Find out the correct horizontal distance:

A 20 m steel tape was standardised on flat ground, at a temperature of 20^oC and at a pull of 15 kg. The tape was used within catenary at a temperature of 30^oC and the pull applied was 10 kg. A cross-sectional area of the tape is 0.02 cm², and its total weight is 400 gm. The Young's modulus of elasticity (E) and coefficient of thermal expansion (α) is 2.1 × 106 kg/cm² and 11 × 10^{- 6} per ^oC respectively. Find out the correct horizontal distance.

Solution

Correction for temperature C_t = α (T_m- T₀) L

= 11 × 10^{- 6}  (30 - 20) × 20

= 0.00220 m

Correction for pull C _p = (P-P₀ / AE) L

=   (10 - 15) × 20/0.02 × 2.1 × 10⁶

=- 0.00238 m

Correction for sag Cs  = l (wl )²/24 p²

=20 (0.4)²/ 24 (10)² = 0.00133

Because correction for sag is always negative so it will be subtracted. Thus, total correction per tape length would be:

0.00220 - 0.00238 - 0.00133 = - 0.00151 m

Correct horizontal distance = 20 - 0.00151 = 19.99849 m.