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INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT
BILHAUR, KANPUR NAGAR -209202
PROJECT REPORT OF THE FINAL YEAR -2013
PROJEC...
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INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT, BARAULI
BILHAUR , KANPUR
SESSION 2014-2015
“DEPARTMENT OF CIVIL ENGINEERIN...
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INDEX
S.no. Title Page No.
1 Acknowledgement 4
2 Certificate 6
3 Objective 8
4
5
Introduction
Survey
10
13
6 Structural ...
=
ACKNOWLEDGEMENT
We want to express our heartiest thanks to our most respected and
honoured head of department of civil e...
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CERTIFICATE
It is certified that the project report entitled “ PRELIMINARY DESIGN
OF WATER TREATMENT PLANT” is submitted...
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OBJECTS
OBJECTIVE
To prepare a project of Preliminary Design Of “ WATER
TREATMENT PLANT”at Kanpur.
The objectives of the...
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“INTRODUCTION”
INTRODUCTION
The primary needs for any living being are the water,air,food,sheltered, for
which water has...
=
ultrapure water, using a combination of active charcoal membranes, and
reverse osmosis filter. Internet sites of these c...
=
A Survey is an inspection of an area where work is proposed, to
gather information for a design or an estimate to comple...
=
The structural elements are those elements which form the
supporting skeleton frame work of the supply scheme. It includ...
=
water. The requisite alum dose shall be mixed in two
polyethylene tanks of 3000L each, which shall be
interconnected. Th...
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(vii) Main distribution tank-The main distribution reservoir will
obtain pumped water from 2nd stage pumped house and
de...
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(1) Design of pre-sedimentation tank- We shall design the tank
to remove particles upto 0.1 mm size, and adopt the
follo...
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Water requirement for the year 2030= 27.0 MLD
Taking 9.00 hrs detention period capacity of tank
=(8.00*27.0)/24= 9000 m3...
=
Water required by the year 2030 = 27.0 MLD = 1125m3/hr
Water lost in desludging = 2%
Design average flow = 1125 * 100/(1...
=
Design flow for filter after accounting for backwash
water3% and washing time 0.5hr
=1125*(1+0.03)*24/(23.5)= 1183.4 m3/...
=
Capacity of clear water reservoir = 27.00*10^6*10^-3 *
8.0/(24) = 9000m^3
Assume water depth as 3.0 m
Plan area required...
=
Unit weight of soil = 16.8 KN/ 𝑚2
Height of wall above the ground level
H = height of wall above the plinth + plinth hei...
=
Here,
D = 160/16.8 ×( 1- sin 30) ^2 ÷ ( 1- sin30) ^2
D = 1.06 m = 1.10 m (say)
Hence adopt depth of foundation = 1.10 m
...
=
Live load on roof slab = 2000KN/ 𝑚2
The slab is simply supported on all four sides with corners not
held down.
Using M15...
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R = (5×0.865×0.404)÷2
R = 0.874
Let,
Over all depth of slab = 150 mm
Assuming,10mm dia. Main bars and 15 mm clear cover
...
=
= 4.150 m
(b) Clear span + effective depth = 4000 + 130 = 4130 mm
= 4.130 m
2. (a) Centre to centre bearing = 3500 + 300...
=
= 1920 N/ 𝑚2
2. Live load = 2000 N/ 𝑚2
Hence , total load = 3750 + 1920 + 2000 = 7670 N
= 7.670 KN
Total load = 7670 N
B...
=
(b) Maximum bending moment on longer span,
Ml = ( Wl× L^2)/8
= 2867 × (4.130^4) / 8
= 6112.76 N-m
Bending moment = 61127...
=
= 156.37 mm = 150 mm (say)
Hence, provide 10 mm diameter bars @ 150 C/C,
Bend up alternate bar at L/7 = 3.630 / 7 = 0.51...
=
= 261.08m 𝑚2
Check –
1. Shear force for shorter span –
Vb = (W × B) ÷ 3
= (7670 × 3.630) ÷ 3
= 9280 N
2. Shear force for...
=
M1 = 3.95 × 10 ^6 N-mm
So,
Development length Ld ≤ [ 1.3 (M1/V) + Lo ]
Anchorage length, Lo = 12 Φ or d (max.)
= 12 × 10...
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But dia. of main bar is 10 Φ, so bars safe in development.
2. Considering longer span of slab –
M1 = (fst×Ast× y × z) ÷ ...
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ESTIMATION
METHOD OF ESTIMATION
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Long wall short wall method
Centre Line
Method
Long Wall
and Short
Wall
Crossing
Method
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Length of long wall
𝑙1=c/c length + width of step
Length of long wall
𝑙1=c/c length + width of step
Length of short wall...
=
Concrete work
Long wall
4 18.37 0.60 0.60 26.45 𝑚3
17.77 +0.6 =
18.39 m
Short wall
3 14.71 0.60 0.60 15.88 𝑚3
15.31 – 0....
