Geotechnical Design of Gabion Wall

7/29/2019 Geotechnical Design of Gabion Wall

1/238

RESEARCH & DEVELOPMENT

REPORT NO. RD 1045

GEOTECHNICAL DESIGN OF GABION WALL

Mainland North Division

Drainage Services Department

Version no. : 3.0

November 2006


2/238


3/238

RD 1045Geotechnical Design of Gabion Wall Mainland North Division, DSD

Table of Content

Page

1. Scope and Qualifications 1

2. General Background 1

3. Design Considerations of Gabion Wall used in River Embankment 2-5

3.1 Treatment of the Foundation of Gabion Wall 2

3.2 Provision of Gabion Aprons 2-5

3.3 Provision of Geotextile Filter 5

4. Construction of Gabion Wall 6-8

4.1 Packing and Assembly 6

4.2 Installation and Filling 6-8

4.3 Gabion Stone Placement 8

4.4 Lid Closing 8

5. Installation of Reno Mattress 8-9

6. Sample Particular Specifications, Method of Measurement and 9

Schedule of Rates for Gabion Wall & Reno Mattress

7. Maintenance Related Considerations and Maintenance Requirements 9-11

8. Reference Documents 11

Appendix

A. Typical Layout of Gabion Wall

B. Design Calculations for Gabion Wall

C. Sample Particular Specifications, Method of Measurement and Schedule of Rates

for Gabion Wall & Reno Mattress


4/238


Date: November 2006 Page 1-

1. Scope and Qualifications

This paper gives technical guidance for the design of gabion wall used in river

embankment. It also stipulates the requirements for Reno Mattress against the local

scouring at the toe of gabion wall.

This paper is not applicable to revetment structures other than the vertical faced

gabion wall structures for the protection of river embankment.

This paper does not take into consideration wave forces or other hydrodynamic

forces arising out of supercritical flow, curvature flow, ship waves etc. acting on the

gabion wall. Therefore, the designer should treat the guidance with great caution

when using the guidance for the design of gabions used for coastal protection and in

engineered channels. If in doubt, the designer should consult engineers with

knowledge/experience on hydrodynamics and suppliers of gabion structures.

This paper assumes that gabion wall would sit on top of good soil foundation.

Before carrying out the design of gabion wall, the designer should ensure that the

foundation of the gabion wall should have been properly investigated.

2. General Background

Gabions are employed for many uses due to their versatility, which includes

hydraulic structures in river training works and in protection works for roads and land

reclamation. The gabions are steel wire cages that vary in size and are designed to

abate the destructive forces of erosion. Gabions are uniquely woven by twisting each

pair of wires one and one half turns continuously providing the inherent strength and

flexibility required. Gabion cages are normally designed to contain quarry run or

river run stones available at the site of erection. Cages are stacked to construct

structures of great durability and flexibility. The formed structure is capable of

carrying stress in biaxial tension. Gabion cages are not merely containers of stone

since each unit is securely connected to each adjacent cage during construction. The

wire mesh is monolithic through the structure in three dimensions, from top to bottom,

end to end, and from outer face to inner face. It is, therefore, apparent that the wire

reinforces the stone filling in tension.

Gabions form flexible structures that can deflect and deform to a certain limit in


5/238



any direction without fracture. It can withstand the movement of ground without

inordinate structure deformation. This attribute enables the gabion structure to be

built with a minimum foundation preparation. Gabion structures behave as

perforated barriers, allowing water to gradually pass through them. This is a

valuable characteristic in that hydrostatic pressure never builds up behind or under the

structure and cause failure to the gabion design. Gabion structures are regarded as

permanent. In the early stages after installation, siltation takes place between the

stone fill promoting vegetation and adding to the permanency of the structure. In

view of the environmentally friendly nature of the gabion construction as compared to

concrete, gabions are becoming more popular in engineering works in river

embankments which demand a natural looking environment with growth of vegetation

and potential for ecological lives.

3. Design Considerations of Gabion Wall used in River Embankment

There is currently no universally accepted method for designing gabion walls.

However, it is suggested in GEOGUIDE 1 Guide to Retaining Wall Design, Second

Edition, that gabion walls should be considered as gravity retaining wall for the

purpose of design.

The detailed design calculations for gabion wall of retaining height ranging from

1m to 4m, used in river embankment are shown in Appendix B.

3.1 Treatment of the Foundation of Gabion Wall

Foundation treatment is important to the stability of gabion wall as weak

foundation may result in bearing failure or soil slip. Since it largely depends on the

soil conditions which may varies significantly for different locations, designers should

consider the requirements of treatment of foundation case by case. If necessary,

rockfill and/or other appropriate measures as determined by the designers should be

adopted to stabilize the formation before placing gabions.

3.2 Provision of Gabion Aprons

Gabion aprons are commonly used to protect the toe of a gabion retaining wall

structure from scour that could cause undermining in channel works applications. It

is recommended that the gabion apron in the form of Reno Mattress, (refer to Section


6/238



5.0) be a minimum of 300 mm in depth. The length of the gabion apron shall extend

beyond the toe of the structure a minimum of 2 times the anticipated depth of scour

formed under the apron. This will ensure that the gabion apron reaches beyond the

outer limit of the anticipated scour hole that may form. For fast-flowing rivers,

designers need to determine the exact depth and extension of Reno Mattress case by

case with the consideration of scouring at river invert during peak flow.

Scour occurs at toe of gabion retaining wall when it obstructs the channel flow.

The flow obstructed by the gabions form a horizontal vortex starting at the upstream

end of the gabions and running along the toe of the gabions, and a vertical wake

vortex at the downstream end of the gabions.

In accordance with Hydraulic Engineering Circular No. 18 Evaluating Scour At

Bridges, Fourth Edition, Froehlich's live-bed scour equation can be used to obtain the

potential depth of scour.

Froehlich's Live-Bed Scour Equation

where:

K1 = Coefficient for shape

Shape Coefficients

Description K1

Vertical-wall 1.00

Vertical-wall with wing walls 0.82

Spill-through 0.55


7/238



Fig. 3.1 Abutment shape

K2 = Coefficient for angle of embankment to flow = ( / 90) 0.13

( < 90 if wall points downstream

> 90 if wall points upstream )

L = Length of active flow obstructed by the wall, m

Ae = Flow area of the approach cross section obstructed by the wall, m2

Fr = Froude Number of approach flow upstream of the wall = Ve/(gya)

1/2

Ve = Qe/Ae, m/s

Qe = Flow obstructed by the wall and approach structure, at peak flow, m3/s

ya = Average depth of flow on the floodplain (Ae/L), m

L = Length of wall projected normal to the flow, m

ys = Scour depth, m

Fig. 3.2 Orientation of embankment angle, , to the flow


8/238



Fig. 3.3 Determination of length of embankment blocking live flow for abutment

scour estimation

Example:

Assume

K1 = 0.82,

K2 = ( 90/ 90) 0.13 = 1L and ya are the base width and retaining height of the gabion wall as shown in

the drawing in Appendix A.

Computed Scour Depth, ys as follow:

Frya L

0.25 0.5 0.75 1 1.5 2

1 1.5 2.01 2.54 2.97 3.35 4.01 4.59

2 2.25 3.78 4.72 5.49 6.15 7.32 8.34

3 2.75 5.45 6.74 7.79 8.71 10.31 11.71

4 3.25 7.10 8.73 10.06 11.22 13.25 15.03

3.3 Provision of Geotextile Filter

The gabion apron will require minimal excavation and grade work. Generally the

gabion apron and gabion block are placed directly on the ground utilizing a geotextile

filter fabric between the gabions and soil interface to prevent leaching of soils

underneath the gabions.


9/238



The drawings inAppendix A show the details of gabion wall of retaining height

ranging from 1m to 4m, used in river embankment.

4. Construction of Gabion Wall

4.1 Packing and Assembly

Packing

(i) For ease of handling and shipping, the gabions are bundled folded flat.

Assembly

(i) Open the bundle and unfold each unit.

(ii) Lift the sides, the ends and the diaphragms of each unit into vertical position.

(iii) Attach the sides of four corners together with locking wire fastener ortying wire

and the diaphragms to the front and back of the gabion.

(iv) The tying operation begins at the top of the cage. The tying wire is laced around

theselvedge through each mesh all the way to the bottom of the cage.

4.2 Installation and Filling

Installation

(i) Empty gabion baskets shall be assembled individually and placed on the

approved surface to the lines and grades as shown or as directed, with the

position of all creases and that the tops of all sides are level.

(ii) All gabion baskets shall be properly staggered horizontally and vertically.

Finished gabion structures shall have no gaps along the perimeter of the contact

surfaces between adjoining units.


