Brief history of Soil reinforcement, What is meant by ‘Reinforced Soil Structures’, What elements to be used as reinforcements, What are the possible field of applications, How to go for the analysis and design of such structures, Portrayal of Basic Mechanism, What is the main purpose of Facing in soil reinforcement, Problem regarding soil reinforcement

1. GARGI MEMORIAL INSTITUTE OF TECHNOLOGY
Campus: Balarampur, Baruipur, Kolkata- 700144
Phone: (033) 3262 9317
Fax: (033) 2433 0113
Mob: 94757 46447, 98300 37240
Email: principal@gmitkolkata.org
PROJECT REPORT ON
SOIL REINFORCEMENT RETAINING STRUCTURES
PRESENTED BY: INDRAJIT SARDAR
VISHAKA PRADHAN
SRIJAN KUMAR MONDAL
BALARAM SAHA
UNDER THE GUIDANCE: PROFESSOR JOY KUMAR MONDAL
SUBJECT: PROJECT PART II SUBJECT CODE: CE-881
DEPARTMENT: CIVIL ENGINEERING SEMESTER: 8TH
COLLEGE NAME: GARGI MEMORIAL INSTITUTE OF TECHNOLOGY

2. Page 1 of 31
PROJECT REPORT ON
SOIL REINFORCEMENT RETAINING STRUCTURES
PRESENTED BY:
NAME OF THE STUDENT: INDRAJIT SARDAR
UNIVERSITY REGISTRATION NO: 142810110091 OF: 2014-2015
UNIVERSITY ROLL NO: 28101314007 SESSION: 2014-2018
NAME OF THE COLLEGE: GARGI MEMORIAL INSTITUTE OF TECHNOLOGY
NAME OF THE STUDENT: VISHAKA PRADHAN
UNIVERSITY REGISTRATION NO: 142810110116 OF: 2014-2015
UNIVERSITY ROLL NO: 28101314030 SESSION: 2014-2018
NAME OF THE COLLEGE: GARGI MEMORIAL INSTITUTE OF TECHNOLOGY
NAME OF THE STUDENT: SRIJAN KUMAR MONDAL
UNIVERSITY REGISTRATION NO: 142810110110 OF: 2014-2015
UNIVERSITY ROLL NO: 28101314024 SESSION: 2014-2018
NAME OF THE COLLEGE: GARGI MEMORIAL INSTITUTE OF TECHNOLOGY
NAME OF THE STUDENT: BALARAM SAHA
UNIVERSITY REGISTRATION NO: 152810120085 OF: 2015-2016
UNIVERSITY ROLL NO: 28101315033 SESSION: 2015-2018
NAME OF THE COLLEGE: GARGI MEMORIAL INSTITUTE OF TECHNOLOGY
UNDER THE GUIDANCE:
PROFESSOR JOY KUMAR MONDAL

3. Page 2 of 31
CONTENTS
Brief history of Soil reinforcement ……………………………….................... 03
What is meant by ‘Reinforced Soil Structures’…………………………………… 03
What elements to be used as reinforcements……………………………………... 03
What are the possible field of applications………………………………………. 04
How to go for the analysis and design of such structures……………………….. 04
Portrayal of Basic Mechanism………………………………………………….... 06
What is the main purpose of Facing in soil reinforcement……………………… 06
Problem regarding soil reinforcement…………………………………...………... 07

4. Page 3 of 31
 Brief history of Soil reinforcement:
Soil Reinforcement: Reinvention
Henri Vidal (1963) coined the term – ‘Reinforced Earth’ (La Terre Armee).Accidental invention
while playing with children in beach building wet sand houses.
 What is meant by ‘Reinforced Soil Structures’?
Basically, the reinforced soil is nothing but soil plus reinforcement and we call it as reinforced
soil and the reinforcement, some of the ancient reinforcement products. That means, I mean ancient
means, some 2000 to 3000 years back, people have used the reinforcement but in the form of tree
branches, grass reeds, straw, the roots of vegetation, bamboo, tree trunks and so on. And even the ancient
past, people built very high towers and very high tall structures using the soil plus some form of
reinforcement.
And the modern reinforcement materials, they are steel, polymeric materials and then of course, the
natural materials like the coir and jute. And reason why, we use reinforcement or we need to use the
reinforcement is, the soil is very strong in compression. See, if you are able to apply pure compression
stress on the soil, it can take any amount of compression. But then unfortunately, because of the poisson’s
ratio of it, if we apply compression in one direction, there is tension in the other direction and the soil is
very weak in tension and starts failing.
 What elements to be used as reinforcements?
The following elements are used as reinforcement in soil;
Strip reinforcement
1. Flexible linear elements
2. Plain, grooved or ribbed
3. Materials
4. Metals
5. Galvanized steel
6. Aluminum-Magnesium alloy
7. Chrome Stainless steel
8. Check for durability against corrosion
9. Bamboo
10. Polymers
11. Glass-fiber reinforced plastics
Sheet reinforcement
1. Galvanized steel, textile fabric or expanded metal
2. Geotextiles Textile fabrics
3. Most common nowadays
4. Porous
5. Permeability in the range of coarse gravel to fine sand
6. Manufacturing
7. Woven from continuous monofilament fibers
8. Non-woven_ Staple fibers laid in random pattern and mechanically entangled
9. Fibers may be bonded or interlocked

