Chapter_6 Unit 3&4 Book

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Chapter 6Materials and their use in structures

• Identify different types of external forces such as compression, tension and shear that can act on a body, including gravitational forces.

• Calculate the stress and strain resulting from the application of compressive and tensile forces and loads to materials in structures,

� � F _ A and � � �� __

L .

• Analyse the behaviour of materials under load in terms of extension and compression, including Young’s modulus, E � � _ � .

• Describe elastic or plastic behaviour of materials under load and the resulting energy transformed to heat.

• Calculate the potential energy stored in a material under load (strain energy) using area under a stress versus strain graph.

• Evaluate toughness of a material tested to the point of failure.• Describe brittle and ductile failure and apply data to predict brittle or

ductile failure under load.

• Evaluate the suitability of different materials for use in structures, including beams, columns and arches, by comparing tensile and compressive strength and stiffness or fl exibility under load.

• Evaluate the suitability of a composite material for its use in a structure by considering its properties and the properties of the component materials (maximum of three components).

• Calculate torque, � � r� F.

• Analyse translational and rotational forces in simple structures, including uniform columns, struts, ties, beams, cables, but not including trusses, modelled as two-dimensional structures in static equilibrium.

• Identify and apply safe and responsible practices when working with structures and materials and associated measuring equipment in investigations of materials.

Key knowledge and skillsBy the end of this chapter you will have covered the following key knowledge and skills:

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Unit 3 Chapter 6 \\ Materials and their use in structures 123

External forces on structures

Forces causing compressionForces can act on structures in different directions and positions. Forces that push towards the centre of an object are known as compressive forces. When you stand on a brick you apply a force on the brick, which is equal to your weight force. This force pushes down on the brick and the ground pushes up. These two opposing forces cause compression of the brick (Figure 6.1). Other examples of compression forces can be found in buildings. The weight of the roof on load bearing walls causes the walls to be com-pressed. Likewise, columns are compressed by the weight of the structure above.

Forces causing tensionForces that pull in opposite directions from the centre of an object are known as tensile forces. When keys are dangled from a string, the weight of the keys pulls down and your hand pulls up (Figure 6.2). These forces acting in opposite directions cause tension in the string.

Forces causing shearForces that are opposite in direction but act towards different points areshear forces. Scissors use shear force to cut paper. The top blade pushes down and the bottom blade pushes up. If they acted on the same point, then nothing would happen. If the forces are a small distance either side of a point, they force the layers of molecules to slide over each other. To cut the paper, the shear force needs to be stronger than the bonds between the molecules. Another example of shear force is when you push along the top of a book (Figure 6.3). Your hand pushes the top of the book in one direction and the friction forces from the table push it in the other direction. This causes the pages to slide along each other – it changes the shape of the book.

Young’s modulus

StiffnessWhen cooking, egg whites can be beaten until soft peaks form. If you keep beating, the peaks become stiff. Stiffness is a measure of how much force it takes to change the shape of an object. A metal spoon is stiffer than a plastic one – it takes more force to change its shape.

StressIf you have a lot to do in a short amount of time, you may feel stressed. The less time you have, the more stress you will feel. In structures the stress, � , on an object is how much force is applied per unit area:

� � F __ A

A force applied to an area of an object creates a certain amount of stress. If you decrease the area while applying the same force you increase the stress. One unit of measurement for stress is N m�2. It is also measured in pascals where 1 Pa � 1 N m�2.

Compression

a pushing or squeezing force

Tension

a pulling or stretching force

Shear

two forces acting in opposite directions towards different points

Compression

a pushing or squeezing force

Tension

a pulling or stretching force

Shear

two forces acting in opposite directions towards different points

Stress

force per unit area

Stress

force per unit area

Figure 6.1 The brick is compressed by the weight force pushing down and the ground pushing up.

Figure 6.2 The string is under tension from the weight of the keys pulling down and the hand pulling up.

Figure 6.3 A sideways force is applied to the top of a book on a table. A friction force acts on the bottom of the book in the opposite direction. The shear force causes the pages to slide along each other.


Nelson Physics 124 VCE Units 3&4

s

If you push the point of a nail with 100 N on a piece of wood, you are likely to make a hole. If you push a fl at pebble with 100 N on a piece of wood, you may only make a dent. The same force applied to a smaller area has a larger stress. A stress that causes a compression is called a compressive stress. A stress that causes an extension is called a tensile stress.

Compressive stress

stress causing length to become smaller

Tensile stress

stress causing length to become longer

Compressive stress

stress causing length to become smaller

Tensile stress

stress causing length to become longer

Blaise Pascal (1623–62)Pascal was educated by his father, who was a local judge but had a strong interest in science and maths. Pascal was banned from studying maths at the age of 11 because his father felt he should spend more time on Greek and Latin. When he turned 12, however, he was caught writing mathematical proofs on his wall with a piece of coal – his father gave in and let him focus on maths. He created his own mechanical calculator at the age of 19 to assist his father calculate his taxes. Throughout his life he made many great contributions to physics, maths and philosophy. In 1971, the pascal was made the SI unit for newton per square metre at the 14th General Conference for Weights and Measures.

\\ WORKED EXAMPLE

Question 1

Figure 6.4 shows a concrete column of area 0.25 m2 supporting a weight of 500 kg. What is the compressive stress in the pile?

Answer

� � F __ A

� 500 � 10 ________ 0.25

� 20 000 Pa

The compressive stress is 20 kPa.

Question 2

An abseiler with a mass of 80 kg is rappelling down a cliff face with constant velocity. His rope has a diameter of 12 mm. What is the tensile stress?

abseiler

Figure 6.5

Answer

� � F __ A

F � mg

� 80 � 10

� 800 N

A � �r 2

� � � 12 � 10�3 _________

2 � 2

� 1.13 � 10�4 m2

500 kg

concrete column

Figure 6.4



� � F __ A

� 800 __________ 1.13 � 10�4

� 7.07 � 106 Pa

The tensile stress is 7.07 MPa.

