Tensile Force

 Forces don’t just change the motion of a body, but can change the size and shape of them
too. This is known as deformation
 Forces in opposite directions stretch or compress a body
o When two forces stretch a body, they are described as tensile
o When two forces compress a body, they are known as compressive

Diagram of tensile and compressive forces

Tensile Strength

 Tensile strength is the amount of load or stress a material can handle until it stretches and
breaks
 Here are some common materials and their tensile strength:

Tensile strength of various materials

Worked example

Cylindrical samples of steel, glass and rubber are each subjected to a gradually increasing tensile
force F. The extensions e are measured and graphs are plotted as shown below.

Correctly label the graphs with the materials: steel, glass, rubber.
Exam Tip

Remember to read the questions carefully in order to not confuse the terms ‘tensile stress’ and
‘tensile strain’.
Extension and Compression
 When you apply a force (load) onto a spring, it produces a tensile force and causes the
spring to extend
 Stretching a spring with a load produces a force that leads to an extension

Hooke’s Law

 If a material responds to tensile forces in a way in which the extension produced is

proportional to the applied force (load), we say it obeys Hooke’s Law
 This relationship between force and extension is shown in the graph below
 Force v extension graph for a spring

 The extension of the spring is determined by how much it has increased in length
 The limit of proportionality is the point beyond which Hooke's law is no longer true
when stretching a material i.e. the extension is no longer proportional to the applied load
o The point is identified on the graph where the line is no longer straight and starts
to curve (flattens out)
 Hooke’s law also applies to compression as well as extension. The only difference is that
an applied force is now proportional to the decrease in length
 The gradient of this graph is equal to the spring constant k. This is explored further in
the revision notes “The Spring Constant”

Worked example
Which graph represents the force-extension relationship of a rubber band that is stretched almost
to its breaking point?

ANSWER: A

 Rubber bands obey Hooke’s law until they’re stretched up to twice their original size or
more - this is because the long chain molecules become fully aligned and can no longer
move past each other
 This is shown by graph A - after the section of linear proportionality (the straight line),
the gradient increases significantly, so, a large force is required to extend the rubber band
by even a small amount
 Graph B is incorrect as the gradient decreases, suggesting that less force is required to
cause a small extension
 Graph C is incorrect as this shows a material which obeys Hooke’s Law and does not
break easily, such as a metal
 Graph D is incorrect as the plateau suggests no extra force is required to extend the
rubber as it has been stretched

Exam Tip
Exam questions may ask for the total length of a material after a load is placed on it and it has
extended. Remember to add the extension to the original length of the material to get its final full
length

Hooke's Law
 A material obeys Hooke’s Law if its extension is directly proportional to the applied
force (load)
 The Force v Extension graph is a straight line through the origin (see “Extension and
Compression”)
 This linear relationship is represented by the Hooke’s law equation

Hooke’s Law

 The constant of proportionality is known as the spring constant k

Worked example

A spring was stretched with increasing load.

The graph of the results is shown below.

What is the spring constant?
Exam Tip

Double check the axes before finding the spring constant as the gradient of a force-extension
graph. Exam questions often swap the load onto the x-axis and length on the y-axis. In this case,
the gradient is not the spring constant but 1 ÷ gradient is.
The Spring Constant
 k is the spring constant of the spring and is a measure of the stiffness of a spring
o A stiffer spring will have a larger value of k
 It is defined as the force per unit extension up to the limit of proportionality (after which
the material will not obey Hooke’s law)
 The SI unit for the spring constant is N m-1
 Rearranging the Hooke’s law equation shows the equation for the spring constant is

Spring constant equation

 The spring constant is the force per unit extension up to the limit of proportionality
(after which the material will not obey Hooke’s law)
 Therefore, the spring constant k is the gradient of the linear part of a Force v Extension
graph
 Spring constant is the gradient of a force v extension graph

Combination of springs

 Springs can be combined in different ways

o In series (end-to-end)
o In parallel (side-by-side)
 Spring constants for springs combined in series and parallel

 This is assuming k1 and k2 are different spring constants

 The equivalent spring constant for combined springs are summed up in different ways
depending on whether they’re connected in parallel or series

Worked example
Three springs are arranged vertically as shown.