10/238



Fig. 4.1 Abutment shape

(Courtesy of and adapted from TerraAqua Gabions)

(iii) All adjoining empty gabion units shall be connected along the perimeter of their

contact surfaces in order to obtain a monolithic structure. All lacing wire

terminals shall be securely fastened.

(iv) All joining shall be made through selvedge-selvedge wire connection;

mesh-mesh wire connection is prohibited unless necessary.

Filling

(i) The initial line of gabion basket units shall be placed on the prepared filter layer

surface and adjoining empty baskets set to line and grade, and common sides

with adjacent units thoroughly laced or fastened. They shall be placed in a

manner to remove any kinks or bends in the mesh and to uniform alignment.

The basket units then shall be partially filled to provide anchorage against

deformation and displacement during the filling operation.


11/238



(ii) Deformation and bulging of the gabion units, especially on the wall face, shall

be corrected prior to additional stone filling. Care shall be taken, when placing

the stone by hand or machine, to assure that the PVC coating on the gabions will

not be damaged if PVC is utilized. All stone on the exposed face shall be hand

placed to ensure a neat compact appearance.

(iii) Gabions shall be uniformly overfilled by about 2540 mm to account for future

structural settlements and for additional layers. Gabions can be filled by any

kind of earth filling equipment. The maximum height from which the stones

may be dropped into the baskets shall be 900 mm.

4.3 Gabion Stone Placement

(i) The stone fill shall be placed into the gabion units in 300 mm lifts. Cells shall be

filled to a depth not exceeding 300 mm at a time. The fill layer should never be

more than 300 mm higher then any adjoining cell.

(ii) Connecting wires shall be installed from the front to back and side to side of

individual cell at each 300 mm vertical interval for gabions of depth exceeding

500 mm.

(iii) The voids shall be minimized by using well-graded stone fill and by hand

placement of the facing in order to achieve a dense, compact stone fill.

4.4 Lid Closing

(i) The lids of the gabion units shall be tightly secured along all edges, ends and

diaphragms in the same manner as described for assembling.

5.0 Installation of Reno Mattress

Basically, the procedure for installation of reno mattress is similar to the

construction of gabion units. Particular attention should be paid to the following :

(i) Mattress units should be placed in proper position so that movement of rockfill

inside the cage, due to gravity or flowing current, is minimal.

Thus, on slopes, Mattresses should be placed with its internal diaphragms at

right angles to the direction of the slope.

On river beds, position the Mattress with the internal diaphragms at right anglesto the direction flow.


12/238



(ii) The Mattresses may be either telescoped or cut to form and tied at required

shape when necessary, for example, when Mattresses are laid on a radius. For

a sharp curve, it may be necessary to cut the Mattress diagonally into triangular

sections and tie the open side securely to an intact side panel.

6. Sample Particular Specifications, Method of Measurement and

Schedule of Rates for Gabion Wall and Reno Mattress

Sample clauses of PS, MM and SoR for gabion wall and Reno Mattress are

shown inAppendix C.

7. Maintenance Related Considerations and Maintenance Requirements

Geoguide 1 (Sections 9.5 and 13) may be referenced for the basis of providing a

general guideline on maintenance of gabion walls.

Generally speaking, maintenance requirements should be duly considered during

both the design stage and during routine inspection after completion of works [Ref.

8.5]. Detailed discussion on the maintenance requirement both in detailed design

stage and routine inspection are beyond the scope of this Technical Report. The

necessary maintenance requirements should be judged on a case-by-case basis.

However, some of the important considerations required to be considered during

detailed design stage and routine inspection are listed below.

Suggested considerations on maintenance requirements to be looked at during

design stage :

The water quality of river/stream would affect the durability of the wireused in the basket. The suitability of the gabion structures to be used in

such river/stream environment should be within manufacturers

recommendation. If necessary, corrosion protection measures should be

applied to wires, such as PVC coated galvanized steel wires;

Gradation of stone aggregates should be based on gabion thickness and grid

size. As a rule of thumb, the size of stone measured in the greatest

dimension should range from 150mm to 300mm. In addition, the smallest

stone size must generally be larger than the wire mesh openings (usually of


13/238



about 100mm); Package of stone aggregates should be manually performed instead of

mechanically performed. The mechanically package can cause unwanted

stress to the net. However, manually packing of stone aggregates should

not be over emphasized. Poorly packed gabions will cause undue

movements as well as excessive abrasion to the PVC coating. To allow for

the settlement of the stone aggregates, an over fill of about 25-40mm is

considered to be adequate;

The strength of the stone aggregates should be durable to resist the impact

from flood flow particularly if the flood flow is violent. The stress created

by the violent flood flow against gabions will lead to the shaking and

mutual thrust of stones inside gabions. If the stones are fragile, the stones

will start to crush into pieces small enough to fall outside the gabion net;

The opening of the gabion net can be torn away by the continuous thrust of

materials carried by runoff (e.g. sable, gravel, and rubble) against iron wires.

When the net opens, the stone filling it up fall out, and the structure loses all

its weight and, consequently, its function; and

Gabions structures are generally composed of superimposed layers ofgabion baskets. Special attention should be paid on gabion structures with a

stepped shape, only a part of the superimposed layer rests on a lower layer

of gabions. The remaining part rests directly on the earthfill. In this case,

the underlying earthfill has to be compacted carefully, and its adherence to

the lower layer of gabions should be ensured before surperimposing the

next layer.

Suggested considerations on maintenance requirements to be looked at during

routine inspection :

A gabion structure needs to be inspected annually and after each flood event.

However, a newly placed gabion structure is recommended to be inspected

for every 3 months or after each rainfall event whichever is the less;

Signs of undercutting or other instability should also be checked;

Any displacement or shifting of the wire baskets should need to be

corrected immediately;


14/238



Checking on the sign of damage or erosion of the river embankment should

be included; and

Checking for the wires of panels/cages for any signs of rusting and wearshould be included.

8. Reference Documents

8.1 U.S. Department of Transportation, Federal Highway Administration,

Hydraulic Engineering Circular No. 18 Evaluating Scour At Bridges, Fourth

Edition, May 2001.

8.2 U.S. Ohio Department of Natural Resources, Division of Water, Water Planning,

Stream Guide, Stream Management Guide No. 15 Gabion Revetments

8.3 U.S. Environmental Department of Naval Facilities Engineering Service Center,

Storm Water Best Management Practices Decision Support Tool #129 Gabions

8.4 Tricardi, Watershed Management Use of Gabions in Small Hydraulic Works

8.5 Geotechnical Engineering Office, Civil Engineering Department, the

Government of the Hong Kong Special Administration Region, GEOGUIDE

1 Guide to Retaining Wall Design, Second Edition, October 2003.


15/238




16/238


Appendix A

Typical Layout of Gabion Wall


17/238


18/238


19/238


20/238


App. A - 1

Drawing Notes:

1. All dimensions are in millimeters unless otherwise specified.

2. Depending on the soil conditions, designers should determine whether any

ground treatment for foundation is required in consideration with sliding,

bearing or soil slip failures.

3. Determination of Potential Scour Depth by Froehlich's Live-Bed Scour Equation

where:

K1 = Coefficient for shape

Shape Coefficients

Description K1

Vertical-wall 1.00

Vertical-wall with wing walls 0.82Spill-through 0.55

K2 = Coefficient for angle of embankment to flow = ( / 90) 0.13

( < 90 if wall points downstream

> 90 if wall points upstream )

L = Length of active flow obstructed by the wall, m

Ae

= Flow area of the approach cross section obstructed by the wall, m 2

Fr = Froude Number of approach flow upstream of the wall = Ve/(gya)1/2

Ve = Qe/Ae, m/s

Qe = Flow obstructed by the wall and approach structure, at peak flow, m3/s

ya = Average depth of flow on the floodplain (Ae/L), m

L = Length of wall projected normal to the flow, m

ys = Scour depth, m


21/238


App. A - 2

Example:

Assume

K1 = 0.82,

K2 = ( 90/ 90) 0.13 = 1

L and ya are the base width and retaining height of the gabion wall as shown in

the sketches in the calculations in Appendix B.

Computed Scour Depth, ys as follow:

FrYa L

0.25 0.5 0.75 1 1.5 2

1 1.5 2.01 2.54 2.97 3.35 4.01 4.59

2 2.25 3.78 4.72 5.49 6.15 7.32 8.34

3 2.75 5.45 6.74 7.79 8.71 10.31 11.71

4 3.25 7.10 8.73 10.06 11.22 13.25 15.03

4. Mesh shall be hexagonal double twist and shall not ravel if damaged. The

dimensions of the hexagon shall be 80 x 100 mm.

5. The gabion mesh shall be formed with 2.7 mm diameter mild steel wires, hot dip

galvanized to BS 443 and further coated with polyvinyl chloride (PVC).

