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Chapter 2

Q. 2.7

Q. 2.7

A compression member constructed from steel pipe has an outside diameter of 90 mm and a cross sectional area of 1580 mm². What axial force P will cause the outside diameter to increase by 0.0094 mm? Given E = 200 GPa and 1/m = 0.3.


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First, we shall calculate longitudinal strain

\varepsilon_{\text {long }}=\frac{\sigma}{E}

But,      \text { stress }=\sigma=\frac{P}{1580}

\begin{gathered} \varepsilon_{\text {long }}=\frac{P}{1580 \times 200 \times 10^3}=3.16 \times 10^{-9} P \\ \frac{1}{m}=\frac{\varepsilon_{\text {lat }}}{\varepsilon_{\text {long }}}, \quad \therefore \quad \varepsilon_{\text {lat }}=\frac{\text { change in diameter }}{\text { original diameter }}=\frac{0.0094}{90}=1.04 \times 10^{-4} \end{gathered}

∴        0.3=\frac{1.04 \times 10^{-4}}{3.16 \times 10^{-9} P}, \quad \therefore \quad P=110014.8  N =110.01  kN

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