Slope project daniel

Slope Stabilizing
Environmental and protective structures
2015-2016
POLITECNICO DI MILANOCivil engineering for risk mitigation
SEYED MOHAMMAD SADEGH MOUSAVI 836154
DANIEL JALILI 832852
Prof. Di Prisco
Prof. Galli
1

Contents
The Most
Critical Circular
Slip Surface
Different water
tables
Geoslope Vs.
Fellenius
Active stabilizing
system
Rigid Square
bearing plates
A deep ground
anchor
Optimize
Position of
system
Passive
stabilizing
system
Global safety factor dependency with
respect to the soil displacement amplitude
FRC Vs. Plain
Concrete
Advantages
Results Observations
2

General information
POLITECNICO DI MILANO
Introduction
The proposed design consists of two
main components:
 Plate element
 Grout anchor
Nailing
Anchored Bulkhead
The Experimental Program
Comparison (Stiffness & Ductility) of FRC Vs. RC
Assumption
No Redundancy effect (in case of Safety)
Civil engineering for risk mitigation
3

High Toughness in case of SFRC (S1) Vs. R/C (P1)
Parameters
Comparable Peak (S0C, S1C & R0B)  Crack Occurs (R0B)
Very low carrying capacity (S0A)
Post tensioning  (S1C: Ductile & R0B: Brittle)
Sudden Break
Observations
1st
2nd
3rd
High Performance of
SFRC (HPSFRC)
Very High local
toughness
Compression contribution
Tension Contribution
S1B (4 strands) P1B (4 strands - PC)
4th
Material characteristics of HPSRFC
Mix design of HPSRFC of the plate
fresh state properties of SCC
4

Parameters
Design Material
(Concrete)
Better Performance in comparison of pozzolan
cement (Very Low Porosity along with fiber & low
permeability  High Durability)
Self Compacted
Concrete
 Concrete Type  C30/C37
 Min Cement  300 kg/m3
 Additive  Avoid early Cracking
Standard
Prescription
Method
Avoiding Contaminant (Controlling
water) & 0.4% (Fiber) of total volume
Water & Fiber
Characteristic
Considering time curing & workability
High Flow
Ability
Hardening state
Crack Tip Opening Displacement (mm)
NominalStress(MPa)
Based on the axial test, the mix design of hardened state:
Average Value is considered (RED Curve)
5

Materials
Dimension
Plate Details
 Square plate (HPSFRC)
 80*80*24 cm (Standard method EN 206)
 Weight of Plate: 4 KN
 Rebar's Diameters: 25 mm
 Tensile Strength of steel: 450 N/mm2
 Opening in the middle of plate to Installation
of Anchor & Head’s attachment.
Ultimate Capacity of the plate
Experimental and FEM analysis
Experimental test of the HPSFRC plate, supported at three
points and subjected to a concentrated load at the center
Observation
Since the experimental tests were not performed until
failure condition, therefore FEM analysis was performed
to check the ultimate load of the plate. As you can see in
the front graphs.
Overestimation Rule
Comparison of the results between the experiment & FEM analysis
Ultimate load of the plate based on FEM analysis based on Hordijk tension
softening & multidirectional fixed crack model
6
Impossibility to predict the
real B.C due to roughness of
the morainic soil surface
Preventing any local
cracking due to
concentrating loads

