Correction per tape length (kgf data, 61)

Reading. "Fully support condition" means the tape rests on the ground throughout the measurement, so there is no sag correction. The correction per tape length is made up of the temperature and pull corrections.

Data

• $l = 30$ m; $T_0 = 15^\circ$C, $T = 20^\circ$C; $P_0 = 10$ kgf, $P = 16$ kgf
• $W = 0.8$ kgf; density of steel $= 0.00786$ kgf/cm$^3$; $E = 2.1\times 10^{5}$ kgf/cm$^2$ (as given); $\alpha = 7.1\times 10^{-7}$/$^\circ$C

Cross-sectional area

$$A = \frac{W}{\text{density}\times l} = \frac{0.8}{0.00786\times 3000} = 0.03393\ \text{cm}^2$$

$AE = 0.03393\times 2.1\times 10^{5} = 7124.7$ kgf.

1. Temperature correction

$$C_t = 7.1\times 10^{-7}\times(20 - 15)\times 30 = +0.000107\ \text{m}$$

2. Pull correction

$$C_p = \frac{(16 - 10)\times 30}{7124.7} = +0.02526\ \text{m}$$

3. Correction per tape length

$$C = 0.000107 + 0.02526 = +0.02537\ \text{m}$$

Answer: correction per tape length $= +0.0254$ m (the true length of one tape length is 30.0254 m).

Note on $E$. The usual value for steel is $2.1\times 10^{5}$ N/mm$^2$ $= 2.1\times 10^{6}$ kgf/cm$^2$, and the printed unit may be a misprint for N/mm$^2$. With that value, $AE = 71{,}247$ kgf, $C_p = +0.00253$ m and the correction per tape length is $+0.00263$ m.

(If the tape were suspended at its ends in the field, a further sag correction of $-\dfrac{W^{2}l}{24P^{2}} = -0.0031$ m would apply.)