HOMEWORK

From Dieter
1-1, 1-2, 1-3, 1-5, 1-7, 1-8, 3-1

Module #1
States of Stress and Strain
READING LIST
DIETER: Ch. 1, pp. 7-17; Ch. 2, pp. 18-20 and 31-36; Ch. 3 pp. 70-76;
Ch. 8 pp. 275-289

Ch. 2 in Meyers & Chawla, 1st ed.

Ch. 2 in Roesler et al.
Ch. 2 in Courtney
Ch. 5 in Nye
Ch. 2 in McClintock and Argon
 Tensile Test
The most common way to assess the mechanical behavior of a material
(strength and ductility)

Load Cell

F
Extensometer
• Used mostly for
metals and
Specimen polymers.
• Not used for
Area, Ao
ceramics except at
very high
• Collect force (or load) vs. temperatures.
displacement (or time).
• We use the resulting F
information to assess Moving
“strength” and Crosshead
“deformability”
 Tensile Test Specimen
Reduced section
2 ¼”
A standard
0.505” diameter ¾” diameter
tensile
2”
specimen Gauge length

Standard Geometries
“Buttonhead” – circular cross-section
“Dogbone” – flat cross-section

ASTM E8 or D638
 Force - Extension Curve
A2

Proportional limit
Elastic limit

A1
Force

d1 L d2

L 1 = L2
V1 ≠ V2
K
xelastic(2)

xelastic(1) Extension

When x ≤ xelastic, F = K x
When x > xelastic, F ≠ K x

As deforming volume changes, the force to deform the

specimen changes. WHY?
 FORCE

ONE ELASTIC ROD

the force required to deform it to failure is F.
(stretch)
F F

Lo L

2F 2F

TWO IDENTICAL ELASTIC RODS, CONNECTED TOGETHER

the force required to deform them to failure is 2F
 The amount of force needed to
deform a solid depends on the
volume or surface area of the body.
 Engineering Stress
F applied load
E  s  
Ao originalcross-sectionalarea
Units for stress and conversion factors

• Load per unit area (“force distribution”)

– PSI: 1 lb/in2 = 6.895 x 10-3 MPa

= 7.032 x 10-4 kg/mm2
= 6.8 x 104 dynes/cm2
– MPa: 1 MPa = 1 MN/m2 = 1 N/mm2 = 1 x 106 N/m2

• An engineering stress, as defined above, represents the

average stress in the object.
As defined above, this represents a normal tensile stress.
 TENSILE STRESS

CONSIDER A SINGLE ELASTIC ROD

the stress required to deform it to failure is
 = F/A
F F

Lo L

2F 2F

IF TWO IDENTICAL RODS ARE JOINED AND DEFORMED

the stress required to deform them to failure is
 = 2F/2A = F/A
 The amount of STRESS needed to
deform a solid does not depend upon
the volume or surface area of the
body.
 Engineering Strain
Li  Lo L change in length
E  e   
Lo Lo initial length
• Length per unit length

– We express strain either as a fraction or as a percentage. Be careful

when doing homework or solving real engineering problems.
– Ex:
• e = 0.02 is the same as 2% strain
• e = 0.10 is the same as 10% strain
• e = 1.00 is the same as 100% strain

• Engineering strain, as defined above, represents the average

linear strain in the object.
 STRAIN IS UNITLESS!

It is a measure of the amount of

distortion (i.e., “deformation”) caused
by the application of a force.
 Engineering Stress-Strain Curve in Tension
ELASTIC PLASTIC
Ultimate tensile
Stress (Force/Area)

Strain strength, u or UTS

Hardening
(σ = Kεn) Necking
Elastic
limit

Engineering

Elastic
Modulus
E (σ = εE)
0.2% Offset yield stress
(o, ys, YS, etc.)

Plastic energy Stored

dissipated elastic strain
(heat) energy

0.2% STRAIN, p=0.002

Strain
Uniform plastic strain Non-uniform
Elastic strain
plastic strain
A*
 Engineering Stress-Strain Curve in Tension
ELASTIC PLASTIC
Ultimate tensile • Elastic deformation
Stress (Force/Area)

Strain strength, u or UTS

Hardening
up to elastic limit.
(σ = Kεn) Necking • Plastic deformation
Elastic after elastic limit.
limit
• Uniform plastic
deformation between
Elastic elastic limit and the
Modulus UTS.
E (σ = εE)
0.2% Offset yield stress • Nonuniform plastic
(o, ys, YS, etc.)
deformation after
Plastic energy UTS.
dissipated Stored
(heat) elastic strain • In tension this non-
energy uniform deformation
Strain is called necking.
0.2% STRAIN, p=0.002

Uniform plastic strain Non-uniform

Elastic strain
plastic strain

B*
 True Stress – True Strain
• Volume is generally conserved during deformation.
Thus “shape changes with deformation.”
► Re-define stress and strain to account for shape change.
F
Ao Ai < Ao
z

Lo

y
x unloaded
F
loaded

• True stress, σT = load / instantaneous area = F/Ai

• True strain,  εT = ???
 L3

L2
Why true strain? F

L1 L1-2 L2-3

• Engineering strain is the average linear strain in the solid.

