Question

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 1

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