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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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 2-
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 3-
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 4-
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 5-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 7-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 8-
(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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 9-
(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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 10-
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;
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 11-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 12-
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Appendix A
Typical Layout of Gabion Wall
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. A - 1
Drawing Notes: 1. All dimensions are in millimeters unless otherwise specified. 2. Depending on the soil conditions, designers should determine whether any
ground treatment for foundation is required in consideration with sliding, bearing or soil slip failures.
3. Determination of Potential Scour Depth by Froehlich's Live-Bed Scour Equation
where: K1 = Coefficient for shape
Shape Coefficients
Description K1
Vertical-wall 1.00
Vertical-wall with wing walls 0.82
Spill-through 0.55
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. A - 2
Example: Assume K1 = 0.82, K2 = ( 90 / 90) 0.13 = 1 L´ and ya are the base width and retaining height of the gabion wall as shown in the sketches in the calculations in Appendix B.
Computed Scour Depth, ys as follow:
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. A - 4
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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
(c) Effective friction angle, φ' = 35 o
(II) Loadings
Dead loads
Self weight of the proposed protection wall, earth pressure and hydrostatic pressure are
taken to be dead load for design. The unit weight of water was taken as 9.81 kN/m3.
Imposed 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
para. 1 Step 7 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, 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]A5, A7 & A9
para. 1 Step 8 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!
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
above foundation (without back batter) (with back batter)
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
Prepared by : NG Chun-ling (AE/TM5)
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
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 4.20 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.)
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.)
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
- Annex A1 page 5 -
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 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
ΣM = 431.62 (kNm/m run)ΣMr = 519.39 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 125.65 + 87.76 - 0.00 = 213.41 kNm/m runRestoring Moment ΣMr = 519.39 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 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
Resisting Force against Sliding, Fr == 102.89 kN/m run > Activating Force Fa = 85.08 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.009 m
Effective Width B' = B - 2 e = 4.20 - 2 x 0.009= 4.18 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 4.18 x 100 = 418.16 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'
= / 418.16 = 35.00 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 30.3= 23.30
14634
- Annex A1 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 4.18 / 100 ) /
( 1 + 4.18 / 100 )= 1.96
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 8508 / ( 14634 + 4.2 x 418.16 x cot 30.3 )= 0.48
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
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 1.03 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
=> qult > qmax = 35.00 kPa => OK! OK!
- Annex A1 page 8 -
Project : Design of Gabion Wall Annex A2
Prepared by : NG Chun-ling (AE/TM5)
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
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 Base width of wall 4.2 mφ'f 35.0 o ( = tan-1((tan φ')/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 4.50 = 6.10Pa2 = 0.271 x 3.00 x 21 x 3.00 / 2 = 25.61Pa3 = 0.271 x 3.00 x 21 x 1.50 = 25.61Pa4 = 0.271 x 11 x 1.50 x 1.50 / 2 = 3.35Pwh = 10 x 1.50 x 1.50 / 2 = 11.25
ΣΗ= 71.92Pah = ΣPai = 60.67
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 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
ΣM = 402.15 (kNm/m run)ΣMr = 490.35 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 104.25 + 88.20 - 0.00 = 192.45 kNm/m runRestoring Moment ΣMr = 490.35 kNm/m run
=> 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
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. 4.5 mTable 8 φ' 35 o Water level (from bottom) 1.5 m
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
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 4.50 = 6.10Pa2 = 0.271 x 3.00 x 21 x 3.00 / 2 = 25.61Pa3 = 0.271 x 3.00 x 21 x 1.50 = 25.61Pa4 = 0.271 x 11 x 1.50 x 1.50 / 2 = 3.35Pwh = 10 x 1.50 x 1.50 / 2 = 11.25
ΣΗ= 71.92Pah = ΣPai = 60.67
- Annex A2 page 3 -
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 Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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 4.00 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
ΣM = 431.62 (kNm/m run)ΣMr = 519.39 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 104.25 + 87.76 - 0.00 = 192.01 kNm/m runRestoring Moment ΣMr = 519.39 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 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 -
Project : Design of Gabion Wall Annex A3
Prepared by : NG Chun-ling (AE/TM5)
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
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. 4.00 mTable 8 φ' 35 o Water level (from bottom) 1.33 m
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.)
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 δb/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 A3 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 4.00 = 6.60Pa2 = 0.330 x 2.67 x 21 x 2.67 / 2 = 24.63Pa3 = 0.330 x 2.67 x 21 x 1.33 = 24.63Pa4 = 0.330 x 11 x 1.33 x 1.33 / 2 = 3.22Pwh = 10 x 1.33 x 1.33 / 2 = 8.89
ΣΗ= 67.96Pah = ΣPai = 59.07
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 = 53.04Pwv = 10 x 1.33 x 3.40 / 2 = -22.67Pav = 0.00 = 0.00
ΣV= 122.41
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 254.11 (kNm/m run)ΣMr = 305.49 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 89.72 + 51.38 - 0.00 = 141.09 kNm/m runRestoring Moment ΣMr = 305.49 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 3.4 / 2 - ( 305.49 - 141.09 ) / 122.41= 0.357m
Arm (m)
Arm (m)
- Annex A3 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 67.96 kN / m
ΣV= 122.41 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 122.41 x 0.58= 71.43 kN/m run > Sliding Force Fa = Σ H= 67.96 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.357 m
Effective Width B' = B - 2 e = 3.40 - 2 x 0.357= 2.69 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 2.69 x 100 = 268.59 m2
Sliding Force Qs = ΣH x L' = 67.96 x 100 = 6796 kNNormal Force Qn = ΣV x L' = 122.41 x 100 = 12241 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 268.59 = 45.58 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 2.69 / 100 ) /
( 1 + 2.69 / 100 )= 1.97
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 6796 / ( 12241 + 4.2 x 268.59 x cot 30.3 )= 0.48
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
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 0.23 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
=> qult > qmax = 45.58 kPa => OK! OK!
- 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
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. 4.00 mTable 8 φ' 35 o Water level (from bottom) 1.33 m
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.)
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.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 4.00 = 6.60Pa2 = 0.330 x 2.67 x 21 x 2.67 / 2 = 24.63Pa3 = 0.330 x 2.67 x 21 x 1.33 = 24.63Pa4 = 0.330 x 11 x 1.33 x 1.33 / 2 = 3.22Pwh = 10 x 1.33 x 1.33 / 2 = 8.89
ΣΗ= 67.96Pah = ΣPai = 59.07
- Annex A3 page 5 -
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 = 53.04Pwv = 10 x 1.33 x 3.40 / 2 = -22.67Pav = 0.00 = 0.00
ΣV= 122.41
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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 3.50 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
ΣM = 276.29 (kNm/m run)ΣMr = 327.41 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 89.72 + 51.12 - 0.00 = 140.84 kNm/m runRestoring Moment ΣMr = 327.41 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 3.4 / 2 - ( 327.41 - 140.84 ) / 122.41= 0.176m
Arm (m)
Arm (m)
((
((( )
))))
- Annex A3 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 = 55.38 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 63.49 kN / m
Resisting Force against Sliding, Fr == 63.49 kN/m run > Activating Force Fa = 55.38 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 = 67.96 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 85.60 kN / m
Resisting Force against Sliding, Fr == 85.60 kN/m run > Activating Force Fa = 67.96 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.176 m
Effective Width B' = B - 2 e = 3.40 - 2 x 0.176= 3.05 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 3.05 x 100 = 304.82 m2
Sliding Force Qs = ΣH x L' = 67.96 x 100 = 6796 kNNormal Force Qn = ΣV x L' = 122.41 x 100 = 12241 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 304.82 = 40.16 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 30.3= 23.30
12241
- Annex A3 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 3.05 / 100 ) /
( 1 + 3.05 / 100 )= 1.97
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 6796 / ( 12241 + 4.2 x 304.82 x cot 30.3 )= 0.47
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
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
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 0.24 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
=> qult > qmax = 40.16 kPa => OK! OK!
- Annex A3 page 8 -
Project : Design of Gabion Wall Annex A4
Prepared by : NG Chun-ling (AE/TM5)
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
1000 Insitu soil Pa2
W3
1000 1333W4 Pa3 Pa4 Pwh
500 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. 4.00 mTable 8 φ' 35 o Water level (from bottom) 1.33 m
γm 1 Base width of wall 3.4 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 A4 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 4.00 = 5.42Pa2 = 0.271 x 2.67 x 21 x 2.67 / 2 = 20.23Pa3 = 0.271 x 2.67 x 21 x 1.33 = 20.23Pa4 = 0.271 x 11 x 1.33 x 1.33 / 2 = 2.65Pwh = 10 x 1.33 x 1.33 / 2 = 8.89
ΣΗ= 57.43Pah = ΣPai = 48.54
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 = 53.04Pwv = 10 x 1.33 x 3.40 / 2 = -22.67Pav = 0.00 = 0.00
ΣV= 122.41
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 74.42 (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
ΣM = 254.11 (kNm/m run)ΣMr = 305.49 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 74.42 + 51.38 - 0.00 = 125.80 kNm/m runRestoring Moment ΣMr = 305.49 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 3.4 / 2 - ( 305.49 - 125.80 ) / 122.41= 0.232m
Geoguide 1 By Middle-third Rule, B/6 = 3.4 / 6 = 0.567m > 0.232m OK! OK!Clause 9.2.4
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
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. 4.00 mTable 8 φ' 35 o Water level (from bottom) 1.33 m
γm 1 Base width of wall 3.4 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 4.00 = 5.42Pa2 = 0.271 x 2.67 x 21 x 2.67 / 2 = 20.23Pa3 = 0.271 x 2.67 x 21 x 1.33 = 20.23Pa4 = 0.271 x 11 x 1.33 x 1.33 / 2 = 2.65Pwh = 10 x 1.33 x 1.33 / 2 = 8.89
ΣΗ= 57.43Pah = ΣPai = 48.54
- Annex A4 page 3 -
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 = 53.04Pwv = 10 x 1.33 x 4.20 / 2 = -22.67Pav = 0.00 = 0.00
ΣV= 122.41
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 74.42 (kNm/m run)
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 62.52W2 30.42 2.50 x 0.10 + 2.425 x 0.995 = 2.66 80.93W3 41.34 1.50 x 0.10 + 2.075 x 0.995 = 2.21 91.49W4 53.04 0.50 x 0.10 + 1.700 x 0.995 = 1.74 92.35Pwv -22.67 3.4 x 2 / 3 x 0.995 = 2.26 -51.12Pav 0.00 3.4 = 3.40 0.00
ΣM = 276.17 (kNm/m run)ΣMr = 327.29 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 74.42 + 51.12 - 0.00 = 125.54 kNm/m runRestoring Moment ΣMr = 327.29 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 3.4 / 2 - ( 327.29 - 125.54 ) / 122.41= 0.052m
Geoguide 1 By Middle-third Rule, B/6 = 3.4 / 6 = 0.567m > 0.052m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(
(( )
))(
( ))
- Annex A4 page 4 -
Project : Design of Gabion Wall Annex A5
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 4.5m Gabion WallChecking of Ultimate Limit State (toe at 1.5m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 1.5m above foundation)
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex A5 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 3.00 = 4.95Pa2 = 0.330 x 2.00 x 21 x 2.00 / 2 = 13.85Pa3 = 0.330 x 2.00 x 21 x 1.00 = 13.85Pa4 = 0.330 x 11 x 1.00 x 1.00 / 2 = 1.81Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 39.46Pah = ΣPai = 34.46
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 = 41.34Pwv = 10 x 1.00 x 2.65 / 2 = -13.25Pav = 0.00 = 0.00
ΣV= 78.79
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 39.70 (kNm/m run)
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
ΣM = 122.88 (kNm/m run)ΣMr = 146.29 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 39.70 + 23.41 - 0.00 = 63.11 kNm/m runRestoring Moment ΣMr = 146.29 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.65 / 2 - ( 146.29 - 63.11 ) / 78.79= 0.269m
Arm (m)
Arm (m)
- Annex A5 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 39.46 kN / m
ΣV= 78.79 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 78.79 x 0.58= 45.97 kN/m run > Sliding Force Fa = Σ H= 39.46 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.269 m
Effective Width B' = B - 2 e = 2.65 - 2 x 0.269= 2.11 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 2.11 x 100 = 211.14 m2
Sliding Force Qs = ΣH x L' = 39.46 x 100 = 3946 kNNormal Force Qn = ΣV x L' = 78.79 x 100 = 7879 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 211.14 = 37.32 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 2.11 / 100 ) /
( 1 + 2.11 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 3946 / ( 7879 + 4.2 x 211.14 x cot 30.3 )= 0.42
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
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 1.01 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
=> qult > qmax = 37.32 kPa => OK! OK!
