Scribd
Upload
2K views4 pages

CE 21 Correction

The document contains multiple examples of calculating corrections that need to be applied when measuring distances outdoors using a steel tape. These corrections account for: 1) Thermal expansion due to changes in temperature between when the tape is calibrated and when used. 2) Increased tension when pulled harder than under standard conditions. 3) Sag of the tape when only supported at the ends instead of along the entire length. An example combining all three corrections is given to calculate the corrected distance between zero and 30 meters on a tape under non-standard outdoor conditions.

Uploaded by

Gintoki
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
2K views4 pages

CE 21 Correction

The document contains multiple examples of calculating corrections that need to be applied when measuring distances outdoors using a steel tape. These corrections account for: 1) Thermal expansion due to changes in temperature between when the tape is calibrated and when used. 2) Increased tension when pulled harder than under standard conditions. 3) Sag of the tape when only supported at the ends instead of along the entire length. An example combining all three corrections is given to calculate the corrected distance between zero and 30 meters on a tape under non-standard outdoor conditions.

Uploaded by

Gintoki
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 4

Christian Emil R.

Pacifico

10/13/2020

CORRECTION DUE TO TEMPERATURE

13. A 30-m steel tape is of standard length at 20°C. If the coefficient of thermal expansion
of steel is 0.0000116/1°C, determine the distance to be laid out using this tape to establish
two points exactly1235.65 m apart when the temperature is 33°C.

Given:

NL= 30m

α= 0.0000116/1°C

To= 20°C

L=1235.65m

T=33°C

Required:

The distance to be laid out, L’

Solution:

CT = L × α × (t – t0).

CT = 1235.65 m × 0.0000116/1°C × (33°C - 20°C)

CT = 0.186 m

L’ = L - CT

L’ = 1235.65m – 0.186m

L’= 1235.46 m

CORRECTION DUE TO TENSION

17. A 30-m steel tape weighing 1.75kg is of standard length under a pull of 4.55 kg,
supported for full length. This tape was used in measuring a line (found to be1371.50 m)
on smooth level ground under a steady pull of 8 kg. Assuming E = 2.05 x 106 kg/cm2 and
that the unit weight of steel is 7.9 x 10-3 kg/cm3, determine the following: cross-sectional
area of the tape, correction for increase in tension for the whole length measured, and the
correct length of the measured line.

Given:

NL = 30 m

P = 8.0 kg E = 2.05 x 106 kg/

P0 = 4.55 kg W = 1.75 kg
ρ = 7.9 x 10-3 kg/cm3
ML = 1371.50 m

Required:

Cross-sectional area of the tape, correction to be applied and correct length of the
line, A, C and CL

Solution:

V = A × L, and ρ × V = W,

( )
, and

( )( )
( )( ⁄ )

CL = ML ± CP

CL = 1371.50 m + 0.026 m

CL = 1371.526 m
CORRECTION DUE TO SAG

19. A 30-m steel tape weighing 0.04 kg/m is constantly supported only at its endpoints,
and used to measure a line with a steady pull of 8.5kg. If the measured length of the line
is 2465.18 m, determine the correct length of the line.

Given:

ML = 2465.18 m NL = 30.00 m

w = 0.04 kg/m P = 8.5 kg

Required:

Correct length of the line, CL

Solution:

( )
∑

( ⁄ ) ( ) ( ⁄ ) ( )

( ) ( )

CL = ML ± CS

CL = 2465.18 m – 0.025 m

CL = 2465.155 m
`

COMBINED CORRECTIONS

22. A 30-m tape weighs 12.5 g/m and has a cross section of 0.022 cm2. It measures
correctly when supported throughout under a tension of 8.0 kg and at a temperature of
20ºC. When used in the field, the tape is only supported at its ends, under a pull of 9.0 kg
and at an average temperature of 28ºC. Determine the distance between the zero ad 30-m
marks.

Given:

P = 9.0 kg P0 = 8.0 kg

A = 0.022 cm2 E = 2.0 x 106 ⁄

t = 28ºC t0 = 20ºC

α = 0.0000116 ºC NL = 30.0 m

Required:

Corrected Length, CL

Solution:

CT = Lα(t - t0)

CT = 30 m(11.6 x 10-6 / ºC)(28ºC -20ºC)

CT = 2.784 x 10-3 m
( )
CP =

CL = L ± Cp ± CT
( )( )
CP =
( )( ⁄ )

CP = 6.818 x 10-4 m

CL = 30.0 m + 6.818 x 10-4 m + 2.784 x 10-3 m

CL = 30.0034658 m or 30.003 m

You might also like