978-0134604657 Chapter 26

subject Type Homework Help
subject Pages 9
subject Words 1450
subject Authors Charles D. Ghilani

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26 VOLUMES
Asterisks indicate problems that have partial answers given in Appendix G.
26.1 What are the principle methods used to determine volumes in surveying?
26.2 Prepare a table of end areas versus depths of fill from 0 to 20 ft by increments of 2 ft for
level sections, a 40-ft-wide level roadbed with side slopes of 2:1.
Area = 2h2 + 40h
Fill Depth (ft)
End Area (ft2)
0
0
2
88
4
192
6
312
8
448
10
600
12
768
14
952
16
1152
18
1368
20
1600
26.3 Similar to Problem 26.2, except use side slopes of 1:1.
Area = 1h2 + 40h
Fill Depth (ft)
End Area (ft2)
0
0
2
84
4
176
6
276
8
384
10
500
12
624
14
756
16
896
(*)
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18
1044
20
1200
Draw the cross sections and compute Ve for the data given in Problems 26.4 through 26.7.
26.4* Two level sections 75 ft apart with center heights 4.8 and 7.2 ft in fill, base width 30 ft,
side slopes 2:1.
26.5 Two level sections of 30-m stations with center heights of 1.44 and 1.58 m. in cut, base
width 10 m, side slopes 1-1/2:1.
26.6 The end area at station 36+00 is 265 ft2. Notes giving distance from centerline and cut
ordinates for station 36+80 are C 3.8/16.6; C 4.9/0; C 5.6/10.2. Base is 20 ft.
707.6 yd3 = 19,110 ft3
x
y
10
0
16.6
3.8
38
0
4.9
81.34
10.2
5.6
0
10
0
0
10
0
0
119.34
End Area @ 36 + 80 = 112.66 ft2
26.7 An irrigation ditch with b = 10 ft and side slopes of 2:1. Notes giving distances from
centerline and cut ordinates for stations 52+00 and 53+00 are C 2.4/10.8; C 2.8; C
4.5/10.4; and C 2.1/12.2; C 3.5; C 3.7/11.2.
290.3 yd3 = 7839 ft3
x
y
+
x
y
+
-5
0
0
-5
0
0
10.8
2.4
-12
0
12.2
2.1
-10.5
0
0
2.8
30.24
29.12
0
3.5
42.7
39.2
10.4
4.5
0
22.5
11.2
3.7
0
18.5
5
0
0
0
5
0
0
0
-5
0
0
-5
0
0
18.24
51.62
32.2
57.7
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copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.
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End areas: 52 + 00 = 75.8 ft2 and 53 + 00 = 81.0 ft2
26.8 Why must cut and fill volumes be totaled separately?
are generally paid for cuts only.
26.9* For the data tabulated, calculate the volume of excavation in cubic yards between stations
10 00
and
15 00.
Station
Cut End Area
(ft )
2
10 00
263
11 00
358
12 00
446
13 00
402
14 00
274
15 00
108
4237.0 yd3
Station
Cut End Area (ft2)
Volume
10 + 00
156
11 + 00
208
674.1
12 + 00
342
1018.5
13 + 00
240
1077.8
14 + 00
198
811.1
15 + 00
156
655.6
4237.0
26.10 For the data listed, tabulate cut, fill, and cumulative volumes in cubic yards between
stations
10 00
and
20 00.
Use an expansion factor of 1.25 for fills.
End Area
(ft )
2
Station
Cut
Fill
10 00
0
11 00
108
12 00
238
13 00
201
14 00
106
14 60
0
0
15 00
102
16 00
138
17 00
205
18 00
147
© 2018 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all
copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.
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19 00
138
20 00
106
End Area
(ft2)
Volumes
Station
Cut
Fill
Cut
(yd3)
Fill
(yd3)
1.25Fill
(yd3)
Cumulative (yd3)
10 + 00
0
11 + 00
108
200.0
200.0
12 + 00
238
640.7
840.7
13 + 00
201
813.0
1653.7
14 + 00
106
568.5
2222.2
14 + 60
0
0
117.8
2340.0
15 + 00
102
75.6
94.4
2245.6
16 + 00
138
444.4
555.6
1690.0
17 + 00
205
635.2
794.0
896.0
18 + 00
147
651.9
814.8
81.2
19 + 00
138
527.8
659.7
-578.5
20 + 00
106
451.9
564.8
-1143.3
26.11 Calculate the section areas in Problem 26.4 by the coordinate method.
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26.13 Determine the section areas in Problem 26.7 by the coordinate method.
See solution to problem 26.7
26.14* Compute
P
C
and
P
V
for Problem 26.4. Is
P
C
significant?
 
