Consider a single crystal of BCC iron oriented such that a tensile stress is applied along a [010] direction. Compute the resolved shear stress along a (110) plane and in a [-1 1 1] durection when a tensile stress of 52 MPa is applied. Assume that the angle between the [010] direction and the vector normal to the slip plane [110] is 45 degrees. Assume that the angle between the [010] direction and the [-1 1 1] direction is 54.7 degrees. Put your final answer in MPa.

τR = 21.248 MPa

Explanation:

Given:

Single Crystal BCC iron

Tensile stress applied along [010] direction

φ = 45º is the angle between (110) plane normal and the [010] direction

λ = 54.7º is the angle between the [010] direction and the [-1 1 1] direction

Required:

Compute the resolved shear stress along the (110) plane and [-1 1 1] direction when  a tensile stress of 52 MPa is applied.

We can apply the equation

τR = σ*cos λ*cos φ

⇒  τR = (52MPa)*(cos 54.7º)*(cos 45º) = 21.248 MPa

answer:

okay

explanation: