Question

A 30 m steel tape was standardised at a temperature of 20o C and under
a pull 5 kg. The tape was used in catenary to fix a distance of 28 m
between two points at 40o 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.

Accepted Answer

ΔL=L1ΑΔT L1​=  FRAC{ U0394L}{ U03B1 U0394T} ​L2​−L1​=L1​ΑΔT L2​=L1​(1+ΑΔT) L1​=  FRAC{L2}{1+ U03B1 U0394T} =  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

Question #297103

Expert's answer

Δl=l₁αΔt

l₁= $\frac{Δl}{αΔt}$ l₂−l₁=l₁αΔt

l₂=l₁(1+αΔt)

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

I₁₌ $\frac{308,000,000}{42,000,000}$ m= 7.33m

Question #297103

Expert's answer

Δl=l1αΔt

l1​=ΔlαΔt\frac{Δl}{αΔt}αΔtΔl​ ​l2​−l1​=l1​αΔt

l2​=l1​(1+αΔt)

l1​=l21+αΔt\frac{l2}{1+αΔt}1+αΔtl2​ = (28×11)106(2.1X106)×20)\frac{(28\times 11) 10^6}{(2.1 X 10^6)\times20)}(2.1X106)×20)(28×11)106​

​I​1=308,000,00042,000,000\frac{308,000,000}{42,000,000}42,000,000308,000,000​ m= 7.33m

Δl=l₁αΔt

l₁= $\frac{Δl}{αΔt}$ l₂−l₁=l₁αΔt

l₂=l₁(1+αΔt)

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

I₁₌ $\frac{308,000,000}{42,000,000}$ m= 7.33m