Properties of Matter
part I
presented by Dr. Alan S. Cornell
P125, School of Physics,
University of the Witwatersrand
Wits 2050
Matter as we know it comes in all shapes an sizes. From
microscopic life-forms to gigantic man-made structures.
While we can classify material objects in all sorts of ways
Solids
Liquids
Gases
Metals
Non-Metals
Steels
Cu-Alloys
Al-Alloys
..
.
etc.
where Solids are objects of stable shape and volume,
Liquids of stable volume, and Gases expand to the
container limit. The treatment of liquids and gases
will largely be covered in our studies of fluids
in these lectures we shall study some of the bulk/generic
properties of matter
Overview of material to be covered
• Equilibrium
• Elasticity: - Stress and Strain
- Plastic deformation and breaking stress
- Young’s, shear and bulk moduli
- Hooke’s law
• Hydrostatics (more on this later)
• Surface tension
• Hydrodynamics (which will be with your next lecturer)
Equilibrium
Equilibrium
body is said to be in equilibrium when it has
A body is said to be in
ero net forces
and
torques.
equilibrium when the net
forces and torques
body in (about
stable
anyequilibrium
axis) is zero
eturns to equilibrium when
A body is in a state of
sturbed.stable equilibrium if it
returns to equilibrium
after
body in unstable
equilibrium
it is disturbed
oves away from equilibrium
A body is in a state of
hen disturbed.
unstable equilibrium if
it moves away from
equilibrium after it is disturbed
In order to solve equilibrium problems
1. Identify all the forces acting on the body, including their directions
and locations
2. Draw a free-body diagram with all the forces indicated on it
3. Write down a vector force equation for each component (eg. in the
x-direction, y-direction etc.) with proper signs, that is:
!
!
!
!
F=
Fx =
Fy =
Fz = 0
The vector sums must be independently equal to zero
4. Select a rotational axis for torque calculations (more on this in your
mechanics courses)
5. Write down the torque equations with proper signs
6. Solve the equations for any unknown quantities
Classifying Load
But what happens to objects when they are placed under
some sort of external forces?
Classifying
Loadof loads materials can be
Firstly, let us classify
the types
subjected
to:
• Normal Load (Axial load) : Load is perpendicular to the
Normal
Load
(Axial): Load that is perpendicular to
supporting
material.
- Tension Load
: As the ends
of material
the supporting
material,
such
as: are pulled apart
to make the material longer, the load is called a tension
- Tension
load.Load: where the ends of the material are pulled
- Compression
: As the ends
of material are pushed in
apart to
make theLoad
material
longer,
to make the material
loadends
is called
- Compression
Load:smaller,
wherethethe
of the material are
a compression load.
pushed together to make the material shorter.
Tension
Compression
Tungsten
35 # 10
Steel
20 # 1010
100
Copper
11 # 1010
Strain
Stress
Brass 2)
9.1 # 1010
0
0.002 0.004 0.006 0.008 0.01
(MN/m
Aluminum
7.0 # 1010
Classifying Load
(cont)
Figure 12.15 Stress-versus-strain
400 Glass
6.5–7.8 # 1010
curve for an elastic solid.
Elastic5.6 # 1010
Breaking
• Shear Load : Tangential load 300 Quartz
limit
Water
—
point
—
200 Mercury
∆x
Elastic
behavior
14 # 10
20 # 1
8.4 # 1010
6#1
10
4.2 # 10
14 # 1
Table1012.1
3.5 # 10
6.1 # 1
10
Typical
Values for
2.5 #
10
7.0 #El1
2.6–3.2 # 1010
5.0–5.5 #
Youn
10
2.6 # 10
2.7
#1
Substance
(N/m
—
0.21 # 1
—
2.8 # 1
Shear (or Tangential) Load: where forces act in one
direction on one side of an object, and in another direction
on the other side. For example:
pulling apart
h
–F
Cargo
Pressure
fs
A
Tungsten
35
F
Steel
20
100
Copper
11
Strain
Young’s
modulus
is typically
used to characterize
str
Brass a rod or wire 9.1
0
0.002
0.004 0.006
0.008 0.01
tension or compression. Note that becauseAluminum
strain is a dimension
7.0
Figure
12.15
Stress-versus-strain
Fixed face
units of force per unit area. Typical values are given in Table 12.1.
