Analysis & Design of the supported deep excavation structures

1. Supported Deep Excavation
• Eng. Khaled Sobhy Emam
• Demonstrator, Faculty of Engineering – Ain Shams University
• Khaled.Sobhy@eng.asu.edu.eg
2020-2021

2. Revision on the calculations of the
Lateral Earth Pressure (L.E.P)

3. Active Side
Passive Side
𝐾𝑎 =
1 − sin∅
1 + sin∅
𝐾𝑝 =
1
𝐾𝑎
𝐾𝑜 = 1 − sin∅

4. Pa1
𝐾𝑎 =
1 − sin∅
1 + sin∅
𝐾𝑝 =
1
𝐾𝑎
𝐾𝑜 = 1 − sin∅
Pa2
𝜎𝑎 = 𝐾𝑎(𝛾𝑒𝑓𝑓. ℎ + 𝑞) − 2𝑐 𝐾𝑎
Pp1
Pp1
𝜎𝑝 = 𝐾𝑝(𝛾𝑒𝑓𝑓. ℎ + 𝑞) + 2𝑐 𝐾𝑝

6. 4.756
8.079
52.285
31.44 125

7. Supporting Deep Excavation

8. Retaining Walls

9. SPW

10. Soldier Piles & Lagging
I-beams

11. RC Piles
1- Adjacent Piles
D
Free
distance
D
Spacing (S) = D + free distance
Ground Water (or) fine sand

13. RC Piles
2- Tangent Piles
Spacing (S) = D
D
Ground Water

14. RC Piles
3- Secant Piles
Spacing (S) = DRC + Dplastic – 2 overlaps

15. Diaphragm Walls

16. What is the recommended shoring system ?!
Top Ground Surface
Loose to medium
sand
Very dense gravely
sand
Gravel
Tangent RC Pile Wall

17. What is the recommended shoring system ?!
Top Ground Surface
Soft clay
G.W.T
Sheet Pile Walls
Medium Sand

18. Top Ground Surface
Bottom of Excavation
Earth Side
Air excavation side
H
L
t
H = Excavation depth
t = Penetration depth

19. Cantilever System

20. Free end
Rigid connection
We need to calculate the required penetration
depth to make the bottom connection totally
fixed and prevent any rotation about it.

21. B.M.D
S.F.D
Z
Mmax
O
C
d
tsafe = d x 1.2 x 1.2
To achieve fixation
F.O.S

22. (C) in the undrained condition is bigger than it in the drained condition, so the critical is to use it in the
drained condition
The load is function in the height, so it’s a first-degree equation … the S.F.D is a second-degree equation,
and B.M.D is a third-degree equation

23. Example
G.W.T
Sand
Silty Clay
𝛾 = 18𝐾 Τ
𝑁 𝑚3
∅ = 34𝑜
𝛾 = 19𝐾 Τ
𝑁 𝑚3
𝐶𝑢 = 30𝐾𝑃𝑎
𝐶′ = 5𝐾𝑃𝑎
∅′ = 20𝑜
2 m
3 m

24. G.W.T
Sand
Silty Clay
𝛾 = 18𝐾 Τ
𝑁 𝑚3
∅ = 34𝑜
𝛾 = 19𝐾 Τ
𝑁 𝑚3
𝐶𝑢 = 30𝐾𝑃𝑎
𝐶′ = 5𝐾𝑃𝑎
∅′ = 20𝑜
1
2
3
4
5
6
2 m
3 m
𝜎𝑎1 = 𝑧𝑒𝑟𝑜
𝜎𝑎2 = 3𝑥8𝑥0.28 = 6.79𝐾𝑃𝑎
𝜎𝑎3 = 3𝑥8𝑥0.49 − 2𝑥5 0.49 = 4.76𝐾𝑃𝑎
d
𝜎𝑎4 = 4.76 + (9𝑑𝑥0.49) = 4.76 + 4.41𝑑
𝜎𝑝5 = 2𝑥5 2.04 = 14.28𝐾𝑃𝑎
𝜎𝑝6 = 14.28 + (9𝑑𝑥2.04) = 14.28 + 18.36𝑑

25. 14.28
4.76
4.76+4.41d
6.79
14.28+18.36d
50
125
50d
10.18
2.21d^2
4.76d
14.28d
9.18d^2
O

