VIPSolutionsGive complete answer with answers of all subparts , A copper wire of initial diameter 6 . 0 mm is drawn through a die to a final diameter of 5 . 0 mm as shown in Figure 3 . The die has an entrance angle of 1 5 ° , and the coefficient of friction is 0 . 0 6 . The metal has a strength coefficient of 5 2 0 MPa and a strain hardening exponent of 0 . 2 0 . The process is carried out at room temperature. Determine: ( a ) Area reduction, ( b ) True strain in drawing, ( c ) Drawing stress, ( d ) Drawing force required.
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Give complete answer with answers of all subparts , A copper wire of initial diameter 6 . 0 mm is drawn through a die to a final diameter of 5 . 0 mm as shown in Figure 3 . The die has an entrance angle of 1 5 ° , and the coefficient of friction is 0 . 0 6 . The metal has a strength coefficient of 5 2 0 MPa and a strain hardening exponent of 0 . 2 0 . The process is carried out at room temperature. Determine: ( a ) Area reduction, ( b ) True strain in drawing, ( c ) Drawing stress, ( d ) Drawing force required.
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Asked by: manish kumar
Created at: 2026-01-10 14:13:21
Asked by: manish kumar
Created at: 2026-01-10 14:13:21
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# Wire Drawing Problem
Given:
- **Initial diameter** (\(d_\)) = 6. mm
- **Final diameter** (\(d_f\)) = 5. mm
- **Entrance angle of die** (\(\alpha\)) = 15°
- **Coefficient of friction** (\(\mu\)) = .06
- **Strength coefficient** (\(K\)) = 520 MPa
- **Strain hardening exponent** (\(n\)) = .20
- **Process temperature:** Room temperature
Let us solve each part step by step.
---
## (a) Area Reduction
**Formula:**
\[
\text{Area reduction, } r = \frac{A_ - A_f}{A_}
\]
Where:
- \(A_ = \frac{\pi}{4} d_^2\) = initial area
- \(A_f = \frac{\pi}{4} d_f^2\) = final area
**Calculations:**
\[
A_ = \frac{\pi}{4} (6.^2) = \frac{\pi}{4} \times 36. = 28.274 \text{ mm}^2
\]
\[
A_f = \frac{\pi}{4} (5.^2) = \frac{\pi}{4} \times 25. = 19.635 \text{ mm}^2
\]
\[
r = \frac{28.274 - 19.635}{28.274} = \frac{8.639}{28.274} = .3056 \approx 30.6\%
\]
---
## (b) True Strain in Drawing
**Formula:**
\[
\epsilon = \ln\left(\frac{A_}{A_f}\right)
\]
**Calculation:**
\[
\epsilon = \ln\left(\frac{28.274}{19.635}\right) = \ln(1.439) = .363
\]
---
## (c) Drawing Stress
The **average flow stress** during drawing for a strain-hardening material:
\[
\bar{\sigma}_f = \frac{K \epsilon^n}{1 + n}
\]
But for wire drawing, the **drawing stress** considering friction and die angle:
\[
\sigma_d = \bar{\sigma}_f \left[ 1 + \frac{\mu}{\tan \alpha} \right]
\]
### Step 1: Calculate average flow stress
\[
\bar{\sigma}_f = \frac{K \epsilon^n}{1 + n}
\]
Plug in the values:
- \(K = 520\) MPa
- \(\epsilon = .363\)
- \(n = .20\)
\[
\epsilon^n = .363^{.20} = e^{.20 \ln(.363)} \approx e^{.20 \times (-1.013)} \approx e^{-.203} \approx .816
\]
\[
\bar{\sigma}_f = \frac{520 \times .816}{1 + .20} = \frac{424.3}{1.20} = 353.6 \text{ MPa}
\]
### Step 2: Calculate drawing stress
\[
\sigma_d = \bar{\sigma}_f \left[ 1 + \frac{\mu}{\tan \alpha} \right]
\]
- \(\mu = .06\)
- \(\tan 15^\circ = .2679\)
\[
\frac{\mu}{\tan \alpha} = \frac{.06}{.2679} = .224
\]
\[
\sigma_d = 353.6 \times (1 + .224) = 353.6 \times 1.224 = 432.6 \text{ MPa}
\]
---
## (d) Drawing Force Required
**Formula:**
\[
F = \sigma_d \cdot A_f
\]
**Calculation:**
\[
F = 432.6 \text{ MPa} \times 19.635 \text{ mm}^2 = 8,494 \text{ N}
\]
---
# **Summary Table**
| Part | Quantity | Value |
|------|------------------|------------------|
| (a) | Area reduction | \(30.6\%\) |
| (b) | True strain | \(.363\) |
| (c) | Drawing stress | \(432.6\) MPa |
| (d) | Drawing force | \(8,494\) N |
---
**Figure 3: Wire Drawing Schematic**
*Figure 3: Wire drawing process with labeled parameters*
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