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You want to examine the possibility of using a polymeric material as an artificial load bearing surface for articulating joints. But, the deformation of this material under loading is a concern. You have been informed that if the material has a poisson's ratio under 0.1, then it can undergo further testing for thsi application. You are given a cylindrical sample that is 10 mm in diameter and 4 mm in height. During compression testing, you apply a strain of 20% in the axial direction, and the diameter of the cylindrical sample increases by 0.4 mm at its widest point.

a. What is the Poissonâs ratio of the sample?
b. Does it pass or fail the screening test?

Sagot :

ajeigbeibraheem

Answer:

Step-by-step explanation:

From the given information:

Original diameter of the sample = 10 mm

Diameter increases by 0.4 mm

It means that the New diameter[tex]= 10+0.4 = 10.4 mm[/tex]

The change in diameter = new diameter - original diameter [tex]= (10.4 - 10) mm[/tex]

Transverse strain =  [tex]\dfrac{change \ in \ diameter }{original \ diameter}[/tex]

= [tex]\dfrac{(10.4 - 10) }{10}[/tex]

= 0.04 mm

Original height = 3 mm

Reduction in the height = 20%

New height = [tex]3 -( 3*20) mm= 2.4 mm[/tex]

Change in height = new height - original height

[tex]= (2.4 - 3) mm[/tex]

Longitudinal strain = [tex]\dfrac{change \ in \ height }{original \ height}[/tex]

[tex]=\dfrac{(2.4 - 3)}{3}[/tex]

= - 0.2 mm

Now;

Poisson’s ratio of sample = [tex]- \dfrac{transverse \ strain }{ longitudinal \ strain}[/tex]

= [tex]\dfrac{- (0.04) }{ (-0.2)}[/tex]

= 0.2

From the given statement, Poisson’s ratio of sample is less than 0.1

However, in the estimation of our data, it is 0.2

This implies that It fails the screening test.

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