By C.K.Cheung
Elasticity
Intermolecular forces:
F/N
RE
r/m
F/N
X/m
0
F = - kx = restoring force
( x = small extension )
(L + L)
F
F
If L is small  F  L ( Hook’s Law )
 L depends on the original length of the wire
 We write F  (
L
)
L
Also, F  A ( cross-sectional area of the wire )
Hence,
F  (A)(
L
)
L

F = E(A)(
L
) ( Hook’s Law for a block )
L
Or in general:
(
F
L
)  E( )
A
L
 E=
stress
strain
1
By C.K.Cheung
stress
PD: plastic deformation
A
P
Fracture point
E
D
Yield point
Proportional limit
Elastic deformation
0
O’

0”
strain
0.5 %
0E : straight line ( i.e. directly proporational )
: wire returns to its original length when stress removed  elastic deformation.
EP : non-linear, if the stress is removed at any point between OP, the curve will be retraced & the material
will return to its original length.
Beyond P : permanent or plastic deformation starts  wire retains some of its extension if stress is removed,
e.g. extended  A, reducing the stress at A, wire contracts along A0’ ( almost // to OE ), OO’ =
permanent extension. If the stress is reapplied, the curve O’AD is followed.
Note:
1/
If PD long  ductile ( slip of atom layers )
2/
If PD short  brittle ( separation of atom layers )
Strength
Deals with how great an applied force a material can withstand before breaking. ( depends on
manner of breakage )
Stiffness
The ability of a solid to resist a force against deformation.
2
By C.K.Cheung
94’ MC
22. A uniform wire of force constant k and Young modulus E is cut into two shorter wires of equal length.
If
they are arranged side by side and treated as a single wire combination, what are the force constant and the
Young modulus for this combination?
Force Constant
A.
k
B.
2k
C.
2k
D.
4k
E.
4k
Young modulus
E
E
2E
E
2E
92’ MC
12. A ductile fracture occurs
(1) after appreciable plastic deformation, by slow crack propagation.
(2)
when dislocations are not free to move.
(3)
after a brittle fracture occurs.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
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By C.K.Cheung
W.D. in stretching a wire
Area = A
(L + L)
F
 F 


A A  L2
 L 
L
L
2
 L 
AE (L)
F (L)
 AEx 


W.D. =  Fdx   
= area under F – x curve
dx 
L 
2L
2L
2
0
0
= energy stored in the wire
1
F ( L )
W .D.
1
2
energy stored per unit volume =
=
=
(stress)( strain)
AL
volume
2
= area under stress – strain curve
-
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By C.K.Cheung
Strength of metal
Metals are , theoretically, very hard because it is difficult to slide one layer of the atoms over the next ( 
Break many bonds simultaneously to move one atomic spacing )
If F removed
F
( Slippage of atom layers )
In reality, due to imperfections in crystals ( even pure ), e.g. dislocations ( = missing of atoms in crystal
lattice )  bonds can be weakened.
Due to dislocation, bond between DC is weakened,  the upper layer slip to the right by breaking
weakened & fewer bonds & moved only a small fraction of the atomic spacing.
Note:
1/ slip is due to movement of dislocations.
2/
dislocations may get “stuck” if they encounter some other kind of imperfections in the crystal ( e.g.
impurity atoms; dislocations ).
E.g. steel = small amount of carbon dissolved in Iron.
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By C.K.Cheung
Work hardening
If a metal is deformed, its resistance to plastic deformation increases. This is called ‘work hardening’
e.g.
Iron bar
Deformed ( cannot easily
be unbent again)
Fatigue
This may cause fracture, occurs when a metal is subjected to a large number of cycles of varying stress
( even if the maximum value of this stress could be applied steadily with completely safety ). Repeated work
hardening causes the material to become more inhomogeneous locally. ( i.e. too many dislocations )
e.g.
In aircraft parts
Creep
Fracture occurs when metals are stressed ( even constant ) for a considerable periods of time at
temperatures high enough to allow easy movement of dislocations, due to the vibratory motion of atoms.
e.g. Unsupported lead ( low melting point ) pipes gradually sag:
mg
6
By C.K.Cheung
Microscopic interpretation of Young’s modulus
F
Assume simple cubic structure
Area = A
r
r
r(r+r)
F
Elastic strain =
r
r
Elastic stress =
nKr
A
n= no. of mol. in 1 molecular plane
k= force constant between each pair
of molecules.
For simple cubic arrangement:
Volume of cube = r3
For each molecule at the corners of the cube, it is equally shared by
Effective volume
occupied by each
molecule = r3
8 adjacent cubes
 number of molecule per cube = 8 x
1
1
8
 effective volume occupied by 1 molecule = r3
 number of molecules in each molecular plane = n =
A r
nkr
k
)
2
stress
A k
A
r
E=


