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Worked Solutions
Chapter 7 Materials and their use in structures
7.1 External forces acting on materials
1 B; the cable is being stretched and is under tension.
2 D; the pole is bent so experiences tensile and compressive forces. It also has shear forces acting
where the pole is planted in the track.
3 A; the stumps are being squeezed between the house and the ground.
4 a The towers are being compressed under their own weight.
b The horizontal force of the wind pushing on the blades acts near the top of the tower. A
horizontal force from the ground is acting at the base of the tower. These forces combine
to create a large shear effect.
5 During bending, the upper surface of the plank is being squeezed and is under compression. The
lower surface is being stretched and is under tension.
6 a Forces are balanced, so: F t = mg = 3.0 × 9.80 = 2.9 N
b The fishing line is stretched by a small amount.
7 a The tightrope is being stretched and is under tension.
b The safety mat is being squashed and is under compression.
8 a The ball is being squashed and is under compression.
b, c
F t F t F t
F b
F c
F c
F c
F h
F c
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Worked solutions
Chapter 7 Materials and their use in structures
9 a Forces are balanced, so: F t = mg = 80 × 9.80 = 780 N
b The tensile force will stretch the rope.
10
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Worked solutions
Chapter 7 Materials and their use in structures
7.2 Stress and strength
1 a Force F = mg = 60 × 9.80 = 590 N
Area A (in m 2 ) = πr 2 = π × 0.0050 2 = 7.9 × 10 –5 m 2
Stress σ = F/A = 590/7.9 × 10 –5 = 7.5 × 10 6 Pa
b Stress σ = F/A = 150/(0.015 2 ) = 6.5 × 10 5 Pa
2 a Forces are balanced, so F t = F g = 900 × 9.80 = 8.8 × 10 3 N
8820
b Stress = F/A = = 1.9 × 10 7 Pa = 19 MPa
2
! ! 0.012
3 Stress = F/A =
4
1.0!
10
! ! 0.00500
2
= 1.26 × 10 8 Pa
4 Compressive strength of steel = 500 MPa (from table)
Area A = πr 2 = π × (0.00500) 2 = 7.85 × 10 –5 m 2
Maximum stress = 5.00 × 10 8 = F max /A
F max = 5.00 × 10 8 × 7.85 × 10 –5 = 3.93 × 10 4 N = 39.3 kN
5 Compressive strength of steel = 5000 MPa (from table)
Maximum force = strength × area = 5.00 × 10 8 × 7.85 × 10 –5 = 39.3 kN
Safety factor of 5 means that the maximum force under operating conditions must be 5 times
lower: 39 300/5 = 7.9 kN
6 a Area A = 3.0 cm 2 = 3.0 × 10 –4 m 2
Maximum compressive stress = strength = 170 MPa (from table)
Maximum compressive force = strength × area = 1.70 × 10 8 × 3.0 × 10 –4 = 5.0 × 10 4 N
b Maximum tensile stress = strength = 130 MPa (from table)
Maximum tensile force = strength × area = 1.30 × 10 8 × 3.0 × 10 –4 = 3.9 × 10 4 N
7 a As the beam bends, the material along the centre is not under stress. This is a neutral
zone.
b As the beam bends, the bottom surface is under tension. Steel is weaker under tension, so
more material is needed here.
8 a F = mg = 1.0 × 10 5 × 9.80 = 9.8 × 10 5 N
Compressive strength = 8.0 × 10 7 Pa
Stress = F/A
5
9.80!
10
Area A = F/stress =
7
8.0!
10
= 1.2 × 10 –2 m 2
A = 1.2 × 10 –2 = πr 2
r = 0.063 m = 6.2 cm
b F = mg = 1.0 × 10 5 × 9.80 = 9.8 × 10 5 N
Compressive strength = 500 MPa (from table)
Stress = F/A
5
9.8!
10
Area A = F/stress =
8
5.0!
10
= 2.0 × 10 –3 m 2
A = 2.0 × 10 –3 = πr 2
r = 0.025 m = 2.5 cm
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Worked solutions
Chapter 7 Materials and their use in structures
9 As this beam bends under the load, the underneath surface of the beam becomes stretched and is
therefore under tension. Since concrete is weaker under tension than under compression, this
surface is where cracks are more likely to appear.
