Lecture 8

Jayant Jain Assistant Professor,

Department of Applied Mechanics, IIT Delhi, Hauz Khas, 110016

Lecture 8 Beyond Elasticity:

Plasticity, yielding and ductility

Recap Strength of perfect crystal: theoretical strength Strength of real crystals: PN stress Dislocations and their role in plastic deformation Dislocations are carriers of plastic deformation: Ease of motion of dislocation decides the strength of crystal Plastic deformation takes place by slip: One of the most important mechanisms of deformation Justified that slip takes place on close packed plane along close packed direction

Slip systems

BCC FCC HCP

Slip

Plane

Slip

direction

Crystal

structure

number of Slip

Systems

Slip system of the three common crystal structure

Why HCP structured materials are less ductile??? Because it has limited number of easy slip systems

Dislocation Movement

Materials: engineering, science, processing and design, 2nd edition Copyright (c)2010 Michael Ashby, Hugh Shercliff, David Cebon

For a dislocation to move, only bonds along the line it moves must be broken this is significantly easier than breaking all of the bonds in the plane In crystals there are preferred planes and directions for which dislocation movement is easier these are called the slip planes and slip directions Slip displacements are tiny however, if a large number of dislocations traverse a crystal, moving on many planes, the material deforms at a macroscopic level

Derive an expression on how applied stress relates to resistance force This resistance force can be anything: It is the one that doesn't want dislocation to traverse through the crystal

Force acting on a dislocation

Materials: engineering, science, processing and design, 2nd edition Copyright (c)2010 Michael Ashby, Hugh Shercliff, David Cebon

Dislocations move if

exceeds f/b

Crystals resist the motion of dislocations with a friction-like resistance f per unit length

Dislocations move from an applied shear stress as they move the upper half of the crystal shifts relative to the lower half by a distance b

Remember you don't apply stress on slip system, you apply stress on crystal which gets resolved onto the system Derive an expression that tells you stress resolved on a given slip system?? If your stress reaches some critical value then only slip will initiate on a given system

Concept of Resolved shear stress

Erich Schmid (1935) discovered that if a crystal is stressed, slip begins when the shear stress on a slip system reaches a critical value, CRSS

10

coscosR

Then slip begins, minimum shear stress to initiate slip: Critical resolved shear stress

Slip occurs first in slip systems oriented close to

( = = 45o) with respect to the applied stress

Critical Resolved Shear Stress

This tells you about the significance of geometry of slip system

coscos Schmid factor

If R = CRSS

In response to an applied tensile or compressive stress, slip in a single crystal begins when the resolved shear stress reaches some critical value, crss.

It represents the minimum shear stress required to initiate slip and is a property of the material that determines when yielding occurs.

11

max)cos(cos

crssy

Critical Resolved Shear Stress

Note crystal orientation can make it easy or hard

coscosR

maximum at = = 45

R = 0

=90

R = /2 =45 =45

R = 0

=90

Critical Resolved Shear Stress

Dependence of YS on orientation factor

Yie

ld s

tress (

MP

a)

Hard orientation

Soft orientation

cos cos

0.1

0.2

0.3

0.4 0.2 0 0.4 0.2 0

Single crystal: Pure Magnesium

A zinc crystal (hcp) is oriented with normal to the basal

plane making an angle of 60 with the tensile axis and the three slip directions x1, x2 and x3 lying on its plane making

angles of 38, 45 and 84, respectively with the tensile axis. If the plastic deformation is first observed at a stress of 2.3

MNm-2, find which of the three slip directions has initiated

slip and at what value of the resolved shear stress?

Question