- 32 -

sunglowcitrineΠολεοδομικά Έργα

15 Νοε 2013 (πριν από 3 χρόνια και 8 μήνες)

110 εμφανίσεις

-

32

-




1.

By

the

collimation

system.



S
tation
point



Chainage



BS



IS



FS


RL

of
collimation
line (HI)



RL



Remark

1

165

3.150



101.235

98.085


2

180


2.245



98.990


3

195


1.125



100.
1
10


4

210

3.125


0.860

103.500

100.375

changed

point


5


225



2.760




100.740


6

240


1.835



101.665


7

255


1.470



102.030


8

270

1.225


1.965

102.760

101.535

Change

point

9

285


2.390



100.370


10

300



3.035


99.725


T
otal

7.500


5.860





Arithmetical

check:



BS

-


FS

=

7.500+5.860

=


1.640

Last

RL
-

1
s
t

RL

=
99.725
-

99.085

=

1.640

(ok)



2.

By

the

rise
-
and
-
fall

system



S
tation
point


Chainage


BS


IS


FS


Rise


Fall


RL


Remark

1

165

3.150





98.085


2

180


2.245


0.905


98.990


3

195


1.125


1.120


100.
1
10


4

210

3.125


0.860

0.265


100.375

Change

point

5

225


2.760


0.365


100.740


6

240


1.835


0.925


101.665


7

255


1.470


0.365


102.030


8

270

1.225


1.965


0.495

101.535

Change

point

9

285


2.390



1.165

100.370


10

300



3.035


0.645

99.725


T
otal=


7.500


5.860

3.945

2.305



-

33

-




Arithmetical

checl:

BS

-

F
S

=

7.500
-
5.860

=

+

1.640


Rise

-

fal
l

=

3.945
-
2.30
5

=

+

1.640


Last

RL
-

1
s
t

RL

=

99.725

-

98.085

=

+

1.640




22

(a)

Define

the

following

terms


(i)

Levelling

(2

marks)


(ii)

Level

surface

(2

marks)


(b
)
Th
e

followin
g

consecutiv
e

reading
s

wer
e

take
n

wit
h

a

leve
l

an
d

a

4
-
mete
r

levelling

sta
f
f

on

a

continuously

sloping

ground

at

common

intervals

of

30m:

0.855

(on

A)

1.545m,

2.335,

3.
1
15,

3.825,

0.455,

1.380,

2.055,

2.855,

3.455,


0.585,

1.015,

1.850,

2.755,

3.845

(on

B).


The

RL

of

A

was

380.500.

Mark

entires

in

a

level

book

and

apply

the

usual
checks.

Determine

the

fradient

of

AB.
(16

marks)

(i
)

Levelling

The

art

of

determining

the

relative

h

eights

of

di
f
ferent

point

on

or

below

the
surface

of

the

earth

is

known

as

levelling.

Thus,

levelling

deals

with

measurement

in

the


vertical

plane.


(ii)

Level surface


Any

surface

parallel

to

the

mean

spheroidal

surface

of

the

earth

in

side

to

be

a
level

surface.

Such

a

surface

is

obviously

curved.

The

water

surface

of

a

still

lake

is

also
considered

to

be

level

surface.


(b)

-

34

-





S
tation
point


Chainage


BS


IS


FS


Rise(+)


Fall

(
-
)


RL


Remark

A

0

0.855





380.500



30


1.545



0.690

379.810



60


2.335



0.790

379.020



90


3.
1
15



0.780

378.240



120

0.455


3.825


0.710

377.530

Change

points


150


1.380



0.925

736.605



180


2.055



0.675

375.930



210


2.855



0.800

375.130



240

0.585


3.455


0.600

374.530

Change

points


270


1.015



0.430

374.100



300


1.850



0.835

373.265



330


2.755



0.905

372.360


B

360



3.845


1.090

371.270


T
otal


1.895


1
1.125

0

9.230





Check:

B
S
-

FS

=

1.895

-

1
1
.

