gabian wall design.pdf

RESEARCH & DEVELOPMENT

REPORT NO. RD 1045

GEOTECHNICAL DESIGN OF GABION WALL

Mainland North Division Drainage Services Department

Version no. : 3.0 November 2006

RD 1045 Geotechnical 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


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



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



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



Fig. 3.1 Abutment shape K2 = Coefficient for angle of embankment to flow = ( θ / 90) 0.13

( θ < 90° if wall points downstream θ > 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, θ, to the flow



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 = 1 L´ 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:

Fr ya 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.



The drawings in Appendix 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 or tying 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 the selvedge 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.



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.



(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 25–40 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 angles

to the direction flow.



(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 in Appendix 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 wire used in the basket. The suitability of the gabion structures to be used in such river/stream environment should be within manufacturer’s 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



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 of

gabion 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;



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 wear

should 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.


Appendix A

Typical Layout of Gabion Wall


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


Shape Coefficients

Description K1

Vertical-wall 1.00


Spill-through 0.55

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





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.


Fr Ya 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 and

3.0 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


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 manufacturer’s 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.


App. A - 4


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)

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) RequirementsGeoguide 1 Gabion MaterialsPara. 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.

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

Para. 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/mm 2 .(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).

Geoguide 1 Infill materialPara. 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 -

(B) Assumptions

Gabion and Infill Materials

Block Size

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

Para. 9.5.1

Mesh Size

8cm x 10cm x 2.7mm

Size of Infill Material = 250mm

Refer to Annex E Critical Velocity for water flow = 6.4 m/sMaccaferri Gabions

ParameterGeoguide 1 Specific gravity of the rock, Gs =

Para. 9.5.2 (1) Porosity of the infill =

Mobilized angle of wall friction, δ =

Backfilling Material behind the existing wall

The properties of backfilling material are assumed to be

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

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

(c) Effective friction angle, φ' = 35 o

Insitu Soil beneath the wall (foundation material)

The properties of insitu soil are assumed to be

(a) Unit weight = 19 kN/m3

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


(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 loadGeoguide 1 5kPa surcharge was assumed on the land side.

Para. 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 -

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

Table 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).

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

Table 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.

Geoguide 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 -

Ultimate Limit Statement (ULS)Refer to Annex A1, A3, 1. Checking Overturning [OK if restoring moment > overturning moment]A5, A7 & A9

para. 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!

Refer to Annex A1, A3, 2. Checking Sliding [OK if resisting force > sliding force]A5, A7 & A9



m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

Refer to Annex A1, A3, 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]A5, A7 & A9



m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

Serviceability Limit Statement (SLS)Refer to Annex A2, A4, 1. Check Overturning and Determine EccentricityA6, A8 & A10 [OK if the resultant force acts within the middle third of the wall base]para. 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 -

Construction Aspects

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

Para. 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

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

para. 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 -

Project : Design of Gabion Wall Annex A1


Checked by :

Subject : Design of 4.5m Gabion WallChecking 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 kPa800 700 1300

1000 W1

Pav

1000 3000W2 Pa1

1000 Pa2

W3 Insitu soil1000

W4 Pa3 Pa4 Pwh 1500500 W5

Toe

Pwv

W1, W2, W3, W4, W5 = Self-weight of the proposed protection wallPa1 = Lateral pressure due to surchargePav = Vertical component of active earth pressurePa2, Pa3, Pa4 = Horizontal component of active earth pressurePwh = Lateral pressure due to groundwaterPwv = Upthrust

Abbreviation Value UnitFill (Behind the Wall) Surcharge behind the wall 5 kN/m2

Geoguide 1, clause 4.3.4, γ 21 kN/m3 Surcharge at the wall 0 kN/m2

Table 6, Table 7 and c' 0 kN/m2 Height of R.W. 4.50 mTable 8 φ' 35 o Water level (from bottom) 1.50 m

γm 1.2 Base width of wall 4.2 mφ'f 30.3 o ( = tan-1((tan φ')/γm ))φcv' 30 o Length of wall 100 mKah 0.330(Note: γm is taken to be 1.2 so as to cater for the unknown ground condition.)

Insitu Soil (Beneath the Wall) Gabions and Infill materialsGeoguide 1, clause 4.3.4, γ 19 kN/m3 Specific gravity of the rock, Gs 2.6Table 6, Table 7 and c' 5 kN/m2 Porosity of the infill 0.4Table 8 φ' 35 o

φcv' 34 o

δb 28.9 o (= 0.85fcv' )γm 1.2φ'f 30.3 o ( = tan-1((tan φ')/γm ))cf' 4.2 kN/m2 ( = c' / γm )

Geoguide 1, clause 5.12 δbf 24.1 o ( = δb / γm )Geoguide 1, clause 6.6 Kp 3.610 (β/φ = 0 δ/φ = 0)Fig. 19 (Note: γm is taken to be 1.2 so as to cater for the unknown ground condition.)

- Annex A1 page 1 -

Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 4.50 = 7.42Pa2 = 0.330 x 3.00 x 21 x 3.00 / 2 = 31.17Pa3 = 0.330 x 3.00 x 21 x 1.50 = 31.17Pa4 = 0.330 x 11 x 1.50 x 1.50 / 2 = 4.08Pwh = 10 x 1.50 x 1.50 / 2 = 11.25

ΣΗ= 85.08Pah = ΣPai = 73.83

Geoguide 1, Clause 5.11.2, Step 2 Vertical component of earth pressure (Pav)Table 14 Based on the design assumption, the mobilised angle of wall friction, δ 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 - runW1 = 26 x 0.6 x 1.30 x 1.00 = 20.28W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04W5 = 26 x 0.6 x 4.20 x 0.50 = 32.76Pwv = 10 x 1.50 x 4.20 / 2 = -31.50Pav = 0.00 = 0.00

ΣV= 146.34

Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment

Pa1 7.42 4.50 / 2 = 2.25 16.70Pa2 31.17 3.00 / 3 + 1.50 = 2.50 77.92Pa3 31.17 1.50 / 2 = 0.75 23.37Pa4 4.08 1.50 / 3 = 0.50 2.04Pwh 11.25 1.50 / 3 = 0.50 5.63

ΣM = 125.65 (kNm/m run)

Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 4.2 - 1.30 / 2 = 3.55 71.99W2 30.42 4.2 - 1.95 / 2 = 3.23 98.10W3 41.34 4.2 - 2.65 / 2 = 2.88 118.85W4 53.04 4.2 - 3.40 / 2 = 2.50 132.60W5 32.76 4.2 - 4.20 / 2 = 2.10 68.80Pwv -31.50 4.2 x 2 / 3 = 2.80 -88.20Pav 0.00 4.2 = 4.20 0.00

ΣM = 402.15 (kNm/m run)ΣMr = 490.35 (kNm/m run)

Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 125.65 + 88.20 - 0.00 = 213.85 kNm/m runRestoring Moment ΣMr = 490.35 kNm/m run

=> ΣMr > ΣMo => OK!

OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV

= 4.2 / 2 - ( 490.35 - 213.85 ) / 146.34= 0.211m

Arm (m)

Arm (m)

- Annex A1 page 2 -

Step 7 Check SlidingSliding Force Fa = ΣH = 85.08 kN / m

ΣV= 146.34 kN / m

Resisting Force against Sliding Fr == Ns x tan φ'f= 146.34 x 0.58= 85.39 kN/m run > Sliding Force Fa = Σ H= 85.08 kN / m

=> OK!

OK!

Step 8 Check Bearing CapacityFrom 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 mEffective Area A' = B' x L'

= 3.78 x 100 = 377.88 m2

Sliding Force Qs = ΣH x L' = 85.08 x 100 = 8508 kNNormal Force Qn = ΣV 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 γ ' B' Nγ sγ iγ tγ gγ + q' Nq sq iq tq gq

For Bearing Capacity Factors,Nq = exp(πtanφ'f) tan2(π/4+φ'f/2)

= exp ( π x tanφ'f ) x tan2( π/4 + φ'f / 2 )= exp ( 3.14 x 0.584 ) x tan2( 0.785 + 30.3 / 2 )= 18.96

Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78

Nγ = 2 x ( Nq + 1 ) x tan φ'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

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 3.78 / 100= 0.98

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 3.78 / 100= 1.02

14634.00

- Annex A1 page 3 -

For Inclination Factors,mi = ( 2 + B' / L' ) /

( 1 + B' / L' )= ( 2 + 3.78 / 100 ) /

( 1 + 3.78 / 100 )= 1.96

Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 8508 / ( 14634 + 4.2 x 377.88 x cot 30.3 )= 0.49

iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.49 ) 2.96

= 0.14iq = ( 1 - Ki ) mi

= ( 1 - 0.49 ) 1.96

= 0.27ic = iq - ( 1 - iq ) / ( Nc x tan φ'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 factorsequal 1, i.e. tc = tγ = tq = 1

gc = gγ = gq = 1

Effective Surcharge q' = γ ' x 0.5= 9 x 0.5= 4.5 kPa

As a result, Bearing Capacity qult

= cf ' Nc sc ic tc gc + 0.5 γ ' B' Nγ sγ iγ tγ gγ + 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 1.02x 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 -

2. Checking of Ultimate Limit State (toe at 0m above foundation)(with back batter 1:10)

750 650 5 kPa800 700 1300

1000 W1

Pav

1000 W2 3000Pa1

1000 W3 Pa2

Insitu soil1000 W4

Pa3 Pa4 Pwh 1500500 W5

ToePwv





γm 1.2 Base width of wall 4.20 mφ'f 30.3 o ( = tan-1((tan φ')/γm ))φcv' 30 o Length of wall 100 mKah 0.330 Back batter, θ = 1 : 10 = 0.10(Note: gm is taken to be 1.2 so as to cater for the unknown ground condition.)


φcv' 34 o

δb 28.9 o (= 0.85fcv' )γm 1.2φ'f 30.3 o ( = tan-1((tan φ')/γm ))cf' 4.2 kN/m2 ( = c' / γm )



ΣΗ= 85.08Pah = ΣPai = 73.83

- Annex A1 page 5 -




ΣV= 146.34

Step 4 Overturning moment of earth pressure about Toe

cos θ = 0.995 tan θ = 0.100

Force (kN/ m) MomentPa1 7.42 4.50 / 2 = 2.25 16.70Pa2 31.17 3.00 / 3 + 1.50 = 2.50 77.92Pa3 31.17 1.50 / 2 = 0.75 23.37Pa4 4.08 1.50 / 3 = 0.50 2.04Pwh 11.25 1.50 / 3 = 0.50 5.63

ΣM = 125.65 (kNm/m run)

Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 4.00 x 0.10 + 3.550 x 0.995 = 3.93 79.71W2 30.42 3.00 x 0.10 + 3.225 x 0.995 = 3.51 106.70W3 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 = 2.59 137.22W5 32.76 0.25 x 0.10 + 2.100 x 0.995 = 2.11 69.27Pwv -31.50 4.2 x 2 / 3 x 0.995 = 2.79 -87.76Pav 0.00 4.2 = 4.20 0.00





= 4.2 / 2 - ( 519.39 - 213.41 ) / 146.34= 0.009m

Arm (m)

Arm (m)

((

(((( )

)))))

- Annex A1 page 6 -

Step 7 Check Sliding

Geoguide 1, Clause 9.2.3, Sliding along soil/structure interfaceFigure 41Mechanism 1 Activating force Fa = ΣH x cos θ - ΣV x sin θ = 70.03 kN / m

Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 75.69 kN / m

Resisting Force against Sliding, Fr == 75.69 kN/m run > Activating Force Fa = 70.03 kN / m

=> OK!