=
Base slab
Long wall 2 50.67 0.90 0.30 27.36 𝑚3
Short wall
2 28.87 0.90 0.30 15.58 𝑚3
Concrete Work
Long wall
4 18.37 0.6...
=
Descriptionof work No. L (m) B
(m)
h/d quantity
Remark
1ST STEP
Long wall 2 50.34 0.57 0.15 11.26 𝑚3
49.77 +0.57 =
50.34...
=
Long wall 5 17.96m
Short wall 3 15.11 0.20 0.62
5
5.66 𝑚3
15.31-0.2
=15.11m
Long wall 1 12.20 0.20 0.62
5
1.525 𝑚3
12+.2...
=
Descriptionof work No. L (m) B
(m)
h/d quantity Remark
Long wall 2 50.0 0.23 3.0 69.00𝑚3
49.77 +0.23 =
50.0 m
Short wall...
=
3 B=
15+.23+.23+.2
3= 15.83
7 PLASTER WORK
Descriptionof work
Measurements
quantity RemarkNo. L (m) B
(m)
h/d
Long wall ...
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Estimation
(Material cost)
S. N0. Items quantity Rate (Rs.) Cost (Rs.)
1 Cement 19850 bags 350/bag 6947500
2 L.Sand 987 ...
=
5 Grit 524 2200/ m3 1152800
6 Steel 15123 Kg 45/ Kg 680535
Total material cost in Rs. Rs. 11733555/-
(Labour cost)
S.No....
=
12mm 2488.97𝑚2
42/𝑚2
104536.74
8 R.C.C. 29.485 510/ m3 15037.35
Total 473521.10
Total cost of project
COST OF PROJECT
To...
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(b). 0.5% of civil work for survey work = Rs. 5609.6/-
(c). 4% of civil work for internal road = Rs. 4487.68/-
(d). 2% o...
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FIG- SHOWING VARIOUS UNIT OF WATER SUPPLY PROJECT
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FIG- SHOWING RAW WATER TANK
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FIG- SHOWING PRE-SEDIMENTATION TANK
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FIG- SHOWING RAW WATER TANK
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final year Project (civil)

  1. 1. = INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT BILHAUR, KANPUR NAGAR -209202 PROJECT REPORT OF THE FINAL YEAR -2013 PROJECT TITLE:- “PRELIMINARY DESIGN OF WATER TREATMENT PLANT” Submitted by AKHILESH PRATAP VERMA ABHISHEK TIWARI ANKIT KUMAR JITENDRA KUMAR AKHILESH SINGH ABHAY PRATAP YADAV ASHAD AGNIHOTRI In partial fulfillment for the award of the degree Of BACHELOR OF TECHNOLOGY IN CIVIL ENGINEERING
  2. 2. = INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT, BARAULI BILHAUR , KANPUR SESSION 2014-2015 “DEPARTMENT OF CIVIL ENGINEERING” PROJECT REPORT UNDER THE ABLE GUIDANCE OF MR.NITIN AGRAWAL MR.K. K. THAKUR ( H.O.D. ) (Lecturer) IITM,Kanpur IITM,Kanpur
  3. 3. = INDEX S.no. Title Page No. 1 Acknowledgement 4 2 Certificate 6 3 Objective 8 4 5 Introduction Survey 10 13 6 Structural Elements 15 7 Design Specification 19 8 Design 25 9 Drawing 37 10 Estimation 39 11 Total cost of project 52
  4. 4. = ACKNOWLEDGEMENT We want to express our heartiest thanks to our most respected and honoured head of department of civil engineering, Mr. NitinAgrawal for his valuable teachings and the suggestions along with guideline which really help me in completing this project. We want to thank our Project Head Mr.Krishna Kant Thakur for his dedication and the valuable time which he has spent with us during the completion of project. We are really thankful to entire faculty of theDepartment of Civil Engineering for the help and support we have got from them in any form. And lastly we want to thanks to all the members of group for all the cooperation and help we have got from them. Thanking You!
  5. 5. = CERTIFICATE It is certified that the project report entitled “ PRELIMINARY DESIGN OF WATER TREATMENT PLANT” is submitted by “, AKHILESH PRATAP VERMA,ABHISHEK TIWARI,ANKIT KUMAR,AKHILESH SINGH,ABHAY PRATAP YADAV,JITENDRA KUMAR,ASHAD AGNIHOTRI” student of Indus Institute of Technology And Management , Kanpur in practical fulfillment for degree in “Civil Engineering from the Board of Technical Education U.P.” during the academic session 2014- 2015. The external examiner has checked and taken the oral viva – voice on the project report. My blessing are always with him for his bright future. MR. NITIN AGRAWAL MR.K. K.THAKUR ( H.O.D. ) ( Lecturer)
  6. 6. = OBJECTS OBJECTIVE To prepare a project of Preliminary Design Of “ WATER TREATMENT PLANT”at Kanpur. The objectives of the project are:-  Carrying out a complete analysis and designing of the main structural elements of an Water Treatment Plant including slabs, columns, and beams.  The structure should be able to accommodate all the machineries as well as all the required equipments needed in the water Treatment.  Use structural software (AutoCAD) to make the plan.  Use IS codes.  Getting real life experience with engineering practices.  Use of all the necessary equipments needed for the survey of the site.  To study the various elements of the structure in detail.  To estimate the cost of material as well as cost of labour along with other indirect included cost incurred in the construction of civil structures.