10 a Reinforced concrete combines the properties of the high tensile strength of steel with the
high compressive strength of concrete to produce a composite material that is strong
under both tension and compression.
b The reinforcing should be located near the top surface of the concrete. As the roots
expand, the concrete will bend upwards and the top surface will be under tension.
Concrete is weak under tension and so reinforcing is needed here.
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Worked solutions
Chapter 7 Materials and their use in structures
7.3 Strain
1 a Equal, tension is 98 N in both
b Thin due to its smaller cross-sectional area c Thin because it will stretch more.
2 Strain ε = Δl/l = 0.000127/0.200 = 6.35 × 10 –4
3 Strain ε = Δl/l
Δl = ε × l = 0.00128 × 4.00 = 5.12 × 10 –3 m = 5.12 mm
4 Strain ε = Δl/l
Δl = ε × l = 4.5 × 10 –4 × 7.00 = 3.15 × 10 –3 m = 3.15 mm
5 Strain ε = Δl/l
Δl = ε × l = 0.00075 × 0.08000 = 6.00 × 10 –5 m = 6.00 × 10 –3 cm
Marks will be: 8.000 + 0.00600 = 8.006 cm apart
6 Strain = 0.001 = 0.1%
7 Strain ε = Δl/l
Δl = ε × l = 0.002 × 1.4 = 0.0028 m = 2.8 mm
8 a
F (kN) ΔL(mm) σ ( × 10 7 N m –2 ) ε (%)
1.0 0.013 1.27 0.0065
2.0 0.025 2.55 0.0125
3.0 0.038 3.82 0.0190
4.0 0.051 5.09 0.0255
5.0 0.063 6.37 0.0315
b
! 7
6.0"
10
c Gradient = rise/run = = 2.0 × 10 11 N m –2
0.0003
d When steel is subjected to different loads, the ratio stress/strain is constant so σ = kε
where k is a constant.
9 From graph; when stress = 35 MPa, strain = 0.018%
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Worked solutions
Chapter 7 Materials and their use in structures
784
10 Stress σ = F/A = = 2.50 × 10 8 Pa
2
! ! 0.00100
Assuming that the stress/strain ratio remains constant;
gradient = rise/run = stress/strain
2.0 × 10 11 8
2.55!
10
=
strain
8
2.55!
10
strain = = 0.0013
11
2.0!
10
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Worked solutions
Chapter 7 Materials and their use in structures
7.4 Young’s modulus
1 a Find stress where non-linear behaviour starts; 4.0 × 10 8 Pa
b Find strain where non-linear behaviour begins; 0.0020
c
8
4.0!
10
Gradient = rise/run =
0.0020
= 8.0 × 10 11 N m –2
d The gradient of a stress-strain graph is Young’s modulus
e Strength = maximum stress = 6.0 × 10 8 Pa
2 a Strain at breaking point = 0.0050
Strain ε = Δl/l
Δl = ε × l = 0.0050 × 1.0 = 0.0050 m = 5.0 mm
b Strain at elastic limit = 0.0020
Strain ε = Δl/l
Δl = ε × l = 0.0020 × 1.0 = 0.0020 m = 2.0 mm
3 a Area A = πr 2 = π × (0.0010) 2 = 3.1 × 10 –6 m 2
1000
Stress σ = F/A = = 3.2 × 10 8 Pa
8
3.1"
10 !
From graph, strain = 1.6 × 10 –3
Strain ε = Δl/l
Δl = ε × l = 1.6 × 10 –3 × 1.0 = 1.6 mm
b Zero; the wire has not been stressed beyond the elastic limit, so will return to its original
length once the force has stopped acting.
4 a ductile
b F t = mg = 153 × 9.80 = 1500 N
1500
Stress σ = F/A =
6
3.1"
10 ! = 4.8 × 10 8 Pa
From graph, strain = 3.0 × 10 –3
Strain ε = Δl/l
3.0 × 10 –3 = Δl/1.0
Δl = 3.0 × 10 –3 m = 3.0 mm
c No, since the material does not behave elastically under this tensile stress.
5000
5 Stress σ = F/A = = 5.00 × 10 7 Pa
4
1.00"
10 !
! 3
0.800"
10
Strain ε = Δl/l =
= 4.00 × 10 –4
2.00
7
5.00"
10
E = stress/strain =
= 1.25 × 10 11 Pa
4
4.00"
10 !