• Doesn’t work if we consider shape change during deformation.

L Li  Lo
e 
Lo Lo

L1 2 L2  L1 
e1 2  
L1 L1  L13 L3  L1
e
 13    e1 2  e2 3
L L  L2  L1 L1
e2 3  2 3  3
L2 L2 
• An incremental displacement results in an infinitesimal strain:

change in length dL
d  
instantaneous length L

• Re-define strain by integrating the infinitesimal displacements

from the initial to the final length.
Lf
dL  Lf 
 T  true strain  
Lo
L
 ln  
 Lo 

 L2  
1 2  ln   
 L1    L3   L2   L3   L3 
 13  ln    1 2   23  ln    ln    ln  
 L3    L1   L1   L2   L1 
 23  ln  
 L2  
 True Stress – True Strain
F
Ao Ai < Ao
z

Lo

y
x unloaded
F
loaded

True stress, T = load/instantaneous area = F/Ai = E(εE + 1)

Lf
L dL  Lf 
True strain,  T  
Li Lo L
 ln    ln   E  1
 Lo 
Use true strain when
structure = Fcn(ε)
 True Stress-Strain Curve in Tension
Ultimate tensile
strength, u or UTS
True
Stress (Force/Area)

• True stress-
strain curve
shifts up and
to the left of
Engineering
engineering
stress-strain
curve.
0.2% Offset yield stress
(o, ys, YS, etc.)

Strain
Definitions and relationships between true and engineering stress and strain

Fundamental
Parameter Definition Before necking After necking
F F F
Engineering stress,  E E  E  E 
Ao Ao Ao

F F F
True stress,  T T  T  T 
Ai Ai Aneck
  E 1   E 

L L L
Engineering strain,  E E  E  E 
Lo Lo Lo

Ao Li Ao
True strain,  T  T  ln  E  ln  T  ln
Amin Lo Aneck
Ao
 ln
Ai
 ln 1   E 
 Remarks about σ and ε
1. For small amounts of deformation (i.e., elastic),
engineering and true stresses and strains are the same.

2. They deviate at high strains.

3. Result: true stress and strain are used in metal working

operations.

4. Most handbook values (used for design) are engineering

stress and strain.
 Shear Stress and Strain

δ Ao

F F

Ao
θ
h

  tan 
h
F

In a shear stress, a force is distributed within a planar area

Shear stresses distort objects

Shear strains are a measure of shear distortion
 This region
experiences
This region
a tensile
experiences a
stress
shear stress

Force
Force

F
Force

F
In general the stress required depends on the area of
interest and/or the direction of loading
F F
F
Fs
A1
Fn
A2
Lo
A3
Ao
A

Ao
F
F F
A B C
Round rod in tension. Non-uniform structure in tension. Area of interest not
Stress is uniform throughout Stress varies with position. perpendicular to load.
the structure. Stress can be re-defined
Tensile stress at 1:  = F/A1 relative to a preferred
Tensile stress:  = F/Ao Tensile stress at 2:  = F/A2 coordinate system.
Tensile stress at 3:  = F/A3
Tensile stress:  = F/Ao
Shear stress:  = Fs/A
Normal stress:  = Fn/A
 Thermal Stresses and Strains
• Most engineering materials expand when heated and
contract when cooled.

• The strain caused by a change of one degree (1°) in

temperature is known as the coefficient of thermal
expansion (α).

