Respuesta :
Answer:
Poisson’s ratio = 0.36
Explanation:
Given that
Diameter ,d= 10 mm
Force ,F= 15000 N
Reduction in diameter ,Δd = -7 x 10⁻³ mm
E= 100 GPa
Longitudinal strain
[tex]\varepsilon_{logn}=\dfrac{P}{AE}[/tex]
[tex]A=\dfrac{\pi d^2}{4}[/tex]
[tex]\varepsilon_{logn}=\dfrac{P}{\dfrac{\pi d^2}{4}\times E}[/tex]
[tex]\varepsilon_{logn}=\dfrac{15000}{\dfrac{\pi \times 10^2}{4}\times 100\times 1000}[/tex]
[tex]\varepsilon_{logn}=0.0019[/tex]
Lateral strain
[tex]\varepsilon_{lat}=\dfrac{\Delta d}{d}[/tex]
[tex]\dfrac{\Delta d}{d}=-\mu\times \varepsilon_{long}[/tex]
[tex]\dfrac{7\times 10^{-3}}{10}=-\mu\times 0.0019[/tex]
μ= 0.36
Poisson’s ratio = 0.36
The value of Poisson's ratio of the given material is; μ = 0.36
What is the poisson's ratio?
We are given;
Diameter; d = 10 mm
Force; F = 15000 N
Change in diameter; Δd = -7 x 10⁻³ mm
Elastic modulus; E = 100 GPa = 100000 mPa
Formula for Longitudinal strain is;
ε = P/AE
where;
P is force
A is area = πd²/4 = π × 10²/4 = 25π
Thus;
ε = 15000/(25π * 100000)
ε = 0.0019
Formula for lateral strain is;
ε * μ = -Δd/d
where μ is poisson ratio. Thus;
μ = (7 x 10⁻³)/(10 * 0.0019)
μ = 0.36
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