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lawyer [7]
3 years ago
6

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 c

olumn, determine the stress in the pipe at the maximum allowable contraction.
Engineering
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

\mathbf{stress \ \sigma = 264.6 \ Mpa}

Explanation:

From the concept of Hooke's  Law,

E =\dfrac{ stress \ \sigma}{ strain \  \varepsilon}

where;

strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}

strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}

strain \ \varepsilon =0.00147368

Recall:

E =\dfrac{ stress \ \sigma}{ strain \  \varepsilon}

stress \ \sigma = E \times { strain \  \varepsilon}

stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147

\mathbf{stress \ \sigma = 264.6 \ Mpa}

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

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