Geoguide 1, Clause 9.2.3, Sliding along a foundation soil surfaceFigure 41 Ω = 0.00 o

Mechanism 2 Activating force Fa = ΣH x cos Ω - ΣV x sin Ω = 85.08 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 102.89 kN / m


=> OK!

OK!




= 4.18 x 100 = 418.16 m2



= / 418.16 = 35.00 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78

Nγ = 2 x ( Nq + 1 ) x tan φ'f= 2 x ( 18.96 + 1 ) x tan 30.3= 23.30

14634

- Annex A1 page 7 -


= 1 + 18.96 / 30.78 x 4.18 / 100= 1.03

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 4.18 / 100= 0.98

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 4.18 / 100= 1.02


( 1 + B' / L' )= ( 2 + 4.18 / 100 ) /

( 1 + 4.18 / 100 )= 1.96


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.48 ) 2.96

= 0.14iq = ( 1 - Ki ) mi

= ( 1 - 0.48 ) 1.96

= 0.27ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.27 - ( 1 - 0.27 ) / ( 30.78 x tan 30.3 )= 0.23

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

5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44

Since inclination of ground slope equal 0o , all ground slope factorsequal 1, i.e. gc = gγ = gq = 1




= 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 1.02x 0.27 x 5.44 x 1

= 175.19 + 332.87 + 130.54= 638.60 kPa


- Annex A1 page 8 -



Checked by :

Subject : Design of 4.5m Gabion WallChecking 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 kPa800 700 1300

1000 W1

Pav

1000 3000W2 Pa1

1000 Pa2

W3 Insitu soil1000

W4 Pa3 Pa4 Pwh 1500500 W5

Toe

Pwv





γm 1 Base width of wall 4.2 mφ'f 35.0 o ( = tan-1((tan φ')/γm ))φcv' 30 o

Kah 0.271


φcv' 34 o

δb 28.9 o (= 0.85φcv' )γm 1φ'f 35.0 o ( = tan-1((tan φ')/γm ))cf' 5.0 kN/m2 ( = c' / γm )

Geoguide 1, clause 5.12 δbf 28.9 o ( = δb / γm )Geoguide 1, clause 6.6 Kp 3.610 (β/φ = 0 δ/φ = 0)Fig. 19

- Annex A2 page 1 -


ΣΗ= 71.92Pah = ΣPai = 60.67




ΣV= 146.34


Pa1 6.10 4.50 / 2 = 2.25 13.72Pa2 25.61 3.00 / 3 + 1.50 = 2.50 64.02Pa3 25.61 1.50 / 2 = 0.75 19.21Pa4 3.35 1.50 / 3 = 0.50 1.68Pwh 11.25 1.50 / 3 = 0.50 5.63

ΣM = 104.25 (kNm/m run)

Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 4.2 - 1.30 / 2 = 3.55 71.99W2 30.42 4.2 - 1.95 / 2 = 3.23 98.10W3 41.34 4.2 - 2.65 / 2 = 2.88 118.85W4 53.04 4.2 - 3.40 / 2 = 2.50 132.60W5 32.76 4.2 - 4.20 / 2 = 2.10 68.80Pwv -31.50 4.2 x 2 / 3 = 2.80 -88.20Pav 0.00 4.2 = 4.20 0.00



=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV

= 4.2 / 2 - ( 490.35 - 192.45 ) / 146.34= 0.064m

Geoguide 1 By Middle-third Rule, B/6 = 4.2 / 6 = 0.700m > 0.064m OK! OK!Clause 9.2.4

Arm (m)

Arm (m)

- Annex A2 page 2 -

2. Checking of Serviceability Limit State (toe at 0m above foundation)(with back batter 1:10)

750 650 5 kPa800 700 1300

1000 W1

Pav

1000 W2 3000Pa1

1000 W3 Pa2

Insitu soil1000 W4

Pa3 Pa4 Pwh 1500500 W5

ToePwv





γm 1 Base width of wall 4.2 mφ'f 35.0 o ( = tan-1((tan φ')/γm ))φcv' 30 o

Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o

δb 28.9 o (= 0.85fcv' )γm 1φ'f 35.0 o ( = tan-1((tan φ')/γm ))cf' 5.0 kN/m2 ( = c' / γm )



ΣΗ= 71.92Pah = ΣPai = 60.67

- Annex A2 page 3 -




ΣV= 146.34


cos θ = 0.995 tan θ = 0.100


ΣM = 104.25 (kNm/m run)

Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 4.00 x 0.10 + 3.550 x 0.995 = 3.93 79.71W2 30.42 3.00 x 0.10 + 3.225 x 0.995 = 3.51 106.70W3 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 = 2.59 137.22W5 32.76 0.25 x 0.10 + 2.100 x 0.995 = 2.11 69.27Pwv -31.50 4.2 x 2 / 3 x 0.995 = 2.79 -87.76Pav 0.00 4.2 = 4.20 0.00




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

Geoguide 1 By Middle-third Rule, B/6 = 4.2 / 6 = 0.700m > -0.137m OK! OK!Clause 9.2.4

Arm (m)

Arm (m)

(( )

))(

(((

))

)

- Annex A2 page 4 -



Checked by :

Subject : Design of 4.5m Gabion WallChecking 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 kPa800 700 1300

1000 W1

Pav

1000 2667W2 Pa1

1000 Insitu soil Pa2

W3

1000 1333W4 Pa3 Pa4 Pwh

500 W5

Toe

Pwv





γm 1.2 Base width of wall 3.4 mφ'f 30.3 o ( = tan-1(tan φ'/γm ))φcv' 30 o Length of wall 100 mKah 0.330(Note: γm is taken to be 1.2 so as to cater for the unknown ground condition.)


φcv' 34 o

δb 28.9 o (= 0.85fcv' )γm 1.2φ'f 30.3 o ( = tan-1(tan δb/γm ))cf' 4.2 kN/m2 ( = c' / γm )


- Annex A3 page 1 -


ΣΗ= 67.96Pah = ΣPai = 59.07



Step 3 Wt. of wall + water uplift kN/m - runW1 = 26 x 0.6 x 1.30 x 1.00 = 20.28W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34W4 = 26 x 0.6 x 3.40 x 1.00 = 53.04Pwv = 10 x 1.33 x 3.40 / 2 = -22.67Pav = 0.00 = 0.00

ΣV= 122.41


Pa1 6.60 4.00 / 2 = 2.00 13.19Pa2 24.63 2.67 / 3 + 1.33 = 2.22 54.72Pa3 24.63 1.33 / 2 = 0.67 16.42Pa4 3.22 1.33 / 3 = 0.44 1.43Pwh 8.89 1.33 / 3 = 0.44 3.95

ΣM = 89.72 (kNm/m run)

Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 3.4 - 1.30 / 2 = 2.75 55.77W2 30.42 3.4 - 1.95 / 2 = 2.43 73.77W3 41.34 3.4 - 2.65 / 2 = 2.08 85.78W4 53.04 3.4 - 3.40 / 2 = 1.70 90.17Pwv -22.67 3.4 x 2 / 3 = 2.27 -51.38Pav 0.00 3.4 = 3.40 0.00





= 3.4 / 2 - ( 305.49 - 141.09 ) / 122.41= 0.357m

Arm (m)

Arm (m)

- Annex A3 page 2 -


ΣV= 122.41 kN / m


=> OK!

OK!




= 2.69 x 100 = 268.59 m2



= / 268.59 = 45.58 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 2.69 / 100= 1.02

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 2.69 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 2.69 / 100= 1.02

12241.33

- Annex A3 page 3 -


( 1 + B' / L' )= ( 2 + 2.69 / 100 ) /

( 1 + 2.69 / 100 )= 1.97


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.48 ) 2.97

= 0.14iq = ( 1 - Ki ) mi

= ( 1 - 0.48 ) 1.97

= 0.28ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.28 - ( 1 - 0.28 ) / ( 30.78 x tan 30.3 )= 0.23


gc = gγ = gq = 1




= 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 1.02x 0.28 x 1 x 1

= 30.60 + 39.85 + 23.84= 94.29 kPa


- Annex A3 page 4 -

2. Checking of Ultimate Limit State (toe at 0.5m above foundation)(with back batter 1:10)

750 650 5 kPa800 700 1300

1000 W1

Pav

1000 W2 2667Pa1

1000 W3 Insitu soil Pa2

1000 W4 1333Pa3 Pa4 Pwh

500 W5

ToePwv





γm 1.2 Base width of wall 3.40 mφ'f 30.3 o ( = tan-1(tan φ'/γm ))φcv' 30 o Length of wall 100 mKah 0.330 Back batter, θ = 1 : 10 = 0.10(Note: gm is taken to be 1.2 so as to cater for the unknown ground condition.)


φcv' 34 o

δb 28.9 o (= 0.85fcv' )γm 1.2φ'f 30.3 o ( = tan-1(tan φ'/γm ))cf' 4.2 kN/m2 ( = c' / γm )



ΣΗ= 67.96Pah = ΣPai = 59.07

- Annex A3 page 5 -




ΣV= 122.41


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 3.50 x 0.10 + 2.750 x 0.995 = 3.08 62.56W2 30.42 2.50 x 0.10 + 2.425 x 0.995 = 2.66 80.97W3 41.34 1.50 x 0.10 + 2.075 x 0.995 = 2.21 91.53W4 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 -51.12Pav 0.00 3.4 = 3.40 0.00





= 3.4 / 2 - ( 327.41 - 140.84 ) / 122.41= 0.176m

Arm (m)

Arm (m)

((

((( )

))))

- Annex A3 page 6 -





=> OK!




=> OK!

OK!