r
r
strain
r
( )
r
r
(
e.g. for steel:
Ar
A
= 2
3
r
r
 E decreases as temp. increases
E ~ 2 x 10 11 Nm –2; r ~ 3 x 10 –10 m
 k = 60N m –1
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By C.K.Cheung
Non-metal ( rubber )
1/
2/
Amorphous ( no repeating regular structure )
Great range of elasticity ( 1000 % strain )
3/
E ~ 104 smaller than most solids & its E increases as the temperature rises.
Structure:
Polymer consisting long chains of isoprene molecules ( C5H8 ) : ~ 10 4 isoprene molecules per chain.
H
CH3
H
H
C
C
C
C
H
H
Repeating unit
The chain coils up itself ( or some cross – linking exist ) when unstretched.
When stretched  chains unwind & become straight  has a more ordered structure than unstretched,
when fully extended (  chains uncoil & straighten ) rubber is strong because the bonds are then stretched
directly.
Stress
Metal
Rubber
Glass
0
Strain
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By C.K.Cheung
When the force is removed, chains coil up again. So, unlike metals, there is no plastic deformation
( actually little ) when rubber is stretched.
stress
Hysteresis loop:
(Loop area x vol.
= energy lost as heat
in one expension –
contraction cycle)
0
~ 10 3
strain
Small permanent set
Rubber with small hysteresis loop area are used in car tyres, otherwise there is appreciable loss of
energy resulting in increased petrol consumption or lower maximum speed.
Rubber with large hysteresis loop area are very valuable as vibration absorbers.
Vibrating
machine
Mechanical energy  internal energy  temp.
Small amount of vibration energy
When temperature increases  greater disorder among the chains ( more linkage )
 Young’s modulus increases.
9
By C.K.Cheung
Properties of solid from Molecular Theory
Energy
P.E.
F=
-
dU
dr
RE
Z
locus of
mean
position
Point of inflexion
Total energy
- EB
Bonding energy
thermal
energy
Note:
1/
RE : equilibrium separation
2/
3/
4/
r > RE  attractive force increases
r < RE  repulsive force increases
r > Z  Force decreases  breaking point of the solid
Thermal expansion:
1/
2/
At temp. = 0 K, molecule stationary at r = RE
At temp. > 0 K , molecules vibrate about RE ( i.e P.E.
K.E. ),
due to the non-symmetry of the U-r curve, the mean position shift to the right  solid expands as T .
The linear expansivities increase with rising temp.
3/
4/
When mean position > Z  F   liquid state
When the thermal energy (i.e. K.E. ) = EB  r = 

