Answer:
[tex]\epsilon=4.16\times 10^{-3}\[/tex]
Explanation:
It is given that,
Dimension of specimen of aluminium, 9.6 mm × 12.9 mm
Area of cross section of aluminium specimen, [tex]A=9.6\times 12.9=123.84\times 10^{-6}\ mm^2[/tex]
[tex]A=123.84\times 10^{-6}\ m^2[/tex]
Tension acting on object, T = 35600 N
The elastic modulus for aluminum is, [tex]E=69\ GPa=69\times 10^9\ Pa[/tex]
The stress acting on material is proportional to the strain. Its formula is given by :
[tex]\epsilon=\dfrac{\sigma}{E}[/tex]
[tex]\sigma[/tex] is the stress
[tex]\epsilon=\dfrac{F}{EA}[/tex]
[tex]\epsilon=\dfrac{35600}{69\times 10^9\times 123.84\times 10^{-6}}[/tex]
[tex]\epsilon=4.16\times 10^{-3}[/tex]
Hence, this is the required solution.
