CIVL4720 Geotechnical Analysis & Design 3. Retaining Wall Jidong Zhao This lecture • Retaining walls – What/why/when to use retaining walls? – Major types – Design steps • Stability check • Se>lement check 2 Retaining walls • Retaining walls – Provide permanent lateral support to verCcal or near-­‐
verCcal slopes of soil. – Types: • Gravity wall, semi-­‐gravity wall, canClever wall and counter-­‐fort retaining wall. 3 Retaining walls • In design a retaining walls, we need to know – The basic soil properCes of soil behind the wall, such as unit weight, fricCon angle, cohesion. – Two major steps in design • Calculate the lateral earth pressure, and check for stability including overturning, sliding and bearing capacity failures – Use lateral earth pressure theory and bearing capacity theory • Check each component of the structure for adequate strength and determine the steel reinforcement of each component 4 ProporGoning retaining walls • Assume the dimensions first – proporConing – And then check the stability, essenCal try-­‐and-­‐error. – Minimum • Top: 0.3m; D: 0.6m • Counter-­‐fort slabs: 0.3m thick spaced at 0.3H-­‐0.7H. 5 Design of retaining walls • Use of lateral earth pressure theories – SimplificaCons and approximaCons • VerCcal plane AB • Rankine’s acCve earth pressure • Ws and Wc counted. ⎛ sin α ⎞
η = 45 + − − sin ⎜
⎟
⎜
⎟
2 2
⎝ sin φ1ʹ′ ⎠
α
φ1ʹ′
−1
6 Design of retaining walls Coulomb’s acCve pressure theory 7 Main steps for stability check 1. Check for overturning about its toe (a) 2. Check for sliding failure along its base (b) 8 Main steps for stability check 3. Check for bearing capacity failure of the base • Bearing capacity failure (c) • Deep seated shear failure (d) 4. Check for excessive se>lement 5. Check for overall stability 9 Stability check -­‐ Overturning CanClever Retaining wall Rankine’s acCve pressure acCng on AB; Rankine’s pasive pressure acCng on the verCcal line at the toe; 1
Pp = K pγ 2 D 2 + 2c2ʹ′ K p D
2
FS(overturning
M
∑
) =
∑M
R
o
⎛ H ʹ′ ⎞
∑ M o =Ph ⎜⎝ 3 ⎟⎠
Ph = Pa cos α
10 Stability check -­‐ Overturning ∑M
R
Pv = Pa sin α
M v = Pv B = Pa sin α B
M1 + M 2 + M 3 + M 4 + M 5 + M v
Pa cos α ( H ʹ′ 3)
M + M 2 + M3 + M 4 + M5
FS(overturning) = 1
Pa cos α ( H ʹ′ 3) − M v
FS(overturning ) =
FS(overturning) ≈ 1.5 : 2.0
11 Stability check -­‐ Overturning Gravity Retaining wall FS(overturning ) =
FS(overturning ) =
M1 + M 2 + M 3 + M 4 + M 5 + M 6 + M v
Pa cos α ( H ʹ′ 3)
M1 + M 2 + M 3 + M 4 + M 5 + M 6
Pa cos α ( H ʹ′ 3) − M v
12 Example 1 • The cross secCon of a canClever retaining wall is shown below. Calculate the factor of safety with respect to overturning. 13 Stability check – Sliding along the base • The factor of safety against sliding FS(sliding
F
∑
) =
∑F
Rʹ′
τ f = σ ʹ′ tan φ2ʹ′ + c2ʹ′
d
Rʹ′ = τ f ( area of cross section )
= τ f ( B ×1) = Bσ ʹ′ tan φ2ʹ′ + Bc2ʹ′
Bσ ʹ′ = sum of the verticle force = ∑V
Rʹ′ = ( ∑V ) tan φ2ʹ′ + Bc2ʹ′
∑ F = ( ∑V ) tan φ ʹ′ + Bc ʹ′ + P
2
Rʹ′
2
p
∑ F = P cos α
d
a
14 Stability check – Sliding along the base FS(sliding )
V ) tan φ ʹ′ + Bc ʹ′ + P
(
∑
=
≥ 1.5
2
2
p
Pa cos α
Without a base key
1
Pp = γ 2 D 2 K p + 2c2ʹ′ D K p
2
Ignore Pp and reduce φ2ʹ′ and c2ʹ′
FS(sliding ) =
( ∑V ) tan ( k φ ʹ′ ) + Bk c ʹ′ + P
1 2
2 2
p
Pa cos α
2 ⎞
⎛ 1
≤
k
,
k
≤
1
2
⎜
⎟
2
3
⎝
⎠
15 Stability check – Sliding along the base • Use a base key (normally some main steel) to increase stability 1
Pp = γ 2 D12 K p + 2c2ʹ′ D1 K p
2
D1 > D
16 Example 1 – cont’d • The cross secCon of a canClever retaining wall is shown below. Calculate the factor of safety against sliding. 17 Main steps for stability check 3. Check for bearing capacity failure of the base • Bearing capacity failure (c) • Deep seated shear failure (d) 18 Stability check – bearing capacity !" !!!!" !!!!!!!!!"
