Answer:
Explanation:
From the given information:
The equation for applied stress can be expressed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]
where;
[tex]\phi[/tex] = angle between the applied stress [100] and [111]
To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system
Using the equation:
[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [111]
[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]
Thus;
[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]
[tex]\phi= 54.74^0[/tex]
To determine [tex]\lambda[/tex] for [tex][1 \overline 1 0][/tex]
where;
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [tex][1 \overline 1 0][/tex]
[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]
Thus;
[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]
[tex]\phi= 45^0[/tex]
Thus, the magnitude of the applied stress can be computed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]
[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]
[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]