- Annex A5 page 4 -
2. Checking of Ultimate Limit State (toe at 1.5m above foundation)(with back batter 1:10)
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
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 Back batter, θ = 1 : 10 = 0.10(Note: gm 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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 3.00 = 4.95Pa2 = 0.330 x 2.00 x 21 x 2.00 / 2 = 13.85Pa3 = 0.330 x 2.00 x 21 x 1.00 = 13.85Pa4 = 0.330 x 11 x 1.00 x 1.00 / 2 = 1.81Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 39.46Pah = ΣPai = 34.46
- Annex A5 page 5 -
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 = 41.34Pwv = 10 x 1.00 x 2.65 / 2 = -13.25Pav = 0.00 = 0.00
ΣV= 78.79
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 39.70 (kNm/m run)
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
ΣM = 133.91 (kNm/m run)ΣMr = 157.20 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 39.70 + 23.29 - 0.00 = 63.00 kNm/m runRestoring Moment ΣMr = 157.20 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.65 / 2 - ( 157.20 - 63.00 ) / 78.79= 0.129m
Arm (m)
Arm (m)
((
(( )
)))
- Annex A5 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 = 31.39 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 41.10 kN / m
Resisting Force against Sliding, Fr == 41.10 kN/m run > Activating Force Fa = 31.39 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 = 39.46 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 57.02 kN / m
Resisting Force against Sliding, Fr == 57.02 kN/m run > Activating Force Fa = 39.46 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.129 m
Effective Width B' = B - 2 e = 2.65 - 2 x 0.129= 2.39 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 2.39 x 100 = 239.14 m2
Sliding Force Qs = ΣH x L' = 39.46 x 100 = 3946 kNNormal Force Qn = ΣV x L' = 78.79 x 100 = 7879 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 239.14 = 32.95 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 30.3= 23.30
7879
- Annex A5 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 2.39 / 100 ) /
( 1 + 2.39 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 3946 / ( 7879 + 4.2 x 239.14 x cot 30.3 )= 0.41
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
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
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 1.01 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
=> qult > qmax = 32.95 kPa => OK! OK!
- Annex A5 page 8 -
Project : Design of Gabion Wall Annex A6
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 4.5m Gabion WallChecking of Serviceability Limit State (toe at 1.5m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 1.5m above foundation)
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
γm 1 Base width of wall 2.65 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 A6 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 3.00 = 4.06Pa2 = 0.271 x 2.00 x 21 x 2.00 / 2 = 11.38Pa3 = 0.271 x 2.00 x 21 x 1.00 = 11.38Pa4 = 0.271 x 11 x 1.00 x 1.00 / 2 = 1.49Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 33.32Pah = ΣPai = 28.32
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 = 41.34Pwv = 10 x 1.00 x 2.65 / 2 = -13.25Pav = 0.00 = 0.00
ΣV= 78.79
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 32.92 (kNm/m run)
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
ΣM = 122.88 (kNm/m run)ΣMr = 146.29 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 32.92 + 23.41 - 0.00 = 56.33 kNm/m runRestoring Moment ΣMr = 146.29 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.65 / 2 - ( 146.29 - 56.33 ) / 78.79= 0.183m
Geoguide 1 By Middle-third Rule, B/6 = 2.65 / 6 = 0.442m > 0.183m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex A6 page 2 -
2. Checking of Serviceability Limit State (toe at 1.5m above foundation)(with back batter 1:10)
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
γm 1 Base width of wall 2.65 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 3.00 = 4.06Pa2 = 0.271 x 2.00 x 21 x 2.00 / 2 = 11.38Pa3 = 0.271 x 2.00 x 21 x 1.00 = 11.38Pa4 = 0.271 x 11 x 1.00 x 1.00 / 2 = 1.49Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 33.32Pah = ΣPai = 28.32
- Annex A6 page 3 -
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 = 41.34Pwv = 10 x 1.00 x 4.20 / 2 = -13.25Pav = 0.00 = 0.00
ΣV= 78.79
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 32.92 (kNm/m run)
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 45.38W2 30.42 1.50 x 0.10 + 1.675 x 0.995 = 1.82 55.22W3 41.34 0.50 x 0.10 + 1.325 x 0.995 = 1.37 56.55Pwv -13.25 2.65 x 2 / 3 x 0.995 = 1.76 -23.29Pav 0.00 2.65 = 2.65 0.00
ΣM = 133.85 (kNm/m run)ΣMr = 157.15 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 32.92 + 23.29 - 0.00 = 56.21 kNm/m runRestoring Moment ΣMr = 157.15 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.65 / 2 - ( 157.15 - 56.21 ) / 78.79= 0.044m
Geoguide 1 By Middle-third Rule, B/6 = 2.65 / 6 = 0.442m > 0.044m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(
(( )
))()
- Annex A6 page 4 -
Project : Design of Gabion Wall Annex A7
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 4.5m Gabion WallChecking of Ultimate Limit State (toe at 2.5m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 2.5m above foundation)
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1.2 Base width of wall 1.95 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex A7 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.00 = 3.30Pa2 = 0.330 x 1.33 x 21 x 1.33 / 2 = 6.16Pa3 = 0.330 x 1.33 x 21 x 0.67 = 6.16Pa4 = 0.330 x 11 x 0.67 x 0.67 / 2 = 0.81Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 18.64Pah = ΣPai = 16.42
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 = 30.42Pwv = 10 x 0.67 x 1.95 / 2 = -6.50Pav = 0.00 = 0.00
ΣV= 44.20
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 12.86 (kNm/m run)
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
ΣM = 47.57 (kNm/m run)ΣMr = 56.02 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 12.86 + 8.45 - 0.00 = 21.31 kNm/m runRestoring Moment ΣMr = 56.02 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.95 / 2 - ( 56.02 - 21.31 ) / 44.20= 0.190m
Arm (m)
Arm (m)
- Annex A7 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 18.64 kN / m
ΣV= 44.20 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 44.20 x 0.58= 25.79 kN/m run > Sliding Force Fa = Σ H= 18.64 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.190 m
Effective Width B' = B - 2 e = 1.95 - 2 x 0.190= 1.57 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.57 x 100 = 157.06 m2
Sliding Force Qs = ΣH x L' = 18.64 x 100 = 1864 kNNormal Force Qn = ΣV x L' = 44.20 x 100 = 4420 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 157.06 = 28.14 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.57 / 100 ) /
( 1 + 1.57 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1864 / ( 4420 + 4.2 x 157.06 x cot 30.3 )= 0.34
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
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 1.01 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
=> qult > qmax = 28.14 kPa => OK! OK!
- Annex A7 page 4 -
2. Checking of Ultimate Limit State (toe at 2.5m above foundation)(with back batter 1:10)
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1.2 Base width of wall 1.95 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.00 = 3.30Pa2 = 0.330 x 1.33 x 21 x 1.33 / 2 = 6.16Pa3 = 0.330 x 1.33 x 21 x 0.67 = 6.16Pa4 = 0.330 x 11 x 0.67 x 0.67 / 2 = 0.81Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 18.64Pah = ΣPai = 16.42
- Annex A7 page 5 -
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 = 30.42Pwv = 10 x 0.67 x 1.95 / 2 = -6.50Pav = 0.00 = 0.00
ΣV= 44.20
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 12.86 (kNm/m run)
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
ΣM = 51.88 (kNm/m run)ΣMr = 60.29 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 12.86 + 8.41 - 0.00 = 21.27 kNm/m runRestoring Moment ΣMr = 60.29 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.95 / 2 - ( 60.29 - 21.27 ) / 44.20= 0.092m
Arm (m)
Arm (m)
((
( )
))
- Annex A7 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 = 14.13 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 23.25 kN / m
Resisting Force against Sliding, Fr == 23.25 kN/m run > Activating Force Fa = 14.13 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 = 18.64 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 33.92 kN / m
Resisting Force against Sliding, Fr == 33.92 kN/m run > Activating Force Fa = 18.64 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.092 m
Effective Width B' = B - 2 e = 1.95 - 2 x 0.092= 1.77 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.77 x 100 = 176.54 m2
Sliding Force Qs = ΣH x L' = 18.64 x 100 = 1864 kNNormal Force Qn = ΣV x L' = 44.20 x 100 = 4420 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 176.54 = 25.04 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 30.3= 23.30
4420
- Annex A7 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.77 / 100 ) /
( 1 + 1.77 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1864 / ( 4420 + 4.2 x 176.54 x cot 30.3 )= 0.33
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
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
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 1.01 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
=> qult > qmax = 25.04 kPa => OK! OK!
- Annex A7 page 8 -
Project : Design of Gabion Wall Annex A8
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 4.5m Gabion WallChecking of Serviceability Limit State (toe at 2.5m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 2.5m above foundation)
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1 Base width of wall 1.95 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 A8 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.00 = 2.71Pa2 = 0.271 x 1.33 x 21 x 1.33 / 2 = 5.06Pa3 = 0.271 x 1.33 x 21 x 0.67 = 5.06Pa4 = 0.271 x 11 x 0.67 x 0.67 / 2 = 0.66Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 15.71Pah = ΣPai = 13.49
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 = 30.42Pwv = 10 x 0.67 x 1.95 / 2 = -6.50Pav = 0.00 = 0.00
ΣV= 44.20
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 10.66 (kNm/m run)
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
ΣM = 47.57 (kNm/m run)ΣMr = 56.02 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 10.66 + 8.45 - 0.00 = 19.11 kNm/m runRestoring Moment ΣMr = 56.02 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.95 / 2 - ( 56.02 - 19.11 ) / 44.20= 0.140m
Geoguide 1 By Middle-third Rule, B/6 = 1.95 / 6 = 0.325m > 0.140m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex A8 page 2 -
2. Checking of Serviceability Limit State (toe at 2.5m above foundation)(with back batter 1:10)
750 650 5 kPa800 700 1300
1000 W1 1333Pav Pa1 Pa2
1000 W2
6671000 W3 Pa3 Pa4 Pwh
Insitu soil1000 W4
500 W5
ToePwv
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1 Base width of wall 1.95 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.00 = 2.71Pa2 = 0.271 x 1.33 x 21 x 1.33 / 2 = 5.06Pa3 = 0.271 x 1.33 x 21 x 0.67 = 5.06Pa4 = 0.271 x 11 x 0.67 x 0.67 / 2 = 0.66Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 15.71Pah = ΣPai = 13.49
- Annex A8 page 3 -
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 = 30.42Pwv = 10 x 0.67 x 4.20 / 2 = -6.50Pav = 0.00 = 0.00
ΣV= 44.20
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 10.66 (kNm/m run)
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 29.25W2 30.42 0.50 x 0.10 + 0.975 x 0.995 = 1.02 31.02Pwv -6.50 1.95 x 2 / 3 x 0.995 = 1.29 -8.41Pav 0.00 1.95 = 1.95 0.00
ΣM = 51.86 (kNm/m run)ΣMr = 60.26 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 10.66 + 8.41 - 0.00 = 19.07 kNm/m runRestoring Moment ΣMr = 60.26 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.95 / 2 - ( 60.26 - 19.07 ) / 44.20= 0.043m
Geoguide 1 By Middle-third Rule, B/6 = 1.95 / 6 = 0.325m > 0.043m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
((( )
))
- Annex A8 page 4 -
Project : Design of Gabion Wall Annex A9
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 4.5m Gabion WallChecking of Ultimate Limit State (toe at 3.5m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 3.5m above foundation)
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.30 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex A9 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
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 = 20.28Pwv = 10 x 0.33 x 1.30 / 2 = -2.17Pav = 0.00 = 0.00
ΣV= 18.11
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 2.02 (kNm/m run)
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
ΣM = 11.30 (kNm/m run)ΣMr = 13.18 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.88 - 0.00 = 3.90 kNm/m runRestoring Moment ΣMr = 13.18 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 13.18 - 3.90 ) / 18.11= 0.137m
Arm (m)
Arm (m)
- Annex A9 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 5.48 kN / m
ΣV= 18.11 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 18.11 x 0.58= 10.57 kN/m run > Sliding Force Fa = Σ H= 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.137 m
Effective Width B' = B - 2 e = 1.30 - 2 x 0.137= 1.03 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.03 x 100 = 102.51 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 18.11 x 100 = 1811 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 102.51 = 17.67 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.03 / 100 ) /
( 1 + 1.03 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1811 + 4.2 x 102.51 x cot 30.3 )= 0.22
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
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 1.01 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
=> qult > qmax = 17.67 kPa => OK! OK!
- Annex A9 page 4 -
2. Checking of Ultimate Limit State (toe at 3.5m above foundation)(with back batter 1:10)
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.30 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
- Annex A9 page 5 -
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 = 20.28Pwv = 10 x 0.33 x 1.30 / 2 = -2.17Pav = 0.00 = 0.00
ΣV= 18.11
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 2.02 (kNm/m run)
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
ΣM = 12.26 (kNm/m run)ΣMr = 14.13 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.87 - 0.00 = 3.89 kNm/m runRestoring Moment ΣMr = 14.13 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 14.13 - 3.89 ) / 18.11= 0.085m
Arm (m)
Arm (m)(( )
)
- Annex A9 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 = 3.65 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 9.65 kN / m
Resisting Force against Sliding, Fr == 9.65 kN/m run > Activating Force Fa = 3.65 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 = 5.48 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 15.99 kN / m
Resisting Force against Sliding, Fr == 15.99 kN/m run > Activating Force Fa = 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.085 m
Effective Width B' = B - 2 e = 1.30 - 2 x 0.085= 1.13 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.13 x 100 = 113.03 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 18.11 x 100 = 1811 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 113.03 = 16.03 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 30.3= 23.30
1811
- Annex A9 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.13 / 100 ) /
( 1 + 1.13 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1811 + 4.2 x 113.03 x cot 30.3 )= 0.21
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
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
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 1.01 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
=> qult > qmax = 16.03 kPa => OK! OK!