6. The PVC coating shall be dark green in colour, has an average thickness of 0.5

mm and nowhere less than 0.4 mm.

7. The diameter of the mild steel lacing wire and selvedge wire shall be 2.2 mm and3.0 mm respectively, galvanized and coated with PVC in a similar way to the

mesh wire.

8. All wires shall be mild steel to BS 1052.

9. The gabion shall be formed from one continuous piece of mesh which includes the

lid.

10. All edges of gabions, diaphragms and end panels shall be mechanically selvedged


22/238


App. A - 3

in such a way as to prevent ravelling of the mesh and to develop the full strength

of the mesh.

11. The gabion shall be divided by diaphragms into cells which length shall not be

greater than 1m.

12. Infill to gabion shall be rock fill material of size 150 mm to 300 mm and shall be

placed in accordance with the manufacturers recommendations.

13. All front and side faces of the gabion wall shall be fixed with hand packed square

stones of approximately 300 x 200 x 200 mm in size.


23/238


App. A - 4


24/238


Appendix B

Design Calculations for Gabion Wall

Annexes

A. Design of 4.5m gabion wall A, A1 A10

B. Design of 3.5m gabion wall B, B1 B8

C. Design of 2.5m gabion wall C, C1 C6

D. Design of 1.5m gabion wall D, D1 D4

Page

E. Stone sizes and critical velocities for gabions E1 E3

(courtesy of and adapted from Maccaferri Gabions)


25/238

Project : Design of 4.5m Gabion Wall Annex A

Prepared by : NG Chun-ling (AE/TM5)

Checked by :

Subject : Design of 4.5m Gabion Wall

Design Statement

Design of 4.5m Gabion Wall

1. Design Data

(I) Materials

(A) Requirements

eoguide 1 Gabion Materials

ara. 9.5.3 (1) (i) They should not be susceptible to attack by fire and ultraviolet light.

(ii) They should be in form of hexagonal woven or square welded.

should be twisted together in pairs through three half turns, i.e. 'double-twisted' to form

the mesh.

that of the wire-mesh to prevent unravelling.

eoguide 1 (v) The gabion base, top and sides should be formed from a piece of mesh. The ends and

ara. 9.5.3 (1) & Fig. 47 diaphragms can be attached to this mesh by helical wires or other methods.

(vi) The mesh can stretch or contract in two directions in its own plane and thus a rectangular

wire-mesh basket filled with rock fragments can deform in any direction.(vii) The wires used for the wire mesh should be mild steel wire to BS 1052 (BSI, 1986b),

with a minimum tensile strength of 350 N/mm2.

(viii) The wires should be at least 2.7mm in diameter and galvanized.

(ix) For hexagonal wire-mesh the wires should be galvanized to BS 443 (BSI, 1990b) before

weaving.

(x) For welded mesh, the mesh panels should be hot dip galvanized to BS 729 (BSI, 1986c)

after welding. The making of panels with galvanized wires welded together is not

recommended as the welds are left unprotected.

(xi) If the soil and water conditions are aggressive, PVC (polyvinylchloride) coating should be

provided to the wires. The PVC coating should be at least 0.5mm thick and should

meet the requirements of BS 4102 (BSI, 1991c).

eoguide 1 Infill material

ara. 9.5.3 (1)

filled or 300mm , whichever is less.

at least be twice the largest dimension of the mesh aperture .

Reference

(iv) The edges of the mesh should be selvaged with wires of a diameter of about 1.5 times

Remarks

Design Statement

(iii) Hexagonal woven wire mesh is mechanically woven in a continuous sheet. The wires

(i) Maximum size of rock should not exceed two-thirds the depth of the gabion to be

(ii) The preferred size is 150mm to 300mm . The smallest dimension of the rock should

- Annex A page 1 -


26/238

(B) Assumptions

Gabion and Infill Materials

Block Size

eoguide 1 The gabions are in modules of 2m x 1m x 1m.

ara. 9.5.1

Mesh Size

8cm x 10cm x 2.7mm

Size of Infill Material = 250mm

efer toAnnex E Critical Velocity for water flow = 6.4 m/s

accaferri Gabions

Parameter

eoguide 1 Specific gravity of the rock, Gs =

ara. 9.5.2 (1) Porosity of the infill =

Mobilized angle of wall friction, =

Backfilling Material behind the existing wall

The properties of backfilling material are assumed to be

eoguide 1 (a) Unit weight = 21 kN/m3

able 8 (b) Effective shear strength, c' = 0 kPa

(c) Effective friction angle, ' = 35 o

Insitu Soil beneath the wall (foundation material)

The properties of insitu soil are assumed to be(a) Unit weight = 19 kN/m

3

(b) Effective shear strength, c' = 5 kPa

(c) Effective friction angle, ' = 35 o

(II) Loadings

Dead loads

Self weight of the proposed protection wall, earth pressure and hydrostatic pressure are

taken to be dead load for design. The unit weight of water was taken as 9.81 kN/m3.

Imposed load

eoguide 1 5kPa surcharge was assumed on the land side.

ara. 7.2.2 and Table 16 (Footpaths isolated from roads, cycle tracks and play areas)

(III) Water level of the pond

It is assumed that the most critical situation should be when the channel is completely dry,

which is taken to be the design case.

The groundwater level behind the proposed gabion wall is assumed to be one-third of the

retaining height.

2.6

0.4

0.0

- Annex A page 2 -


27/238

eoguide 1 Para. 6.2.2, Proposed retaining walls have no restraint against translation and rotation about the base,

able 20 and Figure 13 hence active state pressure is assumed and compaction-induced lateral pressure is not

considered.

2. Design Reference and Codes

Design Code

1. Geotechnical Engineering Office (1993), Geoguide 1 - Guide to Retaining Wall Design,

Second Edition.

Design Methodology

In accordance with Geoguide 1, the structures would be designed for both the ultimate limit

state (ULS) and the serviceability limit state (SLS).

eoguide 1, clause 4.3.4, Partial safety factor approach stipulated under Geoguide 1, clause 4.3.4 is adopted. The

able 6 and Table 7 minimum factors of safety recommended in Geoguide 1 are adopted.

Per meter run of the proposed retaining walls is considered for simplicity.

eoguide 1, clause 9.5.2 Limit state checks would be carried out at selected planes through the gabion wall, ignoring

the resistance contributed by the cage material and the connections between the cages.

For stepped walls, stability checks would be carried out at each major change in section

shape.

3. Checking the Stability of the Protection Wall

4.5m Gabion Wall

- Annex A page 3 -


28/238

Ultimate Limit Statement (ULS)

efer to Annex A1, A3, 1. Checking Overturning [OK if restoring moment > overturning moment]

5, A7 & A9

ra. 1 Step 6 Height of Toe Stability Stability

above foundation (without back batter) (with back batter)

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

efer to Annex A1, A3, 2. Checking Sliding [OK if resisting force > sliding force]

5, A7 & A9

ra. 1 Step 7 Height of Toe Stability Stability


m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

efer to Annex A1, A3, 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]

5, A7 & A9

ra. 1 Step 8 Height of Toe Stability Stabilityabove foundation (without back batter) (with back batter)

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

Serviceability Limit Statement (SLS)

efer to Annex A2, A4, 1. Check Overturning and Determine Eccentricity

6, A8 & A10 [OK if the resultant force acts within the middle third of the wall base]

ra. 1 Step 6

Height of Toe Stability Stability


m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

For details of calculations, please refer to the Annex A1 to A10.

2.5

0.0

0.5

0.0

0.5

1.5

3.5

0.0

0.5

1.5

3.5

2.5

3.5

0.0

0.5

1.5

2.5

1.5

2.5

3.5

- Annex A page 4 -


29/238

Construction Aspects

eoguide 1 (i) Horizontal internal bracing wires should be fitted between the outer and inner faces at about

ara. 9.5.4 300mm centres in woven mesh gabions which are deeper than 500mm.

(ii) The lids of the gabions should meet the top edges of the sides and ends when closed,

without leaving any gaps.

(iii) The mesh of the lids should be tied down to the tops of any diaphragms provided, as well as

to the tops of the sides and ends.

(iv) Whenever possible, the vertical joints between the units should be staggered in adjacent

courses.

Drainage provisions

eoguide 1 (i) A geotextile filter would be provided under the base and behind the rear face of the gabion

ra. 9.5.5 wall to prevent migration of fines from the backfill into the coarse rock infill.

(ii) Drainage layers at the rear face are normally not warranted. However, a drainage layer of

adequate permeability would be provided at the base of the wall to guard against erosion

of the foundation material.

References

1. Geotechnical Engineering Office (1993), Geoguide 1 - Guide to Retaining Wall Design, Geoguide 1

Second Edition.