6 a The higher the value of Young’s modulus, the stiffer the material.
b Least flexible is tungsten as it has highest E value, then steel and finally aluminium is
most flexible.
7 Strain ε = 0.010
E = σ/ε
Stress σ = E × ε = 7.0 × 10 10 × 0.010 = 7.0 × 10 8 Pa
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Worked solutions
Chapter 7 Materials and their use in structures
F/A = 7.0 × 10 8
F = 7.0 × 10 8 × (π × 0.0050 2 ) = 5.5 × 10 4 N
8 a Stress at elastic limit is: σ = 1.0 × 10 8 Pa
E = σ/ε
8
1.0!
10
ε = σ/E =
10
1.6!
10
= 6.25 × 10 –3
Strain ε = Δl/l
6.25 × 10 –3 = Δl/0.40
Δl = 0.0025 m = 2.5 mm
b Tension, since the tensile strength for bone is lower than the compressive strength for
bone (from the table).
c As we age, calcium loss causes the cross-sectional area of our bones to decrease. For any
given load, therefore, the stress that the bones will experience will be greater and they
will be more likely to fracture (so eat your dairy products!)
9 Strain ε = Δl/l = 0.00030/0.40 = 0.00075
E = σ/ε
Stress σ = E × ε = 1.6 × 10 10 × 0.00075 = 1.2 × 10 7 Pa
10 a C, D; they break at the elastic limit.
b A; it has the steepest gradient.
c D; it has the smallest gradient.
d B; it withstands the greatest stress before failing.
e A; it has the greatest plastic region.
f A; it has the greatest ability to deform without breaking.
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Worked solutions
Chapter 7 Materials and their use in structures
7.5 Strain energy and toughness
1 a strain = 0.20% = 2.0 × 10 –3
Work done per cubic metre = area under stress-strain graph = 0.5 × 2.0 × 10 –3 × 4.0 × 10 8
= 4.0 × 10 5 J/m 3
b
c
Strain energy = area under graph × volume of material = 4.0 × 10 5 × 4.0 × 10 –5 = 16 J
It is in the elastic region of the graph, so will resume its original length and no energy will
be transformed into heat.
2 a Area under graph up to strain of 4.0 × 10 –3 = (1.0 × 10 8 × 1.0 × 10 –3 ) × 13.5 squares =
1.35 × 10 6 J/m 3
b Strain energy = area under graph × volume of material = 1.35 × 10 6 × 4.0 × 10 –5 = 54 J
c It is past the elastic limit, so the alloy will be permanently deformed and will heat up as it
is stretched.
3 a Point B is the fracture point, so the alloy will break.
b Area under graph up to strain of 5.0 × 10 –3 = (1.0 × 10 8 × 1.0 × 10 –3 ) × 19 squares = 1.9 ×
10 6 J/m 3
c Strain energy = area under graph × volume of material = 1.9 × 10 6 × 4.0 × 10 –5 = 76 J
4 In elastic deformation, all the strain energy is returned by a material as it regains its initial
dimensions after the strain is removed. For plastic deformation, some of the strain energy will
be transformed into heat energy in altering the atomic structure of the material. Consequently,
the material in this situation never regains its original dimensions.
5 a Stress σ = F/A = 1200/(π × 0.00200 2 ) = 9.55 × 10 7 Pa
b This stress is under the elastic limit, so graph is straight line through origin.
E = stress/strain
Strain = stress/E = 9.55 × 10 7 /2.00 × 10 11 = 4.78 × 10 –4
Area under stress-strain graph = 0.5 × 4.78 × 10 –4 × 9.55 × 10 7 = 2.28 × 10 4 J/m 3
c Strain energy = area under graph × volume of material = 2.28 × 10 4 × (0.500 × π ×
0.00200 2 ) = 0.14 J
6 No, the rod fails at a point past the elastic limit where the behaviour of the material is nonlinear.
A stress–strain graph is needed.
7 Need to find: area under graph for P/area under graph for Q = 4.5 squares/3.0 squares = 1.5
8 a Find area under graph for P at failure = 9.5 squares × 1.0 × 10 8 × 1.0 × 10 –3 = 9.5 × 10 5
J/m 3
b Find area under graph for Q at failure = 6.5 squares × 1.0 × 10 8 × 1.0 × 10 –3 =
6.5 × 10 5 J/m 3
9 Material P is tougher because it can absorb a greater amount of strain energy per volume before
failing (indicated by area under stress–strain graph).
10 a Material P since it has a greater value for Young’s modulus (indicated by gradient).
b Material P since it experiences a greater stress value prior to failing.
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Worked solutions
Chapter 7 Materials and their use in structures
7.6 Forces in balance: Translational equilibrium
1 For a body to be in translational equilibrium, the vector sum of the forces acting on it must be
zero. This situation applies to any stationary body, or one moving with constant velocity.
Options A, B and D are therefore examples of translational equilibrium.
2 a If your mass is 60 kg for example: F t = mg = 60 × 9.80 = 590 N (400–10 000 N)
b If the car’s mass is 1000 kg for example: F t = mg = 1000 × 9.80 = 9.8 × 10 3 N (7000–
12 000 N)
c 10 L of water has a mass of 10 kg: F t = mg = 10 × 9.80 = 98 N (100 N)
3 Total downwards force = 600 + 850 + (50 × 9.80) = 1940 N
Assuming weight is evenly distributed, upwards force provided by each cable is: F c = 1940/4 =
485 N
4 Total downwards force on beam = (1000 + 1500 + 2000 + 500) × 9.80 = 4.9 × 10 4 N
Total upwards force on beam = F X + F Y = 4.9 × 10 4 N
F Y = 4.9 × 10 4 – 2.0 × 10 4 = 2.9 × 10 4 N upwards
5 Total working load = 5000 × 9.80 + 20,000 × 9.80 = 245,000 N
Working load for each pillar = 245 000/2 = 123 000 N
Given safety factor of 8, each pillar must be able to withstand maximum load of: 8 × 123 000 =
9.8 × 10 5 N upwards
6 Downwards force on sign = 5.0 × 9.80 = 49 N
Each vertical component of the tension forces in wires is equal to 24.5 N.
F t = 24.5/cos40° = 32 N
7 Adding the 3 vectors and forming a force triangle gives: F g = 2 × 40 × sin40° = 51 N (or use
sine rule)
Mass m = 5.2 kg
8 Total downwards force on wire is : F g = 90 × 9.80 = 880 N
Each side of the cable is therefore providing an upwards force of 440 N.
Using a force triangle gives: F t = 440/sin5° = 5.0 kN
9 a Adding the forces gives a right angled force triangle with a vertical weight force of 980
N. Using this, it can be determined that:
F A = 980/sin60° = 1.13 × 10 3 N
F B = 980/tan60° = 565 N
b Cable A is more likely to break because it is under greater stress.
c Stress σ(A) = F/A = 1.13 × 10 3 /1.5 × 10 –6 = 750 MPa
Stress σ(B) = F/A = 565/1.5 × 10 –6 = 380 MPa
d No, the stress in each wire is lower than the tensile strength of steel (820 MPa).
10 Adding the forces gives a right-angled force triangle with a vertical weight force of 1000 N.
Using this, it can be determined that:
F A = 980/sin30° = 1.9 × 10 3 N
F B = 980/tan30° = 1.7 × 10 3 N
Stress σ(A) = F/A = 1.9 × 10 3 /1.5 × 10 –6 = 1300 MPa
Stress σ(B) = F/A = 1.7 × 10 3 /1.5 × 10 –6 = 1100 MPa
These stresses are both greater than the tensile strength of steel, so both cables would fail.
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Chapter 7 Materials and their use in structures
7.7 Torque
1 a The axis of rotation is the spindle. The size of the lever arm is approximately 3 cm.
b The axis of rotation is the front wheel axle. The size of the lever arm is usually around
1 m.
c The axis of rotation at the end of the tweezers. The size of the lever arm is usually a few
centimetres.
d The axis of rotation is where the screwdriver is in contact with the rim of the tin. The size
of the lever arm is usually around 15–30 cm.
2 a The magnitude of the torque produced by a given force is proportional to the length of the
lever arm. By pushing the door at the handle, rather than the middle, the length to the
lever arm is increased.
b A crowbar can be used to generate a large torque because the force can be applied at a
large distance from the pivot.
3 Torque provided by Renee is: τ = F × r = 392 × 2.25 = 882 N m
4 Torque must exceed 400 N m, so: F × r = 400
F × 1.6 = 400
F = 400/1.6 = 250 N
5 a Weight force of load = mg = 2500 × 9.80 = 24 500 N
τ = F × r = 24 500 × 20 = 4.9 × 10 5 N m
b The crane will have a counterweight providing an opposite torque.
