VOLUME
Earthwork
operations involve the determination of volumes of material that is to be
excavated or embanked in engineering project to bring the ground surface to a
predetermined grade. Volumes can be determined via cross-sections, spot levels
or contours. For computation of the volume of earth work, the sectional areas
of the cross-section which are taken transverse to the longitudinal section
during profile leveling ate first calculated. Again, the cross-sections may be
different types, namely: (a) Level (b) Two-level, (c) Three-level, (d)
Side-hill two level or Part cut-part fill and (e) Multi-level.
After
calculation of cross-sectional areas, the volume of earth work is calculated
by:
i.
The trapezoidal rule or average end
area rule.
ii.
The prismoidal rule.
1) The prismoidal rule gives the correct volume directly.
2) The trapezoidal rule does not give the correct volume. Prismoidal
correction should be applied for this purpose. This correction is always
subtractive.
3) Cutting is denoted by a positive sign and filling by a negative sign.
A. Level Section
When
the ground is level along the transverse direction
Cross-sectional
area = (b + b + 2sh) x h
2
= (b + sh)h
B. Two-Level Section
When
the ground surface has a transverse slope:
PB = b/2
Bx = sh₁
b₁ = b/2 + sh₁ …………………………………………. (a)
Ee = (h₁ – h)
b₁ = n x Ee = n(h₁ - h) …………………………………….…….(b)
From (a) and (b), b/2 + sh₁ = n(h₁ - h)
Or h₁(n – s) = n(h + b/2n)
Or h₁ = n x
(h + b/2n) ….…(1)
(n – s)
From (2) and (a), b₁ = b + ns x
(h + b/2n) ……(2)
2 (n – s)
h₂ = n x
(h - b/2n) . ……(3)
(n + s)
b₂ = b +
ns
x (h - b/2n)
……(4)
2 (n + s)
Area ABCDE = ∆DOE + ∆COE - ∆AOB
= 1 OE X Dd + 1 OE
x Ce - 1 AB x OP
2 2 2
Here,
OE = OP + PE = b + h
2s
Dd = b₂
Ce = b₁
AB = b
OP = b
2s
2s 2s 2 2s
= ½{( b + h)(b₁ + b₂) – b²} ……(5)
2s 2s
C. Three-Level Section
When
the transverse slope is not uniform:
Area
ABCOD = ∆DOP + ∆COP + ∆DAP + ∆BCP
= 1 x h x b₂ + 1
x h x b₁ + 1 x b x h₂ + 1 x b x h₁
2 2 2 2
2 2
Area = {1/2(b₁ + b₂) + b/4(h₁ + h₂)}
Here h₁ = OP + Oe = h + b₁/n₁
h₂ = OP – ef = h – b₂/n₂
Deduction of formula for b₂ and b₁
b₂ = AP + AK = b/2 + sh₂
s
Also
b₂ = ef x n₂ = (h – h₂)n₂
or, h₂ = hn₂ – b₂ … (2)
n₂
From (1) and (2)
{b – (b/2)} = hn₂ – b₂
s₂ n₂
b₂n₂ – bn₂/2 = hn₂s - b₂s
b₂(n₂ + s) = n₂(sh + b/2) =
n₂s(h + b/2s)
b₂ = n₂s x (h + b/2s)
n₂ + s
Similarly,
b₁ = n₁s x (h + b/2s)
n₁ - s
D. Side-hill two level or Part cut-part
fill
When
the ground surface has a transverse slope, but the slope of the ground cuts the
formation level partly in cutting and partly in filling, the following method
is adopted:
Here, h = n x (b/2n + h)
n – s
2 (n – s)
Then h₂ and b₂ are deduced as follows:
b₂ = b + AA'
= b + sh₂ …
(i)
2 2
Again b = EE' = O'E' x n = (h +
h₂)n … (ii)
From
(i) and (ii), b + sh₂ = (h +
h₂)n
2
Or h₂(n – s) = b
– hn = n b – h
2 2n
h₂ = n x (b/2n - h)
n – s
From
(i) b₂ = b
+ ns x (b/2n - h)
2 (n – s)
Area in Cutting:
Area
of ∆PBC,
A₁ = 1 x PB x
h₁
2
Here, PB = OB + OP = b + nh
2
A₁ = 1
x b + nh n x b
+ h
2 2 n – s 2n
= 1 b + nh 1 x b + nh
2 2 n – s 2
= 1 {(b/2) + nh}²
2 n – s
Area in Filling:
Area
of ∆APE,
A₂ = 1 x PA x h₂
2
Here, PA = b - nh
2
A₂ = 1 x b
- nh n x b
- h
2 2 n – s 2n
= 1 b - nh 1 x
b - nh
2 2
n – s 2
= 1 {(b/2) - nh}²
2 n – s
In the
above case, the side slopes for cutting and filling are assumed to be equal.
But in actual practice, the side slope of cutting is different from that of
filling. Let the side slop of cutting be s₁ : 1.
Then,
b₁ = b + ns₁ x
(h + b/2n)
2 (n – s₁)
Area in cutting,
A₁ = 1 {(b/2) + nh}²
2 n – s
E. Multi-level Section
|
Left |
Centre |
Right |
|
±h₂/2 ±h₁/2 |
±h/0 |
±h₃/b₃ ±h₄/b₄ |
A
positive sign in the numerator denotes a cut, and a negative sign indicates a
fill.
The
denominator denotes corresponding horizontal distance from the centre. Starting
from the centre (E) and running outwards to the right and left, the coordinates
of the vertices are arranged, irrespective of algebraic sign, in determinant
form:
A
G F E
D C B
b/2 b₁
b₂ 0 b₄
b₅ b/2
The sum of products of the coordinates joined by the
solid line is
∑ P =
h₃ x 0 + h₄ x b₃ + 0 x b₄ + h₁ x 0 + h₂ x b₁ + 0 x b₂
The sum of products of the coordinates joined by the
dotted line is
∑ Q = h
x b₃ + h₃ x b₄ + h₄ x (b/2) + h x b₁ + h₁ x b₁ + h₂ x (b/2)
Area = ½ (∑ P - ∑ Q)
(Next post
on “FORMULA FOR CLCULATION OF VOLUME”)
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