Azizi - Applied Analysis in Geotechnics-Solution Manual

APPLIED ANALYSES IN GEOTECHNICS: SOLUTIONS MANUAL

APPLIED ANALYSES IN GEOTECHNICS

SOLUTIONS MANUAL Fethi Azizi [email protected]

2m

A

h = 14m

2m

B

2m

C

datum

2m

D

2m 2m

2m

12m

10m

8m

Non fissured rock

1

6

4 2


Chapter 1 Problem 1.1 a)

w=

e Gs

=

b)

w=

e−A (1+e) Gs

c)

× 1 = 1.93Mg/m 3 bulk density : ρ = s1+e ρ w = 2.7×1.252 1.75 bulk unit weight: γ = ρ.g = 9.81 × 1.93 = 18.9 kN/m 3

0.75 2.7

= 28%

=

0.75−0.04×1.75 2.7

= 25.2%

G (1+w)

Þ

G +e

s saturated density : ρ sat = 1+e ρ w = 2.7+0.75 × 1 = 1.97Mg/m 3 Þ 1.75 saturated unit weight : γ sat = g.ρ sat = 9.81 × 1.97 = 19.3kN/m 3

Problem 1.2 - the bulk unit weight :

ùγ γ = G s éë 1+w 1+e û w

- the saturated unit weight :

γ sat =

Hence :

γ sat γ

=

1 é 1 1+w ë

+

e ù Gs û

=

G s +e γ 1+e w

1 1.25

1 ö × æè 1 + 2.7 ø ≈ 1.1

Problem 1.3 a) the water content : w = b) the void ratio : e =

Mw Ms

=

1320−1075 1075

= 22.8%

Vv Vs

- the volume of solids : V s =

Ms G s .ρ w

=

1075 2.7×1

= 398.1cm 3

- the volume of voids : V v = V − V s = π × therefore : e =

387.3 398.1

= 0.97

2

10 2 4

× 10 − 398.1 = 387.3cm 3


c) degree of saturation : S = d) air content : A =

e−wG s 1+e

=

0.228×2.7 0.97

0.97−0.228×2.7 1.97

e) bulk density : ρ =

M V s +Vv

=

f) dry density : ρ d =

ρ 1+e

1.68 1.97

=

=

wG s e

1320 785.4

= 63.3%

= 18%

= 1.68g/cm 3

= 1.37g/cm 3

Problem 1.4 Subscripts s and f refer to soil and filter respectively. From the graph, it is seen that : d 10s = 0.105mm, d 15s = 0.12mm, d 30s = 0.2mm, d 60s = 0.32mm, d 85s = 0.85mm, d max s = 4mm. a) coefficient of curvature : C u = coefficient of gradation : C g =

d 60 d 10

=

0.32 0.105

d 50s = 0.28mm

≈3

d 230 0.2 2 = = 1.2 d 10 d 60 0.105 × 0.32

b) coefficient of permeability : k ≈ d 210 = 0.105 2 = 1.1 × 10 −2 cm/s. c) refer to figure:

Percentage passing

100 80 60 40 20 0 0.01

0.1

1

10

100

Particle size (mm) M

C

F

silt F: fine,

M

C

sand M: medium,

3

F

gravel C: coarse

M

C

cobbles


Problem 1.5 a) the soil plasticity index : I p = A × % clay = 0.72 × 0.23 = 16.56% the liquid limit : w L = I p + w P = 16.56 + 16 = 32.56% Thus : w L = 32.5%, I p = 16.5% Þ clay of low plasticity b) the liquidity index : I L =

w−wP IP

w = ILIP + wP = 0.75 × 0.165 + 0.16 = 28.4%

Þ

Problem 1.6 w op =

1 é G (1 Gs ë s

ρ

w − A) ρd max − 1ùû =

1 2.7

1 × æè 2.7 × [1 − 0.03] × 1.65 − 1 öø ≈ 22%

Problem 1.7 1.8

ρ d (Mg/m 3 ) 1.7

1.6

1.5 0.10

0.15

0.20

0.25

0.30

moisture content

a) From the graph : ρ d max = 1.71 Mg/m 3 ,

w op ≈ 16%

b) the bulk density : ρ = RC.ρ d max (1 + w op ) = 0.95 × 1.71 × 1.16 = 1.88Mg/m 3 . c) the air content : ρ df

A = 1 − ρ wG s (1 + G s w op ) = 1 −

0.95×1.71 2.7

4

× (1 + 2.7 × 0.16) ≈ 13.8%


The degree of saturation : S= 1−A = 1 − 0.138 = 65.2% 0.138 1 + G sAwop 1 + 0.16×2.7 The void ratio : e =

wop G s S

=

0.16×2.7 0.652

Therefore the porosity : n =

e 1+e

= 0.66

≈ 0.4

5


Chapter 2 Problem 2.1 The maximum capillary rise : h c =

C e.d 10

=

1.5×10 −6 0.94×10 −6

= 15.95m

Whence a suction pressure : u = −h c .γ w = −159.5kN/m 2 The capillary saturation rise : h s = h c d 1060 = 15.95 × d

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/m 2 The maximum increase in effective stress at that depth is such that : σ max = 1.2 × 67 = 80.4kN/m 2 Accordingly: 80.4 = 18.5x + 10(5 − x)

Þ

x=

30.4 8.5

= 3.58m

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

Problem 2.3 a) the porwater pressure : u A = 10 × 6 = 60kN/m 2 the total pressure : σ A = 3 × 10 + 3 × 20 = 90kN/m 2 the effective pressure : σ A = σ A − u A = 90 − 60 = 30kN/m 2 b) the porewater pressure : u A = 10 × 3 = 30kN/m 2 the total pressure : σ A = 3 × 20 = 60kN/m 2 the effective pressure : σ A = σ A − u A = 60 − 30 = 30kN/m 2

6


c) the porewater pressure : u A = 2 × 10 = 20kN/m 2 the total pressure : σ A = 1 × 17 + 2 × 20 = 57kN/m 2 the effective pressure : σ A = σ A − u A = 57 − 20 = 37kN/m 2 d) 3m

kN/m 2 40

80

40

kN/m 2

80

40

σ

u

σ

80

2

2

2 4

kN/m 2

4

σ

σ

4

σ

u

u

σ

Problem 2.4 The depth at which ∆σ v = 0.05σ o is in all probability larger than z w = 5m. under these conditions, the effective vertical stress due to self weight is : σ o = γ.z w + (γ sat − γ w )(z − z w ) = 9z w + 10z = 45 + 10z. so that : ∆σ v = 0.05[45 + 10z] = 2.25 + 0.5z But ∆σ v is due to the (three) line loads, and using the superposition principle, it is seen that at a given depth z, ∆σ v beneath the centre of the loaded surface is such that : 2Q 1 ∆σ v = π

z3 æx2 + z2 ö è 1 ø

2

2Q 2 +2× π

z3 æx2 + z2 ö è 2 ø

where, with reference to the figure : x 1 = 0,

2

x 2 = 3.5m .

Thus, equating the two expressions, and rearranging:

7


3.534.z + 0.785.z 2 = 2000 +

2400 æ 12.25 + 1 ö è z2 ø

2

Þ

z ≈ 72.6m

Problem 2.5 a) stresses at point A due to the vertical component of the (line) load: 12 z 3 - vertical stress : σ A = 2F. cos π R4 since R = z = 1m

Þ

σ A = 2 × 620π× cos 12 ≈ 386 kN/m 2

- shear stress : τ xz = 0 The stresses at A due to the horizontal component of the force are zero. Therefore, at A the vertical and shear stresses generated by the inclined load are respectively : σ A = 386kN/m 2 , τ xz = 0 b) at B, the stresses due to the vertical component of the load are: - vertical stress : 12 z 3 = 2 × 620 × cos 12 × 1 σ B = 2F cos = 15.4kN/m 2 π π 2 2 2 R4 (2 + 1 ) - shear stress: 12 z 2 .x = 2 × 620 × cos 12 × 1 × 2 = 30.9k/m 2 τ xz = 2F cos 2 π π R4 (2 2 + 1 2 ) The stresses due to the horizontal component of the load are : 12 z 2 .x = 2 × 620 × cos 12 × 2 × 1 = 6.6kN/m 2 σ v = 2F sin π π 25 R4 12 x 2 .z = 2 × 620 × sin 12 × 2 2 × 1 = 13.1kN/m 2 τ xz = 2F sin π π 25 R4 Whence, trough superposition :

8


- the total vertical and shear stresses at B are respectively : σ B = 15.4 + 6.6 = 22kN/m 2 τ xz = 30.9 + 13.1 = 44kN/m 2

Problem 2.6 2.5m 2.5 3

1

1 H

3 2m

3H

B

1

2 2.5H

The vertical stress at B is evaluated using superposition principle. The uniform load applied by 1 is : q = γH = 21H kN/m 2 . 2.5m

a) vertical stress at B due to 1: σ v1 = π [α + sin α cos (α + 2β)] q

H

From the opposite figure: 2m

tan β = − 1.25 Þ α = −2β = 1.18rd. 2

β

B

Thus : σ v1 = 0.642q. 2.5H

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

H 1.25

ö = 0.558rd β = tan −1 æè 1.25 2 ø α1 =

ùû tan −1 éë 1.25+2.5H 2

2m

− 0.558 rd.

B

using equation 2.35a: σ v2 =

q (1.25+2.5H) é 2π ë 1.25H

α 1 − 0.899 ùû

9

β

α1


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

3H H

ùû tan −1 éë 1.25+3H 2

α2 =

− 0.558 rd 1.25

whence, according to equation 2.35a: σ v3 =

q (1.25+3H) é 2π ë 1.5H

α2

α 2 ùû − 0.899

β

2m B

Adding the three stress contributions at B, then equating the outcome to 150kN/m 2 and substituting for q = 21H , it follows that : 150 = 13.482H + 3.342H [

(1.25+2.5H) 1.25H (1.25+3H) 1.5H

ö + tan −1 æè 1.25+2.5H 1.25H ø ö − tan −1 æè 1.25+3H 2 ø

1.023 H

− 4.034 ]

So that through trial and error : H ≈ 7.3m

Problem 2.7 The tank radius being : a = 10m . The vertical stress increase at A (at a depth z = 1.2m ) is such that ∆σ A = 5kN/m 2 . On the other hand : ∆σ A = q.I σ

Þ

Iσ =

Because z = 1.2m follows that : z a

but :

r a

=

Þ

= 0.12,

x+a a

= 1+

5 250 z a

= 0.02. =

I σ = 0.02 x a

Þ

1.2 10

= 0.12 . From the charts of figure 2.20, it

Þ

r a

≈ 2.5

x = 1.5a = 15m.

10


Problem 2.8

2

Because of the nature of Fadum's method, the following vertical stresses are evaluated using the superposition principle. The irregular shape of foundation makes it inevitable to add some artificial areas; in which case, care must be taken so as to subtract the stress generated by any of the added areas.

D

4

1

6

F 5

B

C

A

3

E

H

G

8

7

Prior to evaluate the stress increase at any of the points, the expression of these stresses are presented in terms of the areas contributing to the overall stress increase. A minus sign indicates that the contribution to the stress of the corresponding area is to be subtracted. Thus, referring to the figure opposite : ∆σ A = A2H7 − A1G7 + A1FE − A2D3 + ABC3 ∆σ B = B5EA + B2H8 − B2DC − B1G8 + B1F5 ∆σ C = CBA3 + C5E3 + CDH8 − C4G8 + C4F5 ∆σ D = DHG4 + D6E3 − D6F4 + D2A3 − D2BC ∆σ E = E6D3 + E5BA − E5C3 + E6H7 − EFG7 ∆σ F = FGH6 + F6D4 + F1AE − F1B5 + F4C5 ∆σ G = GHD4 + G1A7 − GFE7 − G1B8 + G4C8 ∆σ H = H2A7 − H6E7 + H6FG − H2B8 + HDC8 For a uniform load q, the above stresses evaluated at a depth z = 1m using Fadum's charts are : ∆σ A ≈ 0.245q, ∆σ B = 0.25q, ∆σ C ≈ 0.724q, ∆σ D ≈ 0.244q ∆σ E ≈ 0.244Q, ∆σ F ≈ 0.728q, ∆σ G ≈ 0.247q, ∆σ H ≈ 0.249q

Problem 2.9 The expressions of stresses are presented in terms of the areas contributing to the overall stress increase at different points. Referring to the figure above and considering the two uniform surface loads separately : - stress increase at K due to q 1 is calculated from the area : ∆σ K (q 1 ) = B5EA

11


- stress increase at K due to q 2 is calculated from the area : ∆σ K (q 2 ) = B8H2 − B8G1 − BCD2 + B5F1 - stress increase at L due to q 1 is calculated from the area : ∆σ L (q 1 ) = E5BA - stress increase at L due to q 2 is calculated from the area : ∆σ L (q 2 ) = E3D6 − E5C3 + E6H7 − EFG7 - stress increase at M due to q 1 is calculated from the area : ∆σ M (q 1 ) = F1AE − F1B5 - stress increase at L due to q 2 is calculated from the area : ∆σ M (q 2 ) = F4C5 + F6D4 + F6HG Thus using the dimensions of different areas in conjunction with Fadum's charts, then applying the influence factors appropriately and adding the contribution of both loads at every point, it follows that : ∆σ K ≈ 0.245q 1 + 0.005q 2 ∆σ L ≈ 0.242q 1 + 0.002q 2 ∆σ M ≈ 0.008q 1 + 0.72q 2

Problem 2.10 Use the scaled foundation according to Newmark method, then place the point at which the stress is to be estimated at the centre of the concentric circles. Whence point M (see the upper foundation in the following figure) and K (the lower foundation). Now, only the number of elements needs be estimated in either case, taking care of the difference in loading. Whence form the figure : ∆σ M ≈ 0.005 [15q 1 + 24q 2 ] ≈ 32.5 kN/m 2 ∆σ K ≈ 0.005 [19q 1 + 10q 2 ] ≈ 19.5kN/m 2

12


z = scale line

Influence factor : I = 0.005

13


Chapter 3 Problem 3.1

zu

A

datum

B

G

C D

A

ze

B

F E

G

C D

Impervious

F E

Please note: the answers given on page 160 of the book, relating to seepage forces and effective stresses for this problem, are erroneous. The hydraulic gradients at different points, and the corresponding angles with respect to the vertical (refer to enlarged figure on the RHS) are such that: i A = 0.21, α A ≈ 1 i B = 0.23, α B ≈ 6 , i C = 0.3, α c ≈ 12 i D = 0.4, α D ≈ 19 , i E = 0.45, α E ≈ 10 , i F = 0.33, α F ≈ 4.5 i G = 0.3, α G ≈ 3 The seepage forces: - on the upstream side : - on the downstream side :

F = A z u γ w i cos α (downward force) F = −A z e γ w i cos α (upward force)

For a unit area A = 1m 2 , it follows that: F A = 0.21 × 0.9 × 10 × cos 1 = 1.89 kN F B = 0.23 × 2.4 × 10 × cos 6 = 5.49 kN F C = 0.3 × 3.7 × 10 × cos 12 = 10.86 kN F D = 0.4 × 4.7 × 10 × cos 19 = 17.77 kN F E = −0.45 × 2.6 × 10 × cos 10 = −11.52kN F F = −0.33 × 2.6 × 10 × cos 4.5 = −8.55 kN F G = −0.3 × 0.6 × 10 × cos 3 = −1.8kN.

14


The effective stresses : - on the upstream side : σ = z u éë γ + γ w i cos α ùû - on the downstream side : σ = z e éë γ − γ w i cos αùû Whence: σ A = 0.9(11 + 0.21 × 10 × cos 1) = 11.8kN/m 2 , σ B = 2.4(11 + 0.23 × 10 × cos 6) = 31.9kN/m 2 σ C = 3.7(11 + 0.3 × 10 × cos 12) = 51.6kN/m 2 σ D = 4.7(11 + 0.4 × 10 × cos 19) = 69.5kN/m 2 σ E = 2.6(11 − 0.45 × 10 × cos 10) = 17.1kN/m 2 σ F = 1.7(11 − 0.33 × 10 × cos 4.5) = 13.1kN/m 2 σ G = 0.6(11 − 0.3 × 10 × cos 3) = 4.8kN/m 2 The porewater pressure : u = h p .γ w and the pressure head are with reference to the figure (note the datum) : h pA = 3.6 − 0.18 − 1.3 = 2.12m Þ u A = 21.2kN/m 2 h pB = 3.6 − 0.36 − 0.18 + 0.2 = 3.26m Þ u B = 32.6kN/m 2 h pC = 3.6 − 2 × 0.36 − 0.18 + 1.6 = 4.3m Þ u C = 43kN/m 2 h pD = 3.6 − 3 × 0.36 − 0.18 + 2.5 = 4.84m Þ u D = 48.4kN/m 2 h pE = 0.72 + 0.18 + 2.6 = 3.5m Þ u E = 35kN/m 2 h pF = 0.36 + 0.18 + 1.7 = 2.24m Þ u F = 22.4kN/m 2 h pG = 0.6 + 0.18 = 0.78m Þ u G = 7.8kN/m 2 Problem 3.2 The equivalent horizontal permeability is such that : Q = T HL .k h = 1 × 0.88 .k h = 0.55k h 1.6 kh =

Q 0.55

=

8.33 0.55

Þ

× 10 −6 = 1.51 × 10 −5 m/s.

d .k On the other hand : k h = Σ di i =

Σ

i

0.2 (k 1 +k 2 +...+k 5 ) 1

k h = 0.2(k 1 + 0.5k 1 + 0.25k 1 + 0.125k 1 + 0.0625k 1 ) = 0.3875k 1 Hence : k 1 =

1.51 0.3875

× 10 −5 = 3.91 × 10 −5 m/s.