Calculations
Ultimate Moment based on a linear softening model
The linear elastic-softening model  estimating the ultimate moment of the plate based on the bending tests.
Two reference values: 𝑓𝐹𝑡𝑢 is the ultimate residual tensile strength in uniaxial tension and is equated as follows.
According to the front figure, two design parameters were defined
𝑙 𝑐𝑠: Kinematic module which is the distance between 2 cracks
(Assumption: Distributed between 2 openings)
𝑘 𝑎: Ranges between 0.362 & 0.378
(Average value of 0.37 can be considered) as shown in the front figure:
ULS state
Concentrated compressive stress
distribution at upper fibers
Crack Opening  2.5 mm
(Linear Distribution of Stress)
𝑤 = 0 → 𝑘 𝑏 𝑓𝑅1
𝑤𝑖1 = 0.5 𝑚𝑚 → 𝑘 𝑎 𝑓𝑅1
𝑤𝑖2 = 2.5 𝑚𝑚 → 𝑓𝐹𝑡2.5
Rotational Equilibrium
Consideration & (Linear
Model)
& considering 𝒌 𝒃
& 𝒇 𝑹𝟑 = 𝟎. 𝟓𝒇 𝑹𝟏
considering the average value from
the experiment (Page 6)
𝑓𝑅1 = 𝑓1,𝑒𝑞 0−0.6 = 12.06 𝑀𝑃𝑎
𝑓𝑅3 = 𝑓1,𝑒𝑞 0.6−3 = 9.76 𝑀𝑃𝑎
𝑓𝐹𝑡𝑠,𝑚 = 0.45𝑓𝑅1,𝑚 = 0.45 × 12.06 = 5.43 𝑀𝑃𝑎
𝑓𝐹𝑡𝑠,2.5𝑚 = 0.5𝑓𝑅3,𝑚 − 0.2𝑓𝑅1,𝑚 = 0.5 × 9.76 − 0.2 × 12.06 = 2.47 𝑀𝑃𝑎
7

Calculations
𝑙 𝑠 = ℎ = 240 𝑚𝑚 (𝑃𝑙𝑎𝑡𝑒 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠)
𝑤 𝑢 = min 2.5 𝑚𝑚; 𝜀 𝑐𝑢 = 0.02𝑙 𝑠 = 2.5 𝑚𝑚
The ultimate moment considering only the fiber reinforced part
The ultimate moment considering only the steel rebar
The total ultimate moment of the plate
Longitudinal & Rotational equilibrium in
the linear model
8

Calculations – Ultimate Bearing Load
Failure Mechanisms
Assumption
Material is rigid perfectly plastic
& Max bending moment:
Mu = 195.5 KN. m/m
Failure Mechanism 1
Failure Mechanism 2
Failure Mechanism 3
External work
Le = P. δ
Internal work
Li = 𝐿1 𝑀 𝑢 𝜃1 + 𝐿2 𝑀 𝑢 𝜃2
= 𝐿1 𝑀 𝑢
𝛿
𝑅1
+ 𝐿2 𝑀 𝑢
𝛿
𝑅2
= 630𝑀 𝑢
𝛿
175
+ 630𝑀 𝑢
𝛿
350
Le=Li & 𝛿
𝑃 = 1058𝐾𝑁 ≈ 108 𝑇𝑜𝑛
External work
Le = P. δ
Internal work
Li = 𝐿1 𝑀 𝑢 𝜃1 + 𝐿2 𝑀 𝑢 𝜃2 + 𝐿3 𝑀 𝑢 𝜃3
=630𝑀 𝑢
𝛿
𝑅1
𝑐𝑜𝑠27° + 754𝑀 𝑢
𝛿
𝑅2
𝑐𝑜𝑠34°
+ 754𝑀 𝑢
𝛿
𝑅3
𝑐𝑜𝑠29°
𝑅1=𝑅2=336, 𝑅3 =265
Le=Li & 𝛿
𝑃 = 1179𝐾𝑁 ≈ 120 𝑇𝑜𝑛
Punching
There is uncertainty  Punching
behavior of plate due to bending
 Cause more cracks around the
opening  Much higher (Pu)
9
Using a factor of 1.6

Calculations – Safety Factor (Geoslope)
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
SAFETYFACTOR
WATER HEIGHT [M]
Factor of Safety
Fellenius Bishop Spencer Janbu
Fellenius Bishop Spencer Janbu
0 5 1.23 1.22 1.22 1.16
3 8 1.23 1.22 1.22 1.16
5 10 1.23 1.22 1.22 1.16
7 12 1.23 1.21 1.21 1.14
9 14 1.17 1.12 1.12 1.05
11 16 1.11 1.01 1.01 0.93
13 18 1.05 0.89 0.9 0.78
15 20 0.97 0.76 0.77 0.62
Safety Factor
Hw [m+asl]Hw [m]
10
Unstable Zones
Unknown water level Safety Factor  1.1 (9m)
Critical Condition
(Geoslope)
Safety Factor  0.972 (16m)
Starting design Point  SF<1.3

Calculations – Safety Factor
Fellenius
hypothesis
Circular slip is
homogenous
No interaction between
slices (Hi=Vi=hi=0)
Slip surface of each slice [m]
𝑧𝑓𝑖 = 𝑧 𝑐 − 𝑅2 − 𝑥𝑖 − 𝑥 𝑐
2
R: Radius of Rotational Failure
𝑥 𝑐=𝑧 𝑐: Radius axis coordination
Slice Weight [m]
𝑊𝑖 = 𝛥𝑧𝑖 + 𝛥𝑧𝑓𝑖 𝛥𝑥𝑖 𝛾𝑠𝑎𝑟
𝛥𝑥𝑖: Slice Width [m]
Half of the base [m]
ai =
Δxi
2cosαi
Inclination angle of the slice [° ]
αi = arctan
Δzfi
Δxi
Average Hydrostatic Pressure
at base of the slice [KN/m]
Ubi =
Δzwi − Δzfi
2
γwater
Δzfi > Δzwi → Ubi = 0
Average Hydrostatic Force at base of the slice [KN/m]
𝑈𝑖 =
𝑈𝑏𝑖 𝛥𝑥𝑖
cos𝛼𝑖
Average Lateral Hydrostatic Force [KN/m] at base of the
slice & as Lateral hydrostatic force for the next slice
𝑈 )𝑖,𝑠(𝐿𝑒𝑓𝑡 𝑜𝑟𝑈 )𝑖,𝑑(𝑅𝑖𝑔ℎ𝑡 =
𝛥𝑧𝑤𝑖 − 𝛥𝑧𝑓𝑖
2
𝛾 𝑤𝑎𝑡𝑒𝑟
Equation along the normal axis of the slice is the total normal force
𝑁𝑖
′
+ 𝑈𝑖 = 𝑊𝑖cos𝛼𝑖 + 𝑈𝑖,𝑠 − 𝑈𝑖,𝑑 sin𝛼𝑖
𝑁𝑖
′
= Effective Normal Force
Failure Criterion based on equilibrium
𝑇𝑖 = 𝑁𝑖
′
tan𝜙′ +
𝑐′𝛥𝑥𝑖
cos𝛼𝑖
Safety Factor based on the global
equilibrium
𝐹S =
𝑅 𝑇𝑖
𝑅 𝑊𝑖sin𝛼𝑖
Global Equilibrium
R Wisinαi + Qicosβi xp,i − xc = R Ti − Qisinβi zc − zp,i
Qi: The magnitude of the external load, βi: Inclination of external Load
xp,i = zp,i: 𝐶𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
Safety Factor based by considering the external load
𝐹𝑆 =
𝑅 𝑇𝑖
𝑅 𝑊𝑖sin𝛼𝑖 + 𝑄𝑖 cos𝛽𝑖 𝑥 𝑝,𝑖 − 𝑥 𝑐 − 𝑠𝑖𝑛𝛽𝑖 𝑧 𝑐 − 𝑧 𝑝,𝑖
In case of External Load
11

Calculations – Safety Factor
0.7
0.8
0.9
1
1.1
1.2
1.3
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
SAFETYFACTOR
WATER HEIGHT [M]
Excel
Geoslope
Ratio
Observation
 Lower value in Excel Vs. Geoslope
 Average ratio between 2 approaches  1.15
(Due to probably by numerical errors of excel)
Critical Safety Factor
(Using)  0.885
12

Design – Active stabilizing system
Assumption
 S=1.5m, Distance of 2 plates in plane axis
 No Group Effect
 Shear action on strands is neglected
 Uniform applied load (Pa) acting on plate 
Multiplying with 1.6 factor
 Interaction curve for 0.8×0.8×0.24 m plate
 Applied load ⦜ Plate
 Ground anchor ⦜ Slope
Active Stabilizing System
• Slope Stability Evaluation based on Intervention
Pu in Critical Condition
• Anchors Design  Peak of linear behavior of
Interaction Curve
Intervention of External load
𝜎 ⇒
𝑃𝐶𝑢𝑟𝑣𝑒 = 1000 𝐾𝑃𝑎
𝑊 = 18 𝑚𝑚
𝑃𝑙𝑎𝑡𝑒 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛
𝑃𝑎 = 1000 0.8 × 0.8 = 640
𝐾𝑁
𝑎𝑛𝑐ℎ𝑜𝑟
𝑆=1.5 𝑚
𝑃𝑎 =
640
1.5
= 427
𝐾𝑁
𝑚′
𝑃𝑎 = 427
𝐾𝑁
𝑚;
 Pre-Stressed Load
Optimizing position of plates (Assumptions)
 Safety factor > 1.3  5 Plates
 The max applied load for each plate  427 KN/m’
 The distance of 2 plates in plane  S= 1.5, 2, 2.5 m
 Horizontal distance = 1,5 m  (No Group
effect)
13

Design – Plate Positions
S=1.5 m
(Best Configuration )
Applied load acting on the center
axis of the plate
Final Analysis
of SF
Assumption: Applied load is
uniform over the plate
Load = 53 KN/m’
over 0.8 m (width) SF = 1.51
14
S=1.5 m
SF=1.52
S=2 m
SF=1.46
S=2.5 m
SF=1.28

Design – Anchors
Anchors Dimensions
Type  Permanent anchors
 Free Length
 Bulb Length
 Grout injection method
Components & Total Length of Anchor
 Lfree  Length of Tendon + Plate Thickness
(240mm) (Anchor Head)
 𝐿𝑡𝑓 =
𝐿𝑓𝑟𝑒𝑒+𝐿𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡
3
 Actual free length
 Lconstraint  The length of the grout
 𝐿𝑡𝑏 =
2
3
× 𝐿𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡  Actual Grout length
 Le=1000mm  Min Addition Length (attachment to
pull out device)
Location
 Avoid bonding in the slip surface  Grout is placed outside
the unstable zone: (Because)
- Avoid failure risk due to interaction with slide body with
small movement
- Increasing in pre-stress action by soil movement
- Partial movement of anchor head by soil movement
 Anchor ⦜ Slope
15
Anchor Characteristics
 Steel Quality  MW 450
 Area of each tendon  861 mm^2
 Nominal diameter of each tendon  38 mm
 Spiral diameter used for grouting  250 mm
 Gout prop diameter  0.27 mm
 Inclination  0°
 Distance between 2 anchors  1.5 m
 Soil Thickness above the grout  7.5 m
 Plate thickness  0.24 m
 Grout Length or Constraint  9 m
 Free Length  7.74 m
 Additional Length  1 m
 Total Length  17.74 m
 7 Tendons & each tendons has 7 wires
 7×38=266 mm  Available diameter

Design – Failure state
Failure consideration  Grout failure  Uncertainty in the load interaction with tendon (Negative
trend in time)
 Grout & Soil interaction  Positive trend in time (Located in
stable zone)
Ultimate load design
 Applied load of each anchor: 640 KN/anchor
 Pre-stress load = 1865 KPa was derived from the
interaction curve = 55% of 2nd peak of linear curve
 The rest capacity of the anchors will take the rest effect
due to soil movements
 Ultimate external load (Qi,u)  796 KN/m
 External Load Applied (Qi)  427 KN/m
 Pre-Stress load per anchor  640 KN/anchor
 Ultimate design load (Pu,d)  1194 KN/anchor
 Vertical Load due to inclination  0
 Vertical Pre-stress load due to inclination  0
 Plate weight  3.84 KN/plate (Negligible)
16
Difficult to know the real behavior
of the anchors  Uncertainty about
1st Failure is in Tendons/Grouts

