circular reservoir 5m (hydroforum.net).pdf

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DRAFT MARCH 2007 134.
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WATER SUPPLY SAMPLE DESIGNS 8B.2CIRCULAR WATER RESERVOIR 5m3

1. OVERVIEW

The following document outlines the design

calculations for a 5m3 circular shaped water reservoirsuitable for community water supply. The structurehas base and roof slabs made from reinforcedconcrete, with stone masonry (or brick masonry iflocally available) walls. The structure is plasteredinternally with three layers of waterproof mortar. Thedesign has a steel access cover placed centrally overthe valve chamber side. Good quality fittings shouldbe used where possible. Where pipes pass throughthe walls, GI pipes should be used fitted with anti-passage rings to the pipes to prevent the path ofwater filtration along the pipe. Calculations are basedon the following design rules

1. An ordinary 1:2:4 concrete mix is not waterprooftherefore a richer 1:1.5:3 mixture (with a cement

dosage of 380 kg/m3) has been used for floorslab calculations. A standard 1:2:4 concrete mix,with a cement dosage of 320 kg/m

3, has been

used for the roof slab calculation.

2. Base slab thickness is determined using theempirical rule below, with a minimum value of10cm.

Slab Thickness = Shortest Span / 40

3. The base slab is designed for the weight of thestructure and water stored pressing downwardsand the reaction of the soil pressing upwards.

Slab Loading = (Weight of Structure + Water)Slab Area

4. The base slab has been designed with sufficientstrength to enable it to span over possible weakpatches of the ground. The bending moment isgreatest at the center of the slab and has beencalculated using..

Maximum BM = (Loading x Span2) / 24

5. The required reinforcement to resist themaximum bending moment has been calculatedassuming mild steel grade I ribbed bars with atensile strength of 1650kg/cm2. The total area ofreinforcement required for one meters width ofslab has been calculated by..

Area = Maximum Bending Moment(Max. Reinf. Tension x Lever Arm)

Where the lever arm can be found by theempirical rule

Lever Arm = 7/8 x Concrete Wall Thickness

6. Base slab reinforcement sizes have beenselected on the basis of providing the smallestdiameter practicable. Load spreading is providedby the lateral reinforcement members, whichhave the same area and spacing as thelongitudinal reinforcement.

7. Roof slab thickness is determined using the

empirical rule below, with a minimum value of10cm.

Slab Thickness = Shortest Span / 40

8. The roof slab is designed for the weight of theslab structure plus a local access loading of200kg/m

2(the weight of three people).

Loading = (Slab Weight + (200kg/m2

x Area))Slab Area

9. In a simply supported slab resting on four walls,the bending moment is greatest at the center ofthe slab. The maximum bending moment hasbeen calculated using..


10. The required reinforcement to resist themaximum bending moment has been calculatedassuming mild steel grade I ribbed bars with amax imum tensile strength of 1650kg/cm

2. The

total area of reinforcement required for onemeters length of wall has been calculated by..

Total Area = Maximum Bending Moment(1650 kg/cm

2x Lever Arm)

Where the lever arm can be found by theempirical rule

Lever Arm = 7/8 x Concrete Slab Thickness

11. Roof reinforcement sizes have been selected onthe basis of providing the smallest diameterpracticable. Load spreading is provided by thelateral reinforcement members, which have thesame area and spacing as the longitudinalreinforcement.

12. Internal plastering consists of three layers usingsikalite waterproofing compound (whereavailable)

Layer #1: 6mm 1:4 SplatterdashLayer #2: 10mm 1:3 rough finishLayer #3: 10mm 1:2 smooth float

2.0 BILL OF QUANTITIES

Item Description Unit Qty

1.0 Local Material

1.1 Gravel m3

1.4

1.2 Cement sack 21

1.3 Sand m3

1.6

1.4 Bricks or Dressed Stone m3

7.28

1.5 Crushed Rock Foundation m3

2.0

1.6 Planks (3.0m x 0.2m) pc 14

1.7 Sikalite Compound kg 16

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2.0 Ironwork

2.1 Iron Reinforcement Bars 8mm x 12m

pc 8.5

2.2 Iron Reinforcement Bars 10mm x 12m

pc 16

2.2 Tying Wire 1mm kg 8

2.3 Nails 6cm kg 8

2.4 Steel Access Cover 50cm x50cm

pc 1

2.5 Steel Internal AccessLadder

pc 1

3.0 Plumbing

3.1 Float Valve 1 pc 1

3.2 GI Pipe 1 x 6m pc 1

3.3 GI Socket 1 pc 23.4 GI Elbow 1 pc 1

3.5 GI Gate Valve pc 2

3.6 Small Filter Mesh Unit 1 pc 2

3.7 TeflonTape roll 4

3.0 JUSTIFYING CALCULATIONS

Item Dimension

Water Depth (m) 1.40 m

Interior Width (m) 2.20 m

Thickness of Hardcore (m) 0.30 m

Width of Apron (m) 0.20 m

Wall Thickness (m) 0.40 m

Wall Height (m) 1.70 m

Freeboard (m) 0.20 m

Sludge Depth (m) 0.10 m

Table 1: Table of Tentative Dimensions

3.1 EXCAVATION

Excavation = [ x (3.6 m / 2)2

] x 0.4 m= 4.1 m

3

Hardcore = [ x (6.0 m / 2)2 ] x 0.2 m= 2.0 m

3

3.2 ROOF SLAB (Dosage 320 kg/m3

- 1:2:4)

Slab Thickness = Shortest Span / 40= 3.0 m / 40= 0.07 m (minimum 0.10 m)

Slab Concrete = [ x (3.0 m / 2)2

] x 0.10 m= 0.71 m

3

Cement = 0.71 m3

x 320 kg/m3

= 227 kg (4.5 bags)

Sand = 0.71 m3

x 0.45= 0.32 m

3

Gravel = 0.71 m3

x 0.9= 0.64 m

3

Slab Loading = (Slab Weight + (200kg/m2

x Area))Slab Area

= [(0.71 m3

x 2500 kg/m3)

+ (200 kg/m3

x 7.1 m2)] / 7.1 m

2

= 450 kg/m2

Therefore for a unit meter width of slab with adistributed loading of 450 kg/m, spanning 2.8 m ..