• The strain caused by temperature change ΔT is:

 t  T

 t  E t
Modes of Deformation Besides Tension

COMPRESSION TORSION
F
T


T
ℓ L
ℓo 

D
Ao
F
Tr
 max  ;
J
T  torque;
r  radius of cylinder
 r4
J  polar moment of inertia 
2
Modes of Flexure/Bend Testing
Deformation
Besides Load
Tension

• Commonly used with brittle materials that behave in a linear elastic manner
(Ex, ceramics and glasses).
• Governed by two equations:

M E M 
 and 
I R I y

• M = applied bending moment,

• I = second moment of inertia of the beam about the neutral plane,
• E = Young’s modulus,
• R = radius of curvature,
•  = tensile or compressive stress
FL3
• y = planar distance from the neutral plane. y 
48 EI
 Flexure/Bend Testing

3-Point 4-Point
F F/2 F/2

H H
L L1 L1
W W

F/2 F/2 F/2 F/2

FLH FLH
 max   max 
8I 4I
WH 3  D4 WH 3  D4
I rectangular  ; I round  I rectangular  ; I round 
12 64 12 64

max is the Modulus of Rupture

Irectangular and Iround represent the moments of inertia for uniform rectangular or
round cross-sections.
Moment Diagrams and Stress Distributions in Modulus of Rupture Tests

compression

Moment
tension

3-Point

compression

Moment

tension
4-Point

Be careful when comparing fracture results from tension, 3-point bending and 4-point
bending tests as the volume of material at maximum stress is higher in 4-point bending.
 4-Point vs. 3-Point Bend Tests
• In four-point bend tests, the material over the inner span is
subjected to a constant stress; whereas in three-point bend
tests, stresses are localized on the top surface.

• Thus a larger volume of material is tested in 4-point bend

tests than in 3-point bend test. As a result, materials flaws
are more likely to reside in the region of maximum stress in
a 4-point bend test.
 FOR THE TIME BEING, WE WILL
CONCENTRATE ON TENSILE AND
COMPRESSIVE MODES OF LOADING
5 minute break
 State of Stress
Most easily described by normal and shear stress components.

Fy 

y
y
Ay y′ x′
x x Ay   Ay / cos 
Fx′
Fy Fy Fy′
 yy 
Ay

Fy

Define relative to feature of interest.

• Normal Stress 

Fy  Fy cos   yy
 y y      yy cos  
2
(1  cos 2 )
Ay  Ay / cos  2

Ay   Ay / cos 
• Shear Stress Fx′
Fy Fy′

Fx Fy sin   yy
 y x     yy sin  cos   sin 2
Ay  Ay / cos  2
y
y′ x′
x
 Total stress (F) resolved onto an oblique plane
F

Fs Fn F sin  F
n    cos 2 
  Fn A  Ao / cos   Ao

Fs F cos  F
s   cos  sin 
A
 
A  Ao / cos   Ao
z Ao

y F     90
x
 States of Stress & Strain
In most service conditions and forming operations,
loading is not uniaxial.

Most engineering structures and materials experience

multiaxial loading.

This means that they will be subject to varying

combinations of
normal AND shear
stresses.

How do we define stress and strain under these

conditions?
 Multiaxial loading is implied
from changing geometries of
real components.
z´
p z Stress states in vary from
x´
y´
point to point throughout the
o y
entire object.
x

We can easily describe the

state of stress using
extensions of what we have
learned already.
 States of Stress
• It is easiest to define the distribution of forces (and thus the
distribution of stresses) relative to a planar area within the
object.

• NORMAL forces or stresses that are  to the plane.

• SHEAR forces or stresses that are  to the plane.

• The relationships that we will develop will allow us to define

the state of stress relative to any coordinate system. Pay
attention! Think!
 F4 F5 • Consider an arbitrary solid
object that has a series of
external forces applied to it.

O • Assume that the object

remains in static equilibrium.
F3
• Forces applied on the exterior
F2 of the solid are balanced by
F1 internal tractions (i.e., internal
forces) that keep the object in
equilibrium (i.e., keep it from
How do we define moving or changing shape).
the state of stress
• Thus the solid is under a
at point O? state of stress.
 z or x3 z or x3
F4 F4
F5 F5

O O
y or x2 y or x2
F3
F = (F4+F5)
x or x1 x or x1
F2
F1
• Define a planar area that passes through the point (it can be placed anywhere)
and establish an orthogonal (3-D) coordinate system passing through the point
where one axis is the normal to the plane and the other two axes lie within the
plane.
• Sum forces such that the resultant force F on the point keeps the body from
moving. [ΔF=0 at equilibrium]
• This is easiest to visualize if you section the body at the plane, ignore one half,
and place a resultant force F on the plane such that it opposes the surface forces
acting on body.
 Resolution of
force F into
normal and x3
shear force
components
F
relative to a
planar area Fn 
O Fs

y
x

• Focusing on the plane and a small area (A) surrounding point O, the
resultant force F can be resolved into normal, Fn, and shear, Fs,
components relative to the plane and our orthogonal coordinate system.

• The components of force can, in turn, be converted into stresses.

