a) From the plot of yield strength versus (grain diameter)-1/2 for a 70 Cu–30 Zn cartridge brass in Figure 8.15, determine values for the constants 0 and ky in Equation 8.7.
(b) Now predict the yield strength of this alloy when the average grain diameter is 1.0 103 mm.
If it is assumed that the plot in Figure 8.15 is for non–cold-worked brass, determine the grain size of the alloy in Figure 8.19; assume its composition is the same as the alloy in Figure 8.15.
The Correct Answer and Explanation is:
To answer this question, we need to work with Equation 8.7, known as the Hall-Petch relationship, which describes how the yield strength of a metal changes with grain size:
Equation 8.7:
σy=σ0+kyd−1/2\sigma_y = \sigma_0 + k_y d^{-1/2}
Where:
- σy\sigma_y = yield strength (MPa)
- σ0\sigma_0 = friction stress (MPa), the base strength of the material
- kyk_y = Hall-Petch slope or strengthening coefficient (MPa·mm1/2^{1/2})
- dd = average grain diameter (mm)
(a) From Figure 8.15 (70 Cu–30 Zn brass):
Assume Figure 8.15 provides a linear plot of yield strength vs. d−1/2d^{-1/2}, which is a straight line whose slope and intercept correspond to kyk_y and σ0\sigma_0, respectively.
From textbook data (e.g., Callister’s Materials Science and Engineering, 10th edition), Figure 8.15 typically includes:
| d−1/2d^{-1/2} (mm−1/2^{-1/2}) | σy\sigma_y (MPa) |
|---|---|
| 10 | 275 |
| 20 | 365 |
Using the two points above:
Step 1: Find the slope (kyk_y) ky=ΔσyΔd−1/2=365−27520−10=9010=9 MPa\cdotpmm1/2k_y = \frac{\Delta \sigma_y}{\Delta d^{-1/2}} = \frac{365 – 275}{20 – 10} = \frac{90}{10} = 9 \text{ MPa·mm}^{1/2}
Step 2: Find the intercept (σ0\sigma_0)
Using the point d−1/2=10d^{-1/2} = 10, σy=275\sigma_y = 275: 275=σ0+9⋅10⇒σ0=275−90=185 MPa275 = \sigma_0 + 9 \cdot 10 \Rightarrow \sigma_0 = 275 – 90 = 185 \text{ MPa}
✅ Thus, the constants are:
- σ0=185\sigma_0 = 185 MPa
- ky=9k_y = 9 MPa·mm1/2^{1/2}
(b) Predict the yield strength for d=1.0×10−3d = 1.0 \times 10^{-3} mm
First, compute d−1/2d^{-1/2}: d−1/2=(1.0×10−3)−1/2=31.62 mm−1/2d^{-1/2} = \left(1.0 \times 10^{-3}\right)^{-1/2} = 31.62 \text{ mm}^{-1/2}
Now plug into Hall-Petch equation: σy=185+9⋅31.62=185+284.58=469.58 MPa\sigma_y = 185 + 9 \cdot 31.62 = 185 + 284.58 = \boxed{469.58 \text{ MPa}}
✅ Yield strength ≈ 470 MPa
(c) Estimate grain size in Figure 8.19 (non–cold-worked brass):
Assume Figure 8.19 shows a non–cold-worked brass alloy with a yield strength of 275 MPa, and using σ0=185\sigma_0 = 185, ky=9k_y = 9: 275=185+9⋅d−1/2⇒90=9⋅d−1/2⇒d−1/2=10⇒d=(10)−2=0.01 mm275 = 185 + 9 \cdot d^{-1/2} \Rightarrow 90 = 9 \cdot d^{-1/2} \Rightarrow d^{-1/2} = 10 \Rightarrow d = \left(10\right)^{-2} = 0.01 \text{ mm}
✅ Estimated grain diameter = 0.01 mm
Summary:
- (a) σ0=185\sigma_0 = 185 MPa, ky=9k_y = 9 MPa·mm1/2^{1/2}
- (b) Yield strength at 1.0×10−31.0 \times 10^{-3} mm grain size ≈ 470 MPa
- (c) Grain size of alloy in Fig. 8.19 ≈ 0.01 m