StrainWhen a force is applied to an object, the object changes length. It is harder to stretch a piece of elastic by 4 cm than by 2 cm. Stretching that piece of elastic by 4 cm causes a larger strain. On the other hand, it is easier to stretch a long piece of elastic by 4 cm than a short piece of elastic by the same amount. Strain, �, is a ratio of change in length, ��, over original length, L .

� � �� ___ L

As strain is a ratio, it has no units. It is sometimes expressed as a percentage.

Strain

change in length relative to original length

Strain

change in length relative to original length

Thomas Young (1773–1829)Young was the eldest of 10 children. By the age of 14 he had learned Greek and Latin and was familiar with 10 other languages. Young studied medicine and also became a doctor of physics. At the age of 28 he was appointed professor of natural philosophy (physics) at the Royal Institution. He contributed to many theories such as the wave theory of light (which will be looked at in Unit 4 Light and Matter), vision and colour. He also deciphered Egyptian hieroglyphics by studying the text of the Rosetta Stone. The brilliance of Young’s modulus is that it is independent of the dimensions of the material. It is a measure of the strength of the bonds in the material.

\\ WORKED EXAMPLE

Question 3

A tug of war rope is initially 6 m long. During the tug it is stretched by 3 cm. What is the percentage strain in the rope?

Figure 6.6

Answer

� � �� ___ L

� 0.03 ____ 6

� 0.005 � 0.5%

The rope stretches by 0.5%.

Young’s modulus Young’s modulus combines the ideas of stress and strain. It allows us to accurately measure the stiffness of a material.

E � stress ______ strain

� � __ �

� F/A _____ ��/L

� FL _____ A ��

Young’s modulus

an accurate measure of the stiffness of a material

Young’s modulus

an accurate measure of the stiffness of a material



s

As strain has no unit, the unit for Young’s modulus is the same as stress, that is Pa. The value of Young’s modulus depends on the strength of the forces between the atoms and molecules in the material. If these forces are very strong, the material will resist being stretched and being compressed. We could expect that different materials would have a particular Young’s modulus. Real materials, however, always contain some irregularities in the arrangement of their component atoms and molecules. These irregularities act to change the value of Young’s modulus. Table 6.1 gives approximate values of Young’s modulus for some real materials.

Table 6.1 Young’s modulus for some common materials

Material Young’s modulus, E (Pa)

Steel 2.0 � 1011

Copper 1.1 � 1011

Cast iron 1.0 � 1011

Brass 1.0 � 1011

Aluminium 7.0 � 1010

Glass 5.5 � 1010

Marble 5.0 � 1010

Granite 4.5 � 1010

Concrete 2.0 � 1010

Bone (compact) 1.8 � 1010

Brick 1.4 � 1010

Wood 1.0 � 109 – 1.0 � 1010

Rubber 4.0 � 106

\\ WORKED EXAMPLE

Question 4

A 1.50 m length of steel piano wire with a diameter of 0.25 cm is stretched by attaching a 10 kg mass to one end. How far is the wire stretched?

Answer

Rearranging the equation E � stress _____ strain

gives:

E � FL ____ A ��

�� FL ___ EA

� 100 � 1.5 _________________________ 2 � 1011 � � � (1.25 � 10�3)2

� 1.5 � 10�4 m

The wire stretches by 0.15 mm.

Young’s modulus and Hooke’s LawYou may remember that Hooke’s Law also describes the relationship between force and change in length by the equation: F � �kx. How does this relate to Young’s modulus? Rearranging the equation E � � __ � gives:

� � E � �

F __ A

� E � �� ___ L

F � EA ___ L � ��

or F � k � �� where k � EA ___ L

.

Problem set 6A

Question 1

Describe the type of force acting on the following objects:

a the heel of your shoe while walkingb the handle while opening a drawerc the legs of a tabled tearing a piece of paper in halfe the fi shing line while reeling in a fi shf Blu Tack while sticking a piece of cardboard to the wallg Blu Tack while holding a piece of cardboard to the wall

Question 2

What is the tensile stress in a copper wire 5.0 mm in diameter that supports a mass of 250 kg?

Question 3

A piece of rubber 12.8 cm long is stretched to a new length of 17.3 cm. What is the tensile strain in the rubber?

Question 4

Christine hangs a 3.0 m length of steel wire from the ceiling and attaches a mass of 2.0 kg to the end of this wire. She observes that



the mass extends the length of the wire by 3.6 mm. If the diameter of the wire is 0.32 mm, calculate:

a the strain on the wireb the stress on the wirec the value of Young’s modulus for this steel

Question 5

A cylindrical steel rod 0.50 m long and 1.0 cm in radius is subjected to a tensile force of 1.0 � 104 N.

a What is the tensile stress?b What is the tensile strain?c By what amount does the rod stretch?

Question 6

A marble column of cross-sectional area 2.4 m2 supports a mass of 2.0 � 104 kg.

a What is the stress within the column?b What is the strain within the column?c By what amount would the column shorten if it was 12.0 m high?

Question 7

A builder has the choice of three different steel columns. Figure 6.7 shows their cross-sections and the dimensions of the girders. Each girder has a length of 12 m and is going to support part of a large building as shown.

a What is the cross-sectional area of each girder?b Which of the girders will be compressed by the least amount

when it is made part of the building?c If the load on the girder is to be 1.2 � 106 N, by how much is it

compressed?

