SURVEYING & LEVELLING

Lecture # 2 Civil Engineering Dept.

CHAINING ON LEVEL GROUND

To chain the line, the leader moves forward by dragging

the chain and by taking with him a ranging rod and arrows.

The follower stands at the starting station by holding the

other end of the chain

When the chain is fully extended, Leader holds the

ranging rod and the ranging process is done with the help

of directions given by the follower

Then the follower holds the zero end of the chain by

touching the station point

The leader stretches the chain and finally place it on the

line. Once done he inserts an arrow in the ground at the

end of the chain

Now the leader moves forward by dragging the chain with

arrows and the ranging rod and the whole process is

repeated again but this time there should be a surveyor on

the starting station to conduct the ranging operation

Chaining (contd.) On Sloping Ground

There are two methods. Direct and Indirect Method

Direct Method

The distance is measured in small horizontal stretches. Say a1, a2

…an. with suitable length of chain or tape. This method is direct one.

Finally the total horizontal distances are added to get the required

distances.

w

x

y

z

c

b

a

Chaining (contd.)

Chaining (contd.)

Indirect Method:

Horizontal distance of the segment is calculated by knowing sloping

length of the segment and angle of inclination of that with horizontal.

If the elevation difference between 2 terminals points and the sloping

distance between 2 terminal points is known the horizontal distance D

can be calculated as

D = (l²-h²)

Chaining (contd.) Indirect method (Vertical angle measured) Clinometer to measure vertical angle

D1 = s1×cosΦ1

The e ui ed ho izontal distance D =∑s cosΦ

D2

Φ1

Φ2

s1

s2

C

B

A

D1 Indirect method (difference in height)

D = √( s 2 - h 2 ) D

h

s A

B C

Indirect method (Hypotensual allowance)

B

A’

C

A

θ

1 chain

BA / = BC = 1 CHAIN

BA = 1 CHAIN × sec θ

AA / = BA - BA / = 1 CHAIN ( sec θ – 1)

TAPE CORRECTIONS

1. TEMPERATURE CORRECTION

This correction is necessary because the length of the

tape or chain may be increased or decrease due to rise

or fall of temperature, respectively, during

measurement

The sign of correction may be positive or negative

Coefficient of thermal expansion for the steel tape can

be assumed to be 11 x 10-6 per degree centigrade, if

not known

Where,

Ct Correction for temperature (m)

Tm Temperature during measurement (oC)

To Temperature at which tape was standardized (oC)

L Length of the tape (m)

Coefficient of thermal expansion (oC-1)

𝐶 = 𝛼 𝑇 − 𝑇 𝐿

2. PULL CORRECTION

Due to the elastic properties of the tape material,

when pull is applied, the strain will vary according to

the variation of applied pull, and hence necessary

correction should be applied

The sign of correction may be positive or negative

Modulus of elasticity of the tape may be assumed to be

2.1 x 106 kg/cm2, if not known

Where,

Cp Pull Correction (m)

Pm Pull applied during measurement (kg)

Po Pull at which tape was standardized (kg)

L Length of the tape (m)

E Modulus of elasticity (kg/cm2)

A Cross-sectional area of tape (cm2)

𝐶 = 𝑃 − 𝑃 𝐿𝐴 𝐸

3. SLOPE CORRECTION 𝐶ℎ = 𝐿 − 𝐿 cos 𝜃

Where,

Ch Slope correction (m)

h Vertical distance between

two points (m)

𝐶ℎ = 𝐿 − 𝐿 − ℎ 𝐶ℎ = ℎ2𝐿

h L

Ch (always negative)

A

B

C

(APPROXIMATE)

(EXACT)

(EXACT)

4. SAG CORRECTION

This correction is necessary when the measurement is

taken with tape in suspension (always negative)

𝐶 = 𝐿 𝐿24 𝑛 𝑃 = 𝐿𝑊24 𝑛 𝑃

Where,

Cs Correction for sag (m)

Pm Pull applied during measurement (kg)

n number of spans

w weight of tape per unit length (Kg/m)

W Total weight of the tape (kg)

5. NORMAL TENSION

The tensions at which the effect of pull is neutralized

by the effect of sag is known as normal tension

At this the elongation due to pull is balanced by the

shortening due to sag. Mathematically, 𝐶 = 𝐶 𝐿𝑊24 𝑛 𝑃 = 𝑃 − 𝑃 𝐿𝐴 𝐸

𝑊24 𝑃 = 𝑃 − 𝑃𝐴 𝐸

Let, Pn be the Normal tension or pull, then

(Considering n = 1)

𝑊 𝐴 𝐸24 = 𝑃 𝑃 − 𝑃

By substituting the values of Po, W, A and E, we get the

final equation of the following form, 𝑃 ± 𝑃 ± 𝐶 = 0

Pn is determined by trial and error method

CHAIN CORRECTIONS

1. CORRECTION FOR INCORRECT LENGTH 𝐿 = 𝐿′𝐿 𝐿

Where,

L Standard/ true length of chain (m)

Lm Measured length of the line (m)

Lt True length of the line (m)

L’ True length error (m)

(Positive when chain is too long and negative when chain is too short)

2. CORRECTION FOR INCORRECT AREA

𝐴 = 𝐿′𝐿 𝐴

Where,

L Standard/ true length of chain (m)

L’ True length error (m)

Am Measured length of the line (m2)

At True area (m2)

Numerical 1

A 20 m chain was used to measure the distance between two

stations, which was found to be 2500 m. Then same distance

when measured with a 30 m chain was found to be 2495 m. If

the 20 m chain was 0.07 m too long, what was the error in the

30 m chain?

𝐻𝑖𝑛𝑡: 𝐿 = 𝐿′𝐿 𝐿

(Ans. 0.17 m)

Numerical 2

A line was measured by a 20 m chain which was accurate

before starting the day’s work. After chaining 900 m, the

chain was found to be 6 cm too long. After chaining a total

distance of 1, 575 m, the chain was found to be 14 cm too long.

Find the true distance of the line.

𝐻𝑖𝑛𝑡: 𝐿 = 𝐿′𝐿 𝐿

(Ans. 1,579.72 m)

Numerical 3

An old map was plotted to a scale of 40 m to 1 cm. Over the

years, this map has been shrinking, and a line originally 20 cm

long is only 19.5 cm long at present. Again the 20 m chain was 5

cm too long. If the present area of the map measured by

planimeter is 125.50 cm2, find true area of the land surveyed.

𝐻𝑖𝑛𝑡: 𝐴 = 𝐿′𝐿 𝐴

(Ans. 21.23 hectares)

CONCLUDED