F
Stress at Point O  lim
A0 A

• Each stress component can be related to our coordinate system.

 z
Conversion
of force F F
Fn  zz
into stress   ij
Fs 
components O  
zx
acting on a zy
plane x y

Fn F cos 
NORMAL:  ii   zz  
A A

Fs cos   F sin   cos 

Fy
d ir.
 zy   32   
ll el to
y- A A A
Fs F sin  Para
SHEAR:  ij  
A A Para
Fx Fs sin   F sin   sin 
llel
to x
-d i r.
 zx   31   
A A A
 RELATIVE TO ANY PLANE,
The state of stress at any point can be defined by
one NORMAL STRESS and
two SHEAR STRESSES.

The normal and shear components can be

conveniently related to an orthogonal coordinate
system.
 Notation for Stresses
 ij ;  ij
In this course we shall refer to our subscripts* as follows:

First Subscript:
Corresponds to the plane that the stress acts upon.

Second Subscript:
Corresponds to the direction that the stress component is
pointed in.

*Many texts reverse the meaning for each subscript.

 Definition of stresses relative to a planar
area and a 3-D coordinate system

z
Fx
Normal stress:  xx 
Ax

xz Fy
xy
Shear stress:  xy 
Ax
xx y
Fz
 xz 
Ax Ax
x
 Sign Convention for Stresses
Normal Stresses:

Positive – tension
Ex., zz = +100 MPa

Negative – compression
Ex., zz = -100 MPa

Shear Stresses:

Positive – acts on (+) face & points in (+) direction

Ex., zy or zy

Negative – acts on (-) face & points in a (+) direction

Ex., z,-y or z,-y

NOTE: With shear stresses, sometimes a  is used rather than a .

Watch your subscripts.
 Stress at a Point

• Normal stress at a point

 ii (i  x, y, z ) normal to area  A at a point is
 Fy
 ii  lim
 A 0  A

• Shear stress at a point

 ij   ij (i  j  x, y, z ) parallel to area  A at a point is

 Fx
 ij   ij  lim
 A 0  A
 The Stress Tensor
z z
Total stress
on plane z
zz
zy
zx
Total stress Total stress yz
on plane x on plane y xz
yy
y xx xy yx y
x x

(a) (b)
Fig. 1-2. A point-size cube in equilibrium: (a) as extracted showing
contact force vectors on forward faces, and (b) stress components.
[Figure adapted from W.A. Backofen, Deformation Processing (Addison
Wesley, Reading, MA, 1972) p. 3]
Definition of stresses about a point relative to a
3-D coordinate system
z
zz
6 normal stress
zy
zx components
yz
xz 12 shear stress
yy
components
xy yx
xx y 18 total components of
stress
Ax
x

Points on the back sides of the parallelepiped

are not drawn for clarity
 The Stress Tensor
• A body is at equilibrium under arbitrary forces.
– No net forces (∑F = 0)
– No net torques / moments (∑M = 0)

z
zz

zx
zy  xx  xy  xz 
 ij   yx  yy  yz 
yz
xz yy
z xy yx   zx  zy  zz 
Oxx y
x  ij   ji
x y
6 independent stress components
The stress tensor is symmetrical
Proof is provided on the next 2 viewgraphs
 At equilibrium, F = 0 and M = 0
(there can be no net force or torque)
z
zz
zy
zx
yz
xz yy
z xy yx
Oxx y
x
x y

M = 0
Summation of moments about each axis Start

z -axis:  xy yz  x   yx xz  y   xy   yx

with

18 Left
with

y -axis:  xz z y  x   zx xy  z   xz   zx Reduce

by -3 =15
x-axis:  zy yx  z   yz xz  y   zy   yz
 (on x-faces):
SUMMATION OF
FORCES ON EACH
 xx dydz    x  x dydz  0   xx    x  x
FACE  xy dydz    x  y dydz  0   xy    x  y
15
These relationships allow -3 =12
us to reduce the number of  xz dydz    x  z dydz  0   xz    x  z
stress components that (on y -faces):
must be specified to define
the state of stress.  yy dxdz    y  y dxdz  0   yy    y  y

F = 0  yx dxdz    y  x dxdz  0   yx    y  x 12

-3 = 9
z  yz dxdz    y  z dxdz  0   yz    y  z
zz
(on z -faces):
zy
zx
yz  zz dxdy    z  z dxdy  0   zz    z  z
xz

z xy yx

yy
 zx dxdy    z  x dxdy  0   zx    z  x
9
Oxx y -3 = 6
x  zy dxdy    z  y dxdy  0   zy    z  y
x y
[18 - 12 = 6 components]
 The Stress Tensor*
 xx  xy  xz   xx  xy  xz   xx  xy  xz 
 ij   yx  yy  yz  or   xy  yy  yz  or     yz 
 yy

  zx  zy  zz    xz  yz  zz      zz 

 xx ,  yy ,  zz , xy , yz , xz

A TENSOR is
a group of numbers that represents a physical quantity.
i.e., what you get when you relate materials properties to
an x,y,z coordinate system.