  7. 7. = “INTRODUCTION” INTRODUCTION The primary needs for any living being are the water,air,food,sheltered, for which water has the greatest importance water is required for various purposes such as drinking, cooking, bathing, washing, watering of lawns and garden or growing of crops for fire fighting, for heating and air-conditioning systems, etc. In the ancient times human required water for drinking, bathing, cooking etc. but with the advancement of civilization the utility of water enormously increased and now such a stage has come that without well- organized public water supply scheme it is impossible to run the present civic file and develop the towns. In ancient times every individual or family was responsible to arrange for their water supplies. There was no collective effort by the whole community for it. But as the community developed it became essential to have public water supply soon the inhabitants realized that their local sources of water supply such as shallow wells, springs, cisterns etc are inadequate to meet the demand of the town, they started to collect the water from distant large sources and conveyed it to the town through aqueducts, canals etc. when the concentration of town increases, it becomes very difficultto locate wells. In addition to these sourcesof water, having good quality was less readily available. These all situations led to the development of public water supply schemes. Water treatment involves science, engineering, business, and art. The treatment may include mechanical, physical, biological, and chemical methods. As with any technology, science is the foundation, and engineering makes sure that the technology works as designed. The appearance and application of water is an art. In terms of business, RGF Environmental, Water Energy Technologies, Aquasana Store, Vitech, Recalyx Industrial SDN BHD and PACE Chemicals ltd are some of many companiesthat offer various processes for water treatment. Millipore, a Fisher Scientific partner, offers many lines of products to produce
  8. 8. = ultrapure water, using a combination of active charcoal membranes, and reverse osmosis filter. Internet sites of these companies offer useful information regarding water. An environmental scientist or consultant matches the service provider, modify if necessary, with the requirement.  Natural Water includes some discussion on hard and soft water. Softening hard water for boiler, cooler, and domestic application is discussed therein. These treatments prepare water so that it is suitable for the applications.  Water Biology deals with water and biology. Drinking water is part of making water suitable for living. Thus, this link gives some considerations to drinking water problems.  There are many different industry types, and waters from various sources are usually treated before and after their applications. Pre- application treatment and water treatment offer a special opportunity or challenge. Only a general consideration will be given to some industrial processes.  General municipal and domestic water treatment converts used water (waste) into environmentally acceptable water or even drinking water. Every urban centre requires such a facility. SURVEY
  9. 9. = A Survey is an inspection of an area where work is proposed, to gather information for a design or an estimate to complete the initial tasks required for an outdoor activity. It can determine a precise location, access, best orientation for the site and the location of obstacles. The type of site survey and the best practices required depend on the nature of the project. Examples of projects requiring a preliminary site survey include urban construction,specialized construction (such as the location for a telescope) and wireless network design. The most important thing under the water supply schemes is the selection of source of water , which should be reliable and have minimum impurities. After the selection of source water, the next step is to construct intake works in order to collect the water and carry to the treatment plants, where this water will be treated. Type of treatment processes directly depend upon the impurities in water at the source and the quality of water required by the consumers. When the water is treated, it is stored in clear water reservoir , from where it is distributed to the consumers. The distribution system will also depend on the elevation of clear water reservoir and the elevavtion of distribution area. STRUCTURAL ELEMENTS
  10. 10. = The structural elements are those elements which form the supporting skeleton frame work of the supply scheme. It include the following things :- (i) Cascade type aerator (ii) Alum dose sand wash water tank (iii) Combined coagulation cum sedimentation tank (iv) Chlorinator (v) Clear water reservoir (vi) Main distribution tank (i) Cascade type aerator-since the raw water does not contain too much colour and odour, only nominal aeration is proposed by constructing an artistic cascade type aerator .This unit will help in main ting the wanted oxygen levels in water, remove dissolved iron and ,manganese, remove CO2 and H2S gases as well as the colour and tastes caused by volatile oils etc. released by algae and other micro-organisms. The raw water reaching the plant will be pumped into the the aerator tank through a 450mm dia pipe and will outflow through 4no. 250mm dia pipes. (ii) Alum dossers and wash water tank- Alumdossers is the appurtenance through which measured quantity of alum is added to the water supply system before of
  11. 11. = water. The requisite alum dose shall be mixed in two polyethylene tanks of 3000L each, which shall be interconnected. The raw water shall first be added in one tank where the alum dose of shallbemixed manually. The alum mixed solution shall flow into the other tank, where a float control will be produced to ensure constant depth of wall (iii) Combined coagulation cum sedimentation tank-It shall be constructed to allow formation of flocks and settlementof particles. (iv) Rapid gravity filter-Filter units shall be constructed to filter the sediment water, as usual,with provision of wash water tank. (v) Chlorinator-Considering the remote area and difficulty in transporting and storing the chlorine gas cylinders, it has been decided to use bleaching powder for disinfection by providing gravity typeof chlorinator. (vi) Clear water reservoir-It shall be constructed to store treated water , from where it will be lifted by 2nd stage pump house to deliver it to main distribution reservoir. This pumping is to bed for 16 hrs.