= 3.05 x 100 = 304.82 m2



= / 304.82 = 40.16 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


12241

- Annex A3 page 7 -


= 1 + 18.96 / 30.78 x 3.05 / 100= 1.02

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 3.05 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 3.05 / 100= 1.02


( 1 + B' / L' )= ( 2 + 3.05 / 100 ) /

( 1 + 3.05 / 100 )= 1.97


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.47 ) 2.97

= 0.15iq = ( 1 - Ki ) mi

= ( 1 - 0.47 ) 1.97

= 0.28ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.28 - ( 1 - 0.28 ) / ( 30.78 x tan 30.3 )= 0.24


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 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 1.02x 0.28 x 5.44 x 1

= 182.00 + 258.46 + 134.51= 574.98 kPa


- Annex A3 page 8 -



Checked by :

Subject : Design of 4.5m Gabion WallChecking 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 kPa800 700 1300

1000 W1

Pav

1000 2667W2 Pa1


W3

1000 1333W4 Pa3 Pa4 Pwh

500 W5

Toe

Pwv





γm 1 Base width of wall 3.4 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o

Kah 0.271


φcv' 34 o

δb 28.9 o (= 0.85φcv' )γm 1φ'f 35.0 o ( = tan-1(tan φ'/γm ))cf' 5.0 kN/m2 ( = c' / γm )


- Annex A4 page 1 -


ΣΗ= 57.43Pah = ΣPai = 48.54




ΣV= 122.41


Pa1 5.42 4.00 / 2 = 2.00 10.84Pa2 20.23 2.67 / 3 + 1.33 = 2.22 44.96Pa3 20.23 1.33 / 2 = 0.67 13.49Pa4 2.65 1.33 / 3 = 0.44 1.18Pwh 8.89 1.33 / 3 = 0.44 3.95






= 3.4 / 2 - ( 305.49 - 125.80 ) / 122.41= 0.232m


Arm (m)

Arm (m)

- Annex A4 page 2 -

2. Checking of Serviceability Limit State (toe at 0.5m above foundation)(with back batter 1:10)

750 650 5 kPa800 700 1300

1000 W1

Pav

1000 W2 2667Pa1

1000 W3 Pa2

Insitu soil1000 W4 1333

Pa3 Pa4 Pwh

500 W5

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o

δb 28.9 o (= 0.85fcv' )γm 1φ'f 35.0 o ( = tan-1(tan φ'/γm ))cf' 5.0 kN/m2 ( = c' / γm )



ΣΗ= 57.43Pah = ΣPai = 48.54

- Annex A4 page 3 -




ΣV= 122.41


cos θ = 0.995 tan θ = 0.100







= 3.4 / 2 - ( 327.29 - 125.54 ) / 122.41= 0.052m


Arm (m)

Arm (m)

(

(( )

))(

( ))

- Annex A4 page 4 -



Checked by :


Reference Remarks


750 650 5 kPa800 700 1300

1000 W1

Pav Pa1 Pa2

1000 2000W2

1000 Insitu soil Pa3 Pa4 Pwh 1000W3

1000W4

500 W5

Toe

Pwv





γm 1.2 Base width of wall 2.65 mφ'f 30.3 o ( = tan-1(tan φ'/γm ))φcv' 30 o Length of wall 100 mKah 0.330(Note: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)


φcv' 34 o


Geoguide 1, clause 5.12 δbf 24.1 o ( = δb / γm )Geoguide 1, clause 6.6 Kp 3.610 (β/φ = 0 δ/φ = 0)Fig. 19 (Note: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)

- Annex A5 page 1 -


ΣΗ= 39.46Pah = ΣPai = 34.46



Step 3 Wt. of wall + water uplift kN/m - runW1 = 26 x 0.6 x 1.30 x 1.00 = 20.28W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42W3 = 26 x 0.6 x 2.65 x 1.00 = 41.34Pwv = 10 x 1.00 x 2.65 / 2 = -13.25Pav = 0.00 = 0.00

ΣV= 78.79


Pa1 4.95 3.00 / 2 = 1.50 7.42Pa2 13.85 2.00 / 3 + 1.00 = 1.67 23.09Pa3 13.85 1.00 / 2 = 0.50 6.93Pa4 1.81 1.00 / 3 = 0.33 0.60Pwh 5.00 1.00 / 3 = 0.33 1.67


Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 2.65 - 1.30 / 2 = 2.00 40.56W2 30.42 2.65 - 1.95 / 2 = 1.68 50.95W3 41.34 2.65 - 2.65 / 2 = 1.33 54.78Pwv -13.25 2.65 x 2 / 3 = 1.77 -23.41Pav 0.00 2.65 = 2.65 0.00





= 2.65 / 2 - ( 146.29 - 63.11 ) / 78.79= 0.269m

Arm (m)

Arm (m)

- Annex A5 page 2 -


ΣV= 78.79 kN / m


=> OK!

OK!




= 2.11 x 100 = 211.14 m2



= / 211.14 = 37.32 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 2.11 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 2.11 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 2.11 / 100= 1.01

7879.00

- Annex A5 page 3 -


( 1 + B' / L' )= ( 2 + 2.11 / 100 ) /

( 1 + 2.11 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.42 ) 2.98

= 0.20iq = ( 1 - Ki ) mi

= ( 1 - 0.42 ) 1.98

= 0.34ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

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


gc = gγ = gq = 1




= 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 1.01x 0.34 x 1 x 1

= 39.36 + 43.21 + 29.34= 111.91 kPa


- Annex A5 page 4 -


750 650 5 kPa800 700 1300

1000 W1

Pav

1000 W2 Pa1 Pa2 2000

1000 W3 Insitu soil Pa3 Pa4 Pwh 1000

1000 W4

500 W5

ToePwv







φcv' 34 o




ΣΗ= 39.46Pah = ΣPai = 34.46

- Annex A5 page 5 -




ΣV= 78.79


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 2.50 x 0.10 + 2.000 x 0.995 = 2.24 45.40W2 30.42 1.50 x 0.10 + 1.675 x 0.995 = 1.82 55.24W3 41.34 0.50 x 0.10 + 1.325 x 0.995 = 1.37 56.56Pwv -13.25 2.65 x 2 / 3 x 0.995 = 1.76 -23.29Pav 0.00 2.65 = 2.65 0.00





= 2.65 / 2 - ( 157.20 - 63.00 ) / 78.79= 0.129m

Arm (m)

Arm (m)

((

(( )

)))

- Annex A5 page 6 -





=> OK!




=> OK!

OK!




= 2.39 x 100 = 239.14 m2



= / 239.14 = 32.95 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


7879

- Annex A5 page 7 -


= 1 + 18.96 / 30.78 x 2.39 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 2.39 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 2.39 / 100= 1.01


( 1 + B' / L' )= ( 2 + 2.39 / 100 ) /

( 1 + 2.39 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.41 ) 2.98

= 0.21iq = ( 1 - Ki ) mi

= ( 1 - 0.41 ) 1.98

= 0.35ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.35 - ( 1 - 0.35 ) / ( 30.78 x tan 30.3 )= 0.31


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.31 x 5.69 x 1+ 0.5 x 9 x 2.39 x 23.30 x 0.99 x

0.21 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.35 x 5.44 x 1

= 232.59 + 278.44 + 164.91= 675.94 kPa


- Annex A5 page 8 -



Checked by :


Reference Remarks


750 650 5 kPa800 700 1300

1000 W1

Pav

1000 Pa1 Pa2 2000W2

1000 Insitu soil Pa3 Pa4 1000W3 Pwh

1000W4

500 W5

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex A6 page 1 -


ΣΗ= 33.32Pah = ΣPai = 28.32




ΣV= 78.79


Pa1 4.06 3.00 / 2 = 1.50 6.10Pa2 11.38 2.00 / 3 + 1.00 = 1.67 18.97Pa3 11.38 1.00 / 2 = 0.50 5.69Pa4 1.49 1.00 / 3 = 0.33 0.50Pwh 5.00 1.00 / 3 = 0.33 1.67






= 2.65 / 2 - ( 146.29 - 56.33 ) / 78.79= 0.183m


Arm (m)

Arm (m)

- Annex A6 page 2 -


750 650 5 kPa800 700 1300

1000 W1

Pav

1000 W2 Pa1 Pa2 2000

1000 W3 Pa3 Pa4 Pwh 1000Insitu soil

1000 W4

500 W5

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 33.32Pah = ΣPai = 28.32

- Annex A6 page 3 -




ΣV= 78.79


cos θ = 0.995 tan θ = 0.100







= 2.65 / 2 - ( 157.15 - 56.21 ) / 78.79= 0.044m


Arm (m)

Arm (m)

(

(( )

))()

- Annex A6 page 4 -



Checked by :


Reference Remarks


750 650 5 kPa800 700 1300

1000 W1

Pav Pa1 Pa2 13331000

W2 6671000 Insitu soil Pa3 Pa4 Pwh

W3

1000W4

500 W5

Toe

Pwv







φcv' 34 o



- Annex A7 page 1 -


ΣΗ= 18.64Pah = ΣPai = 16.42



Step 3 Wt. of wall + water uplift kN/m - runW1 = 26 x 0.6 x 1.30 x 1.00 = 20.28W2 = 26 x 0.6 x 1.95 x 1.00 = 30.42Pwv = 10 x 0.67 x 1.95 / 2 = -6.50Pav = 0.00 = 0.00

ΣV= 44.20


Pa1 3.30 2.00 / 2 = 1.00 3.30Pa2 6.16 1.33 / 3 + 0.67 = 1.11 6.84Pa3 6.16 0.67 / 2 = 0.33 2.05Pa4 0.81 0.67 / 3 = 0.22 0.18Pwh 2.22 0.67 / 3 = 0.22 0.49


Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 1.95 - 1.30 / 2 = 1.30 26.36W2 30.42 1.95 - 1.95 / 2 = 0.98 29.66Pwv -6.50 1.95 x 2 / 3 = 1.30 -8.45Pav 0.00 1.95 = 1.95 0.00





= 1.95 / 2 - ( 56.02 - 21.31 ) / 44.20= 0.190m

Arm (m)

Arm (m)

- Annex A7 page 2 -


ΣV= 44.20 kN / m


=> OK!

OK!




= 1.57 x 100 = 157.06 m2



= / 157.06 = 28.14 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 1.57 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.57 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.57 / 100= 1.01

4420.00

- Annex A7 page 3 -


( 1 + B' / L' )= ( 2 + 1.57 / 100 ) /

( 1 + 1.57 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.34 ) 2.98

= 0.29iq = ( 1 - Ki ) mi

= ( 1 - 0.34 ) 1.98

= 0.44ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.44 - ( 1 - 0.44 ) / ( 30.78 x tan 30.3 )= 0.41


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.41 x 1 x 1+ 0.5 x 9 x 1.57 x 23.30 x 0.99 x

0.29 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.44 x 1 x 1

= 53.38 + 48.12 + 38.17= 139.67 kPa


- Annex A7 page 4 -


750 650 5 kPa800 700 1300

1000 W1

Pav Pa1 Pa2 13331000 W2

6671000 W3 Insitu soil Pa3 Pa4 Pwh

1000 W4

500 W5

ToePwv







φcv' 34 o




ΣΗ= 18.64Pah = ΣPai = 16.42

- Annex A7 page 5 -




ΣV= 44.20


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 1.50 x 0.10 + 1.300 x 0.995 = 1.44 29.26W2 30.42 0.50 x 0.10 + 0.975 x 0.995 = 1.02 31.03Pwv -6.50 1.95 x 2 / 3 x 0.995 = 1.29 -8.41Pav 0.00 1.95 = 1.95 0.00





= 1.95 / 2 - ( 60.29 - 21.27 ) / 44.20= 0.092m

Arm (m)

Arm (m)

((

( )

))

- Annex A7 page 6 -





=> OK!




