Question

A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction (a) If the applied stress if 0.45 MPa, what will be the resolved shear stress, tRss, along the [101] direction within the (11 T) plane? (b) What tensile stress is required to produce a critical resolved shear stress, TcRss of 0.242 MPa

Answer 1

Given:

Applied stress, $\sigma = 0.45 MPa$

Critical Resolved Stress,  $T_{cRss}= 0.242 MPa$

Solution:

a). According to the question, orientation of tensile load is along [1 1 1],

$\sigma = 0.45 MPa$

Now, for resolved shear stress,  $\t_{Rss}$ along [1 0 1] within (1 1 T)

let  '$\theta$' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

$cos\theta$ = $\frac{1\times 1 + 1\times 0 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+0^{2}+1^{2}}}}$

$cos\theta$ = $\sqrt{\frac{2}{3}}$

let  '$\phi$' be the angle between [1 1 1] and [1  1 1], then by coordinate formula:

$cos\phi = \frac{1\times 1 + 1\times 1 + 1\times 1}{\sqrt{1^{2}+1^{2}+1^{2}\sqrt{1^{2}+1^{2}+1^{2}}}}$

$cos\phi$ = 1

Now, for the resolved components along [1 0 1]

$\t_{Rss}$  = $\sigma cos\theta cos\phi$

$\t_{Rss}$ = $0.45\times 10^{6}\times \sqrt{\frac{2}{3}}\times 1$ = 0.3673 MPa

b).  For required tensile stress to produce  $T_{cRss}= 0.242 MPa$:

$\sigma _{1} = \frac{T_{cRss}}{cos\theta cos\phi }$

$\sigma _{1} = \frac{0.242\times 10^{6}}{\sqrt{\frac{2}{3}}\times 1}$

$\sigma _{1} = 0.2964 MPa$