1. CE 208
Quantity Surveying
(Estimation of an Underground Water Reservoir)
Md. Sazedul Islam
Lecturer
Department of Civil Engineering
Ahsanullah University of Science and Technology
2. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 2
Worked Out Problem
3. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 3
Worked Out Problem
Plan of the Underground Water Reservoir
Concrete Cover = 3in
4. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 4
Worked Out Problem
Section A-A
5. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 5
Estimation of BFS
Item Description Length (ft) Width (ft) Height/ thickness (ft) Area (ft2
)
1. 3 inch Brick Flat Soling 16 11 - 176
6. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 6
Estimation of Concrete
Item Description Length (ft) Width (ft) Height/ thickness (ft) Volume (cft)
2. Base Slab 16 11 1 176
3. Wall 2 x (9+14) 1 11 – 1 – Τ
5
12 440.83
4. Cover Slab 15 10 ൗ
5
12 62.5
Total = 679.33
7. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 7
1. 3 inch BFS [one layer]
Size of one brick = 9.5 in x 4.5 in x 2.75 in
Area of one brick = 9.5 in x 4.5 in = 42.75 in2
No of bricks =
176 × 144
42.75
= 593 Nos.
Sand volume required per 10 sqm area of BFS = 0.1 cum
Volume of sand =
0.1 × 176 × 0.30482
10
= 0.1635 m3 = 5.77 cft
Quantity of materials
4.5 in
2.75 in
8. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 8
2. Reinforced Cement Concrete (1:2:4)
Final volume (hard concrete) = 679.33 cft (From Table)
Initial volume (before mixing) = 679.33 x 1.5 = 1018.995 cft
Mix ratio = 1:2:4
Cement = 1018.995 x
1
7
= 145.57 cft = 117 bags
[One bag cement = 50 kg = 1.25 cft]
Sand = 1018.995 x
2
7
= 291.14 cft
Brick chips/ Khoa = 1018.995 x
4
7
= 582.28 cft
No. of bricks =
300×582.28
35.315
= 4947 Nos.
[1m3= 35.315 ft3 brick chips required 300 Nos. of full size brick]
Quantity of materials
9. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 9
Total volume of cement = 117 bags
Total volume of sand = [5.77 + 291.14] = 297 cft
Total number of bricks = [593 + 4947] = 5540 Nos.
Materials Quantity
Cement 117 bags
Sand 297 cft
Brick 5540 Nos.
Table: Summary of Quantity of Materials:
Quantity of materials
10. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 10
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Base
Slab
#6 @5” c/c 2 ×
16′ × 12 − 3" − 3"
5
+ 1 = 78 11′
−
3"
12
−
3"
12
+ 2 × 10.5 ×
6
8 × 12
= 11.813 921.414
11 ft
11. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 11
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Base
Slab
#5 @6”c/c 2 ×
11′ × 12 − 3" − 3"
6
+ 1 = 44 16′
−
3"
12
−
3"
12
+ 2 × 10.5 ×
5
8 × 12
= 16.594 730.136
11 ft
12. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 12
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Wall #3 @5”c/c 2 ×
11′ × 12 − 5" − 3" − 3"
5
+ 1 = 52 46 2392
11 ft
13. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 13
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Wall
#4 @5”c/c
(Outside)
46′ × 12
5
+ 1 = 112 11′
−
5"
12
−
3"
12
−
3"
12
+
4"
12
+ 2 × 10.5 ×
4
8 × 12
= 11.292 1264.704
11 ft
14. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 14
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Wall
#4 @5”c/c
(Inside)
46′ × 12
5
+ 1 = 112 11′
−
5"
12
−
3"
12
−
3"
12
+
6"
12
+ 2 × 10.5 ×
4
8 × 12
= 11.458 1283.296
11 ft
15. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 15
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Cover
Slab
#6 @5”c/c
(Bottom)
15′ × 12 − 3" − 3"
5
+ 1 = 36 10′
−
3"
12
−
3"
12
+ 2 × 10.5 ×
6
8 × 12
= 10.813 389.268
11 ft
16. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 16
Estimation of Reinforcement
Item Bar Designation No. Length (ft) Total Length (ft)
Cover
Slab
#6 @5”c/c
(Top)
10′ × 12 − 3" − 3"
5
+ 1 = 24 15′
−
3"
12
−
3"
12
+ 2 × 10.5 ×
6
8 × 12
= 15.813 379.512
11 ft
17. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 17
Table: Calculation of Weight of Reinforcement
Bar Length (ft)
Additional
(2%)
Final Length
(ft)
Weight/Length
(lb/ft)
Weight (lb)
#3 bar 2392 47.840 2439.84 0.376 918
#4 bar 2548 50.960 2598.96 0.668 1737
#5 bar 730.136 14.603 744.739 1.043 777
#6 bar 1690.194 33.804 1724.00 1.502 2590
Reinforcement
18. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 18
Report
B
A
1’
6”
1
1
19. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 19
Report
#7 @6” c/c
#6 @5” c/c
#7 @7” c/c
#4 @6” c/c
#5 @6” c/c
5”
1’
B
20. 6 February 2022
MD. SAZEDUL ISLAM, LECTURER, DEPARTMENT OF CIVIL ENGINEERING, AUST. 20
