1. The Engineering of Foundations
In this lecture
Design of Retaining Wall Examples
Sheet Piles Walls
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The Eng. of Foundations

2. Retaining Walls
Check the stability of the retaining wall shown in the Fig.
qall = 2.5 kg/m2
0.3 m
q =10 kN/m2 W3 W2 W1
Sand
γ= 18kN/m3
ϕ=30 0
5.8 m P1
0.8 m P2 W4
1.2 m Pp W5
0.7 m C
1.5 m
3 m
Original soil
γ= 19 kN/m3, c = 35 kPa
ϕ=34 0
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The Eng. of Foundations

3. Retaining walls Solution Calculate earth pressure coefficient Ka, Kp
Ka = (1-sinφ) / (1+sin φ) = 0.33, Kp = 3.5
Identify the forces that affect the stability of the retaining wall calculate the active and passive
At z = 0 m the lateral stress is σ = q Ka = 10 * 0.33 = 3.33 kN/m2
Then, P1 = 7 * 3.33 = kN/m
At z = 7 m, P2 = 0.5 (18)*0.33* (7*7) = 147 kN/m
Passive earth pressure is
σp = γ z Kp
At z = 1.2 m, Pp= 0.5 * 19* 3.5* (1.2*1.2) = kN/m
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The Eng. of Foundations

4. Design of Retaining Wall
Organize a table
Calculate overturning moment = P1 * P2 * 7/3 = 424 kN.m
Fs = MR / M0 = 1355 / 424 = 3 > 1.5 OK
Area
Weigh/unit length
Moment arm from C
Moment about C (kN.m) 1
18 * 6.3 * 3 = 340.2 3
1020.6 2
6.3 * 0.3 * 22 = 41.6
1.35
56.13
0.5 * 0.4 * 6.3 * 22 = 27.7
1.07
29.70 4
19 * 0.5 * 0.8 = 7.6
0.40
2.88 5
4.5 * 0.7 * 22 = 69.3
2.25
155.93
Surcharge
10 * 3 = 30 90
∑ V = 516 kN
∑ MR = 1355 kN.m
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The Eng. of Foundations

5. Design of retaining walls
Check for sliding
Sliding forces = P1 + P2 = 170 kN/m
Resisting forces = R + Pp
R = B * 0.67 * tan 2/3 φ = tan 23
= kN/m
Fs = / 170 = 2 OK
Check bearing capacity
Calculate the moment of all forces about C
1355 – 424 = ∑ V X X = 931 / 516 = 1.8 m
Therefore , e = 4.5 / 2 – 1.8 = 0.45 < B/6 OK
The pressure under the base of the retaining wall is calculated as
= V / B (1 ± 6e / B)
= (516/ 4.5) (1 ± ((6 * 0.45) / 4.5))
= 184 kN/m2 (1.84 kg/cm2), 46 kN/m2
qu < qall OK.
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The Eng. of Foundations

6. Design of retaining walls
Example 2: For the cross section of the retaining wall shown in the Fig. calculate the factor of safety with respect of overturning and sliding
1.5 ft
1.06 ft
100 W2 W1 W4
Sand
γ = 117 Ib/ft3
C = 0
ϕ =30 0
18 ft Pv Pa Ph
4 ft W5
4 ft Pp W3
2.75 ft C
4 ft
2.5 ft
6 ft
Original soil
γ= 107 Ib/ft3, c = 900 Ib/ft2
ϕ=18 0
28/2/2014
The Eng. of Foundations

7. Design of retaining walls
Checking for overturning
Total H = tan 100
= ft
Pa = 1/2 γH2 Ka, For φ = 300 , α = 100 , the value of Ka is 0.355
Pa = 0.5 (117)(21.81) 2 (0.355) / 1000 = 9.88 K/ft
Pv = Pa sin 100 = K/ft
Ph = Pa cos100 = 9.73 K/ft
Area
Weight K/ft
Moment arm from C (ft)
Moment about C (K.ft) 1
1.5*18*0.15 = 4.05
5.75
23.29 2
0.5*1*18*0.15 = 1.35
4.67
6.3 3
12.5*2.75*0.15 = 5.16
6.25
32.23 4
(( )/2) (6)(0.117) = 13.01
9.6
124.8
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The Eng. of Foundations

8. Design of retaining walls
The overturning moment (M0)
= Ph (H / 3) = 9.73 (21.81 / 3) =70.74 K.ft/ft
Fs = MR / M0 = / = 2.93 > 2 OK
Check for sliding
Fs (sliding) = (∑V tan (K1φ) + B K2 C2 +Pv+ Pp ) / Pa cos α , let K1 = K2 = 2/3
= ( tan (0.67 * 18) + (12.5 * 0.67 * 0.9) (1.622)/9.73
= 1.67 > 1.5 OK
Area
Weight K/ft
Moment arm from C (ft)
Moment about C (K.ft)
Pv = 1.715
12.5
21.44
∑ V =
∑ MR =
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The Eng. of Foundations

9. Sheet piles walls (c) (b) (a) Fig.1
They can be defined as a group of piles made of timber, steel or concrete set close together to resist lateral pressure as from earth or water. They are often used for water front structures. The construction of the sheet piles does not usually require dewatering of the site.
Sheet Piles Construction Method
Three construction procedures are mainly used to install sheet piles walls . The sheets are driven into the ground for a certain depth of embedment. This type of sheet piles is referred to as cantilever sheet piles as shown in the Fig.1 (a). If the cantilever pile, as in (b) is provided with anchors they are classified as anchored sheet piles. The braced sheet piles are other type of the piles as shown in (c).
Wales
Anchor
Sheet pile
Dredge line
Dead man
Struts
(c)
(b)
(a)
Fig.1
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The Eng. of Foundations

10. Sheet piles walls Types of sheet piles based on materials they made of
Timber sheet piles
This type of piles used for short spans, light lateral loads and commonly
for temporary structures. They are fabricated in forms shown in the
Fig. 2 the disadvantage of this type is that they require caps and if they are drive into hard soil layer they tend to split.
Fig.2
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The Eng. of Foundations

11. Sheet piles walls Reinforced concrete sheet piles
This type of piles is used to provide lateral support for large spans. The shortcoming of this type is that they require more driving efforts because of their size. Their transportation cost is high due to their weight. This type of sheet piles is manufactured in standard size as shown in Fig. 3.
Fig.3
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The Eng. of Foundations

12. Sheet Piles Walls Steel sheet piles
They are the most common type because of these advantages:
High resistance to driving stress
Relatively light weight
Can be reused several times
Long service life
Easy to increase the pile length
They are available in different shapes
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The Eng. of Foundations

13. Sheet Piles Walls
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The Eng. of Foundations

14. Sheet Piles Walls
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The Eng. of Foundations