6 a τ = F × r = 9.8 × 0.50 = 4.9 N m
b τ = F × r = 9.8 × 1.0 = 9.8 N m
c τ = F × rsinθ = 9.8 × 1.0 × sin30° = 4.9 N m
7 The long pole with the heavy weight at the end can be used to provide a large counter-torque
should the performer overbalance. Only a small movement of the pole is need to balance the
torque produced by the performers body when they overbalance.
8 It will not work because the centre of mass lies outside the base of support. To make the design
work, the bench or supports should be moved so that the centre of gravity is between the
supports. . Otherwise, he could use bolts to attach the beam to the left-hand support.
9 a The weight of the suitcase will produce a torque about a pivot point around the base of
the spine which will tend to rotate your torso to the right. You compensate for this by
leaning to the left or extending your left arm.
b Force = mg = 14 × 10 = 140 N m
Assume that you are holding the suitcase approximately 0.5 m from spine.
τ = F × r ≈ 140 × 0.50 ≈ 70 N m
10 a Weight of skip: F g = mg = 3500 × 9.80 = 3.4 × 10 4 N
b The effective lever arm remains at 15 m throughout, so the torque does not change.
c τ = F × rsinθ = 3.43 × 10 4 × 25 × sin37° = 5.1 × 10 5 N m clockwise about the pivot.
(or simply find: τ = F × r = 3.43 × 10 4 × 15 = 5.1 × 10 5 N m clockwise about the pivot.)
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Worked solutions
Chapter 7 Materials and their use in structures
7.8 Structures in translational and rotational equilibrium
1 Torque produced by first child is: τ = F × r = 200 × 1.2 = 240 N m
Torque produced by second child is: τ = F × r = mg × 1.5 = 240 N m
mg = 240/1.5 = 160 N
mass m = 16 kg
2 To generate less torque, the adult should sit near the pivot. The child should sit at end of beam.
3 = (19.6 × 0.25) + (49 × 0.50) = 29 N m
4 a Torque of cable acting around axis is: τ = F t × 10sin30° = 5 × F t
b F t (v) = F t sin30°, F t (h) = F t cos30°
c Assume that each cable supports half the weight of the bridge i.e. 3430 N
Taking torques around the axis: (F t sin30° × 10) – (3430 × 5.0) = 0
F t = 17 200/5.0 = 3440 = 3.4 × 10 3 N
5 Call support forces F L and F R .
Total of downwards forces = (100+150+200+50) × 9.80 = 4900 N
Total of upwards forces: F L + F R = 4900 N
Taking torques around left-hand end:
–(980 × 3.0) – (490 × 5.0) – (1470 × 7.0) – (1960 × 8.5) + (F R × 10) = 0
F R × 10 = 32,300
F R = 3200 N = 3.2 kN up
F L = 4900 – 3200 = 1700 N = 1.7 kN up
6 The beam and the train have a weight of 49 000 N each.
Taking torques around end X: –(4.9 × 10 4 × d) – (4.9 × 10 4 × 10) + (30 600 × 20) = 0
d = 2.5 m from end X or 17.5 m from end Y.
7 a τ = F × rsinθ = 157 × 2.4 × sin25° = 160 N m
b τ = F × rsinθ = 490 × 1.2 × sin25° = 250 N m
c τ = F × rsinθ = 490 × 3.6 × sin25° = 750 N m
8 a The vector sum of all the forces is zero i.e. F X + F Y +F g = 0
Taking the magnitudes of these forces gives: F X + F Y = 900 × 9.80 = 8,820 N
b (F Y × 1.8) – (8820 × 2.0) = 0
c F Y = 17 640/1.8 = 9,800 N = 9.8 kN up
d –(F X × 1.8) + (8820 × 0.2) = 0
e F X = 1760/1.8 = 980 N down
Y is being squashed beneath the beam and so is under compression. X is exerting a
downwards force on the beam and so is under tension.
9 a Taking magnitudes of forces: F L + F R = 686 + 196 = 882 N
Taking torques around left-hand end: (686 × 1.50) + (196 × 2.50) – (F R × 5.00) = 0
F R = 1520/5.00 = 300 N up
F L = 882 – 300 = 580 N up
b Taking torques around left-hand end: (686 × d) + (196 × 2.50) – (325 × 5.00) = 0
d = 1135/686 = 1.7 m from left-hand end.