R = ∑V + Pa cos α
(
)
M net = ∑ M R −∑ M o
qmin = qheel
M net
∑V
V M y
∑
q=
±
B
e = − CE
2
qmax = qtoe =
CE = X =
net
A
∑V +
e ( ∑V )
( B )(1)
I
B
2 = ∑V ⎛1 + 6e ⎞
⎜
⎟
B ⎝
B ⎠
3
⎛ 1 ⎞
⎜ ⎟ ( B )
⎝ 12 ⎠
V ⎛ 6e ⎞
∑
=
1−
⎜
B ⎝
e 1
≥ , qmin ≤ 0
B 6
⎟
B ⎠
Needs re-proportioning
19 Stability check – bearing capacity 1
qu = c2ʹ′ N c Fcd Fci + qN q Fqd Fqi + γ 2 Bʹ′Nγ Fγ d Fγ i
2
Fcd = 1 + 0.4
Bʹ′ = B − 2e
q = γ 2D
(
Fqd = 1 + 2 tan φ2ʹ′ 1 − sin φ2ʹ′
o
⎛ ψ ⎞
Fci = Fqi = ⎜1 − o ⎟
⎝ 90 ⎠
2
⎛ Pa cos α ⎞
ψ = tan ⎜⎜
⎟⎟
V
⎝ ∑
⎠
o
−1
FS( bearing capacity) =
)
2
D
Bʹ′
D
Bʹ′
Fγ d = 1
⎛ ψ o ⎞
Fγ i = ⎜1 −
⎟
⎜ φ ʹ′o ⎟
⎝
2 ⎠
2
Fγ s = Fqs = Fcs = 1
qu
qmax
20 Example 2 • The cross secCon of a canClever retaining wall is shown below. Calculate the factor of safety with respect to bearing capacity. 21 Example 2 -­‐ soluGon 22 Example 3 A gravity retaining wall is shown below. Use δ’=2φ1’/3 and Coulomb’s acCve earth pressure theory. Determine: a. FS against overturning b. FS against sliding c. the pressure on the soil at the toe and heel. 23 Example 3 -­‐ soluGon 24 Other issues – Joints and Drainages • Joints – ConstrucCon Joints (a) – ContracCon joints (b) – Expansion joints (c) 25 Other issues – Joints and Drainages • Drainage from the backfill – Saturated soil (due to rainfall or other condiCons) may increase the pressure on the wall • Create instability problem – Drainage types • Weep holes (a)/ Perforated drainage pipes (b) Minimum
diameter 0.1m
Filter needed to
prevent wash
26 Other issues – Joints and Drainages • Drainage – Weep hole/perforated pipe – Need filter to prevent wash up of sand/clog – Factors on the choice of filter materials – the grain size distribuCon of backfill materials be such that • The soil to be protected is not washed into the filter • The excessive hydrostaCc pressure head is not created in the soil with a lower hydraulic conducCvity (Terzaghi and Peck, 1976)
F: Filter; B: Base.
