ1 − γw

γsatcos α

cos β. cos (α−β)(1 − zwz )

Substituting for β = 16 , α = 8 , F = 1.3, z = 6m, φ = 24, then rearranging, it emerges that :γsat = 21kN/m3, c = 7kN/m3

F = 0.994 + 0.128zw zw = 2.39m

Problem 7.2

In the short term, the factor of safety of the slope without piles is :

F = cuR2αW.d = 50×122×1.57

196×20×2.5 = 1.15

To increase the factor of safety to 1.8, n piles need be used whereby :

F = R.cu

Wd [L + 9n.a] n = 19a

W.d.FR.cu

− L

= 19×0.15 ×

3920×2.5×1.6

12×50 − 12 × 1.57 = 5.4

Hence .n = 6


61

Problem 7.3

From the geometry of the figure, the area of thefailing soil mass is :

.W ≈ 2010kN/m

The factor of safety in theshort term is thence :

F = 65×1.448×182

2010×7.5 ≈ 2

Note 83 = 1.448rd

Problem 7.4

Referring to the figureopposite, the potential failingsoil mass is subdivided intofive slices of equal width

From the geometryb ≈ 4.2m.of the figure, and using aunit weight , theγ = 20kN/m3

following weights can beobtained:

slice A B C D Eweight w 260.4 630 754.8 516.6 180.6(kN/m)

Applying Bishop routine using :φ = 23 , c = 7kN/m2


62

AB

C

D

E

56

35.5

18.47−7

clay fill

7.5m

10.7m

18m

W

O

45

70

20

13

slice α w. sin α (kN/m) w tan φ + c b (kN/m)A 56 215.9 139.9B 35.5 365.8 296.8C 18.4 238.3 349.8D 7 63 248.7E -7 -22 106.1

The calculations yield : Σ w sin α = 861 kN/m.Now, starting with a factor of safety , and using the parameter :F = 1

, the first set of iterative calculations is as follows:mα = cos α + tanφ sinαF

slice mα1

mαc b + w tan φ

(kN/m)

A 0.911 153.4B 1.06 279.8C 1.08 323.05D 1.04 238.1E 0.94 112.7

Thus : , Σ 1mαc b + w tan φ

= 1107 kN/mand the calculated factor of safety is different from theF = 1107

861 = 1.28selected value . Accordingly, another iteration using an initial factorF = 1

leads to the following :F = 1.28

slice mα = cos α + tanφ sinα1.28

1mαc b + w tan φ

(kN/m)

A 0.834 167.7B 1.007 294.8C 1.053 332D 1.03 240.8E 0.94 112.8

Whence : Σ 1mαc b + w tan φ

= 1148 kN/m

and the calculated factor of safety : F = 1148861 = 1.33

Another iteration is needed using a factor F = 1.33 :


63

slice mα = cos α + tanφ sinα1.33

1mαc b + w tan φ

(kN/m)

A 0.82 169.8B 0.999 297C 1.05 333.3D 1.03 241.1E 0.95 111.2

Accordingly : Σ 1mαc b + w tan

= 1152.4kN/m

yielding a factor of safety : .F = 1152.4861 = 1.34

Problem 7.5

The slope is now waterlogged; hence the saturated unit weight applies. Consequently, the following quantity must beγsat = 21kN/m3

used:Σ w sin α = 861 × 21

20 ≈ 904 kN/m.

From the previous figure, the following estimates of average porewaterpressure in each slice can be obtained:

slice w(kN/m) u(kN/m2) (w − ub) tan φ + c b (kN/m)A 273.4 31 90.2B 661.5 75 176.5C 792.5 94 198.2D 542.4 61 150.9E 189.6 23 68.9

The iterative calculations can now start in earnest. Applying initially afactor o safety , it follows that :F = 1


64

slice mα = cos α + tanφ sinα1

1mα(w − ub) tan φ + c b (kN/m)

A 0.94 99B 1.06 166.5C 1.08 183.5D 1.044 144.5E 0.94 73.3

Whence : , yielding a factor ofΣ 1mα(w − ub) tan φ + c b = 666.8kN/m

safety : F = 666.8

904 = 0.74

This value being different from , hence a new iteration is neededF = 1using F = 0.74 :

slice mα = cos α + tan sinα0.74

1mα(w − ub) tan φ + c b (kN/m)

A 1.035 87.1B 1.147 153.9C 1.13 175.4D 1.062 142.1E 0.923 74.6

Σ 1mα(w − ub) tan φ + c b = 633.1kN/m F = 633.1

904 = 0.7

Were another iteration to be undertaken using an initial , theF = 0.7outcome would be a calculated value . Therefore, for all practicalF = 0.69purposes .F = 0.7


65

Problem 7.6

The five slices of saturated soil have a combined total weight such that : Considering the porewater pressure values estimatedΣ w sin α = 904kN/m.

in the previous problem, and taking into account the geometry of elements( for each element) and the soil characteristics ( , theb = 4.2m c = 7kN/m2)following calculations can be made :

slice w(kN/m) u(kN/m2) c bcos α +

w cos α − u.bcos α

tan φ (kN/m)

A 273.4 31 18.6B 661.5 75 100.5C 792.5 94 173.6D 542.4 61 148.6E 189.6 23 68.2

Accordingly : Σ

c bcos α +

w cos α − ubcos α

tan φ

= 509.5kN/m

Using Fellenius method, a factor of safety implies :F = 1.7

1.7 = 509.5Σ w sin α−Σ T Σ T = 904 − 509.5

1.7 = 604.3kN/m

Since the horizontal tensile force developed by each geotextile layer is, the number of layers required is thus: T = 150kN/m

n = Σ T150 = 604.3

150 ≈ 4


66

Chapter 8

Problem 8.1

a- Referring to thefigure, it is seen thatthe work done bythe weight of soil islimited to the areaabgf since the workdone in abc cancelsout that done in bde.Accordingly, workdone by the soilweight is :

δEs = δωcos 45.γH2 [H + 2B] = δω

2γH2

2 1 + 2BH

The work done by the external loads and is:qo qu

δEq = δω2

(qo[H + B] − quB)

The internal work done on the slip plans :

δWsp = δω 2 .cu(H + B ) = δω2

2cu(H + B)

The internal work done within the slip fan : δWsf = δω2

πcuB

Therefore, the total external work is :

δE = δω2γH2

21 + 2B

H + qo(H + B) − quB

Similarly, the total internal work :

δW = δω2

cu[πB + 2(H + B)]


67

B

H

ba

cd

e

gf

qo qo

qu

45

cu

δω

Equating both expressions, then rearranging, it follows that :

qu = qo1 + H

B + γH2

2

1B + 2

H − 2cu

1 + π

2 + HB

b- in the long term, theangles depicted in theopposite figure are such

that: , α = π4 + φ

2 β = π4 − φ

2

Moreover :δωb = δωexp

π2 tan φ

The work done by theexternal loads is :

δE = qo.C.δω. cos π4 + φ

2 − qu.B.δωbsin

π4 + φ

2

On the other hand, it is seen that :

C = 2r1cos π4 + φ

2 + H

tan

π4 +φ

2

B = 2r2. cos π4 − φ

2 = 2r2. sin

π4 + φ

2

Therefore :

C = B. r1

r2. tan

π4 +φ

2

1 + H

2r1. sin

π4 +φ

2

Also : r1r2 = exp

−π2 tan φ


68

H

B

C

qo qo

qu

φ

φ

α

α β

45 − φ2

δω

δωb

r1 r2

thus substituting for C in the expression of then equating it to zero, itδEfollows that:

qusin

π4 + φ

2 exp

π2 tan φ

= qor1r2

cos

π4 +φ

2

tan

π4 +φ

2

1 + H

2r1sin

π4 +φ

2

Now substituting for the ratio and rearranging:r1r2

qu = qoexp −πtanφ

tan2

π4 +φ

2

1 + H

2r1sin

π4 +φ

2

But 2r1sin

π4 + φ

2 = B. r1

r2 = B exp −π2 tan φ

whence :

qu = qoexp −πtanφ

tan2

π4 +φ

2

1 + H

B exp

π2 tan φ

and therefore :

HB = qu

qo tan2

π4 + φ

2 exp

π2 tan

− exp −

π2 tan φ


69

Problem 8.2

Referring to the figure above, the expression of the work done by theexternal loads is :

δE = quB.δωcos π4 + φ

2 − qoC.δωexp

π2 tan φ

sin

π4 + φ

2

Because of the logarithmic spiral nature of the slip fan, it follows that :

r2r1 = δωb

δω = exp

π2 tan φ

From the geometry of the figure :