5. Page 4 of 31
Anchor reinforcement
1. Flexible linear elements with distortions at the end.
 What are the possible fields of applications?
1. Mostly as retaining structures
2. Advantages over conventional type of walls
3. More economical if wall heights are large and subsoil is poor
4. Can be rapidly constructed
5. Require simple equipment for construction
6. Flexible structures
7. Soil forms bulk of their volume
8. Greater ability to withstand differential settlement than the rigid retaining walls
9. Large base-to-height ratio
10. Foundation stress distribution s nearly uniform
11. Less stress concentration at the toe
12. Permits construction of geotechnical structures on poor and difficult subsoil conditions.
13. Economical 25-50% saving in cost.
 How to go for the analysis and design of such structures?
In the reinforced earth wall two type of stability checked:
External stability
For failure
A) Calculation of Earth pressure(Static)
Earth pressure due to backfill soil weight
Earth pressure due to surcharge
Earth pressure due to cohesion
Earth pressure due to water
B) Calculation of moments of all static forces about the toe of retaining wall
C) Calculation of dynamic increments of forces
Above Ground water table
1. Case-I (For positive Av)
2. Case-II (For negative Av)
Below Ground water table
1. Case-I (For positive Av)
2. Case-II (For negative Av)
D) Calculation of dynamic incremental forces due to backfill and surcharge
Due to backfill
Case-I (For positive Av)
Case-II (For negative Av)
Due to additional dynamic moment
Case-I (For positive Av)
Case-II (For negative Av)

6. Page 5 of 31
Due to surcharge additional forces and moments
Free moments due to wall fill
Static forces/moments due to wall fill
Static forces/moments due to surcharge
Incremental dynamic force/moment for wall fill
Incremental dynamic force/moment due to surcharge
Factor of safety
E) Sliding stability
Static case
Seismic case
F) Overturning stability
Check against bearing
If fails,
Two remedies
i) Either increase surcharge, but that will also increase surcharge on the backfill which
develops more lateral stresses.
ii) Increase the length of the reinforcement.
For this particular case we will increase the reinforcement.
Revised calculation for wall fill
i) Factor of safety
ii) Sliding stability
iii) Overturning stability
iv) Check against bearing
a) Static case
b) Seismic case
Internal stability
Rupture Failure
Check for tension under static case
Case I- When hi is 4m
Case II- When hi is 6m
Check for tension under dynamic case
Case I- When hi is 4m
Case II- When hi is 6m
Wedge/Pullout failure
Static case
Case I- When hi is 4m
Case II- When hi is 6m
Seismic case
Case I- When hi is 4m
Case II- When hi is 6m
Total Self weight of soil wedge
Force equilibrium of wedge
Critical angle of inclination

7. Page 6 of 31
Anchorage length of reinforcement at any depth for seismic condition
Total length of reinforcement required
 Portrayal of Basic Mechanism
1. Primarily developed for cohesion less soils
2. Carries tensile stress
3. Suppression of anisotropic lateral suppression or strain
4. Cohesion
5. Bond between adjacent particles
6. Electro-static forces
7. Cementation
8. Prohibits free movement
9. Results in increased shear strength
10. Concept of pseudo-cohesion
11. Particles are tied to each other
12. The tie provides a pseudo-bond between the particles
13. Result in enhanced shear strength
 What is the main purpose of Facing in soil reinforcement?
Facing in reinforced soil structures
Required for vertical or near-vertical structures
Main purpose
1. Retains the soil between the reinforcement in the immediate vicinity to the edge of the structure
2. Does not affect the overall stability of the structure affect the local stability
3. Should be able to adopt to deformations without distortions and introduction of stresses materials
4. Galvanized steel, Stainless steel, Aluminum, Bricks, Precast concrete panels,
5. Precast concrete slabs, Geotextiles, Geogrids, Plastics, Glass-reinforced plastics, Timber
6. Metal and precast concrete panels are mostly used
7. Ease in handling and assembling
Metal facing
1. Mild steel, Galvanized steel or Aluminum
2. Same property as the reinforcement strips
3. Facing is semi-elliptical
4. Continuous horizontal joint along one edge
5. Holes are provided for bolting of reinforcing elements
6. Very flexible
7. Can adapt to significant deformation
Concrete Panel Facing
1. Cruciform shaped.
2. Vertical dowel-groove system to accommodate other adjacent panels.
3. Dowels allow for restricted lateral and rotational movement.
4. Renders the entire facing structure as flexible.

8. Page 7 of 31
PROBLEM:
Design the retaining structure as given in the following figure:
RT =16.5
qallowable= 200kpa
q=18kn
μ= 0.5
γw= 10kn
δ= 0.67×φ=0.67×35° = 23.45°
δ′ = 0.335×φ =0.335×35° = 11.72°
Zone factor (v) = 0.36
∴αh = 0.18
∴αv = 0.09
SOLUTION:
We will be analyzing according to IS-1893-1984
EXTERNAL STABILITY CHECK
For failure,
θ = 45°–
∅
= 45°–
°
= 26°
Now,
x=6×tan θ
= 6×tan26°
= 2.96 meter

9. Page 8 of 31
A. CALCULATION OF EARTH PRESSURE (STATIC)
1. Earth pressure due to backfill soil weight
Now for above the water table
P=ka×r×z
Where, ka= co-efficient of active earth pressure under static condition above water table.
For above the water table,
∴ k =
cos ∅
cosδ 1 +
(∅ δ) ∅
δ
cos∅= cos35° = 0.819
cosδ= cos23.45° = 0.917
sin(∅+δ)=sin(35°+0.67×35°) = 0.852
sin∅= sin35° = 0.573
∴ K =
cos ∅
cosδ 1 +
(∅ δ) ∅
δ
=
(0.819)
0.917 1 +
( . × . )
.
= 0.244
Now for below the water table
δ′ =
. °×φ
= 0.335×φ = 0.335×35° = 11.72°
cos∅ = cos35° = 0.819
cosδ = cos11.72° = 0.979
sin(∅+δ) = sin(35°+11.72°) = 0.728
sin∅ = sin35° = 0.573
Now,
∴ K′ =
cos ∅
cosδ 1 +
(∅ δ) ∅
δ