=> OK!

OK!




= 1.77 x 100 = 176.54 m2



= / 176.54 = 25.04 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


4420

- Annex A7 page 7 -


= 1 + 18.96 / 30.78 x 1.77 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.77 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.77 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.77 / 100 ) /

( 1 + 1.77 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.33 ) 2.98

= 0.31iq = ( 1 - Ki ) mi

= ( 1 - 0.33 ) 1.98

= 0.45ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.45 - ( 1 - 0.45 ) / ( 30.78 x tan 30.3 )= 0.42


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.42 x 5.69 x 1+ 0.5 x 9 x 1.77 x 23.30 x 0.99 x

0.31 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.45 x 5.44 x 1

= 312.71 + 305.23 + 213.12= 831.07 kPa


- Annex A7 page 8 -



Checked by :


Reference Remarks


750 650 5 kPa800 700 1300

1000 W1

Pav Pa1 Pa2 13331000


W3

1000W4

500 W5

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex A8 page 1 -


ΣΗ= 15.71Pah = ΣPai = 13.49




ΣV= 44.20


Pa1 2.71 2.00 / 2 = 1.00 2.71Pa2 5.06 1.33 / 3 + 0.67 = 1.11 5.62Pa3 5.06 0.67 / 2 = 0.33 1.69Pa4 0.66 0.67 / 3 = 0.22 0.15Pwh 2.22 0.67 / 3 = 0.22 0.49






= 1.95 / 2 - ( 56.02 - 19.11 ) / 44.20= 0.140m


Arm (m)

Arm (m)

- Annex A8 page 2 -


750 650 5 kPa800 700 1300

1000 W1 1333Pav Pa1 Pa2

1000 W2

6671000 W3 Pa3 Pa4 Pwh

Insitu soil1000 W4

500 W5

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 15.71Pah = ΣPai = 13.49

- Annex A8 page 3 -




ΣV= 44.20


cos θ = 0.995 tan θ = 0.100







= 1.95 / 2 - ( 60.26 - 19.07 ) / 44.20= 0.043m


Arm (m)

Arm (m)

((( )

))

- Annex A8 page 4 -



Checked by :


Reference Remarks


750 650 5 kPa800 700 1300

1000 W1 Pa2 667Pav Pa1 333

1000 Pa3 Pa4 Pwh

W2

1000 Insitu soilW3

1000W4

500 W5

Toe

Pwv







φcv' 34 o



- Annex A9 page 1 -


ΣΗ= 5.48Pah = ΣPai = 4.93



Step 3 Wt. of wall + water uplift kN/m - runW1 = 26 x 0.6 x 1.30 x 1.00 = 20.28Pwv = 10 x 0.33 x 1.30 / 2 = -2.17Pav = 0.00 = 0.00

ΣV= 18.11


Pa1 1.65 1.00 / 2 = 0.50 0.82Pa2 1.54 0.67 / 3 + 0.33 = 0.56 0.86Pa3 1.54 0.33 / 2 = 0.17 0.26Pa4 0.20 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06


Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 1.3 - 1.30 / 2 = 0.65 13.18Pwv -2.17 1.3 x 2 / 3 = 0.87 -1.88Pav 0.00 1.3 = 1.30 0.00





= 1.3 / 2 - ( 13.18 - 3.90 ) / 18.11= 0.137m

Arm (m)

Arm (m)

- Annex A9 page 2 -


ΣV= 18.11 kN / m


=> OK!

OK!




= 1.03 x 100 = 102.51 m2



= / 102.51 = 17.67 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 1.03 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.03 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.03 / 100= 1.01

1811.33

- Annex A9 page 3 -


( 1 + B' / L' )= ( 2 + 1.03 / 100 ) /

( 1 + 1.03 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.22 ) 2.99

= 0.48iq = ( 1 - Ki ) mi

= ( 1 - 0.22 ) 1.99

= 0.62ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.62 - ( 1 - 0.62 ) / ( 30.78 x tan 30.3 )= 0.60


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.60 x 1 x 1+ 0.5 x 9 x 1.03 x 23.30 x 1.00 x

0.48 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.62 x 1 x 1

= 76.85 + 51.77 + 52.94= 181.56 kPa


- Annex A9 page 4 -


750 650 5 kPa800 700 1300

1000 W1 Pa2 667Pav Pa1 333

1000 W2 Pa3 Pa4 Pwh

1000 W3 Insitu soil

1000 W4

500 W5

ToePwv







φcv' 34 o




ΣΗ= 5.48Pah = ΣPai = 4.93

- Annex A9 page 5 -




ΣV= 18.11


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 20.28 0.50 x 0.10 + 0.650 x 0.995 = 0.70 14.13Pwv -2.17 1.3 x 2 / 3 x 0.995 = 0.86 -1.87Pav 0.00 1.3 = 1.30 0.00





= 1.3 / 2 - ( 14.13 - 3.89 ) / 18.11= 0.085m

Arm (m)

Arm (m)(( )

)

- Annex A9 page 6 -





=> OK!




=> OK!

OK!




= 1.13 x 100 = 113.03 m2



= / 113.03 = 16.03 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


1811

- Annex A9 page 7 -


= 1 + 18.96 / 30.78 x 1.13 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.13 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.13 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.13 / 100 ) /

( 1 + 1.13 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.21 ) 2.99

= 0.50iq = ( 1 - Ki ) mi

= ( 1 - 0.21 ) 1.99

= 0.63ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.63 - ( 1 - 0.63 ) / ( 30.78 x tan 30.3 )= 0.61


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.61 x 5.69 x 1+ 0.5 x 9 x 1.13 x 23.30 x 1.00 x

0.50 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.63 x 5.44 x 1

= 444.90 + 317.81 + 292.72= 1055.43 kPa


- Annex A9 page 8 -



Checked by :


Reference Remarks


750 650 5 kPa800 700 1300

1000 W1 Pa2 667Pav Pa1 333

1000 Pa3 Pa4 Pwh

W2

1000 Insitu soilW3

1000W4

500 W5

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex A10 page 1 -


ΣΗ= 4.61Pah = ΣPai = 4.05




ΣV= 18.11


Pa1 1.35 1.00 / 2 = 0.50 0.68Pa2 1.26 0.67 / 3 + 0.33 = 0.56 0.70Pa3 1.26 0.33 / 2 = 0.17 0.21Pa4 0.17 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06






= 1.3 / 2 - ( 13.18 - 3.55 ) / 18.11= 0.118m


Arm (m)

Arm (m)



750 650 5 kPa800 700 1300

1000 W1 Pa2 667Pav Pa1 333

1000 W2 Pa3 Pa4 Pwh

1000 W3

Insitu soil1000 W4

500 W5

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 4.61Pah = ΣPai = 4.05





ΣV= 18.11


cos θ = 0.995 tan θ = 0.100







= 1.3 / 2 - ( 14.12 - 3.54 ) / 18.11= 0.066m


Arm (m)

Arm (m)

(( )

)


Project : Design of 3.5m Gabion Wall Annex B


Checked by :


Design Statement


1. Design Data

(I) Materials




the mesh.









weaving.










Reference


Remarks

Design Statement




- Annex B page 1 -

(B) Assumptions


Block Size


Para. 9.5.1

Mesh Size

8cm x 10cm x 2.7mm
















(II) Loadings

Dead loads









retaining height.

2.6

0.4

0.0

- Annex B page 2 -



considered.


Design Code


Second Edition.

Design Methodology









shape.


3.5m Gabion Wall

- Annex B page 3 -

Ultimate Limit Statement (ULS)Refer to Annex B1, B3, 1. Checking Overturning [OK if restoring moment > overturning moment]B5 & B7



m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

Refer to Annex B1, B3, 2. Checking Sliding [OK if resisting force > sliding force]B5 & B7



m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

Refer to Annex B1, B3, 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]B5 & B7



m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

Serviceability Limit Statement (SLS)Refer to Annex B2, B4, 1. Check Overturning and Determine EccentricityB6 & B8 [OK if the resultant force acts within the middle third of the wall base]para. 1 Step 6



m OK! OK!

m OK! OK!

m OK! OK!

m OK! OK!

For details of calculations, please refer to the Appendix B1 to B8.

0.0

0.5

1.5

2.5

1.5

2.5

0.0

0.5

1.5

0.0

0.5

1.5

2.5

2.5

0.0

0.5

- Annex B page 4 -









courses.

Drainage provisions






References


Second Edition.

- Annex B page 5 -

Project : Design of Gabion Wall Annex B1


Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1

Pav

1000 2333W2 Pa1


W3

500 1167W4 Pa3 Pa4 Pwh

Toe

Pwv

W1, W2, W3, W4 = Self-weight of the proposed protection wallPa1 = Lateral pressure due to surchargePav = Vertical component of active earth pressurePa2, Pa3, Pa4 = Horizontal component of active earth pressurePwh = Lateral pressure due to groundwaterPwv = Upthrust






φcv' 34 o



- Annex B1 page 1 -


ΣΗ= 52.75Pah = ΣPai = 45.95




ΣV= 154.39


Pa1 5.77 3.50 / 2 = 1.75 10.10Pa2 18.85 2.33 / 3 + 1.17 = 1.94 36.66Pa3 18.85 1.17 / 2 = 0.58 11.00Pa4 2.47 1.17 / 3 = 0.39 0.96Pwh 6.81 1.17 / 3 = 0.39 2.65







= 2.95 / 2 - ( 297.22 - 95.21 ) / 154.39= 0.167m

Arm (m)

Arm (m)

- Annex B1 page 2 -


ΣV= 154.39 kN / m


=> OK!

OK!




= 2.62 x 100 = 261.69 m2



= / 261.69 = 59.00 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 2.62 / 100= 1.02

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 2.62 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 2.62 / 100= 1.02

15439.17

- Annex B1 page 3 -


( 1 + B' / L' )= ( 2 + 2.62 / 100 ) /

( 1 + 2.62 / 100 )= 1.97


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.30 ) 2.97

= 0.34iq = ( 1 - Ki ) mi

= ( 1 - 0.30 ) 1.97

= 0.49ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.49 - ( 1 - 0.49 ) / ( 30.78 x tan 30.3 )= 0.46


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.02 x 0.46 x 1 x 1+ 0.5 x 9 x 2.62 x 23.30 x 0.99 x

0.34 x 1 x 1 + 4.5 x 18.96 x 1.02x 0.49 x 1 x 1

= 59.86 + 92.06 + 42.26= 194.17 kPa


- Annex B1 page 5 -


650 600 5 kPa600 1100

1000 W1

Pav

1000 W2 2333Pa1

1000 W3 Insitu soil Pa2

500 W4 1167Pa3 Pa4 Pwh

ToePwv







φcv' 34 o




ΣΗ= 52.75Pah = ΣPai = 45.95

- Annex B1 page 6 -




ΣV= 85.36


cos θ = 0.995 tan θ = 0.100








= 2.95 / 2 - ( 208.97 - 95.04 ) / 85.36= 0.140m

Arm (m)

Arm (m)

((

((( )

))))

- Annex B1 page 7 -





=> OK!