9
1
d
.

2
i
f
3
f
0
e
=
r
e
-
n
9
t

.
o
2
f
3
l
0
e
v
el

3
h
3
9
o
6
r
i
0
zontal

distance

Rise

-

fall

=

0

-

9.230

=

-

9.230


Last

RL
-

1
s
t

RL

= 371.270
-

380.500 =
-

9.230



Falling

gradient

of

AB

=



=

=

(i.e 1 in 39)




23

(a)

Define

the

following

terms.


(i)

Datum

surface

(or)

line
(2marks)


(ii)

Reduced

level

(2

marks)


(b
)
The

following

consecutive

readings

were

taken

with

a

level

and

a

4

metre

levelling
sta
f
f

on

continuously

sloping

ground

at

a

common

interval

of

30m.

0.585

on

A,

0.936,

1.953,

2.846,

3.644,

3.938,

0.962,

1.035,

1.689,

2.534,

3.844,

0.956,


1.579,

3.016,

on

B.

-

35

-




The

elevation

of

A

was

520,

450.

Make

up

a

level

book

and

apply

the

usual

checks.
Determine

the

gradient

of

the

line

AB.
(16

marks)

Solution:


(i)

Datum

surface

or

line


This

is

an

imaginary

level

surface

or

level

line

from

which

the

vertical

distances
of

di
f
ferent

points

(above

or

below

this

line)

are

measured.

(ii
)
Reduced

level

(RL)


The

vertical

distance

of

a

point

above

or

below

the

datum

line

is

known

as

the
reduced

level

(RL)

of

that

point.

The

RL

of

a

point

may

be

positive

or

negative

according
as

the

point

is

above

or

below

the

datum.




(b)





S
tation


Distance

Readings



Collimation


Reduced

Level



Remarks

Back
sight

Inter
sight

Fore
sight

A





















B

0

30


60


90

120


150


180

210


240


270

300


330

0.585









0.962








0.956



0.936


1.953


2.846

3.644




1.035

1.689


2.534




1.579











3.938








3.844




3.016

521.035









518.059








515.171

520.450

520.099


519.082


518.189

517.391


517.097


517.024

516.370


515.525


514.215

513.592


512.155











Chang
e

point








Chang
e

point


Arithmetical
check

2.503


10.798



-

8.295


-

8.295

-

36

-




R.L

of

A

=

520.450


R.L

of B

= 521.155


Di
f
ference

=

-

8.295


There

is

a

fall

of

8.925

metres

from

A

to

B.

The

distance

AB

=

330

metres.


Gradient

of

the

line

AB =


i.e.

1 in 39.8 (falling)




24

(a)

Define

the

following

terms.


(i)

Foresigh

reading

(2

marks)


(ii)

Intermediates

sights

reading.

(2

marks)


(b)

Following

consecutive

readings

were

taken

with

a

dumpy

level:


0.894,

1.643,

2.896,

3.016,

0.954,

0.692,

0.582,

0.251,

1.532,

0.996,

2.135.

The

instrument

was

shifted

after

the

fourth

and

the

eighth

readings.

The

first

reading
was

taken

on

the

sta
f
f

held

on

the

bench

mark

of

R.L

820.765.


Rule

out

a

page

of

a

level

field

book

and



8.
2
t
h
9
e
5

a
=
bov
e
1
re
a
=
di
n
g
1
s.

Calculate

the

reduced
enter

33
0

39.7
7

39.8

levels

of

the

points

and

show

the

usual

checks.


What

is

the

di
f
ference

of

level

between

the

first

and

last

points?

Using

the

collimation
system.

(16

marks)

Solution:


(i)

Foresigh

reading

(FS)


It

is

the

last

sta
f
f

reading

in

any

set

up

of

the

instrument

and

indicates

the

shifting
of the latter .

(ii)

Intermediates

sights

reading

(IS)


It

is

any

other

sta
f
f

reading

between

the

BS

and

FS

in

the

same

set

up

of

the
instrument.

