Give step-by-step solution with explanation and final answer:
Act like a helpful tutor and exlain me :Required information
‘Compute the true stress and the logarithmic strain using the data given be

Question

Give step-by-step solution with explanation and final answer:
Act like a helpful tutor and exlain me :Required information
‘Compute the true stress and the logarithmic strain using the data given below and plot the results on log-log paper. Then,
ind he plastic strength cooficient 7, and th siran-sirengthening exponent m. The yield srength and he uma
Stent aftr the specimen has hod 70 percent cold work The orginal damete of the test specmen s 0.506 in. and the
originallengih, L., - 201. The following data was obtained for the sastc and plastic sites.
For the elastic state
I CN
CT 1
1
1
EI 1
For the plastic state
Pod
CE ——
What ar the yield an the utimt tanslie siength folowing 20% cold working? Assume th tense sirengh prior to cod working 5
856 kp.
The yield ond the uitmata tense strengths are
5'= Cw
Sm est

VIPSolutions · Accepted Answer

Let's break down the problem step by step:

---

## **Step 1: Given Data**

- **Original diameter, $ d_0 $ = 0.506 in**
- **Original length, $ L_0 $ = 2.0 in**
- **Tensile strength prior to cold working, $ S_u = 85.6 $ ksi**
- **20% cold work**
- **Tables provide:**
  - For elastic state: $ P_i $ (load), $ \Delta l $ (elongation)
  - For plastic state: $ P_i $ (load), $ A_i $ (instantaneous area)

---

## **Step 2: Calculate Original Area, $ A_0 $**

$$
A_0 = \frac{\pi d_0^2}{4} = \frac{\pi (0.506)^2}{4} \approx 0.201 \text{ in}^2
$$

---

## **Step 3: Cold Work Calculation**

**% Cold Work =**  
$$
\%CW = \frac{A_0 - A_f}{A_0} \times 100
$$
Given 20% cold work, solve for $ A_f $:

$$
0.20 = \frac{A_0 - A_f}{A_0}
\implies A_f = A_0 (1 - 0.20) = 0.201 \times 0.80 = 0.1608 \text{ in}^2
$$

---

## **Step 4: Find Yield and Ultimate Strengths After Cold Work**

### **Empirical Relationships:**

For most metals:
- **Yield Strength after CW:**
  $$
  S_y' = S_y + \Delta S_{cw}
  $$
- **Ultimate Strength after CW:**
  $$
  S_u' = S_u + \Delta S_{cw}
  $$
Where $ \Delta S_{cw} $ is the increase in strength due to cold work.

**Empirical equation:**
$$
S = S_0 (1 + k \cdot \%CW)
$$
But here, we're told just to assume the ultimate tensile strength prior to cold work is 85.6 ksi.

### **Ultimate Strength After Cold Work**

A common approximate relationship:
$$
S_u' = \frac{S_u}{1 - \text{Fraction Cold Work}}
$$
$$
S_u' = \frac{85.6}{1 - 0.20} = \frac{85.6}{0.80} = 107 \text{ ksi}
$$

### **Yield Strength After Cold Work**

The yield strength usually increases by a similar factor:
$$
S_y' \approx \frac{S_{y,0}}{1 - \text{Fraction Cold Work}}
$$
If not given, you can estimate it as 0.6–0.7 times the ultimate strength for many steels:

$$
S_{y,0} \approx 0.6 \times 85.6 = 51.36 \text{ ksi}
$$
$$
S_y' \approx \frac{51.36}{0.80} = 64.2 \text{ ksi}
$$

---

## **Step 5: Final Answers**

$$
\boxed{S_y' = 64.2 \text{ ksi}}
$$

$$
\boxed{S_u' = 107 \text{ ksi}}
$$

---

### **Explanation**

- **Cold working** makes the material stronger by reducing its area and increasing dislocation density.
- The **yield and ultimate strengths** *increase* by the reciprocal of the remaining area fraction.
- This method is a standard estimate; for more accurate results, you'd need to use the actual stress-strain data and plot it.

---

## **Summary Table**

| Property                  | Before CW | After 20% CW |
|---------------------------|-----------|--------------|
| Yield Strength ($S_y$)  | ~51.36 ksi| 64.2 ksi     |
| Ultimate Strength ($S_u$)| 85.6 ksi | 107 ksi      |

---

If you have any further questions about the stress-strain plotting or about the equations, let me know!

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