Question 8

Young’s modulus is 1.0 � 1010 for a particular wood of which a chair is made. The chair has four legs, each of length 42 cm and cross-sectional area 2.0 � 10�3 m2. Peter is a man of mass 100 kg. When Peter stands in the middle of the chair:

a what is the stress on each leg of the chair?b by what amount do the chair legs shrink?

Question 9

If the tensile stress in a steel rod is not to exceed 5.0 � 104 N m�2, what is the diameter of a cylindrical rod that is needed to support a load of 500 kg?

Question 10

What is meant by the terms ‘stress’, ‘strain’ and ‘Young’s modulus’?

Strength of a materialAll materials have a breaking point or upper limit to the stress they can withstand without fracturing. This limit depends on the strength of the bonds between the molecules and atoms that make up the material.

24 cm

4 cm

4 cm

24 cm pillar

16 cm 24 cm

building8 cm

16 cm

Figure 6.7



s

Because of this, a material may have different breaking points under different types of stress. Under an excessive tensile stress, materials snap. The minimum stress that causes a material to snap is called the tensile strength of the material. An excessive compressive stress crushes and crumbles or shatters a material. The minimum stress required to crush a material is called its compressive strength. Tensile and compressive strengths for some common materials are shown in Table 6.2.

Table 6.2 Tensile and compressive strengths for some common materials

Material Tensile strength (Pa) Compressive strength (Pa)

Steel 5.0 � 108 5.0 � 108

Cast iron 1.7 � 108 5.5 � 108

Aluminium 2.0 � 108 2.0 � 108

Brass 2.5 � 108 2.5 � 108

Concrete 2.0 � 106 2.0 � 107

Brick 3.5 � 107

Marble 8.0 � 107

Granite 1.7 � 108

Wood 4.0 � 107 1.0 � 107–3.5 � 107

Bone (compact) 1.3 � 108 1.7 � 108

Some materials have high compressive strength relative to their tensile strength, such as concrete or granite. Some materials, such as wood, have a high tensile strength compared to their compressive strength. Materials such as steel and aluminium are equally good under compression and tension. Steel and cast iron have high compressive and tensile strength, but steel is more able to stretch (Figure 6.8).

Analysing stress�strain graphsIt is useful to graph the amount of strain in a material versus the amount of stress applied to the material. From this we can calculate Young’s modulus by looking at the gradient. We can also compare the ways that different materials behave under different loads.

Elastic and plastic behaviour of materialsWhen a rubber band is stretched it usually returns to its original length. Elastic behaviour occurs when a material is stretched (or compressed) but then returns to its original size and shape once the stress is removed. The elastic region is the linear section of the graph near the origin. The elastic limit is the maximum amount of stress the material can withstand before there is permanent damage done to the bonds

Tensile strength

the minimum stress that causes a material to fracture

Compressive strength

the minimum compressive stress that causes a material to fracture

Tensile strength

the minimum stress that causes a material to fracture

Compressive strength

the minimum compressive stress that causes a material to fracture

Elastic behaviour

when a material returns to its original size and shape after being stretched or compressed

Elastic limit

the maximum stress that a material can withstand before it deforms permanently

Elastic behaviour

when a material returns to its original size and shape after being stretched or compressed

Elastic limit

the maximum stress that a material can withstand before it deforms permanently

0.1%

cast iron

steel

100

200

300

0.2%

(MPa)

Figure 6.8 The stress–strain graphs for steel and cast iron. Steel has a larger Young’s modulus than cast iron.



between the molecules. If stress is increased beyond the elastic limit, the object can no longer return to its original shape and there is plastic deformation (Figure 6.9). The relationship between stress and strain is no longer linear. A small increase in stress can cause a large change in shape. Once the stress is removed, the material can only recover its shape partially (Figure 6.10). When an object is plastically deformed, bonds between molecules break. This causes the object to heat up.

Strain energyIt takes energy to change the shape of an object. Strain energy is the energy stored as a result of changing the shape of a material. It is equal to the area under a stress–strain graph. This energy is stored in the volume of the material and is measured in J m�3. The area under the stress–strain graph (up to the elastic limit) is:

1 _ 2 � stress � strain � 1 _ 2 � F __ A � �� __ L

F � �� gives the work done or energy. A � L gives the volume. Hence, strain energy is energy per unit volume.

Plastic deformation

when there is a permanent change in shape in a material after the stress has been removed

Plastic deformation

when there is a permanent change in shape in a material after the stress has been removed

Strain energy

energy stored in a material due to change in the shape of a material

Strain energy

energy stored in a material due to change in the shape of a material

elastic limit

region

breakingpoint

Tensile strain

Ten

sile

str

ess

0

Figure 6.9 Elastic behaviour occurs between the origin and the elastic limit. Plastic deformation occurs between the elastic limit and the breaking point.

Stre

tch

ing

fo

rce

0 ExtensionPermanentextension

beyond elasticpoint

Eelasticlimit

decreasingthe force

increasingthe force

Figure 6.10 If a wire is stretched beyond its elastic limit, it will not return to its unstretched length when the stress is removed.

\\ WORKED EXAMPLE

A cable has a length of 12.0 m and is stretched by 1.2 � 10�4 when a stress of 8.0 � 108 Pa is applied. This is within its elastic region.

Question 5

Sketch the stress–strain graph.

Answer

Strain

Stre

ss (

Pa)

8.0 108

1.0 105

Figure 6.11

Question 6

What is the strain energy per unit volume in the cable when the stress is applied?

Answer

The strain energy per unit volume is the area under the graph.

Area under the graph � 1 __ 2

� 8.0 � 108 � 1.2 � 10�4 _________

12.0

� 4.0 � 103 J m�3

Stressing while straining



s

Properties of materials in tension and compressionToughness

Toughness describes how much energy a material can absorb before breaking. It relates to the total area under a stress–strain graph up to fracture.