- Annex A9 page 8 -
Project : Design of Gabion Wall Annex A10
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 4.5m Gabion WallChecking of Serviceability Limit State (toe at 3.5m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 3.5m above foundation)
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1.30 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 A10 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
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 = 20.28Pwv = 10 x 0.33 x 1.30 / 2 = -2.17Pav = 0.00 = 0.00
ΣV= 18.11
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 1.67 (kNm/m run)
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
ΣM = 11.30 (kNm/m run)ΣMr = 13.18 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.88 - 0.00 = 3.55 kNm/m runRestoring Moment ΣMr = 13.18 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 13.18 - 3.55 ) / 18.11= 0.118m
Geoguide 1 By Middle-third Rule, B/6 = 1.3 / 6 = 0.217m > 0.118m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex A10 page 2 -
2. Checking of Serviceability Limit State (toe at 3.5m above foundation)(with back batter 1:10)
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1.3 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
- Annex A10 page 3 -
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 = 20.28Pwv = 10 x 0.33 x 4.20 / 2 = -2.17Pav = 0.00 = 0.00
ΣV= 18.11
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 1.67 (kNm/m run)
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 14.12Pwv -2.17 1.3 x 2 / 3 x 0.995 = 0.86 -1.87Pav 0.00 1.3 = 1.30 0.00
ΣM = 12.25 (kNm/m run)ΣMr = 14.12 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.87 - 0.00 = 3.54 kNm/m runRestoring Moment ΣMr = 14.12 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 14.12 - 3.54 ) / 18.11= 0.066m
Geoguide 1 By Middle-third Rule, B/6 = 1.3 / 6 = 0.217m > 0.066m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(( )
)
- Annex A10 page 4 -
Project : Design of 3.5m Gabion Wall Annex B
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion Wall
Design Statement
Design of 3.5m 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 B 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
(c) Effective friction angle, φ' = 35 o
(II) Loadings
Dead loads
Self weight of the proposed protection wall, earth pressure and hydrostatic pressure are
taken to be dead load for design. The unit weight of water was taken as 9.81 kN/m3.
Imposed 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 B 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
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
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!
Refer to Annex B1, B3, 2. Checking Sliding [OK if resisting force > sliding force]B5 & B7
para. 1 Step 7 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!
Refer to Annex B1, B3, 3. Checking Bearing Capacity [OK if bearing capacity > bearing pressure]B5 & B7
para. 1 Step 8 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!
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
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!
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 -
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 B page 5 -
Project : Design of Gabion Wall Annex B1
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion WallChecking of Ultimate Limit State (toe at 0m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 0m above foundation)
650 600 5 kPa600 1100
1000 W1
Pav
1000 2333W2 Pa1
1000 Insitu soil Pa2
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
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. 3.50 mTable 8 φ' 35 o Water level (from bottom) 1.17 m
γm 1.2 Base width of wall 2.95 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex B1 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 3.50 = 5.77Pa2 = 0.330 x 2.33 x 21 x 2.33 / 2 = 18.85Pa3 = 0.330 x 2.33 x 21 x 1.17 = 18.85Pa4 = 0.330 x 11 x 1.17 x 1.17 / 2 = 2.47Pwh = 10 x 1.17 x 1.17 / 2 = 6.81
ΣΗ= 52.75Pah = ΣPai = 45.95
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88W4 = 26 x 0.6 x 2.95 x 0.50 = 92.04Pwv = 10 x 1.17 x 2.95 / 2 = -17.21Pav = 0.00 = 0.00
ΣV= 154.39
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 61.36 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 2.95 - 1.10 / 2 = 2.40 41.18W2 26.52 2.95 - 1.70 / 2 = 2.10 55.69W3 35.88 2.95 - 2.30 / 2 = 1.80 64.58W4 92.04 2.95 - 2.95 / 2 = 1.48 135.76Pwv -17.21 2.95 x 2 / 3 = 1.97 -33.84Pav 0.00 2.95 = 2.95 0.00
ΣM = 263.38 (kNm/m run)ΣMr = 297.22 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 61.36 + 33.84 - 0.00 = 95.21 kNm/m runRestoring Moment ΣMr = 297.22 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.95 / 2 - ( 297.22 - 95.21 ) / 154.39= 0.167m
Arm (m)
Arm (m)
- Annex B1 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 52.75 kN / m
ΣV= 154.39 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 154.39 x 0.58= 90.09 kN/m run > Sliding Force Fa = Σ H= 52.75 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.167 m
Effective Width B' = B - 2 e = 2.95 - 2 x 0.167= 2.62 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 2.62 x 100 = 261.69 m2
Sliding Force Qs = ΣH x L' = 52.75 x 100 = 5275 kNNormal Force Qn = ΣV x L' = 154.39 x 100 = 15439 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 261.69 = 59.00 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 2.62 / 100 ) /
( 1 + 2.62 / 100 )= 1.97
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 5275 / ( 15439 + 4.2 x 261.69 x cot 30.3 )= 0.30
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
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 0.46 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
=> qult > qmax = 59.00 kPa => OK! OK!
- Annex B1 page 5 -
2. Checking of Ultimate Limit State (toe at 0m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1
Pav
1000 W2 2333Pa1
1000 W3 Insitu soil Pa2
500 W4 1167Pa3 Pa4 Pwh
ToePwv
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
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. 3.50 mTable 8 φ' 35 o Water level (from bottom) 1.17 m
γm 1.2 Base width of wall 2.95 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 3.50 = 5.77Pa2 = 0.330 x 2.33 x 21 x 2.33 / 2 = 18.85Pa3 = 0.330 x 2.33 x 21 x 1.17 = 18.85Pa4 = 0.330 x 11 x 1.17 x 1.17 / 2 = 2.47Pwh = 10 x 1.17 x 1.17 / 2 = 6.81
ΣΗ= 52.75Pah = ΣPai = 45.95
- Annex B1 page 6 -
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88W4 = 26 x 0.6 x 2.95 x 0.50 = 23.01Pwv = 10 x 1.17 x 2.95 / 2 = -17.21Pav = 0.00 = 0.00
ΣV= 85.36
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 61.36 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 3.00 x 0.10 + 2.400 x 0.995 = 2.69 46.10W2 26.52 2.00 x 0.10 + 2.100 x 0.995 = 2.29 60.69W3 35.88 1.00 x 0.10 + 1.800 x 0.995 = 1.89 67.83W4 23.01 0.25 x 0.10 + 1.475 x 0.995 = 1.49 34.34Pwv -17.21 2.95 x 2 / 3 x 0.995 = 1.96 -33.68Pav 0.00 2.95 = 2.95 0.00
ΣM = 175.30 (kNm/m run)ΣMr = 208.97 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 61.36 + 33.68 - 0.00 = 95.04 kNm/m runRestoring Moment ΣMr = 208.97 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.95 / 2 - ( 208.97 - 95.04 ) / 85.36= 0.140m
Arm (m)
Arm (m)
((
((( )
))))
- Annex B1 page 7 -
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 = 43.96 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 43.98 kN / m
Resisting Force against Sliding, Fr == 43.98 kN/m run > Activating Force Fa = 43.96 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 = 52.75 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 62.10 kN / m
Resisting Force against Sliding, Fr == 62.10 kN/m run > Activating Force Fa = 52.75 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.140 m
Effective Width B' = B - 2 e = 2.95 - 2 x 0.140= 2.67 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 2.67 x 100 = 266.94 m2
Sliding Force Qs = ΣH x L' = 52.75 x 100 = 5275 kNNormal Force Qn = ΣV x L' = 85.36 x 100 = 8536 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 266.94 = 31.98 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 30.3= 23.30
8536
- Annex B1 page 8 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 2.67 / 100 ) /
( 1 + 2.67 / 100 )= 1.97
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 5275 / ( 8536 + 4.2 x 266.94 x cot 30.3 )= 0.51
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
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
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 0.21 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
=> qult > qmax = 31.98 kPa => OK! OK!
- Annex B1 page 9 -
Project : Design of Gabion Wall Annex B2
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion WallChecking of Serviceability Limit State (toe at 0m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 0m above foundation)
650 600 5 kPa600 1100
1000 W1
Pav
1000 2333W2 Pa1
1000 Insitu soil Pa2
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
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. 3.50 mTable 8 φ' 35 o Water level (from bottom) 1.17 m
γm 1 Base width of wall 2.95 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 B2 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 3.50 = 4.74Pa2 = 0.271 x 2.33 x 21 x 2.33 / 2 = 15.49Pa3 = 0.271 x 2.33 x 21 x 1.17 = 15.49Pa4 = 0.271 x 11 x 1.17 x 1.17 / 2 = 2.03Pwh = 10 x 1.17 x 1.17 / 2 = 6.81
ΣΗ= 44.56Pah = ΣPai = 37.75
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88W4 = 26 x 0.6 x 2.95 x 0.50 = 92.04Pwv = 10 x 1.17 x 2.95 / 2 = -17.21Pav = 0.00 = 0.00
ΣV= 154.39
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 50.89 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 2.95 - 1.10 / 2 = 2.40 41.18W2 26.52 2.95 - 1.70 / 2 = 2.10 55.69W3 35.88 2.95 - 2.30 / 2 = 1.80 64.58W4 92.04 2.95 - 2.95 / 2 = 1.48 135.76Pwv -17.21 2.95 x 2 / 3 = 1.97 -33.84Pav 0.00 2.95 = 2.95 0.00
ΣM = 263.38 (kNm/m run)ΣMr = 297.22 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 50.89 + 33.84 - 0.00 = 84.74 kNm/m runRestoring Moment ΣMr = 297.22 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.95 / 2 - ( 297.22 - 84.74 ) / 154.39= 0.099m
Geoguide 1 By Middle-third Rule, B/6 = 2.95 / 6 = 0.492m > 0.099m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex B2 page 2 -
2. Checking of Serviceability Limit State (toe at 0m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1
Pav
1000 W2 2333Pa1
1000 W3 Pa2
Insitu soil500 W4 1167
Pa3 Pa4 Pwh
ToePwv
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
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. 3.50 mTable 8 φ' 35 o Water level (from bottom) 1.17 m
γm 1 Base width of wall 2.95 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 3.50 = 4.74Pa2 = 0.271 x 2.33 x 21 x 2.33 / 2 = 15.49Pa3 = 0.271 x 2.33 x 21 x 1.17 = 15.49Pa4 = 0.271 x 11 x 1.17 x 1.17 / 2 = 2.03Pwh = 10 x 1.17 x 1.17 / 2 = 6.81
ΣΗ= 44.56Pah = ΣPai = 37.75
- Annex B2 page 3 -
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88W4 = 26 x 0.6 x 2.95 x 0.50 = 92.04Pwv = 10 x 1.17 x 2.95 / 2 = -17.21Pav = 0.00 = 0.00
ΣV= 154.39
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 50.89 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 3.00 x 0.10 + 2.400 x 0.995 = 2.69 46.08W2 26.52 2.00 x 0.10 + 2.100 x 0.995 = 2.29 60.67W3 35.88 1.00 x 0.10 + 1.800 x 0.995 = 1.89 67.82W4 92.04 0.25 x 0.10 + 1.475 x 0.995 = 1.49 137.36Pwv -17.21 2.95 x 2 / 3 x 0.995 = 1.96 -33.68Pav 0.00 2.95 = 2.95 0.00
ΣM = 278.25 (kNm/m run)ΣMr = 311.92 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 50.89 + 33.68 - 0.00 = 84.57 kNm/m runRestoring Moment ΣMr = 311.92 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.95 / 2 - ( 311.92 - 84.57 ) / 154.39= 0.002m
Geoguide 1 By Middle-third Rule, B/6 = 2.95 / 6 = 0.492m > 0.002m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(
(( )
))(
( ))
- Annex B2 page 4 -
Project : Design of Gabion Wall Annex B3
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m 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)
650 600 5 kPa600 1100
1000 W1
Pav Pa1 Pa2
1000 2000W2
1000 Insitu soil Pa3 Pa4 Pwh 1000W3
500W4
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
γm 1.2 Base width of wall 2.30 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex B3 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 3.00 = 4.95Pa2 = 0.330 x 2.00 x 21 x 2.00 / 2 = 13.85Pa3 = 0.330 x 2.00 x 21 x 1.00 = 13.85Pa4 = 0.330 x 11 x 1.00 x 1.00 / 2 = 1.81Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 39.46Pah = ΣPai = 34.46
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88Pwv = 10 x 1.00 x 2.30 / 2 = -11.50Pav = 0.00 = 0.00
ΣV= 68.06
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 39.70 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 2.3 - 1.10 / 2 = 1.75 30.03W2 26.52 2.3 - 1.70 / 2 = 1.45 38.45W3 35.88 2.3 - 2.30 / 2 = 1.15 41.26Pwv -11.50 2.3 x 2 / 3 = 1.53 -17.63Pav 0.00 2.3 = 2.30 0.00
ΣM = 92.11 (kNm/m run)ΣMr = 109.75 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 39.70 + 17.63 - 0.00 = 57.34 kNm/m runRestoring Moment ΣMr = 109.75 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.3 / 2 - ( 109.75 - 57.34 ) / 68.06= 0.380m
Arm (m)
Arm (m)
- Annex B3 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 39.46 kN / m
ΣV= 68.06 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 68.06 x 0.58= 39.71 kN/m run > Sliding Force Fa = Σ H= 39.46 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.380 m
Effective Width B' = B - 2 e = 2.30 - 2 x 0.380= 1.54 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.54 x 100 = 154.01 m2
Sliding Force Qs = ΣH x L' = 39.46 x 100 = 3946 kNNormal Force Qn = ΣV x L' = 68.06 x 100 = 6806 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 154.01 = 44.19 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.54 / 100 ) /
( 1 + 1.54 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 3946 / ( 6806 + 4.2 x 154.01 x cot 30.3 )= 0.50
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
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 1.01 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
=> qult > qmax = 44.19 kPa => OK! OK!