Design - Grout
 Brice’s Theory (Equilibrium equation)
Limitation between 50%-67%
𝑃𝑢,𝑑 = 0.5 𝑡𝑜 0.67 𝐿 𝑐𝑜𝑛𝑠𝑡𝑎𝑖𝑛𝑡 𝜋∅𝜏 𝑏
∅: Grout Diameter
𝜏 𝑏: Grout Shear Length
 Alpha Method for estimating the 𝝉 𝒃
𝜏 𝑏 = 𝛼 𝑠 𝑞 𝑐,𝑎𝑣𝑒𝑟𝑎𝑔𝑒
𝛼 𝑠 = 0.02  Specific type of anchor
𝑞 𝑐,𝑎𝑣𝑒𝑟𝑎𝑔𝑒: Average value of CPT test
 𝛾𝑠𝑎𝑡 = 18
𝐾𝑁
𝑚3 & ∅′
= 35° ⇒ 𝑞 𝑐 = 25 𝑀𝑃𝑎
 The Max limit of 𝑞 𝑐
= 𝟏𝟓 𝑴𝑷𝒂 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛: 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑠𝑜𝑖𝑙 𝑛𝑜𝑡 𝐺𝑟𝑜𝑢𝑡
Partial factor (Both reduce total up to 56%)
 Ksi-factor  x=0.75
 Grout material  𝛾 𝑀,𝑏 = 1.35
𝜏 𝑏,𝑑 =
𝜉
𝛾 𝑀,𝑏
𝛼 𝑠 𝑞 𝑐,𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 0.56 × 0.02 × 15 = 160 𝐾𝑃𝑎
Grout unity check
- Uncertainty of uniform circumference shear distribution  Grout length reduced up to 1/3
- As I mentioned before, Actual free length  𝐿𝑡𝑓 =
𝐿𝑓𝑟𝑒𝑒+𝐿𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡
3
& Actual Grout length
 𝐿𝑡𝑏 =
1
2
𝑡𝑜
2
3
× 𝐿𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡
Shear force based on 𝒒 𝑐
CPT Quantity  1 x  0.75
𝛾 𝑀,𝑏 = 1.35 𝛼 𝑠 = 0.02
𝑞 𝑐,𝑔 = 15000 𝐾𝑃𝑎 𝜏 𝑏,𝑑 = 160 𝐾𝑃𝑎
Carrying load of grout
𝐹𝑟,𝑔𝑟𝑜𝑢𝑡,𝑑 = 𝐿 𝑔𝑟𝑜𝑢𝑡 𝜋∅𝜏 𝑏,𝑑 = 9 × 3.14 × 0.27 × 610
= 𝟏𝟐𝟐𝟏 𝐾𝑁 > 1194 (𝑃𝑢, 𝑘)
Grout Design
Grout Quantity  1 𝐹𝑠,𝑔𝑟𝑜𝑢𝑡,𝑑 (factor 1)  1194 KN
𝐹𝑟,𝑔𝑟𝑜𝑢𝑡,𝑑 (CPT)  1221 Max unity  0.98
Grout Check  Satisfied
17