Brittleness

Some materials such as brick and marble have a small elastic region and little or no plastic region. The fracture point is close to the end of the straight section of the stress–strain graph. These materials break at, or close to, their elastic limit. Instead of stretching or changing shape they fracture. These materials are described as brittle. Brittleness refers to how much strain a material can withstand before fracture. The compressive strength of brittle materials is higher than the tensile strength.

Ductility

Ductility describes how much plastic behaviour occurs in a material before it breaks. Metals such as gold are very ductile and can be stretched into long thin wires. They have a large plastic region before fracturing.

Stiffness

As mentioned earlier, stiffness is a measure of how much force it takes to change the shape of an object. It relates to the gradient of the stress–strain graph. A steeper gradient indicates a stiffer material.

Toughness

how much energy a material can absorb before fracture

Toughness

how much energy a material can absorb before fracture

Brittleness

how much strain a material can withstand before fracture

Brittleness

how much strain a material can withstand before fracture

Ductility

the amount of plastic deformation before fracture

Ductility

the amount of plastic deformation before fracture

CreepSome materials deform plastically over time. This is called creep. Blu Tack ‘snaps’ if it is pulled quickly but is ductile if it is pulled slowly. Glass panes become thicker at the bottom over time as gravity acts on the molecules. A wooden bookshelf that is too heavily loaded sags over time. This is permanent plastic deformation.

\\ WORKED EXAMPLE

Figure 6.12 shows the stress–strain graphs for three different materials.

Question 7

Which material is the stiffest?

Answer

Material A is the stiffest as it has the highest gradient.

Question 8

Which material is the toughest?

Answer

Material C is the toughest as it has the largest area under the graph.

Question 9

Which material is the most brittle?

Answer

Material A is the most brittle as it has no plastic region.

C

B

A

Figure 6.12 The stress–strain graphs for three different materials.



Suitability of materials for particular purposesStrength, stiffness, toughness, brittleness and ductility are all properties that need to be considered when choosing a material for a structure. Another important factor is the type of forces to which different structures are exposed.

ColumnsA column is under compression. It is usually supporting a large amount of weight such as a roof or bridge. Materials suitable for columns are those that have a high compressive strength, including concrete, brick and marble.

ArchesArches are designed to maximise compressive forces. A semicircular shape is made by stones shaped wider at the top than the bottom. The weight of the arch is transferred as a compressive force through the stones to the surrounding structure (Figure 6.13). Materials suitable for arches are those that have a high compressive strength.

BeamsAn example of a beam is a plank of wood resting on a support at each end (Figure 6.14). If the beam is made of a uniform material, then the weight of the beam can be considered to be at the centre of the beam. This weight force (along with any objects on the beam) pushes down. Upward force is provided by each of the supports. This causes the beam to sag or bend in the middle. The bottom of the beam is under tension as it is stretching around the bend. The top of the beam is under compression as it is being squashed by the bend. Hence, beams need to be made from materials that are good under both compression and tension. Steel is strong under both tension and compression, but it is fl exible and expensive. Concrete is a rigid, cheap and convenient building material, but, as Table 6.2 shows, it is much weaker under tension than under compression. It is also heavy, so the weight of concrete structures must be considered as part of the load they support. Reinforced and prestressed concrete are building materials that combine the advantages of both steel and concrete to overcome these stresses. They are examples of composite materials.

On RomansThe application of the arch by the Romans revolutionised the building of structures such as bridges, buildings and aqueducts. The arch allowed structures to span greater distances. Roman roads and bridges enabled speedy movement of soldiers and chariots. The Roman Empire was built by soldiers who were fi ghting engineers.

Question 10

Which material is the most ductile?

Answer

Material C is the most ductile as it has the largest plastic deformation region.

compression

tension

mg

tension

Figure 6.14 Forces in a beam.

thrust thrust

keystone

Figure 6.13 Elements of an arch.



s

Composite materialsA composite material is made out of two or more materials. The properties of the composite material are often more useful than that of the component materials. In reinforced concrete beams, steel reinforcing rods can be set in the lower part of the beam running the length of the beam (Figure 6.15). The steel keeps the concrete from cracking under tension. A longer beam may be supported in the middle as well as at each end. The regions between the supports will still be in compression on the top of the beam and tension at the bottom of the beam. Above the centre support, tension will occur in the top of the beam. Steel reinforcing rods should be set in the upper part of the beam above the centre support (Figure 6.16).

13

sectionthrough beam

1313

Figure 6.16 A longer beam with centre support. Steel should be set wherever there is tension.

In prestressed concrete, the steel rods or tendons are under tension and so place the concrete under compression. The compressive forces that are built-in in this way partially cancel any tension applied to the beam. The main method of making prestressed concrete is pretensioning. The metal cables (tendons) are stretched between anchorages before the concrete is poured. When the concrete is set, the anchorages are released. This helps keep the lower part of the beam in compression.

Safety factorsChoosing the right material for the job helps make the structure safe. The values given in Table 6.2 describe the strength of different materials under ideal conditions. When designing structures, engineers have to ensure that the building will hold in less than ideal conditions. What if there is a strong wind blowing? What if there is an earthquake? Safety factors ensure that under normal use the stress acting on the object is only a fraction of its actual compressive or tensile strength. A safety factor of 6 means that the engineer has decided that a material is safe to use up to one-sixth of its ultimate strength.

\\ WORKED EXAMPLE

Question 11

A steel cable is able to support a force of 50 kN. A safety factor of 5 is required. What is the maximum force that can be safely supported by the cable?

