Chapter 5 Dislocations of Plastic Deformation

CHAPTER 5 – DISLOCATIONS OF PLASTIC DEFORMATION

5.3 There are three slip systems on an fcc octahedral plane. Assume a 2 MPa tensile stress is applied along the [100] direction of a gold crystal, whose critical resolved shear stress is 0.91 MPa. Demonstrate quantitatively that measurable slip will not occur on any of three slip systems in the (111) plane as a result of this applied stress.

Solution:

The slip systems associated with the (111) plane in a fcc metal are ( This problem requires the use of the Schmid equation:

Where is the angle between the stress axis, [100], and the pole of the slip plane and the angle between the stress axis, and one of the three slip directions. Both can be evaluated with the aid of Eq. 5.3, which is:

To find for the slip system (111) with a [100] stress axis we have:

A similar calculation, for this slip system, yields: so that we have:

In the same manner we find that for the slip system (111)

and for the (111) system:

Note that the resolved shear stress on all three of the slip systems is less than the critical resolved shear stress of 0.91 MPa. As a consequence, the applied stress will not cause slip on any of the slip systems.

5.5 Deformation twins are also able to form along the {111} planes of fcc crystals, as a result of the application of a shear stress across this type of plane. In twinning, the shear directions are .

(a) Prove, using Eq. 5.3, that the are directions that lie in the (111) plane.

(b) Determine the Schmid factors for the (111) twinning systems if a tensile stress is applied along the [711] direction.

Solution:

(a) For a direction to lie in a plane, the dot product of the direction and the pole of the plane must equal zero. Thus, consider the dot product of the direction, and the [111] pole;

and for the other system:

Consequently, we may conclude that both of these directions lie in the (111) plane.

(b) First consider the Schmid factor for the system. The cosine of the angle between the pole of the twinning plane, [111], and the [711] stress axis is given by:

and the cosine of the angle between the [711] stress axis, and the twinning shear direction is:

And the Schmid factor is:

If we let represent the angle between the stress axis and the twinning shear direction in the twin system, it may be shown that:

So that the Schmid factor for the twinning system is

Finally, it may be shown that the Schmidt factor for the twinning system, when the stress axis is along [711] is:

5.8 A 10 mm diameter zinc crystal has a longitudinal axis that makes an angle of 85 degrees with the pole of the basal plane, and a 7 degree angle with the closest slip direction in the basal plane. If the critical resolved shear stress of zinc is 0.20 MPa, at what axial load would the crystal begin to deform by basal slip: (a) newtons, (b) kilograms force?

Solution:

The Schmid equation is:

In this equation the only unknown is , the applied tensile stress. Solving for and substituting into this equations for , the critical resolved shear stress, and the values of the angles yields:

Note: 2.31 MPa = . Let p represent the load. Then, since 10 mm = 0.01 m, the cross-section area of the specimen will be . The load p is accordingly:

(a)

(b)

5.11 The total line length of the dislocations in a 4 cm by 4 cm TEM photograph of a metal foil, taken at a magnification of 25,000X, is 400 cm. The foil imaged by the picture had a thickness of 300 nm. Determine the disloaction density in the foil.

Solution:

The total area of the photograph is 0.04 m by 0.04 m. This corresponds to an area:

The volume of the foil, V, is equal to this area multiplied by the foil thickness, and is . The length of the dislocations in this volume is:

The dislocation density, p, may now be calculated:

5.17 Johnston and Gilman have reported that in a grown LiF crystal, subjected to a constant stress of the dislocation velocity at 249.1 K was and at 227.3 K the velocity was They also observed that their data suggested an Arrhenius relationship between the dislocation velocity and the absolute temperature, so that one might write where v is the dislocation velocity, A a constant of proportionality, Q an effective activation energy in J per mole, and R the international gas constant (8.314 J per mole K). Use the velocity versus temperature data of Johnston and Gilman, given above, to determine Q and A for their LiF crystal, stressed at 1100 . Note that Q may be obtained using the relation:

Once Q has been obtained, A may be determined by substitution into the Arrhenius equation.

Solution:

Given:

Now solving the above equation for Q, and substituting these data into this equation yields:

Next to obtain A use:

and solve for A:

5.19 A typical cross-headed speed in a tensile machine is 0.5 cm per minute.

(a) What is the nominal engineering strain rate imposed by this cross-head speed on a typical engineering tensile specimen with a 5 cm gage length?

(b) Estimate the dislocation velocity that would be obtained with this strain rate in an iron specimen, with a dislocation density of . Assume that the Burgers vector of iron is 0.248 nm.

(c) If a very slow tensile test at a strain rate of is used, what dislocation velocity would be expected in the specimen?

Solution:

(a)

(b)

Where is the strain rate, the dislocation density, b the Burgers vector, and v the dislocation density.

Solving for v, and substituting the given values of p, b, and v give:

(c)

5.21 A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40 mm. After fracture, the length of the gage section was found to be 50 mm, the reduction in area 90 percent, and the load at fracture 1000 N. Compute:

(a) The specimen elongation.

(b) The engineering fracture stress.

(d) The true strain at the neck.

Solution:

(a) The specimen elongation is

(b) The engineering fracture stress, , equals the fracture load, divided by the original specimen area, , or:

where , thus:

The true stress at fracture is equal to the load, P, divided by or:

(d) The true stress at fracture is given by the equation:

5.23 The slope, m, of the curve drawn through the data points in Fig. 5.35 is approximately equal to using the units in the figure. Compute the fractional increase in the dislocation density that would correspond to an increase in flow stress from 588 to 784 MPa (use the titanium data of Jones and Conrad).

Solution:

Since the data in Fig. 5.35 are given in units, first convert the given stresses, 588 and 784 MPa into . This may be done by dividing by the factor 9.81. Thus, 588 MPa = 60 and 784 MPa = 80 . Next use the equation relating the stress to the square root of the dislocation density.

Now solve for .

where is the intercept of the straight line, through the data in Fig. 5.35, with the stress axis. The value of is 39 and Accordingly for the 60 stress we have:

A similar calculation for the dislocation density at the 80 stress yields:

The ratio of these two dislocation denstities is: