Sheet Pile Wall Design and Construction: A Practical Guide for Civil Engineer...
final year Project (civil)
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. =
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. =
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. =
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. =
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. =
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. =
“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. =
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. =
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. =
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. =
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. =
(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. =
(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. =
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. =
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. =
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. =
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. =
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. =
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. =
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. =
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. =
= 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. =
= 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. =
(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. =
= 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. =
= 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. =
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. =
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
30. =
Long wall short wall method
Centre Line
Method
Long Wall
and Short
Wall
Crossing
Method
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. =
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. =
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. =
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. =
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. =
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. =
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
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. =
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. =
(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/-
