# Submitted to the siwes coordinator, department of civil engineering

 Page 6/6 Date conversion 14.06.2018 Size 400.6 Kb.

## Analysis of Structure

FEMBA = (w1l2 /12)+(w2l2/20) ={(5.303 * 0.4752)/12}+{(3.006 * 0.4752)/20}= 0.134kNm

FEMAB = (-w1l2 /12)-(w2l2/30)={(-5.303 * 0.4752)/12}-{(3.006 * 0.4752)/30}= -0.122kNm

FEMDC = (w1l2 /12)+(w2l2/30) = 0.122kNm

FEMCD = (-w1l2 /12)-(w2l2/20) = -0.134kNm

FEMCB = (wl2 /12) = (5.912 * 0.652)/12 = 0.208kNm

FEMBC = (-wl2 /12) = (-5.912 * 0.652)/12 = -0.208kNm

## FIGURE B2: Typical Load Distribution on Drainage Channel

MAB = FEMAB + {(2EI/0.475) (2A + B)}= -0.122 + 8.421EIA + 4.211EIB

MBA = FEMBA + {(2EI/0.475) (A + 2B)}= 0.134 + 4.211EIA + 8.421EIB

MBC = FEMBC + {(2EI/0.65) (2B + C)}= -0.208 + 6.154EIB + 3.077EIC

MCB = FEMCB + {(2EI/0.65) (B + 2C)}= 0.208 + 3.077EIB + 6.154EIC

MCD = FEMCD + {(2EI/0.475) (2C + D)}= -0.134 + 8.421EIC + 4.211EID

MDC = FEMDC + {(2EI/0.475) (C + 2D)}= 0.122 + 4.211EIC + 8.421EID

Conditions for Equilibrium:

MAB = MBC = 0

MBA + MBC = 0

MCB + MCD = 0

MAB = 0: 8.421EIA + 4.211EIB = 0.122 (a)

MBC = 0: 4.211EIC + 8.421EID = -0.122 (b)

MBA + MBC = 0: 4.211EIA + 14.575EIB + 3.077EIC = 0.074 (1)

MCB + MCD = 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,

MAB = 0, MBA = 0.204kNm, MBC = -0.204kNm,

MCB = 0.204kNm, MCD = -0.204kNm, MDC = 0.

These moments are illustrated on Figure B3. From the figure, it can be seen that they are end-moments. 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:

Mmax = -0.204 – (0.5 * 5.303 * 0.4752) – {0.5 * 3.006 * 0.475 * (2 * 0.475/3)}= -1.03kNm

For Member BC:

About point B, Mmax = -0.204 + 0.204 – (0.5 * 5.912 * 0.652) = -1.25kNm

For Member DC:

Mmax = 0.204 + (0.5 * 5.303 * 0.4752) + {0.5 * 3.006 * 0.475 * (2 * 0.475/3)}= 1.03kNm

## Design Procedure

Cover = 25mm, effective depth, d = 75 – 25 – 10 = 40mm,

fcu = 25N/mm2, fy = 425N/mm2

## FIGURE B3: Resultant End-Moments of Drainage Channel

Wall

K = Mmax ÷ (bd2fcu) = 1.03E6Nmm ÷ (1000mm * 402mm2 * 25N/mm2) = 0.026

Z = d{0.5 + [0.25 – (K/0.9)]} Z = 0.95d

Ast = 1.03E6Nmm ÷ (0.95 * 425N/mm2 * 0.95 * 40mm) = 67mm2

Asmin = 0.0013bh = 0.0013 * 1000mm * 75mm = 97.5mm2

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 (Asprov. = 393mm2).

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/mm2

c = 0.8fcu = 4N/mm2

Since  < c, shear is OK and no shear reinforcement is needed.

#### Base

K = Mmax ÷ (bd2fcu) = 1.25E6Nmm ÷ (1000mm * 402mm2 * 25N/mm2) = 0.0313

That implies Z = 0.95d.