Glass
6.5–7
curve for an elastic solid.
(a)
(a) that for a fixed applied force, the change in length is proporti
Quartz
5.6
length and (b) that the force necessary to produce a given strain
Water are in accord w
the cross-sectional area. Both of these observations
Mercury
For∆xrelatively small stresses, the bar will return
to its initial lengt
removed. TheAelastic limit of a substance is defined as the maxim
F
F
be applied to the substance before it becomes permanently deform
turnhto its initial length. It is possible to exceed the elastic limit of a
ing a sufficiently large stress, as seen in Figure 12.15. Initially, a
Young’s modulus is ty
curve is a straight line. As the stress increases, however, the cu
tension
or compressio
straight
line. When the stress exceeds the
elastic limit,
the object
–F
Fixed face
units after
of force
per isuni
torted and does not return to its original shape
the stress
rem
(a)
(a) that
for a fixed ap
(b)
is increased
material
ultimately
breaks.
compare
materials
iseven
byfurther,
howthethey
respond
Elastic
behavior
A good Active
wayFigure
to 12.16 (a) A shear
length and (b) that t
deformation
in
which
a
rectangular
to different types of loads.
the cross-sectional are
Shear Modulus: Elasticity of Shape
block is distorted by two forces of
magnitude
but opposite
We willequal
need
some
sorts of measures
for how they do this.
For relatively small
Another type of deformation occurs when an object is subjected to a
directions applied to two parallel
faces. (b) A book under shear stress.
At the Active Figures link
removed. The elastic
of its faces while the opposite face is held fixed by another force (Fig
F If the object
be applied
to the
subs
in this case is called a shear stress.
is originally
a rectan
Stress and Strain
So to start us off with the elastic properties of solids, where
in reality objects will be deformed by the action of external
forces (though internal forces resist this deformation).
The deformation of solids can be understood in terms of
Stress and Strain
Stress and Strain, where:
Stress: AIn quantity
proportional
tohave
themeasures.
force causing
order to compare
materials, we must
the deformation,
that is, the load per unit area
• Stress : load per unit Area
F
F
ó
=
σ=
A
A
F : load applied in Newtons
where F: load applied in Newtons
A : cross sectional area in m!
A: the cross-sectional area!in
m2in Pa
: stress
σ: stress in P a
A
F
F
12.4 Elastic Properties of Solids
Except for our discussion about springs in earlier chapters, we have assumed that objects remain rigid when external forces act on them. In reality, all objects are deformable. That is, it is possible to change the shape or the size (or both) of an object
by applying external forces. As these changes take place, however, internal forces in
the object resist the deformation.
We shall discuss the deformation of solids in terms of the concepts of stress and
strain. Stress is a quantity that is proportional to the force causing a deformation;
more specifically, stress is the external force acting on an object per unit cross-sectional area. The result of a stress is strain, which is a measure of the degree of deformation. It is found that, for sufficiently small stresses, strain is proportional to
stress; the constant of proportionality depends on the material being deformed and
on the nature of the deformation. We call this proportionality constant the elastic
modulus. The elastic modulus is therefore defined as the ratio of the stress to the resulting strain:
Strain: A measure of the degree of deformation, as
measure by the elongation of a material from its
original length
(cont)
∆L
Strain =
L0
ss and Strain
tion of a
n
stress
(12.5) of the material in m
modulus
where ∆L = Elastic
L−
L0!: strain
the elongation
The
modulusunloaded
in general relates what
is done to a solid object
(a force isof
ap- the material (m)
L0elastic
: the
(original)
length
plied) to how that object responds (it deforms to some extent).
We consider three types of deformation and define an elastic modulus for each:
Note
that Strain
hasoriginal
no dimensions.
material
to the
length
1. Young’s modulus, which measures the resistance of a solid to a change in its
length
Lo
e
2. Shear modulus, which measures the resistance to motion of the planes within a
solid parallel to each other
Li
∆L
3. Bulk modulus, which measures the resistance of solids or liquids to changes in
their volume
A
Young’s Modulus: Elasticity in Length
L
Consider a long bar of cross-sectional area A and initial length Li that is clamped at
one end, as in Figure 12.14. When an external force is applied perpendicular to the
cross section, internal forces in the bar resist distortion (“stretching”), but the bar
reaches an equilibrium situation in which its final length Lf is greater than Li and in
which the external force is exactly balanced by internal forces. In such a situation, the
bar is said to be stressed. We define the tensile stress as the ratio of the magnitude of
the external force F to the cross-sectional area A. The tensile strain in this case is defined as the ratio of the change in length !L to the original length Li. We define
Young’s modulus by a combination of these two ratios:
elongation (m)
unloaded(original) length of a material (m)
strain (m/m)
F
Active Figure 12.14 A long bar
clamped at one end is stretched by
an amount !L under the action of
a force F.
At the Active Figures link
at http://www.pse6.com, you
can adjust the values of the
applied force and Young’s
modulus to observe the
change in length of the bar.
Uniaxial (Tension or Compression)
Forces
and
Distortions
Uniaxial
(Tension
or Compressive)
Forces and Distortions
To recap, for a material under a uniaxial force (the
definitions
be for/ more
general loads):
F isabove
+ve –can
Tension
Elongation
If F is positive the material will be under tension
is –ve – Compressive / Shortening
(that is,Felongated)
If F is negative the material will be under a comUniaxial force
F
pressive
forceStress
(that is,
Uniaxial
= shortened)
=
cross sectional area A
Uniaxial force
F
Uniaxial
Stress
=
=
-2L-2]=[ML-1T-2]
[!] =[stress]=[F/A]=[MLT
cross − sectional
area
A
−2 −2
−1 −2
[σ]
=
[stress]
=
[F/A]
=
[M
LT
L
]
=
[M
L
-2
-3
Units: Nm = Pascal = Pa or Nm m = Joules Tm-3 ]
That is, it has units: N m−2 = Pascal = P a or
N.m/m3 = Joules m−3
extension
L ! L0
extension ÄL ∆L
Longitudin
al Strain
= =
= ==
Longitudinal
Strain
original
length
L0 L0 L0
original
length
Strain
hashas
no no
units
or or
dimensions
Strain
units
dimensions
Tension testing machines
A36 Steel
Stress and Strain Diagram
Stress-Strain Diagrams
As we saw from the movie, a plot of stress vs. strain gives
us the behaviour of the material to applied forces, where
Stress-Strain
these diagrams
are different forDiagram
different materials.
ultimate
tensile
strength
3
Slope=
E
! UTS
yield
strength
!y
E
=
Eε
ó =E
å
σ
=E = ó
å
ε
Strain
Hardening
2
Stress (F/A)
σ
Plastic
Region
Elastic
Region
1
E=
necking
óy
å2 ! å1
4
Strain (
Fracture
5
Elastic region
slope=Young’s(elastic) modulus
yield strength
Plastic region
ultimate tensile strength
strain hardening
fracture
!) (e/Lo)
Before continuing with our definitions let us look at each
element in this plot and what it represents in the movie we
saw:
• Elastic Region (Points 1-2)
- The material will return to its original shape after the
material is unloaded (like a rubber band)
- The stress is linearly proportional to the strain in
this region
σ
σ = E Strain or E =
Strain
where σ = Stress
E is the Elastic modulus (Young’s Modulus)
• Yield Strength (Point 2): a point at which permanent deformation occurs. If it is passed, the material will
no longer return to its original length.
• Plastic Region (Point 2-3)
- If the material is loaded beyond the yield strength,
the material will not return to its original shape after
unloading
- It will have some permanent deformation
- If the material is unloaded at Point 3, the curve will proceed from Point 3 to Point 4. The slope will be the same
as the slope between Point 1 and 2
- The distance between Point 1 and 4 indicates the amount
of permanent deformation
• Strain Hardening
- If the material is loaded again from Point 4, the curve
will follow back to Point 3 with the same Elastic Modulus
(slope)
- The material now has a higher yield strength of Point 4
- Raising the yield strength by permanently straining the
material is called Strain Hardening
• Tensile Strength (Point 3)
- The largest value of stress on the diagram is called the
Tensile Strength or the Ultimate Tensile Strength
- It is the maximum stress which the material can support
y all solid without
materials
respond to a small
breaking
d force by• changing
length.5)
Fracture (Point
- If the material is stretched beyond Point 3, the stress
esponsedecreases
takes three
progressive
as necking
and non-uniform deformation occurs
as the force
increases,
- Fracturing
will finally occur at Point 5.
Elasticity
ic region –
ortional deformation
ic region – permanent
mation
re - fracture
Poisson’s ratio
Poisson’s
Ratio
Note
that longitudinal strain does
not occur in isolation, there is
Longitudinal Strain does not occur in isolation,
also some thinning (or widening)
there is also a thinning (widening) of the
specimen
whichspecimen.
is called transverseThis
Strain is called
of the
Transverse Strain
!b b - b0
Transverse Strain =
b0
=
b0
∆b
b − b0
Transverse Strain =
=
b0
ÄL Lb!0 L
Longitudinal Strain =
L
=
0
L
∆L
where Longitudinal Strain =
.
L0
Strain
( "b / b )
It Transverse
is therefore
convenient
to= define
ratio:s Ratio
=Const. = the
! = Poisson'
0
0
0
Longitudinal Strain
( "L / L )
0
Transverse Strain
(∆b/b0 )
Poisson s Ratio =
=−
=ν
Longitudinal Strain
(∆L/L0 )
!
where ν is a constant.
Young’s, shear and bulk moduli
As seen earlier in the stress vs. strain plots, there is a
region where (the elastic region) where stress is
proportional to strain, this constant of proportionality is
the Elastic Modulus
stress
Elastic Modulus ≡
strain
There are three types of elastic modulus we shall look at
here:
1. Young’s modulus: which is a measure of the elasticity
in length
2. Shear modulus: which is a measure of the elasticity in
the plane
3. Bulk modulus: which is a measure of the elasticity in
volume
Tensile deformation again
In the case of tensile deformation, like in the movie, in
the elastic region, the elastic modulus is called the Young’s
!"#$%&"'(")*+,-.%*#
modulus, where
9,$)"*0$('0#1$)"#$#-'0)*($5+2
)"#$:+4,.;0$5+24-
T ensile Stress
F/A
F Li
Y =
=
=
T ensile Strain
∆L/Li
A∆L
where some common values are
Y = 107 − 108 P a for rubber,
and Y = 2 × 1011 P a for steel.
) *&+%'* !"#*++
( "
) *&+%'* !"#$%&
!"#$%&
!"#$%&'()"*#+#'
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( " '!( !
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Shear deformation
In the same way, when we have deformation caused by a
force acting parallel to one face of an object whilst an
opposing force acts on another face in the opposite
direction
01',*%&'()*+,-.)/
(eg holding it in place), we can define shear stress and strain
A"##B1+,*#()*+((#,-.#()*,/-"#
as:
F
Shear stress =
$"
F
%7(.' %&'(
A
where F is the tangential to the area A
φ #
! &.09(0&/.# : 2';( +
,%
of the face being sheared
!
x
&
Shear strain =
h
where x is the horizontal distance that
6
φ
%7(.' %&'./0 !
%
7
the sheared face moves and h is the
height of the object. The Shear
modulus is then defined as:
F/A
shear stress
=
S=
shear strain
x/h
S E C T I O N 1 2 . 4 • Elastic Properties of Solids
a single subforces occurs
f deformation
ess is defined
the area A of
y in more det "P ! "F/A,
in is equal to
Equation 12.5,
ulk modulus,
(12.8)
number. This
ses a decrease
375
Note that that shear modulus or coefficient of rigidity can
also be called the torsion modulus or modulus of rigidity
and is usually between Y /3 to Y /2 (Y is the Young’s
modulus). This is for solid materials only!
Bulk modulus: Volume Elasticity
Finally, when the stress is a uniform pressure on all sides
Bulk modulus
of
the object, and the resulting deformation is a volume
change, we speak of bulk stress and bulk strain.
Vi
alues in a difsted. The reci-
ulus. However,
a liquid does
ensile force is
oor. The fric-
F
A
∆V
Bulk Strain = V olume Strain =
V0
∆p
Bulk M odulus = B = −
∆V /V0
Bulk Stress = P ressure = P =
F
Vi + ∆V
Active Figure 12.17 When a solid
is under uniform pressure, it undergoes a change in volume but no
change in shape. This cube is compressed on all sides by forces nor-
Note that the negative sign is inserted to ensure that B is
a positive number, as an increase in pressure (∆P ) causes
a decrease in volume (negative ∆V ) and vice versa.
In some texts the reciprocal of the bulk modulus is given,
this quantity is known as the compressibility.
Hooke’s law
As most of you should know, the work done in moving a
particle a distance x for a constant force F is
W = Fx
But this is only true if F is a constant. In cases where F is
a function of x
! xf
W =
F (x)dx
xi
One common example of this is for the case of a block
moving on a frictionless surface which is connected to a
spring. If the spring is either stretched or compressed a
small distance from its unstretched (equilibrium) position,
the force exerted on the block can be expressed as:
S ECTI O N 7.4 • Work Done b
F = −kx
Fs is negative.
x is positive.
where x is the position of the block
relative to the equilibrium position (x =
0) and k is a positive constant called
the force constant or the spring constant. This force law for a spring is
known as Hooke’s law, where k is a
measure of the stiffness of the spring.
In this case, the work done in moving a
particle from position xi to xf while attached to a spring with spring constant
k is:
! xf
! xf
1
W =
F dx =
(−kx)dx = k(x2i −x2f )
2
xi
xi
As we shall see in the next section, the
spring constant k can be related to the
modulus for elasticity.
x
x
x=0
(a)
Fs = 0
x=0
x
x=0
(b)
Fs is positive.
x is negative.
x
x
x=0
(c)
Fs
Area = –1 kx 2max
2
kx max
x
0
x max
Fs = –kx
(d)
Active Figure 7.10 The force exerted by a spring on a block varies with the block’s
position x relative to the equilibrium position x ! 0. (a) When x is positive (stretched
spring), the spring force is directed to the left. (b) When x is zero (natural length of
the spring), the spring force is zero. (c) When x is negative (compressed spring), the
spring force is directed to the right. (d) Graph of Fs versus x for the block–spring
At th
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you can o
motion fo
32$/+,1&#-#%45
Elastic energy
∆*
?
@
>
>
>(%
/,
4( !
+ (,
+ (,
/,
2
/ !
!
+
#3&
!
3
!
4
3
(
In the above)1(2
we stretch
a rod which has
)*+,-./01-)2'-2
3(2
2 Young’s modulus:
F/A
F/A
H/&(;!"I('!1&(*1()-"&-#/*12(-/&("!'(*1#"&$)&)(-/&(7!-&1-*$%(
=
Y =
&1&"2,(!<(-/&("!'>(H/*)(*)(#$%%&'(*#(,)$0&*-*+1.>
∆!/L
x/L
"6 is related to
"6the spring constant:
where the force
$
%
5 !
+ #3&73 ! 4 (
*
373 !
%
4( 6
YA *
YA
F (x) =
x = kx , k =
L
L
$
8$1 ! 4( 6 %
%
The work done in stretching this rod increases the potential
energy of the rod. This is called the elastic energy
! x
! x
1 2
1 2
W =
F (x)dx = k
xdx = kx ⇒ Eelastic = kx
2
2
0
0
That is, the energy stored in the material is also expressible
as:
1 Y Ax2
1 " x #2
Estored =
= Y
(AL)
2 L
2
L
1
=
Y (strain)2 V
2
The energy density stored within the material is then:
Estored
1
1 (stress)2
2
u≡
= Y (strain) =
V
2
2
Y