Answer:
For any material if ∈ is the axial strain then the lateral strain is given by -μ∈ is the lateral strain in the object
Where,
μ is the poisson's ratio of the material
The longitudinal strain is calculated as follows
[tex]\varepsilon _{axial}=\frac{\Delta length}{Length_{original}}\\\\\therefore \varepsilon _{axial}=\frac{0.7}{800}=8.75\times 10^{-4}[/tex]
Thus the lateral strain becomes
[tex]\varepsilon _{lateral}=-\mu\varepsilon _{axial}\\\\\varepsilon _{lateral}=-0.27\times 8.75\times 10^{-4}=-2.36\times 10^{-4}[/tex]
now by definition of lateral strain we have
[tex]\varepsilon _{lateral}=\frac{\Delta diameter}{diameter_{original}}\\\\\Rightarrow \Delta Diameter=-2.36\times 10^{-4}\times 32=-7.56\times 10^{-3}\\\\D_{f}-D_{i}=-7.56\times 10^{-3}\\\\D_{f}=32-7.56\times 10^{-3}=31.992mm[/tex]
By hookes law the stress developed due to the given strain is given by
[tex]\sigma =\varepsilon _{axial}E[/tex]
Applying values we get
[tex]\sigma =8.75\times 10^{-4}\times 150\times 10^{9}\\\\\sigma =131.25MPa[/tex]
Thus the force is calculated as
[tex]Force=\sigma \times Area\\\\Force=131.25\times 10^{6}\times \frac{\pi (32\times 10^{-3})^{4}}{4}\\\\Force=105.55kN[/tex]
