These all information is updated by Rajnish kumar

1. BY- RAJNISH KUMAR

2. Calculation of cement, sand, aggregates use in column :-
 Size of column from ground level to ceiling level is 9" × 12" × 10′
& the grade of concrete use in column is m 20 grade.
 Quantities of cement :-
Formula for calculating cement in concrete is = concrete volume × 𝑝𝑎𝑟𝑡 𝑜𝑓
𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑠𝑢𝑚
𝑜𝑓 𝑟𝑎𝑡𝑖𝑜 × 𝑑𝑟𝑦 𝑣𝑜𝑙𝑢𝑚𝑛
sum of ratio = 1: 1.5: 3 = 1 + 1.5 + 3 = 5.5
concrete volume= l× 𝑏 × ℎ = 0.75′
× 1′
× 10′
= 7.5 𝑐𝑢. 𝑓𝑡 𝑤𝑒𝑡 𝑣𝑜𝑙𝑢𝑚𝑒
Dry volume= 7.5 cu.ft × 1.54 = 11.55𝑐𝑢. 𝑓𝑡
 Cement= (1 ÷ 5.5) × 11.5 = 2.1𝑐𝑢. 𝑓𝑡
 = 2.1 ÷ 1.23 = 1.70 𝑏𝑎𝑔𝑠 50 𝑘𝑔 𝑏𝑎𝑔 𝑣𝑜𝑙𝑢𝑚𝑒 = 1.23𝑐𝑢. 𝑓𝑡
 = 𝑡𝑜𝑡𝑎𝑙 𝑐𝑒𝑚𝑒𝑛𝑡 = 1.70 × 50 = 85 𝑘𝑔 𝑐𝑒𝑚𝑒𝑛𝑡
 Sand = (1.5 ÷ 5.55) × 11.5 = 3.15𝑐𝑢. 𝑓𝑡.
 Aggregate = 3 ÷ 5.5 × 11.55 = 6.29𝑐𝑢. 𝑓𝑡

3. There are four column in my house
 The quantities of cement, sand & aggregates use in one column is: 85
kg,3.15𝑐𝑢𝑏. 𝑓𝑡 & 6.29𝑐𝑢. 𝑓𝑡
 So quantities of cement ,sand & aggregate for 4 column =
85𝑘𝑔 × 4
= 540 kg
‘’ ‘’ sand ‘’ =
3.15 × 4
=
12.6𝑐𝑢. 𝑓𝑡
‘’ ‘’ aggregate ‘’ =
6.29 × 4
=
25.16𝑐𝑢. 𝑓𝑡

4. Calculation of quantities of cement, sand & aggregates use in beam
in a room, there are two types of beams, named PB1& PB2
LENGTH =2.7 M, WIDTH=0.25 M & DEPTH= 0.35 M.
VOLUME = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑖𝑔ℎ𝑡
= 2.7𝑚 × 0.25𝑚 × 0.35m
= 0.236 𝑐𝑢𝑚
No.of PB1 =2
Total volume of PB1= 2× 0.236 = 0.472 𝑐𝑢𝑚
PB2:- LENGTH =3.1M, WIDTH=0.25M, DEPTH=0.35M
VOLUME = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑖𝑔ℎ𝑡
= 3,1𝑚 × 0.25𝑚 × 0.35𝑚 = 0.271 𝑐𝑢𝑚 𝑛𝑜. 𝑜𝑓 𝑃𝐵2 = 2
TOTAL VOLUM OF PB2 =2 × 0.271 = 0.54 𝐶𝑈𝑀

5. Total volume of beam=total volume of pb1+total volume of pb2
 0.472+0.542= 1.014 cum
 So, total quantities of concrete in four beams
=1.014 cum
Dry volume of concrete = 1.014× 1.54 𝑑𝑟𝑦 𝑐𝑜 − 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑓𝑜𝑟 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 1.54
=1.562cum
& grade of concrete use in beam is = m25= 1:1:2
sum of all part=1+1+2=4
Quantity of cement :-
In cum.= dry volume of concrete × 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝑠𝑢𝑚 𝑜𝑓 𝑟𝑎𝑡𝑖𝑜
1.562 ×
1
4
= 0.391𝑐𝑢𝑚
Density of cement =1440kg/cum
in kg =0.391× 1440 = 563.04 𝑘𝑔
weight of one bag cement= 50 kg

6. No.of bag of cement=563.04/50=12 bags
 QUANTITY OF SAND =
1.562×
1
4
= 0.391𝐶𝑈𝑀
= 1 cum =35.31 cu.ft
so in cu.ft .= 0.391× 35.31 = 13.806 𝑐𝑢. 𝑓𝑡
QUANTITIES OF AGGREGATES :-
= Dry volume of concrete× 𝑝𝑎𝑟𝑡 𝑜𝑓
𝑎𝑔𝑔𝑟𝑒𝑔𝑎𝑡𝑒𝑠
𝑠𝑢𝑚 𝑜𝑓 𝑟𝑎𝑡𝑖𝑜
= 1.562×
2
4
= 0.781
= 0.781× 35.31 = 27.577 𝑐𝑢. 𝑓𝑡

7. RESULT :- ( for casting for beam )
DESCRIPTION OF
ITEM
QUANTITY UNIT
CEMENT 12
Bags
SAND 13.806 Cu.ft
AGGREGATE 27.577 Cu.ft

8. Quantity survey for brick work :-
 For 200mm wall:-
 Length= 2m, width=0.2m, height =4m
 volume of brick work =2m × 0.2𝑚 × 4m=1.6 cum
 Use standard size of brick = 190× 90 × 90𝑚𝑚
 Use mortar thickness = 10mm
 volume of one brick with mortar = 200× 100 × 100
 in m = 0.2m × 0.1𝑚 × 0.1𝑚 =
0.002𝑚
no of brick = (volume of brick work in one wall/volume of one
brick with mortar)
= 1.620/0.002=810 Nos.

9. Volume of mortar:-
 volume of mortar = volume of brickwork –(No.of bricks ×
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑟𝑖𝑐𝑘 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑚𝑜𝑟𝑡𝑎𝑟)
 = 1.6-(810× 0.19 × 0.09 × 0.09) = 0.373 𝑐𝑢𝑚. 𝑓𝑡
 Dry co-efficient of mortar -1.33
 So dry volume of mortar= 0.496 cum
 use mortar of 1:4(cement : sand)
 Quantity of cement :-
 dry volume of mortar × 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝑠𝑢𝑚 𝑜𝑓 𝑝𝑎𝑟𝑡
×
𝑑𝑒𝑛𝑠𝑖𝑡𝑦𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 ÷ 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑔 𝑐𝑒𝑚𝑒𝑛𝑡
0.496×
1
5
× 1440 ÷ 50= 3 Bags

10. Quantity of sand :-
 Dry volume of mortar × 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
𝑠𝑢𝑚 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑎𝑟𝑡𝑠
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑎𝑛𝑑
 =0.496×
4
5
× 35.31 = 14.011 𝑐𝑢. 𝑓𝑡
Description of item Quantity UNIT
Bricks 810 Nos
Cement 3 Bags
Sand 14.011 Cuft

11. For 100 mm wall :-
 area of brickwork = 𝑙𝑒𝑛𝑔𝑡ℎ × ℎ𝑖𝑔ℎ𝑡
 4 × 3 = 12 𝑆𝑞𝑚
Area of one brick with mortar= 0.2× 0.1 = 0.02 𝑠𝑞𝑚
Area of one brick with mortar= 0.2× 0.1 𝑠𝑞𝑚
No.of bricks = area of brick work/area of one brick with mortar
= 12/0.02 = 600 no’s
volume of mortar = volume of brick work – (number of bricks ×
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑟𝑖𝑐𝑘 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑚𝑜𝑟𝑡𝑎𝑟 )
= (12× 0.1 ) – (600× 0.19 × 0.09 × 0.09)
= 0.2766 cum
Dry volume of mortar=0.2766× 1.33 = 0. 367878 𝑐𝑢𝑚
Mortar = 1:4

12. Quantity of Cement :-
 0.3678 ×
1
5
= 0.1356
 0.07356 × 1440 = 264.816 kg
264.816÷ 50 = 5.296 bags
 QUANTITY OF SAN :- 0.3678×
4
5
× 35.31 = 10.3896
 RESULT:-
DESCRIPTION OF ITEM QUANTITY UNIT
BRICKS 465 Nos.
CEMENT 7 bags
UNIT 32.542 Cuft

13. 

14. Length of wall = 3.1m, hight of wall=3m & thickness of plaster =0.015m
volume of mortar used in plaster = 3.1× 3 × 0.015 = 0.139𝑐𝑢𝑚
Dry co- efficient for mortar= 1.33
Dry volume of mortar =0.139× 1.33 = 0.185 𝑐𝑢𝑚
use mortar = 1:4( cement : sand)
QUANTITY OF CEMENT:-
= 0.185× (
1
5
) × (1440/50) = 2 bags
QUANTITY OF SAND:-
0.185×
4
5
× 35.31 = 5.226 𝐶𝑈. 𝐹𝑇

15. Rr
 old surface cover 350 sq. Ft area per gallon. There for quantity of paint used
for old surface = 472.78/25 = 1.35 gallon =1.35× 4.35 = 6.13 liters

16. Quantity surveying paint work :-
room size 10.2 ft.× 8.9𝑓𝑡 = 38.2 𝑓𝑡
total wall length = 10.2 + 8.9 + 10.2 + 8.9=38.2 ft.
Wall hight =10 ft.
Total area of wall painting = total wall length × ℎ𝑖𝑔ℎ𝑡
= 3.2 × 10 = 382 𝑠𝑞. 𝑓𝑡
CELLING PAINT WORK :-
Ceiling area = 10.2 × 8.9 = 90.78 𝑠𝑞. 𝑓𝑡.
So, total area for painting work =celling panting× 𝑤𝑎𝑙𝑙 𝑝𝑒𝑛𝑡𝑖𝑛𝑔
=90.78+382=472.87squ.ft
1)new surface cover 200 sq.ft . Area per gallon.
Therefor, quantity of paint for new surface =472.78/200= 2.36
gallon,
1 gallon =4.54 liters
2.36× 4.54 = 10.71 liters

17. Details about lap length
 LAP LENGTH- over lapping length or lap length is provided for maintain continuity of
bar in order to safely transfer the stress from one bar to another bar
 Note – as per is cod 456:2000 over lapping length should be not less than 75mm
generally we take 24d-50d where “D” is Día of bar
 Development depth-development length is provided to transfer the load from one
building compartment to the. Other development length is known as anchorage length
 It is denoted by “ld”
ld = ∅
𝜎
4𝑗𝑏𝑑
, ∅ = 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑏𝑎𝑟, 𝜎 =
𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑏𝑎𝑟 𝑎𝑡 𝑐𝑜𝑛𝑠𝑖𝑑𝑟𝑒𝑎𝑡𝑖𝑜𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 , 𝑗𝑏𝑑 = 𝑑𝑒𝑠𝑖𝑔𝑛 𝑏𝑜𝑢𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠
Note = as per is code 456: 2000 development length can be calculated as given above
generally we take 300mm to 50d where “d” is dia of bar