Springs P,Q and O are identical and have spring constant k. Spring R has spring constant
4k.What is the increase in the overall length of the arrangement when a force W is applied as
shown?
Exam Tip

The equivalent (or effective) spring constant equations for combined springs work for any
number of springs e.g. if there are 3 springs in parallel k1 , k2 and k3 , the equivalent spring
constant would be keq = k1 + k2 + k3 .

Stress, Strain & the Young Modulus

Stress

 Tensile stress is the applied force per unit cross sectional area of a material

Stress equation

 The ultimate tensile stress is the maximum force per original cross-sectional area a wire
is able to support until it breaks

Strain

 Strain is the extension per unit length

 This is a deformation of a solid due to stress in the form of elongation or contraction
 Note that strain is a dimensionless unit because it’s the ratio of lengths
 Strain equation

Young’s Modulus

 The Young modulus is the measure of the ability of a material to withstand changes in
length with an added load ie. how stiff a material is
 This gives information about the elasticity of a material
 The Young Modulus is defined as the ratio of stress and strain

Young Modulus equation

 Its unit is the same as stress: Pa (since strain is unitless)

 Just like the Force-Extension graph, stress and strain are directly proportional to one
another for a material exhibiting elastic behaviour
 A stress-strain graph is a straight line with its gradient equal to Young modulus

 The gradient of a stress-strain graph when it is linear is the Young Modulus

Worked example

A metal wire that is supported vertically from a fixed point has a load of 92 N applied to the
lower end.

The wire has a cross-sectional area of 0.04 mm2 and obeys Hooke’s law.

The length of the wire increases by 0.50%.What is the Young modulus of the metal wire?

A. 4.6 × 107Pa B. 4.6 × 1012 Pa C. 4.6 × 109 Pa D. 4.6 × 1011 Pa

Exam Tip

To remember whether stress or strain comes first in the Young modulus equation, try thinking of
the phrase ‘When you’re stressed, you show the strain’ ie. Stress ÷ strain.
Young's Modulus Experiment
 To measure the Young’s Modulus of a metal in the form of a wire requires a clamped
horizontal wire over a pulley (or vertical wire attached to the ceiling with a mass
attached) as shown in the diagram below
  A reference marker is needed on the wire. This is used to accurately measure the
extension with the applied load
 The independent variable is the load
 The dependent variable is the extension

Method

1. Measure the original length of the wire using a metre ruler and mark this reference point
with tape
2. Measure the diameter of the wire with micrometer screw gauge or digital calipers
3. Measure or record the mass or weight used for the extension e.g. 300 g
4. Record initial reading on the ruler where the reference point is
5. Add mass and record the new scale reading from the metre ruler
6. Record final reading from the new position of the reference point on the ruler
7. Add another mass and repeat method

Improving experiment and reducing uncertainties:

 Reduce uncertainty of the cross-sectional area by measuring the diameter d in several

places along the wire and calculating an average
 Remove the load and check wire returns to original limit after each reading
 Take several readings with different loads and find average
 Use a Vernier scale to measure the extension of the wire

Measurements to determine Young’s modulus

1. Determine extension x from final and initial readings

Example table of results:

Table with additional data

2. Plot a graph of force against extension and draw line of best fit

3. Determine gradient of the force v extension graph

4. Calculate cross-sectional area from:

5. Calculate the Young’s modulus from:

Exam Tip

Although every care should be taken to make the experiment as reliable as possible, you will be
expected to suggest improvements in producing more accurate and reliable results (e.g. repeat
readings and use a longer length of wire)
Elastic & Plastic Deformation
 Elastic deformation: when the load is removed, the object will return to its
original shape
 Plastic deformation: when the load is removed, the object will not return
to its original shape or length. This is beyond the elastic limit
 Elastic limit: the point beyond which the object does not return to its
original length when the load is removed
 These regions can be determined from a Force-Extension graph:

