Answer:
[tex]\epsilon = 3.958\times 10^{-3}[/tex]
Step-by-step explanation:
The Young's Module of Aluminium is [tex]E = 69\times 10^{9}\,Pa[/tex]. The axial stress on the specimen is:
[tex]\sigma = \frac{F}{A_{t}}[/tex]
[tex]\sigma = \frac{35500\,N}{(0.01\,m)\cdot (0.013\,m)}[/tex]
[tex]\sigma = 2.731\times 10^{8}\,Pa[/tex]
The strain is derived of following expression:
[tex]\epsilon = \frac{\sigma}{E}[/tex]
[tex]\epsilon = \frac{2.731\times 10^{8}\,Pa}{69\times 10^{9}\,Pa}[/tex]
[tex]\epsilon = 3.958\times 10^{-3}[/tex]
The resulting strain from the specimen of aluminum is 3.958 × 10^-3.
It should be noted that the Young modules of aluminum will be:
= 69 × 10^9 Pa
The axial stress is given as:
= 35500 / (0.01)(0.014)
= 2.731 × 10^8 Pa
The strain will now be:
= (2.731 × 10^8 Pa) / (69 × 10^9 Pa)
= 3.958 × 10^-3.
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