FIGURE B2: Typical Load Distribution on Drainage Channel
M_{AB} = FEM_{AB} + {(2EI/0.475) _{} (2_{A} + _{B})}= 0.122 + 8.421EI_{A }+ 4.211EI_{B }
M_{BA} = FEM_{BA} + {(2EI/0.475) _{} (_{A} + 2_{B})}= 0.134 + 4.211EI_{A }+ 8.421EI_{B }
M_{BC} = FEM_{BC} + {(2EI/0.65) _{} (2_{B} + _{C})}= 0.208 + 6.154EI_{B }+ 3.077EI_{C }
M_{CB} = FEM_{CB} + {(2EI/0.65) _{} (_{B} + 2_{C})}= 0.208 + 3.077EI_{B }+ 6.154EI_{C }
M_{CD} = FEM_{CD} + {(2EI/0.475) _{} (2_{C} + _{D})}= 0.134 + 8.421EI_{C }+ 4.211EI_{D}
M_{DC} = FEM_{DC} + {(2EI/0.475) _{} (_{C} + 2_{D})}= 0.122 + 4.211EI_{C }+ 8.421EI_{D }
Conditions for Equilibrium:
M_{AB} = M_{BC} = 0
M_{BA} + M_{BC} = 0
M_{CB} + M_{CD} = 0
M_{AB} = 0: 8.421EI_{A }+ 4.211EI_{B }= 0.122 (a)
M_{BC} = 0: 4.211EI_{C }+ 8.421EI_{D }= 0.122 (b)
M_{BA} + M_{BC} = 0: 4.211EI_{A }+ 14.575EI_{B }+ 3.077EI_{C} = 0.074 (1)
M_{CB} + M_{CD} = 0: 3.077EI_{B }+ 14.575EI_{C }+ 4.211EI_{D} = 0.074 (2)
From equation (a), 8.421EI_{A }= 0.122  4.211EI_{B}
Hence, 4.211EI_{A }= 0.061 – 2.106EI_{B} (3)
Also, from equation (b), 8.421EI_{D }= 0.122  4.211EI_{C}
Hence, 4.211EI_{D }= 0.061 – 2.106EI_{C } (4)
Substitute equation (3) into equation (1), and equation (4) into equation (2):
12.469EI_{B }+ 3.077EI_{C }= 0.013 (5)
3.077EI_{B }+ 12.469EI_{C }= 0.013 (6)
Solving equations (5) and (6) simultaneously gives:
EI_{B }= 0.001384
EI_{C }= 0.001384
From equation (3), EI_{A }= 0.01379
EI_{D }= 0.01379
Hence,
M_{AB} = 0, M_{BA} = 0.204kNm, M_{BC} = 0.204kNm,
M_{CB }= 0.204kNm, M_{CD }= 0.204kNm, M_{DC }= 0.
These moments are illustrated on Figure B3. From the figure, it can be seen that they are endmoments. Hence, it is important to determine the maximum moment that would act along the span of each member, about a point, using the principles of static equilibrium.
For Member AB:
About point A,
M_{max }= 0.204 – (0.5 _{*} 5.303 _{* }0.475^{2}) – {0.5 _{*} 3.006 _{* }0.475 _{*} (2 _{* }0.475/3)}= 1.03kNm
For Member BC:
About point B, M_{max }= 0.204 + 0.204 – (0.5 _{*} 5.912 _{* }0.65^{2}) = 1.25kNm
For Member DC:
About point D,
M_{max }= 0.204 + (0.5 _{*} 5.303 _{* }0.475^{2}) + {0.5 _{*} 3.006 _{* }0.475 _{*} (2 _{* }0.475/3)}= 1.03kNm
Design Procedure
Cover = 25mm, effective depth, d = 75 – 25 – 10 = 40mm,
f_{cu} = 25N/mm^{2}, f_{y }= 425N/mm^{2}
FIGURE B3: Resultant EndMoments of Drainage Channel
Wall
K = M_{max} ÷ (bd^{2}f_{cu}) = 1.03E6Nmm ÷ (1000mm _{*} 40^{2}mm^{2} _{* }25N/mm^{2}) = 0.026
Z = d{0.5 + [0.25 – (K/0.9)]} Z = 0.95d
A_{st} = 1.03E6Nmm ÷ (0.95 _{* }425N/mm^{2} _{*} 0.95 _{* }40mm) = 67mm^{2 }
A_{smin} = 0.0013bh = 0.0013 _{* }1000mm _{* }75mm = 97.5mm^{2}
Design specifications allow 10mm as the least bar diameter for ease of construction. Hence, balancing with the provisions of clause 3.12.11.2 in BS 8110, provide Y10@200mm c/c (A_{sprov. }= 393mm^{2}).
Also, provide Y10@200mm for distribution.
Shear
V = (5.303kN/m _{* }0.475m) + (½ _{*} 0.475m _{* }3.006kN/m) = 3.233kN
= V/(bd) = 3.233E3N/(1000mm _{* }40mm) = 0.040N/mm^{2}
_{c} = 0.8f_{cu} = 4N/mm^{2}
Since < _{c}, shear is OK and no shear reinforcement is needed.
Base
K = M_{max} ÷ (bd^{2}f_{cu}) = 1.25E6Nmm ÷ (1000mm _{*} 40^{2}mm^{2} _{* }25N/mm^{2}) = 0.0313
That implies Z = 0.95d.
Hence, A_{st} = 1.25E6Nmm ÷ (0.95 _{* }425N/mm^{2} _{*} 0.95 _{* }40mm) = 81.5mm^{2 }
A_{smin} = 0.0013bh = 0.0013 _{* }1000mm _{* }75mm = 97.5mm^{2}
Provide Y10@150mm (main and distribution bars respectively) in accordance with the provisions of clause 3.12.11.2.3.
Detailing and BarBending Schedule
The detailed drawing for the drainage channel is illustrated in Figure B4. From this, the barbending schedule for the entire channel’s length (including the irregular section) is produced as shown on Table B1.
Figure B4: Detailing for a typical crosssection of the drainage channel
Table B1 Bar Bending Schedule of Reinforcement for the Drainage Channel
APPENDIX C
DESIGN PROCESS FOR THE CANTILEVER RETAINING WALL
Load Estimation
(a) Retaining Wall’s Backfill Load

