A composite cylinder has inner radius a, outer radius b, and interface radius c (Figure P1...

A composite cylinder has inner radius a, outer radius b, and interface radius c (Figure P11.8). Initially, the outer radius of the inner cylinder is larger than the inner radius of the outer cylinder by an amount δ. Show that after assembly the shrink-fit pressure is

where E is the modulus of elasticity of the cylinders and δ/c is the shrinkage factor. Hint: The increase of the inner radius of the outer cylinder plus the decrease in the outer radius of the inner cylinder produced by PS must be equal to δ. (See the solution of Example 11.3.)

FIGURE P11.8

Example 11.3: Stresses in a Composite Cylinder

Let the cylinder in Example 11.1 be a composite cylinder made by shrinking an outer cylinder onto an inner cylinder. Before assembly, the inner cylinder has im1er and outer radii of a = 10 mm and ci = 25.072 mm, respectively. Likewise, the outer cylinder has inner and outer radii of co = 25.000 mm and b = 50 mm, respectively. Determine the stress components σrr and σθθ at r = a = 10 mm, r = 25 mm, and r = h =50 mm for the composite cylinder. For assembly purposes, the inner cylinder is cooled to a uniform temperature T1 and the outer cylinder is heated to a uniform temperature T2to enable the outer cylinder to slide freely over the inner cylinder. It is assumed that the two cylinders will slide freely if we allow an additional 0.025 mm to the required minimum difference in radii of 0.072 mm. Determine how much ti1e temperature (in degrees Celsius) must be raised in the outer cylinder above the temperature in the inner cylinder to freely assemble the two cylinders. α = 0.0000117/°C.

After the composite cylinder has been assembled, the change in stresses caused by the internal pressure p1 = 300 MPa is the same as for the cylinder in Example 11.1. These stresses are added to the residual stresses in the composite cylinder caused by shrinking the outer cylinder onto the inner cylinder.

Solving for ps, we obtain

ps = 189.1 MPa

The pressure ps produces stresses (residual or shrink-fit stresses) in the nonpressurized composite cylinder. For the inner and outer cylinders, the residual stresses and at the inner and outer radii are given by Eqs. 11.20 and 11.21. For the inner cylinder, p1 = 0, p2 = ps, a = 10 mm, and b = 25 mm. For

the outer cylinder, p1 = ps, p2 = 0, a = 25 mm, and b =50 mm. The residual stresses are found to be

The stresses in the composite cylinder after an internal pressure. of 300 MPa has been applied are obtained by adding these residual stresses to the stresses calculated m Example 11.1. Thus, we find

A comparison of the stresses for the composite cylinder with those for the solid cylinder in Example 11.1 indicates that the stresses have been changed greatly. The determination of possible improvements in the design of the open-end cylinder necessitates consideration of particular criteria of failure

(see Section 11.4)

To have the inner cylinder slide easily into the outer cylinder during assembly, the difference in temperature between the two cylinders is given by the relation

since for uniform temperatures T1 , T2, we have σrr. = σθθ = σzz = 0 in each cylinder, and since then Eqs. 11.2 and 11.4 yield ∈θθ = u/r = αΔT, where r = c0 = c1.

Example 11.1: Stresses in a Hollow Cylinder

A thick-wall cylinder is made of steel (E = 200 GPa and V = 0.29), has an inside diameter of 20 mm, and has an outside diameter of 100 mm. The cylinder is subjected to an internal pressure of 300 MPa. Determine the stress components σrr and σθθat r = a = 10 mm, r = 25 mm, and r = b = 50 mm.

Solution:

The external pressure p2 = 0. Equations 11.20 and 11.21 simplify to

Substitution of values for r equal to 10 mm, 25 mm, and 50 mm, respectively, into these equations yields the following results:

 (11.20)

 (11.21)

 (11.25)

 (11.20)

 (11.21)

 (11.2)

 (11.4)