15


Problem 3.3 The equivalent vertical permeability is such that : Q = A. HL .k v = (1.6 × 0.8) × 0.88 .k v = 1.126k v 1.6 kv =

2.78 1.126

Þ

× 10 −6 = 2.47 × 10 −6 m/s.

But: d k v = Σd /ki =

Σ

i

i

therefore : k v1 =

1 0.2

æ1 è k v1

2.47 0.1613

+

1 0.5k v1

+

1 0.25k v1

+

1 0.125k v1

1 ö + 0.0625k v1 ø

= 0.1613k v1

× 10 −6 = 1.53 × 10 −5 m/s.

Problem 3.4 The permeability of the isotropic layer being : k = [15.1 × 2.47] 1/2 × 10 −6 m/s = 6.11 × 10 −6 m/s. The new tank length : 0.5 k 1/2 ö = 0.647m L = L. éë k hv ùû = 1.6 × æè 2.47 15.1 ø The hydraulic gradient in the vertical direction : i v =

0.88 1

= 0.88

Whence the quantity of vertical flow: Q = L × 0.8k.i v = 0.647 × 0.8 × 6.11 × 0.88 × 10 −6 = 2.78 × 10 −6 m 3 /s

16


Problem 3.5 The approximate solution is as per the following figure

Impervious

Problem 3.6 12.5

15

r5

7.5 15

15

r6

r7

12.5

r8

r9

r4

15

45 m

r 10

r3

r 11

7.5

r2

r1

r 12

55 m

- The radii at point A: r 1 = 1.5m,

r 7 = 43.5m ,

r 3 = r 11 = (6 2 + 27.5 2 )

r 2 = r 12 = (1.5 2 + 15 2 )

1/2

1/2

= 28.14m

r 4 = r 10 = (21 2 + 27.5 2 )

1/2

= 45.3m

r 6 = r 8 = (15 2 + 43.5 2 )

r 5 = r 9 = (27.5 2 + 36 2 )

17

= 15.07m 1/2

1/2

= 34.6m

= 46.01m


- the radii at point B:

r1 = (

1/2 1.5 2 + 6 2

r2 = (

1.5 2

)

r4 = (

7.5

r3

15

r2

)

1/2

= 21.05m

12.5

r6

r5

r7 r8

15

r1

r9

7.5

= 44.87m

)

15

45 m

= 36.03m

1/2 11 2 + 43.5 2

15

r4

= 6.18m

1/2 + 21 2

r 3 = (1.5 2 + 36 2 )

12.5

r 12

r 10

r 11 55 m

r 5 = (26 2 + 43.5 2 )

1/2

= 50.68m

r 6 = (43.5 2 + 41 2 )

1/2

= 59.78m

r 7 = (36 2 + 53.5 2 )

1/2

= 64.48m

r 8 = (53.5 2 + 21 2 )

1/2

= 57.47m

r 10 = (1.5 2 + 41 2 )

1/2

= 41.03m

r 12 = (1.5 2 + 11 2 )

1/2

= 11.1m

r 9 = (53.5 2 + 6 2 )

1/2

r 11 = (1.5 2 + 26 2 )

= 53.83m

1/2

= 26.04m

- the radii at point C: 12.5

7.5

r2

15

r1

15

r3

15

r4

12.5

r5 r6 r7

45 m

r8

15

r9

r 12

7.5

r 11

r 10 55 m

r 2 = r 12 = (1.5 2 + 15 2 )

1/2

r 4 = r 10 = (26 2 + 22.5 2 ) r 6 = r 8 = (15 2 + 53.5 2 )

= 15.07m

1/2

1/2

r 3 = r 11 = (11 2 + 22.5 2 )

= 34.38m

1/2

= 25.04m

1/2

= 46.77m

r 5 = r 9 = (22.5 2 + 41 2 )

= 55.56m

r 1 = 1.5m,

18

r 7 = 53.5m


- the drawdown at A is thus: H − hA = dA = dA =

q 2πDk

4.6×10 −2 2π×15×1.9×10 −3

12

R ln éë r Σ j=1

[ ln

900 1.5

o j

ùû

900 900 900 900 + 2 ln 15.07 + 2 ln 28.14 + 2 ln 34.6 + 2 ln 45.3 +

900 900 2 ln 46.01 + ln 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

900 6.18

900 900 900 900 + ln 21.05 + ln 36.03 + ln 44.87 + ln 50.68 +

900 900 900 900 900 900 900 ln 59.78 + ln 64.48 + ln 57.47 + ln 53.83 + ln 41.03 + ln 26.04 + ln 11.1

=

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

900 1.5

900 900 900 900 + 2 ln 15.07 + 2 ln 25.04 + 2 ln 34.38 + 2 ln 46.77 +

900 900 2 ln 55.56 + ln 53.5

=

4.6×10 −2 30π×1.9×10 −3

]

× 42.57 = 10.93m

19

]


Problem 3.7

12.5

= r 3 = r 7 = r 9 = 27.04m = r 8 = 22.5m = r 6 = r 10 = r 12 = 31.32m = r 11 = 27.5m

r 12

15

and

15

12

R Σ r j=1

o j

r1

r2

r3

r4

r 11 r 10

r6

r9

r8

r7

45 m

7.5

H 2 − h 2D =

é q Thus : d D = H − ê H 2 − πk ë with :

12.5

r5

- the drawdown at D (unconfined flow) is : dD = H − hD

15

7.5

- the radii at point D: r1 r2 r4 r5

15

n

R ln r Σ j=1

o j

q πk

55 m n

R ln r Σ j=1

o j

ù ú û

700 700 700 700 ùû = 38.79 = éë 4 ln 27.04 + 2 ln 22.5 + 4 ln 31.32 + 2 ln 27.5 −3

d D = 20 − éë 20 2 − 8.33×10 × 38.79 ùû = 8.05m 4π×10 −4 - point E is in a similar position than point A in the previous problem (i.e. the same radii apply) : 12

R Σ r j=1

o j

=

[ln

700 1.5

700 700 700 700 + 2 ln 15.07 + 2 ln 28.14 + 2 ln 34.6 + 2 ln 45.3 +

700 700 2 ln 46.01 + ln 43.5

] = 39.96 −3

× 39.96 ùû ≈ 8.38m and : d E = 20 − éë 20 2 − 8.33×10 4π×10 −4 - point F is in a similar position than point B in the previous problem: 12

R Σ r j=1

o j

=

[ln

700 6.18

700 700 700 700 700 + ln 21.05 + ln 36.03 + ln 44.87 + ln 50.68 + ln 59.78 +

700 700 700 700 700 700 ln 64.48 + ln 57.47 + ln 53.83 + ln 41.03 + ln 26.04 + ln 11.1

Whence :

−3

d F = 20 − éë 20 2 − 8.33×10 × 36.75 ùû = 7.50m 4π×10 −4

20

]

= 36.75


Problem 3.8

14m

A Non fissured rock

According to the figure, there are 15 equipotential drops, so that the porewater pressure at A is : ö ≈ 117kN/m 2 u A = γ w h A = 10 × æè 14 − 2.5 × 14 15 ø as opposed to : u A = 10 × (14 − 2.75 × 147) = 85kN/m 2 with a drainage blanket as per the following figure. 8m 2.5 1

3 14m

1

A Non fissured rock

21

drainage blanket


Problem 3.9 From the opposite figure, it is seen that : - the exit hydraulic gradient estimated from element a : ie ≈

0.39 0.96

C L element a 2m

= 0.41

- the factor of safety against piping 1 F = 0.41 = 2.4 - the flownet contains 10 flow channels (both sides count) and 14 equipotential drops. Thus, the quantity of flow : Nf

q = H. N d .k = 5.5 ×

10 .k 14

impermeable

= 1.96k (m 3 /s/m)

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

C L

u u σ

σ 100

60

0

u, σ

60

(kN/m 2 )

22

100


Problem 3.10

b CL

With reference to the opposite figure, it is seen that for the (circular) cofferdam of the previous problem : d1 T1

=

5.5 11

= 0.5,

d2 T2

= 0.5,

T2 b

=

11 6

H

= 1.833

d1 T1

d2 T2

Hence the form factors from figure 3.45 : φ 1 ≈ 1,

φ 2 ≈ 1.38 .

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

impervious

φ

h 2 = 1.3H φ1 +φ2 2 = 1.3 × 5.5 × 1.38 = 4.145m 2.38 whence an average exit hydraulic gradient : i eav =

h2 l

=

4.145 5.5

= 0.75

The quantity of flow being : 1 k q = 0.8H φ1 +φ k = 0.8 × 5.5 × 2.38 = 1.85k (m 3 /s per metre of perimeter) 2

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

23


Problem 3.11

y

Referring to the figure opposite, it is seen that the sheet pile in the case of problem 3.11 is characterised by the following : h u = 5.5m, h d = 0, H = 5.5m L = 5.5m, T = D = 11m

hu

H

A zu

hd B

T

ze

F

Hence the parameter : ξ=

D

L

H é ln ê TL + è 2 −1 öø L ë æ T2

0.5 ù

éD ú +ln ê L + è 2 −1 öø L û ë æ D2

0.5 ù

x

ú û

Impervious

=

5.5 2 ln éë 2+ 3 ùû

≈ 2.09m

The velocity potential can now be calculated at points A, F, and B : 0.5 2 é ù Φ A = ξ ln ê TL + æè TL 2 − 1 öø ú = 2.09 ln æè 2 + 3 öø = 2.752m ë û

ΦF = 0 0.5 2 é ù Φ B = −ξ ln ê DL + æè DL 2 − 1 öø ú = −2.09 ln æè 2 + 3 öø = −2.752m ë û

The velocity potential, total head, and hydraulic gradient at a depth y on the upstream side along AF are respectively : 0.5 éy y2 ù Φ y = ξ ln ê L + æè L 2 − 1 öø ú ë û

h = Φ y − Φ A + h u + T = Φ y + 13.748 (m) iy =

Φ A −Φ y T−y

=

2.752−Φ y 11−y

24


The porewater pressure and effective vertical stress along AF are thereafter evaluated as follows : u = γ w (h − y) σ = z u .[γ sat − γ w (1 − i y )] Note that y in the above relationships is measured from the top of the impermeable soil layer. Thus along AF: point A

point F

y(m) Φ y (m)

h(m) i y

11 10 9 8 7 5.5

16.50 16.27 16 15.67 15.26 13.75

2.752 2.518 2.248 1.924 1.51 0

u (kN/m 2 )

0.234 0.252 0.276 0.310 0.50

55 63 70 77 82.5 82.5

σ (kN/m 2 ) 0 13.3 27 41 56.4 88

The velocity potential, total head, and hydraulic gradient at a depth y on the downstream side along BF are respectively : 0.5 éy y2 ù Φ y = −ξ ln ê L + æè L2 − 1 öø ú ë û

h = Φ y − Φ B + h d + D = (Φ y + 13.752) m iy =

Φ y −Φ B D−y

=

Φ y +2.752 11−y

and the porewater pressure and vertical effective stress : u = γ w (h − y) σ = z e [γ sat − γ w (1 + i y )] Whence the corresponding numerical values:

25


point B

point F

y(m) Φ y (m) 11 -2.752 10 -2.518 9 -2.248 8 -1.924 7 -1.51 5.5 0

h(m) 11 11.23 11.5 11.8 12.24 13.75

26

iy u(kN/m 2 ) 0 0.234 12.3 0.252 25 0.276 38 0.31 52.4 0.50 82.5

σ (kN/m 2 ) 0 8.66 17 24.7 31.6 33


Chapter 4

0.9

Problem 4.1

eo

e Schmertmann method yields a preconsolidation pressure of :

0.70

σ p ≈ 900 kN/m 2 , whence : OCR =

900 200

0.50

= 4.5 0.30

100 σ vo

1000

10000 σ v (kN/m 2 )

0

∆e σ p ≈ 900 kN/m 2 0.2

0.9

Problem 4.2

σp

e

Casagrande method leads to :

0.70

σ P ≈ 1, 180 kN/m corresponding to : OCR = 2

1180 200

= 5.9 0.50

Note that, when applicable, Schmertmann method is more reliable. 0.30 100

1000

10000

σ v (kN/m 2 )

27


Problem 4.3

0

0.5

∆h (mm) 1.0 1.5 2.0 0

10

20

30

40

t (min) From the graph : t 90 ≈ 9 min Therefore : cv =

T 90 .d 2 t 90

=

0.848×10 −4 81×60

= 1.74 × 10 −8 m 2 /s

or

c v ≈ 0.55 m 2 /year

For a degree of vertical consolidation U = 0.8 , the corresponding time factor is : T v ≈ 0.54 . Accordingly, the actual time of consolidation in the laboratory is : t 80 =

T v80 .d 2 cv

=

0.54 0.55

× 10 −4 = 9.8 × 10 −5 years

On site, the clay layer thickness being 8m and is drained on both sides; hence H = 4m. Moreover, the relationship between consolidation times t l in the laboratory and t f in the field is such that : 2

t f = t l æè Hd öø Þ

2

t f80 = 9.8 × 10 −5 × æè 104−2 öø ≈ 15.7 years

28


Problem 4.4 a- The load can be assumed to have been applied instantaneously for a period of 5 months, at the end of which U = 90% was achieved in the vertical direction. Accordingly : t 90 =

T v90 H 2 cv

=

0.848×4 2 2

= 6.785 years

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

cvt H2

=

2 42

× 125 = 0.052

Þ

U v ≈ 28%

The radial and vertical degrees of consolidation are related to the overall degree as follows : 1 − U = (1 − U v )(1 − U r ) Therefore : (1 − 0.9) = (1 − U r )(1 − 0.28)

Þ

U r = 86.1%

The number of sand drains needed to achieve such degree of radial consolidation in a time t = 5 months is estimated using the following procedure : the drain radius being r w = 32.5mm , first select a curve corresponding to a ratio n 1 on figure 4.24. Next read on the selected curve the time factor T r corresponding to a degree of radial consolidation U r = 0.881 , then calculate 0.5 c .t 0.5 1 . n 2 = r1w éë Thr ùû = 0.0325 × éë 3.1 × 125 ùû Tr Compare thereafter n 2 to n 1 : if different, restart the same calculation using the curve corresponding to n 2 in figure 4.24 (pencil in an intermediate curve if the need arises). The results are as follows : n 1 = 20 n 1 = 23

Þ Þ

T r = 2.3 T r = 2.4

Þ Þ

n 2 = 23.06 n 2 = 22.6

29


Hence the number of drains needed : n = 23 n=

re rw

r e = 23 × 0.0325 ≈ 0.75m .

