Post on 17-Jun-2021
Luma Khalid
E-mail : luma.k@coeng.uobaghdad .edu.iq
Engineering Surveying 3rd Stage
Volumes
Volume of Pyramid
β’ V=1
3 A.h
Volume
β’ Vc1-2=L1-2(π΄1:π΄2
2)
β’ Vc2-3=L2-3(π΄2:π΄3
2)
β’ Vc3-1f=1
3 A3.L3-1
β’ Vf1-3c=1
3 A1.L3-1
β’ Vf1-2=L1-2(π΄1:π΄2
2)
β’ Vf2-3=L2-3(π΄2:π΄3
2)
β’ Total Volume C = L(π΄1:π΄π
2+ π΄2 + π΄3 + β― + π΄π β 1)
β’ Total Volume F = L(π΄1:π΄π
2+ π΄2 + π΄3 + β― + π΄π β 1)
1cut
2cut
3cut
1fill
2fill
3fill
L1-2
L2-3
L3-1
L2-3
L1-2
Volume
β’ Vc1-2=L1-2(π΄1:π΄2
2)
β’ Vc2-3=L2-3(π΄2:π΄3
2)
β’ Vc3-4=L3-4(π΄3:π΄4
2)
β’ Vc4-1=1
3 A4.L1-2
β’ Vf1-3c=1
3 A1.L3-4
β’ Vf1-2=L1-2(π΄1:π΄2
2)
β’ Vf2-3=L2-3(π΄2:π΄3
2)
1cut
2cut
3cut
1fill
2fill
3fill
L1-2
L2-3
L3-4
L2-3
L1-2
4cut
Example
β’ Compute the total volumes between the following cross sections have width (b=14 m)
Sta. Cross - Section Slope
4+20 π?
20.00 ,
π1.80
0.00 ,
π?
?
Ground has uniform slope: (1:sf)= (1:3.0)
5+38 π?
? ,
π1.90
0.00 ,
π?
?
Ground from left to right has (1:n1)= up slope(1:15)
(1:n2)=up slope(1:10),(1:s)=(1:3)
6+75 π?
9.00 ,
π 0.25
0.00 ,
π 1.2
?
Ground has uniform slope: (1:n)= (1:12)
8+09 π 2.75
? ,
π 1.1
4.00 ,
π 2.30
0.00,
π 1.75
3.20 ,
π ?
16.50 (1:sc)= (1:3.0)
Station 4+20
β’1
3 =
ΞY13
β’ ΞY=4.33 m
β’1
π =
2.53320
β’ n= 7.89 Μ΄8
β’ A= (h+π
2π )2(
π2. π
π2;π 2) - π2
4π
β’ A= (1.8+14
2β3)2(
82. 3
82;32) - 142
4β3
β’ Af=43.31m2
n 1 1:n
7 7 C.L
1 3
1 3 1.8
20
ΞX1=13
ΞY1
?= 2.533
2.533
20
Station 5+38
β’ A=1
2β +
π
2π
2(
π1π
π1;π +
π2π
π2:π ) β
π2
4π
β’ A=1
21.9 +
14
2β3
2(
15β3
15;3+
10β3
10:3) β
142
4β3
β’ Af= 37.95 m2
1:10
1:15
7 7 C.L
1 3
1 3 1.9
-
+
Station 6+75
β’1
π =
0.25d
β’ d= 3 m.
β’1
π =
Y9
β’ Y=0.75
β’ hf= 0.75-0.25= 0.5 m.
1:12
7 7
C.L
1
Sc
1 Sf
1.2
hf
wc 9
d 0.25
1:12 hf
9
d 0.25
Y
Station 6+75
β’ Ac=βπ
2(
π
2+ π)
β’ Ac=1.2
2(7 + 3)
β’ Ac = 6 m2
β’ Af=βπ
2
π
2β π
β’ Af =0.5
27 β 3
β’ Af = 1 m2
Station 8+09
D
C B
A
E
F
G
C.L
X - axis
Y - axis
16.5 ??? 3.2 4
ΞY
1.75 2.3
1.1 2.75
1
π =
ΞY16.5β7
1
3 =
ΞY16.5β7
ΞY= 3.17 m.
1
3 = 2.75
ΞX
ΞX= 8.25 m.
???=7+8.25= 15.25 m.
Station 8+09
β’ A=(8.25,0)
β’ B=(0,2.75)
β’ C=(11.25,1.1)
β’ D=(15.25,2.3)
β’ E=(18.45,1.75)
β’ F=(31.75,3.17)
β’ G=(22.25,0)
D
C B
A
E
F
G
C.L
X - axis
Y - axis
16.5 15.25 3.2 4
3.17
1.75 2.3
1.1 2.75
Station 8+09
N E
N1 E1
N2 E2
N3 E3
N4 E4
N5 E5
N6 E6
N7 E7
N1 E1
N E
0 8.25
2.75 0
1.1 11.25
2.3 15.25
1.75 18.45
3.17 31.75
0 22.25
0 8.25
2A =I 82.506I
A c =41.25 m2
Computation volume
β’ Vf1-2=L1-2(π΄1:π΄2
2)
β’ Vf=118(43.31:37.95
2) +137(
37.95:1
2) +134(
1
3)
β’ Vf=7506.39m3
β’ Vc1-2=L1-2(π΄1:π΄2
2)
β’ Vc=134(41.25:6
2) +137(
6
3)
β’ Vc=3439.95m3
Approximate Method for Computing Volume From the Profile
β’ dc(in m.)=π΄π
πΏπ
β’ ac=dc (bc+sc.dc)
β’ Vc= ac*Lc
β’ df(in m.)=π΄π
πΏπ
β’ af=df (bf+sf.df)
β’ Vf= af*Lf
β’ 80Ω Ψ«Ψ§Ω Ψ§ΩΩΨͺΨ§Ψ¨ Ψ΅
Volume of a Barrow Pit or Volume from Spotheights
β’ Vc= A(square)(ββ1:2ββ2:3ββ3:4ββ4
4)
β’ Vc=A(triangle)(ββ1:2ββ2:3ββ3:4ββ4:5ββ5:6ββ6:7ββ7:8ββ8
3)
