Question

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [101]direction and is initiated at an applied tensile stress of 13.9 MPa (2020 psi), compute the critical resolved shear stress

Answer 1

`\mathbf{\tau_c =5.675 \ MPa}`

Step-by-step explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [
`\bar 1`01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

`cos \lambda = \Big [(d_1d_2+e_1e_2+f_1f_2)/(√((d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) )) \Big]`

where;

`[d_1\ e_1 \ f_1]` = directional indices for tensile stress

`[d_2 \ e_2 \ f_2]` = slip direction

replacing their values;

i.e
`d_1` = 0 ,
`e_1` = 0
`f_1` = 1 &
`d_2` = -1 ,
`e_2` = 0 ,
`f_2` = 1

`cos \lambda = \Big [((0* -1)+(0* 0) + (1* 1) )/(√((0^2+0^2+1^2)+((-1)^2+0^2+1^2) )) \Big]`

`cos \ \lambda = (1)/(√(2))`

Also, to find the angle
`\phi` between the stress [001] & normal slip plane [111]

Then;

`cos \ \phi = \Big [(d_1d_3+e_1e_3+f_1f_3)/(√((d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) )) \Big]`

replacing their values;

i.e
`d_1` = 0 ,
`e_1` = 0
`f_1` = 1 &
`d_3` = 1 ,
`e_3` = 1 ,
`f_3` = 1

`cos \ \phi= \Big [ \frac{ (0 * 1)+(0 * 1)+(1 * 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]`

`cos \phi= \frac{1} {√(3) }`

However, the critical resolved SS(shear stress)
`\mathbf{\tau_c}` can be computed using the formula:

`\tau_c = (\sigma )(cos \phi )(cos \lambda)`

where;

applied tensile stress
`\sigma =` 13.9 MPa

∴

`\tau_c =13.9* ( (1)/(√(2)) )( (1)/(√(3)))`

`\mathbf{\tau_c =5.675 \ MPa}`