= [450 kg/m x (3.0 m)2

] / 8= 506.3 kg.m

Total Area = Maximum Bending Moment(1650 kg/cm2

x Lever Arm)

Where

Lever Arm = 7/8 x Concrete Slab Thickness= 7/8 x 10 cm= 8.75 cm

Rebar Area = 506.3 kg.m / (1650 x 0.0875 m)(per m width) = 3.5 cm

2

Assume 8 mm reinforcement bars available with area0.5 cm

2

Bars per m = 3.5 cm2

/ 0.5 cm2

Width = 7 bars

Rebar Spacing = 1 m / 7 bars= every 14 cm

Assume that load spreading (longitudinal)reinforcement will also consist of 8mm reinforcementbars spaced every 14 cm. A 1 m

2area of slab will

require the following amount of reinforcement..

Reinforcing Bar = (7 x 1.0 m) x 2= 14 m / m

2

Therefore for the entire roof slab we will require..

Reinforcing Bar = 14 m / m2

x [ x (3.0 m / 2)2

]

= 99 m (8.5 bars)= 99 m x 0.38 kg/m= 37.6 kg

Tying Wire = 37.6 kg x 5%= 2 kg

Shuttering = [ x (3.0 m / 2)2

] / (3.0 m x 0.2 m)= 12 planks

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3.3 BRICK / STONE MASONRY WALLS

Wall Volume = [( x 1.52

) x 1.70 m]

- [( x 1.12

) x 1.70 m]= 5.6 m

3

Brick/Stone Vol = 5.6 m3

x 1.3= 7.28 m

3

Mortar Volume = 5.6 m3

x 0.23= 1.3 m

3

Cement (1:3) = 1.3 m3

x 117kg= 152 kg

Vol of Sand = 1.3 m3

x 0.243= 0.32 m

3

3.4 BASE SLAB (Dosage 380 kg/m3

- 1:1.5:3)

Slab Thickness = Shortest Span / 40= 3.4 / 40

= 0.085 m (minimum 0.10 m)

Slab Concrete = [ x (3.4 m / 2)2

] x 0.10 m= 0.9 m

3

Cement = 0.9 m3

x 380 kg/m3

= 342 kg (7 bags)Sand = 0.9 m

3x 0.42

= 0.38 m3

Gravel = 0.9 m3

x 0.84= 0.76 m

3

Slab Loading = (Weight of Structure + Water)Slab Area

Where the total weight of the structure can be foundby..

Total Weight = Floor + Roof + Walls + Water= ((0.76 m

3+ 0.71 m

3+ 5.6 m

3)

x 2,500 kg / m3) + 5,000 kg

= 22,675 kg

Base Loading = 22,675 kg / 9.1 m2

= 2,492 kg/m2

Therefore, for a unit meter width of slab with adistributed loading of 2,292 kg/m, spanning 3.4 m ..


= ((2,292 kg/m x (3.4 m)2

) / 24

= 1,104 kg.m

Longit. Rebar = Maximum Bending Moment(per m width) (Max. Reinf. Tension x Lever Arm)

Where

Lever Arm = 7/8 x Concrete Slab Thickness= 7/8 x 10 cm= 8.75 cm

Longit. Rebar = 1,104 kg.m / (1650 x 0.0875 m)(per m width) = 7.6 cm

2

Assume 10 mm reinforcement bars available witharea 0.79 cm

2

Longit. Rebar = 7.6 cm2

/ 0.79 cm2

(per m width) = 9.6 (10 bars)

Longit. Rebar = 1 m / 10 barsSpacing = every 10 cm

Lat. Bar Area = Longitudinal Reinforcement Area(per m length) = 7.6 cm

2

Therefore lateral reinforcement will also consist of 10x 10 mm reinforcement bars spaced every 10 cm. A1 m

2area of slab will require the following amount of

reinforcement..

Reinforcing Bar = (10 x 1.0 m) x 2= 20 m / m

2

Therefore for the entire floor slab we will require..

Reinforcing Bar = 20 m / m2

x [ x (3.4 m / 2)2

]= 182 m (16 bars)= 182 m x 0.59 kg/m= 107 kg

Tying Wire = 107 kg x 5%= 6 kg

3.5 INTERNAL PLASTERING

Plaster Area = [ x 1.12

] + (2 x 1.1 x 1.7 m)= 15.6 m

2

Splatterdash = 15.6 m

2

x 2.735 kgCement = 42.7 kg (6mm 1:4)Vol. of Sand = 15.6m

2x 0.00766

= 0.12 m3

Rough Finish = 15.6 m2

x 7.34 kgCement = 114.5 kg (12mm 1:3)Vol. of Sand = 15.6 m

2x 0.01541

= 0.24 m3

Smooth Float = 15.6 m2

x 9.79 kgCement = 152.7 kg (12mm 1:2)Vol. of Sand = 15.6 m

2x 0.01371

= 0.21 m3

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