References:A text book of SURVEYING – M. Shahjahan, M.A. Aziz

Surveying – B.C. Punmia

Prepared by: Md. Aminul Islam, Lecturer,Dept. of CEE,SUST.

Introduction Calculation of area and volumes Chain Surveying Plane table Surveying Tacheometry or Stadia Surveying Introduction of photogrammetry

CHAIN SURVEYINGErrors in Chaining:If the length of the chain used in measuring length of the line

is not equal to the true length of the designated length, the measured length of the line will not be correct and suitable correction will have to be applied. The error may also be either cumulative or compensating.Cumulative error:Cumulative errors are those which may either go on

increasing or decreasing when a chain is shorter or longer than its standard length.

Cumulative error:When the chain is too short, the measured length of the

line is too great i.e., greater than its true length and therefore, the error is positive and the correction is negative. Again when the chain is too long, the measured distance

will be less, the error is negative and the correction is positive.

Compensating errors:Compensating errors are those which cancel one another

and finally their total effect remains approximately same. While stretching a chain one may pull it less than the

standard pull of that chain.

Compensating errors:

Again one may stretch it with a greater pull than the standard one. As a result the measured length in the former case will be higher and in the later case will be less.But when these two lengths are added, the two errors will

compensate each other.Errors and mistakes may arise form:1. Erroneous length of chain or tape (Cumulative + or -)2. Bad ranging (Cumulative +)3. Careless holding and marking (Compensating ±)4. Bad straightening (Cumulative +)

Errors and mistakes may arise form:

5. Non-horizontality (Cumulative +)6. Sag in chain (Cumulative +)7. Variation in temperature (Cumulative + or -)8.Variation in pull ( Compensating ±,Cumulative + or -)9. Personal mistake Displacement of arrows Miscounting chain length Misreading Erroneous reading

Details from Book: Punmia

Correction

Correction in length area and volume:Let,Le be the incorrect length of the chain and Lc be the corrected length of the chain.The correct distance,

The correct area,

The correct volume,

Example: The road from Dhaka to Mirpur is actually 25320 ft. long. This distance was measured by an Engineers defective chain and was found to be 25270 ft. How much correction does the chain need?

Solution:Correct distance, L = 25320 ftMeasured lenghth = 25270 ftCorrected length of the chain, Lc = 100 ftIncorrect length of the chain = Le

measured incorrect length

Correction = Le - Lc = 0.197 ft. (should be shortened)

Example: The length and width of a plot of land were measured by an Engineer's chain exactly 100 ft. in length at the beginning. But it was found to be 100.3 ft. long at the end of the survey work. The area of the plot drawn to a scale 1 inch = 100 ft. was 25.60 sq. inches. What was the true area of the plot?

Solution:Corrected length of the chain, Lc = 100 ftIncorrect length, Le = 100.3 ft

From the scale on the map 1 in2=1002 =10000 sq. ft.

Area of the plot= 25.70*10000=257000sq.ft Ans.

Example: The length, width and depth of a pond were measured 63.8 ft by an incorrect Gunter's chain. The volume of the pond was calculated to be 1,60,000 cft. The chain was tested at the end of the measurement of the tank. 65.8

Correction for temperature: If the temperature in field is more than the temperature at which the tape or

chain was standardized, the length of the tape increases, measured distance becomes less and therefore the correction becomes additive.

On the other hand it is negative.

The temperature correction is given by,

Where,∝ = co-efficient of thermal expansion.Tm =Mean temperature in the field during measurement.To =The standard temperature for the tape.L =Measured length. The co-efficient of thermal expansion of steel varies from 5.5x10-6 to6.85x10-6 per degree oF. The sign of the correction is plus or minus according

as Tm greater or less than To . The steel tapes are generally standardized at65°F.

𝐶𝐶𝑡𝑡 =∝ (𝑇𝑇𝑚𝑚 − 𝑇𝑇0)𝐿𝐿

Example: A distance of 1840 ft. was measured with a steel tape which was exactly 100 ft, long at 65°F. The temperature during measurement in the field was 85°F. Calculate the actual length of the line. Take the co-efficient of thermal expansion of tape =6.25xl0-6 per I'-'F.

First Method :Temperature correction for each tape length Ct= ∝((Tm –To)L =6.25 x10-6(85-65) 100 =0.012 ft. (positive)Length tape at 85°F= 100+0.012 = 100.012 ft.

Second Method :Total correction = 6.25x10-6 (85—65)x1840 = 0.230 ft.True length of the line= 1840+0.230 = 1840.230 ft.

Correction for pull or tension: If the pull applied during measurement is more than the pull at which the tape was

standardized, the length of the tape increases, measured distance becomes less and the correction is positive.

Similarly, if the pull is less, the length of the tape decreases, measured distance becomes more and the correction is negative.

If Cp is the correction for pull then,

Where L= length of tape, A=cross-sectional area of tape, Ff = pull applied in the field,Fs = pull at standardization, andE= Young's Modulus of Elasticity (for steel, E=30x106psi). Since the effect of pull on

tape is to make the measured length too short, the correction is always positive.

Example: A steel tape of 100 ft. length, standardized at 25 lb. pull, was used in the field with a pull of 35 lbs. The cross-sectional area of the tape is 0.025 sq. inch. Take the value of Young's Modulus of Elasticity for steel, 30xl06

psi. Calculate the correction for excess pull.

Correction for sag: When the tape is stretched on supports between two points it takes the form of

horizontal catenary. The horizontal distance will be less than the distance along the curve. Therefore, the

correction for sag is always negative. If Cs is the correction for sag, then,

Where, W=wt. of the tape in lb.,L=length of the tape in ft. and Ff = pull applied in the field in lb.Since the effect of sag on tape is to make the measured length too large. The

correction is always negative.

Example: A steel tape of 100 ft. length weighing 1.2lbs. was pulled with a force of 20 lbs. in the field to measure a certain distance. Calculate the correction for sag.

ProblemA steel tape 20m long standardized at 55ºF with a pull of 10Kg was used for measuring a base line. Find the correction per meter length, if the temperature at that time of measurement was 80ºF and the pull exerted was 16Kg. Weight of 1 cubic cm of steel = 7.86 gm. Weight of tape = 0.8Kg and E = 2.109x106 Kg/cm2. Co-efficient of

expansion of tape per one foot is 6.25 x10-6 .Solution:1. Correction for temperature, Ct= ∝((Tm –To) L

=6.25 x10-6(80-55) 20 =0.0031 m. (Additive)

2.Correction for pull,

Weight of tape, A(20x100)7.86x 10-3 kg = 0.8 kgA=0.051 sq.cm

Total correction= Ct +Cp-CS

=+0.0031+0.00112-0.00208=+0.00214m. Ans.

Thank You