Answer:
The answer is below
Explanation:
The poison ratio is given by the formula:
[tex]v=-\frac{\epsilon_x}{\epsilon_z} =-\frac{\epsilon_y}{\epsilon_z} \\\\Where\ \epsilon_y=transverse\ strain, \epsilon_z=longitudinal\ strain,v=poison\ ratio\\ \\The\ transverse\ strain(\epsilon_x)=\epsilon_y=\frac{change\ in\ diameter}{initial\ diameter}=\frac{\Delta d}{d_o} =\frac{-0.0105\ mm}{32\ mm} \\=-0.000328\\\\v=-\frac{\epsilon_x}{\epsilon_z} \\\\\epsilon_z=\frac{-\epsilon_x}{v} =\frac{-(-0.000328)}{0.34}\\ \\\epsilon_z=9.65*10^{-4}[/tex]
We then locate the strain of 9.65 * 10⁻⁴ on the stress-strain curve, this gives a stress of 68.9 MPa
The stress required to cause a reduction of 0.0105mm on the alloy is 68.9MPa
Data Given;
The transverse strain of the cylindrical alloy is the ratio between the reduction in diameter to the actual diameter
[tex]E^x = \frac{\delta d}{d}\\ E^x = \frac{-0.0105}{32}\\ E^x = -0.00033\\ E^x = -3.3*10^-^4[/tex]
The longitudinal strain is the ratio between the transverse strain to the Poisson's ratio.
[tex]E^y = \frac{-E^x}{v}\\ E^y = \frac{3.3*10^-^4}{0.34}\\ E^y = 0.00097 = 9.7 *10^-^4[/tex]
Using the strain-stress graph of a cylindrical alloy, we would find the strain of 0.00097 around the stress of 68.9MPa
Learn more on strain-stress here;
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