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Question 3.6: A nylon [E = 2,500 MPa; ν = 0.4] bar is subjected to an axia...

A nylon [E = 2,500 MPa; ν = 0.4] bar is subjected to an axial load that produces a normal stress of σ. Before the load is applied, a line having a slope of 3:2 (i.e., 1.5) is marked on the bar as shown in Figure P3.6. Determine the slope of the line when σ = 105 MPa.

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From the given stress and elastic modulus, compute the longitudinal strain in the bar:

\varepsilon_{\text {long }}=\frac{\sigma}{E}=\frac{105  MPa }{2,500  MPa }=0.04200  mm / mm

Use Poisson’s ratio to calculate the lateral strain:

\varepsilon_{\text {lat }}=-v \varepsilon_{\text {long }}=-(0.4)(0.04200  mm / mm )=-0.01680  mm / mm

Before deformation, the slope of the line is

\text { slope }=\frac{3}{2}=1.500

After deformation, the slope of the line is

\text { slope }=\frac{3(1-0.01680)}{2(1+0.04200)}=\frac{2.950}{2.084}=1.415

 

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