INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT
Bilhaur Kanpur- 209202
A
Project report On
Automobile Workshop
“
”
In partial fulfillment for the award award of the degree Of BACHELOR OF TECHNOLOGY In CIVIL ENGINEERING
Submitted to:
Under Guidance of:
Mr. S.K. Mishra
Mr. Shiv Mohan Dubey
(H.O.D. Of civil engineering)
(Lecturer,, Dept. Of Civil Engineering) (Lecturer
INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT
Bilhaur Kanpur- 209202
A
Project report On
Automobile Workshop
“
”
In partial fulfillment for the award award of the degree Of BACHELOR OF TECHNOLOGY In CIVIL ENGINEERING
Submitted to:
Submitted by:
Mr.. S.K. Mr S.K . Mishra Mis hra
Balvant Chauhan
(H.O.D. Of civil engineering)
Vijay Bahadur Bind Vikram I.P I.P.. Singh Si ngh
Under Guidance of:
Mr. Shiv Mohan Dubey (Lecturer,, Dept. Of Civil Engineering) (Lecturer
Vivek Kumar Vivek Kr. Kr. Singh Si ngh Vivek Yadav
Name of Project
Automobil A utomobile e Workshop Workshop (Maruti Suzuki workshop at Jarib Cahuki, Kanpur)
INDEX
S.No.
Title
Page No.
1
Certificate
1
2
Object
2
3
Survey
3
4
Drawing
4
5
Estimation
6
6
Rate of cost
7
OBJECT
To prepare a project of AUTOMOBILE AUTOMOBILE WORKSHOP WORKSHOP (Maruti Suzuki workshop) in the plot of 50m X 30m at Kanpur.
The objectives of the project are:-
Carrying out a complete analysis and designing of the main structural elements of an Automobile Workshop including slabs, columns, and beams. The structure should be able to accommodate all the machineries as well as all the required equipments equipments needed in the auto mobile work shop. Use structural software (AutoCAD) (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.
INTRODUCTION
BUILDING A building can be defined as a structure which consists of walls, floors, and roofs, erected to provide covered space for different uses such as
Residence Education Business Storage Workshop Worship
TYPES OF BUILDING Buildings may be divided into following parts:(i) (ii)
Residential apartment building. Office building.
(i)
Residential Residential apartment building Apartment buildings are multi-story buildings where three or more residences are contained within one structure.
(ii)
Office building The primary purpose of an office building is to provide a workplace and working environment. environment.
AUTOMOBILE AUTOMOBILE WORK WORK SHOP:The automobile workshop is a civil structure which is basically work place for repairing and servicing of the automobiles. Or we may saygarage) is a repair “An automobile repair shop (also known as a garage) shop where automobiles are repaired by auto mechanics and electricians.”
STRUCTURAL ELEMENTS
The structural elements are those elements which form the supporting skeleton skeleton frame work work of the building. building. It include the following things :(i) (ii) (iii) (iv)
Foundation Column Beam Slab
(i)
Foundation- There main function is to transmit the building load to the ground safely.
(ii)
Column- There main function is to transmit the beam load to the foundation and to resist the lateral loads. It also hold the building in straight position.
(iii)
Beam- there main function is to hold the slab load and to transmit them to the column.
(iv)
Slab- there main function is to support the load resting on it and to transfer these load to the beam.
Detail specifications
The main items of work in a load bearing wall type of construction may be given as:– (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii)
Excavation P.C.C Foundation Masonry in foundation Filling in plinth Damp proof course Flooring Masonry work in super structure Doors and window Roof Plaster White washing
(i)
Excavation:- a trench of desired length, width and depth is excavated in order to prepare a foundation for the building.
(ii)
Plain Cement Concrete: - PCC is a construction material generally used as binding materials and is composed of cement, (commonly Portland cement) and a nd
other cementations materials such as fly ash and slag cement, aggregate water, and chemical admixtures.
(iii)
Foundation: – It is necessary to provide a bed of lean concrete under the main walls. The load of super structure is distributed by foundation to the ground.