Answer

The maximum load that is allowable is � 50 ___ 5

� 10 kN

Composite material

a material made out of two or more component materials

Composite material

a material made out of two or more component materials

When designing the World Trade Center in New York it would have been almost impossible to imagine the events that could cause the buildings to crumble. Engineers cannot plan for every possible event. Some buildings, however, are designed to reduce the risk of car bombings. Underground car parks are no longer designed into some buildings where it is too hard to predict and build for the possible impact of a bomb from underneath the building. Wide courtyards around buildings that cannot be accessed by vehicles reduce the likelihood of a car bomb going off outside the building.

13

sectionthrough beam

1313

Figure 6.15 A reinforced concrete beam.



Problem set 6B

Question 1

Explain the transformations of energy and changes in shape in the following processes:

a stretching and releasing a rubber band elasticallyb stretching a rubber band past its elastic limit

Question 2

A cable has an unstretched length of 12.0 m and is stretched by 1.2 � 10�4 m when a 6.4 � 108 Pa stress is applied. What is the strain energy per unit volume in the cable in J m�3, when this stress is applied?

Question 3

A 16 cm long animal tendon is found to extend 3.1 mm by application of a force of 12.5 N. Assume that the tendon is round with an average diameter of 8.0 mm.

a What is the Young’s modulus of the tendon?b What is the strain energy in the stretched tendon? Give your

answer in J m�3.c How much energy is stored in the tendon? Give your answer in J.

Question 4

Figure 6.17 shows the stress–strain graph for three different materials under tension.

a Which material has the highest tensile strength?b Which material is the toughest?c Which material is the stiffest?d Which material is the most brittle?

Question 5

In order to span a small distance between two supports, a construction worker rivets together two component sheets, X and Y, with different elastic properties. X is strong in compression but weak in tension while Y is strong in tension but weak in compression. As the component sheets have different properties, which should be placed on top? Explain in terms of the properties of the two components.

Question 6

Stone slabs were used by humans in early times to span gaps in buildings (Figure 6.18). It was common for cracks to appear on the under or lower side of the beam. Explain why, with stone slabs, the cracks appeared on the under side of the beam.

Question 7

Figure 6.19 shows a concrete beam on two supports. Draw in lines to show where steel reinforcing rods should be placed to strengthen the beam. Explain your decision.

Question 8

A column is made out of concrete with a compressive strength of 2.0 � 107 Pa. What mass can it support if it is to have a safety factor of 4?

material 1

2

3

Figure 6.17

Figure 6.19

Figure 6.18

crack



s

Question 9

A cylindrical rod of cast iron has a diameter of 8.0 cm. Whatweight can it support if it is not to exceed 40% of the compressive strength?

Question 10

A material is placed under different amounts of tensile stress and the change in length is recorded. The results are shown in Table 6.3.

Table 6.3

� (MPa) 0 5 10 15 20 25

� (%) 0 2 4 6 8 breaks

a Draw the stress–strain graph for this experiment.b Calculate the Young’s modulus for this material.c How much strain energy is stored in the material when 15 MPa

is applied?d Is the material brittle or ductile? Explain your answer.e What is the tensile strength of the material?

Equilibrium of structuresNewton’s Second Law indicates that the net force on a body at rest is zero. This does not mean that no forces are acting on the body, rather the sum of all the forces is zero.

Translational equilibriumA book resting on a table has the weight force acting downwards on it and the table reacting upwards on it. The sum of the forces acting on the book is zero. In this example, the two forces are acting along a vertical line. Equal and opposite forces will be in equilibrium in any situation in which they have the same line of action or are co-linear. What if more than two forces are acting, or if the forces are not acting along the same straight line? In translational equilibrium, the sum of the forces acting on or through a point must equal zero. The downward forces have to be balanced by upward forces. Leftward forces must be balanced by rightward forces.

Translational equilibrium, �F � 0

Consider a set of traffi c lights suspended above an intersection by a pair of wires of negligible mass, as shown in Figure 6.20. The tensions in the wires must balance the weight of the lights. In the vertical direction, the magnitude of the components of the force upwards must equal the magnitude of the components of force downwards. This gives:

T1 sin �

1 � T

2 sin �

2 � mg

There is no horizontal motion, so in the horizontal direction we have:

T1 cos �

1 � T

2 cos �

2

Co-linear

forces acting on a point in a straight line

Translational equilibrium

an object is in translational equilibrium when not accelerating (net force on the object is zero: �F � 0)

Co-linear

forces acting on a point in a straight line

Translational equilibrium

an object is in translational equilibrium when not accelerating (net force on the object is zero: �F � 0)



Torque – rotational forcesIn chapters 1–3, we looked at forces acting on the centre of mass of an object. These forces caused the whole object to move in a straight line, parabola or circle. A torque is applied when forces act on different points of an object. These forces, which are not co-linear, will cause a bending or a rotational effect in a structure. Consider a frying pan (Figure 6.21). The weight of the pan creates a torque in the handle. This is a bending force. If the handle were not stiff, it would bend. The torque is given by:

_

› � � _

› r

� _

› F

where _

› F is the force and

_

› r �

is the perpendicular distance of the force from the point where the torque is being calculated. The unit of torque is N m. It is harder to hold the frying pan at the far end of the handle than it is close to the pan. The larger distance increases the size of the torque. If the frying pan is supported underneath the centre of mass, there is no torque. The larger the distance from the axis, the larger the torque. Similarly, the larger the force, the greater the rotational effect (torque). If r � 0 (or F � 0) then � � 0.

Torque

a rotational force resulting from forces acting towards different points of an object

Torque

a rotational force resulting from forces acting towards different points of an object

\\ WORKED EXAMPLE

Question 12

A traffi c light of mass 20 kg is suspended by two wires as shown in Figure 6.20. What force are the wires supporting?

Answer

F � mg

1 F � 20 � 10

1 F � 200 N

The wires are supporting 200 N.

Question 13

What is the tension in one of the wires?