=> OK!

OK!




= 2.67 x 100 = 266.94 m2



= / 266.94 = 31.98 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


8536

- Annex B1 page 8 -


= 1 + 18.96 / 30.78 x 2.67 / 100= 1.02

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 2.67 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 2.67 / 100= 1.02


( 1 + B' / L' )= ( 2 + 2.67 / 100 ) /

( 1 + 2.67 / 100 )= 1.97


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.51 ) 2.97

= 0.12iq = ( 1 - Ki ) mi

= ( 1 - 0.51 ) 1.97

= 0.25ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.25 - ( 1 - 0.25 ) / ( 30.78 x tan 30.3 )= 0.21


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.02 x 0.21 x 5.69 x 1+ 0.5 x 9 x 2.67 x 23.30 x 0.99 x

0.12 x 5.44 x 1 + 4.5 x 18.96 x 1.02x 0.25 x 5.44 x 1

= 153.86 + 185.79 + 117.53= 457.18 kPa


- Annex B1 page 9 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1

Pav

1000 2333W2 Pa1


W3

500 1167W4 Pa3 Pa4 Pwh

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex B2 page 1 -


ΣΗ= 44.56Pah = ΣPai = 37.75




ΣV= 154.39


Pa1 4.74 3.50 / 2 = 1.75 8.30Pa2 15.49 2.33 / 3 + 1.17 = 1.94 30.12Pa3 15.49 1.17 / 2 = 0.58 9.04Pa4 2.03 1.17 / 3 = 0.39 0.79Pwh 6.81 1.17 / 3 = 0.39 2.65






= 2.95 / 2 - ( 297.22 - 84.74 ) / 154.39= 0.099m


Arm (m)

Arm (m)

- Annex B2 page 2 -


650 600 5 kPa600 1100

1000 W1

Pav

1000 W2 2333Pa1

1000 W3 Pa2

Insitu soil500 W4 1167

Pa3 Pa4 Pwh

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 44.56Pah = ΣPai = 37.75

- Annex B2 page 3 -




ΣV= 154.39


cos θ = 0.995 tan θ = 0.100







= 2.95 / 2 - ( 311.92 - 84.57 ) / 154.39= 0.002m


Arm (m)

Arm (m)

(

(( )

))(

( ))

- Annex B2 page 4 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1

Pav Pa1 Pa2

1000 2000W2


500W4

Toe

Pwv







φcv' 34 o



- Annex B3 page 1 -


ΣΗ= 39.46Pah = ΣPai = 34.46




ΣV= 68.06


Pa1 4.95 3.00 / 2 = 1.50 7.42Pa2 13.85 2.00 / 3 + 1.00 = 1.67 23.09Pa3 13.85 1.00 / 2 = 0.50 6.93Pa4 1.81 1.00 / 3 = 0.33 0.60Pwh 5.00 1.00 / 3 = 0.33 1.67







= 2.3 / 2 - ( 109.75 - 57.34 ) / 68.06= 0.380m

Arm (m)

Arm (m)

- Annex B3 page 2 -


ΣV= 68.06 kN / m


=> OK!

OK!




= 1.54 x 100 = 154.01 m2



= / 154.01 = 44.19 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 1.54 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.54 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.54 / 100= 1.01

6806.00

- Annex B3 page 3 -


( 1 + B' / L' )= ( 2 + 1.54 / 100 ) /

( 1 + 1.54 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.50 ) 2.98

= 0.13iq = ( 1 - Ki ) mi

= ( 1 - 0.50 ) 1.98

= 0.25ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.25 - ( 1 - 0.25 ) / ( 30.78 x tan 30.3 )= 0.21


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.21 x 1 x 1+ 0.5 x 9 x 1.54 x 23.30 x 0.99 x

0.13 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.25 x 1 x 1

= 27.43 + 20.37 + 21.82= 69.62 kPa


- Annex B3 page 4 -


650 600 5 kPa600 1100

1000 W1

Pav

1000 W2 Pa1 Pa2 2000


500 W4

ToePwv







φcv' 34 o




ΣΗ= 39.46Pah = ΣPai = 34.46

- Annex B3 page 5 -




ΣV= 68.06


cos θ = 0.995 tan θ = 0.100








= 2.3 / 2 - ( 119.21 - 57.25 ) / 68.06= 0.240m

Arm (m)

Arm (m)

((

(( )

)))

- Annex B3 page 6 -





=> OK!




=> OK!

OK!




= 1.82 x 100 = 182.09 m2



= / 182.09 = 37.38 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


6806

- Annex B3 page 7 -


= 1 + 18.96 / 30.78 x 1.82 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.82 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.82 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.82 / 100 ) /

( 1 + 1.82 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.49 ) 2.98

= 0.14iq = ( 1 - Ki ) mi

= ( 1 - 0.49 ) 1.98

= 0.27ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.27 - ( 1 - 0.27 ) / ( 30.78 x tan 30.3 )= 0.23


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.23 x 5.69 x 1+ 0.5 x 9 x 1.82 x 23.30 x 0.99 x

0.14 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.27 x 5.44 x 1

= 166.41 + 140.94 + 124.99= 432.34 kPa


- Annex B3 page 8 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1

Pav

1000 Pa1 Pa2 2000W2


500W4

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex B4 page 1 -


ΣΗ= 33.32Pah = ΣPai = 28.32




ΣV= 68.06


Pa1 4.06 3.00 / 2 = 1.50 6.10Pa2 11.38 2.00 / 3 + 1.00 = 1.67 18.97Pa3 11.38 1.00 / 2 = 0.50 5.69Pa4 1.49 1.00 / 3 = 0.33 0.50Pwh 5.00 1.00 / 3 = 0.33 1.67






= 2.3 / 2 - ( 109.75 - 50.55 ) / 68.06= 0.280m


Arm (m)

Arm (m)

- Annex B4 page 2 -


650 600 5 kPa600 1100

1000 W1

Pav

1000 W2 Pa1 Pa2 2000


500 W4

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 33.32Pah = ΣPai = 28.32

- Annex B4 page 3 -




ΣV= 68.06


cos θ = 0.995 tan θ = 0.100







= 2.3 / 2 - ( 119.16 - 50.47 ) / 68.06= 0.141m


Arm (m)

Arm (m)

(

(( )

))()

- Annex B4 page 4 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1

Pav Pa1 Pa2 13331000


W3

500W4

Toe

Pwv







φcv' 34 o



- Annex B5 page 1 -


ΣΗ= 18.64Pah = ΣPai = 16.42




ΣV= 38.01


Pa1 3.30 2.00 / 2 = 1.00 3.30Pa2 6.16 1.33 / 3 + 0.67 = 1.11 6.84Pa3 6.16 0.67 / 2 = 0.33 2.05Pa4 0.81 0.67 / 3 = 0.22 0.18Pwh 2.22 0.67 / 3 = 0.22 0.49







= 1.7 / 2 - ( 42.28 - 19.29 ) / 38.01= 0.245m

Arm (m)

Arm (m)

- Annex B5 page 2 -


ΣV= 38.01 kN / m


=> OK!

OK!




= 1.21 x 100 = 120.96 m2



= / 120.96 = 31.43 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 1.21 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.21 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.21 / 100= 1.01

3801.33

- Annex B5 page 3 -


( 1 + B' / L' )= ( 2 + 1.21 / 100 ) /

( 1 + 1.21 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.40 ) 2.99

= 0.22iq = ( 1 - Ki ) mi

= ( 1 - 0.40 ) 1.99

= 0.36ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.36 - ( 1 - 0.36 ) / ( 30.78 x tan 30.3 )= 0.33


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.33 x 1 x 1+ 0.5 x 9 x 1.21 x 23.30 x 1.00 x

0.22 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.36 x 1 x 1

= 42.29 + 27.49 + 31.17= 100.94 kPa


- Annex B5 page 4 -


650 600 5 kPa600 1100

1000 W1

Pav Pa1 Pa2 13331000 W2


500 W4

ToePwv







φcv' 34 o




ΣΗ= 18.64Pah = ΣPai = 16.42

- Annex B5 page 5 -




ΣV= 38.01


cos θ = 0.995 tan θ = 0.100








= 1.7 / 2 - ( 45.95 - 19.25 ) / 38.01= 0.148m

Arm (m)

Arm (m)

((

( )

))

- Annex B5 page 6 -





=> OK!




=> OK!

OK!




= 1.40 x 100 = 140.44 m2



= / 140.44 = 27.07 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


3801

- Annex B5 page 7 -


= 1 + 18.96 / 30.78 x 1.40 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.40 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.40 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.40 / 100 ) /

( 1 + 1.40 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.39 ) 2.99

= 0.23iq = ( 1 - Ki ) mi

= ( 1 - 0.39 ) 1.99

= 0.38ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.38 - ( 1 - 0.38 ) / ( 30.78 x tan 30.3 )= 0.34


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.34 x 5.69 x 1+ 0.5 x 9 x 1.40 x 23.30 x 0.99 x

0.23 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.38 x 5.44 x 1

= 251.91 + 183.78 + 176.46= 612.14 kPa


- Annex B5 page 8 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1

Pav Pa1 Pa2 13331000


W3

500W4

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex B6 page 1 -


ΣΗ= 15.71Pah = ΣPai = 13.49




ΣV= 38.01


Pa1 2.71 2.00 / 2 = 1.00 2.71Pa2 5.06 1.33 / 3 + 0.67 = 1.11 5.62Pa3 5.06 0.67 / 2 = 0.33 1.69Pa4 0.66 0.67 / 3 = 0.22 0.15Pwh 2.22 0.67 / 3 = 0.22 0.49






= 1.7 / 2 - ( 42.28 - 17.08 ) / 38.01= 0.187m


Arm (m)

Arm (m)

- Annex B6 page 2 -


650 600 5 kPa600 1100

1000 W1 1333Pav Pa1 Pa2

1000 W2

6671000 W3 Pa3 Pa4 Pwh

Insitu soil500 W4

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 15.71Pah = ΣPai = 13.49

- Annex B6 page 3 -




ΣV= 38.01


cos θ = 0.995 tan θ = 0.100







= 1.7 / 2 - ( 45.93 - 17.05 ) / 38.01= 0.090m


Arm (m)

Arm (m)

((( )

))

- Annex B6 page 4 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1 Pa2 667Pav Pa1 333

1000 Pa3 Pa4 Pwh

W2

1000 Insitu soilW3

500W4

Toe

Pwv







φcv' 34 o



- Annex B7 page 1 -


ΣΗ= 5.48Pah = ΣPai = 4.93




ΣV= 15.33


Pa1 1.65 1.00 / 2 = 0.50 0.82Pa2 1.54 0.67 / 3 + 0.33 = 0.56 0.86Pa3 1.54 0.33 / 2 = 0.17 0.26Pa4 0.20 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06







= 1.1 / 2 - ( 9.44 - 3.36 ) / 15.33= 0.154m

Arm (m)

Arm (m)

- Annex B7 page 2 -


ΣV= 15.33 kN / m


=> OK!