(iv)
Masonry in foundation: – Load bearing wall type of construction can have walls made out of bricks or stone. Hence specifications for these items of work comprise of brick masonry in cement mortar.
(v)
Filling in plinth:- Back filling of earth of done till the plinth level. This prevents the surrounding water to entre in the building because the plinth level is quiet higher than that of the ground level.
(vi)
Damp proof course (DPC): – Damp proof course is provided provided at plinth level of the building check the rise of moisture from ground to super structure. The specifications for this item of work depend upon the intensity of rainfall in the area, the depth of sub-soil water-table and quality. For this work 35 mm thick cement concrete M15 is provided. provided.
(vii)
Flooring :– flooring is provided to have a hard, durable smooth, easy to clean, impervious and dust free surface and the materials used in flooring should be strong enough to resist its wear and tear that take place during used. It is normally to provide provide a layer of base concrete under ground floor over 15 cm layer of sand over over earth filling in plinth.
(viii)
Masonry work in super structure:- The outer walls of 230mm thickness and the inner walls of 115mm are
constructed to provide privacy, safety and strength to the structure along with providing providing the real look to building. (ix)
Doors and windows: – Doors are provided to perform the function of access to a building and for intercommunication between different rooms. Generally Generally normal height of door is 2100mm and width is 900mm for residential building.
(x)
Roof: – Depending upon the climatic conditions, building can have a sloping roof or a flat roof. Flat roof can be in the form of cast – in situ RCC slabs or precast units having minimum thickness 100 mm.
(xi)
Plaster: - Coating of the fare faced wall by the 12mm thick mortar of cement and sand, and a nd rough faced walls by 20mm thick mortar.
(xii)
White washing: - Washing of plastered walls by the proper mixture of the lime and water is done to provide the base for the paints which is supposed to be used in the building.
S.no 1 2 3 4 5 6 7 8 9 10
Operation Survey, Survey, Design and Layout Construction of foundation construction of boundary wall Construction of super structure Roofing Flooring Fixing of doors and windows frame Plastering Carpentry work White washing of walls and painting
Time (in days) 3 5 3 11 5 2 4 4 4 3
BAR CHART
0
S. no.
Description of activity
1
Survey design and layout
2
Construction of foundation
3
construction of boundary wall
4
Construction of super structure
5
Roofing
6
Fixing of doors and windows frame
5
10
15
20
25
30
35
40
45
7
8
9
10
Plastering
Flooring Carpentry work White washing and painting
Surveying
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.
design
1. Design of wall foundation – Let us assuming that – Angle of repose of soil = 30* Bearing capacity of soil = 160 KN/ 2 Unit weight of soil = 16.8 KN/ 2 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/ 2000 N/ 2 = 2 KN/ 2
2- Depth of foundation – By Ramekin’s formula ,
D = p/y × (1- sin α)^2 ÷(1-sinα)^2
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 foundation = 1.10 m
3- Load calculation – Self weight weight of wall per metre = L× B×H = 1×0.3×3.45×19.2 = 17.28 KN/m Width of foundation = 2×t + 30 = 2×30 + 30 = 90 cm = 0.9 m
4- Design of roof slab – Size of room room = 4.0 m × 3.5 m (clear inside dimension) dimension) Thickness of wall supporting slab = 300 mm = 0.3 00 m Live Live load on roof slab = 2000KN/ 20 00KN/ 2 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 2 Fst = 140 N/m 2 , m =19
Neutral axis factor , k = x ÷ d = (m ×fck×d)÷ ×fck×d)÷ (mfck + fst) k = (19×5) ÷(19×5 + 140) = 0.404 Hence k = 0.404 Lever arm factor factor , j = (1- k)÷3 j = (1- 0.404)÷ 3 = 0.065