Þ

So that for a square arrangement, the spacing between drains is such that : r e = 0.564S

Þ

S = 1.32m

Problem 4.5 a- the average immediate elastic settlement can be estimated in the B.q following way : S i = α 1 α 2 E u Figure 4.28 yields : H/B = 0.6, Whence: S i = 0.2 × 0.9 ×

10×160 12000

D f /B = 0.15

Þ

α 1 ≈ 0.2,

α 2 ≈ 0.9

= 0.024m

b- the clay layer being only drained on the top side, therefore the time needed to achieve U v = 90% in situ is estimated as follows : t f = t l éë 106−2 ùû tl =

T v90 .d 2 cv

=

2

where laboratory consolidation time is :

0.848 2.1

× 10 −4 = 0.403 × 10 −4 years

Thus: t f = 0.403 × 10 −4 × 1036−4 = 14.51 years c- an overall degree of consolidation U = 0.9 is to be achieved in a time t = 1.5years , and accordingly, the corresponding degree of vertical consolidation is such that : T v = c v Ht 2 = 2.1 × 1.5 = 0.0875 36

Þ

U v ≈ 50%

Now the degree of radial consolidation : 1 − U = (1 − U r )(1 − U v )

Þ

(1 − 0.9) = (1 − U r )(1 − 0.5)

30

Þ


U r = 80%. Thus estimating the number of sand drains needed to achieve such a degree in the knowledge that : c h = 3.2m 2 /year, t = 1.5years, r w = 0.075m : n 1 = 20

Þ

T r ≈ 1.7

n 1 = 22

Þ T r ≈ 1.75

Þ Þ

n2 =

1 0.075

ö × æè 3.2×1.5 1.7 ø

0.5

= 22.4

n 2 ≈ 22

Whence : n = 22 = r we Þ r e = 22 × 0.075 = 1.65m so that for a square 1 arrangement, the spacing between drains is : S = 0.564 r e = 2.92m r

Problem 4.6 a- the stress induced by the loaded foundation will affect the settlement calculations up to a depth of 3B = 30m . Hence the entire clay layer is affected. Select sublayers of a maximum thickness B/2 = 5m , and therefore 2 sublayers each with a thickness of 3m will be adequate. Next, estimate the effective vertical stress due to self weight of soil beneath the foundation, and the initial void ratio (from the curve provided) at the middle of each sublayer (points A and B): point A B

depth (m) 1.5 4.5

sand

A 1.5m

B

clay

4.5m

impermeable rock

σ vo (kN/m 2 ) e o 59 0.845 149 0.83

The increase in effective vertical pressure at both points is estimated as follows : ù é ù é 1 1 ú ê ú ∆σ A = q. êê 1 − ≈ 156kN/m 2 ú 3/2 ú = 160 × ê 1 − 3/2 æ 1+[ a ] 2 ö ú æ 1+ é 5 ù 2 ö ú ê ê z û ë ø ø 1.5 è è û ë û ë

31


é ∆σ B = 160 × êê 1 − ê ë

1 æ 1+ é 5 ù 2 ö è ë 4.5 û ø

3/2

ù úú ≈ 112 kN/m 2 ú û

The preconsolidation pressure at both points is : σ pA = 325kN/m 2 ,

σ pB = 375kN/m 2

Accordingly : point ∆σ v (kN/m 2 ) A 156 B 112

σ v + ∆σ v (kN/m 2 ) σ p (kN/m 2 ) 215.5 < 325 261.3 < 375

Apply equation 4.48 to estimate the consolidation settlement at the surface of each layer : 1 ù = 0.108m S c1 = 0.12 × 3 × 1+0.86 log éë 215.5 59 û 1 ù = 0.048m S c2 = 0.12 × 3 × 1+0.838 log éë 261.3 149 û

so that the total consolidation settlement (without correction) is : S c = 0.108 + 0.048 = 0.156m This value has to be corrected for soil type and foundation size. According to figure 4.35 : H/B = 0.6, A = 0.25 Þ µ = 0.62 . Whence the corrected consolidation settlement : S c = 0.62 × 0.156 = 0.097m. The total settlement of the flexible foundation consists of adding the elastic settlement calculated in the previous problem : S = 0.097 + 0.024 = 0.121m. c- the creep settlement is estimated at the top of each sublayer : S s1 =

Cα 1+e o

log t 21 = t

0.015 1+0.86

log 3.5 = 4.4 × 10 −3 m

32


S s2 =

0.015 1+0.838

log 3.5 = 4.4 × 10 −3 m

leading to a total magnitude of creep settlement between t 1 = 2years and t 2 = 7yeras of : S s = 8.8 mm

33



300

q (kN/m 2 )

The measured results yield the following :

200

q (kN/m 2 ) p (kN/m 2 ) 64 71.5 129 143 193 214 257 286

100

0 0

100

200

300

p (kN/m ) 2

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

6 sin φ 3−sin φ

Þ

3M ùû = sin −1 éë 2.7 ù ≈ 23 φ = sin −1 éë 6+M 6.9 û

Problem 5.2 120

The graph in the figure opposite σ (kN/m 2 ) yields a slope corresponding to an average maximum angle of 90 friction : φ max ≈ 45 60

Whence the angle of friction at the critical density :

30

φ c = φ max − υ = 45 − 10 = 35 0 0

30

60

90

τ max (kN/m ) 2

34

120


Problem 5.3 a- the time needed for a given degree U of porewater pressure dissipation is estimated as follows : t=

h2 (1−U) η.c v

In this case, U = 0.97, η = 3 (sample drained on both sides without radial drainage), c v = 2.7m 2 /year = 5.14 × 10 −6 m 2 / min, h = 38mm. Therefore : 38 2 ×10 −6 t 97 = (1−0.97)×3×5.14×10 −6 = 3121 min (≈ 52h)

The shear speed is such that :

v = ε 1 . t 97l =

0.003×76 3121

= 7.3 × 10 −5 mm/ min

consequently, the speed used (6 × 10 −4 mm/ min) is in this case inadequate b- were the sample to be drained radially as well as vertically, then η = 40.4 and : t 97 =

38 2 ×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=

0.003×76 231.8

= 9.83 × 10 −4 mm/ min ,

meaning that the minimum shear speed of 6 × 10 −4 mm/ min is very adequate

35


The figure opposite indicates that were the sample to drained only on both sides, the measurements are not representative of soil behaviour up to a shear strain of 2% beyond which a minimum degree of porewater pressure dissipation of 95% is achieved

100

U(%) 80 v = 6 × 10 −4 mm/ min

60

drainage : both ends only

40

radial + both ends

20 0 0

2

ε s (%)

4

6

Problem 5.4 a- first calculate the different quantities as follows : σ 3f = æè σ 3 − u f öø (kN/m 2

σ 1f = æè q f + σ 3f öø kN/m 2

104 176.3

224 379.3

p f = 13 é σ 1f + 2σ 3f ù kN/m 2 ë û

144 244

Because the clay is normally consolidated, c = 0 and at failure, the increments of stresses æè ∆p , ∆q öø are related through the slope M : M=

∆q f ∆p f

=

203−120 244−144

= 0.83

Moreover, M is related to the angle of shearing resistance as follows : 3M ùû = sin −1 éë 3×0.83 ù = 21.4 φ = sin −1 éë 6+M 6.83 û

36


b- at failure, the stresses are such that : τf =

qf 2

cos φ ,

σf =

τf tan φ

Whence : σ 3 (kN/m 2 )

τ f (kN/m 2 )

200 300

55.9 94.5

σ f (kN/m 2 ) 142.5 241.1

Problem 5.5 The stress paths of the two tests are depicted in the figure both in terms of effective stress path (ESP) and total stress paths (TSP). The difference between TSP and ESP in each test corresponds to the porewater pressure. 200 ESP

q(kN/m 2 )

B

100 A

TSP

0 0

100

200

300

p (kN/m 2 )

At failure, the porewater pressure parameter A is estimated as follows : uf 12 - sample A: A f = q f = − 145 = −0.08 : heavily overconsolidated clay 105 - sample B : A f = 195 = 0.54 : lightly overconsolidated clay.

37


Chapter 6 Problem 6.1 a) The stress-strain relationship equation 6.42 : é δε v ù éê K1 0 ùú éê δp ê ú=ê 1 úê ë δε s û êë 0 3G úû êë δq K=

δp δε v

=

50 10 −2

in the elastic range is represented by ùú úú . û

Whence the bulk modulus :

= 5MN/m 2 . δq

60 −3 . Now that Moreover, it is seen that : 3G = δε s Þ δε s = 3×10 4 = 2 × 10 δε v and δε s are known, use both equations 6.36 & 6.37 :

δε v = δε 1 + 2δε 3 = 10 −2 δε s = 23 [δε 1 − δε 3 ] = 2 × 10 −3

Þ

δε 1 = 5.3 × 10 −3 ,

δε 3 = 2.35 × 10 −3

b) Under undrained conditions, δε v = 0 and according to equation 6.37 : δε s = δε 1 . Thus for the applied increment of deviator stress, it follows 30 −3 that : δq = 3Gδε s = 3Gδε 1 Þ δε 1 = 3×10 . 4 = 10 The increment of excess porewater pressure generated by δq is such that : δq δu = 3 = 10 kN/m 2 . c) If the drainage were opened, then when all excess porewater pressure has dissipated, the effective mean pressure would have increased by exactly the same amount : δp = δu = 10 kN/m 2 . Hence an increment of volumetric strain : δε v =

δp K

=

10 5×10 3

= 2 × 10 −3 .

q (kN/m 2 )

80

40

d) See figure opposite.

0 200

240

280

p (kN/m 2 )

38


Problem 6.2 a) For the stress path to be within the elastic wall, it suffices to show that p 2 = 350 kN/m 2 is smaller than the mean effective pressure at yield p y . Assuming the drained loading was pursued until the yield curve was met, it is clear that yielding occurs at the point of intersection of the yield curve and the effective stress path. Thus equating the yield curve equation 6.20 to that of the effective stress path 6.4, it follows that : py = po

whence:

p1 po

+

M2 é qy ù M 2 + êê úú ê py ú ë û

qy 3p o

py = p1 +

and

2

M2

=

é q y ê ê p + qy ë 1 3

M 2+ ê

ù ú ú ú û

2

Þ

qy −

1215 é qy 0.81+ êê ê 300+ q y ë 3

ù ú ú ú û

2

qy 3

+ 900 = 0.

The solution can be obtained by trial and error : q y ≈ 199 kN/m 2 Þ p y = 366.3 kN/m 2 . It is therefore seen that p 2 = 350 kN/m 2 < p y = 366.3 kN/m 2 , meaning that the loading applied is well within the elastic wall. b) The increment of mean effective pressure that generated δε v = 7 × 10 −3 is in fact : δp = p 2 − p 1 = 350 − 300 = 50 kN/m 2 ; accordingly :

K=

δp δε v

=

50 7

× 10 3 = 7.14 MN/m 2 .

On the other hand, the corresponding increment of deviatoric stress is such that : δq = 3 æè p 2 − p 1 öø = 150 kN/m 2 ; and therefore : G =

δq 3δε s

=

50 4.2

× 10 3 = 11.9 MN/m 2 .

c) The specific volume corresponding to the point where: p 2 = 350 kN/m 2 , q 2 = 150 kN/m 2 is calculated using the elastic wall

39


equation :

p v 2 = v o − κ ln æè 2 öø po

where v o

is the specific volume

corresponding to p o , calculated from the NCL equation : v o = Γ + (λ − κ)ln 2 − λ ln p o = 2.4 + 0.1 ln 2 − 0.15 ln 500 = 1.537 . ö = 1.555. Whence : v 2 = 1.537 − 0.05 ln æè 350 500 ø

Accordingly, if, from the

point with the co-ordinates æè v 2 , p 2 , q 2 öø , an undrained loading is applied, then the mean effective stress remains constant up to the yielding point (remember that within the elastic wall, an undrained loading implies δp

δε v = K = 0 Þ p = p 2 = constant). Moreover, the undrained loading means that the specific volume remains constant : v = v 2 . Consequently, substituting for p and v into the yield curve equation 6.20, it follows that : 350 500

=

0.81 0.81+η 2

Þ

η=

q 350

= 0.5891

Þ

q = 206.2 kN/m 2 .

Therefore the point where the yield curve is met has the co-ordinates : v 2 = 1.555, q 2 = 206.2kN/m 2 , p 2 = 350 kN/m 2 . Note that the same value q 2 can be found by substituting for p' and v into 1/2 Γ−v 2 −λ ln p the Roscoe surface equation 6.18a : q = Mp 2 éê 2 exp æè λ−κ 2 öø − 1 ùú . ë û The corresponding porewater pressure generated by the undrained loading at yield is estimated from the difference between the total and the effective mean pressures at yield. The total mean pressure is such that : p 2 = 350 + 206.2−150 = 368.7 kN/m 2 ; whence : 3 u = p 2 − p 2 = 368.7 − 350 = 18.7 kN/m 2 . d) At failure, the specific volume is constant v f = v 2 = 1.555 since the loading is undrained. Thus, using the CSL equations 6.7 & 6.8, it follows that at failure : æ ö è Γ−v f ø (2.4−1.555) p f = exp λ = exp 0.15 = 279.6 kN/m 2 , q f = Mp f = 0.9 × 279.6 = 251.6 kN/m 2 .

40


The total mean pressure at failure is calculated as follows : p f = 350 + 251.6−150 = 383.9 kN/m 2 , 3 whence the porewater pressure at failure : u f = p f − p f = 383.9 − 279.6 = 104.3 kN/m 2 .

e) Refer to the figure. 1.7

v NCL 1.6 A

B,C

D O 1.5

CSL

1.4 300

q (kN/m ) 2

D C

200

CSL

B

100

A

0 0

100

200

O

300

400

p (kN/m ) 2

41

500

600


Problem 6.3 a) Because the clay sample was not unloaded, the undrained stress path is on the Roscoe surface. Moreover, undrained conditions imply that the specific volume remains constant. Its value can be found from the NCL equation : v o = Γ + (λ − κ)ln 2 − λ ln p o , with the quantity p o = σ 3 = 220 kN/m 2 . Thus : v o = 1.8 + 0.05 ln 2 − 0.1 ln 220 = 1.295. Now the Roscoe surface equation is used to find p 1 corresponding to q 1 = 74 kN/m 2 and v 1 = 1.295 : æ Γ−v −λ ln p ö ù é 1 1ø è q 1 = Mp 1 ê 2 exp − 1ú λ−κ û ë

0.9 ëé 2 exp 10.1 − p 2 ùû

1/2

1/2

− 74 = 0 .

Þ

Hence

p 1 ≈ 204.8 kN/m 2 .

b) The total mean pressure corresponding to q 1 = 74 kN/m 2 is such that : p 1 = 220 + 743 = 244.7 kN/m 2 . Accordingly, the pore water pressure is: u = 244.7 − 204.8 = 39.9 kN/m 2 . c) Once all excess pore water pressure has dissipated, the mean effective stress will have increased by the quantity u, while the deviator stress remains constant : p 2 = 204.8 + 39.9 = 244.7 kN/m 2 ,

q 2 = 74 kN/m 2 .

The corresponding specific volume is then calculated from the Roscoe surface equation 6.18b : éæ q ö 2 1 ù v 2 = Γ − λ ln p − (λ − κ)ln ê + 2ú Þ ë è Mp 2 ø û

42


é 74 1.8 − 0.1 ln 244.7 − 0.05 ln ê æè 0.9×244.7× ë d)

δp K

δε v =

=

39.9 7

2

2 ö + 1ù 2ú ø û

Þ

v 2 = 1.279.

× 10 −3 = 5.7 × 10 −3 .

e) The loading being drained, hence at failure : pf =

3p o 3−M

=

3×220 3−0.9

= 314 kN/m 2 ,

also, at failure the CSL equations apply : q = Mp f = 0.9 × 314 = 283 kN/m 2 , v f = Γ − λ ln p f = 1.8 − 0.1 ln 314 = 1.225.

f) Refer to figure. 1.4

NCL

v O

1.3

CSL

A B C

1.2 1.1 300

C

q (kN/m ) 2

200

CSL 100 A

B

O

0 0

100

200

300

p (kN/m 2 )

43

400


Problem 6.4 a) the sample was unloaded isotropically until the mean effective stress reached a value p such that : OCR =

po p

= 1.5

Þ

p =

600 1.5

= 400 kN/m 2 .

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

implying that within the elastic wall : δε v = K = 0 Þ p = constant. Accordingly, the mean effective stress at the point where the yield curve is met is p 1 = 400 kN/m 2 . The corresponding specific volume can be p calculated from the elastic wall equation : v 1 = v o − κ ln æè 1 öø , where the po quantity v o represents the specific volume under p o and is estimated from the NCL equation : v o = Γ + (λ − κ)ln 2 − λ ln p o = 1.8 + 0.05 ln 2 − 0.1 ln 600 = 1.195. Whence : v 1 = 1.195 − 0.05 ln 46 = 1.215. Now that v 1 and p 1 are known, use the yield curve equation to determine the corresponding value of q 1 : p1 po

=

M2 2 q M 2 + æè 1 öø p1

Þ

q 1 ≈ 254.5 kN/m 2 .

Note that an identical q 1 value could have been found using the Roscoe surface equation 618a since the point of intersection belongs to both yield and Roscoe surfaces.

b) The undrained loading is pursued until the mean effective stress reaches p 2 = 360 kN/m 2 . Obviously the corresponding specific volume remains constant (undrained load) and thus v 2 = v 1 = 1.215. The deviator stress can now easily be found from the Roscoe surface equation 6.18a :

44


æ Γ−v −λ ln p ö ù é 2 2ø è q 2 = Mp 2 ê 2 exp − 1ú λ−κ û ë

q 2 = 0.9 × 360 éë 2 exp

1/2

(1.8−1.215−0.1 ln 360) 0.05

− 1 ùû

0.5

≈ 300 kN/m 2 .

The total mean pressure corresponding to q 2 is such that : p2 = po +

q2 3

= 400 + 300 = 500 kN/m 2 . 3

Thus, the excess porewater pressure generated under p 2 is as follows : u = p 2 − p 2 = 500 − 360 = 140 kN/m 2 .

c) Once all excess pore water pressure has dissipated, the total mean pressure become an effective pressure, while the deviator stress remains constant : p 3 = p 2 = 500 kN/m 2 and q 3 = q 2 = 300 kN/m 2 . The corresponding specific volume decreases because of the water expelled from within the clay sample, and its magnitude can be calculated from the Roscoe surface equation 618b : ù éæ q ö 2 v 3 = Γ − λ ln p 3 − (λ − κ)ln ê ç 3 ÷ + 12 ú û ë è Mp 3 2 ø é v 3 = 1.8 − 0.1 ln 500 − 0.05 ln ê æè 300 ë 0.9×500

2

2 ö + 0.5 ù ≈ 1.195. ú ø û

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

3p o 3−M

=

3×400 3−0.9

= 571.4 kN/m 2 .