- Annex A page 5 -


30/238

Project : Design of Gabion Wall Annex A1


Checked by :


Checking of Ultimate Limit State (toe at 0m above foundation)

Reference Remarks

1. Checking of Ultimate Limit State (toe at 0m above foundation)

750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 3000

W2 Pa1

1000 Pa2W3 Insitu soil

1000

W4 Pa3 Pa4 Pwh 1500

500 W5

Toe

Pwv

W1, W2, W3,W4,W5 = Self-weight of the proposed protection wall

Pa1 = Lateral pressure due to surcharge

Pav = Vertical component of active earth pressure

Pa2, Pa3, Pa4 = Horizontal component of active earth pressure

Pwh = Lateral pressure due to groundwaterPwv = Upthrust

Abbreviation Value Unit

Fill (Behind the Wall) Surcharge behind the wall 5 kN/m2

eoguide 1, clause 4.3.4, 21 kN/m3

Surcharge at the wall 0 kN/m2

able 6, Table 7 and c' 0 kN/m2

Height of R.W. 4.50 m

able 8 ' 35o

Water level (from bottom) 1.50 m

m 1.2 Base width of wall 4.2 m

'f 30.3o ( =tan

-1((tan ')/m ))

cv' 30o

Length of wall 100 m

Kah 0.330

(Note: m is taken to be 1.2 so as to cater for the unknown ground condition.)

Insitu Soil (Beneath the Wall) Gabions and Infill materials


Specific gravity of the rock, Gs 2.6


Porosity of the infill 0.4

able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1.2

'f 30.3o ( =tan

-1((tan ')/m ))

cf' 4.2 kN/m2 ( = c' / m )

eoguide 1, clause 5.12 bf 24.1o ( = b / m )

eoguide 1, clause 6.6 Kp 3.610 (/ = 0 / = 0)

g. 19 (Note: m

is taken to be 1.2 so as to cater for the unknown ground condition.)

- Annex A1 page 1 -


31/238

Step 1 Sliding Force (kN/m - run)

Pa1 = 0.330 x 5.00 x 4.50 = 7.42

Pa2 = 0.330 x 3.00 x 21 x 3.00 / 2 = 31.17

Pa3 = 0.330 x 3.00 x 21 x 1.50 = 31.17

Pa4 = 0.330 x 11 x 1.50 x 1.50 / 2 = 4.08

Pwh = 10 x 1.50 x 1.50 / 2 = 11.25

= 85.08

Pah = Pai = 73.83

oguide 1, Clause 5.11.2, Step 2 Vertical component of earth pressure (Pav)

ble 14 Based on the design assumption, the mobilised angle of wall friction, is taken as 0.

Thus, the vertical component of earth pressure (Pav) acting on the proposed protection wall is considered as

= 0.00 kN/m - run

Step 3 Wt. of wall + water uplift kN/m - run

W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

W5 = 26 x 0.6 x 4.20 x 0.50 = 32.76

Pwv = 10 x 1.50 x 4.20 / 2 = -31.50

Pav = 0.00 = 0.00

V= 146.34

Step 4 Overturning moment of earth pressure about Toe

Force (kN/ m) Moment

Pa1 7.42 4.50 / 2 = 2.25 16.70

Pa2 31.17 3.00 / 3 + 1.50 = 2.50 77.92

Pa3 31.17 1.50 / 2 = 0.75 23.37

Pa4 4.08 1.50 / 3 = 0.50 2.04

Pwh 11.25 1.50 / 3 = 0.50 5.63M = 125.65 (kNm/m run)

Step 5 Moment of Vertical Force about Toe


W1 20.28 4.2 - 1.30 / 2 = 3.55 71.99

W2 30.42 4.2 - 1.95 / 2 = 3.23 98.10

W3 41.34 4.2 - 2.65 / 2 = 2.88 118.85

W4 53.04 4.2 - 3.40 / 2 = 2.50 132.60

W5 32.76 4.2 - 4.20 / 2 = 2.10 68.80

Pwv -31.50 4.2 x 2 / 3 = 2.80 -88.20

Pav 0.00 4.2 = 4.20 0.00

M = 402.15 (kNm/m run)

Mr = 490.35 (kNm/m run)

Step 6 Check Overturning and Determine Eccentricity

Overturning Moment Mo = 125.65 + 88.20 - 0.00 = 213.85 kNm/m run

Restoring Moment Mr= 490.35 kNm/m run

=> Mr > Mo => OK!

OK!

Eccentricity e = B / 2 - ( Mr - Mo ) / V

= 4.2 / 2 - ( 490.35 - 213.85 ) / 146.34

= 0.211m

Arm (m)

Arm (m)

- Annex A1 page 2 -


32/238

Step 7 Check Sliding

Sliding Force Fa = H = 85.08 kN / m

V= 146.34 kN / m

Resisting Force against Sliding Fr =

= Ns x tan 'f= 146.34 x 0.58

= 85.39 kN/m run > Sliding Force Fa = H= 85.08 kN / m

=> OK!

OK!

Step 8 Check Bearing Capacity

From above, Eccentricity e = 0.211 m

Effective Width B' = B - 2 e= 4.20 - 2 x 0.211

= 3.78 m

Effective Length L' = L' = 100 m

Effective Area A' = B' x L'

= 3.78 x 100 = 377.88 m2

Sliding Force Qs = H x L' = 85.08 x 100 = 8508 kN

Normal Force Qn = V x L' = 146.34 x 100 = 14634 kN

Thus, at the level just below 500mm rock fill,

Bearing Pressure qmax = Qn / A'

= / 377.88 = 38.73 kPa

To calculate Bearing Capacity qult ,

qult = cf'Nc sc ic tc gc + 0.5 ' B' Nsitg + q' Nq sq iq tq gq

For Bearing Capacity Factors,

Nq = exp(tan'f) tan2(/4+'f/2)

= exp ( x tan'f ) x tan2( /4 + 'f / 2 )

= exp ( 3.14 x 0.584 ) x tan2( 0.785 + 30.3 / 2 )

= 18.96

Nc = ( Nq - 1 ) x cot 'f= ( 18.96 - 1 ) x cot 30.3

= 30.78

N= 2 x ( Nq + 1 ) x tan 'f

= 2 x ( 18.96 + 1 ) x tan 30= 23.30

For Shape Factors,

sc = 1 + Nq / Nc x B' / L'

= 1 + 18.96 / 30.78 x 3.78 / 100

= 1.02

s = 1 - 0.4 x B' / L'

= 1 - 0.4 x 3.78 / 100

= 0.98

sq = 1 + tan 'f x B' / L'

= 1 + tan 30.3 x 3.78 / 100

= 1.02

14634.00

- Annex A1 page 3 -


33/238

For Inclination Factors,

mi = ( 2 + B' / L' ) /

( 1 + B' / L' )

= ( 2 + 3.78 / 100 ) /

( 1 + 3.78 / 100 )

= 1.96Ki = Qs / ( Qn + cf' x A' x cot 'f )

= 8508 / ( 14634 + 4.2 x 377.88 x cot 30.3 )

= 0.49

i = ( 1 - Ki )mi + 1

= ( 1 - 0.49 )2.96

= 0.14

iq = ( 1 - Ki )mi

= ( 1 - 0.49 )1.96

= 0.27

ic = iq - ( 1 - iq ) / ( Nc x tan 'f )

= 0.27 - ( 1 - 0.27 ) / ( 30.78 x tan 30.3 )= 0.22

Since tilting of wall and inclination of ground slope both equal 0o

, all tilt and ground slope factors

equal 1, i.e. tc = t = tq = 1

gc = g = gq = 1

Effective Surcharge q' = ' x 0.5

= 9 x 0.5

= 4.5 kPa

As a result, Bearing Capacity qult

= cf ' Nc sc ic tc gc + 0.5 ' B' Nsitg + q' Nq sq iq tq gq= 4.17 x 30.78 x 1.02 x 0.22 x 1 x 1

+ 0.5 x 9 x 3.78 x 23.30 x 0.98 x

0.14 x 1 x 1 + 4.5 x 18.96 x 1.02

x 0.27 x 1 x 1

= 29.50 + 52.76 + 23.17

= 105.42 kPa

=> qult > qmax = 38.73 kPa => OK! OK!

- Annex A1 page 4 -


34/238

2. Checking of Ultimate Limit State (toe at 0m above foundation)

(with back batter 1:10)

750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 W2 3000

Pa1

1000 W3 Pa2

Insitu soil

1000 W4

Pa3 Pa4 Pwh 1500

500 W5

Toe

Pwv





Pwh = Lateral pressure due to groundwater

Pwv = Upthrust







able 8 ' 35o



'f 30.3o ( =tan

-1((tan ')/m ))

cv' 30o


Kah 0.330 Back batter, = 1 : 10 = 0.10

(Note: gm is taken to be 1.2 so as to cater for the unknown ground condition.)






able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1.2

'f 30.3

o

( =tan

-1

((tan ')/m ))cf' 4.2 kN/m

2 ( = c' / m )



g. 19 (Note: m is taken to be 1.2 so as to cater for the unknown ground condition.)