-

37

-






S
tation

S
tation



Reduced

Level



Remarks

Back
sight

Inter
sight

Fore
sight


Rise


Fall


1


2


3


4


5


6


7


8


9


0.894






0.954






1.532




1.643


2.896




0.692


0.582




0.996








3.016






0.251




2.135










0.262


0.
1
10


0.331


0.536




0.749


1.253


0.120










1.139


820.765


820.016


818.763


818.643


818.905


819.015


819.346


819.882


818.743


B.M
C.P
C.P


Arithmetic

Check

3.380


5.402

1.239

3.261


-

2.022


-

2.022

-

2.022





The

di
f
ference

of

level

between

the

first

and

last

points

=

820.765

-

818.143

=

2.022,
which

indicates

that

there

is

a

fall

form

the

1
s
t

point

to

the

last

point.




25

(a)

Define

the

following

terms.
(i)

Bench
-

marks
(2 marks)

(ii)

Backsight

reading

(2

marks)


(b)

The

following

consecutive

readings

were

taken

with

a

levelling

instrument

of
intervals

of

20m.

2.375,

1.730,

0.615,

3.450,

2.835,

2.070,

1.835,

0.985,

0.435,

1.630,

2.255

and

3.630m.
The

instrument

was

shifted

after

the

fourth

and

eight

readings.

The

last

reading

was
taken

on

a

B.M

of

R.L
1
10.200m.

Find

the

R.Ls

of

all

the

points.

Using

Rise

and

Fall
system.

(16

marks)

Solution:


(a)

(i)

Bench
-
marks

(BM)

-

38

-




S
tation


Chainage

(m)


BS


IS



FS



Rise



Fall


RL


Remarks


0


20


40


60


80


100


120


140


160


180

2.375






2.835






0.435



1.73


0.615




2.07


1.835




1.63


2.255







3.45






0.985






3.63



0.645


1.
1
15




0.765


0.235


0.85







2.835








1.195


0.625


1.375

1
12.62


1
13.265


1
14.38


11
1.545


1
12.31


1
12.545


1
13.395


1
12.2


11
1.575


1
10.2




C.P
C.P
B.M


5.645


8.065

3.61






These

are

fixed

points

or

marks

of

known

RL

determined

with

reference

to

the
datum

line.

These

are

very

important

marks.

They

serve

as

reference

points

for

finding
the RL of new points or for conducting levelling opreations
in projects involving roads
railways

etc.

(ii)

Backsight

reading

(BS)


Thi
s

i
s

th
e

firs
t

sta
f
f

readin
g

take
n

i
n

an
y

se
t

u
p

o
f

th
e

instrumen
t

afte
r

th
e

levelling

has

been

perfectly

done.

This

reading

is

always

taken

on

a

point

of

known

RL

i.e

on
bench
-
mark

or

change

points.




(b)

Rise

and

Full

system




























T
otal



Check




BS

-

F
S

=

5.645

-

8.065

=

-
2.42


Rise

-

Fall

=

3.61

-

6.03

=

-
2.42


Last

RL
-

1
s
t

RL

=

1
10.2

-

1
12.62

=

-
2.42

(ok)

-

39

-




S
tation


Chainage

(m)



BS


IS



FS



Rise



Fall



RL


Remarks


140


160


180


200


220


240


260


280


300

3.15




3.86






0.47



2.245




20125


0.76




1.935


3.225





1.125






2.235






3.89



0.905


1.12


1.735


1.365











1.475


1.465


1.29


0.665

103.565


104.47


105.59


107.325


108.69


107.215


105.75


104.46


103.795





CP






CP


7.48


7.25

5.125

4.895




26

(a)

Define

the

following

terms.
(i)

Change

point

(2

marks)

(ii)

Height

of

instrument

(2

marks)


(b)

The

following

successive

readings

were

taken

with

a

dumpy

level

along

a

chain
line

at

common

intervals

of

20m.

The

first

reading

was

taken

on

a

chainage

140m.

