1. ( BDA 3033 ) Chapter 5: Thick Cylinder **Compound Cylinders**
SOLID MECHANICS II
( BDA 3033 )
Chapter 5: Thick Cylinder
**Compound Cylinders**

2. Topics
Introduction
Lame’s Equation
Composite (Compound) Cylinders

3. Compound (Composite ) Cylinder
A compound cylinder is made by press- fitting one or more jackets around an inner cylinder. The purpose is that the outer cylinder puts pressure on the outside of the inner cylinder. This means that the bore is put into compression
In a compound cylinder, the outer cylinder is having inner diameter smaller than outer diameter of inner cylinder
The inner cylinder is shrink fit to outer cylinder by heating and cooling
On cooling the contact pressure is developed at the junction of two cylinders

4. Compound Cylinder (Contd.)
It introduces compressive tangential stress at the inner cylinder and Tensile stresses at outer cylinder.
Inner cylinder subjected to external pressure and outer cylinder subjected to internal pressure

5. Compound Cylinder (Contd.)
The method of solution for compound cylinders constructed from similar materials is to break the problem down into three separate effects:
Shrinkage pressure only on the inside cylinder
Shrinkage pressure only on the outside cylinder
Internal pressure only on the complete cylinder

6. Compound Cylinder (Contd.)
For each of the resulting load conditions there are two known values of radial stress which enable the Lame’s constant to be determined in each case.

7. Compound Cylinder (Contd.)
Consider the thick compound cylinder as shown in Figure:- ri rf r0
Outer cylinder
R1 = Inner radius of inner cylinder
Rc = Outer radius of inner cylinder (common)
R2 = Outer radius of outer cylinder
P = Radial pressure at the junction of
the two cylinder
Inner cylinder

8. Compound Cylinder (Contd.)

9. Compound Cylinder (Contd.)
Values of Constants A1, A2,B1 and B2 can be found if the radial pressure is known
Hoop stress can also be found by using relative expressions
If the fluid pressure is admitted inside the compound shell, it will be resisted by both the shells
The resultant shell will be the algebraic some of initial stresses and
Those due to fluid pressure

10. Problem 3
A compound cylinder is made by shrinking a tube of 160 mm internal diameter and 20 mm thick over another tube of 160 mm external diameter and 20 mm thick. The radial pressure at the common surface, after shrinking is 80 kgf/cm2. Find the final stress setup across the section when the compound cylinder is subjected to an internal fluid pressure of 600 kgf/cm2 Solution:

11. Solution (Contd.)
Applying Lame’s equation for inner cylinder without fluid pressure

12. Solution (Contd.)

13. Solution (Contd..)
Now, apply the Lame’s equation for the inner cylinder only after the fluid under pressure of 600 kg/cm2 admitted.

14. Solution (Contd..)

15. Difference in Radius due to Shrinking
The radial pressure is same for inner radius of outer tube and outer radius of inner tube Difference in radius is given as
Tensile strain at the shell is
Increase in inner radius
Decrease in outer radius
Subscript 2 =outer tube
Subscript 1 =inner tube

16. Problem 4
A steel cylinder of 300 mm external diameter is to be shrunk to another steel cylinder of 150 mm internal diameter. After shrinking the diameter at the junction is 250 mm and radial pressure at the juction is 28 N/mm2. Find the original difference in radii at the junction. Take E=2x105 N/mm2
[Answer: mm]