OK!




= 0.79 x 100 = 79.25 m2



= / 79.25 = 19.34 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 0.79 / 100= 1.00

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.79 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.79 / 100= 1.00

1532.67

- Annex B7 page 3 -


( 1 + B' / L' )= ( 2 + 0.79 / 100 ) /

( 1 + 0.79 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.26 ) 2.99

= 0.40iq = ( 1 - Ki ) mi

= ( 1 - 0.26 ) 1.99

= 0.55ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.55 - ( 1 - 0.55 ) / ( 30.78 x tan 30.3 )= 0.52


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.00 x 0.52 x 1 x 1+ 0.5 x 9 x 0.79 x 23.30 x 1.00 x

0.40 x 1 x 1 + 4.5 x 18.96 x 1.00x 0.55 x 1 x 1

= 67.24 + 33.46 + 46.88= 147.58 kPa


- Annex B7 page 4 -


650 600 5 kPa600 1100

1000 W1 Pa2 667Pav Pa1 333

1000 W2 Pa3 Pa4 Pwh

1000 W3 Insitu soil

500 W4

ToePwv







φcv' 34 o




ΣΗ= 5.48Pah = ΣPai = 4.93

- Annex B7 page 5 -




ΣV= 15.33


cos θ = 0.995 tan θ = 0.100








= 1.1 / 2 - ( 10.24 - 3.36 ) / 15.33= 0.101m

Arm (m)

Arm (m)(( )

)

- Annex B7 page 6 -





=> OK!




=> OK!

OK!




= 0.90 x 100 = 89.87 m2



= / 89.87 = 17.05 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


1533

- Annex B7 page 7 -


= 1 + 18.96 / 30.78 x 0.90 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.90 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.90 / 100= 1.01


( 1 + B' / L' )= ( 2 + 0.90 / 100 ) /

( 1 + 0.90 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.25 ) 2.99

= 0.42iq = ( 1 - Ki ) mi

= ( 1 - 0.25 ) 1.99

= 0.56ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.56 - ( 1 - 0.56 ) / ( 30.78 x tan 30.3 )= 0.54


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.54 x 5.69 x 1+ 0.5 x 9 x 0.90 x 23.30 x 1.00 x

0.42 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.56 x 5.44 x 1

= 393.18 + 214.04 + 261.54= 868.76 kPa


- Annex B7 page 8 -



Checked by :


Reference Remarks


650 600 5 kPa600 1100

1000 W1 Pa2 667Pav Pa1 333

1000 Pa3 Pa4 Pwh

W2

1000 Insitu soilW3

500W4

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex B8 page 1 -


ΣΗ= 4.61Pah = ΣPai = 4.05




ΣV= 15.33


Pa1 1.35 1.00 / 2 = 0.50 0.68Pa2 1.26 0.67 / 3 + 0.33 = 0.56 0.70Pa3 1.26 0.33 / 2 = 0.17 0.21Pa4 0.17 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06






= 1.1 / 2 - ( 9.44 - 3.02 ) / 15.33= 0.131m


Arm (m)

Arm (m)

- Annex B8 page 2 -


650 600 5 kPa600 1100

1000 W1 Pa2 667Pav Pa1 333

1000 W2 Pa3 Pa4 Pwh

1000 W3

Insitu soil500 W4

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 4.61Pah = ΣPai = 4.05

- Annex B8 page 3 -




ΣV= 15.33


cos θ = 0.995 tan θ = 0.100







= 1.1 / 2 - ( 10.24 - 3.01 ) / 15.33= 0.078m


Arm (m)

Arm (m)

(( )

)

- Annex B8 page 4 -

Project : Design of 2.5m Gabion Wall Annex C


Checked by :


Design Statement


1. Design Data

(I) Materials




the mesh.









weaving.










Reference


Remarks

Design Statement




- Annex C page 1 -

(B) Assumptions


Block Size


Para. 9.5.1

Mesh Size

8cm x 10cm x 2.7mm
















(II) Loadings

Dead loads









retaining height.

0.0

2.6

0.4

- Annex C page 2 -



considered.


Design Code


Second Edition.

Design Methodology









shape.


2.5m Gabion Wall

- Annex C page 3 -

Ultimate Limit Statement (ULS)Refer to Annex C1, C3, 1. Checking Overturning [OK if restoring moment > overturning moment]& C5



m OK! OK!

m OK! OK!

m OK! OK!

Refer to Annex C1, C3, 2. Checking Sliding [OK if resisting force > sliding force]& C5



m OK! OK!

m OK! OK!

m OK! OK!

Refer to Annex C1, C3, 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]& C5



m OK! OK!

m OK! OK!

m OK! OK!

Serviceability Limit Statement (SLS)Refer to Annex C2, C4, 1. Check Overturning and Determine Eccentricity& C6 [OK if the resultant force acts within the middle third of the wall base]para. 1 Step 6



m OK! OK!

m OK! OK!

m OK! OK!

For details of calculations, please refer to the Appendix C1 to C6.

1.5

0.0

0.0

0.5

1.5

0.5

1.5

0.0

0.5

0.0

0.5

1.5

- Annex C page 4 -









courses.

Drainage provisions






References


Second Edition.

- Annex C page 5 -

Project : Design of Gabion Wall Annex C1


Checked by :


Reference Remarks


500 5 kPa500 1000

1000 W1

Pav Pa1 Pa2

1000 1667W2


Toe

Pwv

W1, W2, W3 = Self-weight of the proposed protection wallPa1 = Lateral pressure due to surchargePav = Vertical component of active earth pressurePa2, Pa3, Pa4 = Horizontal component of active earth pressurePwh = Lateral pressure due to groundwaterPwv = Upthrust






φcv' 34 o



- Annex C1 page 1 -


ΣΗ= 28.09Pah = ΣPai = 24.62




ΣV= 93.07


Pa1 4.12 2.50 / 2 = 1.25 5.15Pa2 9.62 1.67 / 3 + 0.83 = 1.39 13.36Pa3 9.62 0.83 / 2 = 0.42 4.01Pa4 1.26 0.83 / 3 = 0.28 0.35Pwh 3.47 0.83 / 3 = 0.28 0.96


Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 2 - 1.00 / 2 = 1.50 23.40W2 23.40 2 - 1.50 / 2 = 1.25 29.25W3 62.40 2 - 2.00 / 2 = 1.00 62.40Pwv -8.33 2 x 2 / 3 = 1.33 -11.11Pav 0.00 2 = 2.00 0.00





= 2 / 2 - ( 115.05 - 34.95 ) / 93.07= 0.139m

Arm (m)

Arm (m)

- Annex C1 page 2 -


ΣV= 93.07 kN / m


=> OK!

OK!




= 1.72 x 100 = 172.14 m2



= / 172.14 = 54.06 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 1.72 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.72 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.72 / 100= 1.01

9306.67

- Annex C1 page 3 -


( 1 + B' / L' )= ( 2 + 1.72 / 100 ) /

( 1 + 1.72 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.27 ) 2.98

= 0.40iq = ( 1 - Ki ) mi

= ( 1 - 0.27 ) 1.98

= 0.54ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.54 - ( 1 - 0.54 ) / ( 30.78 x tan 30.3 )= 0.52


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.52 x 1 x 1+ 0.5 x 9 x 1.72 x 23.30 x 0.99 x

0.40 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.54 x 1 x 1

= 66.76 + 71.05 + 46.59= 184.41 kPa


- Annex C1 page 5 -


500 5 kPa500 1000

1000 W1

Pav

1000 W2 Pa1 Pa2 1667


ToePwv







φcv' 34 o




ΣΗ= 28.09Pah = ΣPai = 24.62

- Annex C1 page 6 -




ΣV= 46.27


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 2.00 x 0.10 + 1.500 x 0.995 = 1.69 26.39W2 23.40 1.00 x 0.10 + 1.250 x 0.995 = 1.34 31.43W3 15.60 0.25 x 0.10 + 1.000 x 0.995 = 1.02 15.91Pwv -8.33 2 x 2 / 3 x 0.995 = 1.33 -11.06Pav 0.00 2 = 2.00 0.00





= 2 / 2 - ( 73.73 - 34.89 ) / 46.27= 0.161m

Arm (m)

Arm (m)

((

(( )

)))

- Annex C1 page 7 -





=> OK!




=> OK!

OK!




= 1.68 x 100 = 167.90 m2



= / 167.90 = 27.56 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


4627

- Annex C1 page 8 -


= 1 + 18.96 / 30.78 x 1.68 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.68 / 100= 0.99

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.68 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.68 / 100 ) /

( 1 + 1.68 / 100 )= 1.98


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.48 ) 2.98

= 0.14iq = ( 1 - Ki ) mi

= ( 1 - 0.48 ) 1.98

= 0.27ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.27 - ( 1 - 0.27 ) / ( 30.78 x tan 30.3 )= 0.23


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.23 x 5.69 x 1+ 0.5 x 9 x 1.68 x 23.30 x 0.99 x

0.14 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.27 x 5.44 x 1

= 169.78 + 133.43 + 127.01= 430.22 kPa


- Annex C1 page 9 -



Checked by :


Reference Remarks


500 5 kPa500 1000

1000 W1

Pav

1000 Pa1 Pa2 1667W2


Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex C2 page 1 -


ΣΗ= 23.70Pah = ΣPai = 20.23




ΣV= 93.07


Pa1 3.39 2.50 / 2 = 1.25 4.23Pa2 7.90 1.67 / 3 + 0.83 = 1.39 10.98Pa3 7.90 0.83 / 2 = 0.42 3.29Pa4 1.04 0.83 / 3 = 0.28 0.29Pwh 3.47 0.83 / 3 = 0.28 0.96


Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 2 - 1.00 / 2 = 1.50 23.40W2 23.40 2 - 1.50 / 2 = 1.25 29.25W3 62.40 2 - 2.00 / 2 = 1.00 62.40Pwv -8.33 2 x 2 / 3 = 1.33 -11.11Pav 0.00 2 = 2.00 0.00




= 2 / 2 - ( 115.05 - 30.87 ) / 93.07= 0.095m

Geoguide 1 By Middle-third Rule, B/6 = 2 / 6 = 0.333m > 0.095m OK! OK!Clause 9.2.4

Arm (m)

Arm (m)

- Annex C2 page 2 -


500 5 kPa500 1000

1000 W1

Pav

1000 W2 Pa1 Pa2 1667


ToePwv





γm 1 Base width of wall 2 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o

Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 23.70Pah = ΣPai = 20.23

- Annex C2 page 3 -




ΣV= 93.07


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 2.00 x 0.10 + 1.500 x 0.995 = 1.69 26.37W2 23.40 1.00 x 0.10 + 1.250 x 0.995 = 1.34 31.42W3 62.40 0.25 x 0.10 + 1.000 x 0.995 = 1.02 63.63Pwv -8.33 2 x 2 / 3 x 0.995 = 1.33 -11.06Pav 0.00 2 = 2.00 0.00




= 2 / 2 - ( 121.43 - 30.81 ) / 93.07= 0.026m


Arm (m)

Arm (m)

(

(( )

))()

- Annex C2 page 4 -



Checked by :


Reference Remarks


500 5 kPa500 1000

1000 W1

Pav Pa1 Pa2 13331000


W3

Toe

Pwv







φcv' 34 o



- Annex C3 page 1 -


ΣΗ= 18.64Pah = ΣPai = 16.42




ΣV= 34.00


Pa1 3.30 2.00 / 2 = 1.00 3.30Pa2 6.16 1.33 / 3 + 0.67 = 1.11 6.84Pa3 6.16 0.67 / 2 = 0.33 2.05Pa4 0.81 0.67 / 3 = 0.22 0.18Pwh 2.22 0.67 / 3 = 0.22 0.49







= 1.5 / 2 - ( 33.15 - 17.86 ) / 34.00= 0.300m

Arm (m)

Arm (m)

- Annex C3 page 2 -


ΣV= 34.00 kN / m


=> OK!

