In Problem 11.38, let the disk have a central hole of radius a = 100 mm. Determine the val...

In Problem 11.38, let the disk have a central hole of radius a = 100 mm. Determine the value of w for which the elastic-plastic interface occurs at the mean radius rP=(a + b)/2= 250 mm.

Problem 11.38

A solid steel disk in a state of plane stress (σzz = 0) has an outer radius b = 400 mm and hm; properties p = 7850 kg/m3, E = 200 GPa, v = 0.29, and Y = 620 MPa. The disk is subjected to an angular velocity ω [rad/s], where ωY ≤ ω ≤ ωP, ωY is the angular velocity at yield, and mp is the angular velocity at which the disk is fully plastic (see Example 11.8). When ω increases beyond ωY, a plastic zone develops at the center and progresses to r = rP. As m increases to ωP, the disk becomes fully plastic, that is, rP → b. Determine the location r = rPof the interface between the elastic and plastic regions as a function of ω. Hint: Continuity of stresses at r = rPrequires that

where at r = rP,  is in the elastic region and  is in the plastic region. Also, the stresses in the elastic region are given by Eqs. 11.49 and 11.50.

Example 11.8: Plastic Deformation of a Rotating Disk

A circular steel disk of inner radius a = 100 mm and outer radius b = 300 mm is subjected to a constant angular velocity ω [rad/s] (Figure E11.8). The steel has material properties Y = 620 MPa, E = 200 GPa, v = 0.29, and ρ = 7.85 × 103 kg/m3. Assume that the disk is in a state of plane stress (σzz = 0) and that yield is governed by the maximum shear-stress criterion. Also, let the disk be traction free at r =a and r= b, and let T = 0.

(a) Determine the angular velocity ωY at which yield in the disk is initiated.

(b) Determine the angular velocity ωPat which the disk is fully plastic; compare ωPto ωY.

FIGURE E11.8

Solution:

(a) Since σzz = 0 and σθθ > σrrfor all values of a and b (see Eqs. 11.49 and 11.50), then the maximum shear-stress criterion is given by

or

Since the disk is traction free at r = a and r = h, the maximum value of σθθ is given by Eq. 11.54 as

With the given data, Eqs. (b) and (c) yield

or

(b) When the angular speed increases beyond ωY, a plastic zone develops at r =a (Figure E11.8). In other words, the region a < r < rpwill be plastic and the region rp < r < b will be elastic, where r = rpis the interface between the plastic and elastic regions. In the plastic region σθθ = Y, but σrr is not known. However, we may obtain σrr from the equilibrium condition (Eq. 11.45). Thus, with σθθ = Y, we have in the plastic region

or, rewriting the left side, we have

Integration yields

For r = a, σrr = 0; so by Eq. (f),

Equations (f) and (g) yield

When the disk is fully plastic, r = rp =b. Then, since σrr = 0 at r = b, Eq. (h) yields with the given data

Comparing ωPto ωY, we have by Eqs. (e) and (i)

In this case the speed at the fully plastic condition is 32% larger than that at yield.

 (11.49)

 (11.50)

 (11.45)