Chapter 8, Problem 8.27P

Interpretation Introduction

(a)

Interpretation:

The value of true strain for extrusion of diameter ranging from $\text{25 mm to 22 mm}$ and $22\text{mm}\text{to}20\text{mm}$ should be determined.

Concept Introduction:

The relationship between true strain and area is given by:

  $\epsilon =\ln \left(\frac{A_0}{A_f}\right)$

  $\begin{array}{l}{A}_0=\left(\frac{\pi }{4}\right)\times d_0^2\\ A_f=\left(\frac{\pi }{4}\right)\times d_f^2\end{array}$

Answer to Problem 8.27P

Value of true strain for the diameter ranging from $\text{25 mm to 22 mm}$ is $0.2556$

Value of true strain for the diameter ranging between is $\text{22 mm to 20 mm}$ is $0.19062$

Explanation of Solution

Initial diameter ${\text{d}}_0$ = $\text{25}\text{mm}$

Final diameter ${\text{d}}_{\text{f}}$ = $\text{22 mm}$

  $\epsilon =\ln \left(\frac{A_0}{A_f}\right)$

The above formula to be used for the calculation of true strain as given in the question is the value of initial diameter, that is, $\text{25 mm}$ and reduced diameter that is final diameter for first case is $\text{22 mm}$ and ${\epsilon }_1$ is taken as the true strain.

On substituting the value of area in the equation of true strain:

  ${\epsilon }_1=\ln \left(\frac{\left(\frac{\pi }{4}\right)\times d_0^2}{\left(\frac{\pi }{4}\right)\times d_f^2}\right)$

On solving:

  ${\epsilon }_1=2\ln \left(\frac{d_0}{d_f}\right)$

Now on substituting the value of initial diameter and final diameter:

  ${\epsilon }_1=2\ln \left(\frac{25}{22}\right)$

  ${\epsilon }_1=2\ln \left(1.1363\right)$

  ${\epsilon }_1=0.2556$

If the diameter is changing from $22mmto20mm,$ true strain is calculated using relationship

  ${\epsilon }_2=2\ln \left(\frac{d_0}{d_f}\right)$

Initial Diameter ${\text{d}}_0$ = $\text{22mm}$

Final Diameter ${\text{d}}_{\text{f}}$ = $\text{20mm}$

The values of area based on diameter is given below in the form of formula

On inserting the values in terms of diameter

  ${\epsilon }_2=2\ln \left(\frac{22}{20}\right)$

  ${\epsilon }_2=2\ln \left(1.1\right)$

  ${\epsilon }_2=0.19062$

Value of true strain for the diameter ranging from $\text{25 mm to 22 mm}$ is $\mathrm{0.2556.}$

Value of true strain for the diameter ranging between is $\text{22 mm to 20 mm}$ is $\mathrm{0.19062.}$

Interpretation Introduction

(b)

Interpretation:

The value of true strain for each given condition is to be determined. Also, the comparison of true strain for each step is to be explained

Concept Introduction:

The relationship between true strain and area is given by

  $\epsilon =\ln \left(\frac{A_0}{A_f}\right)$

  $\begin{array}{l}{A}_0=\left(\frac{\pi }{4}\right)\times d_0^2\\ A_f=\left(\frac{\pi }{4}\right)\times d_f^2\end{array}$

Answer to Problem 8.27P

Value of true strain for the diameter ranging from $25\text{mm}\text{to}20\text{mm}$ is $0.4462$

The value of true strain is $0.4462$ same for diameter ranging from $\text{25 mm to 22 mm}$ and $\text{22 mm to 20 mm}$ in comparison with $\text{25}\text{mm}\text{to}\text{20}\text{mm}\text{.}$

Explanation of Solution

The above formula to be used for determination of true strainis given as

Initial Diameter ${\text{d}}_0$ = $\text{25 mm}$

Final Diameter ${\text{d}}_{\text{f}}$ = $\text{20 mm}$

On inserting the values of initial and final diameter:

  $\begin{array}{l}\epsilon =2\ln \left(\frac{d_0}{d_f}\right)\\ \epsilon =2\ln \left(\frac{25}{20}\right)\\ \epsilon =2\ln \left(1.25\right)\\ \epsilon =0.4462\end{array}$

The value of true strain is $0.4462$ for the diameter ranging from $\text{25}\text{mm}\text{to}\text{20}\text{mm}$

Sum of the true strain of diameter ranging from $\text{25 mm to 22 mm}$ and $\text{22 mm to 20 mm}$ is $\mathrm{0.4462.}$

On comparing the sum of true strain the value remains the same for diameter ranging from $\text{25}\text{mm}\text{to}\text{20}\text{mm}$ and from $\text{25 mm to 22 mm}$ and $\text{22 mm to 20 mm}$.