Question

 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). 

Answer 1

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)