Pa1 = 0.330 x 5.00 x 4.50 = 7.42

Pa2 = 0.330 x 3.00 x 21 x 3.00 / 2 = 31.17

Pa3 = 0.330 x 3.00 x 21 x 1.50 = 31.17

Pa4 = 0.330 x 11 x 1.50 x 1.50 / 2 = 4.08

Pwh = 10 x 1.50 x 1.50 / 2 = 11.25

= 85.08Pah = Pai = 73.83

- Annex A1 page 5 -


35/238




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

W5 = 26 x 0.6 x 4.20 x 0.50 = 32.76

Pwv = 10 x 1.50 x 4.20 / 2 = -31.50

Pav = 0.00 = 0.00

V= 146.34


cos = 0.995 tan = 0.100


Pa1 7.42 4.50 / 2 = 2.25 16.70

Pa2 31.17 3.00 / 3 + 1.50 = 2.50 77.92

Pa3 31.17 1.50 / 2 = 0.75 23.37

Pa4 4.08 1.50 / 3 = 0.50 2.04

Pwh 11.25 1.50 / 3 = 0.50 5.63

M = 125.65 (kNm/m run)



W1 20.28 4.00 x 0.10 + 3.550 x 0.995 = 3.93 79.71

W2 30.42 3.00 x 0.10 + 3.225 x 0.995 = 3.51 106.70

W3 41.34 2.00 x 0.10 + 2.875 x 0.995 = 3.06 126.49W4 53.04 1.00 x 0.10 + 2.500 x 0.995 = 2.59 137.22

W5 32.76 0.25 x 0.10 + 2.100 x 0.995 = 2.11 69.27

Pwv -31.50 4.2 x 2 / 3 x 0.995 = 2.79 -87.76

Pav 0.00 4.2 = 4.20 0.00

M = 431.62 (kNm/m run)

Mr = 519.39 (kNm/m run)




=> Mr > Mo => OK!

OK!Eccentricity e = B / 2 - ( Mr - Mo ) / V

= 4.2 / 2 - ( 519.39 - 213.41 ) / 146.34

= 0.009m

Arm (m)

Arm (m)

(

(

((

(

( )

)

)

))

)

- Annex A1 page 6 -


36/238


oguide 1, Clause 9.2.3, Sliding along soil/structure interface

gure 41

echanism 1 Activating force Fa = H x cos - V x sin = 70.03 kN / m

Resisting force Fr= (V x cos - H x sin ) x tan b = 75.69 kN / m

Resisting Force against Sliding, Fr =

= 75.69 kN/m run > Activating Force Fa = 70.03 kN / m

=> OK!

oguide 1, Clause 9.2.3, Sliding along a foundation soil surface

gure 41 = 0.00o


Resisting force Fr= (V x cos - H x sin ) x tan ' + c' l = 102.89 kN / m



=> OK!

OK!



Effective Width B' = B - 2 e

= 4.20 - 2 x 0.009

= 4.18 m



= 4.18 x 100 = 418.16 m2





= / 418.16 = 35.00 kPa


qult

= cf

'

Nc

sc

ic

tc

gc

+ 0.5 ' B' N

s

i

t

g

+ q' Nq

sq

iq

tq

gq



= exp ( x tan'f ) x tan2( /4 + 'f / 2 )

= exp ( 3.14 x 0.584 ) x tan2( 0.785 + 30.3 / 2 )

= 18.96

Nc = ( Nq - 1 ) x cot 'f= ( 18.96 - 1 ) x cot 30.3

= 30.78

N= 2 x ( Nq + 1 ) x tan 'f= 2 x ( 18.96 + 1 ) x tan 30.3

= 23.30

14634

- Annex A1 page 7 -


37/238

For Shape Factors,

sc = 1 + Nq / Nc x B' / L'

= 1 + 18.96 / 30.78 x 4.18 / 100

= 1.03

s = 1 - 0.4 x B' / L'

= 1 - 0.4 x 4.18 / 100= 0.98

sq = 1 + tan 'f x B' / L'

= 1 + tan 30.3 x 4.18 / 100

= 1.02


mi = ( 2 + B' / L' ) /

( 1 + B' / L' )

= ( 2 + 4.18 / 100 ) /

( 1 + 4.18 / 100 )

= 1.96

Ki = Qs / ( Qn + cf' x A' x cot 'f )

= 8508 / ( 14634 + 4.2 x 418.16 x cot 30.3 )

= 0.48

i = ( 1 - Ki )mi + 1

= ( 1 - 0.48 )2.96

= 0.14

iq = ( 1 - Ki )mi

= ( 1 - 0.48 )1.96

= 0.27

ic = iq - ( 1 - iq ) / ( Nc x tan 'f )

= 0.27 - ( 1 - 0.27 ) / ( 30.78 x tan 30.3 )

= 0.23

For Tilt Factors,tc = tq - ( 1 - tq ) / ( Nc x tan 'f )

5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )

5.69

t = ( 1 - tan 'f )2

= ( 1 - 5.71 tan 30.3 )2

5.44

tq = t

= 5.44

Since inclination of ground slope equal 0o

, all ground slope factors

equal 1, i.e. gc = g = gq = 1

Effective Surcharge q' = ' x 0.5= 9 x 0.5

= 4.5 kPa


= cf ' Nc sc ic tc gc + 0.5 ' B' Nsitg + q' Nq sq iq tq gq= 4.17 x 30.78 x 1.03 x 0.23 x 5.69 x 1

+ 0.5 x 9 x 4.18 x 23.30 x 0.98 x

0.14 x 5.44 x 1 + 4.5 x 18.96 x 1.02

x 0.27 x 5.44 x 1

= 175.19 + 332.87 + 130.54

= 638.60 kPa


- Annex A1 page 8 -


38/238



Checked by :


Checking of Serviceability Limit State (toe at 0m above foundation)

Reference Remarks

1. Checking of Serviceability Limit State (toe at 0m above foundation)

750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 3000

W2 Pa1

1000 Pa2W3 Insitu soil

1000

W4 Pa3 Pa4 Pw 1500

500 W5

Toe

Pwv












able 8 ' 35o


m 1 Base width of wall 4.2 m

'f 35.0o ( =tan

-1((tan ')/m ))

cv' 30o

Kah 0.271






able 8 ' 35o

cv' 34o

b 28.9o (= 0.85cv' )

m 1

'f 35.0o ( =tan

-1((tan ')/m ))

cf' 5.0 kN/m2 ( = c' / m )



g. 19

- Annex A2 page 1 -


39/238


Pa1 = 0.271 x 5.00 x 4.50 = 6.10

Pa2 = 0.271 x 3.00 x 21 x 3.00 / 2 = 25.61

Pa3 = 0.271 x 3.00 x 21 x 1.50 = 25.61

Pa4 = 0.271 x 11 x 1.50 x 1.50 / 2 = 3.35

Pwh = 10 x 1.50 x 1.50 / 2 = 11.25

= 71.92

Pah = Pai = 60.67




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

W5 = 26 x 0.6 x 4.20 x 0.50 = 32.76Pwv = 10 x 1.50 x 4.20 / 2 = -31.50

Pav = 0.00 = 0.00

V= 146.34



Pa1 6.10 4.50 / 2 = 2.25 13.72

Pa2 25.61 3.00 / 3 + 1.50 = 2.50 64.02

Pa3 25.61 1.50 / 2 = 0.75 19.21

Pa4 3.35 1.50 / 3 = 0.50 1.68

Pwh 11.25 1.50 / 3 = 0.50 5.63

M = 104.25 (kNm/m run)



W1 20.28 4.2 - 1.30 / 2 = 3.55 71.99

W2 30.42 4.2 - 1.95 / 2 = 3.23 98.10

W3 41.34 4.2 - 2.65 / 2 = 2.88 118.85

W4 53.04 4.2 - 3.40 / 2 = 2.50 132.60

W5 32.76 4.2 - 4.20 / 2 = 2.10 68.80

Pwv -31.50 4.2 x 2 / 3 = 2.80 -88.20

Pav 0.00 4.2 = 4.20 0.00

M = 402.15 (kNm/m run)

Mr = 490.35 (kNm/m run)




=> Mr > Mo => OK!