  12. 12. = (vii) Main distribution tank-The main distribution reservoir will obtain pumped water from 2nd stage pumped house and deliver it by gravity to service reservoirs of towns and villages for 18 hours per day. DESIGN SPECIFICATION
  13. 13. = (1) Design of pre-sedimentation tank- We shall design the tank to remove particles upto 0.1 mm size, and adopt the following general parameters. General parameters- 1. Overflow rate = 20 to 80 m3/m2/day 2. Minimum side water depth = 2.5m 3. Detention time = 0.5 to 3 hrs 4. Side slopes =10% from sides 5. Longitudinal slope =1% in rectangular tank 6. Settling velocity = 0.1mm Hydraulic Design Water required by the year 2030 =27.00 MLD =27.0*1000/24=1125m3/hr pumping is for 16 hrs, thus discharge = 1125*24/16=1687.5m3/hr Water loss in desludging = 2% Design average flow = 1687.5*100/(100-2)=1721.9388m3/hr Assume detention period = 1.5 hrs Effective storage of sedimentation tank= 1727.9388*1.5=2582.91m3 Assuming effective depth = 3.0 m Area of tank required =2582.91/3.0=860.97 m2 Assume L:B = 3:1 3B*B = 860.97 B= 16.941m L= 3*16.941=50.823 Provide a tank of size 50.823m*16.941m*3.5m (2) Design of raw water tank- A raw water storage tank of 8.0 hrs detention period is proposed.
  14. 14. = Water requirement for the year 2030= 27.0 MLD Taking 9.00 hrs detention period capacity of tank =(8.00*27.0)/24= 9000 m3 Provide water depth = 3.0 m Plan area of tank required 9000/3 = 3000 m2 Assume a ratio of L:B =2.5 : 1 2.5B*B= 3000 B=35.00M L= 87.00 M Provide a tank of size 87m*35m*3.5m (3) Design of cascade aerator- water requirement for the year 2030 = 27.00 MLD Average water requirement Q = 27*1000000/(24*3600*1000)m3/s Q = .3125m^3/s for broad crest Q=1.65B*H^1.5 Taking width B= 4 m H=13cm (4) Design of sedimentation tank- Overflow rate = 15 to 30 m3/d/m2 Minimum side water depth = 2.5 m Detention period for coagulated water = 2 to 4 hrs Weir loading = 300 m3/d/m2 Side slopes = 10 % Longitudinal slopes = 1% Settling velocity = 0.02 m
  15. 15. = Water required by the year 2030 = 27.0 MLD = 1125m3/hr Water lost in desludging = 2% Design average flow = 1125 * 100/(100 – 2)= 1147.95 m3/hr Assume a total detention period of 2.6 hrs Effective storage of sedimentation tank = 1147.95 * 2.6 = 2984.69m3 Assume effective depth = 3.0 m Area of the tank required A = 2984.69/3 = 994.89 m2 Assume L/B = 3 :1 3B*B = 994.89 m2 B = 18.2 m L = 54.6 m Provide a tank of size 54.6m * 18.2m* 3.0m (5) Design of rapid gravity filter- A-required flow of filtered water = 27.00 MLD B-quantity of back water used= 3% of filter output C-time lost during back washing = 30 min D-design rate of filtration =5.4 m3/m2/hr E-length width ratio = 1.25 F-size of perforation=9mm Filter water required =27.00 MLD Filtered water required per hour = 27.00*1000000/(24*10000)= 1125 m3/hr
  16. 16. = Design flow for filter after accounting for backwash water3% and washing time 0.5hr =1125*(1+0.03)*24/(23.5)= 1183.4 m3/hr Plan area of filter required = 1183.4/5.4 = 219.14 m2 Generally provide two filter units each of 112m2 Provide 2 units, each 12.5 m * 9m size (6) Designof chlorinator- Two polythene tanks,each of 3000L capacity, shall be installed to mix clear water with the requisite amount of bleaching powder, which water will flow by gravity into clear water tank to have contact period of more than ½ hr, as to cause disinfection. Normal dose of chlorine to be taken = 0.3 ppm for a contact period of 30 min Average daily demand of water of the year 2030 = 27.0 MLD Chlorine required per day =0.3*27.00*10^6/(10^6)kg= 8.1 kg since chlorine content in bleaching powder is 30% it means that 30 kg of chlorine in 100 kg of bleaching powder. Bleaching powder required per day = (8.1 *100)/(30)= 27 kg Annual consumption of bleaching powder = 27* 365 = 9855kg (7) Design of clear water reservoir- Treated water stored in clear water reservoir = 8.00 MLD Since pumping from clear water reservoir distribution tank is also 16 hrs, detention period of 8 hrs is provided in clear water reservoir to ensure 24 hrs supply capacity of clear water reservoir.
