1.3.4 Behaviour of Springs
and Materials
Objective
Describe how deformation is caused by a
force in one direction and can be tensile or
comprehensive
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Deformation
Can be caused by tensile or compressive forces
 Tensile


cause tension
stretching forces
 Compressive


cause compression
squeezing forces
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Deformation
two equal and opposite tensile
forces stretching a wire
two equal and opposite compressive
forces squeezing a spring
Objective
Describe the behaviour of springs and wires
in terms of force, extension, elastic limit,
Hooke’s Law and the force constant – i.e.
force per unit extension or compression
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Definitions
 Force (F)

applied to a spring or wire in tension or
compression
 Extension (x)

the change in length of a material when
subjected to a tension, measured in metres
 Elastic Limit

the point at which elastic deformation becomes
plastic deformation
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Definitions
 Elastic Deformation

when the deforming force is removed, the object
will return to it’s original shape

eg rubber band, spring (usually)
 Plastic Deformation

when the deforming force is removed, the object
will not return to it’s original shape

eg Plasticine, Blutack
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Hooke’s Law
 When tension is plotted against extension, a
straight line graph denotes elastic deformation
 This is summarised by Hooke’s Law:
‘The extension of a body is proportional to the
force that causes it’
or as a formula:
F = kx
where F = Force
x = extension
k = force/spring
constant
Tension /N
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Hooke’s Law
F
Extension /mm
x
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Force/Spring Constant
 F = kx
 Expressed in newtons per metre
 How much force is required per unit of
extension

eg 5 N mm-1 means a force of 5 N causes an
extension of 1 mm
 Can only be used when the material is
undergoing elastic deformation
Objective
Determine the area under a force against
extension (or compression) graph to find
the work done by the force
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Work Done
 Extension produced by tension F is x
 Work done to reach this extension is the area
under the graph
work done = area of triangle
= ½Fx
Objective
Select and use the equations for elastic
potential energy, E = ½Fx and ½kx2
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Elastic Potential Energy
 As work has been done to stretch the wire, the
wire then stores Elastic Potential Energy
 This also applies to compression forces
 For elastic deformation, the elastic potential
energy equals work done:
E =
½Fx
as F = kx then
E
=
½kx2
Objective
Define and use the terms stress, strain,
Young modulus and ultimate tensile
strength (breaking stress)
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Stretching Materials
 One way of describing the property of a
material is to compare stiffness
 In order to calculate stiffness, two
measurements need to be made:
 strain
 stress
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Stretching Materials
 Strain is the fractional increase in the length of
a material
Strain =
extension (m)
original length (m)
 Stress is the load per unit cross-sectional area
of the material
Stress (Nm-2) =
force (N)
cross-sectional area (m2)
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Young Modulus
 To calculate stiffness, calculate the ratio of
stress to strain:
Young Modulus (Nm-2) = stress
strain
or
E = stress
strain
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Young Modulus
Hooke’s Law Region
Elastic limit
Limit of proportionality
stress
gradient = Young modulus
strain
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Ultimate Tensile Stress
 Stiffness tells us about the elastic behaviour of
a material (Young modulus)
 Strength tells us how much stress is needed to
break the material
 The amount of stress supplied at the point at
which the material breaks is called the ultimate
tensile stress of the material
Objective
Describe an experiment to determine the
Young modulus of a metal in the form of a
wire
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Young Modulus Practical
wooden blocks
cardboard bridges
wire
marker
rule
load
Young modulus practical
Objective
Define the terms elastic deformation and
plastic deformation of a material
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Definitions
 Elastic Deformation

when the deforming force is removed, the object
will return to it’s original shape

eg rubber band, spring (usually)
 Plastic Deformation

when the deforming force is removed, the object
will not return to it’s original shape

eg Plasticine, Blutack
Objective
Describe the shapes of the stress against
strain graphs for typical ductile, brittle and
polymeric materials
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Ductile
 Will stretch beyond
it’s elastic limit
 Will deform
plastically
 Can be shaped by
stretching,
hammering, rolling
and squashing
 Examples include
copper, gold and
pure iron
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Brittle
 Will not stretch
beyond it’s elastic
limit
 Will deform
elastically
 Will shatter if you
apply a large
stress
 Examples include
glass and cast iron
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Polymeric
Will perform differently depending on the
molecular structure and temperature
 Can stretch beyond it’s elastic limit
 Can deform plastically
 Can be shaped by stretching, hammering,
rolling and squashing
 Examples include polythene
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Polymeric
 Cannot stretch
beyond it’s elastic
limit
 Can deform
elastically
 Can shatter if you
apply a large stress
 Examples include
perspex
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Summary
 All materials show elastic behaviour up to the
elastic limit
 Brittle materials break at the elastic limit
 Ductile materials become permanently
deformed beyond the elastic limit
 Polymeric materials can show either
characteristics, depending on the molecular
structure and temperature
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Questions
 Physics 1 – Chapter 8


SAQ 1 to 9
End of Chapter questions 1 to 4