Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer 1

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is
`\sigma = 10. 655 MPa`

Step-by-step explanation:

From the question we are told that

The critical yield resolved shear stress is
`\sigma = 2.9Mpa`

First we obtain the angle
`\lambda` between the slip direction [121] and [111]

`\lambda = cos^(-1) [((u_1 u_2 + v_1 v_2 + w_1 w_2)/(√(u_1^2 + v_1 ^2+ w_1^2))√(( u_2^2 + v_2^2 + w_2 ^2)) ) ]`

Where
`u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2` are the directional indices

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

`= cos^(-1) [(2)/(√(6) √(3) ) ]`

`= 61.87^0`

Next is to obtain the angle
`\O` between the direction [121] and [101]

`\O = cos^(-1) [((u_1 u_3 + v_1 v_3 + w_1 w_3)/(√(u_1^2 + v_1 ^2+ w_1^2))√(( u_3^2 + v_3^2 + w_3 ^2)) ) ]`

Substituting 1 for
`u_1` , 2 for
`v_1` , 1 for
`w_1` , 1 for
`u_2`, 0 for
`v_2`, and 1 for
`w_2`

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

`\O = cos^(-1) [(2)/(√(6) √(2) ) ]`

`= 54.74 ^o`

The stress is mathematically represented as

`\sigma = (\tau_c)/(cos \O cos \lambda )`

`= (2.9)/(cos 54.74^o cos 61.87^o)`

`= (2.9)/(0.2722)`

`\sigma = 10. 655 MPa`