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# Answer to Question #128735 in Mechanics | Relativity for Ariyo Emmanuel

Question #128735
(3) A 20 cm long steel tube of 15 cm internal diameter and 1.0cm thickness is surrounded by a brass tube of the same length and thickness. The tubes carry an axial load of 150 kN. Estimate the load carried by each.
Es = 21 x 106 N/cm2 Eb = 10 x 106 N/cm2.
1
2020-08-17T09:20:40-0400

Given:

Length L: 20 cm

Steel tube

Internal diameter "d_s": 15 cm

External diameter "D_s": 17 cm

Es = 21x106 N/cm2

Brass tube

Internal diameter "d_b": 17 cm

External diameter "D_b": 19 cm

Eb = 10x106 N/cm2

solution

Area of steel and brass tubes

"A_{s}=\\frac{\\pi}{4}(D^{2}_{s}-d^{2}_{s}=50.27cm^{2}"

"A_{b}=\\frac{\\pi}{4}(D^{2}_{b}-d^{2}_{b})=56.55cm^{2}"

Under the same load strain "\\epsilon" in steel equals the strain in brass:

"\\frac{\\sigma_{s}}{E_{s}}=\\frac{\\sigma_{b}}{E_{b}}"

"\\sigma_{s}=\\sigma_{b}\\frac{E_{s}}{E_{b}}"

"P_{s}+P_{b}=P"

"\\sigma_{s}A_{s}+\\sigma_{b}A_{b}=P"

"\\sigma_{b}(\\frac{E_{s}}{E_{b}}A_{s}+A_{b})=P"

Load carried by brass and steel respectively:

"P_{b}=\\sigma_{s}A_{b}=\\frac{PA_{b}}{\\frac{E_{s}}{E_{b}}A_{s}+A_{b}}=\\frac{PE_{b}A_{b}}{E_{s}A_{s}+E_{b}A_{b}}=52.3kN"

"P_{s}=\\sigma_{s}A_{s}=\\frac{PA_{s}}{\\frac{E_{s}}{E_{b}}A_{s}+A_{b}}=\\frac{PE_{s}A_{s}}{E_{s}A_{s}+E_{b}A_{b}}=97.7kN"

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