10. Page 9 of 31
=
(0.819)
0.979 1 +
( . × . )
.
= 0.250
Now,
P1a= ×(ka×γ1×4) = ×(0.244×19×4) = 37.088kn/m
P1b = ka×γ1×4×2 = 37.088kn/m
P2 =(k′a–ka)×γ1×4×2 = 0.912kn/m
P3 = ×(k′a×γw×2×2) = 0.5×0.250×10×2×2 = 5.0 kn/m
2. Earth pressure due to surcharge( )
P4 = q×ka×6 = 26.352 kn/m
P5 = q×(k′a–ka)×2 = 0.216 kn/m
3. Earth pressure due to cohesion( )
P6 = 2× C2× k ×6 = 11.855 kn/m
P7 = 2× C2× k′ – k ×2 = 0.0482 kn/m
4. Earth pressure due to water
P = ×γw×2×2 = 20 kn/m
Now,
P = P1a+ P1b+ P2+P3
=37.088+37.088+0.912+5
=80.088 kn/m
P =P6+ P7
=11.855+0.048
=11.903 kn/m
P =P4+ P5
= (26.352+0.216) kn/m
= 26.568 kn/m
P =P
= 20 kn/m
Total active lateral thrust
P =P –P +P +P
= 80.088-11.903+26.568+20
=114.753 kn/m
B. Calculation of moments of all static forces about the toe of retaining wall
Mar = P1a×(2+4/3)+P1b×1+P2×1+P3×2/3
= (37.088×3.33)+(37.088×1)+(0.912×1)+(5×2/3)
=164.836 kn/m
Maq= P4×3+P5×1
= 26.352×3+0.216×1
= 79.272 kn/m

11. Page 10 of 31
Mac = P6×3+P7×1
= 11.855×3+0.048×1
= 35.613 kn/m
Maw = P8×2/3
= 20×2/3
= 13.33 kn/m
Total moment due to static forces
MT(st)= Mar+ Maq– Mac+ Maw
= 164.836+79.272–35.613+13.33
= 221.825 kn/m
C. Calculation of dynamic increments of forces
We will analysis by pseudo static condition. We will introduce a new dynamic earth pressure co-efficient.
According to IS-1893 code,
Kad,which is a function of (φ,δ,λ)
Where, λ=tan-1 α
±α
λ(+αv) = tan-1 α
α
= tan-1 .
.
= 9.37°
λ(–αv) = tan-1 α
–α
= tan-1 .
.
= 11.18°
Now,
K =
(1 ± α )cos (∅ − λ − α)
(cosλ)(cos α) cos(δ + α + λ) 1 +
(∅ δ) (∅ λ)
(α ) (δ α λ)
Above the ground water table
Case–1
For (+ v) condition
λ= 9.37° and α=0, i=0
cos(∅ − λ − α)= cos(35°–9.37°) = 0.9016
cos(δ + α + λ)= cos(23.45°+9.37°) = 0.840
sin(∅ + δ)= 0.8521
sin(∅ − i − λ)= 0.432
cosλ= cos9.37°= 0.986
K ( α ) =
(1 + α )cos (∅ − λ − α)
(cosλ)(cos α) cos(δ + α + λ) 1 +
(∅ δ) (∅ λ)
(α ) (δ α λ)
=
(1 + 0.09)0.9016²
(0.986)(0.840) 1 +
( . )( . )
( . )
= 0.386

12. Page 11 of 31
Case–2
For (– v) condition
λ= 11.18° and α=0, i=0
cos(∅ − λ − α)= cos(35°–11.18°) = 0.915
cos(δ + α + λ)= cos(23.45°+11.18°) = 0.822
sin(∅ + δ)= 0.8521
sin(∅ − i − λ)= 0.4038
cosλ= cos11.18°= 0.981
K (–α ) =
(1– α )cos (∅ − λ − α)
(cosλ)(cos α) cos(δ + α + λ) 1 +
(∅ δ) (∅ λ)
(α ) (δ α λ)
=
(1– 0.09)0.915²
(0.981)(0.822) 1 +
( . )( . )
( . )
= 0.348
Below the ground water table
λ = tan − 1
γ
γ
×
α
1 ± α
γ =20kn/m3
γ =(γ –γ ) = (20–10) = 10kn/m3
Now for (+ ),
λ(+α ) = tan − 1
γ
γ
×
α
1 + α
= tan − 1
20
10
×
0.18
1 + 0.09
= 18.277
Now for (– ),
λ(– α ) = tan − 1
γ
γ
×
α
1– α
= tan − 1
20
10
×
0.18
1 − 0.09
= 21.584
Case–1
For (+ ) condition,
λ= 18.27° and α=0, i=0
cos(∅ − λ − α)= cos(35°–18.27°) = 0.957

13. Page 12 of 31
cos(δ + α + λ)= cos(23.45°+18.27°) = 0.867
sin(∅ + δ)= 0.726
sin(∅ − i − λ)= 0.288
cosλ= cos18.27°= 0.949
K′ ( α ) =
(1 + α )cos (∅ − λ − α)
(cosλ)(cos α) cos(δ + α + λ) 1 +
(∅ δ) (∅ λ)
(α ) (δ α λ)
=
(1 + 0.09)0.957²
(0.949)(0.867) 1 +
( . )( . )
( . )
= 0.546
Case–2
For (– ) condition,
λ= 21.58° and α=0, i=0
cos(∅ − λ − α)= cos(35°–21.58°) = 0.9726
cos(δ + α + λ)= cos(23.45°+21.58°) = 0.837
sin(∅ + δ)= 0.725
sin(∅ − i − λ)= 0.232
cosλ= cos21.58°= 0.9299
K′ ( α ) =
(1 − α )cos (∅ − λ − α)
(cosλ)(cos α) cos(δ + α + λ) 1 +
(∅ δ) (∅ λ)
(α ) (δ α λ)
=
(1 − 0.09)0.9726²
(0.929)(0.837) 1 +
( . )( . )
( . )
= 0.527
D. Calculation of dynamic incremental forces due to backfill and surcharge
Calculation of dynamic incremental forces due to backfill
Paγi= ×γ2×(Kad–Ka) ×
– ²
×( H + 2H )+ (K′ad–K′a) ×
( )²
×[3γ2(H– H )+ γb2Hw]
Where, Paγi= dynamic passive earth pressure in kg/m length of wall.
For (+ ) condition,
Kad= 0.386
Ka= 0.244
K′ad= 0.546
K′a= 0.250
γ2= 18
γ =20kn/m3
γ =(γ –γ ) = (20–10) = 10kn/m3
H= 6m
Hw= 2m

