Question

A single crystal in the form of a cube has two possible slip planes. (1) and (2). as shown in the sketch below. Each of these planes has two possible slip directions, (a) and (b), as is also shown. The angle between the tensile load axis and slip plane normal for both slip planes is 450 40 <plane (1)> <plane (2)> a) If a stress is applied to this crystal in the direction noted, specify which particular (eg. plane (1), direction (a)) slip system(s) will be active at the beginning of plastic deformation. b) If the critical resolved shear stress is 7 MN/m2, calculate the tensile stress at which plastic deformation initiates. c) Looking normal to slip plane (1), four distinct types of dislocations - A. B, C and D- are observed as shown below. Dislocations A and B are parallel to slip direction (a) whereas C and Dare parallel to slip direction (b). Dislocation A and C are observed to glide during plastic deformation dislocations B and D do not. Construct a table with headings as shown below-that lists characteristics of the four dislocation types. Explain your reasoning: (Dislocation character' refers to the dislocation as edge, screw, or mixed). (a) D A B or Dislocation Character Dislocation Type e.g. A) Slip Direction (6) d) Suppose the tensile axis were parallel to direction (b) in slip plane (1). Briefly describe the tensile stress-strain behavior of the single crystal for this condition.

Answer 1

Given: Angle between tensile load axis and slip plane normal for both planes is 45 degree.

Slip will initiate if the Schmid factor (m) exceeds a certain value. Schmid factor (m) is given by dot product of applied tensile load axis direction with shear stress directions. Let us consider that the tensile load is applied in the direction [010].

Therefore, m = cos $\theta$ . cos $\lambda$ , where cos $\theta$ is the angle between tensile axis direction [010] and normal to the plane
(011)
and cos $\lambda$ is the angle between tensile axis direction [010] and slip direction.

(a)
Plane (1)

<Plane (1)>: Slip Plane is
(011)
and Slip Directions are
011
and
| 100
.

In this case, we have two slip directions. Now Schmid factor, m = cos $\theta$ . cos $\lambda$ .

cos $\theta$ = Angle between tensile axis [010] and = 45 degree

(011)

=

2

cos $\lambda$ = Angle between tensile axis [010] and  
011
and
| 100
.

(i) cos $\lambda$ =
0+1*1+0*1 V0+1+00+ (1)2 +12
=
2

Thus, Schmid factor, m = cos $\theta$ . cos $\lambda$ = 0.49

(ii) cos $\lambda$ = = 0

0*-1+0+0*0 V0+1+0/C-1)2 +0+ 0

Thus, Thus, Schmid factor, m = cos $\theta$ . cos $\lambda$ . = 0

Hence, in this case, slip will initiate on slip plane
(011)
in the slip direction  
011
.

  

Plane (2)

<Plane (2)>: Slip Plane is
(011
and Slip Directions are
| 100
and $\left [ 01\overline{1} \right ]$

In this case, we again have two slip directions. Now Schmid factor, m = cos $\theta$ . cos $\lambda$ .

cos $\theta$ = Angle between tensile axis [010] and = 45 degree

(011)

=

2

cos $\lambda$ = Angle between tensile axis [010] and  
| 100
and $\left [ 01\overline{1} \right ]$ .

(i) cos $\lambda$ = = 0

0本 -1+1+0+0 v0+1P 400-12 +0+0

Thus, Schmid factor, m = cos $\theta$ . cos $\lambda$ = 0

(ii) cos $\lambda$ =
00+1+1+0% -1 0+1 +0. (0 +12 -12
=
2

Thus, Schmid factor, m = cos $\theta$ . cos $\lambda$ . = 0.49.

Hence, in this case, slip will initiate on slip plane
(011)
in the slip direction  $\left [ 01\overline{1} \right ]$.

(b) Given: Critical resolved shear stress is 7 MN/m2

The formula for Critical resolved shear stress (CRSS) is given by

CRSS =
1.7
, where $\sigma$ = applied stress and m = Schmid factor.

Hence,    = 14.28 MN /m2

CRSS 0 = m

(c)

Dislocation Type	Slip Direction	Dislocation character
A	(a)	edge
B	(b)	screw
C	(a)	edge
D	(b)	screw

(d) If tensile axis were parallel to direction (b) in slip plane (1), then no slip will occur and therefore, no deformation will be observed in the single crystal.

Thank you. I have tried to answer the questions.