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Worked solutions
Chapter 7 Materials and their use in structures
10 a i The three forces acting on the left-hand cantilever are the weight of the beam
(3920 N), the force from pillar A (F A ) acting downwards and the force from pillar
B (F B ) acting upwards
F A + 3920 = F B
Taking torques around the left-hand end of the left cantilever:
(F B × 1.2) – (3920 × 1.5) = 0
F B = 5880/1.2 = 4.9 kN up
So: F A = 980 N down
By symmetry: F C = 4.9 kN up and F D = 980 N down
ii The force that the pedestrian F P exerts on the end of the cantilever is 350 N down.
F A + 4000 + 343 = F B
Taking torques around the left-hand end of the left cantilever:
(F B × 1.2) – (4000 × 1.5) – (343 × 3.0)= 0
F B = 7030/1.2 = 5860 = 5.9 kN up
So: F A = 1520 N = 1.5 kN down
By symmetry: F C = 5.9 kN up and F D = 1.5 kN down
b As the woman walks from A to B, the force acting in pillar A decreases and the force
acting in B increases. When the woman passes point B and continues on to point P, the
forces in both A and B increase, in order to produce a greater torque to counterbalance the
increase in torque as she moves to point P.
11 The weight of the beam is 49 N.
Taking torques around the left-hand end of the beam: (F t sin45° × 1.2) – (49 × 0.9) = 0
F t = 44.1/0.85 = 52 N
12 a F h = 784sin30° = 390 N
F v = 784cos30° = 680 N
b Taking torques around base of post: (F t × 1.0) – (392 × 0.85) = 0
F t = 330 N
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Worked solutions
Chapter 7 Materials and their use in structures
Chapter review
1 Strength = maximum stress prior to failure = 350 × 10 6 = 350 MPa so D is the correct option.
2 Elastic limit = stress at end of non-linear region = 300 × 10 6 N m –2 so C is correct.
3 E = gradient of elastic region = 300 × 10 6 /1.5 × 10 –3 = 2.0 × 10 11 N m –2 so C is correct.
4 First, find area up to elastic limit X.
Area = ½ × 1.5 × 10 –3 × 300 × 10 6 = 225 000 J m –3
Energy = area × volume of wire = 225 000 × (1.5 × 2.0 × 10 –6 ) = 0.675 J so D is correct.
5 First find area up to point Y.
Area (28 squares) = 28 × 50 × 10 6 × 1.0 × 10 –3 = 1 400 000 J m –3
Energy = area × volume = 1 400 000 × (1.5 × 2.0 × 10 –6 ) = 4.2 J so E is correct.
6 This is plastic behaviour so B.
7 Weight F g = 7.5 × 9.8 = 74 N, distance r = 3.0 m
Torque τ = (74cos65°) × 3.0 = 94 N m so E is correct.
8 Torque τ = F × r = (60 × 9.8 × cos65°) × 2.0 = 500 N m so A is correct.
9 Balanced forces so F t = mg = 4500 × 9.8 = 4.4 × 10 4 N m = 44 kN. B is the correct option.
10 The forces are again balanced so F t = mg. B is correct.
11 a E = stress/strain = F/A / Δl/l =
1000
! ! 0.0010
0.00049 0.20
2
= 1.3 × 10 11 Pa
b % strain = Δl/l × 100 = 0.49/200 × 100 = 0.25%
c Young’s modulus depends only on the properties of the material itself whereas the spring
constant depends on the nature of the material as well as its length and thickness.
12 a Stress at elastic limit = F/A = 2.0 × 10 6 Pa
5
1.0!
10
2
!r
= 2.0 × 10 6
r 2 5
1.0!
10
=
6
! ! 2.0!
10
= 0.0159
r = 0.125 m = 12.5 cm
b
6
2.0!
10
Strain at elastic limit = stress/E =
10
2.0!
10
= 1.0 × 10 –4
Area under graph to elastic limit = 0.5 × 2.0 × 10 6 × 1.0 × 10 –4 = 100 J m –3
Energy = area under graph × volume of material = 100 × 4.0 = 400 J
c Concrete is much stronger under compression than tension. A column is always under
compression, whereas a beam can be under both compression and tension.
d The concrete is poured over a lattice of steel rods and sets around them. This results in the
formation of a composite material called reinforced concrete, which is suitable for use in
beams and flooring.
