Question

A compound cylinder is formed by shrinking a tube of 200mm outside and 150mm inside diameter on to one of 150mm outside and 100mm inside diameter. Owing to shrinkage the radial stress at the common surface is 20 MN/m². If this cylinder is now subjected to an internal pressure of 100 MN/m², what is the magnitude and position of the maximum hoop stress?

Answer 1

Solution;

Shrinkage only-outer tube;

$r=0.1,\sigma_r=0,r=0.075,\sigma_r=-20$

$0=A-\frac{B}{0.1^2}=A-100B$

$-20=A-\frac{B}{0.075^2}=A-177.78B$

From both equations;

$-20=100B-177.78B=-77.78B$

$B=0.2571$

$A=25.71$

Therefore hoop stressses are;

At r=0.1,$\sigma_H=A+100B=51.42MN/m^2$

At r=0.075;

$\sigma_H=A+177.78B=71.42MN/m^2$

Shrinkage only-inner tube;

$r=0.05,\sigma_r=0$

$r=0.075,\sigma_r=-20$

$0=A-\frac{B}{0.05^2}=A-400B$

$-20=A-\frac{B}{0.075^2}=A-177.78B$

From both equations;

$-20=400B-177.78B=222.22B$

$B=-0.09$

$A=-36$

Therefore ,hoop stresses are;

At r=0.05;

$\sigma_H=A+400B=-72MN/m^2$

At r=0.075;

$\sigma_H=A+177.78B=-52MN/m^2$

Considering internal pressure only on the complete cylinder;

At r=0.05,$\sigma_r=-100$

At r=0.1,$\sigma_r=0$

$0=A-100B$

$-100=A-400B$

From both equations;

$-100=100B-400B=-300B$

$B=0.333$

$A=33.33$

The hoop stresses are ;

At r=0.05;

$\sigma_H=A+400B=165.33MN/m^2$

At r=0.075;

$\sigma_H=A+177.78B=92MN/m^2$

At r=0.1;

$\sigma_H=A+100B=66.66MN/m^2$

The resultant hoop stress from combined shrinkage and internal pressure is;

Outer tube;

At r=0.1;

$\sigma_H=66.66+51.42=118.0MN/m^2$

At r=0.075;

$\sigma_H=92+71.42=163.42MN/m^2$

Inner tube;

At r=0.05;

$\sigma_H=165.33-72=93.33MN/m^2$

At r=0.075;

$\sigma_H=92-52=40MN/m^2$

Maximum hoop stress;

$\sigma_{H_{max}}=163.42MN/m^2$ At 0.075m of the outer tube