estimation of two room building

1. Estimating Costing and Valuation
Diploma –Vth Semester
Topic: Detail Estimate
(centre line and long wall short wall method)
Subject Code:3350604
Prepared by:
Y A .Challawala
Civil Engineering Department
Government Polytechnic Dahod

2. Exercise-2
Estimate the quantities of following items of a two roomed building given in Fig.
a. Earthwork in excavation in foundation trench
b. Lime concrete in foundation
c. First class brick work in 1:6 cement mortar in foundation and plinth
d. 2.5 cm thick DPC (1:2:4) with water proofing compound
e. 1st class brick work in cement mortar superstructure
Using (a) Centre line method (b) Long wall short wall method
D1=1.2 x 2.1 m
W1=1.2 x 1.2m
W1
W1D1
D1

3. (a) Centre line Method
Using the above image, we can first find the total length of Centre line.
Centre to Centre length of wall A = 6.3+ (1/2×0.30) + (1/2 x 0.30) = 6.60 m
Centre to Centre length of wall B = 4+ (1/2 x 0.30) + (1/2×0.30) = 4.30 m
Total length of Centre line = 6.6 +6.6 +4.3+ 4.3+4.3=26.1m
Net Centre Line = Total centre line – 1/2 ( no of junction x width of junction)
After finding out the total length of centre line, now find the quantity of the various
items which are used in construction.
W1
W1
D1
D1

4. a. Earthwork in excavation in foundation trench
Net Centre Line = Total centre line – 1/2 ( no of junction x width of junction)
= 26.1-1 /2 (2 x 0.9 ) =25.2
Total quantity of earth work = L x B x H m3
= 25.2 m x 0.9 x 0.9
Total quantity =20.412 m3
b. Lime concrete in foundation:
Net Centre Line = 26.1-1 /2 (2 x 0.9 ) =25.2m
Total quantity of earth work = L x B x H m3
= 25.2 x 0.9 x 0.3
Total quantity =6.804 m3

5. c. First class brick work in 1:6 cement mortar in foundation and plinth
Net Centre Line = Total centre line – 1/2 ( no of junction x width of junction)
Net Centre Line (60 cm Layer) =26.1-1 /2 (2 x 0.6 ) =25.5m
Net Centre Line (50 cm Layer) =26.1-1 /2 (2 x 0.5 ) =25.6m
Net Centre Line (40 cm Layer) =26.1-1 /2 (2 x 0.4 ) =25.7m
Brick work for (60 cm Layer)= 25.5 x 0.6 x 0.3 = 4.59m3
Brick work for (50 cm Layer)= 25.6 x 0.5 x 0.3 = 3.84m3
Brick work for (40 cm Layer)= 25.7 x 0.4 x 0.57 = 5.911 m3
Total Quantity = 14.341 m3

6. d. 2.5 cm thick DPC (1:2:4) with water proofing compound
Net Centre Line (40 cm Layer) =26.1-1 /2 (2 x 0.4 ) =25.7m
Total quantity of D.P.C work= L x B m2
= 25.7 x 0.4
Total quantity =10.28 m3
e. 1st class brick work in cement mortar superstructure
Net Centre Line (30 cm Layer) =26.1-1 /2 (2 x 0.3 ) =25.8m
Total quantity of 1st class brick work work = L x B x H m3
= 25.8 x 0.3 x 3.5
Total quantity =27.09m3

7. Sr.
No. Description
No
.
Leng
th
M.
Brea
dth
M.
Height
M.
Qty. Total Unit
1
Earthwork in
excavation
1 25.2 0.9 0.9 21.141 20.412
Cu.
m
2
Lime Concrete in
Foundation
1 25.2 0.9 0.3 7.047 6.804
Cu.
m.
3
Brickwork in
foundation and Plinth
(a) 60 cm layer 1 25.5 0.6 0.3 4.59
(b) 50 cm layer 1 25.6 0.5 0.3 3.84
(c) 40 cm layer 1 25.7 0.4 0.57 5.911
14.341
Cu.
m
Measurement Sheet

