Respuesta :
Answer:
The original length of the specimen [tex]l_{o} = 104.7 mm[/tex]
Explanation:
Original diameter [tex]d_{o}[/tex] = 30 mm
Final diameter [tex]d_{1}[/tex] = 30.04 mm
Change in diameter Δd = 0.04 mm
Final length [tex]l_{1}[/tex] = 105.20 mm
Elastic modulus E = 65.5 G pa = 65.5 × [tex]10^{3}[/tex] M pa
Shear modulus G = 25.4 G pa = 25.4 × [tex]10^{3}[/tex] M pa
We know that the relation between the shear modulus & elastic modulus is given by
[tex]G = \frac{E}{2(1 + \mu)}[/tex]
[tex]25.5 = \frac{65.5}{2 (1 + \mu)}[/tex]
[tex]\mu = 0.28[/tex]
This is the value of possion's ratio.
We know that the possion's ratio is given by
[tex]\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }[/tex]
[tex]{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}[/tex]
[tex]{\frac{change \ in \ length}{l_{o} } =[/tex] 0.00476
[tex]\frac{l_{1} - l_{o} }{l_{o} } = 0.00476[/tex]
[tex]\frac{l_{1} }{l_{o} } = 1.00476[/tex]
Final length [tex]l_{o}[/tex] = 105.2 m
Original length
[tex]l_{o} = \frac{105.2}{1.00476}[/tex]
[tex]l_{o} = 104.7 mm[/tex]
This is the original length of the specimen.