The
R.L

of

the

second

change

point

was

107.215m.

The

instrument

was

shifted

after

the
third

and

seventh

readings.

Calculated

the

RLs,

of

all

the

points;

3.150,

2.245,

3.860,

2.125,

0.760,

2.235,

0.470,

1.953,

3.225,

and

3.890m.

Using

Rise
and

Fall

method.

(16

marks)

Solution:


(a)

(i)

Change

point

(CP)


This

point

indicates

the

shifting

of

the

instrument.

At

this

point,

an

FS

is

taken
from

one

setting

and

a

BS

from

the

next

setting.

(ii)

Height of
instrument

(HI)


Whe
n

th
e

levellin
g

instrumen
t

i
s

propert
y

levelle
d

th
e

R
L

o
f

th
e

lin
e

o
f

collimation

is

known

as

the

height

of

the

instrument.

This

is

obtained

by

adding

the

BS

reading

to
the

RL

of

the

BM

or

CP

on

which

the

sta
f
f

reading

was

taken.



(b)

Rise

and

Fall

method


























T
otal

-

40

-




Check




BS

-

F
S

=

7.48

-

7.25

=

0.23


Rise

-

Fall

=

5.125

-

4.895

=

0.23


Last

RL
-

1
s
t

RL

=

103.795

-

103.565

=

0.23

(ok)





27

(a)

Define

the

following

terms.
(i)

Horizontal

plane

(
2

marks)

(ii)

Line

of

the

collimation

(2

marks)


(b
)
In

fly

levelling

from

a

BM

of

R.L

140.605,

the

following

readings

were

observed;
Backsight

-

1.545,

2.695,

1.415,

2.925

Foresight

-

0.575,

1.235,

0.595

From

the last position of the
instrument, six pegs at 20m intervals are to be set out on a
uniformly

rising

gradient

of

1

in

50,

the

first

peg

is

to

have

an

RL

of

144.000.

Find

the
sta
f
f

readings

and

RLs

of

the

pegs.

(16

ma
r



k
1
s
)

50

Solution


(a)

(i)

Horizontal

plane


Any

plane

tangential

to

the

level

surface

at

any

point

is

known

as

the

horizontal
plane.

It

is

perpendicular

to

the

plumb

line

which

indicates

the

direction

of

gravit
y
.

(ii)

Line

of

the

collimation


It

is

an

imaginary

line

passing

through

the

intersection

of

the

cross
-
hairs

at

the
diaphragm

and

the

optical

centre

of

the

object

glass

and

its

continuation.

It

is

also

known
as

the

line

of

sight.




(b
)
For

pegs,

rising

gradiend

=

1

in

50

=


H

= 50


H

= 20

V

= 1


V

= ? (




×

20 = 0.4m)


RL

of

peg

(2)

=

144

+

0.4

=

144.4


RL

of peg

(3)

=

144.4

+

0.4

=

144.8

and

so on.

-

41
-












rface

with

the

ground

surface

is

known

as

the
so

be

difined

as

a

line

passing

through

points



dicates

that

all

the

points

on

this

line

have

an






S
tation


Distance


BS


IS


FS


HI


RL


Remark








0

20

40

60

80

100

1.545

2.695

1.415

2.925







2.78

2.38

1.98

1.58

1.18



0.575

1.235

0.595









0.78

142.15

144.27

144.45

146.78

146.78

140.605

141.575

143.035

143.855

144

144.4

144.8

145.2

145.6

146

BM
CP
CP
CP

pe
g
-

1
pe
g
-

2
pe
g
-

3
pe
g
-

4
pe
g
-

5
pe
g

-

6


8.58


3.185





Check

BS

-

FS

=

8.58

-

3.185

=

5.395

Last

RL
-

1
s
t

RL

= 146
-

140.605 = 5.395 (ok)



28

Define

the

terms

"contour

line",

"contour

interval"

and

"horizontal

equiralent".


(20

marks)
Solution:

1.

