A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the va...
Question 1 Report
A metallic bar 50 cm long has a uniform cross-sectional area of 4.0 cm\(^2\). If a tensile force of 35 kN produces an extension of 0.25 mm, calculate the value of Young's modulus
Answer Details
Definition. Young's modulus is the ratio of tensile stress to tensile strain within the elastic limit:
\[ Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{F\,L}{A\,e} \]Convert the data.
- Original length \(L = 50\ \text{cm} = 0.50\ \text{m}\)
- Cross-sectional area \(A = 4.0\ \text{cm}^{2} = 4.0\times10^{-4}\ \text{m}^{2}\)
- Force \(F = 35\ \text{kN} = 35\,000\ \text{N}\)
- Extension \(e = 0.25\ \text{mm} = 2.5\times10^{-4}\ \text{m}\)
Substitute.
\[ Y = \frac{35\,000\times0.50}{4.0\times10^{-4}\times2.5\times10^{-4}} = \frac{17\,500}{1.0\times10^{-7}} \] \[ Y = 1.75\times10^{11}\ \text{Pa} \]Young's modulus of the bar is \(1.75\times10^{11}\ \text{Pa}\) (N m\(^{-2}\)).
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