Answer:
The the elongated length is [tex]\Delta L = 0.4 \ mm[/tex]
The change in diameter is [tex]\Delta d = - 0.0136\ mm[/tex]
Explanation:
From the question we are told that
The diameter of the cylindrical bar is [tex]d = 21.0 \ mm = \frac{21}{1000} = 0.021 \ m[/tex]
The length of the cylindrical bar is [tex]L= 210 \ mm = 0.21 \ m[/tex]
The force that deformed it is [tex]F = 46800 \ N[/tex]
Elastic modulus is [tex]E = 60.9 \ GPa = 60.9 *10^{9}Pa[/tex]
The Poisson's ratio is [tex]\mu = 0.34[/tex]
Generally elastic modulus is mathematically represented as
[tex]E = \frac{\sigma }{\epsilon}[/tex]
Where
[tex]\epsilon[/tex] is the strain which is mathematically represented as
[tex]\epsilon = \frac{L}{\Delta L}[/tex]
Where [tex]\Delta L[/tex] is the elongation length
[tex]\sigma[/tex] is the stress on the cylinder which is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
Where F is the force and
A is the area which is calculated as
[tex]A = \frac{\pi} {4} d^2[/tex]
Substituting values
[tex]A = \frac{\pi}{4} * (0.021)[/tex]
[tex]A = 0.000346 \ m^2[/tex]
So the stress is
[tex]\sigma = \frac{46800}{0.000346}[/tex]
[tex]\sigma = 1.35 *10^{8} \ N \cdot m^2[/tex]
Thus the elastic modulus is
[tex]E = \frac{1.35 *10 ^{8}}{\frac{\Delta L}{L} }[/tex]
making [tex]\Delta L[/tex] the subject
[tex]\Delta L = \frac{EL}{1.35 *10^{8}}[/tex]
Substituting values
[tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{1.35 *10^{8}}[/tex]
[tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{60.9*10^{9}}[/tex]
[tex]\Delta L = 0.0004 \ m[/tex]
Converting to mm
[tex]\Delta L = 0.0004 * 1000[/tex]
[tex]\Delta L = 0.4 \ mm[/tex]
Generally the poisson ratio is mathematically represented as
[tex]\mu = - \frac{\frac{\Delta d }{d} }{\frac{\Delta L }{L} }[/tex]
The negative sign indicate a decrease in diameter as a result of the force
making [tex]\Delta d[/tex] the subject
[tex]\Delta d = - \mu * \frac{\Delta L }{L } * d[/tex]
Substituting values
[tex]\Delta d = - 0.34 * \frac{0.0004 }{0.210 } * 0.021[/tex]
[tex]\Delta d = - 1.36 *10^{-5} \ m[/tex]
Converting to mm
[tex]\Delta d = - 0.0136\ mm[/tex]
