Answer to Question #297103 in Civil and Environmental Engineering for himi

Question

A 30 m steel tape was standardised at a temperature of 20^o C and under a pull 5 kg. The tape was used in catenary to fix a distance of 28 m between two points at 40^o C and under a pull of 5 kg. Given that the cross-sectional area of the tape = 0.02 cm², total weight 470 g. Young’s modulus of steel = 2.1 X 10⁶ kg/cm, and coefficient of linear expansion = 11 x10^-6per ^oC, (a) find the correct distance between the points, and (b) find the value of pull for which the measured distance would be equal to the correct distance.

Answer 1

Δl=l₁αΔt

l₁​=$\frac{Δl}{αΔt}$ ​l₂​−l₁​=l₁​αΔt

l₂​=l₁​(1+αΔt)

l₁​=$\frac{l2}{1+αΔt}$ = $\frac{(28\times 11) 10^6}{(2.1 X 10^6)\times20)}$

​I​₁₌$\frac{308,000,000}{42,000,000}$ m= 7.33m

=7.33m