- Annex B3 page 4 -
2. Checking of Ultimate Limit State (toe at 0.5m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1
Pav
1000 W2 Pa1 Pa2 2000
1000 W3 Insitu soil Pa3 Pa4 Pwh 1000
500 W4
ToePwv
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
γm 1.2 Base width of wall 2.30 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 3.00 = 4.95Pa2 = 0.330 x 2.00 x 21 x 2.00 / 2 = 13.85Pa3 = 0.330 x 2.00 x 21 x 1.00 = 13.85Pa4 = 0.330 x 11 x 1.00 x 1.00 / 2 = 1.81Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 39.46Pah = ΣPai = 34.46
- Annex B3 page 5 -
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88Pwv = 10 x 1.00 x 2.30 / 2 = -11.50Pav = 0.00 = 0.00
ΣV= 68.06
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 39.70 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 2.50 x 0.10 + 1.750 x 0.995 = 1.99 34.15W2 26.52 1.50 x 0.10 + 1.450 x 0.995 = 1.59 42.22W3 35.88 0.50 x 0.10 + 1.150 x 0.995 = 1.19 42.84Pwv -11.50 2.3 x 2 / 3 x 0.995 = 1.53 -17.55Pav 0.00 2.3 = 2.30 0.00
ΣM = 101.67 (kNm/m run)ΣMr = 119.21 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 39.70 + 17.55 - 0.00 = 57.25 kNm/m runRestoring Moment ΣMr = 119.21 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.3 / 2 - ( 119.21 - 57.25 ) / 68.06= 0.240m
Arm (m)
Arm (m)
((
(( )
)))
- Annex B3 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 = 32.46 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 35.21 kN / m
Resisting Force against Sliding, Fr == 35.21 kN/m run > Activating Force Fa = 32.46 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 = 39.46 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 49.30 kN / m
Resisting Force against Sliding, Fr == 49.30 kN/m run > Activating Force Fa = 39.46 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.240 m
Effective Width B' = B - 2 e = 2.30 - 2 x 0.240= 1.82 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.82 x 100 = 182.09 m2
Sliding Force Qs = ΣH x L' = 39.46 x 100 = 3946 kNNormal Force Qn = ΣV x L' = 68.06 x 100 = 6806 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 182.09 = 37.38 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 30.3= 23.30
6806
- Annex B3 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.82 / 100 ) /
( 1 + 1.82 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 3946 / ( 6806 + 4.2 x 182.09 x cot 30.3 )= 0.49
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
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
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 1.01 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
=> qult > qmax = 37.38 kPa => OK! OK!
- Annex B3 page 8 -
Project : Design of Gabion Wall Annex B4
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m 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)
650 600 5 kPa600 1100
1000 W1
Pav
1000 Pa1 Pa2 2000W2
1000 Insitu soil Pa3 Pa4 1000W3 Pwh
500W4
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
γm 1 Base width of wall 2.30 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 B4 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 3.00 = 4.06Pa2 = 0.271 x 2.00 x 21 x 2.00 / 2 = 11.38Pa3 = 0.271 x 2.00 x 21 x 1.00 = 11.38Pa4 = 0.271 x 11 x 1.00 x 1.00 / 2 = 1.49Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 33.32Pah = ΣPai = 28.32
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88Pwv = 10 x 1.00 x 2.30 / 2 = -11.50Pav = 0.00 = 0.00
ΣV= 68.06
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 32.92 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 2.3 - 1.10 / 2 = 1.75 30.03W2 26.52 2.3 - 1.70 / 2 = 1.45 38.45W3 35.88 2.3 - 2.30 / 2 = 1.15 41.26Pwv -11.50 2.3 x 2 / 3 = 1.53 -17.63Pav 0.00 2.3 = 2.30 0.00
ΣM = 92.11 (kNm/m run)ΣMr = 109.75 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 32.92 + 17.63 - 0.00 = 50.55 kNm/m runRestoring Moment ΣMr = 109.75 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.3 / 2 - ( 109.75 - 50.55 ) / 68.06= 0.280m
Geoguide 1 By Middle-third Rule, B/6 = 2.3 / 6 = 0.383m > 0.280m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex B4 page 2 -
2. Checking of Serviceability Limit State (toe at 0.5m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1
Pav
1000 W2 Pa1 Pa2 2000
1000 W3 Pa3 Pa4 Pwh 1000Insitu soil
500 W4
ToePwv
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
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. 3.00 mTable 8 φ' 35 o Water level (from bottom) 1.00 m
γm 1 Base width of wall 2.3 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 3.00 = 4.06Pa2 = 0.271 x 2.00 x 21 x 2.00 / 2 = 11.38Pa3 = 0.271 x 2.00 x 21 x 1.00 = 11.38Pa4 = 0.271 x 11 x 1.00 x 1.00 / 2 = 1.49Pwh = 10 x 1.00 x 1.00 / 2 = 5.00
ΣΗ= 33.32Pah = ΣPai = 28.32
- Annex B4 page 3 -
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52W3 = 26 x 0.6 x 2.30 x 1.00 = 35.88Pwv = 10 x 1.00 x 2.95 / 2 = -11.50Pav = 0.00 = 0.00
ΣV= 68.06
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 32.92 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 2.50 x 0.10 + 1.750 x 0.995 = 1.99 34.13W2 26.52 1.50 x 0.10 + 1.450 x 0.995 = 1.59 42.20W3 35.88 0.50 x 0.10 + 1.150 x 0.995 = 1.19 42.83Pwv -11.50 2.3 x 2 / 3 x 0.995 = 1.53 -17.55Pav 0.00 2.3 = 2.30 0.00
ΣM = 101.62 (kNm/m run)ΣMr = 119.16 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 32.92 + 17.55 - 0.00 = 50.47 kNm/m runRestoring Moment ΣMr = 119.16 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2.3 / 2 - ( 119.16 - 50.47 ) / 68.06= 0.141m
Geoguide 1 By Middle-third Rule, B/6 = 2.3 / 6 = 0.383m > 0.141m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(
(( )
))()
- Annex B4 page 4 -
Project : Design of Gabion Wall Annex B5
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion WallChecking of Ultimate Limit State (toe at 1.5m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 1.5m above foundation)
650 600 5 kPa600 1100
1000 W1
Pav Pa1 Pa2 13331000
W2 6671000 Insitu soil Pa3 Pa4 Pwh
W3
500W4
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1.2 Base width of wall 1.70 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex B5 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.00 = 3.30Pa2 = 0.330 x 1.33 x 21 x 1.33 / 2 = 6.16Pa3 = 0.330 x 1.33 x 21 x 0.67 = 6.16Pa4 = 0.330 x 11 x 0.67 x 0.67 / 2 = 0.81Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 18.64Pah = ΣPai = 16.42
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52Pwv = 10 x 0.67 x 1.70 / 2 = -5.67Pav = 0.00 = 0.00
ΣV= 38.01
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 12.86 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 1.7 - 1.10 / 2 = 1.15 19.73W2 26.52 1.7 - 1.70 / 2 = 0.85 22.54Pwv -5.67 1.7 x 2 / 3 = 1.13 -6.42Pav 0.00 1.7 = 1.70 0.00
ΣM = 35.85 (kNm/m run)ΣMr = 42.28 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 12.86 + 6.42 - 0.00 = 19.29 kNm/m runRestoring Moment ΣMr = 42.28 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.7 / 2 - ( 42.28 - 19.29 ) / 38.01= 0.245m
Arm (m)
Arm (m)
- Annex B5 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 18.64 kN / m
ΣV= 38.01 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 38.01 x 0.58= 22.18 kN/m run > Sliding Force Fa = Σ H= 18.64 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.245 m
Effective Width B' = B - 2 e = 1.70 - 2 x 0.245= 1.21 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.21 x 100 = 120.96 m2
Sliding Force Qs = ΣH x L' = 18.64 x 100 = 1864 kNNormal Force Qn = ΣV x L' = 38.01 x 100 = 3801 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 120.96 = 31.43 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.21 / 100 ) /
( 1 + 1.21 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1864 / ( 3801 + 4.2 x 120.96 x cot 30.3 )= 0.40
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
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 1.01 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
=> qult > qmax = 31.43 kPa => OK! OK!
- Annex B5 page 4 -
2. Checking of Ultimate Limit State (toe at 1.5m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1
Pav Pa1 Pa2 13331000 W2
6671000 W3 Insitu soil Pa3 Pa4 Pwh
500 W4
ToePwv
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1.2 Base width of wall 1.70 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.00 = 3.30Pa2 = 0.330 x 1.33 x 21 x 1.33 / 2 = 6.16Pa3 = 0.330 x 1.33 x 21 x 0.67 = 6.16Pa4 = 0.330 x 11 x 0.67 x 0.67 / 2 = 0.81Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 18.64Pah = ΣPai = 16.42
- Annex B5 page 5 -
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52Pwv = 10 x 0.67 x 1.70 / 2 = -5.67Pav = 0.00 = 0.00
ΣV= 38.01
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 12.86 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 1.50 x 0.10 + 1.150 x 0.995 = 1.29 22.20W2 26.52 0.50 x 0.10 + 0.850 x 0.995 = 0.90 23.75Pwv -5.67 1.7 x 2 / 3 x 0.995 = 1.13 -6.39Pav 0.00 1.7 = 1.70 0.00
ΣM = 39.56 (kNm/m run)ΣMr = 45.95 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 12.86 + 6.39 - 0.00 = 19.25 kNm/m runRestoring Moment ΣMr = 45.95 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.7 / 2 - ( 45.95 - 19.25 ) / 38.01= 0.148m
Arm (m)
Arm (m)
((
( )
))
- Annex B5 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 = 14.75 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 19.85 kN / m
Resisting Force against Sliding, Fr == 19.85 kN/m run > Activating Force Fa = 14.75 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 = 18.64 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 29.26 kN / m
Resisting Force against Sliding, Fr == 29.26 kN/m run > Activating Force Fa = 18.64 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.148 m
Effective Width B' = B - 2 e = 1.70 - 2 x 0.148= 1.40 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.40 x 100 = 140.44 m2
Sliding Force Qs = ΣH x L' = 18.64 x 100 = 1864 kNNormal Force Qn = ΣV x L' = 38.01 x 100 = 3801 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 140.44 = 27.07 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 30.3= 23.30
3801
- Annex B5 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.40 / 100 ) /
( 1 + 1.40 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1864 / ( 3801 + 4.2 x 140.44 x cot 30.3 )= 0.39
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
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
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 1.01 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
=> qult > qmax = 27.07 kPa => OK! OK!
