Chapter 6, Problem 6.11P

To determine

The total tensile proportional limit stress.

Answer to Problem 6.11P

The tensile proportional limit stress is $49\text{MPa}$ .

Explanation of Solution

Given:

The critical shear stress of niobium is $20\times {10}^6\text{Pa}$ .

The temperature is $295\text{K}$ .

Formula Used:

Write the expression for the angle between normal to $\left[011\right]$ slip plane and $\left[001\right]$ crystal direction.

  $\varphi ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]$ …… (I)

Here, $\varphi$ is the angle between the normal to the slip plane and direction of the crystal orientation, $u_1$ is the $x$ -component of vector 1, $v_1$ is $y$ -component of vector 1 and $w_1$ is the $z$ -component of vector 1, $u_2$ is the $x$ -component of vector 2, $v_2$ is $y$ -component of vector 2 and $w_2$ is the $z$ -component of vector 2.

Write the expression for the angle between $\left[111\right]$ slip direction and $\left[001\right]$ crystal direction.

  $\lambda ={\cos }^{-1}\left[\frac{u_1{u}_2+v_1{v}_2+w_1{w}_2}{\sqrt{\left(u_1^2+v_1^2+w_1^2\right)\left(u_2^2+v_2^2+w_2^2\right)}}\right]$ …… (II)

Here, $\lambda$ is the angle between slip direction and the direction of crystal orientation.

Write the expression for relation between resolved shear stress and tensile proportional limit stress.

  ${\tau }_{\text{CRSS}}={\sigma }_{\text{y}}\cos \lambda \cos \varphi$ …… (III)

Here, ${\tau }_{\text{CRSS}}$ is the resolved shear stress and ${\sigma }_{\text{y}}$ is the tensile proportional limit stress.

Calculation:

Substitute $0$ for $u_1$ , $1$ for $v_1$ , $1$ for $w_1$ , $0$ for $u_2$ , $0$ for $v_2$ and $1$ for $w_2$ in equation (I).

  $\begin{array}{l}\varphi ={\cos }^{-1}\left[\frac{\left(0\right)\left(0\right)+\left(1\right)\left(0\right)+\left(1\right)\left(1\right)}{\sqrt{\left(0^2+1^2+1^2\right)\left(0^2+0^2+1^2\right)}}\right]\\ ={\cos }^{-1}\left[\frac{1}{\sqrt{2\left(1\right)}}\right]\\ ={\cos }^{-1}\left[\frac{1}{\sqrt{2}}\right]\\ =45°\end{array}$

Substitute $1$ for $u_1$ , $1$ for $v_1$ , $1$ for $w_1$ , $0$ for $u_2$ , $0$ for $v_2$ and $1$ for $w_2$ in equation (II).

  $\begin{array}{c}\lambda ={\cos }^{-1}\left[\frac{\left(1\right)\left(0\right)+\left(1\right)\left(0\right)+\left(1\right)\left(1\right)}{\sqrt{\left(0^2+1^2+1^2\right)\left(0^2+0^2+1^2\right)}}\right]\\ ={\cos }^{-1}\left[\frac{1}{\sqrt{3\left(1\right)}}\right]\\ ={\cos }^{-1}\left[\frac{1}{\sqrt{3}}\right]\\ =54.74°\end{array}$

Substitute $45°$ for $\varphi$ , $20\times {10}^6\text{Pa}$ for ${\tau }_{\text{CRSS}}$ and $54.74°$ for $\lambda$ in equation (III).

  $\begin{array}{c}20\times {10}^6\text{Pa}={\sigma }_{\text{y}}\cos \left(54.74°\right)\cos \left(45°\right)\\ {\sigma }_{\text{y}}=\frac{20\times {10}^6\text{Pa}}{\cos \left(54.74°\right)\cos \left(45°\right)}\\ =\frac{20\times {10}^6\text{Pa}}{\left(0.577\right)\left(0.707\right)}\\ =\left(4.90\times {10}^7\text{Pa}\right)\left(\frac{{10}^{-6}\text{MPa}}{1\text{Pa}}\right)\end{array}$

Further simplify the above,

  $\begin{array}{c}{\sigma }_{\text{y}}=\left(\left(4.90\times {10}^7\text{Pa}\right)\left(\frac{{10}^{-6}\text{MPa}}{1\text{Pa}}\right)\right)\\ =49\text{MPa}\end{array}$

Conclusion:

Thus, the tensile proportional limit stress is $49\text{MPa}$ .