j = 0.065 Coefficient of moment of resistance, resistance, R = ( fck × j × k )÷2 R = (5×0.865×0.404)÷2
R = 0.874 Let , Over all depth of slab = 150 mm Assuming , 10mm dia. Main bars and and 15 mm clear cover cover Hence , Effective Effective depth of slab, slab, d = overall 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 Therefore slab will be designed as two way slab and effective span shall be smaller of the following – 1. (a) Centre to to centre bearing bearing = 4000 + 300/2 = 4150 mm = 4.150 m (b) Clear span + effective depth = 4000 + 130 = 4130 mm = 4.130 m 2. (a) Centre Centre to centre bearing = 3500 + 300/2 = 3650 mm
= 3.650m (b) Clear span + effective effective depth = 3500 + 130 = 3630 mm = 3.630 m Hence, effective effective span lx (shorter) = 3.630 m Effective Effective span ly (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/ 2 (b) Weight Weight of 100 mm thick lime concrete = 0.100 × 19200 = 1920 N/ 2 2. Live load = 2000 N/ 2 Hence , total load = 3750 + 1920 + 2000 = 7670 N = 7.670 KN Total load = 7670 N
By Ramkine’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) Maximum bending moment on shorter span, Mb = (Wb × B^2) ÷ 8 = 4803 × (3.630)^2 ÷ 8 = 7911.08 N-m
Bending moment = 7911081 N-mm (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 of resistance ) ÷ ( 0.874 × 1000))^0.5 d = ( ( 7911081) ÷ ( 0.874 ×1000))^0.5 = 95.13 mm d = 100 mm (say) < 130 mm Therefore Therefore overall overall depth of slab = 150 mm and effective 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 = 156.37 mm = 150 mm (say) Hence , provide 10 mm diameter bars @ 150 C/C , Bend up alternate alternate bar at L/7 = 3.630 / 7 = 0.518 mm = 520 mm from the centre of bars.
Area of reinforceme reinforcement nt along the span (Ly) (Ly) perpendicular perpendic ular 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 Therefore centre to centre spacing = 190 mm Bendup alternate bars @ L /7 = (4.130) ÷ 7 = 590 mm Actual area of provided , Ast = (1000 × 78.54) ÷ 300 = 261.08m 261.08m 2
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 depth , from table = k × τ* = 1.30 × 0.21 = 0.273 N/ 2 But 0.273 > 0.077 ; 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 M1 = 3.95 × 10 ^6 N-mm
So, Development 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 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) 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.
METHOD OF ESTIMATION ESTIMATION
METHOD OF ESTIMATION ESTIMATION
Centre Line Method
Long Wall and Short Wall
Crossing Method
Long wall short wall method
Length of long wall 1 =c/c
length + width of step
Length of short wall 2 =c/c
length -width of step
1- Excavation work Measurements Description of work
No.
quantity
L (m)
B (m)
h/d
50.67
0.90
0.75 68.40 3
2
28.87
0.90
0.75 38.97 3
4
18.37
0.60
0.60 26.45 3
3
14.71
0.60
0.60 15.88 3
Short wall Long wall
1
12.6
0.60
0.60 4.53 3
Short wall
2
7.05
0.60
0.60 5.07 3
Other wall
1
18.54
0.6
0.60 6.67 3
Earth work (boundary work (boundary walls) Long wall 2
Short wall
remark
49.77+0.9 = 50.67 m 29.77 – 0.9 = 28.87 m
Earth work (inner work (inner walls) Long wall
17.77 +0.6 = 18.39 m 15.31 – 0.6 = 14.71 m 12 + .6 =12.60m 7.65 – 0.6 = 7.05m
166 3
Total
2-
P.C.C. .C.C . WORK IN FOUNDATION
Measurements Description of work
P.C.C. (boundary walls) Long wall
quantity
No.
L (m)
B (m)
h/d
2 2
50.67 28.87
0.90 0.90
0.30 0.30
27.36 3 15.58 3
4
18.37
0.60
0.15
6.61 3
3
14.71
0.60
0.15
3.97 3
1 2 1
12.6 7.05 18.54
0.60 0.60 0.6
0.15 0.15 0.15
1.14 3 1.26 3 1.67 3 57.6 3
Short wall P.C.C. (inner walls) Long wall Short wall Long wall Short wall Other wall Total
remark
3-
BRICK MASSONARY IN FOUNDATION FOUNDATION Measurements
Description of work
1ST STEP Long wall
No.