  17. 17. = Capacity of clear water reservoir = 27.00*10^6*10^-3 * 8.0/(24) = 9000m^3 Assume water depth as 3.0 m Plan area required =9000/3=3000 m^2 Assume a ratio of L:B of 2.5B*B=3000 B=35m L= 87m Thus , provide a tank of size 87m * 35m *3.0m (8) Design of main distribution tank- water requirement for the year 2030 =27.00 MLD Since the water will come in the tank in 16 hrs pumping from clear water reservoir and it will go to the service reservoirs in 18 hrs, a storage of 2 hrs is required on the tank. Capacity of tank required = 27.00*3*10^3/(24)=3375m3 Provide depth of water = 3.0 m Plan area of tank required =3375/3=1125m2 Assume L:B = 2.5 :1 2.5B*B=1125m^2 , B=21m, L=53m, ..provide a covered tank of size 53m*21m*3.5m as per detail DESIGN 1. Design of base slab- Let us assuming that – Angle of repose of soil = 30* Bearing capacity of soil = 160 KN/ 𝑚2
  18. 18. = Unit weight of soil = 16.8 KN/ 𝑚2 Height of wall above the ground level H = height of wall above the plinth + plinth height from ground level = 3+ 0.45 = 3.45 m Over all depth of roof slab D = 150 mm Live load on roof slab = 2000N/ 𝑚2 = 2 KN/ 𝑚2 2- Depth of base slab– By Ramekin’sformula, D = p/y × (1- sinα)^2 ÷(1-sinα)^2
  19. 19. = Here, D = 160/16.8 ×( 1- sin 30) ^2 ÷ ( 1- sin30) ^2 D = 1.06 m = 1.10 m (say) Hence adopt depth of foundation = 1.10 m 3- Load calculation – Self weight of wall per metre = L× B×H = 1×0.3×3.45×19.2 = 17.28 KN/m Width of slab = 2×t + 30 = 2×30 + 30 = 90 cm = 0.9 m 4- Design of roof slab – Size of room = 4.0 m × 3.5 m (clear inside dimension) Thickness of wall supporting slab = 300 mm = 0.3 00 m
  20. 20. = Live load on roof slab = 2000KN/ 𝑚2 The slab is simply supported on all four sides with corners not held down. Using M15 grade of concrete and mild steel Fe415. Design constants – For M15 grade of concrete and Fe 415 grade reinforcement then, fck = 5 N /m 𝑚2 Fst = 140 N/m 𝑚2 , m =19 Neutral axis factor, k = x÷ d = (m ×fck×d)÷ (mfck + fst) k = (19×5) ÷(19×5 + 140) = 0.404 Hence k = 0.404 Lever armfactor, j = (1- k)÷3 j = (1- 0.404)÷ 3 = 0.065 j = 0.065 Coefficient of moment of resistance, R = ( fck× j × k )÷2
  21. 21. = R = (5×0.865×0.404)÷2 R = 0.874 Let, Over all depth of slab = 150 mm Assuming,10mm dia. Main bars and 15 mm clear cover Hence, Effective depth of slab d = overall depth – clear cover – 0.5 × dia. of main bars = 150 – 0.5 ×10 = 130 mm Effective depth, d = 130 mm Length of room = width of room L = 8 , B = 3.5 Hence, L/B = 4/3.5 = 1.1228 < 2 Therefore slab will be designed as two way slab and effective span shall be smaller of the following – 1. (a) Centre to centre bearing = 4000 + 300/2 = 4150 mm
  22. 22. = = 4.150 m (b) Clear span + effective depth = 4000 + 130 = 4130 mm = 4.130 m 2. (a) Centre to centre bearing = 3500 + 300/2 = 3650 mm = 3.650m (b) Clear span + effective depth = 3500 + 130 = 3630 mm = 3.630 m Hence,effective span lx (shorter) = 3.630 m Effective sanely (longer) = 4.130 m Effective span = 3.630 m Load calculations – 1.(a) Due to self weight of 150 mm thick slab = 0.15 ×25000 = 3750 N/ 𝑚2 (b) Weight of 100 mm thick lime concrete = 0.100 × 19200
