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While supporting load P, the maximum contraction permitted in a 4.75-m long pipe column is 7 mm. Given that E =180 GPa for the column, determine the stress in the pipe at the maximum allowable contraction.

Answer :

ajeigbeibraheem

Answer:

[tex]\mathbf{stress \ \sigma = 264.6 \ Mpa}[/tex]

Explanation:

From the concept of Hooke's  Law,

[tex]E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}[/tex]

where;

[tex]strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}[/tex]

[tex]strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}[/tex]

[tex]strain \ \varepsilon =0.00147368[/tex]

Recall:

[tex]E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}[/tex]

[tex]stress \ \sigma = E \times { strain \ \varepsilon}[/tex]

[tex]stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147[/tex]

[tex]\mathbf{stress \ \sigma = 264.6 \ Mpa}[/tex]

Thus, the stress in the pipe at the maximum allowable contraction = 264.6 Mpa

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