Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36 [tex]E = 200000 MPa[/tex]
The strain of the bar when subjected to 150 MPa is
[tex]\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075[/tex]
Therefore, if the bar elongates by 1.35 mm, then the original length L would be:
[tex]\epsilon = \frac{\Delta L}{L}[/tex]
[tex]L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm[/tex] or 1.8m