14. Page 13 of 31
Paγi= ×γ2×(Kad–Ka) ×
– ²
×( H + 2H )+ (K′ad–K′a) ×
( )²
×[3γ2(H– H )+ γb2Hw]
= ×18×(0.386–0.244) ×
– ²
×( 6 + 2 × 2)+ (0.546–0.250) ×
( )²
×[3×18 (6– 2)+ 10×2]
= 57.365
For (– ) condition,
Kad= 0.348
Ka= 0.244
K′ad= 0.527
K′a= 0.250
γ2= 18
γ =20kn/m3
γ =(γ –γ ) = (20–10) = 10kn/m3
H= 6m
Hw= 2m
Paγi= ×γ2×(Kad–Ka) ×
– ²
×( H + 2H )+ (K′ad–K′a) ×
( )²
×[3γ2(H– H )+ γb2Hw]
= ×18×(0.348–0.244) ×
– ²
×( 6 + 2 × 2)+ (0.527–0.250) ×
( )²
×[3×18 (6– 2)+ 10×2]
= 46.750
Additional dynamic moment
Maγi= ×γ2×(Kad–Ka) ×
– ²
×(H + 2H × H + 3H )+ (K′ad–K′a) ×
( )
×[4γ2(H– H )+ γb2Hw]
For (+ ) condition,
Kad= 0.386
Ka= 0.244
K′ad= 0.546
K′a= 0.250
γ2= 18
γ =20kn/m3
γ =(γ –γ ) = (20–10) = 10kn/m3
H= 6m
Hw= 2m
Maγi= ×γ2×(Kad–Ka) ×
– ²
×(H + 2H × H + 3H )+ (K′ad–K′a) ×
( )
×[4γ2(H– H )+ γb2Hw]
=
18 × (0.386– 0.244) × (6 − 2)
4 × 6
× (6 + 2 × 6 × 2 + 3 × 2 ) +
(0.546 − 0.250) × 2
4 × 6
× [4 × 18 × (6 − 2) + 10 × 2]
= 153.077
For (+ ) condition,
Kad= 0.348
Ka= 0.244
K′ad= 0.527
K′a= 0.250
γ2= 18

15. Page 14 of 31
γ =20kn/m3
γ =(γ –γ ) = (20–10) = 10kn/m3
H= 6m
Hw= 2m
Maγi= ×γ2×(Kad–Ka) ×
– ²
×(H + 2H × H + 3H )+ (K′ad–K′a) ×
( )
×[4γ2(H– H )+ γb2Hw]
=
18 × (0.348– 0.244) × (6 − 2)
4 × 6
× (6 + 2 × 6 × 2 + 3 × 2 ) +
(0.527 − 0.250) × 2
4 × 6
× [4 × 18 × (6 − 2) + 10 × 2]
= 118.294
Calculation of dynamic incremental forces due to surcharge additional forces and moments
P = q (K − K )
(H − H )
H
+ K′
− K′
H
H
M =
2
3
×
q
H
× [(K − K ) × (H − H ) + (K′
− K′
) × H ]
For (+∝ )condition,
P ( ∝ ) = q (K − K )
(H − H )
H
+ K′
− K′
H
H
= 18[(0.386– 0.244) ×
6 − 2
6
+ (0.546 − 0.250) ×
2
6
= 17.184
For (−∝ )condition,
P ( ∝ ) = q (K − K )
(H − H )
H
+ K′
− K′
H
H
= 18[(0.348– 0.244) ×
6 − 2
6
+ (0.527 − 0.250) ×
2
6
= 13.308
For (+∝ )condition,
M =
2
3
×
q
H
× [(K − K ) × (H − H ) + (K′
− K′
) × H ]
=
2
3
×
18
6
× [(0.386 − 0.244) × (6 − 2 ) + (0.546 − 0.250) × 2 ]
= 63.808
For (−∝ )condition,
M =
2
3
×
q
H
× [(K − K ) × (H − H ) + (K′
− K′
) × H ]
=
2
3
×
18
6
× [(0.348 − 0.244) × (6 − 2 ) + (0.527 − 0.250) × 2 ]
= 47.696
Free moments due to wall fill (Restoring forces)
Static forces-moments due to wall fill
Total weight of the wall = L[γ (H − H ) + γ H ]
L= reinforcement length= 5m

16. Page 15 of 31
∴ W = 5[19(6 − 2) + 21 × 2]
= 590 KN
Effective weight of the wall = L[γ (H − H ) + γ H ]
∴ W′
= 5[19(6 − 2) + (21 − 10) × 2]
= 490 KN
∴ M = W ×
L
2
=
590 × 5
2
= 1475 KN ∙ m
∴ M′
= W ×
L
2
=
490 × 5
2
= 1225 Kn ∙ m
Static forces–moments due to surcharge
∴ Static force Q= qL= 18× 5 = 90 KN
∴ Static moment = M = Q ×
L
2
=
90 × 5
2
= 225 Kn ∙ m
Incremental dynamic force–moment for wall fill
Horizontal seismic force= P = W ×∝ = 106.2 KN
Vertical seismic force= P = W × (±∝ ) = ±53.1 KN
M =
L
2
[γ (H − H ) + γ H ] ×∝
=
5
2
[19(6 − 2 ) + 21 × 2 ] × 0.18
= 311.4 KN
M = ± ∝ × W ×
L
2
= ±
53.1 × 5
2
= ±132.75 KN ∙ m
Incremental dynamic forces–moments due to surcharge
P = ∝ × Q = 0.18 × 90 = 16.2 KN
P = ±∝ × Q = ±0.09 × 90 = ±8.1 KN
M = P × H = 16.2 × 6 = 97.2 KN ∙ m
M = P ×
L
2
= ±20.25 KN ∙ m
Factor of safety
Sliding stability
For the static case
F ( ) =
μ W′
+ Q
P
≥ 2
Now, F ( ) =
μ W′
+ Q
P
=
0.5(490 + 90)
114.753
= 2.527
∴ 2.527 > 2 Hence ok.