- Annex B5 page 8 -
Project : Design of Gabion Wall Annex B6
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion WallChecking of Serviceability Limit State (toe at 1.5m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 1.5m above foundation)
650 600 5 kPa600 1100
1000 W1
Pav Pa1 Pa2 13331000
W2 6671000 Insitu soil Pa3 Pa4 Pwh
W3
500W4
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1 Base width of wall 1.70 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 B6 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.00 = 2.71Pa2 = 0.271 x 1.33 x 21 x 1.33 / 2 = 5.06Pa3 = 0.271 x 1.33 x 21 x 0.67 = 5.06Pa4 = 0.271 x 11 x 0.67 x 0.67 / 2 = 0.66Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 15.71Pah = ΣPai = 13.49
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52Pwv = 10 x 0.67 x 1.70 / 2 = -5.67Pav = 0.00 = 0.00
ΣV= 38.01
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 10.66 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 1.7 - 1.10 / 2 = 1.15 19.73W2 26.52 1.7 - 1.70 / 2 = 0.85 22.54Pwv -5.67 1.7 x 2 / 3 = 1.13 -6.42Pav 0.00 1.7 = 1.70 0.00
ΣM = 35.85 (kNm/m run)ΣMr = 42.28 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 10.66 + 6.42 - 0.00 = 17.08 kNm/m runRestoring Moment ΣMr = 42.28 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.7 / 2 - ( 42.28 - 17.08 ) / 38.01= 0.187m
Geoguide 1 By Middle-third Rule, B/6 = 1.7 / 6 = 0.283m > 0.187m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex B6 page 2 -
2. Checking of Serviceability Limit State (toe at 1.5m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1 1333Pav Pa1 Pa2
1000 W2
6671000 W3 Pa3 Pa4 Pwh
Insitu soil500 W4
ToePwv
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1 Base width of wall 1.7 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.00 = 2.71Pa2 = 0.271 x 1.33 x 21 x 1.33 / 2 = 5.06Pa3 = 0.271 x 1.33 x 21 x 0.67 = 5.06Pa4 = 0.271 x 11 x 0.67 x 0.67 / 2 = 0.66Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 15.71Pah = ΣPai = 13.49
- Annex B6 page 3 -
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 1.10 x 1.00 = 17.16W2 = 26 x 0.6 x 1.70 x 1.00 = 26.52Pwv = 10 x 0.67 x 2.95 / 2 = -5.67Pav = 0.00 = 0.00
ΣV= 38.01
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 10.66 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 1.50 x 0.10 + 1.150 x 0.995 = 1.29 22.18W2 26.52 0.50 x 0.10 + 0.850 x 0.995 = 0.90 23.74Pwv -5.67 1.7 x 2 / 3 x 0.995 = 1.13 -6.39Pav 0.00 1.7 = 1.70 0.00
ΣM = 39.54 (kNm/m run)ΣMr = 45.93 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 10.66 + 6.39 - 0.00 = 17.05 kNm/m runRestoring Moment ΣMr = 45.93 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.7 / 2 - ( 45.93 - 17.05 ) / 38.01= 0.090m
Geoguide 1 By Middle-third Rule, B/6 = 1.7 / 6 = 0.283m > 0.090m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
((( )
))
- Annex B6 page 4 -
Project : Design of Gabion Wall Annex B7
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion WallChecking of Ultimate Limit State (toe at 2.5m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 2.5m above foundation)
650 600 5 kPa600 1100
1000 W1 Pa2 667Pav Pa1 333
1000 Pa3 Pa4 Pwh
W2
1000 Insitu soilW3
500W4
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.10 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex B7 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
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 1.10 x 1.00 = 17.16Pwv = 10 x 0.33 x 1.10 / 2 = -1.83Pav = 0.00 = 0.00
ΣV= 15.33
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 2.02 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 1.1 - 1.10 / 2 = 0.55 9.44Pwv -1.83 1.1 x 2 / 3 = 0.73 -1.34Pav 0.00 1.1 = 1.10 0.00
ΣM = 8.09 (kNm/m run)ΣMr = 9.44 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.34 - 0.00 = 3.36 kNm/m runRestoring Moment ΣMr = 9.44 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.1 / 2 - ( 9.44 - 3.36 ) / 15.33= 0.154m
Arm (m)
Arm (m)
- Annex B7 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 5.48 kN / m
ΣV= 15.33 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 15.33 x 0.58= 8.94 kN/m run > Sliding Force Fa = Σ H= 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.154 m
Effective Width B' = B - 2 e = 1.10 - 2 x 0.154= 0.79 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.79 x 100 = 79.25 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 15.33 x 100 = 1533 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 79.25 = 19.34 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.79 / 100 ) /
( 1 + 0.79 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1533 + 4.2 x 79.25 x cot 30.3 )= 0.26
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
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 1.00 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
=> qult > qmax = 19.34 kPa => OK! OK!
- Annex B7 page 4 -
2. Checking of Ultimate Limit State (toe at 2.5m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1 Pa2 667Pav Pa1 333
1000 W2 Pa3 Pa4 Pwh
1000 W3 Insitu soil
500 W4
ToePwv
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.10 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
- Annex B7 page 5 -
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 1.10 x 1.00 = 17.16Pwv = 10 x 0.33 x 1.10 / 2 = -1.83Pav = 0.00 = 0.00
ΣV= 15.33
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 2.02 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 0.50 x 0.10 + 0.550 x 0.995 = 0.60 10.24Pwv -1.83 1.1 x 2 / 3 x 0.995 = 0.73 -1.34Pav 0.00 1.1 = 1.10 0.00
ΣM = 8.91 (kNm/m run)ΣMr = 10.24 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.34 - 0.00 = 3.36 kNm/m runRestoring Moment ΣMr = 10.24 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.1 / 2 - ( 10.24 - 3.36 ) / 15.33= 0.101m
Arm (m)
Arm (m)(( )
)
- Annex B7 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 = 3.92 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 8.12 kN / m
Resisting Force against Sliding, Fr == 8.12 kN/m run > Activating Force Fa = 3.92 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 = 5.48 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 13.53 kN / m
Resisting Force against Sliding, Fr == 13.53 kN/m run > Activating Force Fa = 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.101 m
Effective Width B' = B - 2 e = 1.10 - 2 x 0.101= 0.90 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.90 x 100 = 89.87 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 15.33 x 100 = 1533 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 89.87 = 17.05 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 30.3= 23.30
1533
- Annex B7 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.90 / 100 ) /
( 1 + 0.90 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1533 + 4.2 x 89.87 x cot 30.3 )= 0.25
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
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
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 1.01 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
=> qult > qmax = 17.05 kPa => OK! OK!
- Annex B7 page 8 -
Project : Design of Gabion Wall Annex B8
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 3.5m Gabion WallChecking of Serviceability Limit State (toe at 2.5m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 2.5m above foundation)
650 600 5 kPa600 1100
1000 W1 Pa2 667Pav Pa1 333
1000 Pa3 Pa4 Pwh
W2
1000 Insitu soilW3
500W4
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1.10 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 B8 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
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 1.10 x 1.00 = 17.16Pwv = 10 x 0.33 x 1.10 / 2 = -1.83Pav = 0.00 = 0.00
ΣV= 15.33
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 1.67 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 1.1 - 1.10 / 2 = 0.55 9.44Pwv -1.83 1.1 x 2 / 3 = 0.73 -1.34Pav 0.00 1.1 = 1.10 0.00
ΣM = 8.09 (kNm/m run)ΣMr = 9.44 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.34 - 0.00 = 3.02 kNm/m runRestoring Moment ΣMr = 9.44 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.1 / 2 - ( 9.44 - 3.02 ) / 15.33= 0.131m
Geoguide 1 By Middle-third Rule, B/6 = 1.1 / 6 = 0.183m > 0.131m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex B8 page 2 -
2. Checking of Serviceability Limit State (toe at 2.5m above foundation)(with back batter 1:10)
650 600 5 kPa600 1100
1000 W1 Pa2 667Pav Pa1 333
1000 W2 Pa3 Pa4 Pwh
1000 W3
Insitu soil500 W4
ToePwv
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1.1 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
- Annex B8 page 3 -
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 1.10 x 1.00 = 17.16Pwv = 10 x 0.33 x 2.95 / 2 = -1.83Pav = 0.00 = 0.00
ΣV= 15.33
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 1.67 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 17.16 0.50 x 0.10 + 0.550 x 0.995 = 0.60 10.24Pwv -1.83 1.1 x 2 / 3 x 0.995 = 0.73 -1.34Pav 0.00 1.1 = 1.10 0.00
ΣM = 8.90 (kNm/m run)ΣMr = 10.24 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.34 - 0.00 = 3.01 kNm/m runRestoring Moment ΣMr = 10.24 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.1 / 2 - ( 10.24 - 3.01 ) / 15.33= 0.078m
Geoguide 1 By Middle-third Rule, B/6 = 1.1 / 6 = 0.183m > 0.078m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(( )
)
- Annex B8 page 4 -
Project : Design of 2.5m Gabion Wall Annex C
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m Gabion Wall
Design Statement
Design of 2.5m 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 C 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
(c) Effective friction angle, φ' = 35 o
(II) Loadings
Dead loads
Self weight of the proposed protection wall, earth pressure and hydrostatic pressure are
taken to be dead load for design. The unit weight of water was taken as 9.81 kN/m3.
Imposed 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.
0.0
2.6
0.4
- Annex C 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
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
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!
Refer to Annex C1, C3, 2. Checking Sliding [OK if resisting force > sliding force]& C5
para. 1 Step 7 Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
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
para. 1 Step 8 Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
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
Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
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 -
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 C page 5 -
Project : Design of Gabion Wall Annex C1
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m Gabion WallChecking of Ultimate Limit State (toe at 0m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 0m above foundation)
500 5 kPa500 1000
1000 W1
Pav Pa1 Pa2
1000 1667W2
500 Insitu soil Pa3 Pa4 Pwh 833W3
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
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. 2.50 mTable 8 φ' 35 o Water level (from bottom) 0.83 m
γm 1.2 Base width of wall 2.00 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex C1 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.50 = 4.12Pa2 = 0.330 x 1.67 x 21 x 1.67 / 2 = 9.62Pa3 = 0.330 x 1.67 x 21 x 0.83 = 9.62Pa4 = 0.330 x 11 x 0.83 x 0.83 / 2 = 1.26Pwh = 10 x 0.83 x 0.83 / 2 = 3.47
ΣΗ= 28.09Pah = ΣPai = 24.62
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40W3 = 26 x 0.6 x 2.00 x 0.50 = 62.40Pwv = 10 x 0.83 x 2.00 / 2 = -8.33Pav = 0.00 = 0.00
ΣV= 93.07
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 23.84 (kNm/m run)
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
ΣM = 103.94 (kNm/m run)ΣMr = 115.05 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 23.84 + 11.11 - 0.00 = 34.95 kNm/m runRestoring Moment ΣMr = 115.05 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2 / 2 - ( 115.05 - 34.95 ) / 93.07= 0.139m
Arm (m)
Arm (m)
- Annex C1 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 28.09 kN / m
ΣV= 93.07 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 93.07 x 0.58= 54.30 kN/m run > Sliding Force Fa = Σ H= 28.09 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.139 m
Effective Width B' = B - 2 e = 2.00 - 2 x 0.139= 1.72 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.72 x 100 = 172.14 m2
Sliding Force Qs = ΣH x L' = 28.09 x 100 = 2809 kNNormal Force Qn = ΣV x L' = 93.07 x 100 = 9307 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 172.14 = 54.06 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.72 / 100 ) /
( 1 + 1.72 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 2809 / ( 9307 + 4.2 x 172.14 x cot 30.3 )= 0.27
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
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 1.01 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
=> qult > qmax = 54.06 kPa => OK! OK!
- Annex C1 page 5 -
2. Checking of Ultimate Limit State (toe at 0m above foundation)(with back batter 1:10)
500 5 kPa500 1000
1000 W1
Pav
1000 W2 Pa1 Pa2 1667
500 W3 Insitu soil Pa3 Pa4 Pwh 833
ToePwv
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
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. 2.50 mTable 8 φ' 35 o Water level (from bottom) 0.83 m
γm 1.2 Base width of wall 2.00 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.50 = 4.12Pa2 = 0.330 x 1.67 x 21 x 1.67 / 2 = 9.62Pa3 = 0.330 x 1.67 x 21 x 0.83 = 9.62Pa4 = 0.330 x 11 x 0.83 x 0.83 / 2 = 1.26Pwh = 10 x 0.83 x 0.83 / 2 = 3.47
ΣΗ= 28.09Pah = ΣPai = 24.62
- Annex C1 page 6 -
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40W3 = 26 x 0.6 x 2.00 x 0.50 = 15.60Pwv = 10 x 0.83 x 2.00 / 2 = -8.33Pav = 0.00 = 0.00
ΣV= 46.27
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 23.84 (kNm/m run)
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
ΣM = 62.68 (kNm/m run)ΣMr = 73.73 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 23.84 + 11.06 - 0.00 = 34.89 kNm/m runRestoring Moment ΣMr = 73.73 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2 / 2 - ( 73.73 - 34.89 ) / 46.27= 0.161m
Arm (m)
Arm (m)
((
(( )
)))
- Annex C1 page 7 -
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 = 23.33 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 23.86 kN / m
Resisting Force against Sliding, Fr == 23.86 kN/m run > Activating Force Fa = 23.33 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 = 28.09 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 35.33 kN / m
Resisting Force against Sliding, Fr == 35.33 kN/m run > Activating Force Fa = 28.09 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.161 m
Effective Width B' = B - 2 e = 2.00 - 2 x 0.161= 1.68 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.68 x 100 = 167.90 m2
Sliding Force Qs = ΣH x L' = 28.09 x 100 = 2809 kNNormal Force Qn = ΣV x L' = 46.27 x 100 = 4627 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 167.90 = 27.56 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 30.3= 23.30
4627
- Annex C1 page 8 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.68 / 100 ) /
( 1 + 1.68 / 100 )= 1.98
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 2809 / ( 4627 + 4.2 x 167.90 x cot 30.3 )= 0.48
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
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
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 1.01 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
=> qult > qmax = 27.56 kPa => OK! OK!
