Calc CL & no. T junction first of all calculate the total centre line of the plan given in the question bt drawing the centre line diagram

Total C = 3 x 5.6 x + 6 x 3.8 + 2 x 4.3 + 2.8 + 1.8 =52.8 m
Note:- see the shape in the figure and count it as 1 segment.
◻ = No. of T junctions marked
No. of T junctions = 10
Step 2:- To calculate what is asked.
Sr No. |
Description of work |
Nos |
L |
B |
H |
Quantity |
Unit |
a) |
Earthquake in excavation for foundation:-,Net length=total CL−No. of T×B2=5.2−10×0.92=48.3 m |
1 |
(calc),48.3 |
0.9,(P.C.C width) |
1.4,[H from G.L] |
L×B×H,60.86 |
m3 |
b) |
P.C.C (1:2:4) for foundation bed Net length = Total CL−No. of T junction×B2=52.8−10×0.92=48.3 m |
1 |
48.3 |
0.9 |
HH of Pcconly,0.2 |
L×B×H,8.69 |
m3 |
c) |
D.P.C (2.5 cm thick) in 1:3:6 (At P.L),Note:- No. height in DPC,Net length =Tc1−NT×B2=52.8−10×0.32=51.3 |
1 |
51.3 |
0.3 |
- |
L×B15.39 |
m3 |
d) |
1st class brickwork in superstructure,Net length = Tc1−NT×B2=52.8−10×0.3251.3 |
1 |
51.3 |
0.3 |
Floor to floor ht.,3 |
L×B×H,46.17 |
m3 |
- |
Total,quantity |
- |
- |
- |
- |
46.17 |
m3 |
- |
Deduction:-,(-), |
- |
- |
- |
- |
- |
- |
- |
1) Door (Analysis fromdiagram & given data) |
- |
- |
- |
- |
L×B×H |
- |
- |
D |
-2 |
1.2 |
0.3 |
2.1 |
-1.512 |
m3 |
- |
D1 |
-2 |
1 |
0.3 |
2.1 |
-1.26 |
m3 |
- |
2) Windows |
- |
- |
- |
- |
L×B×H |
- |
- |
W1 |
-1 |
2 |
0.3 |
1.3 |
-0.7 |
m3 |
- |
W2 |
-4 |
1.5 |
0.3 |
1.3 |
-2.34 |
m3 |
- |
V |
-2 |
0.6 |
0.3 |
0.9 |
-0.32 |
m3 |
- |
O |
-2 |
1.2 |
0.3 |
2.1 |
-1.5 |
m3 |
- |
- |
- |
- |
- |
- |
38.53 |
- |
Note:- Brickwork in super structure possess deductions. Deductions are done because of doors & windows present in the wall Information given in the diagram helps us for the deduction.