8. Sr.
No.
Description No.
Lengt
h M.
Bread
th M.
Height
M.
Qty. Total Unit
3 D.P.C 1 25.7 0.4 -- 10.44 10.28 Sq.m
4
1st class brick work in
cement mortar
superstructure
1 25.8 0.3 3.5 27.405 27.09
Cu.
m.
Deduction for Doors and
Windows
D1 2 1.2 0.3 2.1 1.512
W1 2 1.2 0.3 1.2 0.864
-2.376 Cu.m
Deduction for Lintels above
doors and windows with 15
cm bearing at each end (20
cm thick)
D1 2 1.5 0.3 0.2 0.18
W1 2 1.5 0.3 0.2 0.18
-0.36 Cu.m
Net Quantity of 1st class brick work in cement
mortar superstructure
24.354 Cu.m

9. (b) Long wall short wall method
Using the above image, we can first find the length of long wall and short wall.
Centre to Centre length of long wall = 5+ (1/2×0.30) + (1/2 x 0.30) = 6.60 m
Centre to Centre length of short wall = 4+ (1/2 x 0.30) + (1/2×0.30) = 4.30 m
After finding out the length of the long wall and short wall, now find the quantity of
the various items which are used in construction.
Room-1
3 x 4 m
Room-2
3 x 4 m
D1
D1
W1
W1

10. a. Earthwork in excavation in foundation trench
Length of long wall = c/c length + width of earthwork
= 6.6+ 0.9= 7.5m
Length of short wall = c/c length -width of earthwork
= 4.3 - 0.9= 3.4m
Total quantity of earth work = L x B x H m3
Long wall = 2 x 7.5 x 0.9 x 0.9 =12.15m3
Short wall = 3 x 3.4 x 0.9 x 0.9 =8.262m3
Total =12.15 + 8.262=20.412 m3
Sr.
No.
Description No.
Length
M.
Breadth
M.
Height
M.
Qty. Total
1
Earthwork in
excavation
Long Wall 2 7.5 0.9 0.9 12.15
Short Wall 3 3.4 0.9 0.9 8.262
20.412

11. b. Lime concrete in foundation:
Length of long wall = c/c length + width of concrete in foundation
= 6.6 + 0.9= 7.5m
Length of short wall = c/c length - width of concrete in foundation
= 4.3 - 0.9= 3.4m
Total quantity of earth work = L x B x H m3
Long wall = 2 x 7.5 x 0.9 x 0.3 = 4.05m3
Short wall = 3 x 3.4 x 0.9 x 0.3 = 2.754m3
Total =4.05 + 2.754=6.804m3
Sr.
No.
Description No.
Length
M.
Breadth
M.
Height
M.
Qty. Total
2
Lime
Concrete in
Foundation
Long Wall 2 7.5 0.9 0.3 4.05
Short Wall 3 3.4 0.9 0.3 2.754
6.804

12. c. First class brick work in 1:6 cement mortar in foundation and plinth
Length of long wall (60 cm Layer)= c/c length + width of First class brick work
= 6.6+ 0.6= 7.2m
Length of short wall (60 cm Layer) = c/c length - width of First class brick work
= 4.3 - 0.6= 3.7m
Length of long wall (50 cm Layer)= c/c length + width of First class brick work
= 6.6+ 0.5= 7.1m
Length of short wall (50 cm Layer) = c/c length - width of First class brick work
= 4.3 - 0.5= 3.8m
Length of long wall (40 cm Layer)= c/c length + width of First class brick work
= 6.6+ 0.4= 7.0m
Length of short wall (40 cm Layer) = c/c length - width of First class brick work
= 4.3 - 0.4= 3.9m
Sr.
No.
Description No.
Length
M.
Breadth
M.
Height M. Qty. Total Unit
3
Brickwork in foundation and
plinth
Long Wall
1st footing 60 cm 2 7.2 0.6 0.3 2.592
2nd footing 50 cm 2 7.1 0.5 0.3 2.13
2nd footing 40 cm 2 7 0.4 0.575 3.22
Short Wall
1st footing 60 cm 3 3.7 0.6 0.3 1.998
2nd footing 50 cm 3 3.8 0.5 0.3 1.71
2nd footing 40 cm 3 3.9 0.4 0.575 2.691
14.341 Cu.m.

