Chapter 2, Problem 2.96P

(a)

To determine

The true strain for the 1112 steel specimen during necking.

The true strain for the annealed 1112 steel specimen during necking.

The elongation of the 1112 steel specimen during necking.

The elongation of the annealed 1112 steel specimen during necking.

Answer to Problem 2.96P

The true strain for the 1112 steel specimen during necking is $0.08$ .

The true strain for the annealed 1112 steel specimen during necking is $0.19$ .

The elongation of the 1112 steel specimen during necking is $8.32\%$ .

The elongation of the annealed 1112 steel specimen during necking is $20.92\%$ .

Explanation of Solution

Given:

The diameter of specimen is $d=20\text{mm}$ .

The length of the specimen is $l_o=25\text{mm}$ .

Formula used:

The expression for the true strain for 1112 steel specimen during necking is given as,

  $\epsilon =n$

Here, $n$ is the exponent for the 1112 steel specimen.

The expression for the true strain for 1112 annealed steel specimen during necking is given as,

  ${\epsilon }_o=n_o$

Here, $n_o$ is the exponent for the 1112 annealed steel specimen.

The expression for the true strain for the 1112 steel specimen is given as,

  $\epsilon =\ln \left(\frac{l}{l_o}\right)$

Here, $l$ is the length of the cold rolled 1112 steel after necking.

The expression for the true strain for the 1112 annealed steel specimen is given as,

  ${\epsilon }_o=\ln \left(\frac{l_e}{l_o}\right)$

Here, $l_e$ is the length of the 1112 annealed steel after necking.

The expression for the elongation of the 1112 steel specimen at that instant is given as,

  $e=\left(\frac{l-l_o}{l_o}\right)\times 100$

The expression for the elongation of the 1112 annealed steel specimen at that instant is given as,

  $e_o=\left(\frac{l_e-l_0}{l_o}\right)\times 100$

Calculation:

The properties of 1112 steel specimen is given as,

The constant is $K=760\text{MPa}$ .

The exponent is $n=0.08$ .

The properties of 1112 annealed steel specimen is given as,

The constant is $K_o=760\text{MPa}$ .

The exponent is $n_o=0.19$ .

The true strain when necking occurs can be calculated as,

  $\begin{array}{c}\epsilon =n\\ \epsilon =0.08\end{array}$

The true strain for 1112 annealed steel specimen during necking can be calculated as,

  $\begin{array}{c}{\epsilon }_o=n_o\\ {\epsilon }_o=0.19\end{array}$

The length of the cold rolled 1112 steel after necking can be calculated as,

  $\begin{array}{c}\epsilon =\ln \left(\frac{l}{l_o}\right)\\ 0.08=\ln \left(\frac{l}{25\text{mm}}\right)\\ e^{0.08}=\ln \left(\frac{l}{25\text{mm}}\right)\\ l=27.08\text{mm}\end{array}$

The elongation of the 1112 steel specimen at that instant can be calculated as,

  $\begin{array}{c}e=\left(\frac{l-l_o}{l_o}\right)\times 100\\ e=\left(\frac{27.08\text{mm}-25\text{mm}}{25\text{mm}}\right)\times 100\\ e=8.32\%\end{array}$

The length of the 1112 annealed steel after necking can be calculated as,

  $\begin{array}{c}{\epsilon }_o=\ln \left(\frac{l_e}{l_o}\right)\\ 0.19=\ln \left(\frac{l_e}{25\text{mm}}\right)\\ l_e=30.23\text{mm}\end{array}$

The elongation of the 1112 annealed steel specimen at that instant can be calculated as,

  $\begin{array}{c}{e}_o=\left(\frac{l_e-l_0}{l_o}\right)\times 100\\ e_o=\left(\frac{30.23\text{mm}-25\text{mm}}{25\text{mm}}\right)\times 100\\ e_o=20.92\%\end{array}$

Conclusion:

Therefore, the true strain for the 1112 steel specimen during necking is $0.08$ .

Therefore, the true strain for the annealed 1112 steel specimen during necking is $0.19$ .

Therefore, the elongation of the 1112 steel specimen during necking is $8.32\%$ .

Therefore, the elongation of the annealed 1112 steel specimen during necking is $20.92\%$ .

(b)

To determine

The ultimate strength for 1112 steel specimen.

The ultimate strength for 1112 annealed steel specimen.

Answer to Problem 2.96P

The ultimate strength for 1112 steel specimen is $573\text{MPa}$ .

The ultimate strength for 1112 annealed steel specimen is $458.13\text{MPa}$ .

Explanation of Solution

Formula used:

The expression for the ultimate strength for 1112 steel specimen is given as,

  $S_s=K{\epsilon }^n$

The expression for the ultimate tensile strength for 1112 steel specimen is given as,

  $UT{S}_s=S_s\times e^n$

The expression for the ultimate strength for 1112 annealed steel specimen is given as,

  $S_a=K_o{\epsilon }_o{}^{n_o}$

The expression for the ultimate tensile strength for 1112 annealed steel specimen is given as,

  $UT{S}_a=S_a{e}^n$

Calculation:

The ultimate strength for 1112 steel specimen can be calculated as,

  $\begin{array}{c}{S}_s=K{\epsilon }^n\\ S_s=760\text{MPa}{\left(0.08\right)}^{0.08}\\ S_s=621\text{MPa}\end{array}$

The ultimate tensile strength for 1112 steel specimen can be calculated as,

  $\begin{array}{c}UT{S}_s=S_s\times e^n\\ UT{S}_s=621\text{MPa}\times e^{0.08}\\ UT{S}_s=573\text{MPa}\end{array}$

The ultimate strength for 1112 annealed steel specimen can be calculated as,

  $\begin{array}{c}{S}_a=K_o{\epsilon }_o{}^{n_o}\\ S_a=760\text{MPa}{\left(0.19\right)}^{0.19}\\ S_a=554\text{MPa}\end{array}$

The ultimate tensile strength for 1112 annealed steel specimen can be calculated as,

  $\begin{array}{c}UT{S}_a=S_a{e}^n\\ UT{S}_a=554\text{MPa}\times e^{0.19}\\ UT{S}_a=458.13\text{MPa}\end{array}$

Conclusion:

Therefore, the ultimate strength for 1112 steel specimen is $573\text{MPa}$ .

Therefore, the ultimate strength for 1112 annealed steel specimen is $458.13\text{MPa}$ .