OK!




= 0.90 x 100 = 89.92 m2



= / 89.92 = 37.81 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 0.90 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.90 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.90 / 100= 1.01

3400.00

- Annex C3 page 3 -


( 1 + B' / L' )= ( 2 + 0.90 / 100 ) /

( 1 + 0.90 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.46 ) 2.99

= 0.16iq = ( 1 - Ki ) mi

= ( 1 - 0.46 ) 1.99

= 0.29ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.29 - ( 1 - 0.29 ) / ( 30.78 x tan 30.3 )= 0.25


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.25 x 1 x 1+ 0.5 x 9 x 0.90 x 23.30 x 1.00 x

0.16 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.29 x 1 x 1

= 32.57 + 14.78 + 25.05= 72.40 kPa


- Annex C3 page 5 -


500 5 kPa500 1000

1000 W1

Pav Pa1 Pa2 13331000 W2


ToePwv







φcv' 34 o




ΣΗ= 18.64Pah = ΣPai = 16.42

- Annex C3 page 6 -




ΣV= 34.00


cos θ = 0.995 tan θ = 0.100








= 1.5 / 2 - ( 36.48 - 17.84 ) / 34.00= 0.202m

Arm (m)

Arm (m)

((

( )

))

- Annex C3 page 7 -





=> OK!




=> OK!

OK!




= 1.10 x 100 = 109.64 m2



= / 109.64 = 31.01 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


3400

- Annex C3 page 8 -


= 1 + 18.96 / 30.78 x 1.10 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.10 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.10 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.10 / 100 ) /

( 1 + 1.10 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.45 ) 2.99

= 0.17iq = ( 1 - Ki ) mi

= ( 1 - 0.45 ) 1.99

= 0.31ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.31 - ( 1 - 0.31 ) / ( 30.78 x tan 30.3 )= 0.27


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.27 x 5.69 x 1+ 0.5 x 9 x 1.10 x 23.30 x 1.00 x

0.17 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.31 x 5.44 x 1

= 198.88 + 106.74 + 144.48= 450.10 kPa


- Annex C3 page 9 -



Checked by :


Reference Remarks


500 5 kPa500 1000

1000 W1

Pav Pa1 Pa2 13331000


W3

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex C4 page 1 -


ΣΗ= 15.71Pah = ΣPai = 13.49




ΣV= 34.00


Pa1 2.71 2.00 / 2 = 1.00 2.71Pa2 5.06 1.33 / 3 + 0.67 = 1.11 5.62Pa3 5.06 0.67 / 2 = 0.33 1.69Pa4 0.66 0.67 / 3 = 0.22 0.15Pwh 2.22 0.67 / 3 = 0.22 0.49






= 1.5 / 2 - ( 33.15 - 15.66 ) / 34.00= 0.236m


Arm (m)

Arm (m)

- Annex C4 page 2 -


500 5 kPa500 1000

1000 W1 1333Pav Pa1 Pa2

1000 W2

667500 W3 Pa3 Pa4 Pwh

Insitu soil

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 15.71Pah = ΣPai = 13.49

- Annex C4 page 3 -




ΣV= 34.00


cos θ = 0.995 tan θ = 0.100







= 1.5 / 2 - ( 36.46 - 15.63 ) / 34.00= 0.137m


Arm (m)

Arm (m)

((( )

))

- Annex C4 page 4 -



Checked by :


Reference Remarks


500 5 kPa500 1000

1000 W1 Pa2 667Pav Pa1 333

1000 Pa3 Pa4 Pwh

W2

500 Insitu soilW3

Toe

Pwv







φcv' 34 o



- Annex C5 page 1 -


ΣΗ= 5.48Pah = ΣPai = 4.93




ΣV= 13.93


Pa1 1.65 1.00 / 2 = 0.50 0.82Pa2 1.54 0.67 / 3 + 0.33 = 0.56 0.86Pa3 1.54 0.33 / 2 = 0.17 0.26Pa4 0.20 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06


Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1 - 1.00 / 2 = 0.50 7.80Pwv -1.67 1 x 2 / 3 = 0.67 -1.11Pav 0.00 1 = 1.00 0.00





= 1 / 2 - ( 7.80 - 3.13 ) / 13.93= 0.165m

Arm (m)

Arm (m)

- Annex C5 page 2 -


ΣV= 13.93 kN / m


=> OK!

OK!




= 0.67 x 100 = 67.01 m2



= / 67.01 = 20.79 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 0.67 / 100= 1.00

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.67 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.67 / 100= 1.00

1393.33

- Annex C5 page 4 -


( 1 + B' / L' )= ( 2 + 0.67 / 100 ) /

( 1 + 0.67 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.29 ) 2.99

= 0.35iq = ( 1 - Ki ) mi

= ( 1 - 0.29 ) 1.99

= 0.50ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.50 - ( 1 - 0.50 ) / ( 30.78 x tan 30.3 )= 0.47


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.00 x 0.47 x 1 x 1+ 0.5 x 9 x 0.67 x 23.30 x 1.00 x

0.35 x 1 x 1 + 4.5 x 18.96 x 1.00x 0.50 x 1 x 1

= 60.95 + 24.82 + 42.92= 128.68 kPa


- Annex C5 page 5 -


500 5 kPa500 1000

1000 W1 Pa2 667Pav Pa1 333

1000 W2 Pa3 Pa4 Pwh

500 W3 Insitu soil

ToePwv







φcv' 34 o




ΣΗ= 5.48Pah = ΣPai = 4.93

- Annex C5 page 6 -




ΣV= 13.93


cos θ = 0.995 tan θ = 0.100



Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 0.50 x 0.10 + 0.500 x 0.995 = 0.55 8.54Pwv -1.67 1 x 2 / 3 x 0.995 = 0.66 -1.11Pav 0.00 1 = 1.00 0.00





= 1 / 2 - ( 8.54 - 3.13 ) / 13.93= 0.112m

Arm (m)

Arm (m)(( )

)

- Annex C5 page 7 -





=> OK!




=> OK!

OK!




= 0.78 x 100 = 77.68 m2



= / 77.68 = 17.94 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


1393

- Annex C5 page 8 -


= 1 + 18.96 / 30.78 x 0.78 / 100= 1.00

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.78 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.78 / 100= 1.00


( 1 + B' / L' )= ( 2 + 0.78 / 100 ) /

( 1 + 0.78 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.28 ) 2.99

= 0.37iq = ( 1 - Ki ) mi

= ( 1 - 0.28 ) 1.99

= 0.52ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.52 - ( 1 - 0.52 ) / ( 30.78 x tan 30.3 )= 0.49


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.00 x 0.49 x 5.69 x 1+ 0.5 x 9 x 0.78 x 23.30 x 1.00 x

0.37 x 5.44 x 1 + 4.5 x 18.96 x 1.00x 0.52 x 5.44 x 1

= 359.54 + 164.16 + 241.26= 764.96 kPa


- Annex C5 page 9 -



Checked by :


Reference Remarks


500 5 kPa500 1000

1000 W1 Pa2 667Pav Pa1 333

1000 Pa3 Pa4 Pwh

W2

500 Insitu soilW3

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex C6 page 1 -


ΣΗ= 4.61Pah = ΣPai = 4.05




ΣV= 13.93


Pa1 1.35 1.00 / 2 = 0.50 0.68Pa2 1.26 0.67 / 3 + 0.33 = 0.56 0.70Pa3 1.26 0.33 / 2 = 0.17 0.21Pa4 0.17 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06






= 1 / 2 - ( 7.80 - 2.78 ) / 13.93= 0.140m


Arm (m)

Arm (m)

- Annex C6 page 2 -


500 5 kPa500 1000

1000 W1 Pa2 667Pav Pa1 333

1000 W2 Pa3 Pa4 Pwh

500 W3

Insitu soil

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 4.61Pah = ΣPai = 4.05

- Annex C6 page 3 -




ΣV= 13.93


cos θ = 0.995 tan θ = 0.100







= 1 / 2 - ( 8.53 - 2.78 ) / 13.93= 0.087m


Arm (m)

Arm (m)

(( )

)

- Annex C6 page 4 -

Project : Design of 1.5m Gabion Wall Annex D


Checked by :


Design Statement


1. Design Data

(I) Materials




the mesh.









weaving.










Reference


Remarks

Design Statement




- Annex D page 1 -

(B) Assumptions


Block Size


Para. 9.5.1

Mesh Size

8cm x 10cm x 2.7mm
















(II) Loadings

Dead loads









retaining height.

0.0

2.6

0.4

- Annex D page 2 -



considered.


Design Code


Second Edition.

Design Methodology









shape.


1.5m Gabion Wall

- Annex D page 3 -

Ultimate Limit Statement (ULS)Refer to Annex D1 & D3 1. Checking Overturning [OK if restoring moment > overturning moment]para. 1 Step 6



m OK! OK!

m OK! OK!

Refer to Annex D1 & D3 2. Checking Sliding [OK if resisting force > sliding force]para. 1 Step 7



m OK! OK!

m OK! OK!

Refer to Annex D1 & D3 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]para. 1 Step 8



m OK! OK!

m OK! OK!

Serviceability Limit Statement (SLS)Refer to Annex D2 & D4 1. Check Overturning and Determine Eccentricitypara. 1 Step 6 [OK if the resultant force acts within the middle third of the wall base]



m OK! OK!

m OK! OK!