,C = 2r2cos π4 − φ

2 + H

tan

π4 −φ

2

B = 2r1cos π4 + φ

2

Hence : C = B. r2r1 tan

π4 + φ

2

1 + H

2r2cos

π4 +φ

2

Accordingly, substituting for C in the expression of external workestablished earlier, then equating it to zero :


70

C

qu

qo

β

βαφ

φ

H

B

δω

δωb

r1r2

qu = qo tan2

π4 + φ

2 exp

πtan φ

1 + H

2r2cos

π4 +φ

2

But : 2r2cos π4 + φ

2 = B. exp

π2 tan φ

Thence:

qu = qo exp πtan φ

tan2

π4 + φ

21 + H

B exp −

π2 tan φ

Problem 8.3

From the figure,it is seen that theexternal work isdone by theweight of the soilblock abcd andby the horizontalforce .P1

The weight ofabcd is:

γBH + H2

2 − H2

2 tan α = γH2

2 2BH + 1 − tan α

Hence the work done by the soil weight and :P1

δE = −P1δω

2+ γH2

2δω2

2BH + 1 − tan α

-the internal work due to energy dissipation along the slip plans :

δWsp = δω.cuH 2 + 2δω B2

.cu

-the internal work done within the slip fan :

δWsf = 2cuB2

π2δω


71

a b

cd

H

B

P1 P1

α α

45 45 45

45

δω

Thus the total internal work :

δW = δω2

.2cuH + B1 + π

2

Equating to and rearranging :δE δW

P1 = γ H2

2 2BH + 1 − tan α − 2cuH

1 + B

H1 + π

2

Problem 8.4

According to Mohr's circles representation of stresses, it is seen that :

qu = qo + γH + 2p1sin φ + x1 + x2 + 2p3sin φ

where :

p1 = qo+γH

1−sinφ

p1

p2 dp

p= o

π4 2 tan φ dθ p2 = p1exp

π2 tan φ

, ,x1 = p2sinφtan 60 x2 = p2sinφ

tan 75 p3 = qu

1+sinφ

Hence : x1 + x2 = 0.845 sin φ exp

π2 tan φ

(qo+γH)

1−sinφ

On substitution for the different quantities in the equation establishedearlier :

qu = (qo+γH)

tan2

π4 −φ

2

1+sinφ +0.845 exp

π2 tanφ

1−sinφ

so that for φ = 22 qu = 10.43(qo + γH)


72


73

AB

C

Dqo + γH 2p1sin φ

x1 x2

2p3 sin φ

qu

120

6075

45

σ

τ

p3

p2

p1

φ

H

A

B C

D

qo

60 75

qu

qo

Problem 8.5

-the effective width of foundation : B = B − 2e = (3 − 2e)m-the net pressure at failure : qnf = 1

2(γsat − γw).B Nγ iγ sγ dγ + γ.D.Nq iq sq dq − γ.D

with the inclination factors :

, iγ = 1 − α

φ

2= 1 − 20

35

2= 0.184 iq =

1 − α90

2= 1 − 20

90

2= 0.605

the bearing capacity factors read from figure 8.26 : φ = 35 Nγ ≈ 43, Nq ≈ 30

the shape factors : since the length L >> B sq = sγ = 1

the depth factors : because , then ξ = DB = 1.2

3 = 0.4 < 1

dq = 1 + ξ tan φ 1 − sin φ

= 1 + 0.4 tan 35(1 − sin 35) = 1.12dγ = 1

On substitution for the different quantities :qnf = 1

2 × 9.8(3 − 2e) × 43 × 0.184 × 1+ 18 × 1.2 × 30 × 0.605 × 1.12 − 18 × 1.2

qnf = 533.8 − 77.54e

The net pressure at foundation base : qn = Rv

B− γ.D

with : , therefore :Rv = 400 cos 20 ≈ 376 kN/m

qn = 376(3−2e) − 21.6

The factor of safety against shear failure being , thence :F = 3

qnfqn = 3 533.8 − 77.54e = 3 ×

3763−2e − 21.6

e2 − 9.22e + 4.3 = 0 e = 0.493m < B6 = 0.5m


74

Problem 8.6

The factor of safety being defined as : F = qunqn = 3

and the ultimate net pressure at foundation level for a rectangularfoundation is : . Accordingly, the bearing capacity factor forqun = cuNc

rec

the rectangular foundation is :

Ncrec = 3.qn

cu = 3×15060 = 7.5

This factor being related to that corresponding to an equivalent squarefoundation as follows :Nc

rec = Ncsq 0.84 + 0.16B

L = Nc

sq(0.84 + 0.16 × 0.8) = 0.958Ncsq

Whence : , and from the charts in figure 8.27 , this valueNcsq = 7.75

corresponds to a ratio , and DB ≈ 1.25 B = 1.6m L = B

0.8 = 2m

Problem 8.7

First, calculate the ratios :eL

L = 0.54 = 0.125, eB

B = 0.33 = 0.1

Using the charts in figures 8.32 & 8.33, it follows that :B1

B ≈ 0.1 B1 = 0.1 × 3 = 0.3m

L1

L ≈ 0.23 L1 = 4 × 0.23 = 0.92m

The equivalent foundation width can now be estimated from equation 8.63:

B = 12(B + B1) + L1

2L(B − B1) = 0.5 × (3 + 0.3) + 0.922×4 × (3 − 0.3) = 2.02m

The effective foundation area is therefore :

A = LB = 4 × 2.02 = 8.08m2

Whence the following ultimate pressure :


75

qu = γsat DNqsq dq iq + γsatB2 Nγ sγ dγ iγ

For an angle , figure 8.26 yields : φ = 39 Nq ≈ 52, Nγ ≈ 92The inclination factors are :

, iq = 1 − β

90

2= 1 − 18

90

2= 0.64 iγ =

1 − βφ

2= 1 − 18

39

2= 0.29

The shape factors :sq = 1 + B

L tan φ = 1 + 2.024 tan 39 = 1.41

sγ = 1 − 0.4BL = 1 − 0.4 × 2.02

4 ≈ 0.8

The depth factors :ξ = D

B = 2.53 = 0.83 < 1 dq = 1 + ξ tan φ

1 − sin φ

= 1 + 0.83 tan 39 × (1 − sin 39) = 1.25dγ = 1

Consequently, the ultimate pressure is :

qu = 20 × 2.5 × 52 × 1.41 × 1.25 × 0.64 + 20 × 2.022 × 92 × 0.8 × 1 × 0.29

= 3364 kN/m2

and the net ultimate pressure :

qnu = qu − γsatD = 3364 − 20 × 2.5 = 3314 kN/m2

The net pressure applied at foundation level is :

qn = R. cos 18A − γsatD = 15000×cos 18

3×4 − 20 × 2.5 = 1139 kN/m2

Hence a factor of safety against shear failure :

F = qnuqn = 3314

1139 = 2.91


76

Chapter 9

Problem 9.1

In the long term, the ultimate loading capacity is :

Qu = AbσvbNq − W + pK tan δ o

Lσv.dz

- the area of pile base : Ab = πB2

4 = 0.567m2,- the effective vertical stress at the pile base :

σvb = 3 × 18.7 + 9.8 × (L − 3) = (9.8L + 26.7) kN/m2,- for an angle , figure 9.18 yields φ = 22 Nq ≈ 12,- the coefficient of earth pressure K =

1 − sin φ OCR =

,(1 − sin 22) 2 = 0.884 pK tan δ = πB × 0.884 × tan 0.75φ

≈ 0.7

- the quantity : o

Lσv.dz = A1 + A2

A1 = 12(18.7 × 3 × 3) = 84.15kN/m,

A2 = (L − 3) × 18.7 × 3 + 12(L − 3)2 × 9.8

= (4.9L2 + 26.7L − 124.2)kN/m.

- the effective weight of pile (weight of steel + weightof soil plug) :

steel cross sectional area (15mm thick wall) :S = π

4 × (0.852 − 0.72) = 0.1826m2

soil plug cross section : A = π ×0.72

4 = 0.385m2

whence an effective weight : W = 0.1826 × [3 × 77 + (L − 3) × 67] + 0.385[18.7 × 3 + 9.8 × (L − 3)]

= (16L + 15.8) kN.

Therefore :Qu = 0.567 × (26.7 + 9.8L) × 12 − 16L − 15.8+0.7 × (84.15 + 4.9L2 + 26.7L − 124.2)


77

3m

L

Z

σv

A1

A2

On substitution for it follows that :Qu = 3, 800kN,L2 + 20.23L − 1067.7 = 0 L = 24.09m.