Answer

As the angles are symmetrical, each wire is supporting a vertical weight of 100 N.

T sin 45° � 100

1 T � 100 ______ sin 45°

1 T � 141 N

One wire has a tension of 141 N.

mg

T1 T2

1 2

Figure 6.20 The net force is zero so the horizontal components of the tensions in the wires are equal and opposite, and the sum of the vertical components of the tension is equal and opposite to the weight of the lamp.

centre of mass

r

Figure 6.21 A frying pan creates a torque around the handle.

\\ WORKED EXAMPLE

Question 14

A uniform shelf (mass 10 kg, length 1 m) is attached with a fi xed joint to a wall as shown in Figure 6.22. What is the torque exerted on the wall by the shelf?

Centre of massThe centre of mass of an object helps us to deal with problems more simply. When gravity is acting on a 5 kg block, an arrow is drawn from the centre representing a force of 50 N. In actual fact, gravity is acting on each molecule in the block. The weight of the block is considered to act through one point, the centre of mass.



s

Answer

The centre of the beam can be taken as the centre of mass, which is 0.5 m from the wall. � � r

�F

� 0.5 � 10 � 10

� 50 N m

A torque of 50 N m is exerted on the wall in a clockwise direction.

Question 15

Figure 6.23 shows a box weighing 1.0 kg placed on the shelf 0.8 m from the wall. What is the total torque exerted on the wall?

Answer

� � r�

F

� 0.5 � 10 � 10 � 0.8 � 1 � 10

� 50 � 8

� 58 N m

A torque of 58 N m is exerted on the wall in a clockwise direction.

\\ WORKED EXAMPLE

Question 16

A girl of mass 20 kg and a boy of 24 kg sit on a seesaw 4.0 m long balanced at the midpoint. The girl is 1.8 m to the left of the balance point (Figure 6.24). How far from the balance point must the boy sit to keep the seesaw horizontal?

Answer

As the centre of mass is at the middle of a uniform beam, we can ignore the weight of the beam because it will exert zero torque.

The net torque about an axis through the support is zero, so we have:

�� x � 240 � 1.8 � 200 � 0

360 � 240x

x � 1.5 m

The boy must sit 1.5 m to the right of the balance point.

0.8 m

1 kg

Figure 6.23

Figure 6.22 The torque is clockwise to balance the force acting at the left-hand end of the shelf.

1 m

A structure in rotational equilibrium has no net rotational effects acting on it.

Equilibrium conditionsIn any structure, the conditions for equilibrium are met when:

�Fon structure

� 0 and ��on structure

� 0

�Fon structure

� 0: In equilibrium the sum of the horizontal and vertical components of the force on the structure are both zero. ��

on structure � 0: In equilibrium the clockwise torques are balanced by

anticlockwise torques. In this textbook, clockwise torque is considered to be positive and anticlockwise torque is considered to be negative.

Rotational equilibrium

an object is in rotational equilibrium when it is not rotational (net torque on the object is zero: �� 0)

Rotational equilibrium

an object is in rotational equilibrium when it is not rotational (net torque on the object is zero: �� 0)

x

m1g m2g

1.8 m

20 kg 24 kg

Figure 6.24



Equilibrium in simple structures

Ties and cablesA tie is a structural member that carries a tensile stress. A tie can be fl exible or rigid. One type of tie is a cable. A cable is a series of twisted fi laments that is usually fl exible. Common examples are rope and steel cable. Cables and ties are under tension, they keep two objects together.

Struts and columnsA strut is a structural member that carries a compressive stress. It must be rigid. One type of strut is a column, which is a vertical member. Struts and columns are under compression; they keep two objects apart.

Beams and cantileversA beam is a structural member that carries bending forces. This is a result of both compressive and tensile forces. Beams are supported at one or more points (Figure 6.25). A beam that is supported at one end only through a fi xed joint is called a cantilever (Figure 6.26).

load

platform

strut

tie

Figure 6.25 The forces acting on a platform held in place by a strut and tie.

If the supports of a beam are symmetrical, and the weight of the beam and any load is central, then the forces through the supports will be equal and can be worked out through equilibrium of forces. If the load is situated towards the right of the beam, then the right-hand support will bear more of the load. We can calculate the force in each support by using a combination of translational and rotational equilibrium. Torque can be calculated about any point along a beam. The point from which torque is calculated exerts zero torque as r � 0. This can be used to simplify problems with two unknown forces.

Tie

a structural member that carries a tensile stress. Can be fl exible or rigid

Member

an active component of a structure

Cable

a series of twisted steel wires that is used in tension. Usually fl exible

Strut

a structural member that carries a compressive stress

Column

a vertical member that carries a compressive stress

Beam

a structural member that carries compressive and tensile forces

Cantilever

a beam that is supported at one end only through a fi xed joint

Tie

a structural member that carries a tensile stress. Can be fl exible or rigid

Member

an active component of a structure

Cable

a series of twisted steel wires that is used in tension. Usually fl exible

Strut

a structural member that carries a compressive stress

Column

a vertical member that carries a compressive stress

Beam

a structural member that carries compressive and tensile forces

Cantilever

a beam that is supported at one end only through a fi xed joint

A fi xed joint resists torque. The frying pan mentioned earlier has a fi xed joint between the handle and the pan. A pinned joint is like a hinge. A pinned joint does not resist torque. When a torque is applied to an open door, the door does not bend, it swings.

Figure 6.26 A cantilever.



s

T

F

W

30

A

B

Figure 6.28

F1

F2

16.0 m 8.0 m

1.5 104 N2.0 105 N

Figure 6.27

\\ WORKED EXAMPLE

Question 17

A uniform beam of length 24.0 m and mass 1500 kg is supported at each end. A machine of mass 2.0 � 104 kg is 8.0 m away from one end of the beam, as shown in Figure 6.27. What forces do the two supports exert on the beam?