Hence, Ast = 1.25E6Nmm ÷ (0.95 * 425N/mm2 * 0.95 * 40mm) = 81.5mm2

Asmin = 0.0013bh = 0.0013 * 1000mm * 75mm = 97.5mm2

Provide Y10@150mm (main and distribution bars respectively) in accordance with the provisions of clause 3.12.11.2.3.

#### Detailing and Bar-Bending Schedule

The detailed drawing for the drainage channel is illustrated in Figure B4. From this, the bar-bending schedule for the entire channel’s length (including the irregular section) is produced as shown on Table B1.

Figure B4: Detailing for a typical cross-section of the drainage channel

Table B1 Bar Bending Schedule of Reinforcement for the Drainage Channel

APPENDIX C
DESIGN PROCESS FOR THE CANTILEVER RETAINING WALL

• Aerator Unit + Contained water 120kN
• Aerator Concrete Unit (24kN/m3 * 4.356m3) 105kN

• Roof (0.7kN/m2 * 193.23m2) 136kN

• Concrete Blockwall (5kN/m2 * 28.85m2) 145kN

• Filter’s Concrete Units (24kN/m3 * 48.58m3) 1166kN

• Davnor Filter Unit (40kN * 2) 80kN

• Live Load (2.5kN/m2 * 227.73m2) 570kN Total 2322kN

Therefore, Surcharge Pressure = (Total Backfill Load)/(Backfill area)

Surcharge Pressure = 2322kN/(18.9m * 12.049m) = 10.2kN/m2

For design purpose, Backfill surcharge pressure = 11kN/m2

• Pumps (3kN * 2 nos.) 6kN

• Pump roof (0.7kN/m2 * 42m2) 30kN

• Pump House’s Concrete Unit (24kN/m3 * 2.96m3) 71kN
• Sandcrete Blockwall (5kN/m2 * 80m2) 400kN

• Treated Water 640kN

• Treated Water tanks 192kN

• Treated Water tanks’ support (24kN/m3 * 5.540m3) 133kN

• Appurtenances 50kN

• Pump House’s slab (24kN/m3 * 80.833m2) 1940kN/m3

• Live Load (2.5kN/m2 * 270m2) 675kN

## Total 4137kN

Frontfill Surcharge Pressure = (Total Frontfill Load)/(Frontfill area)

Frontfill surcharge pressure = 4137kN/(9.146m * 29.436m) = 15.37kN/m2

For design purpose, frontfill surcharge pressure = 16kN/m2

## Analysis of Structure

• Bulk Unit Weight of Soil,  = 19kN/m3

• Allowable bearing pressure = 60kN/m2
• Angle of internal friction,  = 30

• Angle of Cohesion = 0 [Being a homogeneous cohesionless soil used land reclamation.]

Active earth pressure coefficient, Ka = tan2 [45 – (/2)]  (1 – sin )/(1 + Sin)

Passive earth pressure coefficient, Kp = 1/ Ka

Total height of retaining wall, H = 1.7m

Ka = 0.333, Kp = 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, PB = Ka * * HB

PB = 0.333 * 19kN/m3 * 1.7m = 10.76kN/m211kN/m2

Frontfill Earth Pressure, PF = 3 * 19kN/m3 * 0.6m = 34.2kN/m2

Backfill Surcharge Pressure, PSB = 0.333 * 11kN/m2 = 3.663kN/m2

## Figure C2 Cross-sectional Load Diagram of the Cantilever Retaining Wall

Frontfill Surcharge Pressure, PSF = 3 * 16kN/m2 = 48kN/m2

### Horizontal Forces

• Backfill Earth Force, HB = ½ * 1.7m * 11kN/m2 * 1m = 9.35kN ( )

• Frontfill Earth Force, HF = ½ * 0.6m * 34.2kN/m2 * 1m = 10.26kN ( )

• Backfill Surcharge Force, HSB = 0.333 * 1.7m * 11kN/m2 * 1m = 6.24kN ( )

• Frontfill Surcharge Force, HSF = 3 * 0.6m * 16kN/m2 * 1m = 28.8kN ( )

### Vertical Forces

• Wall = 24kN/m3 * 0.3m * 1.4m * 1m = 10.08kN ( )
• Base = 24kN/m3 * 0.3m * 1.8m * 1m = 12.96kN ( )