Aerator Unit + Contained water 120kN

Aerator Concrete Unit (24kN/m^{3 }_{* }4.356m^{3}) 105kN

Roof (0.7kN/m^{2} _{* }193.23m^{2}) 136kN

Concrete Blockwall (5kN/m^{2} _{* }28.85m^{2}) 145kN

Filter’s Concrete Units (24kN/m^{3} _{* }48.58m^{3}) 1166kN

Davnor Filter Unit (40kN _{* }2) 80kN

Live Load (2.5kN/m^{2} _{* }227.73m^{2}) 570kN Total 2322kN
Therefore, Surcharge Pressure = (Total Backfill Load)/(Backfill area)
Surcharge Pressure = 2322kN/(18.9m _{* }12.049m) = 10.2kN/m^{2}
For design purpose, Backfill surcharge pressure = 11kN/m^{2}
(b) Retaining Wall’s Frontfill Load

Pumps (3kN _{* }2 nos.) 6kN

Pump roof (0.7kN/m^{2} _{* }42m^{2}) 30kN

Pump House’s Concrete Unit (24kN/m^{3} _{* }2.96m^{3}) 71kN

Sandcrete Blockwall (5kN/m^{2} _{* }80m^{2}) 400kN

Treated Water 640kN

Treated Water tanks 192kN

Treated Water tanks’ support (24kN/m^{3} _{* }5.540m^{3}) 133kN

Appurtenances 50kN

Pump House’s slab (24kN/m^{3} _{* }80.833m^{2}) 1940kN/m^{3} ^{ }

Live Load (2.5kN/m^{2} _{* }270m^{2}) 675kN
Total 4137kN
Frontfill Surcharge Pressure = (Total Frontfill Load)/(Frontfill area)
Frontfill surcharge pressure = 4137kN/(9.146m _{* }29.436m) = 15.37kN/m^{2}_{ }
For design purpose, frontfill surcharge pressure = 16kN/m^{2}
The various loads listed above can be seen from Figure C1. The figure gives an outline of the retaining site and shows the locations of various features relative to the retaining wall.
Analysis of Structure

Bulk Unit Weight of Soil, = 19kN/m^{3}

Allowable bearing pressure = 60kN/m^{2}

Angle of internal friction, = 30

Angle of Cohesion = 0 [Being a homogeneous cohesionless soil used land reclamation.]
Active earth pressure coefficient, K_{a} = tan^{2} [45 – (/2)] (1 – sin )/(1 + Sin)
Passive earth pressure coefficient, K_{p }= 1/ K_{a}
Total height of retaining wall, H = 1.7m
K_{a }= 0.333, K_{p }= 3.000
The loading diagram of the retaining wall is illustrated in Figure C1. From the figure, it can be seen that the load diagrams have different shapes for different loads. Moreover, it clearly shows the backfill and frontfill sides.
Hence, Backfill Earth Pressure, P_{B} = K_{a} _{* } _{* }H_{B }
P_{B} = 0.333 _{* }19kN/m^{3 }_{* }1.7m = 10.76kN/m^{2} 11kN/m^{2}
Frontfill Earth Pressure, P_{F} = 3 _{* }19kN/m^{3 }_{* }0.6m = 34.2kN/m^{2}
Backfill Surcharge Pressure, P_{SB }= 0.333 _{* }11kN/m^{2 }= 3.663kN/m^{2}
Figure C1: Outline of the Retaining Wall Site
Frontfill Surcharge Pressure, P_{SF }= 3 _{* }16kN/m^{2 }= 48kN/m^{2}
Horizontal Forces

Backfill Earth Force, H_{B} = ½ _{*} 1.7m _{*} 11kN/m^{2 }_{* }1m = 9.35kN ( )