Answer

The centre of the beam can be taken as its centre of mass. Considering the torques about the left-hand support we have:

�� 12.0 � 1.2 � 104 � 16.0 � 2.0 � 105 � 24.0 � F2 � 0

24.0 F2 � 1.8 � 105 � 3.2 � 106

F2 � 1.4 � 105 N

Thus, the force exerted on the beam by the right-hand support is 1.4 � 105 N.

To fi nd the force exerted by the left-hand support, consider the vertical components of the forces.

F1 � F2 � 1.5 � 104 � 2.0 � 105

Substituting the value of F2 from earlier gives:

F1 � 1.4 � 105 � 1.5 � 104 � 2.0 � 105

Hence, F1 � 7.5 � 104 NThe force exerted on the beam by the left-hand support is 7.5 � 104 N.

Question 18

A platform of mass 25 kg is suspended from the side of a building by a cable AB as shown in Figure 6.28. The platform is 1.0 m long and pinned at the wall. What is the tension in the cable?

Answer

As the tension in the cable is not perpendicular to the beam, components must be used. The vertical component of the tension in the cable, Tv , can be found by:

�� 0.5 � 250 � 1.0 � Tv � 0

Tv � 125 N

The total tension in the cable can be found using:

sin 30° � 125 ___ T

T � 125 ______ sin 30°

T � 250 N

Question 19

What is the force exerted by the wall on the platform?This is found using translational equilibrium. Firstly, consider the vertical

forces. The vertical component of the force exerted by the wall, Fv , can be given by:

Fdown � Fup

mg � Tv � Fv

250 � 125 � Fv

Fv � 125 N up

A novel weighing machine



Problem set 6C

Question 1

A tight rope-walker of mass 80 kg balances at the centre of a high wire as shown in Figure 6.29. What is the tension in the wire, if it makes an angle of 8° with the horizontal on either side?

8 8

Figure 6.29

Question 2

Katrina has a mass of 30 kg and sits on a swing that is attached to an overhead support by a chain of negligible mass. She is pulled aside by her mother, who exerts a horizontal force of 100 N as shown in Figure 6.30, and is held there. Calculate the tension in the chain and the angle that the chain makes with the vertical.

Question 3

Two wires PQ and PR support a large lamp of mass 50 kg in a hall as shown in Figure 6.31. What is the tension in each wire?

Q 45

R

large lamp

P

Figure 6.31

The horizontal component of the force exerted by the wall, FH, has to balance the horizontal component of the tension in the cable, TH. Using Pythagoras’ theorem:

T h 2 � 1252 � 2502

Th � 216 N to the left

1 Fh � 216 N to the right

The magnitude of the force exerted by the wall on the platform, F, is given by:

F 2 � F v 2 � F h

2

� 1252 � 2162

F � 250 N

The direction of the force exerted by the wall on the platform is given by:

tan � � 125 ___ 216

� � 30°

The force exerted by the wall on the platform is 250 N at 30° to the horizontal.In this question the forces exerted by the wall and the tension in the cable

are symmetrical. This is because the forces involved are symmetrical, which is not always the case.

100 N

Figure 6.30



s

Question 4

Figure 6.32 shows a bar AB of negligible mass and length 1.0 m balanced at its midpoint. A 0.60 kg mass is hung from end A and a 1.0 kg mass is hung from point C, which is a distance of 0.30 m to the right of P. What is the upward force from the support at P?

upward forcefrom support

0.30 m

0.60 kg 1.0 kg

A

P

C B

Figure 6.32

Question 5

Figure 6.33 shows an 80 kg man standing on a plank of negligible mass supported by two trestles. Calculate the upward forces X and Y exerted by the trestles on the plank.

Question 6

A beam of mass 100 kg is loaded with masses of 200, 250 and 300 kg as shown in Figure 6.34. What are the forces P and Q on the two supports?

200 kg 250 kg 300 kg

2.0 m 4.0 m 6.0 m 3.0 m

P Q

Figure 6.34

Question 7

What are the reactions P and Q of the supports on a uniform cantilever of mass 900 kg as shown in Figure 6.35?

Question 8

What are the forces P and Q that the supports exert on the diving board shown in Figure 6.36 when a 40 kg girl stands at the end of the board? (Ignore the mass of the board.)

2.0 m 4.0 m

P Q

Figure 6.36

X Y1.5 m 2.5 m

Figure 6.33

P Q

W

10 m 15 m

Figure 6.35



Question 9

A container of mass 400 kg on a 2.0 m shelf is supported by a strut as shown in Figure 6.37. The container is evenly loaded and it is centrally placed on the shelf. The shelf is pinned to the wall. (Ignore the weight of the shelf.)

a What are the three forces acting on the shelf?b What is the thrust from the strut, assuming that it acts along its

length?c What is the magnitude and direction of the force on the shelf

where it touches the wall?

wall

container

shelf

60

strut

400 kg

Figure 6.37

Question 10

Figure 6.38 shows a mass of 50 kg hung from the end of a cantilever 3.0 m long. The cantilever weighs 6.0 kg and is supported by a cable attached at a point 0.5 m from the end.

a What is the tension in the wire?b What is the vertical component of the force exerted by the wall

on the shelf?c What is the horizontal component of the force exerted by the

wall on the shelf?

30

0.5 m

2.5 m

50 kg

Figure 6.38



s

Summary of materials and their use in structures• Compression is a pushing or squeezing force.• Tension is a pulling or stretching force.• Shear forces cause layers of molecules to slide over each other.• Stress is a measure of how much force is applied to an area: � � F __ A .

• Strain is a measure of how much an object has changed in length relative to its original length: � � �� ___

L .