- Annex C1 page 9 -
Project : Design of Gabion Wall Annex C2
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m Gabion WallChecking of Serviceability Limit State (toe at 0m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 0m above foundation)
500 5 kPa500 1000
1000 W1
Pav
1000 Pa1 Pa2 1667W2
500 Insitu soil Pa3 Pa4 833W3 Pwh
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
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. 2.50 mTable 8 φ' 35 o Water level (from bottom) 0.83 m
γm 1 Base width of wall 2.00 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 C2 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.50 = 3.39Pa2 = 0.271 x 1.67 x 21 x 1.67 / 2 = 7.90Pa3 = 0.271 x 1.67 x 21 x 0.83 = 7.90Pa4 = 0.271 x 11 x 0.83 x 0.83 / 2 = 1.04Pwh = 10 x 0.83 x 0.83 / 2 = 3.47
ΣΗ= 23.70Pah = ΣPai = 20.23
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40W3 = 26 x 0.6 x 2.00 x 0.50 = 62.40Pwv = 10 x 0.83 x 2.00 / 2 = -8.33Pav = 0.00 = 0.00
ΣV= 93.07
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 19.76 (kNm/m run)
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
ΣM = 103.94 (kNm/m run)ΣMr = 115.05 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 19.76 + 11.11 - 0.00 = 30.87 kNm/m runRestoring Moment ΣMr = 115.05 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 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 -
2. Checking of Serviceability Limit State (toe at 0m above foundation)(with back batter 1:10)
500 5 kPa500 1000
1000 W1
Pav
1000 W2 Pa1 Pa2 1667
500 W3 Pa3 Pa4 Pwh 833Insitu soil
ToePwv
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
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. 2.50 mTable 8 φ' 35 o Water level (from bottom) 0.83 m
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
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.50 = 3.39Pa2 = 0.271 x 1.67 x 21 x 1.67 / 2 = 7.90Pa3 = 0.271 x 1.67 x 21 x 0.83 = 7.90Pa4 = 0.271 x 11 x 0.83 x 0.83 / 2 = 1.04Pwh = 10 x 0.83 x 0.83 / 2 = 3.47
ΣΗ= 23.70Pah = ΣPai = 20.23
- Annex C2 page 3 -
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40W3 = 26 x 0.6 x 2.00 x 0.50 = 62.40Pwv = 10 x 0.83 x 2.00 / 2 = -8.33Pav = 0.00 = 0.00
ΣV= 93.07
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 19.76 (kNm/m run)
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
ΣM = 110.37 (kNm/m run)ΣMr = 121.43 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 19.76 + 11.06 - 0.00 = 30.81 kNm/m runRestoring Moment ΣMr = 121.43 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 2 / 2 - ( 121.43 - 30.81 ) / 93.07= 0.026m
Geoguide 1 By Middle-third Rule, B/6 = 2 / 6 = 0.333m > 0.026m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(
(( )
))()
- Annex C2 page 4 -
Project : Design of Gabion Wall Annex C3
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m 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)
500 5 kPa500 1000
1000 W1
Pav Pa1 Pa2 13331000
W2 667500 Insitu soil Pa3 Pa4 Pwh
W3
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1.2 Base width of wall 1.50 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex C3 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.00 = 3.30Pa2 = 0.330 x 1.33 x 21 x 1.33 / 2 = 6.16Pa3 = 0.330 x 1.33 x 21 x 0.67 = 6.16Pa4 = 0.330 x 11 x 0.67 x 0.67 / 2 = 0.81Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 18.64Pah = ΣPai = 16.42
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40Pwv = 10 x 0.67 x 1.50 / 2 = -5.00Pav = 0.00 = 0.00
ΣV= 34.00
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 12.86 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.5 - 1.00 / 2 = 1.00 15.60W2 23.40 1.5 - 1.50 / 2 = 0.75 17.55Pwv -5.00 1.5 x 2 / 3 = 1.00 -5.00Pav 0.00 1.5 = 1.50 0.00
ΣM = 28.15 (kNm/m run)ΣMr = 33.15 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 12.86 + 5.00 - 0.00 = 17.86 kNm/m runRestoring Moment ΣMr = 33.15 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.5 / 2 - ( 33.15 - 17.86 ) / 34.00= 0.300m
Arm (m)
Arm (m)
- Annex C3 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 18.64 kN / m
ΣV= 34.00 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 34.00 x 0.58= 19.84 kN/m run > Sliding Force Fa = Σ H= 18.64 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.300 m
Effective Width B' = B - 2 e = 1.50 - 2 x 0.300= 0.90 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.90 x 100 = 89.92 m2
Sliding Force Qs = ΣH x L' = 18.64 x 100 = 1864 kNNormal Force Qn = ΣV x L' = 34.00 x 100 = 3400 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 89.92 = 37.81 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.90 / 100 ) /
( 1 + 0.90 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1864 / ( 3400 + 4.2 x 89.92 x cot 30.3 )= 0.46
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
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 1.01 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
=> qult > qmax = 37.81 kPa => OK! OK!
- Annex C3 page 5 -
2. Checking of Ultimate Limit State (toe at 0.5m above foundation)(with back batter 1:10)
500 5 kPa500 1000
1000 W1
Pav Pa1 Pa2 13331000 W2
667500 W3 Insitu soil Pa3 Pa4 Pwh
ToePwv
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1.2 Base width of wall 1.50 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 2.00 = 3.30Pa2 = 0.330 x 1.33 x 21 x 1.33 / 2 = 6.16Pa3 = 0.330 x 1.33 x 21 x 0.67 = 6.16Pa4 = 0.330 x 11 x 0.67 x 0.67 / 2 = 0.81Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 18.64Pah = ΣPai = 16.42
- Annex C3 page 6 -
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40Pwv = 10 x 0.67 x 1.50 / 2 = -5.00Pav = 0.00 = 0.00
ΣV= 34.00
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 12.86 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.50 x 0.10 + 1.000 x 0.995 = 1.14 17.85W2 23.40 0.50 x 0.10 + 0.750 x 0.995 = 0.80 18.63Pwv -5.00 1.5 x 2 / 3 x 0.995 = 1.00 -4.98Pav 0.00 1.5 = 1.50 0.00
ΣM = 31.50 (kNm/m run)ΣMr = 36.48 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 12.86 + 4.98 - 0.00 = 17.84 kNm/m runRestoring Moment ΣMr = 36.48 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.5 / 2 - ( 36.48 - 17.84 ) / 34.00= 0.202m
Arm (m)
Arm (m)
((
( )
))
- Annex C3 page 7 -
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 = 15.15 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 17.65 kN / m
Resisting Force against Sliding, Fr == 17.65 kN/m run > Activating Force Fa = 15.15 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 = 18.64 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 26.09 kN / m
Resisting Force against Sliding, Fr == 26.09 kN/m run > Activating Force Fa = 18.64 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.202 m
Effective Width B' = B - 2 e = 1.50 - 2 x 0.202= 1.10 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.10 x 100 = 109.64 m2
Sliding Force Qs = ΣH x L' = 18.64 x 100 = 1864 kNNormal Force Qn = ΣV x L' = 34.00 x 100 = 3400 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 109.64 = 31.01 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 30.3= 23.30
3400
- Annex C3 page 8 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.10 / 100 ) /
( 1 + 1.10 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1864 / ( 3400 + 4.2 x 109.64 x cot 30.3 )= 0.45
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
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
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 1.01 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
=> qult > qmax = 31.01 kPa => OK! OK!
- Annex C3 page 9 -
Project : Design of Gabion Wall Annex C4
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m 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)
500 5 kPa500 1000
1000 W1
Pav Pa1 Pa2 13331000
W2 667500 Insitu soil Pa3 Pa4 Pwh
W3
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1 Base width of wall 1.50 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 C4 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.00 = 2.71Pa2 = 0.271 x 1.33 x 21 x 1.33 / 2 = 5.06Pa3 = 0.271 x 1.33 x 21 x 0.67 = 5.06Pa4 = 0.271 x 11 x 0.67 x 0.67 / 2 = 0.66Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 15.71Pah = ΣPai = 13.49
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40Pwv = 10 x 0.67 x 1.50 / 2 = -5.00Pav = 0.00 = 0.00
ΣV= 34.00
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 10.66 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.5 - 1.00 / 2 = 1.00 15.60W2 23.40 1.5 - 1.50 / 2 = 0.75 17.55Pwv -5.00 1.5 x 2 / 3 = 1.00 -5.00Pav 0.00 1.5 = 1.50 0.00
ΣM = 28.15 (kNm/m run)ΣMr = 33.15 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 10.66 + 5.00 - 0.00 = 15.66 kNm/m runRestoring Moment ΣMr = 33.15 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.5 / 2 - ( 33.15 - 15.66 ) / 34.00= 0.236m
Geoguide 1 By Middle-third Rule, B/6 = 1.5 / 6 = 0.250m > 0.236m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex C4 page 2 -
2. Checking of Serviceability Limit State (toe at 0.5m above foundation)(with back batter 1:10)
500 5 kPa500 1000
1000 W1 1333Pav Pa1 Pa2
1000 W2
667500 W3 Pa3 Pa4 Pwh
Insitu soil
ToePwv
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
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. 2.00 mTable 8 φ' 35 o Water level (from bottom) 0.67 m
γm 1 Base width of wall 1.5 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 2.00 = 2.71Pa2 = 0.271 x 1.33 x 21 x 1.33 / 2 = 5.06Pa3 = 0.271 x 1.33 x 21 x 0.67 = 5.06Pa4 = 0.271 x 11 x 0.67 x 0.67 / 2 = 0.66Pwh = 10 x 0.67 x 0.67 / 2 = 2.22
ΣΗ= 15.71Pah = ΣPai = 13.49
- Annex C4 page 3 -
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.50 x 1.00 = 23.40Pwv = 10 x 0.67 x 2.00 / 2 = -5.00Pav = 0.00 = 0.00
ΣV= 34.00
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 10.66 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.50 x 0.10 + 1.000 x 0.995 = 1.14 17.84W2 23.40 0.50 x 0.10 + 0.750 x 0.995 = 0.80 18.62Pwv -5.00 1.5 x 2 / 3 x 0.995 = 1.00 -4.98Pav 0.00 1.5 = 1.50 0.00
ΣM = 31.49 (kNm/m run)ΣMr = 36.46 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 10.66 + 4.98 - 0.00 = 15.63 kNm/m runRestoring Moment ΣMr = 36.46 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.5 / 2 - ( 36.46 - 15.63 ) / 34.00= 0.137m
Geoguide 1 By Middle-third Rule, B/6 = 1.5 / 6 = 0.250m > 0.137m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
((( )
))
- Annex C4 page 4 -
Project : Design of Gabion Wall Annex C5
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m Gabion WallChecking of Ultimate Limit State (toe at 1.5m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 1.5m above foundation)
500 5 kPa500 1000
1000 W1 Pa2 667Pav Pa1 333
1000 Pa3 Pa4 Pwh
W2
500 Insitu soilW3
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.00 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex C5 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 2.02 (kNm/m run)
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
ΣM = 6.69 (kNm/m run)ΣMr = 7.80 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.11 - 0.00 = 3.13 kNm/m runRestoring Moment ΣMr = 7.80 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 7.80 - 3.13 ) / 13.93= 0.165m
Arm (m)
Arm (m)
- Annex C5 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 5.48 kN / m
ΣV= 13.93 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 13.93 x 0.58= 8.13 kN/m run > Sliding Force Fa = Σ H= 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.165 m
Effective Width B' = B - 2 e = 1.00 - 2 x 0.165= 0.67 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.67 x 100 = 67.01 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 13.93 x 100 = 1393 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 67.01 = 20.79 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.67 / 100 ) /
( 1 + 0.67 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1393 + 4.2 x 67.01 x cot 30.3 )= 0.29
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
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 1.00 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
=> qult > qmax = 20.79 kPa => OK! OK!