L (m)
B (m)
h/d
quantity remark
2
50.34
0.57
0.15
11.26 3
2
29.20
0.57
0.15
4.99 3
4
18.17
0.40
0.15
4.36 3
Long wall Short wall Long wall
3 1
14.91 12.40
0.40 0.40
0.15 0.15
2.68 3 0.74 3
Short wall Other wall
2 1
7.25 18.54
0.40 0.40
0.15 0.15
0.87 3 1.12 3
2
50.23
0.46
0.30
13.86 3
2
29.31
0.46
0.30
8.10 3
4
18.09
0.30
1.62 3
Long wall Short wall
3
15.01
0.30
Long wall
1
12.30
0.30
Short wall
2
7.35
0.30
Other wall
1
18.54
0.30
0.07 5 0.07 5 0.07 5 0.07 5 0.07 5
2
50.12
0.35
Short wall
49.77 +0.57 = 50.34 m 29.77- 0.57 = 29.20m 17.76 +.4 = 18.17m 15.31-0.4 =14.91m 12+.4 =12.4m 7.65-0.4 = 7.25m
2ND STEP Long wall
Short wall
1.01 3 0.276 3 0.331 3
49.77 +0.46 = 50.23 m 29.77- 0.46 = 29.31m 17.76 +.3 = 18.09m 15.31-0.3 =15.01m 12+.3 =12.3m 7.65-0.3= 7.35m
0.413
3ND STEP Long wall
0.40
14.943
49.77 +0.35 = 50.12 m
2
29.42
0.35
0.40
8.23 3
4
17.96
0.20
8.98 3
Long wall Short wall
3
15.11
0.20
Long wall
1
12.20
0.20
Short wall
2
7.45
0.20
Other wall
1
18.54
0.20
0.62 5 0.62 5 0.62 5 0.62 5 0.62 5
Short wall
1.525 3 1.86 3
7.65-0.2= 7.45m
2.1383 94.333
TOTAL
4-
5.66 3
29.77- 0.35 = 29.42m 17.76 +.2 = 17.96m 15.31-0.2 =15.11m 12+.2 =12.2m
D.P.C. WORK Measurements
Description of work
Long wall
quantity
No.
L (m)
B (m)
h/d
2 2
50.12 29.42
0.35 0.35
---------
35.08 2 20.59 2
4
17.96
0.20
-----
14.37 2
Short wall Long wall
remark
Short wall Long wall Short wall Other wall
3 1 2 1
15.11 12.20 7.45 18.54
0.20 0.20 0.20 0.20
-----------------
TOTAL
9.06 2 2.44 2 2.98 2 3.70 2 88.247 2
5-
MASSONARY WORK IN SUPER STRUCTURE Measurements
Description of work
quantity
remark
3.0
69.003
0.23
3.0
40.76 3
0.11 5 0.11 5 0.11 5 0.11 5 0.11 5
4.75
38.57 3
4.75
24.90 3
4.75
6.618 3
4.75
8.232 3
49.77 +0.23 = 50.0 m 29.77- 0.23 = 29.54m 17.76 +..115 = 17.65m 15.310.115=1 5.19m 12+.115 =12.115m 7.65-0.115= 7.535m
4.75
10.123
No.
L (m)
B (m)
h/d
2
50.0
0.23
2
29.54
4
17.65
Long wall Short wall
3
15.19
Long wall
1
Short wall
2
12.11 5 7.353
Other wall
1
18.54
Long wall
Short wall
TOTAL
198.217
6-
3
R.C.C. WORK Measurements
Description of work
RCC WORK
No.
1
L (m)
18.62
B (m)
15.8
h/d
0.10
quantity
remark
29.485
L= 17.76+.30+.3 +.23= 18.626
6
7-
3
B= 15+.23+.23+. 23= 15.83
STEEL ROOF TRUSS
STEEL IN ROOF TRUSS
8-
3
@880 Kg/truss with Purling and 5280 Kg accessories
6
PLASTER WORK Measurements
Description of work
Long wall Short wall
No.