12 27
p
26.15 Calculate
P
C
and
P
V
for Problem 26.7. Would
P
C
be significant in rock cut?
26.16 From the following excerpt of field notes, plot the cross section on graph paper and
superimpose on it a design template for a 30-ft wide level roadbed with fill slopes of
1-1/2:1 and a subgrade elevation at centerline of 3250.26 ft. Determine the end area
graphically by counting squares.
HI = 3246.99
20 00 Lt
5.2
50
4.8
22
6.6
0
5.9
12
7.0
30
8.1
50
404.1 ft2
-27.22
-22
0
12
30
30.44
15
-15
3242.12
3242.19
3240.39
3241.09
3239.99
3239.97
3250.26
3250.26
x
y
+
0
3240.4
38884.7
12
3241.1
0
97232.7
30
3240.0
38879.9
98629.9
30.44
3240.0
97199
48599.5
15
3250.26
98942.5
48754
15
3250.26
48753.9
88462
27.22
3242.1
48632
71327
22
3242.19
88242
0
0
3240.4
71289
75613.0
74804.7
Area = 0.5|75,613.0 − 74,804.7|=404.14
26.17 For the data of Problem 26.16, determine the end area by plotting the points in a CAD
package, and listing the area.
404.1 ft2
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26.18 For the data of Problem 26.16, calculate slope intercepts, and determine the end area by
the coordinate method.
26.19 From the following excerpt of field notes, plot the cross section on graph or in a CAD
program and superimpose on it a design template for a 30-ft wide level roadbed with cut
slopes of 1-1/2:1 and a subgrade elevation of 240.88 ft. Determine the end area
graphically by counting squares.
HI = 252.66 ft
46 00 Lt
6.0
50
7.9
27
5.5
10
4.9
0
6.6
24
6.5
50
232.8 ft2
Left side intercept
Grade slope: 247.16244.76
1727 = −0.141176
Grade elevation @ −15 is 247.16 − 0.141176(5)=246.454 ft
26.20 For the data of Problem 26.19, calculate slope intercepts and determine the end area by
the coordinate method.
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26.21* Complete the following notes and compute
e
V
and
.
P
V
The roadbed is level, the base is
26.22 Similar to Problem 26.21, except the base is 24 ft.
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12(27)(4.9 − 3.6)(59.5 − 66.3)= −2.7 yd3
26.23 Calculate
e
V
and
P
V
for the following notes. Base is 24 ft.
12 90
C6.4
43.6
C3.6
0
C5.7
40.8
12 30
C3.1
30.4
C4.9
0
C4.3
35.2
Ve = 477.4 yd3; Vp = 472.9 yd3
12 +
90
12 +
30
x
y
+
x
y
+
0
3.6
146.9
0
4.9
172.5
40.8
5.7
0
68.4
35.2
4.3
0.0
51.6
12
0
0
0.0
12
0
0.0
0.0
12
0
0
0.0
12
0
0.0
0.0
43.6
6.4
76.8
0.0
30.4
3.1
37.2
0.0
0
3.6
156.96
0
4.9
149.0
233.76
215.3
186.2
224.1
w1
84.4
w2
65.6
Area
224.5
Area
205.1
Ve = 12,889.2 ft3 = 477.4 yd3
 
3
60 4.9 3.6 65.6 84.4 4.5 yd
12 27
P
C  
26.24 Calculate
,,
eP
VC
and
P
V
for the following notes. The base in fill is 20 ft and base in cut
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15
0
0
0.0
18.3
22
330.0
0.0
20.1
3.4
51
0.0
0
0
0.0
0.0
0
2
40.2
330.0
0.0
91.2
12.0
Cut area
51.6
Cut area
165
Fill area
4
Fill area
15
46 + 00
fill
45 + 00
fill
6
0
0.0
0
0
0
13
2
12
20.0
14.5
3
0
30
10
0
0
0.0
10
0
0
0
6
0
0
0
0
0
12
20.0
0
30
ft3
yd3
Cut Volume
10830
401.1
Fill Volume
950
35.2
3
100 0 2.0 18.3 20.1 1.1 yd
Elevation (ft)
860
870
880
890
900
910
Area (ft2)
1370
1660
2293
2950
3550
4850
3.1136 ac-ft
Contour
Area
Volume
(ft3)
860
1370
870
1660
0.34780
880
2293
0.45374
890
2950
0.60181
900
3550
0.74610
910
4850
0.96419
3.11364
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26.26
Elevation (ft)
1015
1020
1025
1030
1035
1040
Area (ft2)
1815
2097
2391
2246
2363
2649
2.6001 ac-ft
Contour
Area
Volume
(ft3)
1015
1815
1020
2097
0.44904
1025
2391
0.51515
1030
2246
0.53225
1035
2363
0.52904
1040
2649
0.57530
2.60078
26.27 State two situations where prismoidal corrections are most significant.
26.28* Distances (ft) from the left bank, corresponding depths (ft), and velocities (ft/sec),
respectively, are given for a river discharge measurement. What is the volume in
3
ft /sec?
0, 1.0, 0; 10, 2.3, 1.30; 20, 3.0, 1.54; 30, 2.7, 1.90; 40, 2.4, 1.95; 50, 3.0, 1.60; 60, 3.1,
1.70; 74, 3.0, 1.70; 80, 2.8, 1.54; 90, 3.3, 1.24; 100, 2.0, 0.58; 108, 2.2, 0.28; 116, 1.5, 0.
419.3 ft3/s
Distance
Depth
Area
Velocity
Vavg
Discharge
0
1.0
0.00
10
2.3
16.5
1.30
0.65
10.73
20
3.0
26.5
1.54
1.42
37.63
30
2.7
28.5
1.90
1.72
49.02
40
2.4
25.5
1.95
1.93
49.09
50
3.0
27.0
1.60
1.78
47.93
60
3.1
30.5
1.70
1.65
50.33
70
3.0
30.5
1.70
1.70
51.85
80
2.8
29.0
1.54
1.62
46.98
90
3.3
30.5
1.24
1.39
42.40
100
2.0
26.5
0.58
0.91
24.12
108
2.2
16.8
0.28
0.43
7.22
116
1.5
14.8
0.00
0.14
2.07
419.3
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copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.
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