= 4.2 / 2 - ( 490.35 - 192.45 ) / 146.34

= 0.064m

eoguide 1 By Middle-third Rule, B/6 = 4.2 / 6 = 0.700m > 0.064m OK! OK!

ause 9.2.4

Arm (m)

Arm (m)

- Annex A2 page 2 -


40/238

2. Checking of Serviceability Limit State (toe at 0m above foundation)


750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 W2 3000

Pa1

1000 W3 Pa2

Insitu soil

1000 W4

Pa3 Pa4 Pw 1500

500 W5

Toe

Pwv






Pwv = Upthrust







able 8 ' 35o



'f 35.0o ( =tan

-1((tan ')/m ))

cv' 30o

Kah 0.271 Back batter, = 1 : 10 = 0.10






able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1

'f 35.0

o

( =tan

-1

((tan ')/m ))cf' 5.0 kN/m

2 ( = c' / m )



g. 19


Pa1 = 0.271 x 5.00 x 4.50 = 6.10

Pa2 = 0.271 x 3.00 x 21 x 3.00 / 2 = 25.61

Pa3 = 0.271 x 3.00 x 21 x 1.50 = 25.61

Pa4 = 0.271 x 11 x 1.50 x 1.50 / 2 = 3.35

Pwh = 10 x 1.50 x 1.50 / 2 = 11.25

= 71.92Pah = Pai = 60.67

- Annex A2 page 3 -


41/238




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

W5 = 26 x 0.6 x 4.20 x 0.50 = 32.76

Pwv = 10 x 1.50 x 4.20 / 2 = -31.50

Pav = 0.00 = 0.00

V= 146.34


cos = 0.995 tan = 0.100


Pa1 6.10 4.50 / 2 = 2.25 13.72

Pa2 25.61 3.00 / 3 + 1.50 = 2.50 64.02

Pa3 25.61 1.50 / 2 = 0.75 19.21

Pa4 3.35 1.50 / 3 = 0.50 1.68

Pwh 11.25 1.50 / 3 = 0.50 5.63

M = 104.25 (kNm/m run)



W1 20.28 4.00 x 0.10 + 3.550 x 0.995 = 3.93 79.71

W2 30.42 3.00 x 0.10 + 3.225 x 0.995 = 3.51 106.70

W3 41.34 2.00 x 0.10 + 2.875 x 0.995 = 3.06 126.49W4 53.04 1.00 x 0.10 + 2.500 x 0.995 = 2.59 137.22

W5 32.76 0.25 x 0.10 + 2.100 x 0.995 = 2.11 69.27

Pwv -31.50 4.2 x 2 / 3 x 0.995 = 2.79 -87.76

Pav 0.00 4.2 = 4.20 0.00

M = 431.62 (kNm/m run)

Mr = 519.39 (kNm/m run)




=> Mr > Mo => OK!


= 4.2 / 2 - ( 519.39 - 192.01 ) / 146.34= -0.137m

eoguide 1 By Middle-third Rule, B/6 = 4.2 / 6 = 0.700m > -0.137m OK! OK!

ause 9.2.4

Arm (m)

Arm (m)

(

( )

)

)((

(

(

)

)

)

- Annex A2 page 4 -


42/238



Checked by :


Checking of Ultimate Limit State (toe at 0.5m above foundation)

Reference Remarks

1. Checking of Ultimate Limit State (toe at 0.5m above foundation)

750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 2667

W2 Pa1

1000 Insitu soil Pa2W3

1000 1333

W4 Pa3 Pa4 Pwh

500 W5

Toe

Pwv












able 8 ' 35o



'f 30.3o ( =tan

-1(tan '/m ))

cv' 30o


Kah 0.330







able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1.2

'f 30.3o ( =tan

-1(tan b/m ))

cf' 4.2 kN/m2 ( = c' / m )



g. 19 (Note: m

is taken to be 1.2 so as to cater for the unknown ground condition.)

- Annex A3 page 1 -


43/238


Pa1 = 0.330 x 5.00 x 4.00 = 6.60

Pa2 = 0.330 x 2.67 x 21 x 2.67 / 2 = 24.63

Pa3 = 0.330 x 2.67 x 21 x 1.33 = 24.63

Pa4 = 0.330 x 11 x 1.33 x 1.33 / 2 = 3.22

Pwh = 10 x 1.33 x 1.33 / 2 = 8.89

= 67.96

Pah = Pai = 59.07




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

Pwv = 10 x 1.33 x 3.40 / 2 = -22.67

Pav = 0.00 = 0.00

V= 122.41



Pa1 6.60 4.00 / 2 = 2.00 13.19

Pa2 24.63 2.67 / 3 + 1.33 = 2.22 54.72

Pa3 24.63 1.33 / 2 = 0.67 16.42

Pa4 3.22 1.33 / 3 = 0.44 1.43

Pwh 8.89 1.33 / 3 = 0.44 3.95M = 89.72 (kNm/m run)



W1 20.28 3.4 - 1.30 / 2 = 2.75 55.77

W2 30.42 3.4 - 1.95 / 2 = 2.43 73.77

W3 41.34 3.4 - 2.65 / 2 = 2.08 85.78

W4 53.04 3.4 - 3.40 / 2 = 1.70 90.17

Pwv -22.67 3.4 x 2 / 3 = 2.27 -51.38

Pav 0.00 3.4 = 3.40 0.00

M = 254.11 (kNm/m run)

Mr = 305.49 (kNm/m run)




=> Mr > Mo => OK!

OK!


= 3.4 / 2 - ( 305.49 - 141.09 ) / 122.41

= 0.357m

Arm (m)

Arm (m)

- Annex A3 page 2 -


44/238



V= 122.41 kN / m


= Ns x tan 'f= 122.41 x 0.58


=> OK!

OK!




= 2.69 m



= 2.69 x 100 = 268.59 m2





= / 268.59 = 45.58 kPa


qult = cf'Nc sc ic tc gc + 0.5 ' B' Nsitg + q' Nq sq iq tq gq



= exp ( x tan'f ) x tan2( /4 + 'f / 2 )

= exp ( 3.14 x 0.584 ) x tan2( 0.785 + 30.3 / 2 )

= 18.96

Nc = ( Nq - 1 ) x cot 'f= ( 18.96 - 1 ) x cot 30.3

= 30.78

N= 2 x ( Nq + 1 ) x tan 'f

= 2 x ( 18.96 + 1 ) x tan 30.3= 23.30

For Shape Factors,

sc = 1 + Nq / Nc x B' / L'

= 1 + 18.96 / 30.78 x 2.69 / 100

= 1.02

s = 1 - 0.4 x B' / L'

= 1 - 0.4 x 2.69 / 100

= 0.99

sq = 1 + tan 'f x B' / L'

= 1 + tan 30.3 x 2.69 / 100

= 1.02

12241.33

- Annex A3 page 3 -


45/238


mi = ( 2 + B' / L' ) /

( 1 + B' / L' )

= ( 2 + 2.69 / 100 ) /

( 1 + 2.69 / 100 )

= 1.97

Ki = Qs / ( Qn + cf' x A' x cot 'f )= 6796 / ( 12241 + 4.2 x 268.59 x cot 30.3 )

= 0.48

i = ( 1 - Ki )mi + 1

= ( 1 - 0.48 )2.97

= 0.14

iq = ( 1 - Ki )mi

= ( 1 - 0.48 )1.97

= 0.28

ic = iq - ( 1 - iq ) / ( Nc x tan 'f )

= 0.28 - ( 1 - 0.28 ) / ( 30.78 x tan 30.3 )

= 0.23




gc = g = gq = 1


= 9 x 0.5

= 4.5 kPa


= cf ' Nc sc ic tc gc + 0.5 ' B' Nsitg + q' Nq sq iq tq gq= 4.17 x 30.78 x 1.02 x 0.23 x 1 x 1

+ 0.5 x 9 x 2.69 x 23.30 x 0.99 x

0.14 x 1 x 1 + 4.5 x 18.96 x 1.02

x 0.28 x 1 x 1

= 30.60 + 39.85 + 23.84

= 94.29 kPa


- Annex A3 page 4 -


46/238



750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 W2 2667

Pa1

1000 W3 Insitu soil Pa2

1000 W4 1333

Pa3 Pa4 Pwh

500 W5

Toe

Pwv






Pwv = Upthrust







able 8 ' 35o



'f 30.3o ( =tan

-1(tan '/m ))

cv' 30o


Kah 0.330 Back batter, = 1 : 10 = 0.10







able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1.2

'f 30.3

o

( =tan

-1

(tan '/m ))cf' 4.2 kN/m

2 ( = c' / m )