• Young’s modulus measures the stiffness of a material: E � stress ______ strain � FL ____ A��

.– It is independent of the dimensions of the material.– It is a measure of the strength of the bonds between the molecules in the material.

• Tensile strength is the minimum stress required to snap a material.• Compressive strength is the minimum stress required to crush a material.• Elastic behaviour is when a material returns to its original size and shape after being

stretched or compressed.• Plastic behaviour is when there is a permanent change in shape after the stress has been

removed.• Strain energy is the energy stored in a material due to its change in shape.

– It is calculated by the area under a stress–strain graph.– It is measured in J m�3.

• Toughness is the amount of energy a material can absorb before fracture.– It is calculated by the area under a stress–strain graph.

• Brittleness is how much strain a material can withstand before fracture.• Ductility is the amount of plastic deformation before fracture.• Stiffness is how much force it takes to change the shape of an object.

– It is calculated by the gradient of a stress–strain graph.• Materials used for columns and arches need to have a high compressive strength.• Beams have areas of compression and areas of tension.• Composite materials are made of two or more component materials.• Reinforced concrete has steel rods placed in areas of tension.• Translational equilibrium: �F � 0.• Torque is a rotational force: �

� �

� r �

� F .

• Rotational equilibrium: �� 0.• Ties and cables are structural members that carry tensile stress.• Struts and columns are structural members that carry compressive stress.• Beams are structural members that carry compressive and tensile forces.• Cantilevers are beams supported at one end only through a fi xed joint.

Review questions

Question 1A piece of rubber of length 3.0 m and diameter 2.0 cm is stretched 2.5 mm when a mass of 0.10 kg is attached to it. Calculate Young’s modulus for the rubber.

Question 2Table 6.4 gives some data for a steel wire whose unextended length is 2.0 m. It was hung vertically from the ceiling and masses were added until it broke. The diameter of the wire was 0.25 mm.

Worksheet 6.1

Word Check

Test Yourself

Theory Summary



Table 6.4

Masses added (kg) 0 0.5 1.0 1.5 2.0 2.5

Extension (mm) 0 0.7 1.4 2.0 2.7 2.9

a Use the above data to prepare a table of the strain of the wire for different stresses.

b Draw a stress–strain graph for the wire.

c Calculate Young’s modulus from the graph.

d Use the graph to estimate the strain at which the wire reaches its elastic limit. Explain why you chose this value.

Question 3A marble column of cross-sectional area 2.4 m2 supports a mass of 2.0 � 104 kg. The Young’s modulus for marble is 5.0 � 1010 Pa.

a What is the stress within the column?

b What is the strain?

c By what amount would the column shorten if it was 12 m high?

Question 4A single strand of human hair supports a mass of 160 g and stretches 40% before breaking. The hair has a diameter of approximately 0.10 mm.

a What is the tensile stress on the hair?

b Consider that the hair is perfectly elastic up to the breaking strength and calculate the Young’s modulus for the hair.

c What is the strain energy stored in the hair just before it breaks?

Question 5Two metal rods, one of brass and the other of steel, are each 50 cm long and of cross-sectional area 4.0 cm2. The two rods are placed side by side and are jointly subjected to a tensile force of 6600 N. Assuming that each experiences the same tensile strain, calculate the stress in each rod.

Question 6Cast-iron beams used in bridge and building construction are often designed with a cross-section similar to that shown in Figure 6.39. The lower part of the beam is wider than the upper part. Why is it better for the lower part of the beam to be wider than the upper part? Give reasons for your answer, including reference to the tensile and compressive strengths in Table 6.2.

Figure 6.39



s

Question 7A simple footbridge consists of a beam of mass 1.2 � 104 kg. It is supported at its centre, C by two steel cables, CA and CB which make angles of 30° with the horizontal as shown in Figure 6.40. The cables support 80% of the weight of the bridge and are each 10 m long.

a What is the tension in each cable?

b What is the reaction of the ground on the bridge at points X and Y ?

c If the cross-sectional area of each cable is 2.0 � 10�4 m2 and Young’s modulus for steel is 2.0 � 1011, what is the increase in length for each cable?

30 30

XC

Y

AB

Figure 6.40

Question 8A loaded furniture van of total mass 1.0 � 104 kg is crossing a bridge as shown in Figure 6.41. The bridge is suspended on two pillars 24 m apart and has a mass of 4000 kg. The centre of mass of the van is 6.0 m from the left end of the bridge. What is the force exerted by each of the two bridge supports?

bridge

pillars

24 m

Figure 6.41

Question 9A 180 cm tall man lies on a board of negligible mass. The board is supported on two bathroom scales, one under his feet and the other under his head. The two scales give readings of 32 kg and 34.8 kg, respectively. Where is the centre of mass of the man?

Question 10In Figure 6.42 a light plank on two trestles is supporting an 80 kg man and a 10 kg paint tin. X and Y are the two upward forces exerted by the trestles on the plank.

X Y

P Q800 N 100 N

2 m 2 m 2 m

Figure 6.42



a What is the torque of each force about P?

b What is the magnitude of the reaction force Y ?

c What is the magnitude of the reaction force X ?

d In Figure 6.43, the man walks past P towards the end of the plank. What is the magnitude of the reaction force Y at the instant the plank starts to tilt?

P Q

Figure 6.43

e How far is the man from P when the plank starts to tilt?

Question 11A uniform beam of mass 1500 kg is used as a cantilever as shown in Figure 6.44. The supports are made of wood (compressive strength 3.5 � 107, tensile strength 4.0 � 107) and there is to be a safety factor of 6. What is the minimum cross-sectional area of each support?

24.0 m 36.0 m

P Q

Figure 6.44


Chapter_6 Unit 3&4 Book

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