In the short term, the ultimate loading capacity is : Qu = Qs + Qb − W- the ultimate shaft friction : Qs = πBLαc u

- the average shear strength : . Whence :c u = (120 + L) kN/m2

Qs = π ×0.85 × 0.35L × (120 + L) = (0.935L2 + 112.15L) kN- the quantity is calculated according to equation 9.11:Qb − W

Qb − W = NccuAb = NcπB2

4 (120 + 2L) = 9π × 0.852

4 × (120 + 2L)

Thus, in the short term :Qu = 0.935L2 + 122.4L + 612.8

so that on substitution for :Qu = 3, 800kN

L2 + 130.9L − 3408.8 = 0 L = 22.26m.

The long term conditions are the most critical, hence the length L ≈ 24.1m.

Problem 9.2

a) Short term ultimate loading capacity calculated as follows :Qu = Qs + Qb − W

- the ultimate shaft friction : . Since , figureQs = Asαcu cu < 40kN/m2

9.14c yields Whence: α ≈ 1.

Qs = 30 × 15 × (0.36 + 0.38) × 2 × 1 = 666kN.

- the quantity .Qb − W ≈ NcAbcu = 9 × 0.36 × 0.38 × 40 = 49.25kN

Therefore: Qu = 666 + 49.25 = 715.25 kN.

b) In the long term, the ultimate loading capacity is :

Qu = AbσvbNq − W + pK tan δ o

Lσv.dz


78

- the pile cross sectional area (including the soil plug) : Ab = 0.36 × 0.38 = 0.1368m2

- the effective overburden pressure at the pile base :σvb = 2 × 19 + 13 × 9 = 155 kN/m2

- the bearing capacity factor from figure 9.18: φ = 23 Nq ≈ 14

- the quantity : pK tan δ = 2 × (0.36 + 0.38) × (1 − sin 23) × tan

3φ4 = 0.28

- the integral : o

Lσv.dz = A1 + A2

A1 = 12 × 2 × 19 × 2 = 38kN/m

A2 = 13 × 38 + 12 × 132 × 9 = 1254.5kN/m

- the submerged weight of pile :W ≈ 22 × 10−3 × (2 × 77 + 13 × 67)+ 0.36 × 0.38 × (2 × 19 + 13 × 9) = 43.75kN.

Therefore :Qu = 0.1368 × 155 × 14 − 43.75 + 0.28 × (38 + 1254.5) ≈ 615 kN.

The long term loading capacity being smaller than the short term, hence:

Qu ≈ 615 kN.

Problem 9.3

Because of the negative skin friction developing at the top of the pile shaft,the allowable loading capacity is calculated from equation 9.24:

Qa = Qb+Qs−1.5Qn

F − W

- the base resistance : Qb = AbσvbNq


79

Z

2m

15m

σv

A1

A2

, Ab = πB2

4 = π0.52

4 = 0.1963m2

,σvb = 3 × 18.6 + 4 × 8.6 + (L − 7) × 9.8 = (9.8L + 21.6) kN/m2

figure 9.18 : .φ = 22 Nq ≈ 12

Whence: Qb = 0.1963 × 12 × (9.8L + 21.6) = (23.1L + 50.9) kN

- the positive skin friction : Qs = pK tan δ o

Lσv.dz

p = πB = π ×0.5 = 1.57m K1 = 1 − sin 24 = 0.593, K2 = 1 − sin 22 = 0.625 δ1 = 3

4 × 24 = 18 δ2 = 34 × 22 = 16.5

the positive skin friction is developed from a depthof 5m below ground,

hence: 5

Lσv.dz = A3 + A4

A3 = (3 × 18.6 + 2 × 8.6) × 2 + 12 × 22 × 8.6

= 163.2kN/m

A4 = (3 × 18.6 + 4 × 8.6)(L − 7) + 12(L − 7)2 × 9.8

= (4.9L2 + 21.6L − 391.3)kN/m

Therefore:Qs = p.(A3.K1. tan δ1 + A4.K2. tan δ2)On substitution for the different calculated quantities, it follows that:

Qs = (1.42L2 + 6.28L − 64.37)kN

- the negative skin friction develops between zero and 5m depth :

Qn = pK tan δ o

5σv.dz = p.K1tan δ1 (A1 + A2)

A1 = 12 × 32 × 18.6 = 83.7kN/m


80

Z

3m

5m

7m

L

σv

A1

A2

A3

A4

A2 = 3 × 18.6 × 2 + 12 × 22 × 8.6 = 128.8kN/m

Thus: Qn = 1.57 × 0.593 × tan 18 × (83.7 + 128.8) = 64.3kN

- the submerged weight of pile:

W = Ab[3 × 24 + (L − 3) × 14] = (2.75L + 5.9) kN.

Accordingly, knowing that the allowable load is : ,Qa = Qb+Qs+1.5Qn

3 − Wthen substituting for the different known quantities, including Qa = 500kNand rearranging, the following equation ensues :

L2 + 15L − 1152 = 0 L ≈ 27.3m

Problem 9.4

Because of the negative skin friction : Qa = Qb+Qs−1.5Qn

3.5 − W


, Ab = π0.522

4 = 0.2124m2

- since the pile base is embedded in sand, the angle used in conjunctionwith figure 9.18 is calculated from equation 9.21a :

φ = φ +402 = 38+40

2 = 39 Nq ≈ 155

- the effective overburden pressure at pile base:

σvb = 5 × 19.6 + 2 × 18.5 + 10 × 9.5 + 1.5 × 9.5 = 244.3kN/m2

Hence: Qb = 0.2124 × 244.3 × 155 ≈ 8043kN

- the positive skin friction :Qs = p.K. tan δ zf

Lσv.dz

pile perimeter : p = πB = 0.52π =1.634m


81

fill material: K1 = 1 − sin 21 = 0.64, δ1 = 0.75 × 21 = 15.75clay : K2 = 1 − sin 22 = 0.625, δ2 = 0.75 × 22 = 16.5sand : K3 = 1.5.(1 − sin 38) = 0.576, δ3 = 0.75 × 38 = 28.5

- the quantity : 15

18.5σv.dz = A4 + A5

A4 = 2 × (5 × 19.6 + 2 × 18.5 + 8 × 9.5) + 1

2 × 22 × 9.5 = 441kN/m

A5 = 1.5 × (5 × 19.6 + 2 × 18.5 + 10 × 9.5) + 1

2 × 1.52 × 9.5 = 355.7kN/m

Whence: Qs = p.[A4K2tan δ2 + A5K3tan δ3]so that on substitution for the different calculatedquantities :

Qs ≈ 315kN

- the negative skin friction develops down to a depth of 15m :

, where: Qn = p.K tan δ o

15σv.dz o

15σv.dz = A1 + A2 + A3

, A1 = 12 × 52 × 19.6 = 245kN/m

A2 = 5 × 19.6 × 2 + 12 × 22 × 18.5 = 233kN/m

A3 = 8 × (5 × 19.6 + 2 × 18.5) + 12 × 82 × 9.5 = 1384kN/m

Therefore: Qn = p.[A1K1tan δ1 + (A2 + A3).K2tan δ2]A straightforward substitution for the different quantities yields:

Qn ≈ 561kN

- the submerged pile weight: .W = 0.2124 × (7 × 24 + 11.5 × 14) ≈ 70kN

Accordingly:

Qa = 315+8043−1.5×5613.5 − 70 ≈ 2077kN.


82

5m

7m

15m

17m

18.5

z

σv

A1

A2

A3

A4

A5

Problem 9.5

The ultimate loading capacity : Qu = Qs + Qb − W

- the shaft friction : Qs = p.k. tan δ o

Lσv.dz

pile perimeter : p = π ×0.45 = 1.414m loose sand :K1 = 1.6 × (1 − sin 32) = 0.752, δ1 = 0.75 × 32 = 24 dense sand:K2 = 1.6 × (1 − sin 41) = 0.55, δ2 = 0.75 × 41 = 30.75

- the integral : o

Lσv.dz = A1 + A2

A1 = 12 × 42 × 16.5 = 132kN/m

A2 = 16.5 × 4 × (L − 4) + 12 × (L − 4)2 × 11

= 5.5L2 + 22L − 176

Whence: Qs = p[A1K1tan δ1 + A2K2tan δ2]

Qs = (2.54L2 + 10.3L − 19.7)kN.