Below the elastic limit, the material exhibits elastic behaviour

Above the elastic limit, the material exhibits plastic behaviour

 Elastic deformation occurs in the ‘elastic region’ of the graph. The extension
is proportional to the force applied to the material (straight line)
 Plastic deformation occurs in the ‘plastic region’ of the graph. The extension
is no longer proportional to the force applied to the material (graph starts to
curve)
 These regions are divided by the elastic limit

Brittle and ductile materials

  Brittle materials have very little to no plastic region e.g. glass, concrete. The
material breaks with little elastic and insignificant plastic deformation
 Ductile materials have a larger plastic region e.g. rubber, copper. The
material stretches into a new shape before breaking

Stress-strain curve for a brittle and ductile material

 To identify these materials on a stress-strain or force-extension graph up to
their breaking point:
o A brittle material is represented by a straight line through the origins
with no or negligible curved region
o A ductile material is represented with a straight line through the origin
then curving towards the x-axis

Worked example
A sample of metal is subjected to a force which increases to a maximum value and
then fractures. A force-extension graph for the sample is shown.
What is the behaviour of the metal between X and Y?A. both elastic and plastic
B. not elastic and not plastic
C. plastic but not elastic
D. elastic but not plastic
ANSWER: C
 Since the graph is a straight line and the metal fractures, the point after X
must be its elastic limit
 The graph starts to curve after this and fractures at point Y
 This curve between X and Y denotes plastic behaviour
 Therefore, the correct answer is C

Exam Tip
Although similar definitions, the elastic limit and limit of proportionality are not the
same point on the graph. The limit of proportionality is the point beyond which the
material is no longer defined by Hooke’s law. The elastic limit is the furthest point a
material can be stretched whilst still able to return to its previous shape. This is at a
slightly higher extension than the limit of proportionality. Be sure not to confuse
them.

Area under a Force-Extension Graph

 The work done in stretching a material is equal to the force multiplied by the distance
moved
 Therefore, the area under a force-extension graph is equal to the work done to stretch
the material
  The work done is also equal to the elastic potential energy stored in the material

Work done is the area under the force - extension graph

 This is true for whether the material obeys Hooke’s law or not
o For the region where the material obeys Hooke’s law, the work done is the area of
a right angled triangle under the graph
o For the region where the material doesn’t obey Hooke’s law, the area is the full
region under the graph. To calculate this area, split the graph into separate
segments and add up the individual areas of each

Loading and unloading

 The force-extension curve for stretching and contraction of a material that has exceeded
its elastic limit, but is not plastically deformed is shown below
  The curve for contraction is always below the curve for stretching
 The area X represents the net work done or the thermal energy dissipated in the
material
 The area X + Y is the minimum energy required to stretch the material to extension e

Worked example

The graph shows the behaviour of a sample of a metal when it is stretched until it starts to
undergo plastic deformation.
What is the total work done in stretching the sample from zero to 13.5 mm extension?

Simplify the calculation by treating the curve XY as a straight line.

Exam Tip

Make sure to be familiar with the formula for the area of common 2D shapes such as a right
angled triangle, trapezium, square and rectangles.
 Elastic Potential Energy
 Elastic potential energy is defined as the energy stored within a material (e.g. in a spring)
when it is stretched or compressed
 It can be found from the area under the force-extension graph for a material deformed
within its limit of proportionality

Worked example

A spring is extended with varying forces; the graph below shows the results.

What is the energy stored in the spring when the extension is 40 mm?
 Calculating Elastic Potential Energy
 A material within it’s limit of proportionality obeys Hooke’s law. Therefore, for a
material obeying Hooke’s Law, elastic potential energy can be calculated using:
 Elastic potential energy can be derived from Hooke’s law

 Where k is the spring constant (N m-1) and x is the extension (m)

Exam Tip

The formula for EPE = ½ kx2 is only the area under the force-extension graph when it is a straight
line i.e. when the material obeys Hooke’s law and is within its elastic limit.