- Annex C5 page 5 -
2. Checking of Ultimate Limit State (toe at 1.5m above foundation)(with back batter 1:10)
500 5 kPa500 1000
1000 W1 Pa2 667Pav Pa1 333
1000 W2 Pa3 Pa4 Pwh
500 W3 Insitu soil
ToePwv
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.00 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
- Annex C5 page 6 -
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 2.02 (kNm/m run)
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
ΣM = 7.43 (kNm/m run)ΣMr = 8.54 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.11 - 0.00 = 3.13 kNm/m runRestoring Moment ΣMr = 8.54 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 8.54 - 3.13 ) / 13.93= 0.112m
Arm (m)
Arm (m)(( )
)
- Annex C5 page 7 -
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 = 4.06 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 7.35 kN / m
Resisting Force against Sliding, Fr == 7.35 kN/m run > Activating Force Fa = 4.06 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 = 5.48 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 12.30 kN / m
Resisting Force against Sliding, Fr == 12.30 kN/m run > Activating Force Fa = 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.112 m
Effective Width B' = B - 2 e = 1.00 - 2 x 0.112= 0.78 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.78 x 100 = 77.68 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 13.93 x 100 = 1393 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 77.68 = 17.94 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 30.3= 23.30
1393
- Annex C5 page 8 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.78 / 100 ) /
( 1 + 0.78 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1393 + 4.2 x 77.68 x cot 30.3 )= 0.28
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
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
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 1.00 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
=> qult > qmax = 17.94 kPa => OK! OK!
- Annex C5 page 9 -
Project : Design of Gabion Wall Annex C6
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 2.5m Gabion WallChecking of Serviceability Limit State (toe at 1.5m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 1.5m above foundation)
500 5 kPa500 1000
1000 W1 Pa2 667Pav Pa1 333
1000 Pa3 Pa4 Pwh
W2
500 Insitu soilW3
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1.00 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 C6 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 1.67 (kNm/m run)
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
ΣM = 6.69 (kNm/m run)ΣMr = 7.80 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.11 - 0.00 = 2.78 kNm/m runRestoring Moment ΣMr = 7.80 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 7.80 - 2.78 ) / 13.93= 0.140m
Geoguide 1 By Middle-third Rule, B/6 = 1 / 6 = 0.167m > 0.140m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex C6 page 2 -
2. Checking of Serviceability Limit State (toe at 1.5m above foundation)(with back batter 1:10)
500 5 kPa500 1000
1000 W1 Pa2 667Pav Pa1 333
1000 W2 Pa3 Pa4 Pwh
500 W3
Insitu soil
ToePwv
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
- Annex C6 page 3 -
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 2.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 1.67 (kNm/m run)
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 8.53Pwv -1.67 1 x 2 / 3 x 0.995 = 0.66 -1.11Pav 0.00 1 = 1.00 0.00
ΣM = 7.43 (kNm/m run)ΣMr = 8.53 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.11 - 0.00 = 2.78 kNm/m runRestoring Moment ΣMr = 8.53 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 8.53 - 2.78 ) / 13.93= 0.087m
Geoguide 1 By Middle-third Rule, B/6 = 1 / 6 = 0.167m > 0.087m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(( )
)
- Annex C6 page 4 -
Project : Design of 1.5m Gabion Wall Annex D
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 1.5m Gabion Wall
Design Statement
Design of 1.5m 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 D 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
(c) Effective friction angle, φ' = 35 o
(II) Loadings
Dead loads
Self weight of the proposed protection wall, earth pressure and hydrostatic pressure are
taken to be dead load for design. The unit weight of water was taken as 9.81 kN/m3.
Imposed 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.
0.0
2.6
0.4
- Annex D 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
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
Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
m OK! OK!
m OK! OK!
Refer to Annex D1 & D3 2. Checking Sliding [OK if resisting force > sliding force]para. 1 Step 7
Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
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
Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
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]
Height of Toe Stability Stability
above foundation (without back batter) (with back batter)
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 -
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 D page 5 -
Project : Design of Gabion Wall Annex D1
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 1.5m Gabion WallChecking of Ultimate Limit State (toe at 0m above foundation)
Reference Remarks
1. Checking of Ultimate Limit State (toe at 0m above foundation)
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
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. 1.50 mTable 8 φ' 35 o Water level (from bottom) 0.50 m
γm 1.2 Base width of wall 1.30 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex D1 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.50 = 2.47Pa2 = 0.330 x 1.00 x 21 x 1.00 / 2 = 3.46Pa3 = 0.330 x 1.00 x 21 x 0.50 = 3.46Pa4 = 0.330 x 11 x 0.50 x 0.50 / 2 = 0.45Pwh = 10 x 0.50 x 0.50 / 2 = 1.25
ΣΗ= 11.10Pah = ΣPai = 9.85
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.30 x 0.50 = 40.56Pwv = 10 x 0.50 x 1.30 / 2 = -3.25Pav = 0.00 = 0.00
ΣV= 52.91
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 5.89 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.3 - 1.00 / 2 = 0.80 12.48W2 40.56 1.3 - 1.30 / 2 = 0.65 26.36Pwv -3.25 1.3 x 2 / 3 = 0.87 -2.82Pav 0.00 1.3 = 1.30 0.00
ΣM = 36.03 (kNm/m run)ΣMr = 38.84 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 5.89 + 2.82 - 0.00 = 8.71 kNm/m runRestoring Moment ΣMr = 38.84 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 38.84 - 8.71 ) / 52.91= 0.080m
Arm (m)
Arm (m)
- Annex D1 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 11.10 kN / m
ΣV= 52.91 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 52.91 x 0.58= 30.87 kN/m run > Sliding Force Fa = Σ H= 11.10 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.080 m
Effective Width B' = B - 2 e = 1.30 - 2 x 0.080= 1.14 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.14 x 100 = 113.92 m2
Sliding Force Qs = ΣH x L' = 11.10 x 100 = 1110 kNNormal Force Qn = ΣV x L' = 52.91 x 100 = 5291 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 113.92 = 46.45 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.14 / 100 ) /
( 1 + 1.14 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1110 / ( 5291 + 4.2 x 113.92 x cot 30.3 )= 0.18
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
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 1.01 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
=> qult > qmax = 46.45 kPa => OK! OK!
- Annex D1 page 4 -
2. Checking of Ultimate Limit State (toe at 0m above foundation)(with back batter 1:10)
300 5 kPa0 1000
1000 W1
Pav Pa1 Pa2 1000500 W2
5000 W3 Insitu soil Pa3 Pa4 Pwh
ToePwv
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
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. 1.50 mTable 8 φ' 35 o Water level (from bottom) 0.50 m
γm 1.2 Base width of wall 1.30 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.50 = 2.47Pa2 = 0.330 x 1.00 x 21 x 1.00 / 2 = 3.46Pa3 = 0.330 x 1.00 x 21 x 0.50 = 3.46Pa4 = 0.330 x 11 x 0.50 x 0.50 / 2 = 0.45Pwh = 10 x 0.50 x 0.50 / 2 = 1.25
ΣΗ= 11.10Pah = ΣPai = 9.85
- Annex D1 page 5 -
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.30 x 0.50 = 10.14Pwv = 10 x 0.50 x 1.30 / 2 = -3.25Pav = 0.00 = 0.00
ΣV= 22.49
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 5.89 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.00 x 0.10 + 0.800 x 0.995 = 0.90 13.97W2 10.14 0.25 x 0.10 + 0.650 x 0.995 = 0.67 6.81Pwv -3.25 1.3 x 2 / 3 x 0.995 = 0.86 -2.80Pav 0.00 1.3 = 1.30 0.00
ΣM = 17.98 (kNm/m run)ΣMr = 20.78 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 5.89 + 2.80 - 0.00 = 8.69 kNm/m runRestoring Moment ΣMr = 20.78 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 20.78 - 8.69 ) / 22.49= 0.113m
Arm (m)
Arm (m)
((
( )
))
- Annex D1 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 = 8.80 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 11.74 kN / m
Resisting Force against Sliding, Fr == 11.74 kN/m run > Activating Force Fa = 8.80 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 = 11.10 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 18.54 kN / m
Resisting Force against Sliding, Fr == 18.54 kN/m run > Activating Force Fa = 11.10 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.113 m
Effective Width B' = B - 2 e = 1.30 - 2 x 0.113= 1.07 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 1.07 x 100 = 107.49 m2
Sliding Force Qs = ΣH x L' = 11.10 x 100 = 1110 kNNormal Force Qn = ΣV x L' = 22.49 x 100 = 2249 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 107.49 = 20.92 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 30.3= 23.30
2249
- Annex D1 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 1.07 / 100 ) /
( 1 + 1.07 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 1110 / ( 2249 + 4.2 x 107.49 x cot 30.3 )= 0.37
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
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
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 1.01 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
=> qult > qmax = 20.92 kPa => OK! OK!
- Annex D1 page 8 -
Project : Design of Gabion Wall Annex D2
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 1.5m Gabion WallChecking of Serviceability Limit State (toe at 0m above foundation)
Reference Remarks
1. Checking of Serviceability Limit State (toe at 0m above foundation)
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
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. 1.50 mTable 8 φ' 35 o Water level (from bottom) 0.50 m
γm 1 Base width of wall 1.30 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 D2 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.50 = 2.03Pa2 = 0.271 x 1.00 x 21 x 1.00 / 2 = 2.85Pa3 = 0.271 x 1.00 x 21 x 0.50 = 2.85Pa4 = 0.271 x 11 x 0.50 x 0.50 / 2 = 0.37Pwh = 10 x 0.50 x 0.50 / 2 = 1.25
ΣΗ= 9.35Pah = ΣPai = 8.10
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.30 x 0.50 = 40.56Pwv = 10 x 0.50 x 1.30 / 2 = -3.25Pav = 0.00 = 0.00
ΣV= 52.91
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 4.88 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.3 - 1.00 / 2 = 0.80 12.48W2 40.56 1.3 - 1.30 / 2 = 0.65 26.36Pwv -3.25 1.3 x 2 / 3 = 0.87 -2.82Pav 0.00 1.3 = 1.30 0.00
ΣM = 36.03 (kNm/m run)ΣMr = 38.84 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 4.88 + 2.82 - 0.00 = 7.69 kNm/m runRestoring Moment ΣMr = 38.84 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 38.84 - 7.69 ) / 52.91= 0.061m
Geoguide 1 By Middle-third Rule, B/6 = 1.3 / 6 = 0.217m > 0.061m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex D2 page 2 -
2. Checking of Serviceability Limit State (toe at 0m above foundation)(with back batter 1:10)
300 5 kPa0 1000
1000 W1 1000Pav Pa1 Pa2
500 W2
5000 W3 Pa3 Pa4 Pwh
Insitu soil
ToePwv
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
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. 1.50 mTable 8 φ' 35 o Water level (from bottom) 0.50 m
γm 1 Base width of wall 1.3 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.50 = 2.03Pa2 = 0.271 x 1.00 x 21 x 1.00 / 2 = 2.85Pa3 = 0.271 x 1.00 x 21 x 0.50 = 2.85Pa4 = 0.271 x 11 x 0.50 x 0.50 / 2 = 0.37Pwh = 10 x 0.50 x 0.50 / 2 = 1.25
ΣΗ= 9.35Pah = ΣPai = 8.10
- Annex D2 page 3 -
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 1.00 x 1.00 = 15.60W2 = 26 x 0.6 x 1.30 x 0.50 = 40.56Pwv = 10 x 0.50 x 1.30 / 2 = -3.25Pav = 0.00 = 0.00
ΣV= 52.91
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 4.88 (kNm/m run)
Step 5 Moment of Vertical Force about Toe Force (kN/ m) MomentW1 15.60 1.00 x 0.10 + 0.800 x 0.995 = 0.90 13.96W2 40.56 0.25 x 0.10 + 0.650 x 0.995 = 0.67 27.24Pwv -3.25 1.3 x 2 / 3 x 0.995 = 0.86 -2.80Pav 0.00 1.3 = 1.30 0.00
ΣM = 38.40 (kNm/m run)ΣMr = 41.20 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 4.88 + 2.80 - 0.00 = 7.68 kNm/m runRestoring Moment ΣMr = 41.20 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1.3 / 2 - ( 41.20 - 7.68 ) / 52.91= 0.016m
Geoguide 1 By Middle-third Rule, B/6 = 1.3 / 6 = 0.217m > 0.016m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
((( )
))
- Annex D2 page 4 -
Project : Design of Gabion Wall Annex D3
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 1.5m 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)
300 5 kPa1000
1000 W1 Pa2 667Pav Pa1 333
500 Pa3 Pa4 Pwh
W2
Insitu soil
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.00 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.)
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 δb/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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
- Annex D3 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 2.02 (kNm/m run)
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
ΣM = 6.69 (kNm/m run)ΣMr = 7.80 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.11 - 0.00 = 3.13 kNm/m runRestoring Moment ΣMr = 7.80 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 7.80 - 3.13 ) / 13.93= 0.165m
Arm (m)
Arm (m)
- Annex D3 page 2 -
Step 7 Check SlidingSliding Force Fa = ΣH = 5.48 kN / m
ΣV= 13.93 kN / m
Resisting Force against Sliding Fr == Ns x tan φ'f= 13.93 x 0.58= 8.13 kN/m run > Sliding Force Fa = Σ H= 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.165 m
Effective Width B' = B - 2 e = 1.00 - 2 x 0.165= 0.67 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.67 x 100 = 67.01 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 13.93 x 100 = 1393 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 67.01 = 20.79 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 30.3= 23.30
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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 -
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.67 / 100 ) /
( 1 + 0.67 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1393 + 4.2 x 67.01 x cot 30.3 )= 0.29
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
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 1.00 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
=> qult > qmax = 20.79 kPa => OK! OK!