Ab = πB2

4 = π0.452

4 = 0.159m2

- using equation 9.21a and figure 9.18: φ = 41+402 = 40.5 Nq ≈ 200

- the effective vertical stress at pile base: σvb = 4 × 16.5 + (L − 4) × 11 = (11L + 22) kN/m2

It follows that : Qb = 0.159 × 200 × (11L + 22) = (349.8L + 700) kN

- the submerged weight of pile : W = 0.159 × (4 × 24 + (L − 4) × 14) = (2.23L + 6.36) kN


83

4m

L

Z

σv

A1

A2

Thus, knowing that the ultimate load is , the substitutingQu = Qs + Qb − Wfor the different quantities including :Qu = 1500kN

L2 + 140.9L − 826 = 0 L = 5.64mProblem 9.6

a) Only skin friction provides resistance to the applied load. The ultimateloading capacity is the smaller of the two following quantities:

Q1 = n.Qu − Wc

Q2 = Qsb − Wc − Ws

n being the number of caissons (12), is the ultimate loading capacityQu

(without base resistance) of a single caisson, is the weight of pilecap,Wc

the friction developed along the block sides, and represents theQsb Ws

submerged weight of the block of soil containing the caissons.

- the weight of pilecap : Wc = 24 × 5 × 41 × 30 = 147, 600 kN- the submerged weight of a single caisson : W = π42

4 × 45 × 14 = 7, 917kN

Thence: Q1 = 12 × (175 × π ×4 × 45 − 7917) − 147600 = 944, 918 kN

- the quantity : Qsb = 2 × (37 + 26).cu.L = 2 × 63 × 400 × 45 = 2268, 000 kN- the submerged weight of the soil block (excluding the weight ofcaissons):Ws = (Abb − nAb).γ × L + nAb.γc × L- the area of the block base : Abb = 37 × 26 = 926m2

- the area of a caisson base: Ab = π42

4 = 12.57m2

- effective unit weight of concrete : .γc = 14kN/m3

Therefore:Ws = (926 − 12 × 12.57) × 12 × 45 + 12 × 12.57 × 14 × 45 = 513, 615kN

and : Q2 = 2268 − 147.6 − 513.6 = 1606 MN.

The ultimate load being the smaller of , and accordingly:Q1 & Q2


84

Qu = Q1 ≈ 945 MN.

The corresponding factor of safety is thence: F = 945500 ≈ 1.9

b) If the end bearing were to be mobilised, then the ultimate loadingcapacity of the group (discarding the contribution due to the pilecap)corresponds to the smaller of the two quantities :Q1 = Q1 + n.Nc.Ab.cu

(notice the units).= 944, 918 + 12 × 6 × 12.57 × 1000 ≈ 1, 850 MN

Q2 = Q2 + Qbb

Qbb = Abb(cuNc + γ.L) = (37 × 26) × (1000 × 6 + 22 × 45 + 20 × 5) ≈ 6, 821MN

Q2 = 1606 + 6821 = 8, 427MN.

being the smaller of the two values, and hence the ultimate load :Q1

, corresponding to a factor of safety : Qu = 1, 850MN F = 1850500 = 3.7

Problem 9.7

The ultimate loading capacity is the smaller of the two quantities:Q1 = nQu + Qc − Wc

Q2 = Qbb + Qsb − nWp − Wc − Ws + Qc

- the ultimate loading capacity of a single pile : Qu = 1.5MN- the loading capacity due to the pilecap :

. Qc = 0.4γ BcNγ (Ac − nAb)

Note : area of pilecap, width of pilecap. Whence: Ac = Bc =Qc = 0.4 × 16.5 × 5 × 3 ×

5 × 5 − 16 × π ×0.452

4 = 2, 223kN


85

and therefore : Q1 = 16 × 1500 + 2223 − 720 = 25, 503kN.

- the base resistance of the block : Qbb = AbbσvbNq + 0.4γ b.Nγ

for , figure 9.36 yields : φ = 41 Nq ≈ 47, Nγ ≈ 22- the cross sectional area of the block : ,Abb = 3.9 × 3.9 = 15.21m2

b being the corresponding width (3.9m).- the effective vertical stress at the block base :

σvb = 4 × 16.5 + 1.64 × 11 ≈ 84kN/m2

Accordingly: Qbb = 15.21 × (47 × 84 + 0.4 × 11 × 3.9 × 22) = 65, 791kN.

- the friction on the sides of the block is :

Qsb = pbo

LK tan δ.σv.dz

loose sand : K1 = 1.6 × (1 − sin 32) = 0.752dense sand : K2 = 1.6 × (1 − sin 41) = 0.55block perimeter :

pb = 2 × (3.9 + 3.9) = 15.6m

- the integral : o

Lσv.dz = A1 + A2

A1 = 12 × 42 × 16.5 = 132kN/m

A2 = 16.5 × 4 × 1.64 + 12 × 1.642 × 11 = 123kN/m

Consequently:

Qsb = pb[A1K1tan 32 + A2K2tan 41]

On substitution for the different values, it follows that : Qsb = 1, 885kN

- the contribution due to the area of pilecap situated outside the block :Qc = 0.4γ BcNγ (Ac − Abb) = 0.4 × 16.5 × 5 × 3 × (52 − 3.92) = 969kN


86

4m

z

5.64m

σv

A1

A2

- the submerged weight of all piles :

nWp = 16 × π0.452

4 × (4 × 24 + 1.64 × 14) ≈ 303kN- the weight of pilecap being : Wc = 720kN- the submerged weight of soil within the block: Ws = (Abb − nAb).γ .L =

52 − 16 × π ×0.452

4 × (16.5 × 4 + 11 × 1.64)

= 1, 887kNWhence the quantity: .Q2 = 65791 + 1885 − 303 − 720 + 969 = 67, 622kN

Q1 < Q2 Qug = Q1 = nQu + Qc − Wc = 25, 503 kN

Allowing for a factor of safety of 3, the allowable loading capacity of thegroup is such that :

Qag = nQu+Qc

3 − Wc = 16×1500+22233 − 720 = 8, 021kN.

Problem 9.8

Were the pilecap contribution to the loading capacity to be discarded, thenaccording to the previous calculations in problem 9.7 :

Q1 = Q1 − Qc = 25503 − 2223 = 23, 280kN

Q2 = Q2 − Qc = 67622 − 969 = 66, 653kN.

Obviously the ultimate loading capacity of the group is in this case :. Hence an allowable load: Qug = Q1 = nQu − Wc = 23280kN

Qag = nQu

3 − Wc = 16×15003 − 720 = 7, 280 kN.


87

Chapter 10

Problem 10.1

a- for a Rankine type analysis to apply, thecriterion of equation 10.59 must be satisfied :

tan−1

LH ≥ η

Using equation 10.58 : η = π4 − φ

2 + 12 [δ − α]

Equation 10.36 yields :α = sin−1

sinδsinφ

= sin−1

sin16sin35

= 28.7

Accordingly : η = 45 − 17.5 + 12 × (16 − 28.7) = 21.1

and : tan−1

LH = tan−1

3.57.5 = 25 > η

The criterion is thence satisfied, and the magnitude of active thrust is notaffected by the wall roughness.

Applying Rankine analysis, the coefficient of active pressure is :

Ka = cos δcos δ− cos2δ−cos2φ

cos δ+ cos2δ−cos2φ= cos 16cos 16− cos216−cos235

cos 16+ cos216−cos235≈ 0.3

leading to an effective active thrust :

where Pa = 12γH1

2Ka H1 = 7.5 + 3.5 tan 16 = 8.5m

Pa = 202 × 0.3 × 8.52 = 216.8kN/m

The corresponding line of action is estimated as follows :

above the wall base.d = 1 + 8.53 = 3.83m


88

0.8m

7.5m

3.5m

1.5m

silty sand

clean dense sand

1m

δ = 16

φ = 35

φ = 38

H1

b- first evaluate the total weight including the that of soil above the wallbase. Referring to the figure and using a concrete unit weight

, it is seen that :γc = 24kN/m3

w = w1 + w2 + w3 + w4

w1 = 0.8 × 7.5 × 24 = 144kN/mw2 = 5 × 1 × 24 = 120kN/mw3 = 7.5 × 3.5 × 20 = 525kN/mw4 = 1

2 × 3.5 × 1 × 20 = 35kN/m

Whence a total weight : w = 824kN/m

The sum of vertical forces behind the wall is :

Σ Fv = w + Pasin δ = 824 + 216.8 sin 16 = 883.8kN/m

Neglecting the passive pressure, the resisting force at the wall base is :

Fr = tan 23φ Σ Fv = 883.8 tan 2

3 × 38 = 418.4kN/m

The sliding force :

Fs = Pacos δ = 216.8 cos 16 = 208.4kN/m

Hence a factor of safety against sliding : F = 418.4208.4 = 2

To calculate the factor of safety agianst shear failure, first the eccentricityof the resultant force applied at the base of wall must be determined. To doso, both restoring and overturning moments with respect to the toe of thewall (point O in the following figure) should be calculated.The point of application of the total weight w is found by taking momentsabout O of the elementary weights calculated previously.w1, w2, ....w4

Referring to figure above, it follows that :

d = 1.1×144+2.5×120+3.25×525+3.83×35824 = 2.79m


89

Accordingly, discarding the passive thrust,the restoring moments are such that :

