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1. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Measurement and its conversion
1 Mile = 1609.75 Meter
1 Meter = 1000 mm
1 Meter = 100 cm
1 Foot = 30 cm
1 Inch = 2.54 cm
1 Foot = 12 inch
1 inch = 25.4 mm
1 foot = 0.0254 m
1 inch = 0.064 m
1 Foot = 300 mm
1 Foot = 1/12 = 0.304 m
1 Meter = 3.281 Feet
1 Meter = 39.37 inch
1 Inch = 0.083 foot
1 Inch = 1/39.37 = 0.025
1 Acre = 2 Jarib
1 Acre = 8 Canal
1 Canal = 20 Marla
1 Marla =272.5 Sq. /ft.
1 inch = 8 sutar
1 ton = 1000 kg
1 ton = 2204 Lbs
1 CFT = 490 Lbs
1 CUM = 7850 kg
1 Lbs = 2.204 kg
1 Meter = 10 Desimeter
1 Dm = 10 cm
1 cm = 10 mm
1 M3
= 35.32 Ft3
1 M2
= 10.76 Ft2
1 Mile = 5280 Ft
1 Mile = 8 Furlong
1 Mile = 1760 Yards
1 Furlong = 220 Yards
1 Yard = 3 Feet
1 inch = 0.083 Ft
2 inch = 0.16 Ft
3 inch = 0.25 Ft
4 inch = 0.33 Ft
5 inch = 0.41Ft
6 inch = 0.50 Ft
7 inch = 0.58 Ft
8 inch = 0.66 Ft
9 inch = 0.75 Ft
10 inch = 0.83 Ft
11 inch = 0.91 Ft
12 inch = 1 Ft
1 Mile = 0.621504 Acre
1 Acre = 0.4047 Hector
1 Canal = 20 Marla
1 Jerab = 4 Canal
1 Muraba = 25 Acre

2. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
1 cm = 0.3937 inch 1 Decigram = 10 Centigram
1 Decagram = 10 Gram
1 Hectogram = 10 Decigram
1 kg = 10 Hectogram
1 kg = 2.204 Pounds
1 pound = 16 Ounce
1 Ton = 20 Maan
1 Ton = 1000 kg
10 cm = 1 decimeter
1 Maan = 50 kg
10 decimeter = 1 m
10 m = 1 Decameter
10 Decameter = 1 Hectometer
10 mm = 1 cm
10 Milliliter = 1 Centiliter
10 Centiliter = 1 Decalitre
10 Deciliter = 1 Liter
10 liter = 1 Decaliter
10 Decaliter = 1 Hectoleter
10 Hectoliter = 1 Kiloliter
100 Square meter = 1 Acre
1000 Sq.m = 1 He acre
100 kg = 1 Cointal
10 Cointol = 1 Ton
10 Hectometer = 1 Km
1 Yard = 0.914 m
1 m = 1.09 Yard
1 Meter Square = 10.76 Sft
1 meter cube = 35.32 Cft
1 Marla = 9 Sarsai
1 Sarsai = 30.25 Sft
1 Canal = 5445 Sft
1 Pound = 0.453 kg
1 kg = 9.81 Newton
1 Kilo newton = 1000 Newton
1 Liter = 0.93 kg
1 Kips = 0.93 kg

3. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
RATE ANALYSIS
I. No of bricks in 1 Cft = 13.5 or 14.2
II. No of bricks in 1 cum = 500 bricks
III. Volume of 1 brick in Foot system = 0.0703 Cft
IV. Volume of one brick in meter system = 0.002 Cum
V. Dry mortar used in masonary work = 30 %
VI. Wet mortar convert to dry mortar then multiplying with 1.27
VII. Wet concrete convert to dry multiply with 1.54
VIII. Volume of 1 cement bag = 1.25 Cft
IX. Volume of one cement bag = 0.035 Cum
X. Weight of steel in 1 Cft = 490 Lbs
XI. Weight of steel in 1 Cum = 7850 kg
XII. Volume of block = 0.3075

4. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
To Find Quantity In A Foot System
Q No 1: Determine The Quantities Of Various Materials Used In 1804 Cft Brick Works In
Cement Mortar Ratio 1:4.
Solution: Quantity Of Bricks = 1805 Cft
Find Bricks Cement And Sand
1: Bricks
Quantity Of Bricks = 1805 Cft
Volume Of One Brick = 0.0703 Cft
No Of Bricks In 1805 Cft = 1805/0.0703 =25675 Bricks
2: Cement
Dry Mortar Used In Masonary = 30 %
Quantity Of Mortar = 1805 x 30/100 = 541.5 Cft
Sum Of Ratio = 1 + 4 = 5
Cement = 1/5 X 541.5 = 108.3 Cft
Convert It To Bags One Bag Volume = 1.25 Cft
No Of Bags In 108.3 Cft = 108.3/1.25 = 86.64 Or 87 Bags
3: Sand
4/5 x 541.5 = 433.2 Cft
Abstract Of Cost
Bricks brick 1 trip = 12’000 Rs
In one dumper 2000 bricks
2000/2000 = 12000/2000 = 6 Rs 1 brick = 6 Rs
No of bricks= 25675 x 6 = 2’14’050 Rs
Cement Cement 1 trip = 80’000
Price of 1 bag = 400 400/400 = 80’000/400 = 200 1 Cft = 200 Rs

5. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
No of Cement Bags = 87 87 x 200 = 17400 Rs
Qs: Determine the Quantities of various material to prepare 10 Cum concrete
Ratio 1:2:4.
Sol: Quantity of Material = 10 Cum
Quantity of Dry Material = 10 x 1.54 = 15.4
Ratio of concrete = 1:2:4
Sum of ratio: 1 + 2 + 4 = 7
Cement: 1/7 x 15.4 = 2.2 Cum One bag = 0.035 cum
2.2/0.035 = 62.86 Bags Sand: 2/7 x 15.4 = 4.4 Cum
Bajri: 4/7 x 15.4 = 8.8 Cum
 ________________________$_______________________________________*
D.P.C
Qs: Quantity of D.p.c = 97 sqm thichkness of d.p.c = 2.5 cm Ratio: 1:2:4 Find
The Quantity Of Various Materials.
Sol: Qty of D.p.c = 97 sqm Change it to volume
Thickness of D.p.c + 2.5 cm Change it to meter
2.5/100 = 0.025 Cum
Volume = 97 x 0.025 2.42 Cum
Convert It to Dry = 2.42 x 1.54 = 3.73 cum
Mixing Ratio: 1:2:4
Sum of Ratio = 1 + 2 + 4 = 7
Cement: 1/7 x 3.73 = 0.532 Cum
1 bag = 0.035
0.532/0.035 = 15.2 Bags
Sand: 2/7 x 3.73 1.06 Cum
Bajjri: 4/7 x 3.73 = 2.13 Cum

6. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Sand: Sand 1 trip = 2000 Rs
Volume Of Tracter Trally = 150 Cft & Volume of dumper = 750 Cft
We Take The Volume of Trally 150/150 = 2000/150 = 13.33 Cft
433.2 x 13.33 = 5774.55 Rs
Meter System
Qs: Determine The Quantities Of Various Materials Used In 81 Cum Brick Work In
Cement Mortar 1:5.
Sol: Quantity Of Brick = 81 Cum Mortar Mixing Ratio = 1:5
No of bricks in one Cum = 500 Volume of one brick = 0.002 Cum
No of bricks = 81/0.002 = 40500 bricks or 500 x 81 = 40500 Bricks
Quantity of Mortar = 81 x 30/100 = 24.3 Cum
Sum of Ratio = 1 + 5 = 6 Cement: 1/6 x 24.3 = 4.05 Cum
Volume of 1 bag = 0.035 No of bags = 4.05/0.035 = 115 bags
Sand: = 5/6 x 24.3 = 20.24 Cum
 _________________________________$_______________________________*
Blocks
Qs: Determine the Quantity of Various materials used in 113 cft block work in
Cement mortar 1:4.
Sol: Quantity of blocks = 113 Cft Mortar Mixing Ratio = 1:4
Blocks: Volume of one block = 0.3075 Cft
N of Blocks in 113 Cft + 113/0.3075 = 368 Blocks
Quantity of Mortar = 113 x 30/100 = 33.9 Cft Sum Of Ratio = 1 + 4 = 5
Cement: 1/5 x 33.9 = 6.78 Cft One bag = 1.25 Cft 6.78/1.25 = 5.24 Bags
Sand: 4/5 x 33.9 = 27.12 Cft
 ______________________$____________________________________*
Concrete Work

7. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Qs: Determine The Quantities of Various Materials to Prepare 100 Cft
Concrete Ratio 1:4:8.
Sol: Quantity of Material = 100 x 1.54 = 154 Cft
Sum of Ratio: 1 + 4 + 8 = 13
Cement: 1/13 x 154 = 11.84 Cft
Convert To Bags 1 bag = 1.25 Cft
11.78/1.25 = 9.47 Bags
Sand: 4/13 x 154 = 47.38 Cft
Bajjri: 8/13 x 154 = 94.76 Cft
Floor
Qs: Prepare Analysis of Rates For a Cocrete Floor Consisting Following Structure.
4 inch Base course of brick ballast 2 inch down size
1 inch under layer of cement concrete Ratio 1:3:6
1.5 inch Topping of Cement concrete 1:2:4
Sol: Brick Ballast = 100 x 0.33 = 33.33 Cft
Under Layer = 1:3:6 1 inch = 1/12 = 0.08 ft.
Qty of under layer = 100 x 0.08 = 8 Cft
Dry Qty of concrete = 8 x 1.54 = 12.32 Cft
Sum of Ratio; 1 + 3 + 6 = 10
Cement: 1/10 x 12.32 = 1.232 Cft Convert to Bag
1 Bag = 1.25 Cft 1.232/1.25 = 0.98 Bags
Sand: 3/10 x 12.32 = 3.09 Cft
Bajjri: 6/10 x 12.32 = 7.39 Cft
Topping: P.c.c = 1:2:4 Thickness = 1.5 inch = 1.5/12 = 0.125 Ft
Qty of Topping = 100 x 0.125 = 12.5 Cft
Dry Qty Of P.c.c = 12.5 x 1.54 = 19.25 Cft

8. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Sum of Ratio = 1 + 2 + 4 = 7
Cement: 1/7 x 19.25 = 2.75 Cft Convert to Bags
1 bag = 1.25 Cft 2.75/1.25 = 2.2 Bags
Sand: 2/7 x 19.25 = 5.5 Cft
Bajjri: 4/7 x 19.25 = 11 Cft
Adding The Cft Of Topping And Sub Grade
Cement + Cement: 0.98 + 2.2 = 3.18 bags
Sand + Sand: 3.09 + 5.5 = 8.59 Cft
Bajjri + Bajjri: 7.39 + 11 = 18.39 Cft
Slab
Qs: Length of the slab = 10 Ft Width of the Slab = 9 Ft
Thickness of the Slab = 0.5 Ft Steel = 1.5 %
Mixing Ratio: 1:2:4 Determine tne Anaylsis of Follwing Materials.
Sol: First Calculate Its Quantuty: 1 x 10 x 9 x 0.5 = 45 Cft
Qty Of Cement x Dry Mortar Ratio: 45 x 1.54 = 69.3 Cft
Sum Of Ratio: 1 + 2 + 4 = 7
Cement: 1/7 x 45 = 6.42 Cft Convert To Cement
1 Bag = 1.25 Cft 6.42/1.25 = 5.13 Bags
Sand: 2/7 x 45 = 12.85 Cft
Bajjri: 4/7 x 45 = 25.71 Cft
Steel: 45 x 1.5/100 = 0.675 1 Cft = 490 Lbs
0.675 x 490 = 330.75 Lbs
Change it To Kilogram
330.75/2.204 = 150.06 Kg
Plaster
Qs: Prepare Analysis Of Rates For 1 inch Thick Plaster 100 Sft Long

9. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Cement Mortar Ratio: 1:3.
Sol: Unit of Rates = 100 Sft
Thickness = 1 inch 1/12 = 0.08 Ft
Wet Volume of Mortar: 100 x 1/12 = 8.33
Dry Volume of Mortar: 8.33 x 1.27 = 10.58 Cft
Sum of Ratio: 1 + 3 = 4
Cement: ¼ x 10.58 = 2.65 Cft Convert To Bags
1 Bag = 1.25 Cft 2.65/1.25 = 2.12 Bags
Sand: ¾ x 10.58 = 7.93 Cft.
Tiles
Qs: Find the Quantity Of Tiles If Length of Roof = 20 Ft & Width = 15 Ft ?
Sol: First Find The Volume Of Tile Volume = L x B x H
Volume Of Tile; L = 1 Ft , B = 6 inch or 0.5 Ft
1 x 0.5 = 0.5
Now Divide The Sum of area of the Roof With the Tile volume.
Total Area of Roof = 20 x 15 = 300 Sft
Volume of one Tile = 0.5 SqFt
300/0.5 = 600 Tiles
Or: 20 x 15 = 300 300 x 2 = 600 Tiles
Because In Every Row One Tile Adjusted So The Number Of Rows Are 2 So We Multiply it with 2.
Bricks
Qs:Toi Find The Brick in a Room L = 20 Ft, B = 15 Ft, Thickness Of Wall = 0.75 Ft
Height = 10 Ft ?
Sol: In Room We Have 2 Long Walls & 2 Short Wall So We Will Cut Of Two Short
Walls Or Two Long Walls Width To find Actual Quantity.
20 + 0.75 + 0.75 = 21.5 Ft

10. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Long Wall: No x L x BThickness
x H = Quantity
Long Wall = 2 x 21.5 x 0.75 x 10 = 322.5 Cft
Short Wall = No x L x B Or Thickness x H = Quantity
Short Wall = 2 x 15 x 0.75 x 10 = 225 Cft
Volume of 1 Brick = 0.0703
Now add Long wall & Short Wall
322.5 + 225 = 547.5 547.5/0.0703 = 7788 or 7800 Bricks
Or 547.5 x 14.2 = 7774.5 or 7800 Bricks
Bars
Qs: Suppose the Web of Bar In Which Have 104 Bars & 10 Ft Long
Find Its Weight in Ton ?
Sol: Thickness Of Bar: 6 Sutar
6/4 = 1.5
104 x 10 x 1.5/2040 = 0.76 Ton
1 Ton = 1000 kg 0.76 x 1000 = 760 Kg.
 __________________________$__________________________________*
Qs: Find The Weight of Bars if L = 20 ft. & B = 10 ft.
Width Wise bar: 5 Length Wise Bar: 10
Sol; L x Length Wise Bar = 20 x 5 = 100
W x Width Wise Bar = 10 x 10 = 100
100 + 100/2204 = 200/2204
0.90 Ton 0.90 x 1000 = 90 kg.
 __________________________$____________________________*
Excavation
QS: To Find Out the amount of Excavation Which Length is 100 Ft
Breadth is 3 Ft & Depth is 4 Ft ?

11. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Sol: Volume = L x B x H
100 x 4 x 3 = 1200 Cft
Cost of one Cubic Feet = 5 Rupees
Total Cost = 1200 x 5 = 6000 Rs
Earth Work
Qs: Workout the Quantity of Earth Work Required for an Embankment 150 m Long
& 10 m Wide at the top side Slope is 2:1 & depth at each 30 m interval are 0.60, 1.20
1.40, 1.60, 1.40, & 1.60.
Sol: Abstract of Quantity
R.D Depth B.D S.d2
B.d + Sd2
Mean
Area
Length Quantity
Cutting Filling
0 0.60 6 0.72 6.72 ______ ______ ____ ____
30 1.20 12 2.88 14.88 10.98 30 _____ 324 Cft
60 1.40 14 3.92 15.92 15.40 30 ____ 462 Cft
90 1.60 16 5.12 21.12 18.52 30 ____ 556 Cft
120 1.40 14 3.92 17.92 19.52 30 ____ 556 Cft
150 1.60 16 5.12 21.12 19.52 30 ____ 586 Cft
Total ____ ____ _____ _____ _____ _____ _____ 2513 Cft
 _______________________________$_______________________________________*
Types Of Roads
1. Reinforce Cement Concrete Road { R.C.C }
2. Plain Cement Concrete Road { P.C.C }

12. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
3. Bitumen Road
4. Asphalt Road
5. Tripple Surface Treatment { T.S.T }
6. Double Surface Treatment { D.S.T }
 __________________________$______________________________________*

13. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Blocks
Qs: Find The Blocks Used in A Room Which L is = 18 Ft, B = 14 Ft H = 10 Ft
Thickness = 9 inch or 0.75 Ft ?
Sol: Long Wall : 18 + 0.75 + 0.75 = 19.5
Qty For Long Wall = No x L x Thickness x H = Quantity
2 x 19.5 x 0.75 x 10 = 292.5 Cft
Volume of One Block = 0.3075 Cft
292.5/0.3075 = 951 Blocks
Now Quantity for Short Wall = No x B x Thickness x H = Quantity
2 x 14 x 0.75 x 10 = 210 Cft
210/0.3075 = 682 Blocks
Total Blocks = 1633 Blocks
 ________________________$________________________________________*
Circle Or Well
Qs: Find the Quantity Of Well Which is in circle form Depth is 4 m.
Sol: D = 4 so we addd the thickness of wall 0.15
4 + 0.15 + 0.15 = 4.30
Circle = π x D 3.14 x 4.30 = 13,50 m
No x L x B x H = quantity 1 x 13.50 x 0.30 x 4 = 16.21 Cum
Volume of one Brick = 0.002 Cum 16.21/0.002 = 8105 bricks
Mortar: 1:5 Sum of Ratio: 1 + 5 = 6
Cement: 1/6 x 4.863 = 0.8105 Cum 0.8105/0.035 = 23.15 Bags
Sand: 5/6 x 4.863 = 4.0525 Cum.
____________________________$________________________________________*
Bars On Meter System

14. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Qs: Find the Weight of Bar Which L = 80 m, Dia of Bar = 2 cm ?
Sol: Formulae: D2
/162.162 x L
Dia Must Be in Milimeter So Convert it to Milimeter
1 cm = 10 mm 2 x 10 = 20 mm L= 80 m
Apply Formulae: (202
)/162.162 x 80 = 197.33 kg
I Convert it to Ton so Divide it on 1000 1 Ton = 1000 kg
197.33/1000 = 0.197 Ton
 _________________________________$______________________________________*
Number of Bars
Qs: Find the Number of Bars in 20 m Long & 10 m Wide beam ?
Sol: Main Bar = 20 cm c/c Dia = 16 mm
Distribution Bar = 10 cm c/c Dia = 12 mm
c/c Means center to center
Main Bar = L of Distribution Bar/Space + 1
Main bar = 20/0.20 + 1 = 101 Bars
Distribution Bar = L of Main Bar/Space + 1
Distribution Bar = 10/0.10 + 1 = 101 Bars
Find Weight = D2
/162.162 x L
Main Bar: Weight = (162
)/162.162 x 20 = 31.57 Kg
Distribution Bar: Weight =(122
)/162.162 x 10 = 8.88 kg
 __________________________________$_______________________________________*
Qs: Find the Weight of Bar When its Length is 150 Ft & Dia is 0.5 inch ?
Sol: 1 inch = 8 sutar so First convert it to sutar
0.5 x 8 = 4 Sutar
Formulae of Weight = D2
/24 x L
Weight = (42
)/24 x 150 = 100

15. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
1 kg = 2.204 Lbs 100 kg = 100/2.204 = 45.37 kg
 _____________________________________$_______________________________________*
Qs: What is project ?
Ans: Work Complete in A Specific Time is Called A Project.
Qs: What is Meant by Concrete ?
Ans: Concrete Means C.S.C
C: Means Cement
S: Means Sand
C: Means Course Aggregate
Concrete Means That Can be made by these Three Materials.
In 1:2:4 Concrete we Use
1 is A Cement 2 is A Sand 4 is A Bajjri.
In this Cocrete 25-30 Liters Water can used Or 5-6 Gallon use.
 _______________________________$_______________________________*
Qs: What Are the Main Step to Construct Building Or Any
Construction Work ?
Ans: There are Five Main Step For Project Construction,
1. Budget
2. Layout
3. Site Selection
4. Survey
5. Starting Work
If We Have Money the we can easily construct Anything.
First we layout the Building For Better Results.
We can Choose Better Site For Construction .

16. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
After Site Selection We Survey this place To get Good Result
In last we start the work sothese five are the main steps
To construct any building road Etc.
 ___________________________$__________________________________*
Road Structure
1. Trees: Trees Provide Oxygen And Absorbs Co2
GAS And Protect Human.
2. Drain: In the side We Make Drain To flow Out The road Water.
3. Yellow Line Shows That Don,t Cross me.
4. Center Short Lines Are For Over take when it become long then don,t overtake.
5. Shoulder are created for Maintenance of vehicle When it Switch Down.
Formulae
Formulae For Long Wall = Length of long wall +
{width of short wall - Wall Thickness}
Length of short wall – [width of short wall – Wall thickness]
Quantity of water in Concrete
Concrete Ratio Quantity of Water

17. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
1:3:6 34 Litre
1:2:4 30 Litre
1:1:2 25 Litre
1:1 ½:3 27 Litre
Cement Quantity = Ratio of Cement/Sum of Ratio x Mortar Ratio
Sand Quantity = Ratio of sand/sum of Ratio x Mortar Ratio
Bricks = Quantity of Wall/volume of brick.
Formulae of striup No in beam = Total length of Beam/Space b/w striuf.
T-Iron No in Roof = Total Length/Space in T-Iron
____________________________$_____________________________________*
Dam
Qs: A Dam Having Breadth 4m and Height 10m ?
Given Data: Breadth of Dam = 4 m
Height of Dam = 10 m
Density of Water = ww = 1000 kg/m3
Density of Masonry = wm = 2000 kg/m3
Required Data: Total Pressure = ?
Resultant Pressure = ?
Position of Resultant Pressure = ?
Sol: P = wh2
/2 putting Values
P = 100 x (10)2
/2 = 50000 kg/m2
R.P = 𝑎2 + 𝑏2
Resultant pressure = 500002 + 4 𝑋 10 𝑋 2000
R.P = 94340 kg/m2
 _____________________________$________________________________*
I. 1 mm = 0.0394 inch
II. 1 cm = 10 mm = 0.3937 inch
III. 1 m = 100 cm = 1.0936 yards
IV. 1 km = 1000 m = 0.6214 mile
V. 1 sq inch = 6.4516 cm2
1 sq ft = 144 sq inches = 0.0929 m2
VI. 1 sq yard = 9 sq feet = 0.8361 m2

18. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
VII. 1 acre = 4840 sq yards = 446.9 m2
VIII. 1 sq mile = 640 acre = 259 hectares.
 ___________________________$__________________________________*
a. Area of Circle = πR2
π = 3.14 R = Radius
b. Volume of Cylinder = A = πR2
H
c. 1 Gallon Of Water = Weight = 8.35 Lbs. Pound
d. 1 Gallon = 231 Cubic inches
e. 1 Cubic Foot = 7.48 Gallons
f. 1 Cubic foot = 1728 Cube Nene
g. Atmospheric Pressure = 14.7 Psi
h. Head Pressure Of Water = 0.434 Psi Per Ft
i. 1 Psi = 2.30 Feet of Head.
 ________________________$________________________________*
Qs: A Steel bar Having Length = 10 Ft Thickness is = 1 inch
Find The Weight?
Sol: L = 10 , D = 1 inch 1 inch = 8 Sutar so 1 x 8 = 8 Sutar
W = D2
/24 x L W = (8)2
/24 x 10 = 6400/24 = 266.66 Lbs.
Lbs. is the unit of pound so change it to kg 1kg = 2.204 Lbs.
266.66/2.204 = 120.98 kg.
 __________________________$__________________________________*
QS: A Steel Bar Having Length = 80 m Dia = 6 cm Find Weight?
Sol: in Meter System Dia must Be in millimeter.
1 cm = 10 mm 6 x 10 = 60 mm
W = D2
/162.162 x L = (60)2
/162.162 x 80 = 288000/162.162 = 1776 kg
W = 1776 kg Conver it to Ton Divide it on 1000
1776/1000 = 1.776 Ton.
 _________________________$___________________________________*
Concrete Cylinder Dimension
Height = 30.48 cm Dia = 15.24 cm
Area of Cylinder = πd2
/4 3.142[15.24]2
/4
182.43 Volume = area x Height
Volume = 182.43 x 30.48 = 5560.46 cm3
Strength = Load/Area Load = 35000
35000/182.43 = 191.85 Cucm2
.
 ___________________$__________________________________*

19. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Quantity of Bridge Pile
Depth of Pile = 24 m Dia of Pile = 0.75 cm
Area = πD2
/4 = 3.14 x [0.75]2
/4 = 0.441 m2
Volume = 0.441 x 24 =10.60 m3
convert to foot
10.60 x 35.32 = 374.5 Wet Qty = 374.5 Cft
Convert it to dry = 374.5 x 1.54 = 576.79
Mixing Ratio = 1:3:6
Sum of Ratio = 1 + 3 + 6 = 10
Cement: 1/10 x576.79 = 57.67 ft3
One bag cement = 1.25 57.67/1.25 = 46.13 Bags
Sand: 3/10 x 576.79 = 173.037 cft
Bajri: 6/10 x 576.79 = 346.074 Cft.
 ________________________$____________________________*
Types of Steel
1. Mild Steel
2. Deformed Steel
3. Tar Steel
Kind of Tar Steel
40 Grade = 40000 Lbs/m2
50 Grade = 50000 Lbs/m2
60 Grade = 60000 Lbs/m2
 _______________________$_____________________________*
Formulaes
Circle = πD2
/2
Triangle = ½[a x b]
Rectangle = a x b
Square = a x b
Trapezoid = a + a2
/2 x b + b2
/2
To Find Percentage of Marks
Obtained marks/100 x total marks = percentage
885/100 x 1100 = 80.45 %
Convert it to Number Again Then opposite the process
80.45 x 1100/100 = 885
 _____________________$_______________________________*
Area and Volume Of Different Shapes
Area of Triangles: Formulae: 1/2 x b x h
Suppose: B = 3 Ft H = 5 Ft
Area = 1/2 x 3 x 5 = 7.5 Sft
Volume of Triangle = ½ x b x h x perpendicular

20. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
H =2, b = 1, L = 5 Ft
1/2 x 2 x 1 x 5 = 5 Cft
Area and Volume of Circle
Area of Circle = Formulae = π.D2
/4 D = 2 Ft
Putting Values = 3.14 x (2)2
/4 =3.142 Sft
D = 2 Radius πD2
/4 = πr2
π (2r)2
/4 = π 4r2
/4 = πr2
πD2
/4 = πr2
Volume of Circle = πd2
/4 x L
Suppose: D = 2 Ft, L = 10 Ft v = πD2
/4 x L
V = 3.14 x (2)2/4 x 10 = 31.42 Cft.
Area of Rectangle: Formulae = B x L
Suppose: B = 4 Ft, L = 8 Ft
4 x 8 = 32 Sft
Area and volume of Trapezoid:
Formulae = Sum of Two Parallel Side/2 x H
Area = 10 + 8/2 x 5 = 45 Sft
Volume = 8 + 10/2 x 5 x 20 = 900 Cft
 _______________________$______________________________*
Bricks in Cubic Meter
Length of Wall = 12 m
Thickness of Wall = 0.2286 m
Height of Wall = 3 m
Volume of Wall = L x T x H
12 x 0.2286 x 3 = 8.2296
We Know That: 1 cum = 500 Bricks
8.2296 X 500 = 4115 Bricks.
 ______________________$______________________________*
To Find Cement Mortars in CFT.
Quantity of Brick Work = 375 Cft
No of Bricks = 5062.5
Ratio of Cement Mortar = 1:4
Sum of Ratio = 1 + 4 = 5
We Know That 30 Cft Dry Mortar Used in 100 Cft
So: 375 x 30/100 =112.5 Cft
 ______________________$______________________________*
Cement Mortar in Cubic Meter
Quantity of Bricks = 8.2296 Cum
No of Bricks = 4114.8

21. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
Ratio of Cement = 1:6
Sum of Ratio = 1 + 6 = 7
We know that 0.30 cum dry Mortar Used in 1 Cum
So: Qty of Mortar = 0.30 x 8.2296/1 = 2.4688 cum
Cement: 1/7 x 2.4688 = 0.3526 cum
0.3526/0.035 = 10.07 bags
Sand: 6/7 x 2.4688 = 2.11 cum.
 ___________________________$____________________________*
Quantity of Plaster
Length of Wall = 20 Ft
Height of Wall = 10 Ft
Thickness of Plaster = 1 inch = 1/12 = 0.083 Ft
Wet Volume of Wall = L x H x T
20 x 10 x 0.083 = 16.6 Cft
Dry Volume of Mortar = 16.6 x 1.27 = 21.082 Cft
Ratio of Mortar = 1:3
Sum of Ratio = 1 + 3 = 4
Qty of cement: ¼ x 21.082 = 5.2705 Cft
5.2705/1.25 = 5 Bags
 _____________________$______________________________*
Quantity of Concrete in F.P.S system
Quantity of Wet Material = 150 Cft
Quantity of Dry Material = 150/100 x 154 = 231 Cft
Ratio of Cement: 1:2:4
Sum of Ratio: 1 + 2 + 4 = 7
Cement: 1/7 x 231 = 32.99 Cft 32.99/1.25 = 26.39 Bags
Sand = 2/7 x 231 = 66 Cft
Bajri: 4/7 x 231 = 132 Cft.
 _______________________$_____________________________*
Find Overlap for Steel
Formulae for overlap
Vertical Overlap: Dia of Steel = 40 x D/12 x 8 = 40d/96
Horizontal Overlap: 50 x D/12 x 8 = 50d/96
Suppose: Dia = 4 inch
Vertical: 40d/12 x 8 = 40 x 4/12 x 8 = 160/96 = 1.66 Ft
Vertical Overlap is = 1 Ft & 8 inch
Horizontal: 50d/12 x 8 = 50 x 4/96 = 200/96 = 2.083 Ft
Horizontal Overlap + 1 Ft & 10 inch.
 _________________________$_____________________________*
Stirrup’s (Rings)
Formulae = L x 12/Space

22. Quantity Notes By Engineer Saqib Imran
Cell No : 0341-7549889
L = 37 Ft Space of Rings c/c = 9 inch
Put Value & Apply Formulae
37 x 12/9 = 49.33 Means 50 Rings.
 _______________________$______________________________*
Length of Rings
Length of Column = 10 Ft
Dia of Column = 1 Ft Concrete cover = 2 inch
No of Bars in Column = 4 Space of Bars = 10 inch
Length of Rings = ?
Formulae: Space of Bar x side of column 3 inch
3 inch are added for hock length
10 x 4 x 3 = 40 + 3 = 43 inch 3 Ft & 7 inches.
 ___________________________$___________________________*
Qs: Core Cutter Was Used to Determined Density for a Road
Site. Weight of the Cutter was 1286 Gram & Total Weight Was
3195 gram. Volume of Core Cutter was 1000 cm3
& Water Content
Was 12%. Determine Bulk Density of This Sample?
Sol: Net of soil in Core Cutter:
W = 3195 – 1286 = 1909 gram
Volume of Core Cutter = V = 1000 c.c
Bulk Density = W/V = 1909/1000 = 1.909 Gm/c.c
M = 12% = 12 x 100 = 0.12
Bulk Density = ϓ/1 + m
Symbol of Bulk Density = ϓ
Bulk Density = 1.909/1 + 0.12 = 1.909/1.12
1.705 gm/c.c.
 ___________________________$_______________________________*
Stair Concrete
Waist & Landing Slab = 6 inch Thick
Tread = 12 inch Riser = 6 inch
Width of Stairs = 4.5 Ft
Formulae: 𝐿2 + 𝐻2
L of Waist Slab = 162 + 82 = 17.89 Ft
Volume of Waist Slab = 17.89 x 0.5 x 4.5 = 40.25 Cft
Volume of Landing = 5 x 4.5 x 0.5 = 11.25 Cft
Volume of Steps = 16 x [1/2 x 1 x 0.5] x 4.5 = 18 Cft
Total Volume = 40.25 + 11.25 + 18 = 69.50 Cft.
 __________________________$__________________________________*