26. 125
50d
10.18
2.21d^2
4.76d
14.28d
9.18d^2
O
෍ 𝑀@𝑜 = 𝑧𝑒𝑟𝑜
⇒ 10.18(1 + 𝑑) + 4.76𝑑(
𝑑
2
) + 2.21𝑑2(
𝑑
3
) + 125(
5
3
+ 𝑑) + 50𝑑(
𝑑
2
) − 14.28𝑑(
𝑑
2
) − 9.18𝑑2(
𝑑
3
) = 𝑧𝑒𝑟𝑜
⇒ −2.35𝑑3 + 20.24𝑑2 + 135.18𝑑 + 218.52 = 𝑧𝑒𝑟𝑜

27. 10.18
2.21d^2
4.76d
14.28d
9.18d^2
125
19.5 m
Z
Any factor fn. (d) let it fn. (Z) and recalculate as following to get the point of the max. B.M
෍ 𝐹𝑥 = 𝑧𝑒𝑟𝑜
⇒ 10.18 + 125 + 4.76𝑍 + 2.21𝑍2 + 125 + 50𝑍 = 9.18𝑍2 + 14.28𝑍
⇒ −6.97𝑍2 + 40.47𝑍 + 135.1 = 𝑍𝑒𝑟𝑜
∴ 𝑍 = 8.17𝑚
Point of zero shear (max. B.M) shall be below the dredge line

28. ∴ 𝑀max = −2.35 8.17 3 + 20.24 8.17 2 + 135.18(8.17) + 218.52 = 1392.4𝐾𝑁. 𝑚
Design the shoring system in the following cases …
SPW and fs = 140:160 MPa Sx = Mmax / fs … and get section from tables
Diaphragm wall Get d from C1 = 3.5 and then get As
Piles
Design as circular columns, but in the case
of secant piles design the RC pile only
In the case of the piles .. Multiply moment x spacing to get the acting moment / pile

29. Anchored Systems

30. We use this system if the depth of excavation is very high, which will cause high lateral deformations for the
wall system if it behaves as cantilever system
The reasons which may push me to use another system instead of the cantilever system are …
1- Very deep excavation height.
2- Very high lateral loads (soil – surcharge – ground water - ……).
3- The cantilever system develops large B.M & needs very big penetration depth.

31. The anchored system may be divided to
Anchored-free system Anchored-fixed system
Free = 1.2d Fixed = 1.44d

32. Free-anchored System

33. Driven
Resisting
Resisting

34. To get the reaction on the anchor, multiply the reaction on
the spacing between them

35. Example
G.W.T
Sand
Silty Clay
𝛾 = 18𝐾 Τ
𝑁 𝑚3
∅ = 34𝑜
𝛾 = 19𝐾 Τ
𝑁 𝑚3
𝐶𝑢 = 30𝐾𝑃𝑎
𝐶′ = 5𝐾𝑃𝑎
∅′ = 20𝑜
2 m
3 m
For the same previous example … determine the max. B.M if it’s anchored-
free system
d

36. 14.28
4.76
4.76+4.41d
6.79
14.28+18.36d
50
125
50d
10.18
2.21d^2
4.76d
14.28d
9.18d^2
FA
a

37. ෍ 𝑀@𝑎 = 𝑧𝑒𝑟𝑜
ቇ
⇒ 125(
2
3
𝑥5 − 2) + 10.81(2) + 50𝑑(0.5𝑑 + 3) + 4.76𝑑(0.5𝑑 + 3) + 2.205𝑑2(
2
3
𝑑 + 3
−14.28𝑑(0.5𝑑 + 3) − 9.18𝑑2(
2
3
𝑑 + 3) = 𝑧𝑒𝑟𝑜
⇒ 𝑑 = 5.82𝑚 → 𝑡 = 1.2𝑑 = 1.2𝑥5.82 ≃ 7𝑚
∵ ෍ 𝐹𝑥 = 𝑧𝑒𝑟𝑜
⇒ 125 + 50𝑥5.82 + 10.81 + 4.76𝑥5.82 + 2.205𝑥5.822 − 𝐹𝐴 − 14.28𝑥5.82 − 9.18𝑥5.822 = 𝑧𝑒𝑟𝑜
∴ 𝐹𝐴 = 139.6𝐾 Τ
𝑁 𝑚

38. To get the point of zero shear, calculate σ 𝑭𝑨 above the dredge line,
and you will get case of the followings …
1- If active force > FA … Point of zero shear is above
the dredge line.
2- Otherwise .. It’ll be below the dredge line

39. 125
291
10.18
74.86
27.7
83.11
310.95
139.6
0.113 m
5.82 m
ቇ
𝑀max = 291(
5.82
2
− 0.113) + 74.86(
2
3
𝑥5.82 − 0.113) + 27.7(
5.82
2
− 0.113
−83.11(
5.82
2
− 0.113) − 310.95(
2
3
𝑥5.82 − 0.113) = −230.41𝐾𝑁. Τ
𝑚 𝑚