17. Page 16 of 31
For the seismic case
F ( ) =
μ(W′
+ Q) ± μ(W + Q) ×∝
P + P γ + P + (W + Q) ×∝
≥ 1.5
Now, F ( ),( ∝ ) =
μ(W′
+ Q) ± μ(W + Q) ×∝
P + P γ + P + (W + Q) ×∝
=
0.5(490 + 90) + 0.5(590 + 90) × 0.09
114.753 + 57.365 + 17.184 + (590 + 90) × 0.18
= 1.02
∴ 1.02 ≱ 1.5 Hence not ok.
Now, F ( ),( ∝ ) =
μ(W′
+ Q) ± μ(W + Q) ×∝
P + P γ + P + (W + Q) ×∝
=
0.5(490 + 90) − 0.5(590 + 90) × 0.09
114.753 + 46.750 + 13.308 + (590 + 90) × 0.18
= 0.872
∴ 0.872 ≱ 1.5 Hence not ok.
Overturning stability
For static case
F ( ) =
M′
+ M
M
≥ 2
Now, F ( ) =
M′
+ M
M
=
1225 + 225
221.825
= 6.536
∴ 6.536 ≥ 2 Hence ok.
For seismic case
F ( ),(±∝ ) =
M′
+ M ± M ± M
M + M + M + M γ + M
≥ 1.5
F ( ),( ∝ ) =
M′
+ M ± M ± M
M + M + M + M γ + M
=
1225 + 225 + 132.75 + 20.25
221.825 + 311.4 + 97.2 + 153.077 + 63.808
= 1.891
∴ 1.891 ≥ 1.5 Hence ok.
F ( ),( ∝ ) =
M′
+ M ± M ± M
M + M + M + M γ + M
=
1225 + 225 − 132.75 − 20.25
221.825 + 311.4 + 97.2 + 118.294 + 47.696
= 1.628
∴ 1.628 ≥ 1.5 Hence ok.

18. Page 17 of 31
Check against bearing
For static case
F =
W′
+ Q
L
+
M × 6
L
=
490 + 90
5
+
221.825 × 6
5²
= 169.238
F =
W′
+ Q
L
−
M × 6
L
> 0
=
490 + 90
5
−
221.825 × 6
5²
= 62.762
∴ 62.762 > 0 Hence ok.
For seismic case
F =
(W′
+ Q) ± (W + Q) ∝
L
+
M × 6
L
F =
(W′
+ Q) ± (W + Q) ∝
L
−
M × 6
L
M = M + M γ + M + M + M
M ( ∝ ) = M + M γ + M + M + M
= 221.825+153.077+63.808+97.2+311.4
= 847.31
M ( ∝ ) = M + M γ + M + M + M
= 221.825+118.294+47.676+97.2+311.4
= 796.415
F ( ∝ ) =
(W′
+ Q) + (W + Q) ∝
L
+
M × 6
L
=
(490 + 90) + (590 + 90) × 0.09
5
+
847.31 × 6
5
= 331.59
F ( ∝ ) =
(W′
+ Q) − (W + Q) ∝
L
+
M × 6
L
=
(490 + 90) − (590 + 90) × 0.09
5
+
796.415 × 6
5
= 294.89
F ( ∝ ) =
(W′
+ Q) + (W + Q) ∝
L
−
M × 6
L
=
(490 + 90) + (590 + 90) × 0.09
5
−
847.31 × 6
5
= –75.114
F ( ∝ ) =
(W′
+ Q) − (W + Q) ∝
L
−
M × 6
L
=
(490 + 90) − (590 + 90) × 0.09
5
−
796.415 × 6
5

19. Page 18 of 31
= –87.37
Two remedies
i) Either increase surcharge, but that will also increase surcharge on the backfill which develops
more lateral stress.
ii) Increase the length of the reinforcement.
For this particular case we will increase the reinforcement length.
For Sliding
Now, F ( ),( ∝ ) =
μ ′ μ( )×∝
P + P γ + P + ×∝ × L
= 1.5
=>
L 0.5
( )
+ 0.5
( )
× 0.09
114.753 + 57.365 + 17.184 +
( )
× 0.18 × L
= 1.5
=> L= 10.36 meter
Now, F ( ),( ∝ ) =
μ ′ μ( )×∝
P + P γ + P + ×∝ × L
= 1.5
=>
L 0.5
( )
− 0.5
( )
× 0.09
114.753 + 46.750 + 13.308 +
( )
× 0.18 × L
= 1.5
=> L= 17.296 meter
Check against bearing
Seismic case
M ( ∝ ) = M + M γ + M + M + M
= 221.825+153.077+63.808+97.2+62.28×L
= 535.91+62.28L
M ( ∝ ) = M + M γ + M + M + M
= 221.825+118.294+47.676+97.2+62.28×L
= 485.015+62.28L
F ( ∝ ) =
(W′
+ Q) + (W + Q) ∝
L
+
M × 6
L
= 1.25 × 200
=
(490 + 90) + (590 + 90) × 0.09
L
+
(535.91 + 62.28L) × 6
L
= 1.25 × 200
= L = 6.150 meter (Taking positive value)
F ( ∝ ) =
(W′
+ Q) − (W + Q) ∝
L
+
M × 6
L
= 1.25 × 200
=
(490 + 90) − (590 + 90) × 0.09
L
+
(485.015 + 62.28L) × 6
L
= 1.25 × 200
= L = 5.63 meter (Taking positive value)
F ( ∝ ) =
(W′
+ Q) + (W + Q) ∝
L
−
M × 6
L
= 1.25 × 200

