0
20kviews
A steel bar 50 mm x 50 mm in section, 3m long is subjected to an axial pull of 20kN. Calculate the change in length and change in side of the bar. Take E = 200 GPa and Poisson's ratio = 0.3.
1 Answer
1
3.4kviews

Data: b=50 mm, d =50 mm, L=3m, P = 20 kN, E = 200 GPa μ = 0.3 Calculate: δL, δb, and δd

$\delta \mathrm{L}=\frac{\mathrm{PL}}{\mathrm{AE}}$

$\delta \mathrm{L}=\frac{20 \times 10^{3} \times 3 \times 10^{3}}{50 \times 50 \times 200 \times 10^{3}}$

$\delta \mathrm{L}=0.12 \mathrm{mm}$

$\begin{aligned} \mu &=\frac{\text {Lateral Strain }}{\text {Linear Strain }} \\ \mu &=\frac{\left(\frac{\delta \mathrm{b}}{\mathrm{b}}\right)}{\left(\frac{\delta \mathrm{L}}{\mathrm{L}}\right)} \\ 0.3 &=\frac{\left(\frac{\delta \mathrm{b}}{50}\right)}{\left(\frac{0.12}{3000}\right)} \\ \delta \mathrm{b} &=6 \times 10^{-4} \mathrm{mm} \end{aligned}$

$\mu=\frac{\left(\frac{\delta \mathrm{d}}{\mathrm{d}}\right)}{\left(\frac{\delta \mathrm{L}}{\mathrm{L}}\right)}$

$0.3=\frac{\left(\frac{\delta \mathrm{d}}{50}\right)}{\left(\frac{0.12}{3000}\right)}$

$\delta \mathrm{d}=6 \times 10^{4} \mathrm{mm}$

Please log in to add an answer.