Using the CSL equations, it is seen that : q f = Mp f = 0.9 × 571.4 = 514.3 kN/m 2 , v f = Γ − λ ln p f = 1.8 − 0.1 ln 571.4 = 1.165.

45


e) See figure

1.22

C

A,B

v 1.20 O

D

1.18

E 1.16 1.1 600 E

q (kN/m 2 ) 400 D CSL

C

B

200

A

O

0 0

200

400

600

800

p (kN/m 2 )

Problem 6.5 a) The specific volume corresponding to p 1 is found from the elastic wall equation : v 1 = v o − κ ln

p1 po

with v o = 1.195 as calculated previously in

6.4. Whence : v 1 = 1.195 − 0.05 ln 46 = 1.215

46


b) the point where yielding occurs corresponds to the intersection between the yield curve and the effective stress path. Thus, equating the two following equations : p2 = p1 +

q2 3

Þ p2 = po

p1 po

+

q2 3p o

=

M2 2 æ q ö M 2 + ç 2q2 ÷ è p1+ 3 ø

M2 æq ö M2 + ç 2 ÷ è p2 ø

2

Substituting for p 1 , p o and M : q2 −

1458 + 1200 = 0 2 æ q2 ö 0.81 + ç q2 ÷ è 400 + 3 ø

so that by trial and error : q 2 ≈ 220.1 kN/m 2 ,

p 2 = 473.4 kN/m 2 .

Inserting these two quantities into the yield curve equation 6.19 yields the corresponding specific volume : v 2 = Γ + (λ − κ)ln

2 po

2 − κ ln p 2 = 1.8 + 0.05 ln 600 − 0.05 ln 473.4 = 1.207.

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

c) Because of the drained nature of loading, it is seen that : q 3 = 3 æè p 3 − p 1 öø Þ q 3 = 3(500 − 400) = 300 kN/m 2 . Since the point with the co-ordinates p 3 , q 3 is on the Roscoe surface, use equation 6.18b to calculate the corresponding value of the specific volume:

47


éæ q3 ê v 3 = Γ − λ ln p 3 − (λ − κ)ln ê çç ç ê è Mp 2 3 ë

2 ù ö ÷÷ + 1 úú 2 ÷ ú ø û

é v 3 = 1.8 − 0.1 ln 473.4 − 0.05 ln ê æè 220.1 ë 0.9×473.4

2

2 ö + 0.5 ù ≈ 1.195 . ú ø û

d) Undrained shear occurs under a constant specific volume, thence at failure : v f = v 3 = 1.195. But failure occurs on the CSL, and therefore using the two corresponding equations : Γ−v f ö = 425 kN/m 2 , p f = exp æè λ öø = exp æè 1.8−1.195 0.1 ø

q f = Mp f = 0.9 × 425 = 382.5 kN/m 2 . The total mean stress at failure is thence calculated as follows : pf = p1 +

qf 3

= 400 + 382.5 = 527.5 kN/m 2 . 3

Therefore the porewater pressure at failure is : u f = p f − p f = 572.5 − 425 = 102.5 kN/m 2 .

e) As per the figure.

48


1.22

A

v

B 1.20 D

O

C

1.18 NCL CSL 1.16 1.1 600

q (kN/m 2 ) D

400

C CSL B 200

A

O

0 0

200

400

p (kN/m 2 )

49

600

800


Problem 6.6 a) Given that v 2 = 1.207, p 2 = 473.4 kN/m 2 , q 2 = 220.1 kN/m 2 , an undrained loading implies a constant specific volume so that at failure : v f = v 2 = 1.207. Also, from the CSL equations : Γ−v f ö = 376.2 kN/m 2 , p f = exp æè λ öø = exp æè 1.8−1.207 0.1 ø

q f = Mp f = 0.9 × 376.2 = 338.5 kN/m 2 . Moreover, the total mean stress at failure is : qf p f = p 1 + 3 = 400 + 338.5 = 512.8 kN/m 2 ; hence a porewater pressure at 3 failure : u f = p f − p f = 512.8 − 376.2 = 136.6 kN/m 2 . 1.22

A

v

C B 1.20

O

The stress path is as per the figure.

1.18

NCL CSL 1.16 1.1

600

q (kN/m 2 ) 400

C CSL B 200

A

p (kN/m 2 )

O

0 0

200

400

50

600

800


Problem 6.7 a) The specific volume corresponding to the mean effective stress p 1 is calculated from the elastic wall equation : v 1 = v o − κ ln

p1 po

. The quantity

v o is obviously determined from the NCL equation : v o = Γ + (λ − κ)ln 2 − λ ln p o = 2 + 0.06 ln 2 − 0.1 ln 550 = 1.411. Whence : v 1 = 1.411 − 0.04 ln 100 = 1.479 . 550 Since an undrained load up to q = 120 kN/m 2 was applied, the specific volume remains constant, and more importantly, the mean effective stress remains equal to p 1 until yielding occurs. If failure occurs under undrained conditions, then the mean effective stress at failure can be calculated from the CSL equation : Γ−v

p h = exp λ 1 = exp 2−1.479 = 183.1 kN/m 2 . Since p h > p 1 , yielding occurs 0.1 on the left hand side of the CSL, i.e. on the Hvorslev surface. The point at which yielding occurs is such that : v 2 = v 1 = 1.479 (undrained loading), and

p 2 = p 1 = 100 kN/m 2 (mean effective stress constant up to the yield curve).

Accordingly, inserting these two quantities into the Hvorslev equation 6.23a, it follows that : Γ−v q 2 = (M − h)exp æè λ 2 öø + h.p 2

ö + 0.75 × 100 = 111.6 kN/m 2 . = (0.95 − 0.75)exp æè 2−1.479 0.1 ø

Note that the same value q 2 is obtained from the yield curve equation 6.25, since the yielding point is common to both Hvorslev and yield surfaces.

51


b) If the undrained load is pursued until q 3 = 120 kN/m 2 , then the specific volume remains constant : v 3 = v 2 = 1.479 , and the mean effective stress can be found from the Hvorslev surface equation 6.23a : Γ−v p 3 = 1h éë q 3 − (M − h)exp λ 3 ùû

Þ

p 3 = 111.2 kN/m 2 .

The corresponding total mean pressure being : p3 = p1 +

q3 3

= 10 +

120 3

= 140 kN/m 2 ,

the porewater pressure generated is thence : u = p 3 − p 3 = 140 − 111.2 = 28.8 kN/m 2 .

c) If the drainage was opened, then the excess pore water pressure starts to dissipate. Accordingly, while the deviator stress remains constant at q = 120 kN/m 2 , both the mean effective pressure and the specific volume vary according to the Hvorslev equation : q−hp 120−0.75p v = Γ − λ ln éê M−h ùú = 2 − 0.1 ln éê 0.2 ùú . û ë û ë

Since failure occurs on the CSL, it is seen that : q f = Mp f and

Þ

pf =

120 0.95

= 126.3 kN/m 2 .

v f = Γ − λ ln p f = 2 − 0.1 ln 126.3 = 1.516.

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

d) See figure.

52


1.55 NCL

v D Hvorslev surface

1.5

CSL

C A,B

elastic line

1.45

C

q (kN/m 2 )

B

D

100

CSL 50

NCL

A

0 0

50

100

53

150

p (kN/m 2 )

200


Problem 6.8 a) Obviously, because of the undrained loading from the yield point, the total elastoplastic volumetric strain at failure is zero. The total shear strain in the elastoplastic range is calculated in the same way used in conjunction with example 6-. The yield point is such that : p 3 = 350 kN/m 2 ,

q 3 = 206.2 kN/m 2 ,

v 3 = 1.555.

Up to failure, the stress path is on the Roscoe surface, and since the volume remains constant, the Roscoe surface equation reduces to a relationship æ Γ−v −λ ln p ö ù é 3 è ø q = Mp ê 2 exp − 1ú λ−κ û ë

between p' and q :

0.5

.

Whence the following calculations : p (kN/m 2 ) 350

q (kN/m 2 ) v 206.2 1.555

330

222.2

1.555

310

235.6

1.555

290

246.7

1.555

279.6

251.6

1.555

p 340

Average q v δq 214.4 1.555 -20

δp 16

320

228.9 1.555

-20

13.4

0.715

300

241.1 1.555

-20

11.1

0.804

4.9

0.875

248.8 249.1 1.555 -10.4

η = q/p 0.63

Next, the average values are used to calculate the increments of elastic volumetric strains : δε ev = κ p 340 320 300 284.8

δp

, and plastic shear strains : δε s = − M2 −η 2 δε ev ; thence : p

v.p

q 214.4 228.9 241.1 249.1

δε ev (10 −3 ) -1.89 -2 -2.14 -1.17

v 1.555 1.555 1.555 1.555

54

2η

δε s (10 −3 ) 5.78 9.57 21 46.1 p


Finally, the cumulative plastic shear strains are such that : q (kN/m 2 ) 206.2 222.2 235.6 246.7 251.6

εs

p

0 5.78 × 10 −3 1.53 × 10 −2 3.63 × 10 −2 8.24 × 10 −2

b) Refer to the figure. 260

q (kN/m 2 ) 240

220

200 0

2

4

6

8 p εs

(10 ) −2

Problem 6.9 The stress path being on the Roscoe surface from the onset of undrained loading. Thus, for the undrained part of the stress path, the volume remains constant (refer to the solution of problem 6.3), and the Roscoe surface equation is used to calculate the value of q for each given value of p'. For the drained stress path, all three quantities v, p' and q are variable, and therefore both the Roscoe surface equation and the effective stress path equation are used in a way that, for a selected value p', the corresponding q value is calculated from the effective stress path equation, then the Roscoe

55


surface equation is used to determine the value of v. Whence the following values : p (kN/m 2 ) 220

q(kN/m 2 ) 0

v 1.295

210

60.9

1.295

204.8

74

1.295

244.7

74

1.279

260

120

1.267

280

180

1.250

p 215

Average q v δp δq η = q/p 30.5 1.295 -10 -60.9 0.142

207.4 67.5 1.295 -5.2

300

240

1.235

314

283

1.225

13.1

0.325

224.8 74

1.287 39.9 0

0.329

252.4 97

1.273 15.3 46

0.384

270

150 1.258

20

60

0.556

290

210 1.242

20

60

0.724

307

261.5 1.230 14

43

0.852

The average quantities are thence used to estimate the increments of δε ev = κ

elastoplastic volumetric strains: δε v = p

λ−κ v.p (M +η 2

2

p 215 207.4 224.8 252.4 270 290 307

é (M 2 − η 2 )δp + 2ηδq ù û )ë

q 30.5 67.5 74 97 150 210 261.5

v δε ev (10 −3 ) 1.295 -1.79 1.295 -0.97 1.287 6.89 1.273 2.38 1.258 2.94 1.242 2.78 1.230 1.85

56

δp

,

p

δε s = p

,

δε v (10 −3 ) 1.79 0.97 5.27 7.39 10 9.63 6.42 p

2η M 2 −η 2

δε v . p

δε s (10 −2 ) 0.06 0.09 0.49 0.86 2.22 4.88 13.0 p


The cumulative volumetric and shear elastoplastic strains are thence as follows : p (kN/m 2 ) q(kN/m 2 ) 220 0 210 60.9 204.8 74 244.7 74 260 120 290 180 300 240 314 283

ε v (10 −2 ) 0 0 0 1.22 2.19 3.49 4.73 5.55

ε s (10 −2 ) 0 0.06 0.15 0.64 1.5 3.72 8.6 21.6

The corresponding graphs are as shown. 350

p (kN/m 2 ) 300

250

200 0

2

4

6

ε v (10 −2 )

q (kN/m 2 ) 240

120

0 0

5

10

57

15

20

ε s (10 −2 )


Problem 6.10 Referring to the solution of problem 6-4, then using a similar approach to the calculations related to the solution of problem 6-9, the following quantities can easily be found (note that up to the yield point, the behaviour is undrained and hence the elastic strains are zero) : p (kN/m 2 )

q(kN/m 2 )

v

Average values

400

245.5

1.215

380

279.9

1.215

360

300

1.215

500

300

1.195

525

375

1.184

550

450

1.174

571.4

514.3

1.165

p 390

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

370

289.9 1.215

-20

430

300

1.205

140

0

0.698

512.5 337.5

1.189

25

75

0.658

537.5 412.5

1.179

25

75

0.767

560.7 482.2

1.169

21.4 64.3 0.860

20.1 0.783

The strains increments corresponding to the average values are thus : p

q

v

390 370 430 512.5 537.5 560.7

267.2 289.9 300 337.5 412.5 482.2

1.215 1.215 1.205 1.189 1.179 1.169

δε ev (10 −3 ) -2.1 -2.2 13.5 2.05 1.97 1.63

58

δε v (10 −3 ) p

2.1 2.2 3.36 7.14 6.8 5.52

δε s (10 −2 ) p

0.84 1.75 1.45 2.49 4.7 13.49


Whence the following cumulative elastoplastic strains : p (kN/m 2 ) 400 380 360 500 525 550 571.4

q(kN/m 2 ) 254.5 279.9 300 300 375 450 514.3

ε v (10 −2 ) 0 0 0 1.69 2.60 3.48 4.20

ε s (10 −2 ) 0 0.84 2.59 4.04 6.53 11.23 24.72

It is seen from the latter table that the total cumulative strains at failure are: ε v = 4.2 × 10 −2 ,

ε s = 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 : δε v =

δp K

=

473.4−400 8200

= 8.95 × 10 −3 .

Once yielding occurs, the elastoplastic strains must be calculated in a similar way to that used previously in conjunction with problem 6-10 : p (kN/m 2 )

q(kN/m 2 )

v

473.4

220.1

1.207

500

300

1.195

475

329.6

1.195

450

356.9

1.195

425

382.5

Average values p q 486.7 260

v δp δq η = q/p 1.201 26.6 79.9 0.534

487.5 314.8 1.195 -25

29.6

0.646

462.5 343.2 1.195 -25

27.3

0.742

437.5 369.7 1.195 -25 25.6

0.845

1.195

59


and the strain increments corresponding to the average values : p 486.7 487.5 462.5 437.5

q 260 314.8 343.2 369.7

δε ev (10 −3 ) 2.27 -2.14 -2.26 -2.39

v 1.201 1.195 1.195 1.195

δε v (10 −3 ) 7.75 1.99 2.26 2.39

δε s (10 −2 ) 1.58 0.65 1.29 4.21

p

p

Whence the following cumulative elastoplastic strains : p (kN/m 2 )

q(kN/m 2 )

473.4 500 475 450 425

ε v (10 −3 )

220.1 300 329.6 356.9 382.5

ε s (10 −2 )

0 10.02 9.87 9.87 9.87

0 1.58 2.24 3.52 7.73

The corresponding graphs are depicted in the figures. 500

p (kN/m 2 ) 450

400 0

4

8

12

ε v (10 −3 )

400

q (kN/m 2 ) 300

200 0

2

4

60

6

8

ε s (10 −2 )


Chapter 7 Problem 7.1 The factor of safety in this case is : F=

+

2c z.γ sat . sin 2β

tan φ tan β

é1 − ë

γw cos α γ sat cos β. cos (α−β) (1

z − zw ) ùû

Substituting for β = 16 , α = 8 , F = 1.3, z = 6m, φ = 24 γ sat = 21kN/m 3 , c = 7kN/m 3 , then rearranging, it emerges that : F = 0.994 + 0.128z w

z w = 2.39m

Þ

Problem 7.2 In the short term, the factor of safety of the slope without piles is : F = c u RW.dα = 2

50×12 2 ×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.c u [L Wd

+ 9n.a]

Þ

n= =

1 é W.d.F 9a ë R.c u 1 9×0.15

− L ùû

× æè 3920×2.5×1.6 − 12 × 1.57 öø = 5.4 12×50

Hence n = 6 .

61


Problem 7.3

O

70

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

18m

20 45 13

W ≈ 2010kN/m .

7.5m

W

10.7m

The factor of safety in the short term is thence : F=

65×1.448×18 2 2010×7.5

clay fill

≈2

Note 83 = 1.448rd

Problem 7.4 Referring to the figure opposite, the potential failing soil mass is subdivided into five slices of equal width b ≈ 4.2m. From the geometry of the figure, and using a unit weight γ = 20kN/m 3 , the following weights can be obtained: slice weight w (kN/m)

A 260.4

A B C

56

D

35.5 E

18.4

−7

7

B 630

C 754.8

D 516.6

Applying Bishop routine using φ = 23 , c = 7kN/m 2 :

62

E 180.6


slice A B C D E

α 56 35.5 18.4 7 -7

w. sin α (kN/m) 215.9 365.8 238.3 63 -22

w tan φ + c b (kN/m) 139.9 296.8 349.8 248.7 106.1

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

tan φ sin α F

slice

mα

A B C D E

0.911 1.06 1.08 1.04 0.94

, the first set of iterative calculations is as follows:

1 é mα ëc

b + w tan φ ùû (kN/m) 153.4 279.8 323.05 238.1 112.7

Thus : Σ m1α éë c b + w tan φ ùû = 1107 kN/m , and the calculated factor of safety F = 1107 = 1.28 is different from the 861 selected value F = 1 . Accordingly, another iteration using an initial factor F = 1.28 leads to the following : slice A B C D E

m α = cos α +

tan φ sin α 1.28

1 é mα ë c

b + w tan φ ùû (kN/m)

0.834 1.007 1.053 1.03 0.94

167.7 294.8 332 240.8 112.8

Whence : Σ m1α éë c b + w tan φ ùû = 1148 kN/m and the calculated factor of safety : F =

1148 861

= 1.33

Another iteration is needed using a factor F = 1.33 :

63


slice

m α = cos α +

A B C D E

tan φ sin α 1.33

1 é mα ëc

b + w tan φ ùû (kN/m)

0.82 0.999 1.05 1.03 0.95

169.8 297 333.3 241.1 111.2

Accordingly : Σ m1α éë c b + w tan ùû = 1152.4kN/m yielding a factor of safety : F =

1152.4 861

= 1.34 .

