Chapter 8, Problem 8.11P

Interpretation Introduction

Interpretation:

The strain hardening coefficient of metal needs to be determined. Whether the metal is FCC, BCC or HCP needs to be determined.

Concept introduction:

Strain hardening coefficient is denoted by 'n' it is a material constant it is used for stress-strain behavior calculated in strain hardening.

Answer to Problem 8.11P

The strain hardening coefficient 'n' for the metal is 0.51. It is in the range of FCC.

Explanation of Solution

Given Information:

The measurements in the plastic region are as follows:

Force (N)	Change in gage length (cm)	Diameter(cm)
16,240	0.6642	1.2028
19,066	1.4754	1.0884
19,273	2.4663	0.9848

Calculation:

The diameter of the metal bar =1.33 cm

Gage length of bar = 3 cm

When force = 16240 N, true stress-strain of the bar is given by

The stress is given by a general case:

  ${\sigma }_T=\frac{F}{A}$ ------------- (1)

Where,

  $\begin{array}{l}{F}_1=\text{Force applied on bar}=1624\text{0 N}\\ A_1=\frac{\pi }{4}{d}_1{}^2\\ d_1=1.202\text{8 cm}\end{array}$

Equation (1) becomes,

  $\begin{array}{l}{\sigma }_T=\frac{16240}{\frac{\pi }{4}{(1.2028\times 10)}^2}\\ {\sigma }_T=143MPa\end{array}$

Engineering strain is given by when change length $(\Delta L)$ is 0.6642 to original length is 3 cm.

  $\xi =\frac{(\Delta L)}{l}$ ---------------- (2)

  $\begin{array}{l}\xi =\frac{0.6642}{3}\\ \xi =0.2214\end{array}$

Strain hardening is given by,

  ${\xi }_T=\ln (1+\xi )$ ------------------ (3)

Where,

  $\begin{array}{l}\xi =Engineeringstrain\\ {\xi }_T=\ln (1+0.2214)\\ {\xi }_T=0.200\end{array}$

Similarly, at f = 19066 N and d =1.0884 cm.

Equation (1) becomes,

  $\begin{array}{l}{\sigma }_T=\frac{19066}{\frac{\pi }{4}{(1.0884\times 10)}^2}\\ {\sigma }_T=205MPa\end{array}$

Now at,

  $\begin{array}{l}\Delta L=1.475\text{4cm}\\ l=\text{3cm}\end{array}$

Equation (2) becomes,

  $\begin{array}{l}\xi =\frac{\Delta L}{L}\\ \xi =\frac{1.4754}{3}\\ \xi =0.4918\end{array}$

Also, true strain is given by at $\xi =0.4918$

  $\begin{array}{l}\ln =\ln (1+\xi )\\ {\xi }_T=\ln (1+0.4918)\\ {\xi }_T=0.400\end{array}$

Similarly, at F=19273, d=0.9848.

From equation (1),

  ${\sigma }_T=253MPa$

Also, at

  $\begin{array}{l}\Delta L=2.4663cm\\ L=3\end{array}$

Equation (2) become,

  $\begin{array}{l}\xi =\frac{2.4663}{3}\\ \xi =0.8221\end{array}$

The true strain is given by,

  $\begin{array}{l}\ln =\ln (1+\xi )\\ {\xi }_T=\ln (1+0.8221)\\ {\xi }_T=0.6\end{array}$

Putting all value in the table,

Sr.no.	Force   $(F_t)(N)$	Engineering strain   $(\xi )\text{(cm/cm)}$	Diameter D (mm)	True stress   $({\sigma }_T)MPa$	True strain   $({\xi }_T)(\text{cm/cm)}$
1	16240	0.2214	12.028	143	0.200
2	19066	0.4918	10.884	205	0.400
3	19273	0.8221	9.848	213	0.600

Strength coefficient formula is given by

  ${\sigma }_T=k{({\xi }_T)}^n$ -------------------- (4)

Here,

  $\begin{array}{l}{\sigma }_T=\text{True stress}\hfill \\ {\xi }_T=\text{True strain}\hfill \\ n=\text{Strain hardening component}\hfill \\ k=\text{Strength coefficient} \hfill \end{array}$

Substitute,

  $\begin{array}{l}{\xi }_T=0.2\\ {\sigma }_T=143\end{array}$

Equation (2) becomes,

  $143=k+{0.2}^n$

Applying natural logarithm on both sides,

  $\begin{array}{l}\ln (143)=\ln (k\times {0.2}^n)\\ \ln (143)=\ln (k)+\ln ({0.2}^0)\\ \ln (143)=\ln (k)+n\times \ln (0.2)\\ 4.9628=\ln (k)-1.609n\end{array}$ ---------------- (a)

Similarly substituting,

  $({\sigma }_T=k{({\xi }_T)}^{n}MPa$

  $\begin{array}{l}{\sigma }_T=253\text{MPa}\text{and}\\ {\xi }_T=0.6\end{array}$

Equation (4) becomes,

  $253=k\times {0.6}^n$

Applying logarithm on both sides,

  $\begin{array}{l}\ln (253)=\ln [(k)\times {0.6}^n)\\ \ln (253)=\ln (k)+\ln (0.6)\\ 5.5334=\ln k-0.5108n\end{array}$ ------------------------ (b)

Equating equation (a) and (b),

  $n=0.51$

Conclusion

Thus, strain hardening coefficient for the metal in FCC is 0.51.