13 Maximum stress = 3.0 × 10 8 Pa from graph
Heinemann Physics 12(3e)
Copyright © Pearson Education Australia 2008
Teacher’s Resource and Assessment Disk
(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263
14

Worked solutions
Chapter 7 Materials and their use in structures
14 a When stress = 2.0 × 10 8 Pa, strain = 2.0 × 10 –3 from graph.
Strain ε = Δl/l
2.0 × 10 –3 = Δl/0.0800
Δl = 1.6 × 10 –4 m = 0.16 mm
b Maximum stress = 3.0 × 10 8 = F/A
F = 3.0 × 10 8 × 1.0 × 10 –4 = 3.0 × 10 4 N
c Stress at elastic limit = 2.0 × 10 8 = F/A
F = 2.0 × 10 8 × 1.0 × 10 –4 = 2.0 × 10 4 N
Safety factor = 5, so maximum force under working operations must be 5 times lower
4
2.0!
10
than force at elastic limit: Maximum force = = 4.0 kN
5
15 Area under graph up to breaking point = 16.4 squares × 0.5 × 10 8 × 1.0 × 10 –3 = 8.2 × 10 5 J m –3
Volume of material = 1.0 × 10 –4 × 0.0800 = 8.0 × 10 –6 m 3
Strain energy = area under graph × volume of material = 8.2 × 10 5 × 8.0 × 10 –6 = 6.6 J
16 a Stiffness is the resistance of the material to being stretched or compressed.
b Rank in order of Young’s modulus: nylon, aluminium, steel
c Steel: E = 2.0 × 10 11 = σ/ε
5
1.0!
10
Strain =
11
2.0!
10
= 5.0 × 10 –7
Strain energy per unit volume = area under graph = 0.5 × 1.0 × 10 5 × 5.0 × 10 –7 =
0.025 J m –3
Aluminium: E = 7.0 × 10 10 = σ/ε
5
1.0!
10
Strain =
10
7.0!
10
= 1.5 × 10 –6
Strain energy per unit volume = area under graph = 0.5 × 1.0 × 10 5 × 1.5 × 10 –6 =
0.071 J m –3
Nylon: E = 7.0 × 10 8 = σ/ε
5
1.0!
10
Strain = = 1.4 × 10 –4
8
7.0!
10
Strain energy per unit volume = area under graph = 0.5 × 1.0 × 10 5 × 1.4 × 10 –4 = 7.0 J m –
3
d
The area under the graph up to the point of fracture is greater for steel than it is for
aluminium.
17 The weight of the globe is 15 N. Let us call the left-hand wire A and the wire on the right B.
Adding the 3 force vectors gives a right-angle triangle. This can be used to determine F A and
F B .
F A = 15/sin40° = 23 N
F B = 15/tan40° = 18 N
18 Taking the magnitudes of the forces: F L + F R = 58.8 + 19.6 + 127.4 + 9 8 = 303.8 N
Taking torques around the left end of the beam:
–(58.8 × 2) – (147 × 5.0) – (198 × 7.0) + (F R × 10.0) = 0
F R = 1540/10.0 = 154 N
F L = 310 – 157 = 150 N
19 a Torque created by load is: τ = F × r = (1.5 × 10 3 × 9.8) × 25 = 3.7 × 10 5 N m
Desired torque created by counterweight is: τ = (20 000 × 9.80) × r = 3.7 × 10 5
Heinemann Physics 12(3e)
Copyright © Pearson Education Australia 2008
Teacher’s Resource and Assessment Disk
(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263
15

Worked solutions
Chapter 7 Materials and their use in structures
b
Location of counterweight is r = 1.9 m from pivot.
This reduces the torque acting on the crane making it less likely to topple over.
20 a τ = F × r = 250 × 1.0 = 250 N m –1
b Y, the force will cause the barrier to bend slightly causing compression at X and tension
at Y. Concrete is weaker under tension, so the barrier is more likely to crack at Y.
c The steel rods are strong under tension and reduce the amount of bending and cracking
that occurs at Y.
Heinemann Physics 12(3e)
Copyright © Pearson Education Australia 2008
Teacher’s Resource and Assessment Disk
(a division of Pearson Australia Group Pty Ltd) ISBN 9781442501263
16