Estimation of Underground Water Reservoir

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