Answer to Question #104400 in Physics for Jake

Question #104400
A tubular steel strut of outer diameter 20 mm and inner diameter 5 mm carries a compressive load of 10 kN. The strut is 1 m long and the modulus of elasticity of the material is 60 GPa. Calculate the induced compressive stress, compressive strain and the change in length that takes place during the loading.
1
Expert's answer
2020-03-03T10:05:29-0500

The induced compressive stress:


"\\sigma=\\frac{F}{A}=\\frac{F}{0.25\\pi(D^2-d^2)}"

"\\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{\\sigma}{Y}"

"\\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:


"\\Delta l=\\epsilon l=(5.7\\cdot 10^{-4})(1)=5.7\\cdot 10^{-4}\\ m=0.57\\ mm"


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Comments

Assignment Expert
16.04.20, 15:57

Dear Tm, because A = \pi R^2 = \pi (D/2)^2 = 0.25 \pi D^2

Tm
15.04.20, 19:58

Why was the area multiplied by 0.25

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