- Annex D3 page 4 -
2. Checking of Ultimate Limit State (toe at 0.5m above foundation)(with back batter 1:10)
300 5 kPa0 1000
1000 W1 Pa2 667Pav Pa1 333
500 W2 Pa3 Pa4 Pwh
0 W3 Insitu soil
ToePwv
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1.2 Base width of wall 1.00 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.)
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: γ m is taken to be 1.2 so as to cater for the unknown ground condition.)
Step 1 Sliding Force (kN/m - run)Pa1 = 0.330 x 5.00 x 1.00 = 1.65Pa2 = 0.330 x 0.67 x 21 x 0.67 / 2 = 1.54Pa3 = 0.330 x 0.67 x 21 x 0.33 = 1.54Pa4 = 0.330 x 11 x 0.33 x 0.33 / 2 = 0.20Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 5.48Pah = ΣPai = 4.93
- Annex D3 page 5 -
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 2.02 (kNm/m run)
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
ΣM = 7.43 (kNm/m run)ΣMr = 8.54 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 2.02 + 1.11 - 0.00 = 3.13 kNm/m runRestoring Moment ΣMr = 8.54 kNm/m run
=> ΣMr > ΣMo => OK!
OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 8.54 - 3.13 ) / 13.93= 0.112m
Arm (m)
Arm (m)(( )
)
- Annex D3 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 = 4.06 kN / m
Resisting force Fr = (ΣV x cos θ - ΣH x sin θ) x tan θb = 7.35 kN / m
Resisting Force against Sliding, Fr == 7.35 kN/m run > Activating Force Fa = 4.06 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 = 5.48 kN / mResisting force Fr = (ΣV x cos Ω - ΣH x sin Ω) x tan φ' + c' l = 12.30 kN / m
Resisting Force against Sliding, Fr == 12.30 kN/m run > Activating Force Fa = 5.48 kN / m
=> OK!
OK!
Step 8 Check Bearing CapacityFrom above, Eccentricity e = 0.112 m
Effective Width B' = B - 2 e = 1.00 - 2 x 0.112= 0.78 m
Effective Length L' = L' = 100 mEffective Area A' = B' x L'
= 0.78 x 100 = 77.68 m2
Sliding Force Qs = ΣH x L' = 5.48 x 100 = 548 kNNormal Force Qn = ΣV x L' = 13.93 x 100 = 1393 kN
Thus, at the level just below 500mm rock fill,Bearing Pressure qmax = Qn / A'
= / 77.68 = 17.94 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 30.3= 23.30
1393
- Annex D3 page 7 -
For Shape Factors,sc = 1 + Nq / Nc x B' / L'
= 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
For Inclination Factors,mi = ( 2 + B' / L' ) /
( 1 + B' / L' )= ( 2 + 0.78 / 100 ) /
( 1 + 0.78 / 100 )= 1.99
Ki = Qs / ( Qn + cf ' x A' x cot φ'f )= 548 / ( 1393 + 4.2 x 77.68 x cot 30.3 )= 0.28
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
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
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 1.00 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
=> qult > qmax = 17.94 kPa => OK! OK!
- Annex D3 page 8 -
Project : Design of Gabion Wall Annex D4
Prepared by : NG Chun-ling (AE/TM5)
Checked by :
Subject : Design of 1.5m 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)
300 5 kPa1000
1000 W1 Pa2 667Pav Pa1 333
500 Pa3 Pa4 Pwh
W2
Insitu soil
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1.00 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271
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 (= 0.85φcv' )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 D4 page 1 -
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.00 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about ToeForce (kN/ m) Moment
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
ΣM = 1.67 (kNm/m run)
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
ΣM = 6.69 (kNm/m run)ΣMr = 7.80 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.11 - 0.00 = 2.78 kNm/m runRestoring Moment ΣMr = 7.80 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 7.80 - 2.78 ) / 13.93= 0.140m
Geoguide 1 By Middle-third Rule, B/6 = 1 / 6 = 0.167m > 0.140m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
- Annex D4 page 2 -
2. Checking of Serviceability Limit State (toe at 0.5m above foundation)(with back batter 1:10)
300 5 kPa0 1000
1000 W1 Pa2 667Pav Pa1 333
500 W2 Pa3 Pa4 Pwh
0 W3
Insitu soil
ToePwv
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
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. 1.00 mTable 8 φ' 35 o Water level (from bottom) 0.33 m
γm 1 Base width of wall 1 mφ'f 35.0 o ( = tan-1(tan φ'/γm ))φcv' 30 o
Kah 0.271 Back batter, θ = 1 : 10 = 0.10
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 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
Step 1 Sliding Force (kN/m - run)Pa1 = 0.271 x 5.00 x 1.00 = 1.35Pa2 = 0.271 x 0.67 x 21 x 0.67 / 2 = 1.26Pa3 = 0.271 x 0.67 x 21 x 0.33 = 1.26Pa4 = 0.271 x 11 x 0.33 x 0.33 / 2 = 0.17Pwh = 10 x 0.33 x 0.33 / 2 = 0.56
ΣΗ= 4.61Pah = ΣPai = 4.05
- Annex D4 page 3 -
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 1.00 x 1.00 = 15.60Pwv = 10 x 0.33 x 1.30 / 2 = -1.67Pav = 0.00 = 0.00
ΣV= 13.93
Step 4 Overturning moment of earth pressure about Toe
cos θ = 0.995 tan θ = 0.100
Force (kN/ m) MomentPa1 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
ΣM = 1.67 (kNm/m run)
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 8.53Pwv -1.67 1 x 2 / 3 x 0.995 = 0.66 -1.11Pav 0.00 1 = 1.00 0.00
ΣM = 7.43 (kNm/m run)ΣMr = 8.53 (kNm/m run)
Step 6 Check Overturning and Determine EccentricityOverturning Moment ΣMo = 1.67 + 1.11 - 0.00 = 2.78 kNm/m runRestoring Moment ΣMr = 8.53 kNm/m run
=> ΣMr > ΣMo => OK!Eccentricity e = B / 2 - ( ΣMr - ΣMo ) / ΣV
= 1 / 2 - ( 8.53 - 2.78 ) / 13.93= 0.087m
Geoguide 1 By Middle-third Rule, B/6 = 1 / 6 = 0.167m > 0.087m OK! OK!Clause 9.2.4
Arm (m)
Arm (m)
(( )
)
- Annex D4 page 4 -
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 2-
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 3-
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 4-
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 5-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 6-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 7-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 8-
(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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 9-
(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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 10-
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;
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 11-
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
Date: November 2006 Page 12-
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. A - 1
Drawing Notes: 1. All dimensions are in millimeters unless otherwise specified. 2. Depending on the soil conditions, designers should determine whether any
ground treatment for foundation is required in consideration with sliding, bearing or soil slip failures.
3. Determination of Potential Scour Depth by Froehlich's Live-Bed Scour Equation
where: K1 = Coefficient for shape
Shape Coefficients
Description K1
Vertical-wall 1.00
Vertical-wall with wing walls 0.82
Spill-through 0.55
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. A - 2
Example: Assume K1 = 0.82, K2 = ( 90 / 90) 0.13 = 1 L´ and ya are the base width and retaining height of the gabion wall as shown in the sketches in the calculations in Appendix B.
Computed Scour Depth, ys as follow:
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
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. A - 4
RD 1045 Design of Gabion Wall Used in River Embankment Mainland North Division, DSD
Appendix C
Sample Particular Specifications,
Method of Measurement
and Schedule of Rates
for Gabion Wall & Reno Mattress
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. C - 1
Particular Specifications Gabion Wall & Reno Mattress 7.45 Submissions of Gabion Wall & Reno Mattress
(1) The following particulars of the proposed material together with the method of construction shall be submitted to the Engineer for approval at least 14 days before first delivery of gabion mesh and Reno Mattress mesh to site:
(a) manufacturer’s literature for the gabion mesh and Reno Mattress mesh, including details of:
- types of material employed; and - recommendations of handling, storage, placing, jointing, fixing and
infilling with rocks.
(b) a certificate for the material showing the manufacturer’s name, the date and place of manufacture and details showing that the material complies with the requirements stated in the Contract, including results of tests required in accordance with the Contract.
7.46 Gabion mesh
(1) The mesh used in fabricating gabion shall be of a proprietary type approved by the Engineer.
(2) Mesh shall be hexagonal double twist and shall not ravel if damaged. The
dimensions of the hexagon shall be 80 x 100 mm. (3) 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). The PVC coating shall be dark green in colour, has an average thickness of 0.5 mm and nowhere less than 0.4 mm. 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. All wires shall be mild steel to BS 1052.
(4) The PVC coating shall conform to the following requirements:
- tensile strength shall not be less than 210 kg/cm2 in accordance with ASTM D412-75;
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. C - 2
- elongation shall not be less than 200% and not greater than 280% in accordance with ASTM D412-75; and
- resistance to abrasion shall not be greater than 0.19g in accordance with ASTM D1242-56(75).
7.47Construction of Gabion Wall
(1) The gabion shall be formed from one continuous piece of mesh which includes
the lid. Tying of mesh shall be in strict accordance with the manufacturer’s recommendation.
(2) The gabion shall be divided by diaphragms into cells which length shall not be
greater than 1m. (3) All edges of gabions, diaphragms and end panels shall be mechanically
selvedged in such a way as to prevent ravelling of the mesh and to develop the full strength of the mesh.
(4) 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.
(5) 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.
7.48Reno Mattress mesh
(1) The mesh used in fabricating Reno Mattress shall be of a proprietary type
approved by the Engineer. (2) Mesh shall be hexagonal double twist and shall not ravel if damaged. The least
nominal dimension of the hexagon opening shall be 64 mm.
(3) The gabion mesh shall be formed with 2.2 mm diameter mild steel wires, hot dip galvanized to BS 443 and further coated with polyvinyl chloride (PVC). The PVC coating shall be dark green in colour, has an average thickness of 0.5 mm and nowhere less than 0.4 mm. The diameter of the mild steel lacing wire and selvedge wire shall be 2.2 mm and 2.7 mm respectively, galvanized and coated with PVC in a similar way to the mesh wire. All wires shall be mild
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. C - 3
steel to BS 1052.
(4) The PVC coating shall conform to the following requirements:
- tensile strength shall not be less than 210 kg/cm2 in accordance with ASTM D412-75;
- elongation shall not be less than 200% and not greater than 280% in accordance with ASTM D412-75; and
- resistance to abrasion shall not be greater than 0.19g in accordance with ASTM D1242-56(75).
7.49Construction of Reno Mattress
(1) The Reno Mattress shall be formed from one continuous piece of mesh
excluding the lid. The lid shall be a separate piece made of same type mesh as basket. Tying of mesh shall be in strict accordance with the manufacturer’s recommendation.
(2) The Reno Mattress shall be uniformly partitioned by diaphragms into internal
cells. The diaphragms shall be secured in position to the base and shall be in strict accordance with the manufacturer’s recommendation.
(3) All edges of Reno mattress, diaphragms and end panels shall be mechanically
selvedged in such a way as to prevent unravelling of the mesh and to develop the full strength of the mesh.
(4) Infill to Reno Mattress shall be rock fill material of size 75 mm to 200 mm
depending on the height of Reno Mattress and shall be placed in accordance with the manufacturer’s recommendations. The size of rock fill shall be such that a minimum of two layers of rock must be achieved when filling the mattress.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. C - 4
Method of Measurement Gabion Wall & Reno Mattress 7.82 Item Description Gabion wall and Reno Mattress 7.83 Measurement The unit of measurement shall be in cubic metre. The measurement of gabion wall and Reno Mattress shall
be the volume calculated by multiplying the cross- sectional area of the gabion unit by the height as shown on Drawings or ordered by the Engineer.
7.84 Item Coverage The items for gabion wall shall include for:
(5) levelling and preparation including but not limited to blinding concrete and/or rockfill;
(6) assembling, placing and typing together in position; (7) staking, tensioning and jointing the units together; (8) providing and fabricating the gabion unit including
cutting and folding mesh to form special units and shapes;
(9) providing, packing and compacting rubble filling; (10) overfilling to allow for settlement; (11) bracing wires and wiring lids after fillings; (12) square stone facing; (13) in the case of watercourses and ponds, work in
and/or dealing with the flow of water; (14) placing additional steel wire mesh on the face of the
gabions; (15) all necessary tests including submitting results to the
Engineer.
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
App. C - 5
Schedule of Rates 0720 Gabion Wall & Reno Mattress Item No. Description Unit Schedule Rate 072001 Gabion wall m3 072002 Reno Mattress m3
To be assessed when incorporating into contract SOR
RD 1045 Geotechnical Design of Gabion Wall Mainland North Division, DSD
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