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  $T_{cRss}= 0.242 MPa$:

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

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

$\sigma _{1} = 0.2964 MPa$

Answer 2

The resolved shear stress along the [101] direction is 0.3673 MPa and the tensile stress is required to produce a critical resolved is 0.2964 MPa.

What is a lattice unit cell?

The symmetrical 3-D structural arrangement of the ions, atoms, or molecules inside a crystalline lattice solid as a point.

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 is 0.45 MPa. Then the resolved shear stress, along the [101] direction within the (11T) plane will be

The θ be the angle between [101] and (111) will be

$\rm cos\ \theta = \dfrac{1*1+1*0+1*1}{\sqrt{1^2+1^2+1^2}\sqer{1^2+0^2+1^2}}\\\\\\cos\ \theta = \sqrt{ \dfrac{2}{3}}$

And ∅ be the angle between [111] and [111], then we have

$\rm cos\ \phi= \dfrac{1*1+1*1+1*1}{\sqrt{1^2+1^2+1^2}\sqer{1^2+1^2+1^2}}\\\\\\cos\ \phi= 1$

Now, for the resolved components along [101], we have

$\rm = \sigma cos \ \theta \ \ cos \ \phi\\\\= 0.45 *10^6*\sqrt{\dfrac{2}{3}}*1= 0.3673 \ \ MPa$

(b) The tensile stress is required to produce a critical resolved shear stress of 0.242 MPa will be

$\rm \sigma _1 = \dfrac{shear\ stress}{cos\ \theta \ cos \ \phi}\\\\\\\sigma _1 = \dfrac{0.242*10^6}{\sqrt{\frac{2}{3}} * 1}\\\\\\\sigma _1 = 0.2964 \ MPa$

More about the crystalline lattice link is given below.

https://brainly.com/question/10951564