Σ Mr = 2.79 × 824 + 5 × 216.8 sin 16

= 2597.8kNm/m

The overturning moment due to thathorizontal component of the active thrust:

Σ Mo = 3.83 × 216.8 cos 16 = 798.2kNm/m

Therefore the eccentricity is as follows :

e = B2 − Σ Mr−Σ Mo

Σ Fv= 5

2 − 2597.8−798.2883.8 = 0.464m

Note that e < B6 = 5

6 = 0.83m

The maximum shear stress induced at the base of wall is :

qmax = Σ Fv

B1 + 6e

B = 883.8

5 × 1 + 6×0.464

5 = 275.2kN/m2

The angle of inclination with respect to the vertical of the resultant force isestimated as follows :

α = tan−1 Σ Fh

Σ Fv

= tan−1

Pa cos 16

Σ Fv

= tan−1

216.8 cos 16

883.8 = 13.2

The effective width of foundation :

B = B − 2e = 5 − 2 × 0.464 = 4.07m

The ultimate pressure is therefore :qu = γDNq sq dq iq + γB

2 Nγ sγ dγ iγ

- the bearing capacity factors for an angle estimated from figureφ = 388.26 are : Nq ≈ 45, Nγ ≈ 80


90

3.83m

O

2.79m

w

- the shape factors : sq = 1 + BL tan φ = 1 + 4.07

15 tan 38 = 1.21

sγ = 1 − 0.4BL = 1 − 0.4 × 0.407

15 = 0.89- the depth factors :ξ = D

B = 15 = 0.2 < 1 dq = 1 + ξ tan φ

1 − sin φ

= 1 + 0.2 × tan 38 × (1 − sin 38) = 1.06 dγ = 1

- the inclination factors :

,iq = 1 − 13.2

90

2= 0.73 iγ =

1 − 13.238

2= 0.43

On substitution for the different quantities :qu = (20 × 1 × 45 × 1.21 × 1.06 × 0.73) +20 × 4.07

2 × 80 × 0.89 × 1 × 0.43 = 2089kN/m2

leading thus to a factor of safety against shear failure :

F = quqmax = 2089

275.2 ≈ 7.6

Problem 10.2

Applying Coulomb analysis, the coefficient of active pressure is :

Ka = cos2φ

cos δ

1+

sin φ +δ sinφ

cos δ

2 = cos235

cos 16 1+ sin (16+35) sin 35

cos 16

2 = 0.247

whence an effective lateral thrust :

Pa = 12 × 20 × 7.52 × 0.247 ≈ 139kN/m


91

Problem 10.3

Using table 10.1 with , theφ = 35 , β = 16 β

φ= 0.46

corresponding coefficient of active pressure can be estimated as follows :

β

φ= 0.4, φ = 35 Ka = 0.291

β

φ= 0.6, φ = 35 Ka = 0.329

A linear interpolation yields :

β

φ= 0.46 Ka ≈ 0.291 + (0.329−0.291)

0.2 × 0.06 = 0.302

Accordingly, the effective active thrust applied to the wall is :

Pa = 12 × 20 × 7.52 × 0.302 ≈ 170kN/m

Problem 10.4

First, determine the appropriate coefficients of active and passive pressurethat apply to the soil. Referring to table 10.1, it is seen that for :

, the following values of the coefficientsλ = 0, β

φ= 13.5

22.5 = 0.6, δ = 23φ

of active and passive pressure are read :

φ = 20 Ka = 0.551, Kp = 3.7φ = 25 Ka = 0.468, Kp = 5.7

Hence : φ = 22.5 Ka = 0.51, Kp = 4.7

The coefficient of active and passive pressure applied to the surcharge areestimated from table 10.2, using linear interpolation. The different anglesused in conjunction with table 10.2 are such :

α = 0, Ω = 90 + 13.5 = 103.5 , δ = 23φ


92

Accordingly, the coefficient of active pressure is estimated as follows :Ω = 100 , φ = 22.5 Ka = 0.394+0.314

2 = 0.354

Ω = 105 , φ = 22.5 Ka = 0.37+0.2892 = 0.33

Therefore for an angle , a straightforward linear interpolationΩ = 103.5yields :

Ka = 0.354 − 3.5 × (0.354−0.33)5 ≈ 0.337

Similarly, the coefficient of passive pressure :

Ω = 100 , φ = 20 Kp = 3+4.182 = 3.59

Ω = 105 , φ = 25 Kp = 3.2+4.542 = 3.87

Whence : Ω = 103.5 , Kp = 3.59 + 3.5 × (3.87−3.59)5 = 3.79

The active thrust normal to the wall is thence :

Pan =

12γH2Ka + q1HKa

cos δ

=

12 × 20 × 152 × 0.51 + 15 × 15 × 0.337 cos 15 ≈ 1182kN/m

and the passive thrust normal to the wall is :

Ppn =

12γHp

2Kp + Kp qp Hp cos δ

=

12 × 20 × 52 × 4.7 + 3.79 × 25 × 5 cos 15 ≈ 1593kN/m.


93

Problem 10.5

a- dry soil :- coefficient of active pressure applied to the soil estimated from table10.1, with : λ = 0, δ = 2

3φ , β

φ= 13

25 ≈ 0.5, φ = 25 Ka = 0.422+0.4682 = 0.445

- coefficient of active pressure applied to both the surcharge and soilcohesion estimated from table 10.2. Thus for : , it is seen that :Ω = 103 , α = 0, δ = 2

3φ , φ = 25

,Ω = 100 Ka = 0.314 Ω = 105 Ka = 0.289

Therefore, a linear interpolation yields :

Ω = 103 Ka = 0.289 + 3 × 0.314−0.2895 = 0.304

The total active pressure normal to the wall :

Pan =

12γH2Ka + qKaH

cos δ − c

tanφH1 − Ka cos δ

=

12 × 19 × 72 × 0.445 cos 16.7 + (40 × 0.304 × 7 × cos 16.7) −

5tan25(1 − 0.304 cos 16.7)

Pan = 198.4 + 81.5 − 7.6 = 272.3kN/m

The point of application of this thrust is found by taking moments aboutthe toe of the wall. Accordingly, is applied at a distance from thePan d1

wall base such that :

d1 =198.4×7

3 +81.5×72 −7.6×7

2272.3 = 2.65m


94

b- waterlogged soil :

In this case, the total active thrust normal to the wall is the sum of theeffective active normal thrust and the thrust due to the water. Applying thesame coefficients of active pressure, it follows that :- the effective active thrust normal to the wall :

Pa =

12 × 9.5 × 72 × 0.445 cos 16.7 + (40 × 0.304 × 7 × cos 16.7)−

5tan25(1 − 0.304 cos 16.7)

Pan = 99.2 + 81.5 − 7.6 = 173.1kN/m

- the normal thrust due to water :

Pw = 12γwH2 = 5 × 72 = 245kN/m

Hence the total thrust normal to the wall :

Pan = 173.1 + 245 = 418.1kN/m

whose point of application, situated a distance above the wall base, isd2

calculated by taking moment about the toe of the wall :

d2 =(99.2+245) ×7

3 +(81.5−7.6) ×72

418.1 = 2.54m


95

Problem 10.6

The distance z in the figure is calculated from equation 10.60 :

z = b.Ka(β)Ka(β)−Ka(0) = b.Ka(12.5)

Ka(12.5)−Ka(0)

Table 10.1 is then used to estimate the corresponding coefficients of activepressure, knowing that : . Hence :λ = 0, δ = 2

3φ , φ = 21

β

φ= 0 Ka(0) = 0.442 − 0.442−0.364

5 = 0.426

β

φ= 12.5

21 ≈ 0.6 Ka(12.5) = 0.551 − 0.551−0.4685 = 0.534

Accordingly : z = 1×0.5340.534−0.426 = 4.94m

Moreover, the three areas in the figure above are such that : A1 = (z − b)2 γ

2Ka(β) = (4.94 − 1)2 × 202 × 0.534 = 83kN/m

A2 = γ[H + b − z][ z − b]Ka (β) = = 20 × (6 + 1 − 4.94) × (4.94 − 1) × 0.534 = 86.7kN/m

A3 = γ2 [H + b − z] 2Ka(0) = 20

2 × (6 + 1 − 4.94)2 × 0.426 = 18.1kN/m


96

b

H

z

O

d

σaslope γ.Ka(0)

slope γ.Ka(β)

β

A1

Pa

δ

A2A3

Therefore, the effective active thrust due to the soil weight is :

P1 = A1 + A2 + A3 = 187.8 kN/m

The effective active thrust normal to the wall induced by cohesion is asfollows:

