The induced compressive stress:
"\\sigma=\\frac{10\\cdot 10^3}{0.25\\pi((20\\cdot 10^{-3})^2-(5\\cdot 10^{-3})^2)}=3.4\\cdot 10^{7}\\ Pa"
Te compressive strain:
"\\epsilon=\\frac{3.4\\cdot 10^{7}}{60\\cdot 10^{9}}=5.7\\cdot 10^{-4}"
The change in length that takes place during the loading:
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Dear Tm, because A = \pi R^2 = \pi (D/2)^2 = 0.25 \pi D^2
Why was the area multiplied by 0.25
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