Pa1 = 0.330 x 5.00 x 4.00 = 6.60

Pa2 = 0.330 x 2.67 x 21 x 2.67 / 2 = 24.63

Pa3 = 0.330 x 2.67 x 21 x 1.33 = 24.63

Pa4 = 0.330 x 11 x 1.33 x 1.33 / 2 = 3.22

Pwh = 10 x 1.33 x 1.33 / 2 = 8.89

= 67.96Pah = Pai = 59.07

- Annex A3 page 5 -


47/238




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

Pwv = 10 x 1.33 x 3.40 / 2 = -22.67

Pav = 0.00 = 0.00

V= 122.41


cos = 0.995 tan = 0.100


Pa1 6.60 4.00 / 2 = 2.00 13.19

Pa2 24.63 2.67 / 3 + 1.33 = 2.22 54.72

Pa3 24.63 1.33 / 2 = 0.67 16.42

Pa4 3.22 1.33 / 3 = 0.44 1.43

Pwh 8.89 1.33 / 3 = 0.44 3.95

M = 89.72 (kNm/m run)



W1 20.28 3.50 x 0.10 + 2.750 x 0.995 = 3.08 62.56

W2 30.42 2.50 x 0.10 + 2.425 x 0.995 = 2.66 80.97

W3 41.34 1.50 x 0.10 + 2.075 x 0.995 = 2.21 91.53

W4 53.04 0.50 x 0.10 + 1.700 x 0.995 = 1.74 92.36Pwv -22.67 3.4 x 2 / 3 x 0.995 = 2.26 -51.12

Pav 0.00 3.4 = 3.40 0.00

M = 276.29 (kNm/m run)

Mr = 327.41 (kNm/m run)




=> Mr > Mo => OK!

OK!

Eccentricity e = B / 2 - ( Mr - Mo ) / V= 3.4 / 2 - ( 327.41 - 140.84 ) / 122.41

= 0.176m

Arm (m)

Arm (m)

(

(

(

(( )

)

)

)

)

- Annex A3 page 6 -


48/238


oguide 1, Clause 9.2.3, Sliding along soil/structure interface

gure 41


Resisting force Fr= (V x cos - H x sin ) x tan b = 63.49 kN / m



=> OK!

oguide 1, Clause 9.2.3, Sliding along a foundation soil surface

gure 41 = 0.00o


Resisting force Fr= (V x cos - H x sin ) x tan ' + c' l = 85.60 kN / m



=> OK!

OK!



Effective Width B' = B - 2 e

= 3.40 - 2 x 0.176

= 3.05 m



= 3.05 x 100 = 304.82 m2





= / 304.82 = 40.16 kPa


qult

= cf

'

Nc

sc

ic

tc

gc

+ 0.5 ' B' N

s

i

t

g

+ q' Nq

sq

iq

tq

gq



= exp ( x tan'f ) x tan2( /4 + 'f / 2 )

= exp ( 3.14 x 0.584 ) x tan2( 0.785 + 30.3 / 2 )

= 18.96

Nc = ( Nq - 1 ) x cot 'f= ( 18.96 - 1 ) x cot 30.3

= 30.78

N= 2 x ( Nq + 1 ) x tan 'f= 2 x ( 18.96 + 1 ) x tan 30.3

= 23.30

12241

- Annex A3 page 7 -


49/238

For Shape Factors,

sc = 1 + Nq / Nc x B' / L'

= 1 + 18.96 / 30.78 x 3.05 / 100

= 1.02

s = 1 - 0.4 x B' / L'

= 1 - 0.4 x 3.05 / 100

= 0.99

sq = 1 + tan 'f x B' / L'= 1 + tan 30.3 x 3.05 / 100

= 1.02


mi = ( 2 + B' / L' ) /

( 1 + B' / L' )

= ( 2 + 3.05 / 100 ) /

( 1 + 3.05 / 100 )

= 1.97

Ki = Qs / ( Qn + cf' x A' x cot 'f )

= 6796 / ( 12241 + 4.2 x 304.82 x cot 30.3 )

= 0.47

i = ( 1 - Ki )mi + 1

= ( 1 - 0.47 )2.97

= 0.15

iq = ( 1 - Ki )mi

= ( 1 - 0.47 )1.97

= 0.28

ic = iq - ( 1 - iq ) / ( Nc x tan 'f )

= 0.28 - ( 1 - 0.28 ) / ( 30.78 x tan 30.3 )

= 0.24

For Tilt Factors,

tc = tq - ( 1 - tq ) / ( Nc x tan 'f )

5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )

5.69

t = ( 1 - tan 'f )2

= ( 1 - 5.71 tan 30.3 )2

5.44

tq = t

= 5.44

Since inclination of ground slope equal 0o

, all ground slope factors

equal 1, i.e. gc = g = gq = 1


= 9 x 0.5

= 4.5 kPa


= cf ' Nc sc ic tc gc + 0.5 ' B' Nsitg + q' Nq sq iq tq gq= 4.17 x 30.78 x 1.02 x 0.24 x 5.69 x 1

+ 0.5 x 9 x 3.05 x 23.30 x 0.99 x

0.15 x 5.44 x 1 + 4.5 x 18.96 x 1.02

x 0.28 x 5.44 x 1

= 182.00 + 258.46 + 134.51

= 574.98 kPa


- Annex A3 page 8 -


50/238



Checked by :


Checking of Serviceability Limit State (toe at 0.5m above foundation)

Reference Remarks

1. Checking of Serviceability Limit State (toe at 0.5m above foundation)

750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 2667

W2 Pa1

1000 Insitu soil Pa2W3

1000 1333

W4 Pa3 Pa4 Pw

500 W5

Toe

Pwv












able 8 ' 35o



'f 35.0o ( =tan

-1(tan '/m ))

cv' 30o

Kah 0.271






able 8 ' 35o

cv' 34o

b 28.9o (= 0.85cv' )

m 1

'f 35.0o ( =tan

-1(tan '/m ))

cf' 5.0 kN/m2 ( = c' / m )



g. 19

- Annex A4 page 1 -


51/238


Pa1 = 0.271 x 5.00 x 4.00 = 5.42

Pa2 = 0.271 x 2.67 x 21 x 2.67 / 2 = 20.23

Pa3 = 0.271 x 2.67 x 21 x 1.33 = 20.23

Pa4 = 0.271 x 11 x 1.33 x 1.33 / 2 = 2.65

Pwh = 10 x 1.33 x 1.33 / 2 = 8.89

= 57.43

Pah = Pai = 48.54




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

Pwv = 10 x 1.33 x 3.40 / 2 = -22.67

Pav = 0.00 = 0.00V= 122.41



Pa1 5.42 4.00 / 2 = 2.00 10.84

Pa2 20.23 2.67 / 3 + 1.33 = 2.22 44.96

Pa3 20.23 1.33 / 2 = 0.67 13.49

Pa4 2.65 1.33 / 3 = 0.44 1.18

Pwh 8.89 1.33 / 3 = 0.44 3.95

M = 74.42 (kNm/m run)



W1 20.28 3.4 - 1.30 / 2 = 2.75 55.77

W2 30.42 3.4 - 1.95 / 2 = 2.43 73.77

W3 41.34 3.4 - 2.65 / 2 = 2.08 85.78

W4 53.04 3.4 - 3.40 / 2 = 1.70 90.17

Pwv -22.67 3.4 x 2 / 3 = 2.27 -51.38

Pav 0.00 3.4 = 3.40 0.00

M = 254.11 (kNm/m run)

Mr = 305.49 (kNm/m run)


Overturning Moment Mo = 74.42 + 51.38 - 0.00 = 125.80 kNm/m runRestoring Moment Mr= 305.49 kNm/m run

=> Mr > Mo => OK!


= 3.4 / 2 - ( 305.49 - 125.80 ) / 122.41

= 0.232m

eoguide 1 By Middle-third Rule, B/6 = 3.4 / 6 = 0.567m > 0.232m OK! OK!

ause 9.2.4

Arm (m)

Arm (m)

- Annex A4 page 2 -


52/238

2. Checking of Serviceability Limit State (toe at 0.5m above foundation)


750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 W2 2667

Pa1

1000 W3 Pa2

Insitu soil

1000 W4 1333

Pa3 Pa4 Pw

500 W5

Toe

Pwv






Pwv = Upthrust







able 8 ' 35o



'f 35.0o ( =tan

-1(tan '/m ))

cv' 30o

Kah 0.271 Back batter, = 1 : 10 = 0.10






able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1

'f 35.0

o

( =tan

-1

(tan '/m ))cf' 5.0 kN/m

2 ( = c' / m )



g. 19


Pa1 = 0.271 x 5.00 x 4.00 = 5.42

Pa2 = 0.271 x 2.67 x 21 x 2.67 / 2 = 20.23

Pa3 = 0.271 x 2.67 x 21 x 1.33 = 20.23

Pa4 = 0.271 x 11 x 1.33 x 1.33 / 2 = 2.65

Pwh = 10 x 1.33 x 1.33 / 2 = 8.89

= 57.43Pah = Pai = 48.54

- Annex A4 page 3 -


53/238




= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04

Pwv = 10 x 1.33 x 4.20 / 2 = -22.67

Pav = 0.00 = 0.00

V= 122.41


cos = 0.995 tan = 0.100


Pa1 5.42 4.00 / 2 = 2.00 10.84

Pa2 20.23 2.67 / 3 + 1.33 = 2.22 44.96

Pa3 20.23 1.33 / 2 = 0.67 13.49

Pa4 2.65 1.33 / 3 = 0.44 1.18

Pwh 8.89 1.33 / 3 = 0.44 3.95

M = 74.42 (kNm/m run)



W1 20.28 3.50 x 0.10 + 2.750 x 0.995 = 3.08 62.52

W2 30.42 2.50 x 0.10 + 2.425 x 0.995 = 2.66 80.93

W3 41.34 1.50 x 0.10 + 2.075 x 0.995 = 2.21 91.49W4 53.04 0.50 x 0.10 + 1.700 x 0.995 = 1.74 92.35

Pwv -22.67 3.4 x 2 / 3 x 0.995 = 2.26 -51.12

Pav 0.00 3.4 = 3.40 0.00

M = 276.17 (kNm/m run)

Mr = 327.29 (kNm/m run)




=> Mr > Mo => OK!