20. Page 19 of 31
=
(490 + 90) + (590 + 90) × 0.09
L
−
(535.91 + 62.28L) × 6
L
= 1.25 × 200
= L = 3.54 meter (Taking the positive value)
F ( ∝ ) =
(W′
+ Q) − (W + Q) ∝
L
−
M × 6
L
= 1.25 × 200
=
(490 + 90) − (590 + 90) × 0.09
L
−
(485.015 + 62.28L) × 6
L
= 1.25 × 200
= L = 3.39 meter (taking the positive value)
Now, we will provide maximum length 17.296 meter ≅ 18 meter.
Revised calculations for wall fill
Total weight of the wall(Ww) = 18[19×4+21× 2] KN = 2124 KN
Effective weight of the wall (W′
) = 18[19×4+11× 2] KN = 1764 KN
∴ M = W ×
L
2
=
2124 × 18
2
= 19116 KN ∙ m
∴ M′
= W ×
L
2
=
1764 × 18
2
= 15876 Kn ∙ m
Static forces–moments due to surcharge
∴ Static force Q= qL= 18× 18 = 324 KN
∴ Static moment = M = Q ×
L
2
=
324 × 18
2
= 2916 Kn ∙ m
Incremental dynamic force–moment for wall fill
Horizontal seismic force= P = W ×∝ = 382.32 KN
Vertical seismic force= P = W × (±∝ ) = ±191.16 KN
M =
L
2
[γ (H − H ) + γ H ] ×∝
=
18
2
[19(6 − 2 ) + 21 × 2 ] × 0.18
= 1121.04 KN
M = ± ∝ × W ×
L
2
= ±0.09
2124 × 18
2
= ±1720.44 KN ∙ m
Incremental dynamic forces–moments due to surcharge
P = ∝ × Q = 0.18 × 324 = 58.32 KN
P = ±∝ × Q = ±0.09 × 324 = ±29.16 KN
M = P × H = 58.32 × 6 = 349.92 KN ∙ m
M = P ×
L
2
= ±262.44 KN ∙ m
Factor of safety
Sliding stability
F ( ) =
μ W′
+ Q
P
≥ 2
Now, F ( ) =
μ W′
+ Q
P
=
0.5(1764 + 324)
114.753
= 9.0978

21. Page 20 of 31
∴ 9.0978 > 2 Hence ok.
For the seismic case
F ( ) =
μ(W′
+ Q) ± μ(W + Q) ×∝
P + P γ + P + (W + Q) ×∝
≥ 1.5
Now, F ( ),( ∝ ) =
μ W′
+ Q + μ(W + Q) ×∝
P + P γ + P + (W + Q) ×∝
=
0.5(1764 + 324) + 0.5(2124 + 324) × 0.09
114.753 + 57.365 + 17.184 + (2124 + 324) × 0.18
= 1.83
∴ 1.83 > 1.5 Hence ok.
Now, F ( ),( ∝ ) =
μ W′
+ Q − μ(W + Q) ×∝
P + P γ + P + (W + Q) ×∝
=
0.5(1764 + 324) − 0.5(2124 + 324) × 0.09
114.753 + 46.750 + 13.308 + (2124 + 324) × 0.18
= 1.51
∴ 1.51 > 1.5 Hence ok.
Overturning stability
For static case
F ( ) =
M′
+ M
M
≥ 2
Now, F ( ) =
M′
+ M
M
=
15876 + 2916
221.825
= 84.71
∴ 84.71 ≥ 2 Hence ok.
For seismic case
F ( ),(±∝ ) =
M′
+ M ± M ± M
M + M + M + M γ + M
≥ 1.5
F ( ),( ∝ ) =
M′
+ M + M + M
M + M + M + M γ + M
=
15876 + 2916 + 262.44 + 1720.44
221.825 + 153.077 + 63.808 + 349.92 + 1121.04
= 10.878
∴ 10.878 ≥ 1.5 Hence ok.
F ( ),( ∝ ) =
M′
+ M − M − M
M + M + M + M γ + M
=
15876 + 2916 − 262.44 − 1720.44
221.825 + 118.294 + 47.696 + 349.92 + 1121.04
= 9.043
∴ 9.043 ≥ 1.5 Hence ok.

22. Page 21 of 31
Check against bearing
For static case
F =
W′
+ Q
L
+
M × 6
L
=
1764 + 324
18
+
221.825 × 6
18²
= 120.107
∴ 120.107 < 200kpa Hence ok.
F =
W′
+ Q
L
−
M × 6
L
> 0
=
1764 + 324
18
−
221.825 × 6
18²
= 118.89
∴ 118.89 > 0 Hence ok.
For seismic case
F =
(W′
+ Q) ± (W + Q) ∝
L
+
M × 6
L
F =
(W′
+ Q) ± (W + Q) ∝
L
−
M × 6
L
M = M + M γ + M + M + M
M ( ∝ ) = M + M γ + M + M + M
= 221.825+153.077+63.808+349.92+1121.04
= 1909.67
M ( ∝ ) = M + M γ + M + M + M
= 221.825+118.294+47.696+349.92+1121.04
= 1858.775
F ( ∝ ) =
(W′
+ Q) + (W + Q) ∝
L
+
M × 6
L
=
(1764 + 324) + (2124 + 324) × 0.09
18
+
1909.67 × 6
18
= 163.604
∴ 163.604 < 250 Hence ok.
F ( ∝ ) =
(W′
+ Q) − (W + Q) ∝
L
+
M × 6
L
=
(1764 + 324) − (2124 + 324) × 0.09
18
+
1858.775 × 6
18
= 138.18
∴ 138.18 < 250 Hence ok.
F ( ∝ ) =
(W′
+ Q) + (W + Q) ∝
L
−
M × 6
L
=
(1764 + 324) + (2124 + 324) × 0.09
18
−
1909.67 × 6
18
= 92.875
∴ 92.875 > 0 Hence ok.