D15/D85: the diameters through which 15% and 85% of the soil will pass
27 Other issues • CalculaCon by acCve earth pressure – Need sufficient yielding – Not yielding sufficiently combined with other factors could result in larger lateral forces • Temperature/Ground water fluctuaCon • Adjustment of soil parCcles due to creep and prolonged rainfall • Tidal changes/heavy wave acCon/traffic vibraCon/earthquake – Especially true for • gravity retaining walls/bridge abutments/heavy structures with large secCon modulus – CauCon and judgment needed in choosing the lateral earth pressure in actual design 28 Sheet pile walls • Sheet pile walls – Build conCnuous walls for waterfront structures – Do not need dewatering – Sheet piles also used for temporary supports such as braced cuts (next chapter) 29 Sheet pile walls • Major types – Materials •
•
•
Wooden sheet piles (for temporary light structures) Precast concrete sheet piles Steel (or aluminum) sheet piles – Nature of support •
•
CanClever Anchored 30 Sheet pile walls • Steel sheet piles – Interlocks of sheet pile secCons (a) thumb-­‐and-­‐finger (b) Ball-­‐and-­‐socket joint 31 ConstrucGon methods • Backfilled structure 1. Dredge the in situ soil in front and back of the proposed structure 2. Drive the sheet piles 3. Backfill up to the level of the anchor and place the anchor system 4. Backfill up to the top of the wall • Dredged structure 1. Drive the sheet pile 2. Backfill up to the level of the anchor and place the anchor system 3. Backfill up to the top of the wall 4. Dredge the front side of the wall 32 CanGlever sheet pile walls • Suitable for walls of moderate height (about 6m) • Design of the all – EsCmate of net lateral pressure distribuCon – RotaCng about O – Zone A – Zone B – Zone C 33 PenetraGng Sandy soils Active pressure at z=L1
Active pressure at z=L1+L2
σ 1ʹ′ = γ L1K a
σ 2ʹ′ = (γ L1 + γ ʹ′L2 ) Ka
For z below the dredge line
σ aʹ′ = (γ L1 + γ ʹ′L2 + +γ ʹ′ ( z − L1 − L2 )) Ka
The passive pressure σ ʹ′p = γ ʹ′ ( z − L1 − L2 ) K p
The active pressure
The net pressure
σ ʹ′ = σ ʹ′p − σ aʹ′ = σ 2ʹ′ − γ ʹ′ ( z − L ) ( K p − Ka )
L = L1 + L2
Zero net pressure location
L3 = z − L =
σ 2ʹ′
γ ʹ′ ( K p − K a )
34 PenetraGng sandy soils In the pressure diagram
σ 3ʹ′ = HB = L4γ ʹ′ ( K p − Ka )
At the bottom of the sheet pile the passive pressure (from right to left)
σ ʹ′p = (γ L1 + γ ʹ′L2 + γ ʹ′D ) K p
The active pressure at the same depth σ aʹ′ = γ ʹ′DK a
The net pressure at the bottom of the sheet pile
σ 4ʹ′ = σ ʹ′p − σ aʹ′
= σ 5ʹ′ + γ ʹ′L4 ( K p − K a )
σ 5ʹ′ = (γ L1 + γ ʹ′L2 ) K p + γ ʹ′L3 ( K p − Ka )
D = L3 + L4
35 PenetraGng sandy soils Stability Condition of the wall
⎧⎪∑ FH = 0
⎨
⎪⎩∑ M B = 0
Horizontal force balance
∑F = 0