Contour

line


The

line

of

intersection

of

a

level

su
contour

line

or

simply

the

contou
r
.

It

can

al
of

equal

reduced

levels.

For

example,

a

contour

of

100m

in


RL

of 100m. Similarl
y
, in a contour of 99m all points have an RL

of 99m, and so on.

A


map

showing

only

the

contour

lines

of

an

area

is

called

a

contour

map.

-

42

-



2.

Contour

interval


Th
e

vertica
l

distanc
e

betwee
n

an
y

tw
o

consecutiv
e

contour
s

i
s

know
n

a
s

a

contour

interval.

Suppose

a

map

includes

contour

lines

of

100m,

98m,

96m

and

so

on.

The
contour

interval

here

is

2m.

This

interval

depends

upon:

(i)

the

nature

of

the

ground

(i.e
whether

flat

or

sleep).

(ii)

the

scale

of

the

map

and

(iii)

the

purpose

of

surve
y
.

Contour

intervals

for

flat

country

are

generally

small,

e.g.

0.25m,

0.50m,

0.75m,
etc.

The

contour

interval

for

a

sleep

slope

in

a

hilly

area

is

generally

greate
r
,

e.g.

5m,

10m,

15m,

etc.


Again

for

a

small
-
scale

map,the

interval

may

be

of

1m,

2m,

3m,

etc

and

for

la
r
ge
scale

map,

it

may

be

of

0.25m,

0.05m,

0.75m,

etc.

It

should

be

remembered

that

the

contour

interval

for

a

particular

map

is

constant.


3.

Horizontal

equivalent


Th
e

horizonta
l

distanc
e

betwee
n

an
y

tw
o

consecutiv
e

contour
s

i
s

know
n

as

horizontal

equivalent.

It

is

not

constant.

It

varies

accordin

to

the

steepness

of

the

ground.
For

steep

slopes,

the

contour

lines

run

close

togethe
r
,

and

for

flatter

slopes

they

are
widely

spaced.



29

Describe

full
y
,

with

sketches,

the

characteristics

of

contours.

(20

marks)
Solution:

Characteristics

of

Contours


1.

In

Fig

the

contour

lines

are

closer

near

the

top

of

a

hill

or

high

ground

and

wide

apart
near

the

foot.

This

indicates

a

very

steep

slope

towards

the

peak

and

a

flatter

slope
towards

the

foot.


-

43

-






2.

In

Fig

the

contour

lines

are

closer

near

the

bank

of

a

pond

or

depression

and

wide
apart

towards

the

centre.

This

indicates

a

steep

slope

near

the

bank

and

a

flatter

slope

at

the

cen
tre.





3.

Uniformly

spaced,

contour

lines

indicate

a

uniform

slope.








4.

Contour

lines

always

form

a

closed

circuit.

But

these

lines

may

be

within

or

outside
the

limits

of

the

map.














5.

Counter

lines

cannot

cross

one

anothe
r
,

except

in

the

case

of

an

overhanging

cli
f
f.

But
the

overlapping

portion

must

be

shown

by

a

dotted

line.



-

44

-






6.

When

the

higher

values

are

inside

the

loop,

it

indicates

ridge

line.

Contour

lines

cross
ridge

at

right

angles.















7.

When

the

lower

values

are

inside

the

loop,

it

indicated

a

valley

line.

Contour

lines
cross

the

valley

line

at

right

angles.







8.A

series

of

closed

contours

always

indicates

a

depression

or

summit.

The

lower

values
being inside the loop indicates a depression and the higher values being inside the loop
indicates

a

summit.





-

45

-



9.

Depression

between

summits

are

called

saddles.






10.

Contour

lines

meeting

at

a

point

indicate

a

vertical

cli
f
f.







30

Describe

the

direct

method

of

contouring

with

sketch.

(20

marks)
Solution:

Direct

Method


There

may

be

two

cases,

as

outlined

belo
w
.