  23. 23. = = 1920 N/ 𝑚2 2. Live load = 2000 N/ 𝑚2 Hence , total load = 3750 + 1920 + 2000 = 7670 N = 7.670 KN Total load = 7670 N By Ramekin’s formula – (A) Weight on shorter span – Wb=( L^4 ×W) ÷ ( L^4 + B^4 ) = (4.130^4) × 7670 ÷ (4.130^4 + 3.630^4) = 4803.38 N (B) Weight on longer span, Wl = W – Wb = 7670 – 4803 Wl = 2867 N Bending moment – (a) Maximumbending moment on shorter span, Mb = (Wb× B^2) ÷ 8 = 4803× (3.630)^2 ÷ 8 = 7911.08 N-m Bending moment = 7911081 N-mm
  24. 24. = (b) Maximum bending moment on longer span, Ml = ( Wl× L^2)/8 = 2867 × (4.130^4) / 8 = 6112.76 N-m Bending moment = 6112760 N-mm Thickness of slab – d = ((moment ofresistance)÷(0.874× 1000))^0.5 d = ((7911081)÷(0.874×1000)) ^0.5 = 95.13 mm d = 100 mm (say) < 130 mm Therefore overall depth of slab = 150mm and Effective depth of slab = 130 mm Area of reinforcement along shorter span, = (Moment of resistance)÷ (fst× j × d) = (7911081) ÷ (140 × 0.865 × 130) = 502 mm^2 By using 10 mm of main steel bars – Area of one bar,Ast = (π × d^2) ÷ 4 = (π × 10^2) ÷ 4 = 78. 5 mm^2 And spacing = ( 1000 × 78.5)÷ 502
  25. 25. = = 156.37 mm = 150 mm (say) Hence, provide 10 mm diameter bars @ 150 C/C, Bend up alternate bar at L/7 = 3.630 / 7 = 0.518 mm = 520 mm from the centre of bars. Area of reinforcement along the span (Ly) perpendicular to the above span – Ast = (Moment of resistance) ÷ { j × (d – d‘ ) × st } = (6112760) ÷ {0.865 × (130 – 10)×140}= 420 mm^2 Centre to centre spacing of 10 mm diameter bars = (78.5×1000) ÷ 420 = 187 mm This should not be more than 3d or 300 mm, so 3 × 120 = 360 , or 300mm, Therefore centre to centre spacing = 190 mm Bend up alternate bars @ L /7 = (4.130) ÷ 7 = 590 mm Actual area of provided,Ast= (1000 × 78.54) ÷ 300
  26. 26. = = 261.08m 𝑚2 Check – 1. Shear force for shorter span – Vb = (W × B) ÷ 3 = (7670 × 3.630) ÷ 3 = 9280 N 2. Shear force for longer span – Vl = [L ÷ B] × W × B ÷[ 2 + ( L ÷ B ) ] = [4.130 ÷ 3.630] × 7670 × 3.630 ÷ [ 2 + (4.130 ÷ 3.630) ] = 10091.31 N Hence, τ = V ÷ (b × d) = 10091 ÷ (1000 × 130) = 0.077 N /mm^2 τ = 0.077 N/mm^2 The permissible shear stress τ * for M15 grade concrete; P = (100Ast) ÷ (b × d) = (100 × 261.08) ÷ (1000×130)= 0.21%And for slab 150 mm over all depth, from table = k × τ* = 1.30 × 0.21 = 0.273 N/ 𝑚2 But 0.273 >0.07; hence safe. Check for development length – 1. Considering , shorter span – M1 = (fst × Ast × x × j × d) ÷ 2 = 140 × 502 × 0.865 × 130 = 3951443 N-mm
  27. 27. = M1 = 3.95 × 10 ^6 N-mm So, Development length Ld ≤ [ 1.3 (M1/V) + Lo ] Anchorage length, Lo = 12 Φ or d (max.) = 12 × 10 or 130 mm = 130 mm Hence, Lo = 130 mm will be taken. Development length = (Φ × fst )÷ ( 4 τ*) = (10 × 140) ÷ (4× 0.6) = 583 mm Ld = 583 mm [ 1.3 (M1/V) + Lo ] =[ 1.3 ( 3.95×10^6) ÷ 9280 ] + 130 = 683 mm Since M1 / V + Lo ≥ Ld 58 Φ ≤ 683 Φ = 683/ 58 = 11.78 mm = 12mm (say)