13. d. 2.5 cm thick DPC (1:2:4) with water proofing compound
Length of long wall = c/c length + width of DPC
= 6.6+ 0.4=7.0 m
Length of short wall = c/c length + width of DPC
= 4.3 – 0.4= 3.9m
Total quantity of earth work = L x B m2
Long wall = 2 x7.0 x 0.4 =5.6m2
Short wall = 3 x 3.9 x 0.4 =4.68m2
Total =5.6 +4.68=10.28 m2
Sr.
No.
Description
No
.
Leng
th M.
Brea
dth
M.
Height
M.
Qty. Total Unit
4
2.5 cm thick DPC
(1:2:4) with water
proofing compound
Long Wall 2 7.0 0.4 - 5.6
Short Wall 3 3.9 0.4 - 4.68
10.28 Sq.m

14. e. 1st class brick work in cement mortar superstructure
Length of long wall = c/c length + width of 1st class brick work
= 6.6+ 0.3= 6.9m
Length of short wall = c/c length - width of 1st class brick work
= 4.3 – 0.3= 4.0m
Total quantity of earth work = L x B x H m3
Long wall = 2 x6.9 x 0.3 x3.5=14.59m3
Short wall = 3 x 4.0 x 0.3 x 3.5=12.6m3
Total =14.59+12.6=27.09 m3
Sr.
No.
Description No.
Lengt
h M.
Bread
th M.
Height
M.
Qty. Total Unit
5
1st class brick work in
cement mortar
superstructure
Long Wall
2 6.9 0.3 3.5 14.59
Short Wall 3 4.0 0.3 3.5 12.6 .
27.09 Cu.m

15. Sr.
No. Description
No
.
Len
gth
M.
Bre
adth
M.
Height
M.
Qty. Total Unit
1
Earthwork in
excavation
Long Wall 2
7.5 0.9 0.9 12.15
Short Wall 3
3.4 0.9 0.9 8.262
20.412 Cu.m
2
Lime Concrete in
Foundation
Long Wall 2 7.5 0.9 0.3 4.05
Short Wall 3
3.4 0.9 0.3 2.754
6.804 Cu.m.
(b) Long wall short wall method

16. Sr.
No. Description
No
.
Leng
th
M.
Brea
dth
M.
Heigh
t M.
Qty. Total Unit
3
Brickwork in
foundation and plinth
Long Wall
1st footing 60 cm 2 7.2 0.6 0.3 2.592
2nd footing 50 cm 2 7.1 0.5 0.3 2.13
2nd footing 40 cm 2 7 0.4 0.575 3.22
Short Wall
1st footing 60 cm 3 3.7 0.6 0.3 1.998
2nd footing 50 cm 3 3.8 0.5 0.3 1.71
2nd footing 40 cm 3 3.9 0.4 0.575 2.691
14.341
Cu.
m.

17. Sr.
No.
Description
No
.
Leng
th M.
Brea
dth
M.
Height
M.
Qty. Total Unit
4
2.5 cm thick DPC
(1:2:4) with water
proofing compound
Long Wall 2 7 0.4 - 5.6
Short Wall 3 3.9 0.4 - 4.68
10.28
Sq.m

18. 5
1st class brick work in
cement mortar
superstructure
Long Wall 2 6.9 0.3 3.5 14.49
Short Wall 3 4.0 0.3 3.5 12.6 .
27.09 Cu.m
Deduction for Doors and
Windows
D1 2 1.2 0.3 2.1 1.512
W1 2 1.2 0.3 1.2 0.864
-2.376 Cu.m
Deduction for Lintels
above doors and windows
with 15 cm bearing at
each end (20 cm thick)
D1 2 1.5 0.3 0.2 0.18
W1 2 1.5 0.3 0.2 0.18
-0.36 Cu.m
Net Quantity of 1st class brick work in
cement mortar superstructure
24.354 Cu.m

19. Thank you
Any Question???

20. Exercise-3:
Estimate the quantities of following items of a two roomed building given in Fig.
a. Earthwork in excavation in foundation trench
b. Lime concrete in foundation
c. First class brick work in 1:6 cement mortar in foundation and plinth
d. 2.5 cm thick DPC (1:2:4) with water proofing compound
e. 1st class brick work in cement mortar superstructure
The dimensions of doors, windows and selves are
Door D = 1.20 m × 2.10 m.
Windows W = 1.00 m × 1.50 m
Shelves S = 1.00 m × 1.50 m
4.20
0.6
0.2
0.2
0.2
0.2
0.2
0.3
30
40
50
60
70
80
L.c.c
1.10
P.L
G.L
1.1 m