= 3.4 / 2 - ( 327.29 - 125.54 ) / 122.41

= 0.052meoguide 1 By Middle-third Rule, B/6 = 3.4 / 6 = 0.567m > 0.052m OK! OK!

ause 9.2.4

Arm (m)

Arm (m)

(

(

( )

)

)(( )

)

- Annex A4 page 4 -


54/238



Checked by :


Checking of Ultimate Limit State (toe at 1.5m above foundation)

Reference Remarks


750 650 5 kPa

800 700 1300

1000 W1

Pav Pa1 Pa2

1000 2000

W2

1000 Insitu soil Pa3 Pa4 Pw 1000W3

1000

W4

500 W5

Toe

Pwv












able 8 ' 35o



'f 30.3o ( =tan

-1(tan '/m ))

cv' 30o


Kah 0.330







able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1.2

'f 30.3o ( =tan

-1(tan b/m ))

cf' 4.2 kN/m2

( = c' / m )




- Annex A5 page 1 -


55/238


Pa1 = 0.330 x 5.00 x 3.00 = 4.95

Pa2 = 0.330 x 2.00 x 21 x 2.00 / 2 = 13.85

Pa3 = 0.330 x 2.00 x 21 x 1.00 = 13.85

Pa4

= 0.330 x 11 x 1.00 x 1.00 / 2 = 1.81

Pwh = 10 x 1.00 x 1.00 / 2 = 5.00

= 39.46

Pah = Pai = 34.46

oguide 1, Clause 5.11.2, Step 2 Vertical component of earth pressure (P av)



= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

Pwv = 10 x 1.00 x 2.65 / 2 = -13.25

Pav = 0.00 = 0.00

V= 78.79



Pa1 4.95 3.00 / 2 = 1.50 7.42

Pa2 13.85 2.00 / 3 + 1.00 = 1.67 23.09

Pa3 13.85 1.00 / 2 = 0.50 6.93

Pa4 1.81 1.00 / 3 = 0.33 0.60

Pwh 5.00 1.00 / 3 = 0.33 1.67M = 39.70 (kNm/m run)



W1 20.28 2.65 - 1.30 / 2 = 2.00 40.56

W2 30.42 2.65 - 1.95 / 2 = 1.68 50.95

W3 41.34 2.65 - 2.65 / 2 = 1.33 54.78

Pwv -13.25 2.65 x 2 / 3 = 1.77 -23.41

Pav 0.00 2.65 = 2.65 0.00

M = 122.88 (kNm/m run)

Mr = 146.29 (kNm/m run)




=> Mr > Mo => OK!

OK!


= 2.65 / 2 - ( 146.29 - 63.11 ) / 78.79

= 0.269m

Arm (m)

Arm (m)

- Annex A5 page 2 -


56/238



V= 78.79 kN / m


= Ns x tan 'f= 78.79 x 0.58


=> OK!

OK!




= 2.11 m



= 2.11 x 100 = 211.14 m2





= / 211.14 = 37.32 kPa


qult = cf'Nc sc ic tc gc + 0.5 ' B' Ns i tg + q' Nq sq iq tq gq



= exp ( x tan'f ) x tan2( /4 + 'f / 2 )

= exp ( 3.14 x 0.584 ) x tan2( 0.785 + 30.3 / 2 )

= 18.96

Nc = ( Nq - 1 ) x cot 'f= ( 18.96 - 1 ) x cot 30.3

= 30.78

N= 2 x ( Nq + 1 ) x tan 'f= 2 x ( 18.96 + 1 ) x tan 30.3

= 23.30

For Shape Factors,

sc = 1 + Nq / Nc x B' / L'

= 1 + 18.96 / 30.78 x 2.11 / 100

= 1.01

s = 1 - 0.4 x B' / L'

= 1 - 0.4 x 2.11 / 100

= 0.99

sq = 1 + tan 'f x B' / L'

= 1 + tan 30.3 x 2.11 / 100

= 1.01

7879.00

- Annex A5 page 3 -


57/238


mi = ( 2 + B' / L' ) /

( 1 + B' / L' )

= ( 2 + 2.11 / 100 ) /

( 1 + 2.11 / 100 )

= 1.98Ki = Qs / ( Qn + cf' x A' x cot 'f )

= 3946 / ( 7879 + 4.2 x 211.14 x cot 30.3 )

= 0.42

i = ( 1 - Ki )mi + 1

= ( 1 - 0.42 )2.98

= 0.20

iq = ( 1 - Ki )mi

= ( 1 - 0.42 )1.98

= 0.34

ic = iq - ( 1 - iq ) / ( Nc x tan 'f )

= 0.34 - ( 1 - 0.34 ) / ( 30.78 x tan 30.3 )= 0.30




gc = g = gq = 1


= 9 x 0.5

= 4.5 kPa


= cf ' Nc sc ic tc gc + 0.5 ' B' Ns i tg + q' Nq sq iq tq gq

= 4.17 x 30.78 x 1.01 x 0.30 x 1 x 1+ 0.5 x 9 x 2.11 x 23.30 x 0.99 x

0.20 x 1 x 1 + 4.5 x 18.96 x 1.01

x 0.34 x 1 x 1

= 39.36 + 43.21 + 29.34

= 111.91 kPa


- Annex A5 page 4 -


58/238



750 650 5 kPa

800 700 1300

1000 W1

Pav

1000 W2 Pa1 Pa2 2000

1000 W3 Insitu soil Pa3 Pa4 Pw 1000

1000 W4

500 W5

Toe

Pwv






Pwv = Upthrust







able 8 ' 35o



'f 30.3o ( =tan

-1(tan '/m ))

cv' 30o


Kah 0.330 Back batter, = 1 : 10 = 0.10







able 8 ' 35o

cv' 34o

b 28.9o

(= 0.85fcv' )

m 1.2

'f 30.3 o ( =tan-1(tan '/m ))

cf' 4.2 kN/m2 ( = c' / m )

eoguide 1, clause 5.12 bf 24.1o

( = b / m )




Pa1 = 0.330 x 5.00 x 3.00 = 4.95

Pa2 = 0.330 x 2.00 x 21 x 2.00 / 2 = 13.85

Pa3 = 0.330 x 2.00 x 21 x 1.00 = 13.85

Pa4 = 0.330 x 11 x 1.00 x 1.00 / 2 = 1.81

Pwh = 10 x 1.00 x 1.00 / 2 = 5.00

= 39.46

Pah = Pai = 34.46

- Annex A5 page 5 -


59/238

oguide 1, Clause 5.11.2, Step 2 Vertical component of earth pressure (P av)



= 0.00 kN/m - run


W1 = 26 x 0.6 x 1.30 x 1.00 = 20.28

W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42

W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34

Pwv = 10 x 1.00 x 2.65 / 2 = -13.25

Pav = 0.00 = 0.00

V= 78.79


cos = 0.995 tan = 0.100


Pa1 4.95 3.00 / 2 = 1.50 7.42

Pa2 13.85 2.00 / 3 + 1.00 = 1.67 23.09

Pa3 13.85 1.00 / 2 = 0.50 6.93

Pa4 1.81 1.00 / 3 = 0.33 0.60

Pwh 5.00 1.00 / 3 = 0.33 1.67

M = 39.70 (kNm/m run)



W1 20.28 2.50 x 0.10 + 2.000 x 0.995 = 2.24 45.40

W2

Geotechnical Design of Gabion Wall

Documents

Transcript of Geotechnical Design of Gabion Wall