P2 = o

z−b ctanφ

cos δ.Ka(β) − 1 dz + z−b

H ctan φ

cos δ.Ka(0) − 1 dz

= ctanφ

cos δ.Ka(β) − 1

[z − b] + (H − z + b) c

tanφcos δ.Ka(0) − 1

The corresponding coefficients of active pressure are estimated from table10.2 as follows :

- : , Ka(β) = Ka(12.5) Ω = 90 + 12.5 = 102.5 , α = 0, δ = 23φ

and φ = 21

hence : Ω = 100, φ = 21 Ka(β) = 0.394 − 0.394−0.3145 = 0.378

Ω = 105 , φ = 21 Ka(β) = 0.37 − 0.37−0.2895 = 0.354

Accordingly, for Ω = 102.5 Ka(12.5) = 0.378+0.3542 = 0.366

- : . From table 10.2 :Ka(0) Ω = 90 , α = 0, δ = 23φ , φ = 21

Ka(0) = 0.447 − 0.447−0.3695 = 0.431

The effective active thrust normal to the wall induced by cohesion istherefore :P2 = 5

tan21 × (0.366 cos 14 − 1) × (4.94 − 1) +

(6 − 4.94 + 1) × 5tan 21 × (0.431 cos 14 − 1)

= −48.7kN/m


97

Therefore, the overall active thrust normal to the wall is :

Pan = P1cos δ+ P2 = 187.8 cos 14 − 48.7 = 133.5kN/m

The point of application of is such that :P1

A1H − z + b + z−b

3 + A2

H+b−z2

+ A3

H+b−z3

= P1.d1

d1 = 03m

The point of application of is : .P2 d2 = 62 = 3m

Accordingly, the point of application of the overall normal thrust is :

d = 1Pan

(d1P1cos δ +d2P2) =

= 1133.5 × (2.03 × 187.8 cos 14 − 3 × 48.7) = 1.68m

Problem 10.7


98

q

W

O

Pa R

ηδ

U1

φ

U2

H − zw

zw

z1β

For the soil parameters : , the angle can beφ = 22 , δ = 23φ = 14.7 η

estimated from equation 10.81:tan

η + δ + φ

1cos η sinη − tan

η + φ

= 1

tan (η + 36.7)1

cos η sin η − tan (η + 22) − 1 = 0

So that by trial and error : η = 38.8

The quantity is such that :z1

z1 = H

tanβ tanη1−tanβ tanη

= 6.5 ×

tan14. tan38.81−tan14. tan38.8

= 1.63m

The quantity C in equation 10.80 can then be calculated :

C = qcos β(H + z1) + γ

2 [z1H + zw(2H − zw)] + 12(γsat − γw) (H − zw)2

= 30cos 14 × (6.5 + 1.63) + 19

2 × (1.63 × 6.5 + 2 × (2 × 6.5 − 2)) +

0.5 × 10.5 × (6.5 − 2)2 ≈ 667.3 kN/m

The effective active thrust can now be calculated from equation 10.79 :

Pa = C tan ηcos η+φ

sin η+φ +δ= 667.3 tan 38.8 × cos (38.8+22)

sin (38.8+22+14.7)

= 270.4 kN/m

Next, the thrust due to water is calculated as follows :

U1 = 12γw(H − zw)2 = 5 × (6.5 − 2)2 = 101.25 kN/m

Accordingly, the total active thrust normal to the wall is :

Pan = U1 + Pacos δ = 101.25 + 270.4 cos 14.7 ≈ 363 kN/m


99

Chapter 11

Problem 11.1

Using the results of example 11.5, calculatedwith a factor of safety on , theF = 1 Kp

corresponding effective pressure diagram isdepicted in the figure below.

The equilibrium of moments with respect to the anchorage point O iswritten as follows :


100

Dstiff clay

2m

dense sand

2m

5m

q = 40 kN/m2

9.2

39.3

[23.7+31.534D]

53.6

60.2

O

[17.8]

D

[23.7]2m

dense sand

stiff clay

[60.2+3.312D]

2m

5m

water level

Fa

[9.2] : effective lateral pressure (kN/m2)

Σ Mo = 0

−9.2 × 2 × 1 − (17.8 − 9.2) × 22 × 2

3 + 17.8 × 5 × 2.5

+(39.3 − 17.8) × 52 × 23 × 5 + 53.6 × (2 + D) ×

5 + 2+D2

+(60.2 + 3.312D − 53.6) ×

2+D2 ×

5 + 23(2 + D)

−23.7 × D × 7 + D

2 − 31.534D × D

2 × 7 + 2D

3 = 0

Thus yielding : D3 + 8.55D2 − 27.16D − 112.91 = 0

Whence a depth of embedment D ≈ 4.22m.

The corresponding net pressure diagram is depicted in the following figure.

The anchorage force can be calculated from the equilibrium equation ofFa

horizontal forces :Σ Fh = 0

Fa = (9.2 + 39.3) × 72 + (53.6 + 60.2) × 2

2 + 36.5 × 1.32 − 82.6 × 2.92

2

≈ 187 kN/m.


101

9.2

39.3 53.6

60.236.51.3 m

2.92m82.6

water level

80 40 0 40 80

O

2m

5m

2m

dense sand

stiff clay

net pressure diagram (kN/m2)

Fa

The distribution of shear forces along the pile is calculated in precisely thesame way as in example 11.5. Referring to the net pressure diagram, it isseen that :z (m) shear force (kN/m)0 02− − (9.2 + 17.8) × 2

2 = −272+ − 27 + 186.7 = 159.77 159.7 − (17.8 + 39.3) × 5

2 = 16.99 16.9 − (53.6 + 60.2) × 2

2 = −96.910.3 − 96.9 − 36.5 × 1.3

2 = −120.613.22 − 120.6 + 82.6 × 2.92

2 ≈ 0

The corresponding shear force is depicted in the following figure where thedepth of zero shear force is .z ≈ 7.45m

The maximum bending moment calculated at depth is, withz = 7.45mreference to the net pressure diagram (note that at depth , the netz = 7.45mpressure is :55 kN/m2)

Mmax = −9.2 × 7 × (72 + 0.45) − (39.3 − 9.2) × 7

2 × (73 + 0.45)

−53.6 × 0.45 × 0.452 − (55 − 53.6) × 0.45

2 × 0.453 + 187 × 5.45

≈ 465kNm/m


102

-80 -40 0 40 80 120 160-120

7.45m

shear force (kN/m)

Now that the above quantities were calculated using the limiting conditions(i.e. a factor of safety on ), the actual characteristics of the pileF = 1 Kp

are such that :- depth of embedment : ; hence a total D = 2 × 4.22 = 5.97 ≈ 6m pile length L = 9 + 6 = 15m,

- pile steel section : Mmax = σaIy = 465kNm/m

Iy = Mmax

σa = 465180 × 103 = 2583cm3/m

The appropriate steel section corresponds to a Larssen LX32 with asection modulus ; the maximum driving length of I

y = 3201 cm3/msuch a pile being 28m, well in excess of the total length required.

Problem 11.2

Writing the equilibrium equation in the upper half in the figure above, itfollows that :

Fa + Fb = (9.2 + 39.3) × 72 + (53.6 + 60.2) × 2

2 + 12 × 36.5 × 1.29

= 307.1kN/m


103

9.2

39.3

28.22D -36.5

53.6

60.236.51.29 m

C

D'

net pressure diagram (kN/m2)

Fa

Fb

The equilibrium of moments with respect to the anchorage point O in theupper half is written as follows:

−9.2 × 2 × 1 − (17.8 − 9.2) × 22 × 2

3 + 17.8 × 5 × 52+

(39.3 − 17.8) × 52 × 2

3 × 5 + 53.6 × 2 × 6+

(60.2 − 53.6) × 22 ×

43 + 5 + 36.5 × 1.29

2 × 7 + 1.29

3 − 8.29Fb = 0

Accordingly :

Fb = 1237.58.29 = 149.3kN/m, Fa = 307.1 − 149.3 = 157.8kN/m

The equilibrium of moments with respect to point C in the lower halfyields :

−149.3D − 1.29 + 28.22D − 36.5

D −1.292

D −1.293

= 0

or D 3 − 3.874D 2 − 26.77D + 38.8 = 0 D ≈ 6.93m

The distribution of shear forces along the pile is calculated in conjunctionwith the net pressure diagram, from which it is seen that :

z (m) shear force (kN/m)0 02− − (9.2 + 17.8) × 22 = −272+ − 27 + 157.8 = 130.85 130.8 − (30.7 + 17.8) × 3