Problem 7.5 The slope is now waterlogged; hence the saturated unit weight γ sat = 21kN/m 3 applies. Consequently, the following quantity must be used: ≈ 904 kN/m. Σ w sin α = 861 × 21 20 From the previous figure, the following estimates of average porewater pressure in each slice can be obtained: slice A B C D E

w(kN/m) u(kN/m 2 ) 273.4 31 661.5 75 792.5 94 542.4 61 189.6 23

(w − ub) tan φ + c b (kN/m) 90.2 176.5 198.2 150.9 68.9

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

64


slice A B C D E

tan φ sin α 1

m α = cos α + 0.94 1.06 1.08 1.044 0.94

1 é m α ë (w

− ub) tan φ + c b ùû (kN/m) 99 166.5 183.5 144.5 73.3

Whence : Σ m1α éë (w − ub) tan φ + c b ùû = 666.8kN/m , yielding a factor of safety : F = 666.8 = 0.74 904 This value being different from F = 1 , hence a new iteration is needed using F = 0.74 : slice A B C D E

sin α m α = cos α + tan0.74 1.035 1.147 1.13 1.062 0.923

1 é m α ë (w

Σ m1α éë (w − ub) tan φ + c b ùû = 633.1kN/m

− ub) tan φ + c b ùû (kN/m) 87.1 153.9 175.4 142.1 74.6 Þ

F=

633.1 904

= 0.7

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

65


Problem 7.6 The five slices of saturated soil have a combined total weight such that : Σ w sin α = 904kN/m. Considering the porewater pressure values estimated in the previous problem, and taking into account the geometry of elements (b = 4.2m for each element) and the soil characteristics (c = 7kN/m 2 ) , the following calculations can be made : slice A B C D E

w(kN/m)

u(kN/m 2 )

273.4 661.5 792.5 542.4 189.6

31 75 94 61 23

c b cos α

u.b ö + æè w cos α − cos α ø tan φ (kN/m)

18.6 100.5 173.6 148.6 68.2

c b + æ w cos α − cosubα öø tan φ ù = 509.5kN/m Accordingly : Σ é cos ë α è û

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

509.5

Σ w sin α−Σ T

Þ

= 604.3kN/m Σ T = 904 − 509.5 1.7

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

604.3 150

≈4

66


Chapter 8 qo

Problem 8.1 f

qo g

45

a- Referring to the figure, it is seen that the work done by the weight of soil is limited to the area abgf since the work done in abc cancels out that done in bde. Accordingly, work done by the soil weight is :

B

δω

H

qu a

b e c d

cu

δE s = δω cos 45.γ H2 [H + 2B] =

δω 2

γ H2 éë 1 + 2

2B ù H û

The work done by the external loads q o and q u is: δE q =

δω 2

(q o [H + B] − q u B)

The internal work done on the slip plans : δW sp = δω 2 .c u (H + B ) =

δω 2

2c u (H + B)

The internal work done within the slip fan : δW sf = Therefore, the total external work is : δE =

δω 2

éγ H2 æ1 + ë 2è

2B ö H ø

+ q o (H + B) − q u B ù û

Similarly, the total internal work : δW =

δω 2

c u [πB + 2(H + B)]

67

δω 2

πc u B


Equating both expressions, then rearranging, it follows that : 2 q u = q o æè 1 + HB öø + γ H2 æè B1 + H2 öø − 2c u æè 1 + π2 + HB öø

b- in the long term, the angles depicted in the opposite figure are such that: α = π4 +

φ 2

, β = π4 −

C

qo

qo α

φ 2

B

45 −

φ 2

H

φ

Moreover : δω b = δω exp æè π2 tan φ öø

δω

qu α r1

The work done by the external loads is : φ δE = q o .C.δω. cos æè π4 + 2 öø − q u .B.δω b sin æè π4 +

On the other hand, it is seen that : φ C = 2r 1 cos æè π4 + 2 öø +

B = 2r 2 . cos æè π4 − Therefore : C = B.

Also :

r1 æ φ ö r 2 . tan ç π4 + 2 ÷ è ø r1 r2

φ 2

H æπ φ ö tan ç 4 + 2 ÷ è ø

ö = 2r . sin æ π + 2 è4 ø

é ê ê1 + ê ë

H æ φ ö 2r 1 . sin ç π4 + 2 ÷ è ø

φ 2

ö ø

ù ú ú ú û

= exp æè − π2 tan φ öø

68

φ 2

ö ø

δω b

β r2

φ


thus substituting for C in the expression of δE then equating it to zero, it follows that: æ φ ö ù cos ç π4 + 2 ÷ é ø ê è φ ö r1 ú H π π æ ö æ q u sin è 4 + 2 ø exp è 2 tan φ ø = q o r 2 æ φ ö ê 1 + ú æ φ ö π π tan ç 4 + 2 ÷ ê 2r 1 sin ç 4 + 2 ÷ ú è ø ë è ø û Now substituting for the ratio qu = qo

exp æè −π tan φ öø æ φ ö tan 2 ç π4 + 2 ÷ è ø

But 2r 1 sin æè π4 +

é ê ê1 + ê ë

φ 2

r1 r2

H æ φ ö 2r 1 sin ç π4 + 2 ÷ è ø

and rearranging: ù ú ú ú û

ö = B. r 1 = B exp æ − π tan φ ö r2 ø è 2 ø

whence : qu = qo

exp æè −π tan φ öø æ φ tan 2 ç π4 + 2 è

ö ÷ ø

é1 + ë

H B

exp æè π2 tan φ öø ù û

and therefore : H B

=

qu qo

tan 2 æè π4 +

φ 2

ö exp æ π tan ö − exp æ − π tan φ ö è2 ø ø è 2 ø

69


Problem 8.2 C

qo

B

β

qu

H α δω

φ

δω b

β

r1

φ

r2

Referring to the figure above, the expression of the work done by the external loads is : φ δE = q u B.δω cos æè π4 + 2 öø − q o C.δω exp æè π2 tan φ öø sin æè π4 +

φ 2

ö ø

Because of the logarithmic spiral nature of the slip fan, it follows that : r2 r1

=

δω b δω

= exp æè π2 tan φ öø

From the geometry of the figure : φ C = 2r 2 cos æè π4 − 2 öø +

H æπ φ ö tan ç 4 − 2 ÷ è ø

r Hence : C = B. r 21 tan æè π4 +

φ 2

φ B = 2r 1 cos æè π4 + 2 öø

,

é ö ê1 + ø êê ë

H æ φ ö 2r 2 cos ç π4 + 2 ÷ è ø

ù ú ú ú û

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

70


é ù ú H ö exp æ π tan φ ö ê 1 + ø êê è æπ φ ö ú ø 2r 2 cos ç 4 + 2 ÷ ú è ø û ë φ ö π π æ æ ö But : 2r 2 cos è 4 + 2 ø = B. exp è 2 tan φ ø Thence:

q u = q o tan 2 æè π4 +

φ 2

q u = q o exp æè π tan φ öø tan 2 æè π4 +

φ 2

ö é1 + øë

H B

exp æè − π2 tan φ öø ù û

Problem 8.3 a

b

45

From the figure, it is seen that the external work is done by the weight of the soil block abcd and by the horizontal force P 1 . The weight of abcd is:

P1 H

δω d

P1

α B

α

c 45

45

45

γ éë BH + H2 − H2 tan α ùû = γ H2 éë 2B + 1 − tan α ùû H 2

2

2

Hence the work done by the soil weight and P 1 : δE = −P 1

δω 2

+ γ H2

2

δω 2B é 2 ë H

+ 1 − tan α ùû

-the internal work due to energy dissipation along the slip plans : δW sp = δω.c u H 2 + 2δω

B 2

.c u

-the internal work done within the slip fan : δW sf = 2c u

B π δω 2 2

71


Thus the total internal work : .2c u é H + B æè 1 + π2 öø ù ë û Equating δE to δW and rearranging : δW =

δω 2

2 P 1 = γ H2 éë 2 HB + 1 − tan α ùû − 2c u H é 1 + HB æè 1 + π2 öø ù ë û

Problem 8.4

According to Mohr's circles representation of stresses, it is seen that : q u = q o + γH + 2p 1 sin φ + x 1 + x 2 + 2p 3 sin φ where : q o +γH

p1 =

1−sin φ

p 2 dp

òp

1

p

x1 =

π

= ò 4 2 tan φ dθ o

p 2 sin φ tan 60

Þ

p 2 = p 1 exp æè π2 tan φ öø

x2 =

p 2 sin φ tan 75

,

p3 =

,

Hence : x 1 + x 2 = 0.845 sin φ exp æè π2 tan φ öø

qu 1+sin φ

(q o +γH) 1−sin φ

On substitution for the different quantities in the equation established earlier : qu =

é 1+sin φ +0.845 exp æè π2 tan φ öø ù ú æ φ ö ê 1−sin φ tan 2 ç π4 − 2 ÷ ë û è ø (q o +γH)

so that for φ = 22

Þ

q u = 10.43(q o + γH)

72


qo

qo

H

qu A

60

D

75

B

C

τ φ

C A q o + γH

B

120

x1 x2

2p 1 sin φ

D

75

60

2p 3 sin φ

45

p1 p2 p3

qu

73

σ


Problem 8.5 -the effective width of foundation : B = B − 2e = (3 − 2e)m -the net pressure at failure : q nf = 12 (γ sat − γ w ).B N γ i γ s γ d γ + γ.D.N q i q s q d q − γ.D with the inclination factors : 2

i γ = æè 1 − α öø = æè 1 − φ

20 ö 35 ø

2

i q = æè 1 −

= 0.184 ,

α ö 90 ø

2

2

ö = 0.605 = æè 1 − 20 90 ø

the bearing capacity factors read from figure 8.26 : φ = 35 Þ N γ ≈ 43, N q ≈ 30 the shape factors : since the length L >> B

Þ

sq = sγ = 1

= 0.4 < 1, then the depth factors : because ξ = DB = 1.2 3 d q = 1 + ξ tan φ æè 1 − sin φ öø = 1 + 0.4 tan 35(1 − sin 35) = 1.12 dγ = 1 On substitution for the different quantities : q nf = 12 × 9.8(3 − 2e) × 43 × 0.184 × 1+ 18 × 1.2 × 30 × 0.605 × 1.12 − 18 × 1.2 q nf = 533.8 − 77.54e The net pressure at foundation base : q n =

Rv B

− γ.D

with : R v = 400 cos 20 ≈ 376 kN/m , therefore : qn =

376 (3−2e)

− 21.6

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

=3

Þ

376 533.8 − 77.54e = 3 × éë 3−2e − 21.6 ùû

e 2 − 9.22e + 4.3 = 0

Þ

e = 0.493m <

74

B 6

= 0.5m

Þ


Problem 8.6 The factor of safety being defined as : F = qunn = 3 and the ultimate net pressure at foundation level for a rectangular foundation is : q un = c u N rec c . Accordingly, the bearing capacity factor for the rectangular foundation is : q

N rec c =

3.q n cu

=

3×150 60

= 7.5

This factor being related to that corresponding to an equivalent square foundation as follows : sq æ sq sq Bö N rec c = N c è 0.84 + 0.16 L ø = N c (0.84 + 0.16 × 0.8) = 0.958N c Whence : N c = 7.75 , and from the charts in figure 8.27 , this value B corresponds to a ratio DB ≈ 1.25 Þ B = 1.6m , and L = 0.8 = 2m sq

Problem 8.7 First, calculate the ratios : eL eB = 0.5 = 0.125, = 0.3 = 0.1 L B 4 3 Using the charts in figures 8.32 & 8.33, it follows that : B1 ≈ 0.1 Þ B 1 = 0.1 × 3 = 0.3m B L1 L

≈ 0.23

Þ

L 1 = 4 × 0.23 = 0.92m

The equivalent foundation width can now be estimated from equation 8.63: L B = 12 (B + B 1 ) + 2L1 (B − B 1 ) = 0.5 × (3 + 0.3) + 0.92 × (3 − 0.3) = 2.02m 2×4

The effective foundation area is therefore : A = LB = 4 × 2.02 = 8.08m 2 Whence the following ultimate pressure :

75


q u = γ sat DN q s q d q i q + γ sat B2 N γ s γ d γ i γ For an angle φ = 39 , figure 8.26 yields : N q ≈ 52, The inclination factors are : 2

β i q = æè 1 − 90 öø = æè 1 −

18 ö 90 ø

2

N γ ≈ 92 2

2

β ö = 0.29 i γ = æè 1 − φ öø = æè 1 − 18 39 ø

= 0.64 ,

The shape factors : s q = 1 + BL tan φ = 1 +

2.02 4

tan 39 = 1.41

s γ = 1 − 0.4 BL = 1 − 0.4 × 2.02 ≈ 0.8 4 The depth factors : ξ = DB = 2.5 = 0.83 < 1 3 dγ = 1

Þ

d q = 1 + ξ tan φ æè 1 − sin φ öø = 1 + 0.83 tan 39 × (1 − sin 39) = 1.25

Consequently, the ultimate pressure is : q u = 20 × 2.5 × 52 × 1.41 × 1.25 × 0.64 + 20 × = 3364 kN/m 2

2.02 2

× 92 × 0.8 × 1 × 0.29

and the net ultimate pressure : q nu = q u − γ sat D = 3364 − 20 × 2.5 = 3314 kN/m 2 The net pressure applied at foundation level is : qn =

R. cos 18 A

− γ sat D =

15000×cos 18 3×4

− 20 × 2.5 = 1139 kN/m 2

Hence a factor of safety against shear failure : F=

q nu qn

=

3314 1139

= 2.91

76


Chapter 9 Problem 9.1 In the long term, the ultimate loading capacity is : L

Q u = A b σ vb N q − W + pK tan δ ò σ v .dz o

- the area of pile base : A b = π B4 = 0.567m 2 , - the effective vertical stress at the pile base : σ vb = 3 × 18.7 + 9.8 × (L − 3) = (9.8L + 26.7) kN/m 2 , - for an angle φ = 22 , figure 9.18 yields N q ≈ 12, - the coefficient of earth pressure K = æè 1 − sin φ öø OCR = (1 − sin 22) 2 = 0.884 Þ pK tan δ = πB × 0.884 × tan æè 0.75φ öø ≈ 0.7, 2

L

- the quantity : ò σ v .dz = A 1 + A 2

σv

o

A 1 = 12 (18.7 × 3 × 3) = 84.15kN/m,

3m

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

A1 A2

= (4.9L 2 + 26.7L − 124.2)kN/m.

L Z

- the effective weight of pile (weight of steel + weight of soil plug) : steel cross sectional area (15mm thick wall) : S = π4 × (0.85 2 − 0.7 2 ) = 0.1826m 2 soil plug cross section : A = π ×

0.7 2 4

= 0.385m 2

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 : Q u = 0.567 × (26.7 + 9.8L) × 12 − 16L − 15.8+ 0.7 × (84.15 + 4.9L 2 + 26.7L − 124.2)

77


On substitution for Q u = 3, 800kN, it follows that : L 2 + 20.23L − 1067.7 = 0 Þ L = 24.09m. In the short term, the ultimate loading capacity is : Q u = Q s + Q b − W - the ultimate shaft friction : Q s = πBLαc u - the average shear strength : c u = (120 + L) kN/m 2 . Whence : Q s = π × 0.85 × 0.35L × (120 + L) = (0.935L 2 + 112.15L) kN - the quantity Q b − W is calculated according to equation 9.11: 2 2 Q b − W = N c c u A b = N c π B4 (120 + 2L) = 9π × 0.85 × (120 + 2L) 4 Thus, in the short term : Q u = 0.935L 2 + 122.4L + 612.8 so that on substitution for Q u = 3, 800kN : L 2 + 130.9L − 3408.8 = 0

L = 22.26m.

