Chapter 2, Problem 10P

(a)

To determine

The yield strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation.

The ultimate strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation.

The percentage increase in yield strength.

The percentage increase in ultimate strength.

Answer to Problem 10P

The yield strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation is $77.545\text{kpsi}$.

The ultimate strength if hot rolled AISI 1212 HR steel undergoes a 20 percent cold-work operation is $76.875\text{kpsi}$.

The percentage increase in yield strength is $176.94\%$.

The percentage increase in ultimate strength is $25\%$.

Explanation of Solution

Write the expression for work load factor and area.

$\frac{A_0}{A_i}=\frac{1}{1-W}$ (I)

Here, the original area is $A_0$, the area after load is $A_i$ and work load factor is $W$.

Write the expression for true strain.

${\epsilon }_i=\ln \frac{A_0}{A_i}$ (II)

Here, the true strain is ${\epsilon }_i$.

Calculate the new yield strength after cold work.

${S^{\prime }}_y={\sigma }_0{\epsilon }^m{}_i$ (III)

Here, the yield strength is ${S^{\prime }}_y$, the strength coefficient is ${\sigma }_0$ and strain strengthening exponent is $m$.

Calculate the new ultimate strength after cold work.

${S^{\prime }}_u=\frac{S_u}{1-W}$ (IV)

Here, the new ultimate strength after clod work is ${S^{\prime }}_u$ and the ultimate strength before cold work is $S_u$.

Calculate the percentage increase in yield strength.

${\text{d}}_y\text{=}\left(\frac{\left({S^{\prime }}_y-S_y\right)}{\left(S_y\right)}\right)\times 100\%$                                              (V)

Here, the percentage increase in yield strength is ${\text{d}}_y$, the yield strength before cold-work is $S_y$.

Calculate the percentage increase in ultimate strength.

${\text{d}}_u=\left(\frac{\left({S^{\prime }}_u-S_u\right)}{\left(S_u\right)}\right)\times 100\%$ (VI)

Here, the percentage increase in ultimate strength is ${\text{d}}_u$.

Conclusion:

Refer to Table A-22 “Results of Tensile Test of Some Metals” for hot rolled AISI 1212 steel.

Obtain the yield strength as $28\text{kpsi}$, the ultimate strength before cold work as $61.5\text{kpsi}$, the strength coefficient as $110\text{kpsi}$ and strain strengthening exponent as $0.24$. The fracture strength ${\epsilon }_f$ is $0.85$.

The steel undergoes $20$ percent cold work operation therefore the value of work load factor $W$ is $0.20$

Substitute $0.20$ for $W$ in the Equation (I).

$\begin{array}{c}\frac{A_0}{A_i}=\frac{1}{1-0.20}\\ =1.25\end{array}$

Substitute $1.25$ for $\frac{A_0}{A_i}$ in the Equation (II).

$\begin{array}{c}{\epsilon }_i=\ln 1.25\\ =0.223\end{array}$

Substitute $0.223$ for ${\epsilon }_i$, $0.24$ for $m$ and $110\text{kpsi}$ for ${\sigma }_0$ in the Equation (III).

$\begin{array}{c}{S^{\prime }}_y=\left(110\text{kpsi}\right){\left(0.233\right)}^{0.24}\\ =\left(110\text{kpsi}\right)\left(0.70496\right)\\ \approx 77.545\text{kpsi}\end{array}$

Thus, the yield strength is $77.545\text{kpsi}$.

Substitute $28\text{kpsi}$ for $S_y$ and $77.545\text{kpsi}$ for ${S^{\prime }}_y$ in Equation (V).

$\begin{array}{c}{\text{d}}_y=\left(\frac{\left(77.545\text{kpsi}-28\text{kpsi}\right)}{\left(28\text{kpsi}\right)}\right)\times 100\%\\ =\left(\frac{49.545\text{kpsi}}{28\text{kpsi}}\right)\times 100\%\\ \approx 176.94\%\end{array}$

Thus, the percentage increase in yield strength is $176.94\%$.

Substitute $61.5\text{kpsi}$ for $S_u$ and $0.20$ for $W$ in the Equation (IV).

$\begin{array}{c}{S^{\prime }}_u=\frac{\left(61.5\text{kpsi}\right)}{\left(1-0.20\right)}\\ =\frac{\left(61.5\text{kpsi}\right)}{\left(0.80\right)}\\ =76.875\text{kpsi}\end{array}$

Thus, the ultimate strength is $76.875\text{kpsi}$.

Substitute $61.5\text{kpsi}$ for $S_u$ and $76.875\text{kpsi}$ for ${S^{\prime }}_u$ in Equation (VI).

$\begin{array}{c}{\text{d}}_u=\left(\frac{\left(76.875\text{kpsi}\right)-\left(61.5\text{kpsi}\right)}{\left(61.5\text{kpsi}\right)}\right)\times 100\%\\ =\left(\frac{15.375\text{kpsi}}{61.5\text{kpsi}}\right)\times 100\%\\ =25\%\end{array}$

Thus, the percentage increase in ultimate strength is $25\%$.

(b)

To determine

The ratio of ultimate and yield strength before cold-work operation.

The ratio of ultimate and yield strength after cold-work operation and explain the ductility of part with the help of result.

Answer to Problem 10P

The ratio of ultimate and yield strength before cold-work operation is $2.20$.

The ratio of ultimate and yield strength after cold-work operation is $1.0$ and after cold working operation steel lost most of its ductility.

Explanation of Solution

Calculate the ratio of ultimate strength to yield strength before cold work.

$r_1=\frac{S_u}{S_y}$                                                           (VII)

Here, the ratio of ultimate strength to yield strength is $r_1$, the ultimate strength before cold work is $S_u$ and yield strength before cold work is $S_y$.

Calculate the ratio of ultimate strength to yield strength after cold work.

$r_2=\frac{{S^{\prime }}_u}{{S^{\prime }}_y}$ (VIII)

Here, the ratio of ultimate strength to yield strength after cold work is $r_2$, the ultimate strength after cold work is ${S^{\prime }}_u$ and the yield strength after cold work is ${S^{\prime }}_y$.

Conclusion:

Substitute $61.5\text{kpsi}$ for $S_u$ and $28\text{kpsi}$ for $S_y$ in the Equation (VII).

$\begin{array}{c}{r}_1=\frac{\left(61.5\text{kpsi}\right)}{\left(28\text{kpsi}\right)}\\ =2.1964\\ \approx 2.196\end{array}$

Thus, the ratio of ultimate strength to yield strength before cold work is $2.196$.

Substitute $76.9\text{kpsi}$ for ${S^{\prime }}_u$ and $76.7\text{kpsi}$ for ${S^{\prime }}_y$ in the Equation (VIII).

$\begin{array}{c}{r}_2=\frac{\left(76.9\text{kpsi}\right)}{\left(76.7\text{kpsi}\right)}\\ =1.0026\\ \approx 1.0\end{array}$

Thus, the ratio of ultimate strength to yield strength after cold work is $1.0$.

From the calculated value of $r_1$ and $r_2$, it is obtained that $r_1>r_2$. So, it can be concluded that the steel loses its difficulty after the cold working.