h2 h1 h2 h2 h1
h2
h2
h1
h1 h3
h1
h2
h2
h2
h4 h4 h4
h4 h4 h4
h4 h4
h2 h2
Example
1. Compute amount of cut needed to level the ground at a level of 30 m.
2. Compute the expected amount of fill to raise the ground at a level of 40 m.
3. Calculate the level at which cut and fill are equal.
30 31 34 36
10
30
31
10
35 34
10 33
12
10
33 31 32
32 36 33
35 34
10 10
Compute amount of cut needed to level the ground at a level of 30 m.
β’ Vc=A(ββ1:2ββ2:3ββ3:4ββ4
4)
β’ ββ1=3+1+6+4=14 m.
β’ ββ2 = 1+0+0+4+2+3+4+5
β’ =19 m.
β’ ββ3 = 0
β’ ββ4 = 2+1+3+6= 12m.
β’ Vc(square)=(10*10)(14:2(19):3(0):4(12)
4)=2500 m3
β’ ββ1=6+4=10 m.
β’ ββ2= 2+3 =5m.
β’ ββ3 = 5m.
β’ Vc=A(ββ1:2ββ2:3ββ3:4ββ4:5ββ5:6ββ6:7ββ7:8ββ8
3)
β’ Vc(triangle)=A(ββ1:2ββ2:3ββ3:0
3)= Β½*10*12 (
10:2(5):3(5):0
3)
β’ Vc(triangle)=700m3
β’ Vc(total)= 2500+700=3200m3
h1=6
10
1
0
h3=5 h1=4 12
10
h2=2
h2=3
h2=0 h1=1 h2=4 h1=6
10
h2=0
h2=1
10
35 h1=4
10 h1=3
12
10
h4=3 h4=1 h2=2
h4=2 h4=6 h2=3
h2=5 h2=4
10 10
Compute the expected amount of fill to raise the ground at a level of 40 m.
β’ Vf=A(ββ1:2ββ2:3ββ3:4ββ4
4)
β’ ββ1=7+9+4+6=26 m.
β’ ββ2 = 9+10+10+6+8+7+6+5=61m.
β’ ββ3 = 0
β’ ββ4 = 8+9+7+4= 28m.
β’ Vf(square)=(10*10)(26:2(61):3(0):4(28)
4)=6500 m3
β’ Vf(triangle)=A*h=(1
2*10*12*
4+8+53
)+=(1
2*10*12*
8+7+53
)
β’ +=(1
2*10*12*
6+7+53
)
β’ Vf(triangle)=1100 m3
β’ Vf(total)= 6500+700=7600 m3
h2=10 h1=9 h2=6 h1=4
10
h2=10
h2=9
10
5 h1=6
10 h1=7
12
10
h4=7 h4=9 h2=8
h4=8 h4=4 h2=7
h2=5 h2=6
10 10
h1=4
10
1
0
h3=5 h1=6 12
10
h2=8
h2=7
Volume from Contour
Vertical section
A1 45
A3 55 A2 50
A4 60
A5 65
50
45
55
60
65
A4 50
A2 60 A1 65
A5 45
A3 55
50
45
55
60
65
Example
β’ The land for a specified area containing a hilltop, where the area confined within the contour lines as following below, find the total Volume to level the ground at a level of 105 m. if you know the total area of the land = (60*40) m2
β’ The scale = 1:1500
Con.elev. 125 120 115 110 105 100 95 90
Area (cm2 2.8 3.6 4.2 5.4 6.2 7.4 8.8 10.2
Solution
Con.elev. 125 120 115 110 105 100 95 90
Area (cm2
2.8 3.6 4.2 5.4 6.2 7.4 8.8 10.2
Area (m2
630 810 945 1215 1395 1665 1980 2295
1cm= 15 m 1cm2 = 225 m2
Total Volume C = hi (π΄1:π΄π
2+ π΄2 + π΄3 + β― + π΄π β 1)
Total Volume f = hi (π΄1:π΄π
2+ π΄2 + π΄3 + β― + π΄π β 1)
90 95
100 105 115
125
120
100
95
105
110
115
Cut
fill fill
90
120
125
Vertical section
β’ Area of land (base) = (60*40)= 2400 m2
β’ A105= 2400-1395= 1005
β’ A100= 2400-1665= 735
β’ A95= 2400-1980= 420
β’ A90= 2400-2295= 105
β’ Total Volume f = 5 (105:1005
2+ 420 + 735)= 8550 m3
β’ Total Volume c = 5 (1395:630
2+ 1215 + 945 + 810)= 19912.5 m3