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Give step-by-step solution with explanation and final answer:v Eames x @ow x ® omar x | ® wont x | sw aes x | © anon x | nme x |X ams x | Bann x |B mean x | + - 8 x
DEERE Cr a—— * (2 BE DO eoommmel
Student question ©Timeleftionsros | Tagthe question Step-by-step > Final solution
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ACualloy (i = 483 GPaand b = 0.256 nm) with grains of 120 ym average diameter has a Mechanical Engineering

yield strength of Oy = 185.22 MPa. The same alloy has a yield strength of 180.9 MPa if the

average diameter of its grains is three times bigger. Sobsblect

Search And Select

a. Determine the parameters of the Hall-Petch equation for this alloy and predict its yield -

strength when the average grain diameter is 2 um.

b. Next, suppose that, through appropriate processing, the grain boundaries of this alloy are

strengthened in such a way that the critical local stress to activate poly - slip in neighbor grains.

is increased. In these conditions the alloy shows a yield strength of 210 MPa when the

average grain diameter d = 120 um. Compute the new yield strength when d = 2 ym.
& EQuEB2suecIREOPCR °@Fa ~ Gla BN enh,

VIPSolutions · Accepted Answer

Let's break down the problem and solve it step by step.

## **Given:**
- Shear modulus, μ = 48.3 GPa
- Burgers vector, b = 0.256 nm
- Grain size $ d_1 = 120 $ μm, $ \sigma_{y1} = 185.22 $ MPa
- Grain size $ d_2 = 3 \times 120 = 360 $ μm, $ \sigma_{y2} = 180.9 $ MPa

The Hall–Petch equation:
$$
\sigma_y = \sigma_0 + k_y d^{-1/2}
$$

---

### **a. Determine the Hall–Petch parameters ($\sigma_0$ and $k_y$) and predict yield strength at $d = 2$ μm**

#### **Step 1: Write equations for both cases**

For $ d_1 = 120 $ μm:
$$
185.22 = \sigma_0 + k_y (120)^{-1/2}
$$

For $ d_2 = 360 $ μm:
$$
180.9 = \sigma_0 + k_y (360)^{-1/2}
$$

#### **Step 2: Convert grain sizes to consistent units (μm is fine)**

$$
(120)^{-1/2} = \frac{1}{\sqrt{120}} = 0.0913
$$
$$
(360)^{-1/2} = \frac{1}{\sqrt{360}} = 0.0527
$$

#### **Step 3: Set up the equations**

$$
185.22 = \sigma_0 + k_y \times 0.0913 \tag{1}
$$
$$
180.9 = \sigma_0 + k_y \times 0.0527 \tag{2}
$$

#### **Step 4: Subtract Eq. (2) from Eq. (1)**

$$
185.22 - 180.9 = k_y (0.0913 - 0.0527)
$$
$$
4.32 = k_y \times 0.0386
$$
$$
k_y = \frac{4.32}{0.0386} = 111.66 \ \text{MPa} \cdot \mu m^{1/2}
$$

#### **Step 5: Find $\sigma_0$ using Eq. (1):**

$$
185.22 = \sigma_0 + 111.66 \times 0.0913
$$
$$
185.22 = \sigma_0 + 10.19
$$
$$
\sigma_0 = 185.22 - 10.19 = 175.03 \ \text{MPa}
$$

---

#### **Step 6: Predict yield strength at $ d = 2 $ μm**

$$
d = 2 \ \mu m \implies d^{-1/2} = \frac{1}{\sqrt{2}} = 0.7071
$$

$$
\sigma_y = 175.03 + 111.66 \times 0.7071 = 175.03 + 78.96 = 254.0 \ \text{MPa}
$$

---

### **b. New yield strength when grain boundaries are strengthened**

Given:  
- New yield strength at $d = 120$ μm is 210 MPa.

Use the same Hall–Petch equation to find new $\sigma_0'$:

$$
210 = \sigma_0' + 111.66 \times 0.0913
$$
$$
210 = \sigma_0' + 10.19
$$
$$
\sigma_0' = 210 - 10.19 = 199.81 \ \text{MPa}
$$

Now, for $d = 2$ μm:
$$
\sigma_y' = 199.81 + 111.66 \times 0.7071 = 199.81 + 78.96 = 278.77 \ \text{MPa}
$$

---

## **Final Answers**

**a.**
- Hall–Petch parameters:  
  - $\sigma_0 = 175.03$ MPa  
  - $k_y = 111.66$ MPa·μm$^{1/2}$  
- Predicted yield strength at $d = 2$ μm:  
  - $\sigma_y = \boxed{254.0\ \text{MPa}}$

**b.**
- New predicted yield strength at $d = 2$ μm:  
  - $\sigma_y' = \boxed{278.77\ \text{MPa}}$

---

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