Þ

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 : Q s = A s αc u . 9.14c yields α ≈ 1. Whence:

Since c u < 40kN/m 2 , figure

Q s = 30 × 15 × (0.36 + 0.38) × 2 × 1 = 666kN. - the quantity Q b − W ≈ N c A b c u = 9 × 0.36 × 0.38 × 40 = 49.25kN . Therefore: Q u = 666 + 49.25 = 715.25 kN. b) In the long term, the ultimate loading capacity is : L

Q u = A b σ vb N q − W + pK tan δ ò σ v .dz o

78


- the pile cross sectional area (including the soil plug) : A b = 0.36 × 0.38 = 0.1368m 2 - the effective overburden pressure at the pile base : σ vb = 2 × 19 + 13 × 9 = 155 kN/m 2 - the bearing capacity factor from figure 9.18: φ = 23 Þ

N q ≈ 14

3φ - the quantity : pK tan δ = 2 × (0.36 + 0.38) × (1 − sin 23) × tan æè 4 öø = 0.28 L

- the integral : ò σ v .dz = A 1 + A 2

σv

o

2m

A 1 = 12 × 2 × 19 × 2 = 38kN/m

A1 A2

A 2 = 13 × 38 + × 13 × 9 = 1254.5kN/m 1 2

2

15m

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

Z

Therefore : Q u = 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: Q u ≈ 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 =

Q b +Q s −1.5Q n F

−W

- the base resistance : Q b = A b σ vb N q

79


A b = π B4 = π 0.54 = 0.1963m 2 , 2

2

σ vb = 3 × 18.6 + 4 × 8.6 + (L − 7) × 9.8 = (9.8L + 21.6) kN/m 2 , φ = 22

Þ figure 9.18 : N q ≈ 12 .

Q b = 0.1963 × 12 × (9.8L + 21.6) = (23.1L + 50.9) kN

Whence:

L

- the positive skin friction : Q s = pK tan δ ò o σ v .dz p = πB = π × 0.5 = 1.57m K 1 = 1 − sin 24 = 0.593, δ 1 = 34 × 24 = 18

K 2 = 1 − sin 22 = 0.625 δ 2 = 34 × 22 = 16.5

the positive skin friction is developed from a depth of 5m below ground, hence:

σv

L

ò 5 σ v .dz = A 3 + A 4

A1

A 3 = (3 × 18.6 + 2 × 8.6) × 2 + 12 × 2 2 × 8.6 = 163.2kN/m A 4 = (3 × 18.6 + 4 × 8.6)(L − 7) +

1 (L 2

− 7) × 9.8 2

= (4.9L + 21.6L − 391.3)kN/m 2

3m 5m 7m

A2 A3 A4

L

Z

Therefore: Q s = p.(A 3 .K 1 . tan δ 1 + A 4 .K 2 . tan δ 2 ) On substitution for the different calculated quantities, it follows that: Q s = (1.42L 2 + 6.28L − 64.37)kN - the negative skin friction develops between zero and 5m depth : 5

Q n = pK tan δ ò o σ v .dz = p.K 1 tan δ 1 (A 1 + A 2 ) A 1 = 12 × 3 2 × 18.6 = 83.7kN/m

80


A 2 = 3 × 18.6 × 2 + 12 × 2 2 × 8.6 = 128.8kN/m Thus:

Q n = 1.57 × 0.593 × tan 18 × (83.7 + 128.8) = 64.3kN

- the submerged weight of pile: W = A b [3 × 24 + (L − 3) × 14] = (2.75L + 5.9) kN. Q +Q +1.5Q n

−W , Accordingly, knowing that the allowable load is : Q a = b s3 then substituting for the different known quantities, including Q a = 500kN and rearranging, the following equation ensues : L 2 + 15L − 1152 = 0

Þ

L ≈ 27.3m

Problem 9.4 Because of the negative skin friction :

Qa =

Q b +Q s −1.5Q n 3.5

−W

- the base resistance : Q b = A b σ vb N q A b = π 0.52 = 0.2124m 2 , 4 2

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

φ +40 2

=

38+40 2

= 39

Þ

N q ≈ 155

- the effective overburden pressure at pile base: σ vb = 5 × 19.6 + 2 × 18.5 + 10 × 9.5 + 1.5 × 9.5 = 244.3kN/m 2 Hence:

Q b = 0.2124 × 244.3 × 155 ≈ 8043kN L

- the positive skin friction :Q s = p.K. tan δ ò σ v .dz zf

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

81


fill material: K 1 = 1 − sin 21 = 0.64, δ 1 = 0.75 × 21 = 15.75 clay : K 2 = 1 − sin 22 = 0.625, δ 2 = 0.75 × 22 = 16.5 sand : K 3 = 1.5.(1 − sin 38) = 0.576, δ 3 = 0.75 × 38 = 28.5 - the quantity : ò

18.5 15

σ v .dz = A 4 + A 5

σv A1

A 4 = 2 × (5 × 19.6 + 2 × 18.5 + 8 × 9.5) + 1 × 2 2 × 9.5 = 441kN/m 2

5m

7m

A 5 = 1.5 × (5 × 19.6 + 2 × 18.5 + 10 × 9.5) + 1 × 1.5 2 × 9.5 = 355.7kN/m 2

15m

A2 A3 A4

17m

Whence: Q s = p.[A 4 K 2 tan δ 2 + A 5 K 3 tan δ 3 ] so that on substitution for the different calculated quantities :

A5 18.5

z

Q s ≈ 315kN - the negative skin friction develops down to a depth of 15m : 15

Q n = p.K tan δ ò σ v .dz ,

where:

o

15

òo

σ v .dz = A 1 + A 2 + A 3

A 1 = 12 × 5 2 × 19.6 = 245kN/m , A 2 = 5 × 19.6 × 2 + 12 × 2 2 × 18.5 = 233kN/m A 3 = 8 × (5 × 19.6 + 2 × 18.5) + 12 × 8 2 × 9.5 = 1384kN/m Therefore: Q n = p.[A 1 K 1 tan δ 1 + (A 2 + A 3 ).K 2 tan δ 2 ] A straightforward substitution for the different quantities yields: Q n ≈ 561kN - the submerged pile weight: W = 0.2124 × (7 × 24 + 11.5 × 14) ≈ 70kN. Accordingly: Qa =

315+8043−1.5×561 3.5

− 70 ≈ 2077kN.

82


Problem 9.5 The ultimate loading capacity : Q u = Q s + Q b − W L

- the shaft friction : Q s = p.k. tan δ ò σ v .dz o

pile perimeter : p = π × 0.45 = 1.414m loose sand : K 1 = 1.6 × (1 − sin 32) = 0.752, δ 1 = 0.75 × 32 = 24 dense sand: K 2 = 1.6 × (1 − sin 41) = 0.55, δ 2 = 0.75 × 41 = 30.75

σv 4m

L


A1 A2

o

L

A 1 = 12 × 4 2 × 16.5 = 132kN/m A 2 = 16.5 × 4 × (L − 4) + 12 × (L − 4) 2 × 11 = 5.5L 2 + 22L − 176 Whence: Q s = p[A 1 K 1 tan δ 1 + A 2 K 2 tan δ 2 ]

Z

Þ

Q s = (2.54L 2 + 10.3L − 19.7)kN. - the base resistance : Q b = A b σ vb N q 2 2 A b = π B4 = π 0.45 = 0.159m 2 4 = 40.5 - using equation 9.21a and figure 9.18: φ = 41+40 2 - the effective vertical stress at pile base: σ vb = 4 × 16.5 + (L − 4) × 11 = (11L + 22) kN/m 2 It follows that :

Þ

N q ≈ 200

Q b = 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


Thus, knowing that the ultimate load is Q u = Q s + Q b − W , the substituting for the different quantities including Q u = 1500kN : L 2 + 140.9L − 826 = 0 Problem 9.6

Þ

L = 5.64m

a) Only skin friction provides resistance to the applied load. The ultimate loading capacity is the smaller of the two following quantities: Q 1 = n.Q u − W c Q 2 = Q sb − W c − W s n being the number of caissons (12), Q u is the ultimate loading capacity (without base resistance) of a single caisson, W c is the weight of pilecap, Q sb the friction developed along the block sides, and W s represents the submerged weight of the block of soil containing the caissons. - the weight of pilecap : W c = 24 × 5 × 41 × 30 = 147, 600 kN - the submerged weight of a single caisson : W = π 44 × 45 × 14 = 7, 917kN 2

Thence: Q 1 = 12 × (175 × π × 4 × 45 − 7917) − 147600 = 944, 918 kN - the quantity : Q sb = 2 × (37 + 26).c u .L = 2 × 63 × 400 × 45 = 2268, 000 kN - the submerged weight of the soil block (excluding the weight of caissons): W s = (A bb − nA b ).γ × L + nA b .γ c × L - the area of the block base : A bb = 37 × 26 = 926m 2 2 - the area of a caisson base: A b = π 44 = 12.57m 2 - effective unit weight of concrete : γ c = 14kN/m 3 . Therefore: W s = (926 − 12 × 12.57) × 12 × 45 + 12 × 12.57 × 14 × 45 = 513, 615kN and :

Q 2 = 2268 − 147.6 − 513.6 = 1606 MN.

The ultimate load being the smaller of Q 1 & Q 2 , and accordingly:

84


Q u = Q 1 ≈ 945 MN. The corresponding factor of safety is thence:

F=

945 500

≈ 1.9

b) If the end bearing were to be mobilised, then the ultimate loading capacity of the group (discarding the contribution due to the pilecap) corresponds to the smaller of the two quantities : Q 1 = Q 1 + n.N c .A b .c u = 944, 918 + 12 × 6 × 12.57 × 1000 ≈ 1, 850 MN (notice the units). Q 2 = Q 2 + Q bb Q bb = A bb (c u N c + γ.L) = (37 × 26) × (1000 × 6 + 22 × 45 + 20 × 5) ≈ 6, 821MN Q 2 = 1606 + 6821 = 8, 427MN. Q 1 being the smaller of the two values, and hence the ultimate load : Q u = 1, 850MN , corresponding to a factor of safety :

F=

1850 500

= 3.7

Problem 9.7 The ultimate loading capacity is the smaller of the two quantities: Q 1 = nQ u + Q c − W c Q 2 = Q bb + Q sb − nW p − W c − W s + Q c - the ultimate loading capacity of a single pile : Q u = 1.5MN - the loading capacity due to the pilecap : Q c = 0.4γ B c N γ (A c − nA b ) . Note : A c = area of pilecap, B c = width of pilecap. Whence: 2 ö = 2, 223kN Q c = 0.4 × 16.5 × 5 × 3 × æè 5 × 5 − 16 × π × 0.45 4 ø

85


and therefore : Q 1 = 16 × 1500 + 2223 − 720 = 25, 503kN. - the base resistance of the block : Q bb = A bb æè σ vb N q + 0.4γ b.N γ öø for φ = 41 , figure 9.36 yields : N q ≈ 47, N γ ≈ 22 - the cross sectional area of the block : A bb = 3.9 × 3.9 = 15.21m 2 , b being the corresponding width (3.9m). - the effective vertical stress at the block base : σ vb = 4 × 16.5 + 1.64 × 11 ≈ 84kN/m 2 Accordingly: Q bb = 15.21 × (47 × 84 + 0.4 × 11 × 3.9 × 22) = 65, 791kN.

- the friction on the sides of the block is : L

Q sb = p b ò K tan δ.σ v .dz o

loose sand : K 1 = 1.6 × (1 − sin 32) = 0.752 dense sand : K 2 = 1.6 × (1 − sin 41) = 0.55 block perimeter : p b = 2 × (3.9 + 3.9) = 15.6m

σv 4m

A1 A2

5.64m

L


z

o

A 1 = 12 × 4 2 × 16.5 = 132kN/m A 2 = 16.5 × 4 × 1.64 + 12 × 1.64 2 × 11 = 123kN/m Consequently: Q sb = p b [A 1 K 1 tan 32 + A 2 K 2 tan 41] On substitution for the different values, it follows that : Q sb = 1, 885kN - the contribution due to the area of pilecap situated outside the block : Q c = 0.4γ B c N γ (A c − A bb ) = 0.4 × 16.5 × 5 × 3 × (5 2 − 3.9 2 ) = 969kN

86


- the submerged weight of all piles : nW p = 16 × π 0.45 × (4 × 24 + 1.64 × 14) ≈ 303kN 4 - the weight of pilecap being : W c = 720kN - the submerged weight of soil within the block: 2 ö × (16.5 × 4 + 11 × 1.64) W s = (A bb − nA b ).γ .L = æè 5 2 − 16 × π × 0.45 4 ø 2

= 1, 887kN Whence the quantity: Q 2 = 65791 + 1885 − 303 − 720 + 969 = 67, 622kN . Q1 < Q2

Þ

Q ug = Q 1 = nQ u + Q c − W c = 25, 503 kN

Allowing for a factor of safety of 3, the allowable loading capacity of the group is such that : Q ag =

nQ u +Q c 3

− Wc =

16×1500+2223 3

− 720 = 8, 021kN.

Problem 9.8 Were the pilecap contribution to the loading capacity to be discarded, then according to the previous calculations in problem 9.7 : Q 1 = Q 1 − Q c = 25503 − 2223 = 23, 280kN Q 2 = Q 2 − Q c = 67622 − 969 = 66, 653kN.

Obviously the ultimate loading capacity of the group is in this case : Q ug = Q 1 = nQ u − W c = 23280kN . Hence an allowable load: Q ag =

nQ u 3

− Wc =

16×1500 3

− 720 = 7, 280 kN.

87



0.8m

δ = 16

a- for a Rankine type analysis to apply, the criterion of equation 10.59 must be satisfied : tan −1 æè HL öø

silty sand

φ = 35

7.5m

≥η φ

Using equation 10.58 : η = π4 − 2 + 12 [δ − α] Equation 10.36 yields : 16 ö = 28.7 α = sin −1 æè sin δ öø = sin −1 æè sin sin 35 ø sin φ

H1

3.5m 1m 1.5m

clean dense sand

φ = 38

Accordingly : η = 45 − 17.5 + 12 × (16 − 28.7) = 21.1 ö = 25 > η and : tan −1 æè HL öø = tan −1 æè 3.5 7.5 ø The criterion is thence satisfied, and the magnitude of active thrust is not affected by the wall roughness. Applying Rankine analysis, the coefficient of active pressure is : K a = cos δ

cos δ− cos 2 δ−cos 2 φ cos δ+ cos 2 δ−cos 2 φ

= cos 16

cos 16− cos 2 16−cos 2 35 cos 16+ cos 2 16−cos 2 35

≈ 0.3

leading to an effective active thrust : P a = 12 γH 21 K a Pa =

20 2

H 1 = 7.5 + 3.5 tan 16 = 8.5m

where

× 0.3 × 8.5 2 = 216.8kN/m

The corresponding line of action is estimated as follows : d = 1 + 8.5 = 3.83m above the wall base. 3

88


b- first evaluate the total weight including the that of soil above the wall base. Referring to the figure and using a concrete unit weight γ c = 24kN/m 3 , it is seen that : w = w1 + w2 + w3 + w4 w1 w2 w3 w4

= 0.8 × 7.5 × 24 = 144kN/m = 5 × 1 × 24 = 120kN/m = 7.5 × 3.5 × 20 = 525kN/m = 12 × 3.5 × 1 × 20 = 35kN/m

Whence a total weight : w = 824kN/m The sum of vertical forces behind the wall is :

Σ F v = w + P a sin δ = 824 + 216.8 sin 16 = 883.8kN/m Neglecting the passive pressure, the resisting force at the wall base is : F r = tan 23 φ Σ F v = 883.8 tan 23 × 38 = 418.4kN/m The sliding force : F s = P a cos δ = 216.8 cos 16 = 208.4kN/m Hence a factor of safety against sliding :

F=

418.4 208.4

=2

To calculate the factor of safety agianst shear failure, first the eccentricity of the resultant force applied at the base of wall must be determined. To do so, both restoring and overturning moments with respect to the toe of the wall (point O in the following figure) should be calculated. The point of application of the total weight w is found by taking moments about O of the elementary weights w 1 , w 2 , ....w 4 calculated previously. Referring to figure above, it follows that : d=

1.1×144+2.5×120+3.25×525+3.83×35 824

= 2.79m

89


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

Σ M r = 2.79 × 824 + 5 × 216.8 sin 16 = 2597.8kNm/m

2.79m

w

The overturning moment due to that horizontal component of the active thrust:

Σ M o = 3.83 × 216.8 cos 16 = 798.2kNm/m

3.83m

O

Therefore the eccentricity is as follows : e=

B 2

−Σ

Σ Mo 5 2597.8−798.2 Σ Fv = 2 − 883.8 = 0.464m

Mr−

Note that e <

B 6

=

5 6

= 0.83m

The maximum shear stress induced at the base of wall is : Fv q max = ΣB æè 1 +

6e ö B ø

=

883.8 5

× æè 1 +

6×0.464 ö 5 ø

= 275.2kN/m 2

The angle of inclination with respect to the vertical of the resultant force is estimated as follows : F P cos 16 cos 16 ö α = tan −1 æè Σ Fhv öø = tan −1 æè a Fv öø = tan −1 æè 216.8 883.8 ø = 13.2

Σ

Σ

The effective width of foundation : B = B − 2e = 5 − 2 × 0.464 = 4.07m The ultimate pressure is therefore : q u = γDN q s q d q i q + γ B2 N γ s γ d γ i γ - the bearing capacity factors for an angle φ = 38 estimated from figure 8.26 are : N q ≈ 45, N γ ≈ 80

