Question

A straight aluminum wire 30m Long is subject to a tensile stress of 70MPa.
1. Determine the total elongation of the wire.
2. What temperature change would produce the same elongation
Take: Elastic modulus = 200GNM^-2 And temperature coefficient = 25 × 10^-6

Answer 1

l=30 m, $\sigma_T=$ 70 MPa,Y=200 GPa

we know that

stress= $Y\times strain$

$\sigma _T= Y \times \frac{\Delta l}{l}$

here , $\Delta l = elongation$

$\Delta l= \frac{70 \times 30}{200}=10.5 mm$

(b) as per the concept of thermal stress

$\Delta l= l\alpha \Delta T$

10.5$\times 10^{-3}$ = $30\times 25\times 10^{-6} \Delta T$

$\Delta T =$ 13.33 $^oC$ This is the temperature difference which will produce above elongation