For details of calculations, please refer to the Appendix D1 to D6.

0.5

0.0

0.5

0.0

0.5

0.0

0.0

0.5

- Annex D page 4 -









courses.

Drainage provisions






References


Second Edition.

- Annex D page 5 -

Project : Design of Gabion Wall Annex D1


Checked by :


Reference Remarks


300 5 kPa1000

1000 W1

Pav Pa1 Pa2 1000500

W2 500Insitu soil Pa3 Pa4 Pwh

Toe

Pwv

W1, W2 = Self-weight of the proposed protection wallPa1 = Lateral pressure due to surchargePav = Vertical component of active earth pressurePa2, Pa3, Pa4 = Horizontal component of active earth pressurePwh = Lateral pressure due to groundwaterPwv = Upthrust






φcv' 34 o



- Annex D1 page 1 -


ΣΗ= 11.10Pah = ΣPai = 9.85




ΣV= 52.91


Pa1 2.47 1.50 / 2 = 0.75 1.86Pa2 3.46 1.00 / 3 + 0.50 = 0.83 2.89Pa3 3.46 0.50 / 2 = 0.25 0.87Pa4 0.45 0.50 / 3 = 0.17 0.08Pwh 1.25 0.50 / 3 = 0.17 0.21







= 1.3 / 2 - ( 38.84 - 8.71 ) / 52.91= 0.080m

Arm (m)

Arm (m)

- Annex D1 page 2 -


ΣV= 52.91 kN / m


=> OK!

OK!




= 1.14 x 100 = 113.92 m2



= / 113.92 = 46.45 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 1.14 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.14 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.14 / 100= 1.01

5291.00

- Annex D1 page 3 -


( 1 + B' / L' )= ( 2 + 1.14 / 100 ) /

( 1 + 1.14 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.18 ) 2.99

= 0.55iq = ( 1 - Ki ) mi

= ( 1 - 0.18 ) 1.99

= 0.67ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.67 - ( 1 - 0.67 ) / ( 30.78 x tan 30.3 )= 0.65


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.01 x 0.65 x 1 x 1+ 0.5 x 9 x 1.14 x 23.30 x 1.00 x

0.55 x 1 x 1 + 4.5 x 18.96 x 1.01x 0.67 x 1 x 1

= 84.28 + 65.24 + 57.62= 207.14 kPa


- Annex D1 page 4 -


300 5 kPa0 1000

1000 W1

Pav Pa1 Pa2 1000500 W2


ToePwv







φcv' 34 o




ΣΗ= 11.10Pah = ΣPai = 9.85

- Annex D1 page 5 -




ΣV= 22.49


cos θ = 0.995 tan θ = 0.100








= 1.3 / 2 - ( 20.78 - 8.69 ) / 22.49= 0.113m

Arm (m)

Arm (m)

((

( )

))

- Annex D1 page 6 -





=> OK!




=> OK!

OK!




= 1.07 x 100 = 107.49 m2



= / 107.49 = 20.92 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


2249

- Annex D1 page 7 -


= 1 + 18.96 / 30.78 x 1.07 / 100= 1.01

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 1.07 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 1.07 / 100= 1.01


( 1 + B' / L' )= ( 2 + 1.07 / 100 ) /

( 1 + 1.07 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.37 ) 2.99

= 0.25iq = ( 1 - Ki ) mi

= ( 1 - 0.37 ) 1.99

= 0.40ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.40 - ( 1 - 0.40 ) / ( 30.78 x tan 30.3 )= 0.37


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.01 x 0.37 x 5.69 x 1+ 0.5 x 9 x 1.07 x 23.30 x 1.00 x

0.25 x 5.44 x 1 + 4.5 x 18.96 x 1.01x 0.40 x 5.44 x 1

= 270.13 + 154.76 + 187.41= 612.30 kPa


- Annex D1 page 8 -



Checked by :


Reference Remarks


300 5 kPa1000

1000 W1

Pav Pa1 Pa2 1000500

W2 500Insitu soil Pa3 Pa4 Pwh

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex D2 page 1 -


ΣΗ= 9.35Pah = ΣPai = 8.10




ΣV= 52.91


Pa1 2.03 1.50 / 2 = 0.75 1.52Pa2 2.85 1.00 / 3 + 0.50 = 0.83 2.37Pa3 2.85 0.50 / 2 = 0.25 0.71Pa4 0.37 0.50 / 3 = 0.17 0.06Pwh 1.25 0.50 / 3 = 0.17 0.21






= 1.3 / 2 - ( 38.84 - 7.69 ) / 52.91= 0.061m


Arm (m)

Arm (m)

- Annex D2 page 2 -


300 5 kPa0 1000

1000 W1 1000Pav Pa1 Pa2

500 W2

5000 W3 Pa3 Pa4 Pwh

Insitu soil

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 9.35Pah = ΣPai = 8.10

- Annex D2 page 3 -




ΣV= 52.91


cos θ = 0.995 tan θ = 0.100







= 1.3 / 2 - ( 41.20 - 7.68 ) / 52.91= 0.016m


Arm (m)

Arm (m)

((( )

))

- Annex D2 page 4 -



Checked by :


Reference Remarks


300 5 kPa1000

1000 W1 Pa2 667Pav Pa1 333

500 Pa3 Pa4 Pwh

W2

Insitu soil

Toe

Pwv







φcv' 34 o



- Annex D3 page 1 -


ΣΗ= 5.48Pah = ΣPai = 4.93




ΣV= 13.93


Pa1 1.65 1.00 / 2 = 0.50 0.82Pa2 1.54 0.67 / 3 + 0.33 = 0.56 0.86Pa3 1.54 0.33 / 2 = 0.17 0.26Pa4 0.20 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06







= 1 / 2 - ( 7.80 - 3.13 ) / 13.93= 0.165m

Arm (m)

Arm (m)

- Annex D3 page 2 -


ΣV= 13.93 kN / m


=> OK!

OK!




= 0.67 x 100 = 67.01 m2



= / 67.01 = 20.79 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78



= 1 + 18.96 / 30.78 x 0.67 / 100= 1.00

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.67 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.67 / 100= 1.00

1393.33

- Annex D3 page 3 -


( 1 + B' / L' )= ( 2 + 0.67 / 100 ) /

( 1 + 0.67 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.29 ) 2.99

= 0.35iq = ( 1 - Ki ) mi

= ( 1 - 0.29 ) 1.99

= 0.50ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.50 - ( 1 - 0.50 ) / ( 30.78 x tan 30.3 )= 0.47


gc = gγ = gq = 1




= 4.17 x 30.78 x 1.00 x 0.47 x 1 x 1+ 0.5 x 9 x 0.67 x 23.30 x 1.00 x

0.35 x 1 x 1 + 4.5 x 18.96 x 1.00x 0.50 x 1 x 1

= 60.95 + 24.82 + 42.92= 128.68 kPa


- Annex D3 page 4 -


300 5 kPa0 1000

1000 W1 Pa2 667Pav Pa1 333

500 W2 Pa3 Pa4 Pwh

0 W3 Insitu soil

ToePwv







φcv' 34 o




ΣΗ= 5.48Pah = ΣPai = 4.93

- Annex D3 page 5 -




ΣV= 13.93


cos θ = 0.995 tan θ = 0.100








= 1 / 2 - ( 8.54 - 3.13 ) / 13.93= 0.112m

Arm (m)

Arm (m)(( )

)

- Annex D3 page 6 -





=> OK!




=> OK!

OK!




= 0.78 x 100 = 77.68 m2



= / 77.68 = 17.94 kPa




Nc = ( Nq - 1 ) x cot φ'f= ( 18.96 - 1 ) x cot 30.3= 30.78


1393

- Annex D3 page 7 -


= 1 + 18.96 / 30.78 x 0.78 / 100= 1.00

sγ = 1 - 0.4 x B' / L'= 1 - 0.4 x 0.78 / 100= 1.00

sq = 1 + tan φ'f x B' / L'= 1 + tan 30.3 x 0.78 / 100= 1.00


( 1 + B' / L' )= ( 2 + 0.78 / 100 ) /

( 1 + 0.78 / 100 )= 1.99


iγ = ( 1 - Ki ) mi + 1

= ( 1 - 0.28 ) 2.99

= 0.37iq = ( 1 - Ki ) mi

= ( 1 - 0.28 ) 1.99

= 0.52ic = iq - ( 1 - iq ) / ( Nc x tan φ'f )

= 0.52 - ( 1 - 0.52 ) / ( 30.78 x tan 30.3 )= 0.49


5.44 - ( 1 - 5.44 ) / ( 30.78 x tan 30.3 )5.69

tγ = ( 1 - ω tan φ'f ) 2

= ( 1 - 5.71 tan 30.3 ) 2

5.44tq = tγ

= 5.44





= 4.17 x 30.78 x 1.00 x 0.49 x 5.69 x 1+ 0.5 x 9 x 0.78 x 23.30 x 1.00 x

0.37 x 5.44 x 1 + 4.5 x 18.96 x 1.00x 0.52 x 5.44 x 1

= 359.54 + 164.16 + 241.26= 764.96 kPa


- Annex D3 page 8 -



Checked by :


Reference Remarks


300 5 kPa1000

1000 W1 Pa2 667Pav Pa1 333

500 Pa3 Pa4 Pwh

W2

Insitu soil

Toe

Pwv






Kah 0.271


φcv' 34 o



- Annex D4 page 1 -


ΣΗ= 4.61Pah = ΣPai = 4.05




ΣV= 13.93


Pa1 1.35 1.00 / 2 = 0.50 0.68Pa2 1.26 0.67 / 3 + 0.33 = 0.56 0.70Pa3 1.26 0.33 / 2 = 0.17 0.21Pa4 0.17 0.33 / 3 = 0.11 0.02Pwh 0.56 0.33 / 3 = 0.11 0.06






= 1 / 2 - ( 7.80 - 2.78 ) / 13.93= 0.140m


Arm (m)

Arm (m)

- Annex D4 page 2 -


300 5 kPa0 1000

1000 W1 Pa2 667Pav Pa1 333

500 W2 Pa3 Pa4 Pwh

0 W3

Insitu soil

ToePwv






Kah 0.271 Back batter, θ = 1 : 10 = 0.10


φcv' 34 o




ΣΗ= 4.61Pah = ΣPai = 4.05

- Annex D4 page 3 -




ΣV= 13.93


cos θ = 0.995 tan θ = 0.100







= 1 / 2 - ( 8.53 - 2.78 ) / 13.93= 0.087m


Arm (m)

Arm (m)

(( )

)

- Annex D4 page 4 -



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



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



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


Shape Coefficients

Description K1

Vertical-wall 1.00


Spill-through 0.55



Fig. 3.1 Abutment shape K2 = Coefficient for angle of embankment to flow = ( θ / 90) 0.13




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



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 = 1 L´ and ya are the base width and retaining height of the gabion wall as shown in the drawing in Appendix A.


Fr ya 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.