# x [m] z [m] x [m] z [m] [m] [m] [m] x [m] z [m] x [m] z [m] [m] [m] [m] [degree]
1 16 5.9 19.6 1.8 5.5 0.24 1 19.6 1.8 25.4 -5.1 9 5.74 14.74 49.4
2 17.5 7.1 21.1 2.9 5.5 0.24 1 21.1 2.9 26.9 -3.9 9 5.74 14.74 49.4
3 19 8.6 22.6 4.5 5.5 0.24 1 22.6 4.5 28.4 -2.4 9 5.74 14.74 49.4
4 20.5 9.7 24.1 5.5 5.5 0.24 1 24.1 5.5 29.9 -1.3 9 5.74 14.74 49.4
5 20.5 9.7 27.8 8 7.5 0.24 1 27.8 8 33.7 1.2 9 7.74 16.74 29.1
Top Grout Bottom Grout L grout L free L anchor
Anchor
Direction
Plate Top Tendon Bottom Tendon L free
Plate
Thickne
Additional
Length
Design - Tendon
7 tendons (d=38 mm)  7 wires  1 strand
Due to 7 wires are bounded  section is not perfectly circle, so the computation of Area of each tendon (d-2t):
𝐴 𝑡𝑒𝑛𝑑𝑜𝑛 =
1
4
𝜋 𝑑2
− 𝑑 − 2𝑡 2
=
1
4
𝜋 382
− 38 − 2 × 10 2
= 𝟗𝟑𝟑 𝑚𝑚2
 fy = 470 N/mm^2  Max tensile force:
𝐹𝑟,𝑡𝑒𝑛𝑑𝑜𝑛,𝑘 = 7 × 𝐴 𝑡𝑒𝑛𝑑𝑜𝑛 𝑓𝑦 = 933 ×
477
1000
= 7 × 445𝐾𝑁 = 𝟑𝟏𝟏𝟓 𝐾𝑁 > 1194 (𝑃𝑢, 𝑘)
Shear force based on 𝒒 𝑐
Tendon Quantity  7 𝐹𝑎,𝑚𝑎𝑥  1194 KN
𝐹𝑎,𝑝𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠  641 KN 𝐹𝑠,𝑣,𝑚𝑎𝑥  0
𝐹𝑠,𝑠𝑡𝑎𝑎𝑓,𝑑  1194 KN
(factor 1)
𝐹𝑦,𝑑 /𝐹𝑟,𝑠𝑡𝑎𝑎𝑓  3115 KN
𝛾 𝑚 = 1
Unity check = 0.38 Tendon check  Satisfied
19
Total Length
Extending the Anchors  Top Part of Grout  Out of the unstable zone  The Longest free length last top Plate (#5)

Design – Anchors Location
Anchors that consist of 7 tendons  for No. 1 to 4 (have shorter free length)
∆𝐿 =
𝑃𝑢,𝑘 × 𝐿
7𝐴𝐸
=
1194 × 5.74
933 × 10−6 × 210 × 106
= 0.005𝑚 ⇒ 5 𝑚𝑚
20
Elongation
Safety Factor = 1.52

Design – Passive stabilizing system
 Solve by Means of increment
20
Analyze  Displacement Approach  Equilibrium between the force acting on the plate & Ground anchors
𝑃𝐵2
= 𝐹𝑎𝑛𝑐ℎ𝑜𝑟
𝑃 = 𝑓 𝑈 − 𝑢 = 𝑓(𝑤)
𝑓(𝑈 − 𝑢)𝐵2
= 𝐹𝑎𝑛𝑐ℎ𝑜𝑟 𝑢
𝜕𝑓
𝜕 𝑈 − 𝑢
𝑑𝑈 − 𝑑𝑢 𝐵2
= 𝐹𝑎𝑛𝑐ℎ𝑜𝑟 𝑢
𝐾 𝑝 𝑑𝑈 − 𝑑𝑢 𝐵2
= 𝐹𝑎𝑛𝑐ℎ𝑜𝑟 𝑑𝑢 ⇛
𝑑𝑈 =
𝐹𝑎𝑛𝑐ℎ𝑜𝑟 𝑑𝑢
𝐾 𝑝 𝐵2
+ 𝑑𝑢 =
𝐹𝑎𝑛𝑐ℎ𝑜𝑟
𝐵2 + 𝐾 𝑝
𝐾 𝑝
𝑑𝑢
 Passive system  1st slope stiffness of the curve is considered
𝐹𝑆 = 𝑓 𝑤
𝜕𝑓
𝜕 𝑈−𝑢
= 𝐾 𝑝 Local Stiffness
 Force is non-linear function of relative displacement
𝐾𝑝,1 =
1018 − 0 𝐾𝑃𝑎
(0.0179 − 0) 𝑚
= 56903
𝐾𝑁
𝑚3
𝐹𝑎𝑛𝑐ℎ𝑜𝑟 =
𝐹𝑟,𝑔𝑟𝑜𝑢𝑡,𝑑
𝑠
=
1221
1.5
= 814
𝐾𝑁
𝑚
𝑑𝑈, 1 =
𝐵2 + 𝐾 𝑝,1
𝐾 𝑝,1
𝑑𝑢 =
814
0.82 + 56903
56903
𝑑𝑢 = 1.022 𝑑𝑢
1st Failure  in the grout  Force of the anchor:
𝑷𝒍𝒂𝒕𝒆 𝑹𝒆𝒍𝒂𝒕𝒊𝒗𝒆 𝑫𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
𝑊 = 𝑑𝑈 − 𝑑𝑢
𝐹𝑎𝑛𝑐ℎ𝑜𝑟 =
𝐹𝑟,𝑡𝑒𝑛𝑑𝑜𝑛,𝑑
𝑠
=
3115
1.5
= 2076
𝐾𝑁
𝑚
𝑑𝑈, 1 = 1.052 𝑑𝑢
Grouts
Tendons
Using the interaction Curve

