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

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

Step-by-step explanation:

From the concept of Hooke's Law,

`E =( stress \ \sigma)/( strain \ \varepsilon)`

where;

`strain \ \varepsilon = (change \ in \ dimension )/(original \ dimension)`

`strain \ \varepsilon = (7 \ mm )/(4.75 * 10^(3) \ mm)`

`strain \ \varepsilon =0.00147368`

Recall:

`E =( stress \ \sigma)/( strain \ \varepsilon)`

`stress \ \sigma = E * { strain \ \varepsilon}`

`stress \ \sigma = 180 * 10^(3) \ Mpa * 0.00147`

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

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