A compound thin cylinder has a common diameter of 100 mm and the inner cylinder has a thickness of 2,5 mm. The radial pressure between the two cylinders is 200 kPa and the difference between the two common diameters before shrinkage was 4,305 x 10-3 mm. Determine (a) the thickness of the outer cylinder; and (b) the resultant hoop stresses in both cylinders if the compound cylinder is subjected to an internal pressure of I80 kPa (E = 200 GPa).
Part a
"\\sigma _h= \\frac{Pd}{2t}"
"\\in_n=\\frac{\\sigma_n}{\\in}-\\frac{Pd}{2 \\in \\sigma}"
"\\frac{\\triangle d}{d}=\\in_h-\\frac{Pd}{2t \\in} \\implies \\triangle d= \\frac{Pd^2}{2t \\in}"
"\\frac{Pd_1^2}{2t_1 \\in} +\\frac{Pd_0^2}{2t_0 \\in} =\\triangle d_1+\\triangle d_0"
"\\frac{P}{2 \\in}[\\frac{100^2}{\\in_1}+\\frac{95^2}{2.5}]=\\frac{4.305}{10^2}"
"[\\frac{100^2}{t_1}+3610]=\\frac{4.305}{10^3}* \\frac{2*200*10^3}{0.02}"
"t_1=\\frac{100^2}{5000}=2 mm"
Thickness of the outer cylinder is 2 mm
Part b
Consdering the inner cylinder
"\\sigma_n=\\frac{Pd}{2t}=\\frac{(180-200)*10^3*95}{2*2.5}=-380 kPa"
Therefore "\\sigma_n = 380 kPa" (Compresive)
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