H
1
1
ʹ′
P − σ 3 L4 + (σ 3ʹ′ + σ 4ʹ′ ) L4 = 0
2
2
AACDE − AEFHB + AFHBG = 0
P = AACDE
Moment balance about B
L5
L4 1
1
ʹ′
ʹ′
ʹ′
P ( L4 + z ) − σ 3 L4 + (σ 3 + σ 4 ) L5 = 0
2
3 2
3
σ 3ʹ′ L4 − 2 P
L5 =
σ 3ʹ′ + σ 4ʹ′
D = L3 + L4
36 PenetraGng sandy soils The following 4-th order equation for L4
L44 + B1 L34 − B2 L24 − B3 L4 − B4 = 0
B1 =
σ 5ʹ′
The actual depth of penetration is
increased by 20%-30%
γ ʹ′ ( K p − K a )
8P
B2 =
γ ʹ′ ( K p − K a )
B3 =
B4 =
6 P ⎡⎣ 2 z γ ʹ′ ( K p − K a ) + σ 5ʹ′ ⎤⎦
γ ʹ′2 ( K p − K a )
2
P [6 z σ 5ʹ′ + 4 P ]
γ ʹ′ ( K p − K a )
2
2
37 PenetraGng sandy soils Calculation of maximum bending moment – point of zero shear
1 2
P = zʹ′ γ ʹ′ ( K p − K a )
2
z ʹ′ =
2P
γ ʹ′ ( K p − K a )
1
zʹ′
2
ʹ′
ʹ′
ʹ′
M max = P ( z + z ) − γ z ( K p − K a )
2
3
Section modulus of sheet pile
S=
M max
σ all
Allowable flexural
stress of the sheet pile
38 Example 4 •
The figure below shows a canClever sheet pile wall penetraCng a granular soil. L1=2m, L2=3m, γ=15.9kN/m3,γsat=19.33kN/m3,φ’=32o. a. Determine the required D. b. For a 30% increase in D, what is the required total length of the sheet piles? c. What is the minimum secCon modulus of the sheet piles? Use σall=172MN/m2. 39 Example 4 Left Blank for Note Taking 40 PenetraGng clay Active pressure at z=L1
Active pressure at z=L1+L2
σ 1ʹ′ = γ L1K a
σ 2ʹ′ = (γ L1 + γ ʹ′L2 ) Ka
For z below the dredge line:
undrained condition
φ =0
Ka = K p = 1
Active pressure above rotation point
σ a = (γ L1 + γ ʹ′L2 + +γ sat ( z − L1 − L2 )) − 2c
σ p = γ sat ( z − L1 − L2 ) + 2c
σ 6 = 4c − (γ L1 + γ ʹ′L2 )
At the bottom
σ 7 = 4c + (γ L1 + γ ʹ′L2 )
41 PenetraGng clay Stability Condition of the wall
∑F = 0
H
∑M = 0
B
AACDE − AEFIB + AGIH = 0
∑F = 0
H
1
P1 − σ 6 D + (σ 6 + σ 7 ) L4 = 0
2
D = L3 + L4
P1 = AACDE
Dσ 6 − P1
L4 =
4c
∑M = 0
B
L4
1
1
2
P1 ( D + z1 ) − σ 5 D + (8c ) L4
=0
2
2
3
Combine the two equations
P1 ( P1 + 12cz1 )
σ 6 D − 2 DP1 −
=0
σ 7 − 2c
2
42 PenetraGng clay Major steps: Step 1: Calculate K a for the granular soil;
Step 2: Calcuate σ 1ʹ′ and σ 2ʹ′ ;
Step 3: Calcualte P1 and z1 ;
Step 4: Calcuate σ 6 and σ 7 ;
Step 5: Solve D;
Step 6: Calcuate L4 ;
Step 7: Draw the pressure distribution diagram;
Step 8: The actucal depth of penetration is Dact = (1.4 : 1.6 ) D.
43 PenetraGng clay Calculation of maximum bending moment
Point of zero shear
P1 − σ 6 zʹ′ = 0
L1 + L2 < z < L1 + L2 + L3
P1
ʹ′
z =
σ6
1
M max = P1 ( z1 + zʹ′ ) − σ 6 zʹ′2
2
Section modulus of sheet pile
S=
M max
σ all
Allowable flexural
stress of the sheet pile
44 Example 5 • For the sheet pile shown below, determine: a. The actual depth of penetraCon use 1.5 Cme of the theoreCcal value. b. The minimum size of sheet pile secCon necessary. Use σall=172MN/m2. 45 Example 5 46