23. Page 22 of 31
F ( ∝ ) =
(W′
+ Q) − (W + Q) ∝
L
−
M × 6
L
=
(1764 + 324) − (2124 + 324) × 0.09
18
−
1858.775 × 6
18
=69.33
∴ 69.33 > 0 Hence ok.
G. Internal stability check
Rupture failure
Check for tension under static case
Case–1
When hi ≤ H–Hw
Since Ti (Tension developed in a reinforcement at hi) depends on Fvi and Sz, maximum tension developed
will be at water table.
hi = 4meter
F = γ + q +
6 × M
L
M = P ×
H − H
3
+
P × 4
6
×
4
2
−
P
6
× 4 ×
4
2
= 37.088 ×
6 − 2
3
+
26.352 × 4
6
×
4
2
−
11.855
6
× 4 ×
4
2
= 68.78 kpa
F = γ + q +
6 × M
L
= (19 × 4 + 18) +
6 × 68.78
18
= 95.273 kpa
T = K × F − 2C × K × S
K =
cos φ
cosδ
×
1
1 +
( )
cosφ = cos38 = 0.788
cosδ = cos(0.67 × 38 ) = 0.902
sinφ = sin38 = 0.615
sin(φ + δ ) = 0.894
∴ K =
cos φ
cosδ
×
1
1 +
( )
=
0.788
0.902
×
1
1 +
. × .
.
= 0.217
T = K × F − 2C × K × S

24. Page 23 of 31
= 0.217 × 95.273 − 2 × 1.5 × √0.217 × S
= 19.276× S
Now, T = R = 16.5
19.276× S = 16.5
∴ S = 0.855meter
Case–2
When hi ≥ H–Hw
Here maximum tension will developed at hi = 6meter
F = [γ (H − H ) + γ H + q] +
6M
L
M = M = 221.825
F = [γ (H − H ) + γ H + q] +
6M
L
= [19(6 − 2) + (21 − 10)2 + 18] +
6 × 221.825
18
= 120.107
T = K′ × F − 2C × K′ × S
K′ =
cos φ
cosδ
×
1
1 +
( )
cosφ = cos38 = 0.788
cosδ′ = cos(0.335 × 38 ) = 0.975
sinφ = sin38 = 0.615
sin(φ + δ′ ) = 0.774
∴ K′ =
cos φ
cosδ′
×
1
1 +
( )
=
0.788
0.975
×
1
1 +
. × .
.
= 0.220
T = K′ × F − 2C × K′ × S
= 0.220 × 120.107 − 2 × 1.5 × √0.220 × S
= 25.016× S
Now, T = R = 16.5
25.016× S = 16.5
∴ S = 0.659meter

25. Page 24 of 31
Check for tension under dynamic case
K =
(1 + α )cos (φ − ψ)
cosψcos(δ + ψ) 1 +
( ) ( )
( )
Where,
ψ(± ) = tan − 1
α
1 ± α
ψ( ) = tan − 1
α
1 + α
= 9.37
ψ ( ) = tan − 1
γ
γ
×
α
1 + α
= 17.49
cos(φ − ψ) = 0.877
sin(φ + δ ) = 0.894
sin(φ − ψ) = 0.478
cos(δ + ψ) = 0.820
cosψ = 0.986
K =
(1 + α )cos (φ − ψ)
cosψcos(δ + ψ) 1 +
( ) ( )
( )
=
(1 + 0.09)0.877²
(0.986)(0.820) 1 +
( . )( . )
( . )
= 0.346
cos(φ − ψ′) = 0.936
sin(φ + δ′) = 0.894
sin(φ − ψ′) = 0.350
cos(δ + ψ′) = 0.731
cosψ′ = 0.953
K′ =
(1 + α )cos (φ − ψ′)
cosψ′cos(δ + ψ′) 1 +
( ) ( )
( )
=
(1 + 0.09)0.936²
(0.953)(0.731) 1 +
( . )( . )
( . )
= 0.529
Case–1
When hi ≤ H–Hw
We will consider hi = 4 meter
F = (γ h + q)(1 + α ) +
6M
L

26. Page 25 of 31
M = M + M + M + α [
γ Lh
2
] + α qLh
M =
γ (K − K )
4H
h (2 × H − h )
=
18(0.346 − 0.217)
4 × 6
4 (2 × 6 − 4)
= 49.536
M =
q(K − K )
3H
h (3 × H − h )
=
18(0.386 − 0.244)
3 × 6
4²(3 × 6 − 4)
= 31.808
α
γ Lh
2
= 0.18 ×
19 × 18 × 4
2
= 492.48
α qLh = 0.18 × 18 × 18 × 4
= 233.28
M = M + M + M + α [
γ Lh
2
] + α qLh
= 68.78 + 49.536 + 31.808 + 492.48 + 233.28
= 875.886
F = (γ h + q)(1 + α ) +
6M
L
= (19 × 4 + 18)(1 + 0.09) +
6 × 875.886
18
= 118.680
T = K × F − 2C × K × S
= (0.346 × 118.680 − 2 × 1.5√0.346 ) × S
= 39.298 × S
Now, T = R = 16.5 × 1.25
= 39.298 × S = 16.5 × 1.25
∴ S = 0.524meter
Case–2
When hi ≥ H–Hw
Here maximum tension will developed at hi = 6meter
T = K′ × F − 2C × K′ × S
F = [γ (H − H ) + q](1 + α ) + [h − (H − H )] × (γ + α γ ) +
6 × M
L
M = α qLh
= 349.92
M = α γ L
h − (H − H )
2
= 136.08

27. Page 26 of 31
M = α × γ (H − H ) × L × h −
H − H
2
= 984.96
M = M + M + M + M + M + M
= 1894.847
M = M + M
a = γ (H − H ) = 72
a = h − (H − H ) = 2
M =
γ (K − K )
4H
× (H − H ) × [2Hh − H − 2HH + 4h H + 3H ]
= 111.456
M =
3 × a (K − K )
H
a a H + (γ a H − a a − a H ) ×
a
2
− (γ a + γ H − a )
a
3
+ γ ×
a
4
= 31.724
M = M + M
= 111.456+31.724
= 143.18
M = M + M
M =
2q(K K )(H − H )
H
Hh −
(H − H )
6
{H + 2H + 3h }
= 53.664
M =
2(K − K )q × a
H
a H −
(a + H )a
2
+
a
3
= 5.218
M = M + M
= 53.664+5.218
= 58.882
F = [γ (H − H ) + q](1 + α ) + [h − (H − H )] × (γ + α γ ) +
6 × M
L
= [19(6 − 2) + 18](1 + 0.09) + [6 − (6 − 2)] × (11 + 0.09 × 21) +
6 × 1894.847
18
= 163.329
T = K′ × F − 2C × K′ × S
= 0.529 × 163.329 − 2 × 1.5√0.529 × S
= 84.21× S
Now, T = R = 16.5 × 1.25
= 84.21 × S = 16.5 × 1.25
∴ S = 0.244meter