90


- the shape factors : s q = 1 + BL tan φ = 1 +

4.07 15

s γ = 1 − 0.4 BL = 1 − 0.4 × - the depth factors : ξ=

D B

=

1 5

= 0.2 < 1

tan 38 = 1.21 = 0.89

0.407 15

d q = 1 + ξ tan φ æè 1 − sin φ öø = 1 + 0.2 × tan 38 × (1 − sin 38) = 1.06 dγ = 1

Þ

- the inclination factors : 2 ö = 0.73 , i q = æè 1 − 13.2 90 ø

2

ö = 0.43 i γ = æè 1 − 13.2 38 ø

On substitution for the different quantities : q u = (20 × 1 × 45 × 1.21 × 1.06 × 0.73) + æè 20 × 4.07 × 80 × 0.89 × 1 × 0.43 öø 2 = 2089kN/m 2

leading thus to a factor of safety against shear failure : F=

qu q max

=

2089 275.2

≈ 7.6

Problem 10.2 Applying Coulomb analysis, the coefficient of active pressure is :

Ka =

cos 2 φ é ê cos δ ê 1+ ê ë

sin æè φ +δ öø sin φ cos δ

ù ú ú ú û

2

=

cos 2 35 cos 16 éê 1+ ë

sin (16+35) sin 35 cos 16

whence an effective lateral thrust : P a = 12 × 20 × 7.5 2 × 0.247 ≈ 139kN/m

91

ù ú û

2

= 0.247


Problem 10.3 Using

table

10.1

φ = 35 , β = 16

with

Þ

β φ

= 0.46 ,

the

corresponding coefficient of active pressure can be estimated as follows : β φ β φ

= 0.4,

φ = 35

Þ

K a = 0.291

= 0.6,

φ = 35

Þ

K a = 0.329

A linear interpolation yields : β φ

= 0.46

Þ

K a ≈ 0.291 +

(0.329−0.291) 0.2

× 0.06 = 0.302

Accordingly, the effective active thrust applied to the wall is : P a = 12 × 20 × 7.5 2 × 0.302 ≈ 170kN/m

Problem 10.4 First, determine the appropriate coefficients of active and passive pressure that apply to the soil. Referring to table 10.1, it is seen that for : β λ = 0, = 13.5 = 0.6, δ = 23 φ , the following values of the coefficients 22.5 φ

of active and passive pressure are read : φ = 20 φ = 25

Þ Þ

K a = 0.551, K a = 0.468,

Hence : φ = 22.5

Þ

K p = 3.7 K p = 5.7

K a = 0.51,

K p = 4.7

The coefficient of active and passive pressure applied to the surcharge are estimated from table 10.2, using linear interpolation. The different angles used in conjunction with table 10.2 are such : α = 0,

Ω = 90 + 13.5 = 103.5 ,

δ = 23 φ

92


Accordingly, the coefficient of active pressure is estimated as follows : Ω = 100 , φ = 22.5 Þ K a = 0.394+0.314 = 0.354 2 Ω = 105 ,

φ = 22.5

Þ

Ka =

0.37+0.289 2

= 0.33

Therefore for an angle Ω = 103.5 , a straightforward linear interpolation yields : (0.354−0.33) K a = 0.354 − 3.5 × ≈ 0.337 5 Similarly, the coefficient of passive pressure : Ω = 100 ,

φ = 20

Þ

Kp =

3+4.18 2

Ω = 105 ,

φ = 25

Þ

Kp =

3.2+4.54 2

Whence : Ω = 103.5 ,

= 3.59 = 3.87

K p = 3.59 + 3.5 ×

(3.87−3.59) 5

= 3.79

The active thrust normal to the wall is thence : P an = æè 12 γH 2 K a + q 1 HK a öø cos δ = æè 12 × 20 × 15 2 × 0.51 + 15 × 15 × 0.337 öø cos 15 ≈ 1182kN/m and the passive thrust normal to the wall is : P pn = æè 12 γH 2p K p + K p q p H p öø cos δ = æè 12 × 20 × 5 2 × 4.7 + 3.79 × 25 × 5 öø cos 15 ≈ 1593kN/m.

93


Problem 10.5 a- dry soil : - coefficient of active pressure applied to the soil estimated from table 10.1, with : β λ = 0, δ = 23 φ , = 13 ≈ 0.5, φ = 25 Þ K a = 0.422+0.468 = 0.445 25 2 φ

- coefficient of active pressure applied to both the surcharge and soil cohesion estimated from table 10.2. Thus for : Ω = 103 , α = 0, δ = 23 φ , φ = 25 , it is seen that : Ω = 100

Þ

K a = 0.314 ,

Ω = 105

Þ

K a = 0.289

Therefore, a linear interpolation yields : Ω = 103

Þ

K a = 0.289 + 3 × 0.314−0.289 = 0.304 5

The total active pressure normal to the wall : P an = é 12 γH 2 K a + qK a H ù cos δ − ë û

c tan φ

H æè 1 − K a cos δ öø

= æè 12 × 19 × 7 2 × 0.445 cos 16.7 öø + (40 × 0.304 × 7 × cos 16.7) − 5 (1 tan 25

− 0.304 cos 16.7)

P an = 198.4 + 81.5 − 7.6 = 272.3kN/m

The point of application of this thrust is found by taking moments about the toe of the wall. Accordingly, P an is applied at a distance d 1 from the wall base such that :

d1 =

198.4× 73 +81.5× 72 −7.6× 72 272.3

= 2.65m

94


b- waterlogged soil : In this case, the total active thrust normal to the wall is the sum of the effective active normal thrust and the thrust due to the water. Applying the same coefficients of active pressure, it follows that : - the effective active thrust normal to the wall : P a = æè 12 × 9.5 × 7 2 × 0.445 cos 16.7 öø + (40 × 0.304 × 7 × cos 16.7)− 5 (1 tan 25

− 0.304 cos 16.7)

P an = 99.2 + 81.5 − 7.6 = 173.1kN/m - the normal thrust due to water : P w = 12 γ w H 2 = 5 × 7 2 = 245kN/m Hence the total thrust normal to the wall : P an = 173.1 + 245 = 418.1kN/m whose point of application, situated a distance d 2 above the wall base, is calculated by taking moment about the toe of the wall :

d2 =

(99.2+245) × 73 + (81.5−7.6) × 72 418.1

= 2.54m

95


Problem 10.6

β

b

slope γ.K a (0)

σa

z A1

Pa δ

slope γ.K a (β)

H A2 A3

d O

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

b.K a (β) K a (β)−K a (0)

=

b.K a (12.5) K a (12.5)−K a (0)

Table 10.1 is then used to estimate the corresponding coefficients of active pressure, knowing that : λ = 0, δ = 23 φ , φ = 21 . Hence : β φ β φ

K a (0) = 0.442 −

=0

Þ

=

≈ 0.6

12.5 21

Þ

Accordingly : z =

0.442−0.364 5

= 0.426

K a (12.5) = 0.551 − 0.551−0.468 = 0.534 5 1×0.534 0.534−0.426

= 4.94m

Moreover, the three areas in the figure above are such that : γ

A 1 = (z − b) 2 2 K a (β) = (4.94 − 1) 2 ×

20 2

× 0.534 = 83kN/m

A 2 = γ [H + b − z][z − b]K a (β) = = 20 × (6 + 1 − 4.94) × (4.94 − 1) × 0.534 = 86.7kN/m γ

A 3 = 2 [H + b − z] 2 K a (0) =

20 2

× (6 + 1 − 4.94) 2 × 0.426 = 18.1kN/m

96


Therefore, the effective active thrust due to the soil weight is : P 1 = A 1 + A 2 + A 3 = 187.8 kN/m The effective active thrust normal to the wall induced by cohesion is as follows: P2 = ò =

z−b

c tan φ

o

c tan φ

æ cos δ.K (β) − 1 ö dz + a è ø

ò z−b tanc φ æè cos δ.K a (0) − 1 öø dz H

é cos δ.K a (β) − 1 ù [z − b] + (H − z + b) c é cos δ.K a (0) − 1 ù ë û û tan φ ë

The corresponding coefficients of active pressure are estimated from table 10.2 as follows : - K a (β) = K a (12.5) : Ω = 90 + 12.5 = 102.5 ,

α = 0,

δ = 23 φ ,

and φ = 21 hence : Ω = 100,

φ = 21

Ω = 105 ,

Þ

K a (β) = 0.394 − 0.394−0.314 = 0.378 5

φ = 21

Þ

K a (β) = 0.37 −

Accordingly, for Ω = 102.5

Þ

K a (12.5) =

- K a (0) :

δ = 23 φ ,

Ω = 90 ,

α = 0,

0.37−0.289 5

0.378+0.354 2

= 0.354

= 0.366

φ = 21 . From table 10.2 :

K a (0) = 0.447 − 0.447−0.369 = 0.431 5 The effective active thrust normal to the wall induced by cohesion is therefore : P 2 = tan521 × (0.366 cos 14 − 1) × (4.94 − 1) + (6 − 4.94 + 1) ×

5 tan 21

× (0.431 cos 14 − 1)

= −48.7kN/m

97


Therefore, the overall active thrust normal to the wall is : P an = P 1 cos δ + P 2 = 187.8 cos 14 − 48.7 = 133.5kN/m The point of application of P 1 is such that : ö + A 2 æ H+b−z ö + A 3 æ H+b−z ö = P 1 .d 1 A 1 æè H − z + b + z−b 3 ø è 2 ø è 3 ø d 1 = 03m The point of application of P 2 is : d 2 =

6 2

= 3m .

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

1 (d 1 P 1 cos δ P an 1 133.5

+ d2P2) =

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

Problem 10.7 q z1

β W

zw

U1 δ

U2

η

R

Pa O

98

φ

H − zw


For the soil parameters : φ = 22 , δ = 23 φ = 14.7 , the angle η can be estimated from equation 10.81: tan æè η + δ + φ öø é cos η1sin η − tan æè η + φ öø ù = 1 ë û tan (η + 36.7) éë cos η1sin η − tan (η + 22) ùû − 1 = 0 So that by trial and error : η = 38.8 The quantity z 1 is such that : tan β tan η tan 14. tan 38.8 ù = 1.63m z 1 = H éë 1−tan β tan η ùû = 6.5 × éë 1−tan 14. tan 38.8 û

The quantity C in equation 10.80 can then be calculated : C= =

q (H cos β 30 cos 14

γ

+ z 1 ) + 2 [z 1 H + z w (2H − z w )] + 12 (γ sat − γ w ) (H − z w ) 2

× (6.5 + 1.63) + 192 × (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 : P a = C tan η

cos æè η+φ öø sin æè η+φ +δ öø

= 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 : U 1 = 12 γ w (H − z w ) 2 = 5 × (6.5 − 2) 2 = 101.25 kN/m Accordingly, the total active thrust normal to the wall is : P an = U 1 + P a cos δ = 101.25 + 270.4 cos 14.7 ≈ 363 kN/m

99


Chapter 11 q = 40 kN/m 2

Problem 11.1 2m

dense sand 5m

2m

Using the results of example 11.5, calculated with a factor of safety F = 1 on K p , the corresponding effective pressure diagram is depicted in the figure below.

stiff clay D

The equilibrium of moments with respect to the anchorage point O is written as follows : 9.2 2m

Fa

[17.8] O

5m

dense sand 39.3

2m

water level 53.6

[23.7]

60.2 D

[23.7+31.534D]

stiff clay

[60.2+3.312D]

[9.2] : effective lateral pressure (kN/m 2 )

100


Σ Mo = 0

Þ

−9.2 × 2 × 1 − (17.8 − 9.2) × 22 × 23 + 17.8 × 5 × 2.5 ö +(39.3 − 17.8) × 52 × 23 × 5 + 53.6 × (2 + D) × æè 5 + 2+D 2 ø ö × æ 5 + 2 (2 + D) ö +(60.2 + 3.312D − 53.6) × æè 2+D 2 ø 3 è ø −23.7 × D × æè 7 + D2 öø − 31.534D × D2 × æè 7 + Thus yielding :

2D ö 3 ø

=0

D 3 + 8.55D 2 − 27.16D − 112.91 = 0

Whence a depth of embedment D ≈ 4.22m. The corresponding net pressure diagram is depicted in the following figure. 9.2 2m

Fa O

5m

dense sand 53.6 water level

39.3 2m

36.5

60.2

1.3 m stiff clay

2.92m 82.6 80

40

0

40

80

net pressure diagram (kN/m 2 )

The anchorage force F a can be calculated from the equilibrium equation of horizontal forces : Σ Fh = 0 Þ F a = (9.2 + 39.3) × 72 + (53.6 + 60.2) × 22 + 36.5 × ≈ 187 kN/m.

101

1.3 2

− 82.6 × 2.92 2


The distribution of shear forces along the pile is calculated in precisely the same way as in example 11.5. Referring to the net pressure diagram, it is seen that : z (m) shear force (kN/m) 0 0 2− − (9.2 + 17.8) × 22 = −27 + 2 − 27 + 186.7 = 159.7 5 7 159.7 − (17.8 + 39.3) × 2 = 16.9 2 9 16.9 − (53.6 + 60.2) × 2 = −96.9 10.3 13.22

− 96.9 − 36.5 × 1.3 2 − 120.6 + 82.6 × 2.92 2

= −120.6 ≈0

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

7.45m

-120

-80

-40

0 40 80 shear force (kN/m)

120

160

The maximum bending moment calculated at depth z = 7.45m is, with reference to the net pressure diagram (note that at depth z = 7.45m, the net pressure is 55 kN/m 2 ) : M max = −9.2 × 7 × ( 72 + 0.45) − (39.3 − 9.2) × 72 × ( 73 + 0.45) −53.6 × 0.45 × 0.45 − (55 − 53.6) × 0.45 × 0.45 + 187 × 5.45 2 2 3 ≈ 465kNm/m

102


Now that the above quantities were calculated using the limiting conditions (i.e. a factor of safety F = 1 on K p ), the actual characteristics of the pile are such that : - depth of embedment : D = 2 × 4.22 = 5.97 ≈ 6m; hence a total pile length L = 9 + 6 = 15m, - pile steel section : M max = σ a yI = 465kNm/m I y

=

M max σa

=

465 180

Þ

× 10 3 = 2583cm 3 /m

The appropriate steel section corresponds to a Larssen LX32 with a section modulus yI = 3201 cm 3 /m ; the maximum driving length of such a pile being 28m, well in excess of the total length required.

Problem 11.2 9.2

Fa

39.3 53.6 36.5 1.29 m

60.2

Fb

D' C

28.22D -36.5

net pressure diagram (kN/m 2 )

Writing the equilibrium equation in the upper half in the figure above, it follows that : F a + F b = (9.2 + 39.3) × 72 + (53.6 + 60.2) × 22 + 12 × 36.5 × 1.29 = 307.1kN/m

103


The equilibrium of moments with respect to the anchorage point O in the upper half is written as follows: −9.2 × 2 × 1 − (17.8 − 9.2) × 22 × 23 + 17.8 × 5 × 52 + (39.3 − 17.8) × 52 × 23 × 5 + 53.6 × 2 × 6+ ö − 8.29F b = 0 (60.2 − 53.6) × 22 × æè 43 + 5 öø + 36.5 × 1.29 × æè 7 + 1.29 2 3 ø Accordingly : Fb =

1237.5 8.29

= 149.3kN/m,

F a = 307.1 − 149.3 = 157.8kN/m

The equilibrium of moments with respect to point C in the lower half yields : ö æ D −1.29 ö = 0 −149.3 æè D − 1.29 öø + æè 28.22D − 36.5 öø æè D −1.29 2 øè 3 ø or

D 3 − 3.874D 2 − 26.77D + 38.8 = 0

Þ

D ≈ 6.93m

The distribution of shear forces along the pile is calculated in conjunction with the net pressure diagram, from which it is seen that : z (m) 0 2− 2+ 5 7 9 10.29 11 13 15.93

shear force (kN/m) 0 − (9.2 + 17.8) × 22 − 27 + 157.8 130.8 − (30.7 + 17.8) × 32 58 − (39.3 + 30.7) × 22 − 11.9 − (60.2 + 53.6) × 22 − 125.8 − 36.5 × 1.29 2 − 149.3 + 20 × 0.71 2 − 142.2 + (20 + 76.4) × 22 − 45.8 + (159 + 76.4) × 2.93 2

104

= −27 = 130.8 = 58 = −11.9 = −125.8 = −149.3 = −142.2 = −45.8 = 299


The corresponding diagram is depicted in the figure opposite, from which zero shear force occurs at depths : z 1 ≈ 6.75m z 2 ≈ 13.65m

6.75m

13.65m

13.65m

-200

-100

0

100

200

300

shear force (kN/m)

Using the net pressure diagram, the maximum bending moment at depth z 1 = 6.75m is calculated as follows : z1 M max = −9.2 × 2 × (1 + 4.75) − (17.8 − 9.2) × 22 × æè 23 + 4.75 öø +157.8 × 4.75 − 17.8 × 4.75 × 4.75 − (38.2 − 17.8) × 2