  28. 28. = But dia. of main bar is 10 Φ, so bars safe in development. 2. Considering longer span of slab – M1 = (fst×Ast× y × z) ÷ 2 = (140 × 420 ×0.865 × 130) ÷ 2 = 3.3 × 10^6 N-m Ld = 583 mm 1.3 M1 + Lo =1.3 (3.3× 10^6 ÷ 100.91) + 130 = 555 mm Now , 58 Φ = 555 mm Φ = 9.58 mm (say = 10 mm)Since used bars are also 10 mm Φ , so , bars are safe DRAWING
  29. 29. = ESTIMATION METHOD OF ESTIMATION
  30. 30. = Long wall short wall method Centre Line Method Long Wall and Short Wall Crossing Method
  31. 31. = Length of long wall 𝑙1=c/c length + width of step Length of long wall 𝑙1=c/c length + width of step Length of short wall 𝑙2=c/c length -width of step 1 Raw water tank- Descriptionof work Measurements quantity RemarkNo. L (m) B (m) h/d Base slab Long wall 2 50.67 0.90 0.75 68.40 𝑚3 49.77+0.9 = 50.67 m Short wall 2 28.87 0.90 0.75 38.97 𝑚3 29.77 – 0.9 = 28.87 m
  32. 32. = Concrete work Long wall 4 18.37 0.60 0.60 26.45 𝑚3 17.77 +0.6 = 18.39 m Short wall 3 14.71 0.60 0.60 15.88 𝑚3 15.31 – 0.6 = 14.71 m Long wall 1 12.6 0.60 0.60 4.53 𝑚3 12 + .6 =12.60m Short wall 2 7.05 0.60 0.60 5.07 𝑚3 7.65 – 0.6 = 7.05m Other wall 1 18.54 0.6 0.60 6.67 𝑚3 Total 166 𝑚3 2 Cascade Aerator- Descriptionof work Measurements quantity RemarkNo. L (m) B (m) h/d
  33. 33. = Base slab Long wall 2 50.67 0.90 0.30 27.36 𝑚3 Short wall 2 28.87 0.90 0.30 15.58 𝑚3 Concrete Work Long wall 4 18.37 0.60 0.15 6.61 𝑚3 Short wall 3 14.71 0.60 0.15 3.97 𝑚3 Long wall 1 12.6 0.60 0.15 1.14 𝑚3 Short wall 2 7.05 0.60 0.15 1.26 𝑚3 Other wall 1 18.54 0.6 0.15 1.67 𝑚3 Total 57.6 𝑚3 3 SedimentationTank- Measurements
  34. 34. = Descriptionof work No. L (m) B (m) h/d quantity Remark 1ST STEP Long wall 2 50.34 0.57 0.15 11.26 𝑚3 49.77 +0.57 = 50.34 m Short wall 2 29.20 0.57 0.15 4.99 𝑚3 29.77- 0.57 = 29.20m Long wall 4 18.17 0.40 0.15 4.36 𝑚3 17.76 +.4 = 18.17m Short wall 3 14.91 0.40 0.15 2.68 𝑚3 Long wall 1 12.40 0.40 0.15 0.74 𝑚3 15.31-0.4 =14.91m Short wall 2 7.25 0.40 0.15 0.87 𝑚3 12+.4 =12.4m Other wall 1 18.54 0.40 0.15 1.12 𝑚3 7.65-0.4 = 7.25m 2ND STEP Long wall 2 50.23 0.46 0.30 13.86 𝑚3 49.77 +0.46 = 50.23 m Short wall 2 29.31 0.46 0.30 8.10 𝑚3 29.77- 0.46 = 29.31m Long wall 4 18.09 0.30 0.07 5 1.62 𝑚3 17.76 +.3 = 18.09m Short wall 3 15.01 0.30 0.07 5 1.01 𝑚3 15.31-0.3 =15.01m Long wall 1 12.30 0.30 0.07 5 0.276 𝑚3 12+.3 =12.3m Short wall 2 7.35 0.30 0.07 5 0.331 𝑚3 7.65-0.3= 7.35m Other wall 1 18.54 0.30 0.07 5 0.41𝑚3 3ND STEP Long wall 2 50.12 0.35 0.40 14.94𝑚3 49.77 +0.35 = 50.12 m Short wall 2 29.42 0.35 0.40 8.23 𝑚3 29.77- 0.35 = 29.42m 4 17.96 0.20 0.62 8.98 𝑚3 17.76 +.2 =
  35. 35. = Long wall 5 17.96m Short wall 3 15.11 0.20 0.62 5 5.66 𝑚3 15.31-0.2 =15.11m Long wall 1 12.20 0.20 0.62 5 1.525 𝑚3 12+.2 =12.2m Short wall 2 7.45 0.20 0.62 5 1.86 𝑚3 7.65-0.2= 7.45m Other wall 1 18.54 0.20 0.62 5 2.138𝑚3 TOTAL 94.33𝑚3 4 Rapid Gravity Filter- Descriptionof work Measurements quantity RemarkNo. L (m) B (m) h/d Long wall 2 50.12 0.35 ----- 35.08𝑚2 Short wall 2 29.42 0.35 ----- 20.59𝑚2 Long wall 4 17.96 0.20 ----- 14.37𝑚2 Short wall 3 15.11 0.20 ----- 9.06𝑚2 Long wall 1 12.20 0.20 ----- 2.44𝑚2 Short wall 2 7.45 0.20 ----- 2.98𝑚2 Other wall 1 18.54 0.20 ----- 3.70𝑚2 TOTAL 88.247𝑚2 5 Chlorinator- Measurements