* Read pages 31-36 in Dieter on the stress tensor

5 minute break
 Tensor Notation

• The rank* of a tensor determines:

– Number of components = 3n where n = rank.

– Number of direction cosines = n

• The direction cosines are required to transform

that physical quantity from one coordinate
system to another.
 Ranking of Tensors
• Zero rank – scalars. Scalars are non-directional.
– Temperature
– Density

• First rank tensors – vectors. Vectors are directional.

– Force
– Area

• Second rank tensors – product of two vectors.

– Stress (force & area)
– Strain (displacement & displacement)

• Fourth rank tensors – product of two second rank

tensors.
– Elastic modulus (stress & strain)
 Transformation of First Rank Tensors
θzz = (zz)
z z
• A vector F can be easily
resolved into axial
components. Fz′
Fz
• Transformation is F
changing from one set of
Fy y
axes to another.
Fx′ Fy′ θyy = (yy)
y'
Fx
x'
θxx = (xx)
x
• Relate material behavior to the symmetry of the materials structure.

• Relate behavior of machine part or specimen to symmetry or shape of

part or applied loads.
 Old Axis Vector F resolved onto xyz

Fi
x y z
New Axis
x cos(xx) cos(xy ) cos(xz )
Vector F
resolved onto Fi y cos(yx) cos(yy ) cos(yz )
x´y´z´
z  cos(z x) cos(z y ) cos(z z )

or

Fi
All components are
x y z related to each other
x axx axy axz through a series of
“direction cosines”
Fi y a yx a yy a yz
z  azx azy azz
 Fx  axx Fx  axy Fy  axz Fz
Fy  a yx Fx  a yy Fy  a yz Fz
Fz  azx Fx  azy Fy  azz Fz

 Fx   axx axy axz   Fx 

 F   a a yy

a yz   Fy 
 y   yx  
 Fz   azx azy azz   Fz 
3
Fi   aij Fj  aij Fj
j 1

Vector F Original vector

transformed to F resolved onto
x,y,z x,y,z
Direction cosines
relating x,y,z to
x,y,z
 Transformation of Second Rank Tensors
• Second rank tensors denote
Plane
relationships between two vectors z
normal Force
(Force and Area). y′
on plane
x′
y
• We have to re-orient both vectors. z′

• Thus, we need two rotation

matrices to re-orient stress. x
3 3
 ij    kl aik a jl   kl aik a jl
k 1 l 1
Corresponds Orientation
to area Corresponds matrix between Two direction
normal to i′ to force in j′ re-oriented force cosines to
direction direction and original force transform
Relation to the Deformation of Single Crystals
Fz
• Possible to apply a single normal z
z′
component of stress (zz).


 y′
• As you will learn later, plastic
deformation depends on the
As
shear stress on a specific slip
system.
x′
Fy Fz cos 
 zy     zz cos  cos 
Az Ao cos 
Ao
y
• A slip system is a specific crystal
x Fz
plane + specific crystal direction.
    90
   is not necessarily 90
 The transformation equations denoted above can be applied
to any first or second rank tensor.

Similar relationships exist for higher rank (order) tensors.

REFERENCES

►J.F. Nye: Physical Properties of Crystals, (Oxford University Press, Oxford,

1985).
• R.E. Newnham, Properties of Materials, (Oxford University Press, Oxford, 2005).
• A. Kelly, G.W. Groves, & P. Kidd: Crystallography and Crystal Defects, Revised
Edition, (John Wiley & Sons, New York, 2000).
• D.R. Lovett: Tensor Properties of Crystals, 2nd ed., (IOP Publishing, Philadelphia,
1999).
• S.M. Edelglass: Engineering Materials Science, (The Ronald Press Company,
New York, 1966).
 If a second rank tensor is symmetric, it is possible to define a
unique set of axes where the tensor will have no
off-diagonal components.

 xx  xy  xz   xx 0 0 
 
 ij   xy  yy  yz   ij   0  yy 0 
 
  xz  yz  zz   0 0  zz 

The new coordinate axes (x′, y′, z′) are called the
principal axes

The tensor components (σx′x′, σy′y′, σz′z′) related to them are

called the principal tensor components.