Design – Passive stabilizing system
21
du dU,1 dW W P1, KPa Pa Qi SF
[mm] [mm] [mm] [mm] [KPa] [KN] [KN/m]
0 0 0 0 0 0 0 0.885
25 26 0.5 0.5 28 18 12 0.896
50 51 1 1.5 85 55 36 0.918
75 77 1.5 2.5 142 91 61 0.941
100 102 2 3.5 199 127 85 0.966
200 204 4 6 341 219 146 1.033
250 255 5 9 512 328 219 1.126
400 408 8 13 740 473 316 1.281
500 510 10 18 1024 656 437 1.521
0 0 0 0 0 0 0 0.885
25 26 1 1 57 36 24 0.907
50 52 2 3 171 109 73 0.953
75 78 3 5 285 182 121 1.005
100 104 4 7 398 255 170 1.062
200 208 8 12 683 437 291 1.239
250 260 10 18 1024 656 437 1.521
Weak Grouts Weak Tendons
Observation
Grouts
Weakening
Tendons
Weakening
500 mm Soil Movement
250 mm Soil Movement
SF = 1.521
Better
Performance
Constrained
Grouts
Weak in Tendons (give better
performance)
Higher carrying load capacity of
tendons than the grouts
Rigid movement of the plate ratio +
Soil Movement
Mobilizing
the Soil
Activate the resistance
force from the anchors
Because
affects

Results – Comparison of Active & Passive
23
0 0 0 18 1018 641 427 1.521 1
25 25 0 18.08 1029 658 439 1.552 1
50 50 0 18.23 1037 664 442 1.562 1
75 75 0 18.45 1050 672 448 1.577 1
100 100 0 18.75 1067 683 455 1.597 1
200 201 1 19.35 1101 705 470 1.639 1
250 251 1 20.1 1144 732 488 1.695 1
400 401 1 21.3 1212 776 517 1.793 1
500 502 1 22.8 1297 830 554 1.933 1
De
Va
Active System
0.8
1
1.2
1.4
1.6
1.8
2
0 50 100 150 200 250 300 350 400 450 500
SAFETYFACTOR
DU, SOIL MOVEMENT [MM]
Active System Passive system - Weak Tendons Passive system - Weak Grout Design Value
𝐾 𝑝,2 =
1865 − 1018 𝐾𝑃𝑎
(0.0198 − 0.0179) 𝑚
= 438860
𝐾𝑁
𝑚3
Weakening in Grouts
𝑑𝑈, 2 =
𝐵2 + 𝐾 𝑝,1
𝐾 𝑝,2
𝑑𝑢 =
814
0.82 + 438860
438860
𝑑𝑢 = 1.003 𝑑𝑢
Weakening in Tendons
𝑑𝑈, 2 = 1.000 𝑑𝑢
Evaluate the behavior of active system after pre-stressed due to
soil movement: (2nd Slope of Linear Curve was considered)

Results – Observation
23
Conclusion
Active System
Intervention at early stage
may Avoid Soil Movement
Passive System
Mobilization of the soil mass
(Resistance Activation)
To reach the
Desired Safety Factor
Need soil movement  200-400mm
 Design Safety Factor = 1.3
The most efficient system  Active System
uncertainty of the failure state of the anchors (weakening in
tendons or grouts) may have risk to have excessive soil movement
Advantages

Results – Observation
25
Thanks for your attention

Slope project daniel

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