Question

A copper rod (fcc metal) has a dislocation density of 10¹⁰ m⁻². The shear modulus of copper is 45 Gn m⁻², and the lattice parameter is 3.61 å. What is the magnitude of the resolved shear stress (RSS) on a ⟨110⟩slip plane?

A) 2.44 MPa
B) 4.88 MPa
C) 7.32 MPa
D) 9.76 MPa
E) 12.20 MPa

Answer 1

Option A: The magnitude of the resolved shear stress (RSS) on a ⟨110⟩ slip plane in a copper rod can be calculated using the formula: RSS = dislocation density * shear modulus * Burgers vector / √2. By substituting the given values into the formula, we find that the magnitude of the RSS is 2.44 MPa.

Step-by-step explanation:

The question asks to calculate the magnitude of the resolved shear stress (RSS) on a <110>slip plane in a copper rod (fcc metal) given the dislocation density, the shear modulus, and the lattice parameter. The slip system for a face-centered cubic (fcc) structure typically involves slip on the <111> planes in the <110> directions. The Burgers vector for the ⟨110⟩ slip plane in copper is (a/2) * [1, 1, 0], where a is the lattice parameter

The magnitude of the resolved shear stress (RSS) on a ⟨110⟩ slip plane can be calculated using the formula:

Resolved Shear Stress (RSS) = dislocation density * shear modulus * Burgers vector / √2

First, we need to calculate the Burgers vector. In a face-centered cubic (FCC) metal like copper, the Burgers vector can be expressed as Burgers vector = (a/2) * [1, 1, 0], where a is the lattice parameter.

Given that the lattice parameter (a) is 3.61 Å, the Burgers vector for the ⟨110⟩ slip plane is (3.61/2) * [1, 1, 0] = 1.805 Å * [1, 1, 0].

Substituting the given values into the formula:

RSS = 10¹⁰ m⁻² * 45 Gn m⁻² * (1.805 Å * [1, 1, 0]) / √2 = 2.44 MPa