Solucionário Hosford

 

Chapter 10

1.        Typical values for the dislocation density in annealed and heavily deformed copper are 107 and 1011  cm-2.

A.        Calculate the average distance between dislocations for both cases. For simplicity, assume all of the dislocations are parallel and in a square pattern.

B.        Use equation 9.11 to calculate the energy/length of dislocation line for both cases assuming the dislocations are screws.

For copper: lattice parameter a = 0.361 nm, E = 110 GPa, υ = 0.30,  ρ  = 8.96 Mg/m3.

Solution: A. d = 1/√ρ ; For ρ = 107, d = 316μm. For ρ = 1011, d = 3.16μm.

B. EL = [Gb2/(4π)]ln(r1/ro)

Taking G = E/[2(1+υ)] = 110GPa/(2x1.3) = 42.3G Pa,

ro = b/4 =  0.090 nm and r1= d/2 =  so

EL = [(42.3x109)( 0.361x10-9)2/(4π)]ln(d/0.180x10-9) = 

      4.39x10-10ln(d/0.180x10-9)

For ρ = 107, EL =4.39x10-10ln(316x10-6/0.180x10-9) = 6.31 x10-9 J/m

For ρ = 1011, EL =4.39x10-10ln(3.16x10-6/0.180x10-9) = 4.29x10-9 J/m.

1. For a typical annealed metal, the yield stress in shear is 10-4G. Using a typical value for b, deduce the typical spacing of a Frank-Read source.

 Solution: τ = 2Gb/d, d = 2Gb/τx10-9m,

d = 2G(0.x10-9m)/(10-4G) = 4 μm 

3.        Several theoretical models predict that the dislocation density, ρ, should increase parabolically with strain, ρ = Cε1/2.  Assuming this and the dependence of τ on ρ shown in Figure 10.4, predict the exponent n in power law approximation of the true stress-true strain curve, 

σ = Kεn.

Solution: From Figure 10.4, τ = A√ρ (where A is a constant) so τ = CAε1/4. Since σ is proportional to τ , σ = C’ε0.25, or n = 0.25.

4.        Orowan showed that the shear strain rate `γ = dγ/dt = ρb»v where ρ is the dislocation density, and »v is the average velocity of the dislocation.  In a tension test, the tensile strain rate, `ε, is approximately half of the shear strain rate,  `γ. (The half assumes shear the Schmid factor is 1/2 which is a bit too high.)  In a typical tension test, the crosshead rate is 0.2 in/min and the gauge section is 2 in. in length.  Estimate »v  in a typical tension test for a typical metal with an initial dislocation density of 1010/cm2.

Solution: Taking `γ = 2(0.2/60 in/s)/2 in = .033s-1, »v =  `γ /(bρ) =

.033/[(1010/cm2)(104cm2/m2)(0.25x10-9)] = 1.32x10-7 m/s = 1.32x10-7 μm/s

5.        Consider the possible intersections of dislocations sketched in Figure 10.18.  In the sketches, dislocations B and D are moving to the right and A and C are moving to the left.

A.        Sketch the nature of the jogs or kinks in each dislocation after they intersect one-another.

1. Which, if any, would leave a trail of vacancies or interstitials?[pic 1][pic 2]

Solution: A

                         [pic 3]

B. Neither

6.        The sketches Figure 10.19 show pairs of dislocations that are about to intersect.  The arrows show the directions of motion (Dislocations A, C and E are moving into the paper, dislocations B and F are moving out of the paper and dislocation D is moving to the left.)

A.        After intersection, which of these dislocations (A, B, C, D, E, or F) would have jogs that would produce point defects if the dislocations continued to move in the direction shown?

B.        For each answer to A, indicate whether the point defects would be vacancies or interstitials.[pic 4]

                                 Figure 10.19.  Sketch for problem 6

Solution: A. e and f

B. interstitials

7.    Consider how dislocation intersections affect the stress necessary to continue slip.  Let the density of screw dislocations be ρ (meters of dislocations/m3) and for simplicity assume that these dislocations are equally divided into three sets of dislocations,  each set being parallel to one of the three orthogonal axes.