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Example · Example 5

Q.A steel wire of original diameter 2.0 mm2.0\ \text{mm} is stretched so that it develops a longitudinal strain of 1.5×10−31.5\times10^{-3}. If careful measurement shows the diameter of the wire decreases by 9.0×10−7 m9.0\times10^{-7}\ \text{m} in the process, find

(a) the lateral strain, and
(b) the Poisson's ratio of steel implied by these measurements.
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Given: original diameter d=2.0 mm=2.0×10−3 md = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}, decrease in diameter Δd=9.0×10−7 m\Delta d = 9.0\times10^{-7}\ \text{m}, longitudinal strain =1.5×10−3= 1.5\times10^{-3}.

  1. Lateral strain:

    Lateral strain=Δdd=9.0×10−72.0×10−3=4.5×10−4\text{Lateral strain} = \frac{\Delta d}{d} = \frac{9.0\times10^{-7}}{2.0\times10^{-3}} = 4.5\times10^{-4}

  2. Poisson's ratio: σ=Lateral strainLongitudinal strain=4.5×10−41.5×10−3=0.30\sigma = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} = \frac{4.5\times10^{-4}}{1.5\times10^{-3}} = 0.30 …

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