2 = 587 58 − (39.3 + 30.7) × 2

2 = −11.99 − 11.9 − (60.2 + 53.6) × 2

2 = −125.810.29 − 125.8 − 36.5 × 1.29

2 = −149.311 − 149.3 + 20 × 0.71

2 = −142.213 − 142.2 + (20 + 76.4) × 2

2 = −45.815.93 − 45.8 + (159 + 76.4) × 2.93

2 = 299


104

The corresponding diagram is depictedin the figure opposite, from which zeroshear force occurs at depths :

z1 ≈ 6.75mz2 ≈ 13.65m

Using the net pressure diagram, the maximum bending moment at depth is calculated as follows :z1 = 6.75m

Mmaxz1 = −9.2 × 2 × (1 + 4.75) − (17.8 − 9.2) × 2

2 ×

23 + 4.75

+157.8 × 4.75 − 17.8 × 4.75 × 4.752 − (38.2 − 17.8) × 4.75

2 × 4.753

≈ 320 kNm/m

Similar calculations yield a maximum bending moment at depth : z2 = 13.65m

Mmaxz2 ≈ −324kNm/m ≈ − Mmax

z1

Problem 11.3

First, calculate the porewater pressure distribution along the pile. UsingMandel analysis, it is seen that in this case :

H = 2m, hu = hd = 0, D = 6.5m, T = 8.5m

Therefore :


105

1 3 . 6 5 m

-200 -100 0 100 200 300

6.75m

13.65m

shear force (kN/m)

ξ = 2

ln

8.53.5 +

8.53.5

2−1

+ln

6.53.5 +

6.53.5

2−1

= 0.723m

ΦA = 0.723 ln

8.53.5 +

8.53.5

2− 1

= 1.11m ΦF = 0

ΦB = −0.723 ln

6.53.5 +

6.53.5

2− 1

= −0.89m

Accordingly, on the active side of the wall, thevelocity potential is such that at any depth y :

Φy = ξ ln

yL +

yL

2− 1

Moreover, the total head is : h = Φy − ΦA + hu + T = (ΦA + 7.39) m

The corresponding porewater pressure is : u = γw(h − y)

and the effective unit weight : ; the hydraulic gradientγy = γsat − γw(1 − iy)

being : iy = ΦA−Φy

T−y

Whence the following results on the active side (behind the wall) [note thedepth y is with respect to the top of the impervious layer] :

y (m) u (kN/m2) i γ (kN/m3)8.5 0 - 107 13.4 0.105 11.055.5 26.3 0.123 11.233.5 38.9 0.222 12.22

On the passive side, the velocity potential at depth y is :

Φy = −ξ ln

yL +

yL

2− 1

and the total head :


106

2.5m

3.5m

2m

3m

L= 3.5m

IMPERVIOUS

H

T

A

B

F

D

q = 50 kN/m2

Fa

h = Φy − ΦB + hd + D = (Φy + 7.39) mThe hydraulic gradient being : i = Φy−ΦB

D−y

and the corresponding porewater pressure andsoil effective unit weight are respectively :

,u = γw(h − y) γ = γsat − γw(1 + i)

Therefore on the passive side :

y (m) u (kN/m2) i γ (kN/m3)6.5 0 - 105.5 11.5 0.15 8.54.5 24.1 0.206 7.943.5 38.9 0.297 7.03

The active and passive coefficients of earthpressure are estimated a) from table 10.1 :- loose sand : β = 0, φ = 30 , δ = 2

3φ Ka = 0.300- dense sand: β = 0, φ = 40 , δ = 2

3φ Ka = 0.202Kp = 12

b) from table 10.2 (coefficients applicable to the surcharge load q)- loose sand: α = 0, Ω = 90 , φ = 30 Ka = 0.304- dense sand: α = 0, Ω = 90 , φ = 40 Ka = 0.206

It is seen that, in this case, the coefficients of active pressure are similarand therefore, those calculated from table 10.1 will be used in conjunctionwith the surcharge q. Accordingly, using the free earth support method, the active and passivepressures normal to the wall can now be calculated (using a factor of safety

on . On the active side, the effective active stresses normal to theF = 1 Kp)wall are calculated as follows :σa = Kacos δσv + qOn the passive side, the effective passive stresses normal to the wall areestimated from the expression :σp = Kpcos δσv


107

IMPERVIOUS

z

z

Whence the following results on the active side (z as per the previousfigure)

z (m) γ (kN/m3) u (kN/m2) σv(kN/m2) σa(kN/m2)

0 19 0 0 14.16− 19 0 114 46.26+ 10 0 114 20.67.5 11.05 13.4 130.5 23.69 11.23 26.3 147.3 26.611 12.22 38.9 171.7 31and on the passive side :

z (m) γ (kN/m3) u (kN/m2) σv(kN/m2) σp(kN/m2)

0 10 0 0 01 8.5 11.5 8.5 91.12 7.94 24.1 16.5 1773 7.03 38.9 23.5 252

The corresponding total and net pressure diagrams are depicted in thefollowingfigures:


108

IMPERVIOUS

300 200 100 100

14.1

46.220.6

3742.3

69.9

102.6201.1290.9

total pressure diagram

(kN/m2)σp + u σa + u

Fa

The factor of safety on the passive side is calculated by taking momentsabout the anchorage point O . Thus referring to the net pressure diagram (positive moments clockwise):Σ Mo = 0

−14.1 × 2.5 × 1.25 − (27.5 − 14.1) × 2.52 × 2.5

3 + 27.5 × 3.5 × 1.75+(46.2 − 27.5) × 3.5

2 × 3.5 × 23 + 20.6 × 2 × (1 + 3.5)+

(42.3 − 20.6) × 22 ×

23 × 2 + 3.5 + 42.3 × 0.49

2 ×

0.493 + 5.5

[−1F 49.7 × 0.51

2 ×

23 × 0.51 + 5.99 + 49.7 × 1 × (0.5 + 6.5)+

(139.7 − 49.7) × 12 ×

23 + 6.5 + 139.7 × 1 × (0.5 + 7.5)+

](221 − 139.7) × 1 ×

23 + 7.5 = 0

Once rearranged :

535.74 − 2532.1F = 0 F = 4.73


109

IMPERVIOUS

100 100

14.1

46.220.6

3742.3

200

50140221

net pressure diagram

0.49m

O

pressure (kN/m2)

Fa

The anchorage force is found from the equilibrium equation of horizontalforces. Once more, referring to the net pressure diagram :

−(14.1 + 27.5) × 2.52 + Fa − (27.5 + 46.2) × 3.5

2 − (20.6 + 42.3) × 22

[−0.492 × 42.3 + 1

4.73 49.7 × 0.512 + (139.7 + 49.7) × 1

2

] +(221 + 139.7) × 12 = 0

thus yielding : Fa = 193.4 kN/m

The distribution of shear forces along the pile is next calculated from thenet pressure diagram :

z (m) shear force (kN/m)0 02.5− − (14.1 + 27.5) × 2.5

2 = −522.5+ − 52 + 193.4 = 141.46 141.4 − (27.5 + 46.2) × 3.5

2 = 12.48 12.4 − (20.6 + 42.3) × 2

2 = −50.58.49 − 50.5 − 42.3 × 0.49

2 = −60.89 − 60.8 + 49.7 × 0.51

2 = −48.110 − 48.1 + (49.7 + 139.7) × 1

2 = 46.611 46.6 + (221 + 139.7) × 1

2 = 226.9

The corresponding shear force diagram is depicted in the following figure,from which the depth corresponding to zero shear force (where themaximum bending moment occurs) is estimated at . Whence thez ≈ 6.3mmaximum bending moment calculated with respect to point A in the figure:

Mz=6.3m = Mmax = −14.1 × 2.5 ×

2.52 + 3.8

−(27.5 − 14.1) × 2.52 ×

2.53 + 3.8 + 193.4 × 3.8 −27.5 × 3.5 ×

3.52 + 0.3

−(46.2 − 27.5) × 3.52 ×

3.53 + 0.3 − 20.6 × 0.3 × 0.3

2 −(25 − 20.6) × 0.32 × 0.3

2

Mmax ≈ 234 kNm/m


110

Problem 11.4

If a factor of safety were applied on , then only the effectiveF = 2 Kp

passive pressures normal to the wall will be affected , and therefore needbe calculated as follows :

, with σp = Kp∗ .σv. cos δ Kp

∗ = Kp

2 = 6

Accordingly, the active pressures being as calculated in the previousproblem, the new values of the passive pressures normal to the wall are :

z (m) γ (kN/m3) u (kN/m2) σv (kN/m2) σp (kN/m2)

0 10 0 0 01 8.5 11.5 8.5 45.62 7.94 24.1 16.5 88.53 7.03 38.9 23.5 126

Hence the corresponding net pressure diagram of the following figure.