Case

I
:

When

the

area

is

oblong

and

cannot

be

controlled

from

a

single

station:


Procedure


1
.

Suppose

a

contour

map

is

to

be

prepared

for

an

oblong

area.

A

temporary

beneh
-

mark

is

set

up

near

the

site

by

taking

fly

level

readings

from

a

permanent

bench
-

mark.

2
.

Th
e

leve
l

i
s

the
n

se
t

u
p

a
t

suitabl
e

statio
n

I

fro
m

wher
e

maximu
m

are
a

ca
n

b
e

covered.


3
.

The

plane

table

is

set

up

at

a

suitable

station

P

from

where

the

above

area

can

be
plotted.

4
.

A

back

sight

reading

is

taken

on

the

TBM.

Suppose

the

RI

of

the

TBM

is

249.500m
and

that

the

BS

reading

is

2.250m.

Then

theRI

of

III

is

251.750m.

If

a

contour

of

-

46
-




250.000m

is

required

the

sta
f
f

reading

should

be

1.750m.

If

a

contour

of

249.000m

is
required

the

sta
f
f

reading

should

be

2.750m

and

so

on.

5
.

The

sta
f
fman

holds

the

sta
f
f

at

di
f
ferent

points

of

the

area

by

moving

up

and

down

or
left

and

right,

until

the

sta
f
f

reading

is

exactly

1.750.

Then

the

points

are

marked

by
pegs.

Suppose,

these

points

are

A,

B,

C,

D.

6
.

A

suitable

point

p

is

selected

on

the

sheet

to

represent

the

station

P
.

Then,

with

the
alidade

touching

p

rays

are

drawn

to

A.B.C

and

D.

The

distances

P
A,

PB,

PC

and

PD
are

measured

and

plotted

to

a

suitable

scale.

In

this

manne
r
,

the

points

a,

b,

c

and

d

of
the

contour

line

of

RI

250.000m

are

obtained.

These

points

are

jointed

to

obtain

the
contour

of

250.000m.








7
.

Similarl
y
,

the

points

of

the

other

contours

are

located.


8
.

When

reqired,

the

levelling

instrument

and

the

plane

table

are

shifted

and

set

up

a
new

position

in

order

to

continue

the

operation

along

the

oblong

area.


Case

II


Procedure


1
.

The

plane

table

is

set

up

at

a

suitable

station

P

from

where

the

whole

are

can

be
commanded.

2
.

A

point

p

is

suitably

selected

on

the

sheet

to

represent

the

station

P
.

Radial

lines

are
then

drawn

in

di
f
ferent

directions.


-

47
-



3
.

A

temporary

bench
-
mark

is

established

near

the

site.

The

level

is

set

up

at

a

suitable
positio
n

I

an
d

a

B
S

readin
g

i
s

take
n

o
n

th
e

TBM
.

Le
t

th
e

H
I

i
n

thi
s

settin
g

b
e

153.250m

so

to

find

the

contour

of

RI

152.000m

a

sta
f
f

reading

of

1.250m

is

required

at

a
particular

point,

so

that

the

RI

of

contour

of

that

point

comes

to

152.000m.

RI

= HI
-

sta
f
f reading


=

153.250

-

1.250

=

152.000m


4
.

The

sta
f
fman

holds

the

sta
f
f

along

the

rays

drawn

from

the

plane

table

station

in

such
a

way

that

the

sta
f
f

reading

on

that

point

is

exactly

1.250.

In

this

manner

points

A,

B,
C,

D

and

E

are

located

on

the

ground,

where

the

sta
f
f

readings

are

exactly

1.250.

5
.

The

distances

P
A,

PB,

PC,

PD

and

PE

are

measured

and

plotted

to

any

suitable

scale.


Thus

the points a, b, c are obtained which are jointed in order to obtain a contour of


152.000.


6
.

The

other

contours

may

be

located

in

similar

fashion.









BY

T
U

(Hmawbi)

dr.ayemyint@gmail.com

09
-
5030281,01
-
620072/620454