Corrections Of Tape

TAPING CORRECTIONS

There are two types of corrections depending upon the type of errors in tape due to the different conditions. 

1. Systematic Errors :

• Slope
• Erroneous tape length
• Temperature
• Tension
• Sag

2. Random Errors :

• Slope 
• Alignment
• Marking & Plumbing
• Temperature
• Tension & Sag

1. Temperature Correction 

It is necessary to apply this correction, since the length of a tape is increased as its temperature is raised, and consequently, the measured distance is too small.  It is given by the formula,

                        C_t = 𝛼 (T_m – T_o)L

Where, 

 C_t = the correction for temperature, in m.

    𝛼= the coefficient of thermal expansion.   T_m = the mean temperature during measurement.     

   T_o = the temperature at which the tape is standardized.  

   L = the measure length in m.

• For Foot Unit :

                        C_t = 6.45×10^-6 (T_m – 68)L

• For Metric Unit :

                        C_t = 1.16×10^-5 (T_m – 20)L

Example on Tape Corrections

Question : A line was measured with a steel tape which was exactly  30m long at 18^oC and found to be 452.343 m.  The temperature during measurement  was 32^oC.  Find the true length of the line.  Take coefficient of expansion of the tape  per ^oC=0.0000117.

            Temperature correction per tape length = C_t

                                  = α  (T_m - T_o) l

Here     l = 30 m:  T_o =18^oC; T_m = 32^oC;

            α = 0.0000117

            ~          C_t = 0.0000117 (32-18) 30

                            = 0.004914 m (+ ve)

Hence the length of the tape at 32^oC = 30 + C_t

                                   = 30 + 0.004914 = 30.004914 m.

            Now true length of a line = L’ / L x its measured length.

L = 30 m: L’ = 30.004914 m; measured length = 452.343 m.

~ True length = 30.004914 / 30 x 452.343 = 452.417 m.

2. Length Correction

It is necessary to calculate the correction in leght of tape to reduce error. If the scale has any error then it increases with the number of tape used.

3. Correction Due To Tension Or Pull

If pull applied while standardising the length of tape and pull applied in the field are different, this
correction is required.
Let, Ps = Standard pull
P = Pull applied in the field
A = Cross-sectional area of the tape
L = Measured length of line.
E = Young’s modulus of the material of tape, then

                              Cp = [(P - Ps)L] / AE

And, 

            

E steel = 21 x 10^5 kg / cm2

E steel = 30 x 10^6 /bs/in2

4. Correction Due To Sag

A downward curve or buldge in a structure caused by weakness or excessive weight or pressure is called Sag. 

Calculate sag correction for a 30 m steel tape under a pull of 80 N, if it is suspended in three equal spans. Unit weight of steel is 78.6 kN/m3. Area of cross-section of tape is 8 mm2.
Solution: Length of each span = 10 m

 

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