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A cylindrical bar of metal having a diameter of 20.9 mm and a length of 203 mm is deformed elastically in tension with a force of 46600 N. Given that the elastic modulus and Poisson's ratio of the metal are 64.4 GPa and 0.33, respectively, determine the following: (a) The amount by which this specimen will elongate in the direction of the applied stress. (b) The change in diameter of the specimen. Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answer :

Answer:

[tex]\delta l[/tex] =  0.428 mm

[tex]\delta d[/tex]  = −0.0145 mm

Explanation:

given data

diameter = 20.9 mm

length = 203 mm

force = 46600 N

elastic modulus = 64.4 GPa

Poisson's ratio  = 0.33

solution

we get here first area of cross section that is

Area = [tex]\frac{\pi }{4}d^2[/tex]     .............1

put here value and we will get

Area = [tex]\frac{\pi }{4}\times 20.9^2[/tex]

Area = 343.06 mm²    

and

now we get here change in direction of applied stress that is

[tex]\delta l[/tex] = [tex]\frac{Pl}{AE}[/tex]     .....................2

put here value and we will get

[tex]\delta l[/tex] =  [tex]\frac{46600\times 203}{343.06\times 64.4 \times 10^3}[/tex]  

[tex]\delta l[/tex] =  0.428 mm

and

we know Poisson ratio will be here express as

m = [tex]- \frac{ \frac{\delta d}{d}}{ \frac{\delta l}{l}}[/tex]       ........................3

0.33 =  [tex]- \frac{ \frac{\delta d}{20.9}}{ \frac{0.428}{203}}[/tex]  

[tex]\delta d[/tex]  = −0.0145 mm

so here change in diameter have -ve sign so diameter is decrease.

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