  36. 36. = Descriptionof work No. L (m) B (m) h/d quantity Remark Long wall 2 50.0 0.23 3.0 69.00𝑚3 49.77 +0.23 = 50.0 m Short wall 2 29.54 0.23 3.0 40.76 𝑚3 29.77- 0.23 = 29.54m Long wall 4 17.65 0.11 5 4.75 38.57 𝑚3 17.76 +..115 = 17.65m Short wall 3 15.19 0.11 5 4.75 24.90 𝑚3 15.310.115=1 5.19m Long wall 1 12.115 0.11 5 4.75 6.618 𝑚3 12+.115 =12.115m Short wall 2 7.353 0.11 5 4.75 8.232 𝑚3 7.65-0.115= 7.535m Other wall 1 18.54 0.11 5 4.75 10.12𝑚3 TOTAL 198.217 𝑚3 6 Clear water reservoir at WTP- Descriptionof work Measurements quantity RemarkNo. L (m) B (m) h/d RCC WORK 1 18.626 15.8 0.10 29.485𝑚3 L= 17.76+.30+.3+. 23= 18.626
  37. 37. = 3 B= 15+.23+.23+.2 3= 15.83 7 PLASTER WORK Descriptionof work Measurements quantity RemarkNo. L (m) B (m) h/d Long wall 4 50.0 ------ 3.0 600𝑚2 Short wall 2 29.54 ------ 3.0 354.48𝑚2 Long wall 4 17.65 ------ 4.75 670.85𝑚2 Short wall 3 15.19 ------ 4.75 433𝑚2 Long wall 1 12.115 ------ 4.75 115.09𝑚2 Short wall 2 7.353 ------ 4.75 139.39𝑚2 Other wall 1 18.54 ------ 4.75 176.13𝑚2 TOTAL 2488.97 𝑚2 8 TOTAL MASSONARYWORK T. MASSONARY WORK 94.339 𝑚3
  38. 38. = Estimation (Material cost) S. N0. Items quantity Rate (Rs.) Cost (Rs.) 1 Cement 19850 bags 350/bag 6947500 2 L.Sand 987 m3 1000/m3 987000 3 C.Sand 829 m3 2100/ m3 1740900 4 Bricks 1124100 5000/1000bricks 224820
  39. 39. = 5 Grit 524 2200/ m3 1152800 6 Steel 15123 Kg 45/ Kg 680535 Total material cost in Rs. Rs. 11733555/- (Labour cost) S.No. Items Quantity Rate Cost (Rs.) 1 Base slab 166 m3 70/ m3 11620 2 P.C.C. 57.6 𝑚3 320/ m3 18432 3 Flouring 1552.37 𝑚2 65/𝑚2 100904.05 4 Brick work 88.46 m3 530/ m3 46883.8 5 Staging for brick work 2488.97𝑚2 16/𝑚2 39823.52 6 Shuttering (slab) 378.22𝑚2 130/𝑚2 49169.69 7 Plastering 15mm 2488.97𝑚2 35/𝑚2 87113.95
  40. 40. = 12mm 2488.97𝑚2 42/𝑚2 104536.74 8 R.C.C. 29.485 510/ m3 15037.35 Total 473521.10 Total cost of project COST OF PROJECT Total cost of project – Total civil works = Rs. 985261.35/- Adding – (a)- 20% of civil work for necessary (electric, water,supply, sanitary fittings ) = Rs. 224381.27/-
  41. 41. = (b). 0.5% of civil work for survey work = Rs. 5609.6/- (c). 4% of civil work for internal road = Rs. 4487.68/- (d). 2% of civil work for work charge establishment = Rs. 3244/- (e). 3% of civil work for contingencies = Rs. 5365.76/- (f). 10% of civil work for contractor profit = Rs. 102190.135/- Net total cost = Rs. 13608387.79/-
  42. 42. = FIG- SHOWING VARIOUS UNIT OF WATER SUPPLY PROJECT
  43. 43. = FIG- SHOWING RAW WATER TANK
  44. 44. = FIG- SHOWING PRE-SEDIMENTATION TANK
  45. 45. = FIG- SHOWING RAW WATER TANK
  46. 46. =
  47. 47. =
  48. 48. =
  49. 49. =
  50. 50. =
  51. 51. =
  52. 52. =
  53. 53. =

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