4.75 2

× 4.75 3

≈ 320 kNm/m Similar calculations yield a maximum bending moment at depth z 2 = 13.65m : 2 M max ≈ −324kNm/m ≈

z

1 − M max

z

Problem 11.3 First, calculate the porewater pressure distribution along the pile. Using Mandel analysis, it is seen that in this case : H = 2m, h u = h d = 0, D = 6.5m, T = 8.5m Therefore :

105


ξ=

2 2 2 æ æ 8.5 ö 8.5 ö æ ö −1 ö÷ ln ç 3.5 + è 3.5 ø −1 ÷ +ln ç 6.5 + æè 6.5 3.5 ø è 3.5 è ø ø

æ Φ A = 0.723 ln ç 8.5 + 3.5 è æ Φ B = −0.723 ln ç 6.5 + è 3.5

= 0.723m

2 æ 8.5 ö − 1 ö÷ = 1.11m Φ F = 0 è 3.5 ø ø 2 æ 6.5 ö − 1 ö÷ = −0.89m è 3.5 ø ø

q = 50 kN/m 2

2.5m

Accordingly, on the active side of the wall, the velocity potential is such that at any depth y : 2 æy ö y Φ y = ξ ln ç L + æè L öø − 1 ÷ è ø Moreover, the total head is : h = Φ y − Φ A + h u + T = (Φ A + 7.39) m

3.5m A H B

2m

3m

D

The corresponding porewater pressure is : u = γ w (h − y) and the effective unit weight : the hydraulic gradient γ y = γ sat − γ w (1 − i y ) ; Φ A −Φ y being : i y = T−y

Fa

T F L= 3.5m

IMPERVIOUS

Whence the following results on the active side (behind the wall) [note the depth y is with respect to the top of the impervious layer] : y (m) 8.5 7 5.5 3.5

u (kN/m 2 ) 0 13.4 26.3 38.9

i

γ (kN/m 3 ) 10 0.105 11.05 0.123 11.23 0.222 12.22

On the passive side, the velocity potential at depth y is : 2 æy ö y Φ y = −ξ ln ç L + æè L öø − 1 ÷ è ø

and the total head :

106


h = Φ y − Φ B + h d + D = (Φ y + 7.39) m Φ y −Φ B The hydraulic gradient being : i = D−y z

and the corresponding porewater pressure and soil effective unit weight are respectively : γ = γ sat − γ w (1 + i)

u = γ w (h − y) ,

Therefore on the passive side : y (m) 6.5 5.5 4.5 3.5

u (kN/m 2 ) 0 11.5 24.1 38.9

i

γ (kN/m 3 ) 10 0.15 8.5 0.206 7.94 0.297 7.03

z

IMPERVIOUS

The active and passive coefficients of earth pressure are estimated a) from table 10.1 : - loose sand : β = 0, φ = 30 , δ = 23 φ Þ K a = 0.300 - dense sand: β = 0,

φ = 40 ,

δ = 23 φ

Þ

K a = 0.202 K p = 12

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

Ω = 90 ,

φ = 40

Þ

K a = 0.206

It is seen that, in this case, the coefficients of active pressure are similar and therefore, those calculated from table 10.1 will be used in conjunction with the surcharge q. Accordingly, using the free earth support method, the active and passive pressures normal to the wall can now be calculated (using a factor of safety F = 1 on K p ) . On the active side, the effective active stresses normal to the wall are calculated as follows : σ a = K a cos δ æè σ v + q öø On the passive side, the effective passive stresses normal to the wall are estimated from the expression : σ p = K p cos δσ v

107


Whence the following results on the active side (z as per the previous figure) z (m) γ (kN/m 3 ) u (kN/m 2 ) σ v (kN/m 2 ) σ a (kN/m 2 ) 0 19 0 − 6 19 0 6+ 10 0 7.5 11.05 13.4 9 11.23 26.3 11 12.22 38.9 and on the passive side :

0 114 114 130.5 147.3 171.7

z (m)

σ v (kN/m 2 ) σ p (kN/m 2 )

0 1 2 3

γ (kN/m 3 ) u (kN/m 2 ) 10 8.5 7.94 7.03

0 11.5 24.1 38.9

14.1 46.2 20.6 23.6 26.6 31

0 8.5 16.5 23.5

0 91.1 177 252

The corresponding total and net pressure diagrams are depicted in the following 14.1 figures: Fa total pressure diagram

20.6

46.2 37 42.3

290.9

102.6 201.1 69.9

300

200

σp + u

100

100 (kN/m 2 )

σa + u

IMPERVIOUS

108


14.1

Fa net pressure diagram

O

46.2

20.6 0.49m

37 42.3

50 221

140 200

100

100 pressure (kN/m 2 )

IMPERVIOUS

The factor of safety on the passive side is calculated by taking moments about the anchorage point O . Thus referring to the net pressure diagram (positive moments clockwise): Σ Mo = 0 Þ −14.1 × 2.5 × 1.25 − (27.5 − 14.1) × 2.5 × 2.5 + 27.5 × 3.5 × 1.75+ 2 3 3.5 2 (46.2 − 27.5) × 2 × 3.5 × 3 + 20.6 × 2 × (1 + 3.5)+ (42.3 − 20.6) × 22 × æè 23 × 2 + 3.5 öø + 42.3 × 0.49 × æè 0.49 + 5.5 öø 2 3 − F1 49.7 × 0.51 × æè 23 × 0.51 + 5.99 öø + 49.7 × 1 × (0.5 + 6.5)+ 2 (139.7 − 49.7) × 12 × æè 23 + 6.5 öø + 139.7 × 1 × (0.5 + 7.5)+

[

(221 − 139.7) × 1 × æè 23 + 7.5 öø = 0

]

Once rearranged : 535.74 − 2532.1 =0 F

Þ

F = 4.73

109


The anchorage force is found from the equilibrium equation of horizontal forces. Once more, referring to the net pressure diagram : −(14.1 + 27.5) × − 0.49 × 42.3 + 2

1 4.73

2.5 2

+ F a − (27.5 + 46.2) × 3.5 − (20.6 + 42.3) × 22 2

[49.7 ×

0.51 2

+ (139.7 + 49.7) × 12

+(221 + 139.7) ×

1 2

] =0

thus yielding : F a = 193.4 kN/m The distribution of shear forces along the pile is next calculated from the net pressure diagram : z (m) 0 2.5 − 2.5 + 6 8 8.49 9 10 11

shear force (kN/m) 0 − (14.1 + 27.5) × 2.5 2 − 52 + 193.4 141.4 − (27.5 + 46.2) × 3.5 2 2 12.4 − (20.6 + 42.3) × 2 − 50.5 − 42.3 × 0.49 2 − 60.8 + 49.7 × 0.51 2 − 48.1 + (49.7 + 139.7) × 12 46.6 + (221 + 139.7) × 12

= −52 = 141.4 = 12.4 = −50.5 = −60.8 = −48.1 = 46.6 = 226.9

The corresponding shear force diagram is depicted in the following figure, from which the depth corresponding to zero shear force (where the maximum bending moment occurs) is estimated at z ≈ 6.3m . Whence the maximum bending moment calculated with respect to point A in the figure: M z=6.3m = M max = −(27.5 − 14.1) ×

2.5 2

−(46.2 − 27.5) ×

3.5 2

−14.1 × 2.5 × æè 2.5 + 3.8 öø 2

× æè 2.5 + 3.8 öø + 193.4 × 3.8 −27.5 × 3.5 × æè 3.5 + 0.3 öø 3 2 × æè 3.5 + 0.3 öø − 20.6 × 0.3 × 0.3 −(25 − 20.6) × 3 2

M max ≈ 234 kNm/m

110

0.3 2

× 0.3 2


6.3 m

A

100

-100

200

shear force (kN/m)

IMPERVIOUS

Problem 11.4 If a factor of safety F = 2 were applied on K p , then only the effective passive pressures normal to the wall will be affected , and therefore need be calculated as follows : Kp σ p = K ∗p .σ v . cos δ , with K ∗p = 2 = 6 Accordingly, the active pressures being as calculated in the previous problem, the new values of the passive pressures normal to the wall are : z (m) 0 1 2 3

γ (kN/m 3 ) 10 8.5 7.94 7.03

u (kN/m 2 )

σ v (kN/m 2 ) σ p (kN/m 2 )

0 11.5 24.1 38.9

0 8.5 16.5 23.5

0 45.6 88.5 126

Hence the corresponding net pressure diagram of the following figure.

111


14.1

Fa net total pressure

46.2

20.6 1m

37 42.3 51.2

95

100

pressure (kN/m 2 )

100

IMPERVIOUS

The actual factor of safety on the passive side is determined from the moment equilibrium equation with respect to the anchorage point : Σ Mo = 0 Þ −14.1 × 2.5 × 1.5 − (27.5 − 14.1) × 2.5 × 2.5 + 27.5 × 3.5 × 3.5 2 3 2 2 (1 +(46.2 − 27.5) × 3.5 × × 3.5 + 20.6 × 2 × + 3.5) 2 3 2 æ +(42.3 − 20.6) × 2 × è 23 × 2 + 3.5 öø + 42.3 × 12 × æè 13 + 5.5 öø − F2 51.2 × 12 × æè 23 + 6.5 öø + 51.2 × 1 × æè 12 + 7.5 öø +(95 − 51.2) × 12 × æè 23 + 7.5 öø = 0

[

]

=0 yielding : 600.4 − 1543.8 F

Þ

F = 2.57

The anchorage force can now be calculated from the equilibrium equation of horizontal forces. Referring to the net pressure diagram, it is seen that : F a = (14.1 + 27.5) × +42.3 × 12 −

2 2.57

2.5 2

[51.2 ×

+ (27.5 + 46.2) × 1 2

3.5 2

+ (51.2 + 95) × 12

Hence : F a = 188.2 kN/m

112

+ (20.6 + 42.3) ×

]

2 2


The distribution of shear forces along the pile is calculated as follows : z (m) shear force (kN/m) 0 2.5 − 2.5 + 6 8 9 10 11

0 − (14.1 + 27.5) × 2.5 2 − 52 + 188.2 136.2 − (27.5 + 46.2) × 3.5 2 7.2 − (20.6 + 42.3) × 22 − 55.7 − 42.3 × 12 − 76.9 + 51.2 × 12 − 51.2 + (95 + 51.2) × 12

= −52 = 136.2 = 7.2 = −55.7 = −76.9 = −51.2 = 21.9

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

6.2 m

100

-100

shear force (kN/m) IMPERVIOUS

The maximum bending moment occurring at depth z = 6.2m is thence : M m = −14.1 × 2.5 × æè 2.5 + 3.7 öø − (27.5 − 14.1) × 2.5 × æè 2.5 + 3.7 öø 2 2 3 +188.2 × 3.7 − 27.5 × 3.5 × æè 3.5 + 0.2 öø − (46.2 − 27.5) × 3.5 × æè 3.5 + 0.2 öø 2 2 3 −20.6 × 0.2 × 0.1 M m = 213.1 kNm/m

113


If M m is to be reduced according to Rowe's method, then the design moment is such that : M c = M m − 12 (M m − M r ) First, the reduced moment M r is calculated as follows : α = 118 = 0.73 so that an interpolation on figure 11.7 yields the following results in which the moment M r corresponds to the product of the maximum bending moment M m calculated previously at the depth of zero shear force times the ratio MMmax read from Rowe's graphs : ρ (m 3 /kN) 0.05 0.07 0.1 0.2 0.4 1

M M max

M r = M m MMmax (kNm/m)

0.98 0.89 0.82 0.68 0.56 0.43

209 190 175 145 119 92 210

M r (kNm/m) 190

The figure opposite indicates that a pile section 6W is nearest to the graph and therefore most appropriate. This pile corresponds to a reduced moment M r = 109.8 kNm/m. Accordingly, the design moment is as follows : M c = M m − 12 (M m − M r ) = 213.1 − 12 × (213.1 − 109.8) = 161.5 kNm/m Hence, select a GSP2 pile (British steel handbook)

170 GS P2

LX8

150

130

6W

110

90 0.05

0.1

0.2

ρ (m 3 /kN)

114

0.4

0.6

1.0


Problem 11.5 The coefficients of active and passive pressure are as follows : - sand : β = 0, λ = 0, δ = 23 φ , φ = 35 . Table 10.1 yileds K as = 0.247 The coefficient of active pressure applicable to the surcharge q is found from table 10.2: Ω = 90 , α = 0, δ = 2 , φ = 35 Þ K as = 0.249 3φ

- clay : table 10.1 yileds the following: β = λ = 0, δ = 23 φ , φ = 25 Þ K ac = 0.364, β = λ = 0,

δ=

2 φ 3

φ = 20

,

Therefore for φ = 23 ,

Þ

K ac = 0.442,

K pc = 2.75 K pc = 3.7

K ac = 0.364 + 35 × (0.442 − 0.364) = 0.41 K pc = 2.75 + 35 × (3.7 − 2.75) = 3.32

The coefficients applicable to both cohesion and surcharge are estimated from table 10.2 : Ω = 90 , α = 0, φ = 25 Þ K ac = 0.369, K pc = 3.554 Ω = 90 ,

α = 0,

φ = 20

Hence for φ = 23 ,

Þ

K ac = 0.447,

K pc = 2.645

K ac = 0.369 + 35 × (0.447 − 0.369) = 0.416 K pc = 2.645 + 35 × (3.554 − 2.645) = 3.19

Both active and passive effective normal stresses are then calculated as follows : - throughout the sand layer : σ a = cos δ s é K as σ v + K as q ù ë û - throughout the clay layer : σ a = cos δ c é K ac σ v + K ac q ù − c cot φ é 1 − K ac cos δ c ù ë û ë û σ p = σ v K pc cos δ c + c cot φ é K pc cos δ c − 1 ù ë û with the angles of wall friction : δ s = 23 × 35 = 23.3 ,

115

δ c = 23 × 23 = 15.3


Whence the following stress values (depth z as per the figure opposite): - active pressures : z (m) 0 2 4.5 − 4.5 + 5.5 7 9 4.5 + D

σ v (kN/m 2 ) 0 37 83.3 83.3 102.3 117.3 137.3 92.3 + 10D

σ a (kN/m 2 ) 22.9 31.3 41.8 61.8 69.3 75.2 83.1 65.4 + 3.95D

q = 100 kN/m 2

z 4.5m sand z

1m D' C

- passive pressure : z (m) 0 1 2.5 4.5 D

σ v (kN/m 2 ) 0 19 34 54 10D + 9

σ p (kN/m 2 ) 39.14 99.9 147.9 211.9 32D + 67.9

Hence the ensuing net pressure diagram.

22.9

sand 0.426m 22.7

41.8

30.6 stiff clay

28.D + 2.54

net pressure (kN/m 2 )

116

stiff clay


The depth of embedment D' is found from the equilibrium equation of moments with respect to point C at the foot of the pile : Σ MC = 0 Þ −22.9 × 4.5 × æè 4.5 + D öø − (41.8 − 22.9) × 4.5 × æè 4.5 + D öø 2 2 3 ö + 30.6 × 0.574 × æ D − 1 + 0.574 ö −22.7 × 0.426 × æè D − 0.426 2 3 ø 2 3 ø è +30.6 ×

æ D −1 ö ø è 2

2

+ æè 28D + 2.54 − 30.6 öø

æ D −1 ö ø è 6

2

=0

So whence rearranged, the following cubic equation ensues : D 3 + 0.28D 2 − 33.85D − 62.4 = 0

Þ

D ≈ 6.45m

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

22.9 29.2

0.426m

sand

33.5 22.7 41.8

30.6 72.6

stiff clay

128.5

183.1 net pressure (kN/m 2 )

117


The distribution of shear forces along the pile can now be calculated using the net pressure diagram : z (m) shear force (kN/m) 0 1.5 3 4.5 4.926 5.5 7 9 10.95

0 − (22.9 + 29.2) × 1.5 2 − 39 − (35.5 + 29.2) ×

1.5 2 − 87.6 − (41.8 + 35.5) × 1.5 2 − 145.6 − 0.426 × 22.7 × 12 − 150.4 + 30.6 × 0.574 × 12 − 141.6 + (30.6 + 72.6) × 12 − 64.2 + (128.5 + 72.6) × 22 136.9 + (128.5 + 183.1) × 1.95 2

= −39 = −87.6 = −145.6 = −150.4 = −141.6 = −64.2 = 136.9 = 440.7

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

sand 7.8 m

stiff clay

-200

0

200

400

shear force (kN/m)

118


The maximum bending moment occurring at the depth z = 7.8m is thus (note that in the net pressure diagram, the pressure corresponding to a depth z = 7.8m is 95kN/m 2 ) : M max = −22.9 × 4.5 × ( 4.5 + 3.3) − (41.8 − 22.8) × 4.5 × ( 4.5 + 3.3) 2 2 3 0.426 2 −22.7 × 2 × ( 3 × 0.426 + 0.574 + 2.3) +30.6 × 0.574 × (2.3 + 0.574 ) + 30.6 × 2.3 × 2.3 2 3 2 2.3 +(95 − 30.6) × 2.3 × 2 3 Whence : M max ≈ −633 kNm/m.

119

Azizi - Applied Analysis in Geotechnics-Solution Manual

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