111

IMPERVIOUS

-100 100 200

6.3 m

A

shear force (kN/m)

The actual factor of safety on the passive side is determined from themoment equilibrium equation with respect to the anchorage point :Σ Mo = 0

−14.1 × 2.5 × 1.5 − (27.5 − 14.1) × 2.52 × 2.5

3 + 27.5 × 3.5 × 3.52

+(46.2 − 27.5) × 3.52 × 2

3 × 3.5 + 20.6 × 2 × (1 + 3.5)

+(42.3 − 20.6) × 22 ×

23 × 2 + 3.5 + 42.3 × 1

2 ×

13 + 5.5

[−2F 51.2 × 1

2 ×

23 + 6.5 + 51.2 × 1 ×

12 + 7.5

] +(95 − 51.2) × 12 ×

23 + 7.5 = 0

yielding : 600.4 − 1543.8F = 0 F = 2.57

The anchorage force can now be calculated from the equilibrium equationof horizontal forces. Referring to the net pressure diagram, it is seen that :

Fa = (14.1 + 27.5) × 2.52 + (27.5 + 46.2) × 3.5

2 + (20.6 + 42.3) × 22

[ ]+42.3 × 12 − 2

2.57 51.2 × 12 + (51.2 + 95) × 1

2

Hence : Fa = 188.2 kN/m


112

IMPERVIOUS

100 100

14.1

46.220.6

3742.3

net total pressure

1m

51.295

pressure (kN/m2)

Fa

The distribution of shear forces along the pile is calculated as follows : z (m) shear force (kN/m)

0 02.5− − (14.1 + 27.5) × 2.5

2 = −522.5+ − 52 + 188.2 = 136.26 136.2 − (27.5 + 46.2) × 3.5

2 = 7.28 7.2 − (20.6 + 42.3) × 2

2 = −55.79 − 55.7 − 42.3 × 1

2 = −76.910 − 76.9 + 51.2 × 1

2 = −51.211 − 51.2 + (95 + 51.2) × 1

2 = 21.9

The corresponding shear force diagram is depicted in the following figurefrom which the depth of zero shear force is :z ≈ 6.2m

The maximum bending moment occurring at depth is thence :z = 6.2mMm = −14.1 × 2.5 ×

2.52 + 3.7 − (27.5 − 14.1) × 2.5

2 ×

2.53 + 3.7

+188.2 × 3.7 − 27.5 × 3.5 ×

3.52 + 0.2 − (46.2 − 27.5) × 3.5

2 ×

3.53 + 0.2

−20.6 × 0.2 × 0.1

Mm = 213.1 kNm/m


113

IMPERVIOUS

-100 100

6.2 m

shear force (kN/m)

If is to be reduced according to Rowe's method, then the designMm

moment is such that :Mc = Mm − 1

2(Mm − Mr)

First, the reduced moment is calculated as follows :Mr

so that an interpolation on figure 11.7 yields the followingα = 811 = 0.73

results in which the moment corresponds to the product of theMr

maximum bending moment calculated previously at the depth of zeroMm

shear force times the ratio read from Rowe's graphs :MMmax

ρ (m3/kN) MMmax

Mr = MmM

Mmax(kNm/m)

0.05 0.98 2090.07 0.89 1900.1 0.82 1750.2 0.68 1450.4 0.56 1191 0.43 92

The figure opposite indicates that apile section 6W is nearest to thegraph and therefore mostappropriate.This pile corresponds to a reducedmoment Mr = 109.8 kNm/m.Accordingly, the design moment isas follows :Mc = Mm − 1

2(Mm − Mr) = 213.1 − 1

2 × (213.1 − 109.8) = 161.5 kNm/m

Hence, select a GSP2 pile(British steel handbook)


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0.4 1.00.2 0.60.05 0.1

90

110

130

170

190

150

210

LX8

6 W

GSP2

Mr (kNm/m)

ρ (m3/kN)

Problem 11.5

The coefficients of active and passive pressure are as follows :- sand : . Table 10.1 yileds β = 0, λ = 0, δ = 2

3φ , φ = 35 Kas = 0.247 The coefficient of active pressure applicable to the surcharge q is found from table 10.2: Ω = 90 , α = 0, δ = 2

3φ, φ = 35 Kas = 0.249

- clay : table 10.1 yileds the following:β = λ = 0, δ = 2

3φ , φ = 25 Kac = 0.364, Kpc = 2.75β = λ = 0, δ = 2

3φ , φ = 20 Kac = 0.442, Kpc = 3.7

Therefore for φ = 23 , Kac = 0.364 + 35 × (0.442 − 0.364) = 0.41

Kpc = 2.75 + 35 × (3.7 − 2.75) = 3.32

The coefficients applicable to both cohesion and surcharge are estimatedfrom table 10.2 :Ω = 90 , α = 0, φ = 25 Kac = 0.369, Kpc = 3.554Ω = 90 , α = 0, φ = 20 Kac = 0.447, Kpc = 2.645

Hence for φ = 23 , Kac = 0.369 + 35 × (0.447 − 0.369) = 0.416

Kpc = 2.645 + 35 × (3.554 − 2.645) = 3.19

Both active and passive effective normal stresses are then calculated asfollows :- throughout the sand layer :

σa = cos δs Kasσv + Kasq

- throughout the clay layer :

σa = cos δc Kacσv + Kacq − c cot φ

1 − Kaccos δc

σp = σvKpccos δc + c cot φ Kpccos δc − 1

with the angles of wall friction : δs = 23 × 35 = 23.3 , δc = 2

3 × 23 = 15.3


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Whence the following stress values (depth zas per the figure opposite):- active pressures :

z (m) σv(kN/m2) σa(kN/m2)0 0 22.92 37 31.3

83.3 41.84.5−

83.3 61.84.5+

5.5 102.3 69.37 117.3 75.29 137.3 83.1

4.5 + D 92.3 + 10D 65.4 + 3.95D

- passive pressure :

z (m) σv(kN/m2) σp(kN/m2)0 0 39.141 19 99.92.5 34 147.94.5 54 211.9D 10D + 9 32D + 67.9

Hence the ensuing net pressure diagram.


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stiff clay

4.5m

D'

1m

sand

z

z

C

q = 100 kN/m2

22.9

41.8

sand

stiff clay

22.730.6

0.426m

net pressure (kN/m2)28.D + 2.54

The depth of embedment D' is found from the equilibrium equation ofmoments with respect to point C at the foot of the pile :Σ MC = 0

−22.9 × 4.5 ×

4.52 + D

− (41.8 − 22.9) × 4.52 ×

4.53 + D

−22.7 × 0.4262 ×

D − 0.4263

+ 30.6 × 0.574

2 × D − 1 + 0.574

3

+30.6 ×D −1

2

2 + 28D + 2.54 − 30.6

D −1

2

6 = 0

So whence rearranged, the following cubic equation ensues :

D 3 + 0.28D 2 − 33.85D − 62.4 = 0 D ≈ 6.45m

The complete net pressure diagram is depicted in the following figure.


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41.8

sand

stiff clay

30.6

0.426m22.7

22.9

72.6

128.5

183.1

29.2

33.5

net pressure (kN/m2)

The distribution of shear forces along the pile can now be calculated usingthe net pressure diagram :z (m) shear force (kN/m)

0 01.5 − (22.9 + 29.2) × 1.5

2 = −393 − 39 − (35.5 + 29.2) × 1.5

2 = −87.64.5 − 87.6 − (41.8 + 35.5) × 1.5

2 = −145.64.926 − 145.6 − 0.426 × 22.7 × 1

2 = −150.45.5 − 150.4 + 30.6 × 0.574 × 1

2 = −141.67 − 141.6 + (30.6 + 72.6) × 1

2 = −64.29 − 64.2 + (128.5 + 72.6) × 2

2 = 136.910.95 136.9 + (128.5 + 183.1) × 1.95

2 = 440.7

The corresponding shear force diagram is shown in the following figure.The depth corresponding to zero shear force is z ≈ 7.8m.


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-200 0 200 400

sand

stiff clay

7.8 m

shear force (kN/m)

The maximum bending moment occurring at the depth is thusz = 7.8m(note that in the net pressure diagram, the pressure corresponding to adepth is ) :z = 7.8m 95kN/m2

Mmax = −22.9 × 4.5 × (4.52 + 3.3) − (41.8 − 22.8) × 4.5

2 × (4.53 + 3.3)

−22.7 × 0.4262 × (2

3 × 0.426 + 0.574 + 2.3) +30.6 × 0.574

2 × (2.3 + 0.5743 ) + 30.6 × 2.3 × 2.3

2

+(95 − 30.6) × 2.32 × 2.3

3

Whence : Mmax ≈ −633 kNm/m.


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Azizi Applied Analysis in Geotechnics Solution Manual

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