Example # 1 Mean Area Method

Workout the quantity of earth work for an embankment 150m long and 10m wide at the top. Side slope is 2:1 and depths at e
are 0.60, 1.2, 1.4, 1.6, 1.4 and 1.6m.

Depth Center Area Area of Total Area Mean Area

Stations
(d) (B.d) Sides (Sd2) (Bd + Sd2) (A1 + A2)/2

0 0.6 6 0.72 6.72

1 1.2 12 2.88 14.88 10.8
2 1.4 14 3.92 17.92 16.4
3 1.6 16 5.12 21.12 19.52
4 1.4 14 3.92 17.92 19.52
5 1.6 16 5.12 21.12 19.52
Total filling or embankment quntity (Cu.m)=

Example # 2 Mean Depth Method

Areas
R.D NSL FRL Depth Mean Depth
B.D
0 251 252 1
30 250.9 251.8 0.9 0.95 9.5
60 250.9 251.6 0.7 0.8 8
90 250.8 251.4 0.6 0.65 6.5
108 0 0 0 0.3 3
120 251.2 250.8 -0.4 -0.2 -2
150 251.4 250.6 -0.8 -0.6 -6
 Example # 1
ide slope is 2:1 and depths at each 30m interval

Quantity
Interval R.D (ft)
Cut Fill
0
30 324 100
30 492 200
30 585.6 212.5
30 585.6 300
30 585.6 400
2572.8 500

Areas Sum of Quantity

Interval
Sd2 Areas Cut Fill

1.805 11.305 30 339.15

1.28 9.28 30 278.4
0.845 7.345 30 220.35
0.18 3.18 18 57.24
0.04 -2.04 12 -24.48
0.36 -6.36 30 -190.8
Total = -215.28 895.14
mple # 1 Abstarct of Quantity

Cross Section Mean Area A.M Length (L) Quantity Qty = A.M x
Area (A) Sft = A1 + A2 / 2 ft L in Cutting (Cft)

65.32
31.52 48.42 100 4842
3.32 17.42 100 1742
0 1.66 12.5 20.75
25.3 12.65 87.5
65.75 45.525 100
110.63 88.19 100
Total = 6604.75
ntity

Quantity Qty = A.M x L

in Filling (Cft)

1106.875
4552.5
8819
14478.375
 Center Line Method

S. No Description Nos Length

1 Excavation 1 51.75
2 P. C. C with lean concrete 1:3:6 1 51.75
3 Brick work
in Foundation with cement mortor 1:3
Step 1 1 51.75
Step 2 1 51.75
Step 3 1 51.75
4 D. P. C with cement concrete 1:2:4 1 51.75
5 Brick work
In Super Structure with cement mortor 1:2 1 51.75
6 Plaster work of outside walls with cement 4 16.2
mortor 1:3 2.5 cm thick
Plaster work of inside walls with cement mortor
7 12 5
1:6 2.5 cm thick
8 White wash 3 coats As per item no. 7

Rate Analysis for center line Method

Rate Analysis for Brick work.
Total Volume of Brick work = 11.643 + 9.315 + 6.986 + 54.337 = 82.281m3
Deducction of mortor:
Volume of mortor = 25/100*82.281 = 20.57m3
Net Brick Work= Total Volume of Brick Work - Volume of mortor = 82.281 - 20.57 = 61.71m3
No. of Bricks = 500/1cum*61.71 = 30855 bricks
Let me include 10% of wastage of bricks:
10% wastage = 10/100*30855 = 3086 bricks
Net Number of Bricks = 30855 + 3086 = 33941 bricks
Now calculate cement and sand:
Volume of wet mortor = 20.57m3
Volume of dry mortor = 20.57*1.27 = 26.1239m3
Ratio of mortor = 1:3
Sum of ratio = 1 + 3 = 4
Volume of Cement = 1/4*26.1239 = 6.531m3
Volume of 1 bag of cement in cum = 0.036m3
No. of bags of cement = 6.531/0.036 = 181.416 say 182 bags
Volume of sand = 3/4*26.1239 = 19.592cum

Bricks = 33941 bricks

Cement = 182 bags
Sand = 19.592cum
Rate Analysis for concrete:
Concrete for PCC:
Volume of wet concrete = 10.867cum
Volume of dry concrete = 10.867*1.54 = 16.7351cum
Ratio of concrete = 1:3:6
Sum of ratio = 1 + 3 +6 = 10
Voulume of cement = 1/10*16.7351 = 1.67351cum
Volume of 1 bag of cement = 0.036cum
No. of bags of cement = 1.67351/0.036 = 46.486 say 47 bags
Voulume of sand = 3/10*16.7351 = 5.0205cum
Volume of Bajri = 6/10*16.7351 = 10.0410cum

Bags of cement = 47 bags

Volume of sand = 50205cum
Volume of Bajri = 10.0410cum

Concrete for DPC:

Wet volume of concrete = 0.7405cum
Dry volume of concrete = 0.7406*1.54 = 1.1403cum
Ratio of concrete = 1:2:4
Sum of ratio = 1 + 2 + 4 = 7
Volume of cement = 1/7*1.1403 = 0.1630cum
No of bags of cement = 0.1630/0.036 = 4.525 say 5 bags
Volume of sand = 2/7*1.1403 = 0.3258cum
Volume of Bajri = 4/7*1.1403 = 0.6516cum

Bags of cement = 5 bags

Volume of sand = 0.3258cum
Volume of Bajri = 0.6516cum

Rate Analysis for Palster:

Palster for outside walls:
Wet volume of mortor = 330.48*0.025 = 8.262cum
Dry volume of mortor = 8.262*1.27 = 10.492cum
Ratio of mortor = 1:3
Sum of ratio = 1 + 3 = 4
Volume of cement = 1/4*8.262 = 2.0655cum
No of bags of cement = 2.0655/0.036 = 57.375 say 58 bags
Volume of sand = 3/4*8.262 = 6.1965cum

Bags of cement = 58 bags

Volume of sand = 6.1965cum

Palster for inside walls:

Wet volume of mortor = 270*0.025 = 6.75cum
Dry volume of mortor = 6.75*1.27 = 8.5725cum
Ratio of mortor = 1:6
Sum of ratio = 1 + 6 = 7
Volume of cement = 1/7*8.5725 = 1.2246cum
No of bags of cement = 1.2246/0.036 = 34.572 say 35 bags
Volume of sand = 6/7*8.5725 = 7.3478cum

Bags of cement = 35 bags

Volume of sand = 7.3478cum

Material Statement
r Line Method

(𝑚)^3
Breadth Height Units Quantity

(𝑚)^3
1.05 1.1 59.77125
1.05 0.2 10.8675

(𝑚)^3
0.75 0.3 11.64375
0.6 0.3 9.315
0.45 0.3 6.98625
0.45 0.0318 0.7405425

0.3 3.5 (𝑚)^3 54.3375

5.1 (𝑚)^2 330.48

4.5 (𝑚)^2 270

(𝑚)^2 270

r center line Method

1m3
l Statement
 Remarks
L.O.E = L.O.W + (W.O.E - T.O.W) Long wall = 5 + 5 + 5 + 0.30 + 0.30 + 0.15 + 0.15 = 15.90 x 2 =
L.O.E = 51 + (1.05 - 0.30) = 51.75m Short wall = 4.50 + 0.15 + 0.15 = 4.80 x 4 =
Length of wall = 31.8 + 19.20 =
31.8m
19.20m
51m
 Bar Bending Schedule For Beam

Overall Length Extra Length Less Cover

S. No Size of bar and in position Actual Length (ft) No. of Bars Total Length (ft) Weight Lbs (kg)
(ft) Hook @8d Bend Laps @45D (ft)
1 Main bars @1" dia. 15 0.25 14.75 6 89 107.6748582231
2 Stirrups #3/8" 4.668 0.25 0.083 4.586 18 83 14.12098298677
Total 121.7958412098

Bar Bending Schedule For Beam

Overall Length Extra Length Less Cover

S. No Size of bar and in position Actual Length (ft) No. of Bars Total Length (ft) Weight Lbs (kg)
(ft) Hook @9d Bend Laps @45D (ft)
1 Main bars @ #6 10 0.5625 0.08333333 10.9583333333333 4 43.833333333333 29.82986767486
Cut length of bent of bars @
2 11.11 0.5625 0.08333333 12.0683333333333 2 24.136666666667 16.42570888469
#6
Total 46.25557655955

Length of straight bar with hook = L/4 =10/4 = 2.5'

Length of inclined part h = p/sin Φ= 16/sin45 = 16/0.707 = 22.62"
Length of angular part b = √h^2 - p^2 = √22.62^2 - 16^2 = √511.664 - 256 = √255.664 = 15.899 say 16"
Total length central straight bar = 10 - 2.5 - 2.5 - 1.33 - 1.33 = 2.34'
Cut length of bent of bars = 2.5 + 1.885' + 2.34' + 1.885' + 2.5' + 2{9(6/8)/12) - 2(0.0833) * 2 = 24.1367'

Bar Bending Schedule For Slab

Overall Length Extra Length Less Cover

S. No Size of bar and in position Actual Length (m) No. of Bars Total Length (ft) Weight Lbs (kg)
(m) Hook @9d Bend Laps @45D (m)
1 Main bars @ 25mm 4 0.02 3.96 48 190.08 732.438131173
2 Distribution bars @ 13mm 7 0.02 6.96 28 194.88 203.0525653831

Total 935.4906965561

Bar Bending Schedule For Slab

Overall Length Extra Length Less Cover

S. No Size of bar and in position Actual Length (m) No. of Bars Total Length (ft) Weight Lbs (kg)
(m) Hook @9d Bend Laps @45D (m)
1 Main bars @ 25mm 4 0.02 3.96 48 190.08 732.438131173
2 Distribution bars @ 13mm 7 0.02 6.96 28 194.88 203.0525653831

Total 935.4906965561