Frontfill Earth Force, H_{F} = ½ _{*} 0.6m _{*} 34.2kN/m^{2 }_{* }1m = 10.26kN ( )

Backfill Surcharge Force, H_{SB} = 0.333 _{*} 1.7m _{*} 11kN/m^{2 }_{* }1m = 6.24kN ( )

Frontfill Surcharge Force, H_{SF} = 3 _{*} 0.6m _{*} 16kN/m^{2 }_{* }1m = 28.8kN ( )
Vertical Forces
Total (V) = 71.49kN
Check for Stability
(a) Sliding
Active horizontal forces causing sliding ( ), H_{a} = 6.24 + 9.35 ≈15.6kN
Passive horizontal forces resisting sliding ( ), H_{p} = 10.26 + 28.8 = 39.06kN
Coefficient of friction, µ = tan Ø = tan 30º = 0.58
Total Resisting forces = µ (V) + H_{p} = (0.58 _{*} 71.49) + 39.06 = 80.52kN
Safety Criteria: (Resisting force/Sliding force) ≥ 1.6
(80.52kN) ÷ (15.6kN) = 5.17 >> 1.6.
Hence, the structure is safe against sliding.
(b) Overturning

Overturning moment (caused by horizontal forces about point A), M_{ov}
M_{ov }= (6.24kN _{*} 0.85m) + (9.35kN _{*} 0.57m) = 10.634kNm
(Note that moments due to frontfill forces are neglected because they are just taken to further stabilise the retaining wall.)

Resisting moment (caused by vertical forces about point A), M_{ra}
M_{ra} = (10.08kN _{*} 0.65m) + (12.96kN _{*} 0.9m) + (26.60kN _{*} 1.3m) + (2.85kN _{*} 0.25m) + (11kN _{*} 1.3m) + (8kN _{*} 0.25m) + (10.26kN _{*} 0.2m) + (28.8kN _{*} 0.3m) = 80.501kNm
(Note that the inclusion of the moment due to the frontfill forces is a way to simulate the worst condition that the structure must be able to resist.)
Safety Criteria: (Resisting Moment) ÷ (Overturning Moment) > 2.0
80.5 ÷ 10.53 = 7.65 > > 2.0
Hence, the structure is safe against overturning.
Bearing Pressure Analysis
P_{1,2} = (N/D) ± (6M/D^{2}) = (N/D) [1 ± (6e/D)]
where N = Total vertical forces
D = Base width
M = Net moment about the base bottom of all forces (due to both horizontal and vertical forces)
e = eccentricity
equilibrium distance, x = (M_{resisting} – M_{overturning}) ÷ N = (80.5 – 10.53) ÷ 71.49
x = 0.979m from A.
e = 0.5D – x = 0.9m – 0.979m = 0.079m
e = 0.08m
e_{max} = D ÷ 6 = 0.3m
e < e_{max}. Hence, middle third rule is satisfied.
P_{1, 2} = (71.49/1.8) [1 ± (6 _{*} 0.08m ÷ 1.8m)]
P_{1} = 50.32kN/m^{2}
P_{2} = 29.11kN/m^{2}
Maximum bearing pressure on soil is 50.32kN/m^{2 }(< the allowable bearing capacity of 60kN/m^{2}).
Bending Reinforcement
Wall
A factor of safety of 1.4 is applied to the horizontal forces on the wall. Then, moments of the horizontal forces are taken about the centreline of the base. This is because cantilever walls, generally, fail by rotation about a point some ways above the toe.
M_{ult} = [9.35kN _{*} ((1.7 ÷ 3) – 0.15) _{*} 1.4] + [1.4 _{*} 6.4 _{*} ((1.7 ÷ 2) – 0.15)]
M_{ult} = 11.77kNm
Design Parameters
Cover = 50mm, f_{y }= 425N/mm^{2 }, f_{cu }= 25N/mm^{2 }, diameter = 20mm,
Link = 10mm, d = 300 – 10 – 10 – 50 = 230mm.
K = M_{ult} ÷ (bd^{2}f_{cu}) = 11.77E6Nmm ÷ (1000mm _{*} 230^{2}mm^{2} _{* }25N/mm^{2}) = 0.0088
Hence, Z = 0.95d
A_{st} = 11.77E6Nmm ÷ (0.95 _{* }425N/mm^{2} _{*} 0.95 _{* }230mm) = 133mm^{2 }per 1m run.
A_{smin} = 0.0013bh = 0.0013 _{* }1000mm _{* }300mm = 390mm^{2} per 1m run.
Provide Y10@200mm c/c longitudinal and transverse on both I.F. and O.F. From experience, being the larger unit, the reinforcement information of the wall is suitable for the base. A trial convinced the trainee!
The reinforcement detailing for the crosssection of the retaining wall is as shown on Figure C3. From this figure, the barbending schedule is extracted as tabulated in Table C1.
Figure C3 Detailing of the Crosssection of the Cantilever Retaining Wall
Table C1 Bar Bending Schedule for the Cantilever Retaining Wall