This document provides examples and steps for calculating earthwork quantities for roads and canals. It includes sample problems showing how to calculate the cross-sectional area of a canal cutting using the width, depth, and side slopes. It also shows how to calculate earthwork quantities using the mid-sectional area method and mean sectional area method by considering the ground level, formation level, height/depth at different stations, and length between stations. Formulas and tables with chainage, height, depth, areas, and quantities are provided for illustrative examples.
8. Sl.
No
Chainage
/ Station
Depth /
Height
Mean Depth
/ Height (d)
Width (B) Area of
central portion
(BXd)
Area of sides
(Sd²)
Total sectional
Area
A = (Bd+Sd²)
Length
(L)
Quantity
Q = A X L
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
Sl.
No
Chainage
/ Station
Depth /
Height
(d)
Width
(B)
Area of
central
portion (BXd)
Total (A1)
sectional Area
(Bd1 + Sd1²)
Total (A2)
sectional
Area
(Bd2+ Sd2²)
Mean sectional
Area
A = (A1+A2)/2
Length
(L)
Quantity
Q = A X L
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
9. Example – 1:
A canal is in cutting and has a width of 4.00 m at bottom of the depth of cutting is 2.00 m and side slopes are
1 ½ : 1, find the area of cross section?
Solution: Given Data: - Width of canal = 4.00m; Depth of the cutting = 2.00 m; Side slopes = 1 ½ : 1 ( S: 1 / n: 1)
Area of c/s of canal = (b+nd)d (or) Bd + Sd²
(4 + 1.5*2)2 (or) 4x2 + 1.5 x 2²
4+3*2 (or) 8 + 6
= 14.00 Sq.m (or) 14.00 Sq.m
10. Example – 2:
A) Mid sectional area method.
A) Mean sectional area method.
Solution:
11. Mid - Sectional Area Method:
Volume of earth work (V) = Area of Mid point X Length
Area of the Midpoint = (b+nd)xd
Where b= 10 m, n = 1.5, d = 1+2+3/3 = 2 m. Am = (10 + 1.5*2)x2 = 26 sq. m
Therefore Volume of earthwork (V) = 26 X 120m = 3120 cu m.
12. Chainage 0 50 100 150 200
R. L of Ground 97.00 96.50 96.00 97.50 98.00
13. Height of the
Bank / Cutting
3.00 3.50 4.00 2.50 2.00
Formation Level 100.00 100.00 100.00 100.00 100.00
R. L of Ground 97.00 96.50 96.00 97.50 98.00
Chainage 0 50 100 150 200
21. Station /
Chainage
Length/
Distance
Height / Depth Mean Height /
Depth (d)
Sloping
Breadth of side
slope d√s²+1
Length in
between stations/
chainage
Total Area of both side
slopes (2X L X d√s²+1)
10 300 2.00 - - - -
11 330 1.20 1.60 3.58 30 214.80
12 360 1.16 1.18 2.63 30 157.80
13 390 0.50 0.83 1.85 30 111.00
14 420 0.78 0.64 1.42 30 85.80
15 450 1.60 1.19 2.65 30 159.60
16 480 0.60 1.10 2.45 30 147.60
17 510 1.20 0.90 2.00 30 120.60
18 540 0.38 0.79 1.77 30 106.20
19 570 0.70 0.54 1.20 30 72.60
20 600 1.10 0.90 2.00 30 120.60
TOTAL QUANTITY 1294.80 sq m
22. Item no Particulars of item Quantity Unit Rate Per Amount in
₹.
1 Earth work in Banking 3513.06 Cu m 345 % cu m 12,120.00
2 Turfing/ pitching for Side slopes 1294.80 Sq m 270 % sq m 3495.96
TOTAL COST in Rs. 16,000.00
23. ₹
Chainage / Station / Distance of the point R. L. of Ground Level
420 118.60
450 119.25
480 119.40
510 118.85
540 118.50
570 117.25
600 116.80
630 117.15
660 117.20
24. Chainage 420 450 480 510 540 570 600 630 660
R.L. of G.L 118.60 119.25 119.40 118.85 118.50 117.25 116.80 117.15 117.20
R.L. of F.L 118.60 118.45 118.30 118.15 118.00 117.85 117.70 117.55 117.40
Height of the bank 0 - - - - 0.60 0.90 0.40 0.20
Depth of the Cutting 0 0.80 1.1 0.7 0.50 - - - -
26. Station /
Chainage
Height / Depth Mean Height / Depth
(d)
Sloping Breadth of
side slope d√s²+1
Length in between
stations/ chainage
Total Area of both side slopes
(2X L X d√s²+1)
420 0.00
450 - 0.80
480 -1.1 Being Cutting - No Turfing is to be taken.
510 -0.70
540 -0.50
Passes 0.00
570 0.60 0.30 0.67 16 21.44
600 0.90 0.75 1.68 30 100.80
630 0.40 0.65 1.45 30 87.00
660 0.20 0.30 0.67 30 40.20
TOTAL QUANTITY 249.54 sq .m
27. Item no Description of item Quantity Unit Rate Per Amount in
₹.
1 Earth work in Cutting 1169.76 Cu m 13.00 Cu m 15,206.88
2 Earthwork in banking 740.28 Cu m 13.00 Cu m 9,623.64
3 Turfing/ pitching for Side slopes 294.54 Sq m 9.00 Sq m 2,650.86
Sundries 518.62
TOTAL COST in Rs. 28,000.00