Let's solve **all subparts (a-d)** step by step using the given data: - **Initial diameter, \( d_ = 6. \) mm** - **Final diameter, \( d_f = 5. \) mm** - **Die semi-angle, \( \alpha = 15^\circ \)** - **Coefficient of friction, \( \mu = .086 \)** - **Strength coefficient, \( K = 520 \) MPa** - **Strain hardening exponent, \( n = .20 \)** - **Room temperature process** --- ## (a) **Area Reduction** Area reduction (\( r \)) is given by: \[ r = \frac{A_ - A_f}{A_} \] where \( A_ \) and \( A_f \) are the initial and final cross-sectional areas. \[ A_ = \frac{\pi}{4} d_^2 = \frac{\pi}{4} (6.)^2 = 28.274 \text{ mm}^2 \] \[ A_f = \frac{\pi}{4} d_f^2 = \frac{\pi}{4} (5.)^2 = 19.635 \text{ mm}^2 \] \[ r = \frac{28.274 - 19.635}{28.274} = \frac{8.639}{28.274} = .3056 \approx \boxed{30.56\%} \] --- ## (b) **True Strain in Drawing** True strain (\( \epsilon \)) in drawing is: \[ \epsilon = \ln \left(\frac{A_}{A_f}\right) \] \[ \epsilon = \ln \left(\frac{28.274}{19.635}\right) = \ln(1.439) = \boxed{.363} \] --- ## (c) **Drawing Stress** The average flow stress for strain hardening materials is given by: \[ \bar{\sigma}_f = \frac{K}{1+n} \epsilon^n \] But for drawing, the drawing stress (\( \sigma_d \)) is: \[ \sigma_d = \bar{\sigma}_f \cdot e^{\left( \frac{\mu}{\tan\alpha} \ln\frac{A_}{A_f} \right)} \] First, compute \(\bar{\sigma}_f\): \[ \bar{\sigma}_f = \frac{K}{1+n} \epsilon^n \] \[ \bar{\sigma}_f = \frac{520}{1+.20} \times (.363)^{.2} \] \[ = \frac{520}{1.2} \times (.363)^{.2} \] \[ = 433.33 \times (.363)^{.2} \] Now, \( (.363)^{.2} = e^{.2 \ln .363} = e^{.2 \times (-1.013)} = e^{-.2026} = .8166 \) \[ \bar{\sigma}_f = 433.33 \times .8166 = 354.04 \text{ MPa} \] Now, calculate the exponent for friction: - \(\mu = .086\) - \(\tan 15^\circ = .2679\) So, \[ \frac{\mu}{\tan\alpha} = \frac{.086}{.2679} = .321 \] \[ \ln\left(\frac{A_}{A_f}\right) = \ln(1.439) = .363 \] Exponent: \[ .321 \times .363 = .116 \] \[ e^{.116} = 1.123 \] Now, drawing stress: \[ \sigma_d = 354.04 \times 1.123 = \boxed{397.8\ \text{MPa}} \] --- ## (d) **Drawing Force Required** Drawing force (\( F \)) is: \[ F = \sigma_d \times A_f \] \[ A_f = 19.635\ \text{mm}^2 = 19.635 \times 10^{-6}\ \text{m}^2 \] \[ F = 397.8\ \text{MPa} \times 19.635\ \text{mm}^2 \] Since \(1\ \text{MPa} = 1\ \text{N}/\text{mm}^2\): \[ F = 397.8 \times 19.635 = \boxed{7,813\ \text{N}} \] --- ## **Summary Table** | Subpart | Calculation | Answer | |--------------|-----------------------------------------------------|---------------------| | (a) Area reduction | \(\frac{A_ - A_f}{A_}\) | 30.56% | | (b) True strain | \(\ln(A_/A_f)\) | .363 | | (c) Drawing stress | See above | 397.8 MPa | | (d) Drawing force | \(\sigma_d \times A_f\) | 7,813 N | --- ### **If you need step-by-step working for any subpart, let me know!**