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Pull test on bar: Poisson's ratio, moduli

Short answer

Compute longitudinal strain from extension and gauge length, lateral strain from diameter change. Poisson's ratio μ = 0.289. Elastic modulus E = 188.6 kN/mm², shear modulus G = 73.2 kN/mm², bulk modulus K = 148.9 kN/mm².

In an experiment, a bar of 30 mm diameter is subjected to a pull of 60 kN. The measured extension on gauge length of 200 mm is 0.09 mm and the change in diameter is 0.0039 mm. Calculate the values of Poisson's ratio and three elastic moduli.

Longitudinal quantities

A=π4(30)2=706.86 mm2σ=PA=60×103706.86=84.883 N/mm2εlong=δLL=0.09200=0.000450E=σε=84.8830.000450=188628 N/mm2=188.6 kN/mm2\begin{aligned} A &= \frac{\pi}{4}(30)^2 = 706.86\ \text{mm}^2\\ \sigma &= \frac{P}{A} = \frac{60\times10^3}{706.86} = 84.883\ \text{N/mm}^2\\ \varepsilon_{long} &= \frac{\delta L}{L} = \frac{0.09}{200} = 0.000450\\ E &= \frac{\sigma}{\varepsilon} = \frac{84.883}{0.000450} = 188628\ \text{N/mm}^2 = 188.6\ \text{kN/mm}^2 \end{aligned}

Poisson's ratio

εlat=δdd=0.003930=0.000130,μ=εlatεlong=0.0001300.000450=0.2889\varepsilon_{lat} = \frac{\delta d}{d} = \frac{0.0039}{30} = 0.000130,\qquad \mu = \frac{\varepsilon_{lat}}{\varepsilon_{long}} = \frac{0.000130}{0.000450} = 0.2889

Other moduli

G=E2(1+μ)=1886282(1+0.2889)=73175 N/mm2K=E3(1−2μ)=1886283(1−2×0.2889)=148917 N/mm2\begin{aligned} G &= \frac{E}{2(1+\mu)} = \frac{188628}{2(1+0.2889)} = 73175\ \text{N/mm}^2\\ K &= \frac{E}{3(1-2\mu)} = \frac{188628}{3(1-2\times0.2889)} = 148917\ \text{N/mm}^2 \end{aligned}

Answer: μ=0.289\mu = 0.289; E=188.6E = 188.6 kN/mm²; G=73.2G = 73.2 kN/mm²; K=148.9K = 148.9 kN/mm².

✓ Answer checked by Claude Opus 5.5 (high effort)

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