40. System Cantilever System Anchored-free system
Penetration depth (d) 13.52 m 5.82 m
Max. B.M 1392.4 KN.m/m 230.41 KN.m/m

41. Fixed-anchored System

42. Active
Passive
Fa
d
Point of zero moment
Given

43. X for zero moment
X for zero moment
QA
F
QA
d

44. X
H
There are two points of zero shear, above and below the dredge line
… so there are two points of max. B.M for the system

45. Example
G.W.T
Sand
Silty Clay
𝛾 = 18𝐾 Τ
𝑁 𝑚3
∅ = 34𝑜
𝛾 = 19𝐾 Τ
𝑁 𝑚3
𝐶𝑢 = 30𝐾𝑃𝑎
𝐶′ = 5𝐾𝑃𝑎
∅′ = 20𝑜
2 m
3 m
For the same previous example … determine the max. B.M if it’s anchored-fixed
system, and the point of zero moment is below the dredge line by 0.5 m
D
0.5 m

46. 14.28
4.76
4.76+4.41d
6.79
14.28+18.36d
50
125
50d
10.18
2.21d^2
4.76d
14.28d
9.18d^2
FA
a
d
0.5 m
D

47. 14.28
4.76
6.79 50
125
10.18
FA
a
0.5 m
6.965
23.46
25
2.38
0.551
7.14
2.295
2 m
3 m
M
M@O = zero … 125x(5/3+0.5)+25x0.25+10.18(3/3+0.5)+2.38x(0.25)+0.551(0.5/3)-Fa(3.5)-7.14(0.25)
-2.295(0.25/3) = zero
Fa = 83.16 KN

48. 14.28
4.76
6.79 50
125
10.18
83.16
a
0.5 m
6.965
23.46
25
2.38
0.551
7.14
2.295
2 m
3 m
QM
σ 𝐹𝑥 = 𝑧𝑒𝑟𝑜 … 83.16+7.14+2.295+Qo-125-25-10.18-2.38-0.551 = zero
QM = 70.516 KN

49. 50
6.965
23.46
D
70.516
23.46+18.36D 6.965+4.41D
50D
2.21D^2
4.76D
14.28D
9.18D^2
O
Moment @ O = Zero … 70.516(D)+4.76D(D/2)+2.21D^2(D/3)+50D(0.5D)-14.28D(D/2)
-9.18D^2(D/3) = Zero
-2.323D^3 +20.24D^2 + 70.516D = Zero
D = 11.38 m … d = 0.5+11.38 = 11.88 m …. t = 1.44 x 11.88 = 17.1 = 17.5 m

50. For The Max. B.M
FOR THE UPPER PART
To get the point of zero shear, calculate σ 𝑭𝑨 above the dredge line, and you will get case of the followings …
1- If active force > FA … Point of zero shear is above
the dredge line.
2- Otherwise .. It’ll be below the dredge line
Active = 125 + 10.18 =135.18 KN > Fa = 83.16 KN …. The point of zero shear is above the dredge line

51. 5Z^2+20Z+20
1.12Z^2
83.16
a
2 m
3 m
Z
Z = 1.97 m
M = -56.6 KN.m/m

52. For the lower Part 70.516
2.21Z^2
4.76Z
14.28Z
9.18Z^2
50Z
Z = 6.38 m
M = 528 KN.m/m
Mmax = 528 KN.m/m

53. Design of Lateral Support of the
Anchored System

54. Wale beam
S
S
Lateral Support
Strut
Tie-back

55. Strut
C = F x S
Design as a compression member and the critical is BUCKLING
Tie-back
T =
𝐹 𝑥 𝑆
𝐶𝑂𝑆 𝛼
Lt = Lf + Lg

56. How to determine total length of the tie back ?!
α = 20:40
45 + ø/2
H
h ≥ 5 m

57. α = 20:40
H
𝑇 = 𝑓
𝑠𝑥𝐴𝑠
𝑇 = 𝜎𝑛tan𝛿𝑥𝜋𝑑𝐿𝑔
𝜎𝑛 = (෍ 𝛾𝑥ℎ𝑐.𝑔 + 𝑞)𝑥cos𝛼
ℎ𝑐.𝑔 = ℎ +
𝐿𝑔
2
sin𝛼
∴ 𝐿𝑔 =. . . . .
h

58. Braced Excavation System

60. In the case of sand

61. In the case of Clay

64. Example