• Backfill Earth = 19kN/m3 * 1m * 1.4m * 1m = 26.60kN ( )

• Frontfill Earth = 19kN/m3 * 0.5m * 0.3m * 1m = 2.85kN ( )

• Backfill Surcharge = 11kN/m2 * 1m * 1m = 11kN ( )

• Frontfill Surcharge = 16kN/m2 * 0.5m * 1m = 8kN ( )

Total (V) = 71.49kN

## Check for Stability

(a) Sliding

Active horizontal forces causing sliding ( ), Ha = 6.24 + 9.35 ≈15.6kN

Passive horizontal forces resisting sliding ( ), Hp = 10.26 + 28.8 = 39.06kN

Coefficient of friction, µ = tan Ø = tan 30º = 0.58

Total Resisting forces = µ (V) + Hp = (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), Mov

Mov = (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), Mra

Mra = (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

P1,2 = (N/D) ± (6M/D2) = (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 = (Mresisting – Moverturning) ÷ 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

emax = D ÷ 6 = 0.3m

|e| < emax. Hence, middle third rule is satisfied.

P1, 2 = (71.49/1.8) [1 ± (6 * 0.08m ÷ 1.8m)]

P1 = 50.32kN/m2

P2 = 29.11kN/m2

Maximum bearing pressure on soil is 50.32kN/m2 (< the allowable bearing capacity of 60kN/m2).

## 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.

Mult = [9.35kN * ((1.7 ÷ 3) – 0.15) * 1.4] + [1.4 * 6.4 * ((1.7 ÷ 2) – 0.15)]

Mult = 11.77kNm

### Design Parameters

Cover = 50mm, fy = 425N/mm2 , fcu = 25N/mm2 , diameter = 20mm,

Link = 10mm, d = 300 – 10 – 10 – 50 = 230mm.

K = Mult ÷ (bd2fcu) = 11.77E6Nmm ÷ (1000mm * 2302mm2 * 25N/mm2) = 0.0088

Hence, Z = 0.95d

Ast = 11.77E6Nmm ÷ (0.95 * 425N/mm2 * 0.95 * 230mm) = 133mm2 per 1m run.

Asmin = 0.0013bh = 0.0013 * 1000mm * 300mm = 390mm2 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 cross-section of the retaining wall is as shown on Figure C3. From this figure, the bar-bending schedule is extracted as tabulated in Table C1.

Figure C3 Detailing of the Cross-section of the Cantilever Retaining Wall

Table C1 Bar Bending Schedule for the Cantilever Retaining Wall

1 NTU means Nephthelometric Turbidity Unit. It is used as a unit in determining the intensity of the turbidity of any water sample.

2 Prestressed concrete is a concrete to which an internal compressive stress has been applied by means of wires or tendons, so that a tensile stress equal to the compressive stress can be applied in service (as live loads) such that the net stress is then zero.

3 Slenderness ratio is the ratio of the effective height of a wall to its thickness, i.e. le/h.

4Adverse partial factor is applied to any loads that tend to produce a more critical design condition at the section considered.

5 Beneficial partial factor is applied to any loads that tend to produce a less critical design condition at the section considered.

6Statical determinate structure is one in which the equations of static equilibrium are sufficient to analyse the structure, i.e. ∑Fx=0, ∑Fy=0, ∑M=0

7 Statically indeterminate structure is one in which the equations of static equilibrium are not sufficient to analyse the structure.

8 FEM are the moments at the support of a fixed section (say, beam) that are produced due to lateral or

9 BS 4466: 1969 – Bending, Dimensioning and Scheduling of Bars for the Reinforcement of Concrete.

10 Latitude is one of the axes of an imaginary coordinate system of the Earth to determine the distance of a point from the equator. It is actually the degree of the angle formed between a line from the point to the centre of the Earth and a line from the equator to the centre of the Earth.

11 Longitude is the other axis in the imaginary coordinate system of the Earth. It is the degree of the angle formed between a line from the point to the centre of the Earth and a line from the prime meridian to the centre of the Earth.