Question

Two metals, A and B, have Stacking Fault Energies of $20\text{ mJ/m}^2$ and $160\text{ mJ/m}^2$ respectively. Which of the following is true regarding their deformation behaviour?

• A. Metal A will cross-slip more easily than Metal B.
• B. Metal B will exhibit a higher rate of work hardening than Metal A.
• C. Metal B will cross-slip more easily than Metal A.
• D. Metal A will not exhibit any plastic deformation.

Hint:

High SFE (e.g., Aluminum) $\rightarrow$ Easy cross-slip $\rightarrow$ Low work hardening rate.
Low SFE (e.g., Brass, Stainless Steel) $\rightarrow$ Difficult cross-slip $\rightarrow$ High work hardening rate.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks how the Stacking Fault Energy (SFE) of two metals affects their plastic deformation behavior, specifically their capacity to undergo cross slip.

Step 2: Key Formula or Approach:
The Stacking Fault Energy (SFE) determines the separation distance ($d$) between two partial dislocations:
\[ d \propto \frac{1}{\text{SFE}} \]
- High SFE $\implies$ Narrow stacking fault (small separation).
- Low SFE $\implies$ Wide stacking fault (large separation).

Step 3: Detailed Explanation:

• For a dislocation to cross slip (move from one slip plane to an intersecting plane), any dissociated partial dislocations must first recombine into a single perfect dislocation.

• Metal B (High SFE $= 160\text{ mJ/m}^2$): Because the SFE is very high, the stacking fault width between the partial dislocations is extremely narrow. It requires very little energy to force these partials to recombine, making cross slip easy and highly frequent.

• Metal A (Low SFE $= 20\text{ mJ/m}^2$): Because the SFE is low, the partial dislocations are widely separated. It is highly unfavorable for them to recombine, which suppresses cross slip. Instead, dislocations are restricted to planar slip, leading to high dislocation density pile-ups and a higher rate of work hardening.

Step 4: Final Answer:
Metal B will cross-slip more easily than Metal A because of its higher Stacking Fault Energy.