gas state.
10
By C.K.Cheung
91’ MC
11. The breaking stress of a steel wire is 5.0  108 N/m².
If the steel wire is replaced by a similar piece which is twice as long,
which of the following statements is/are true?
(1) The extension when the longer wire breaks is the same as for the shorter wire.
(2) The work done in stretching the longer wire to the breaking point is the same as for the shorter wire.
(3) The stress needed to break the longer wire is 5.0  108 N/m².
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
12.
Stress
Stretching
Contracting
Strain
A suspended fibre was stretched by an increasing load attached to the bottom end.
reducing the load.
A stress-strain graph was obtained as shown.
Then it was allowed to contract by slowly
Which one of the following conclusions may be deduced
from the graph?
(1) All the work done in stretching the fibre is converted into potential energy.
(2) More work is done in stretching than is recovered in contracting.
(3) The temperature of the fibre rises after it has been stretched and allowed to contract for a few times.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
11
By C.K.Cheung
92’ MC
8.
U
H
O
K
Y
X
r
Z
W
The diagram above shows the variation of the potential energy U between neighbouring atoms in a solid with the separation r
between them .
A.
OH > HK
B.
OH > XZ
C.
KY > YW
D.
HK > XY
E.
YZ > XY
Which of the following features of the curves best explains why the solid expands on heating?
91’MC
14. The graph shows how the potential energy of a pair of ions varies with the distance between them.
potential energy
+
A
0
_
distance
between ions
B
C
Which of the following arguments about the points A, B and C marked on the graph is/are correct?
(1) From A to B the potential energy falls with distance, so here the net force between the ions must be pushing them apart.
(2) At C the potential energy curve is a minimum, so a supply of energy is needed either to increase or decrease the distance
between ions.
(3) At B, the potential energy is zero, so here any repulsive force between the ions must be zero.
A.
(1), (2) and (3)
B.
(1) and (2) only
C.
(2) and (3) only
D.
(1) only
E.
(3) only
12
By C.K.Cheung
15.
x
x + x
x
In an idealized atomic model of the material of a wire, each atom is in equilibrium at a distance x from its nearest neighbours,
both in its own layer and in the layer above or below.
There are n atoms per unit area within each layer.
If the force required
to increase the separation between two atoms from x to (x + x) is (kx), what is the longitudinal stress in the wire?
A.
x/x
B.
k x
C.
nk x/x²
D.
nk x
E.
k/x
91’ Essay
2. (a) Briefly distinguish between the different types of strongly attractive forces which bond atoms of
materials together.
(6 marks)
(b) Taking into consideration the resistance of solids to deformation by external forces, sketch the expected
variations of (1) the interatomic force and (2) the potential energy against the separation of two atoms in
a solid. (4 marks)
(c) (i)
Sketch the expected variations of stress against strain for (1) a copper wire, (2) a rubber band and
(3) a glass fibre in a Young modulus experiment, the loading being increased to just before the
materials break.
(ii)
Briefly account for the different behaviour of the materials.
(6 marks)
90’ MC
8.
Two wires X and Y of the same length and of the same elastic metal are each stretched to the same tension.
X is half that of wire Y.
A.
1 : 1.
B.
1 : 2.
C.
1 : 4.
The diameter of wire
The ratio of the elastic potential energy stored in wire X to that stored in wire Y is
13
By C.K.Cheung
D.
2 : 1.
E.
4 : 1.
89’ IIB
8.
A1
A2
B1
B2
x
x+ x
x
Figure 8.1
In a typical solid model, layers of atoms are arranged in a cubic square lattice array with each atom at an
equilibrium distance x from its nearest neighbours, both in its own layer and in the layers above or below.
Suppose a long steel wire, with many layers, is stretched a little so that each layer is now x + x form those
above or below it, as illustrated in Figure 8.1.
(a) What is the elastic strain produced?
(1 mark)
(b) Assume that the deformation is elastic, and that the binding force holding the pair of atoms Ai and Bi (i =
1, 2, ...) together when the wire is stretched can be considered as acting like a spring.
(i)
If this ‘spring constant’ is k, what is the force between the atoms Ai and Bi ? (1 mark)
(ii)
If each layer contains N atoms, what is the total force between pairs of atoms in adjacent planes? (1
mark)
(iii) Determine the elastic stress acting between the two layers of atoms.
(3 marks)
(c) Use the information in (a) and (b) to determine an expression for the Young modulus of the solid. (2
marks)
(d) It is known that for steel, Young modulus is 2  1011 N/m2 and that the interatomic spacing is 0.30 nm.
(i)
Estimate a value for k.
(1 mark)
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By C.K.Cheung
(ii)
Suppose that a steel wire breaks at a tensile tress of 109 N/m2, estimate the increase in distance, x,
between layers of atoms before breaking occurs.
(2 marks)
88’ MC
10.
stress
wire 2
wire 1
wire 3
strain
Three wires of different material, but of the same length and cross-sectional area are stretched until they break.
stress/strain curves are shown in the figure above.
Their
If E1, E2 and E3 represent the energy required to break wire 1, wire 2 and
wire 3 respectively, which of the following is correct?
A.
E1 > E2 > E3
B.
E3 > E2 > E1
C.
E2 > E1 > E3
D.
E2 > E3 > E1
E.
E3 > E1 > E2
11.
potential
energy
0
r0
separation r
-E
The graph above shows how the potential energy between molecules of a substance varies with their separation.
Which of the
following is an INCORRECT inference from the graph?
A. No resultant force acts on each molecule when r = r0.
B. The force between
molecules is repulsive when r < r0.
C. The average separation between the molecules is r0.
D. The energy required to separate two molecules completely is E.
E. The larger the value of E, the higher is the melting point of the substance.
15