L (m)
B (m)
h/d
4 2
50.0 29.54
-----------
3.0 3.0
quantity
remark
600 2 354.48 2
Long wall
4
17.65
------
4.75
670.85 2
Short wall Long wall Short wall
3 1 2
15.19 12.11 5 7.353
-----------
4.75 4.75
433 2 115.09 2
------
4.75
139.39 2
Other wall
1
18.54
------
4.75
176.13 2
TOTAL
2488.97 2
9-
FLOORING WITH TERA COTA STONE Measurements
Description of work
No.
L (m)
B (m)
h/d
quantity
remark
L = 50.0-
FLOOR AREA
-----
49.54
29.5 4
------
1463.41
2X.23 = 49.54m
2
B = 30 – 2X.23= 29.54 m No. Of terracotta @0.362 /stone
1463.41/0.36= 4065 Nos.
10- DEDUCTIONS Measurements Description of work
DOOR (D1) DOOR (D2) DOOR (D3) DOOR (D4) DOOR (D5) VENTILATOR(V1) TOTAL
No.
L (m)
B (m)
h/d
2 4 1 1 1 1
0.70 0.90 2 3 2 0.6
0.23 0.23 0.23 0.23 0.23 0.23
2.10 2.10 2.10 2.10 2.10 0.6
quantity
remark
0.6763 1.7393 0.9663 1.4493 0.9663 0.0833 5.8793
11- SKIRTING AND DADO Measurements Description of work
SKIRTING SKIRTING
No.
L (m)
B (m)
h/d
-----
158.1 6
------ 0.20
quantity
remark
31.63 2
2x(49.54+29. 54)=158.16m
DADO TOTAL
2
18
------
1.50
54 2 85.63 2
4x(3+6)=18m
12- TOTAL TOTAL MASSONARY MASS ONARY WORK 94.339 – 5.879 = 88.963
T. MASSONARY WORK
13- TOTAL TOTAL NO. OF TERA COTTA COTTA STONE @0.362 /stone
T. NO. STONE
88.96/0.36 =247.11
14- NO. OF ASBESTOR SHEET NEEDED @3.152 /sheet
T. NO. ASBESTOS ASB ESTOS SHEET
453.846/3.15 = 144.07
MATERIAL COST S. N0.
Items
quantity
Rate (Rs.)
Cost (Rs.)
1
Cement
770 bags
350/bag
269500
2
L.Sand
77
m3
1000/m 3
77000
3
C.Sand
29
m3
2100/ m3
60900
4
Bricks
4100
5000/1000bricks
20500
5
Grit
27
2200/ m3
59400
6
Steel
5280 Kg
45/ Kg
237600
7
Asbestos sheet
145 sheet
Total material cost in Rs.
Rs. /-
LABOUR COST
S.No.
Items
Quantity
Rate
Cost (Rs.)
1 2 3 4 5
Earth work P.C.C. Flouring Brick work Staging for brick work Shuttering (slab)
166 m3
2488.97 2
70/ m3 320/ m3 65/2 530/ m3 16/2
11620 18432 100904.05 46883.8 39823.52
378.222
130/2
49169.69
2488.97 2
35/2
87113.95
42/2 510/ m3
104536.74 15037.35 473521.10
6 7
Plastering 15mm
57.6 3
1552.37 2
88.46 m3
2
8
12mm 2488.97 29.485 R.C.C. Total
COST OF PROJECT
Total cost of project projec t – Total civil works = Rs. 1182954.97 Adding – (a). 20% of civil work for necessary (electric, water,supply, sanitary fittings ) = Rs. 236590.97 (b). 0.5% of civil work for survey work = Rs. 5913.27 (c). 4% of civil work for internal road = Rs. 47318.19 (d). 2% of civil work for work charge establishment = Rs. 23659.86 (e). 3% of civil c ivil work for contingencies = Rs. 35488.64 (f). 10% of civil c ivil work for contractor profit = Rs. 118295.50
Net total cost =
Rs. 1650220.62
(only