28. Page 27 of 31
Wedge / Pullout failure
Static case
Wedge failure can occur at any of the plane throughout the depth of the wall.
We have to find out
i) Critical value of β
ii) Critical value of Z
From the calculation it has been established
β = 45° −
φ
2
Case–1
When Z ≤ H − H , Z = 4
W =
1
2
γ Z tanβ
Case–2
When Z > − H , Z = 6
W =
1
2
{γ [2Z − (H − H )] × (H − H ) + γ [Z − (H − H )] }tanβ
For both case
W = a tanβ
T =
(a + q )tanβ
tan(β + φ )
For the first case a =
1
2
γ Z =
1
2
× 19 × 4 = 152kn/m
For the second case a =
1
2
{γ [2Z − (H − H )] × (H − H ) + γ [Z − (H − H )] }
= 326kn/m
Now for first case
T =
(a + q )tanβ
tan(β + φ )
=
(152 + 18 × 4)tan(45° − 19°)
tan(45° − 19° + 38°)
= 53.21kn/m
T =
T
h
at, Z = 4

29. Page 28 of 31
h =
Z
Minimum S
− 1 =
4
0.2
− 1
T =
T
h
= 2.80kn
For the second case
T =
(a + q )tanβ
tan(β + φ )
=
(326 + 18 × 6)tan(45° − 19°)
tan(45° − 19° + 38°)
= 103.241kn/m
T =
T
h
at, Z = 6
h =
Z
Minimum S
− 1 =
6
0.2
− 1
T =
T
h
= 3.560kn
L =
T × FOS
2α tanφ × F
F = 19 × 0.2 + 18 = 21.8kpa
L =
T × FOS
2α tanφ × F
= 0.298meter
L
Ztanβ
=
5.8
6
=>
L
6 × 0.487
=
5.8
6
∴ L = 2.8246meter
L + L = 2.8246 + 0.298
= 3.1226< 18 Hence OK
Seismic case
Effective self-weight of soil wedge=W
Case–1
Z ≤ H − H
W =
1
2
γ Z tanβ
= a tanβ
Case–2
W = a tanβ

30. Page 29 of 31
a =
1
2
{γ [2Z − (H − H )] × (H − H ) + γ [Z − (H − H )] }
Total self-weight of soil wedge
Case–1
Z ≤ H − H
W = a tanβ, a =
1
2
γ Z
Case–2
Z ≥ H − H
a =
1
2
{γ [2Z − (H − H )] × (H − H ) + γ [Z − (H − H )] }
Force equilibrium of wedge
ΣH = 0
T − W α − Qα = R s(φ + β)
ΣV = 0
Rsin(φ + β) = W + W α + Qα + Q
We will consider only +α or it will add up to the weight.
Solving,
T =
a tanβ
tan(φ + β)
+ a tanβ
a = (a + q )α
a = a + a α + (1 + α )q
Critical angle of inclination
β = β = tan
−2a ± 4a − 4 a − a
2(a − a )
a = a cos(2φ ) − a + a sin(2φ )
a = a sin(2φ ) − a cos(2φ )
∴ T =
a tanβ
tan(φ + β )
+ a tanβ ≤ 1.25 × R N
Anchorage length of reinforcement at any depth for seismic condition
L =
T × FOS
2αtanφ × F
Total length of reinforcement required
L = (Z − h )tanβ + L
From the excel sheet it is seen that,
T Occurs for Z= 6meter
T = =
.
= 5.49 < 1.25 × 16.5 Hence safe
FOS = 1.5
α = 0.5
F = 21.8 kpa
β = 33.01° (Average of least and highest value of β)

31. Page 30 of 31
L =
T × FOS
2αtanφ × F
=
5.49 × 1.5
2 × 0.5 × tan38° × 21.8
= 0.48 meter
L = (Z − h )tanβ + L
= (6 − 0.2)tan33.01° + 0.48
= 4.24
∴ 4.24 < 18 .

32. Z(m) a3 a4 a5 a6 a7 a8 ß tanß Tmax(Kn/m) ni Tmax/ni(Kn/m)
0.50 2.38 2.38 2.05 12.40 -7.40 11.53 32.96 0.65 4.11 2.00 2.05
1.00 9.50 9.50 4.95 29.98 -17.90 27.88 32.96 0.65 9.92 4.00 2.48
1.50 21.38 21.38 8.71 52.73 -31.49 49.04 32.96 0.65 17.46 7.00 2.49
2.00 38.00 38.00 13.32 80.66 -48.17 75.02 32.96 0.65 26.71 9.00 2.97
2.50 59.38 59.38 18.79 113.77 -67.95 105.81 32.96 0.65 37.67 12.00 3.14
3.00 85.50 85.50 25.11 152.06 -90.81 141.42 32.96 0.65 50.34 14.00 3.60
3.50 116.38 116.38 32.29 195.52 -116.77 181.85 32.96 0.65 64.74 17.00 3.81
4.00 152.00 152.00 40.32 244.16 -145.82 227.09 32.96 0.65 80.84 19.00 4.25
4.50 191.38 192.63 49.25 297.00 -177.17 276.18 32.99 0.65 98.47 22.00 4.48
5.00 233.50 238.50 59.13 353.07 -210.06 328.18 33.07 0.65 117.43 24.00 4.89
5.50 278.38 289.63 69.95 412.35 -244.46 383.07 33.17 0.65 137.74 27.00 5.10
6.00 326.00 346.00 81.72 474.86 -280.39 440.85 33.29 0.66 159.38 29.00 5.50