UNIT 1 - BONE
Dr M J Dolan
Department of Orthopaedic & Trauma Surgery
University of Dundee
Dr T Drew
Department of Orthopaedic & Trauma Surgery
University of Dundee
SECOND EDITION
Edited by
Dr T Drew
Department of Orthopaedic & Trauma Surgery
University of Dundee
Illustrations by
Mr I Christie
Published by
Distance Learning Section
Department of Orthopaedic & Trauma Surgery
University of Dundee
Second Edition published 2005: ISBN 1-903562-49-X
ISBN 978-1-903562-49-9
First Edition published 1995: ISBN 1-899476-54-7
Copyright © 2005 University of Dundee. All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,
mechanical, photocopying, and recording or otherwise, without prior permission from the publisher.
The University of Dundee is a Scottish Registered Charity, No SC015096.
UNIT 1 - BONE
CONTENTS
1. COMPOSITION AND STRUCTURE
2. BASIC MECHANICAL PROPERTIES
2.1 Stress
2.2 Strain
2.3 Stress-strain Curves
2.4 Young’s Modulus
2.5 Typical Values
3. TYPES OF LOADING
3.1 Shear Loading
3.2 Bending Loading
3.3 Torsional Loading
3.4 Combined Loading
4. INFLUENCE OF MUSCLE ACTIVITY
5. REMODELLING
6. FATIGUE FRACTURES
7. AGE-RELATED CHANGES
SUMMARY
SAQ ANSWERS
END OF UNIT EXERCISE
UNIT 1 - BONE
OBJECTIVES
On completing your study of this unit you should be able to:
1. Describe the material composition and structural organisation of compact and
cancellous bone.
2. Explain how the material composition and structural organisation of bone affects its
strength.
3. Define the terms: “tensile loading”, “compressive loading”, “stress and strain”.
4. Compare and contrast the basic mechanical properties of bone with everyday
materials.
5. Identify and describe the different types of mechanical loads that act on bones in the
human body.
6. Explain the importance of cross-sectional shape in bending and torsion loading.
7. Describe the influence of muscle action on the distribution of stress within bone.
8. Quote Wolff’s law.
9. Describe the influence of exercise and weightlessness on bone mineralisation and
strength.
10. Define a fatigue fracture and explain how they arise.
11. Explain why excessive muscle fatigue may contribute to a fatigue fracture.
12. Describe the influence of age-related changes on the biomechanics of bone.
13. Discuss the differences between age-related changes in men and women.
UNIT 1 - BONE
Bone tissue is a specialised type of connective tissue. It is one of the hardest tissues in
the human body, only dentine and enamel in teeth are harder. Bones, which are made of
bone tissue, perform a number of important functions. They support the structures of the
body, protect delicate structures such as the heart and lungs, and act as lever arms for
movement.
BONE TISSUE
There are a wide range of different bone sizes and shapes. These are grouped into four
different types according to their structure: long bones, short bones, flat bones and
irregular bones. Each one is adapted to perform its specific function; for example the
skull bones are flat to contain and protect the brain whilst the limb bones are long to act
as lever arms for movement.
In this unit we will be examining the basic mechanical properties of bone tissue and
how the structure of bone is adapted to support the structures of the body and allow
movement. In particular we will look at how bones are designed to prevent fractures.
But first the composition and structure of bone will be reviewed.
1. COMPOSITION AND STRUCTURE
Bone tissue is composed of bone cells called osteocytes, a non-cellular organic
component and an inorganic component.
OSTEOCYTES
The non-cellular organic component consists of very strong collagen fibres embedded
in a jelly-like matrix called ground substance. The collagen fibres account for 95% of
the non-cellular component of bone and for about 25 to 30% of the dry weight of bone.
Collagen fibres are flexible but resist stretching.
COLLAGEN FIBRES
The inorganic component consists mainly of the minerals calcium and phosphate.
These are in the form of crystals of calcium phosphate which are deposited within the
matrix. It is the high content of this inorganic component that gives bone its
characteristic hardness and rigidity. The inorganic component accounts for 65 to 70% of
the dry weight of bone.
GROUND SUBSTANCE
INORGANIC
COMPONENT
CALCIUM PHOSPHATE
SAQ 1
(a) What are bone cells called?
(b) Which component of bone gives it its characteristic hardness and
rigidity?
Bones generally consist of two very different types of bone tissues:
 compact bone
 cancellous bone
Compact bone forms the outer layer of bones and has a dense structure. It is also called
cortical bone.
Cancellous bone forms the inner part of short, flat and irregular bones. In long bones it
lines the inner surfaces and makes up the greater part of the metaphyses and epiphyses.
Cancellous bone has a mesh like structure which gives it its other common name spongy bone. The spaces in the mesh contain red bone marrow.
The relative quantities of compact and cancellous bone vary between bones and even
within an individual bone according to the functional requirements. This variation can
easily be observed in the femur. The head and greater trochanter are covered only in a
very thin layer of compact bone yet the shaft has a much thicker layer (Figure 1). Notice
the pattern in the cancellous bone, the cells tend to align themselves in the directions
that will best support the load.
Unit 1 - Bone 1
COMPACT BONE
CANCELLOUS BONE
cancellous bone
compact
bone
compact bone
compact bone
FIGURE 1. SECTION THROUGH HEAD OF A FEMUR.
In compact bone, the basic structural unit is the haversian system arranged
longitudinally in columns of around 200 m in diameter (Figure 2). In these units, the
bone tissue is arranged in layers called lamellae forming concentric cylinders around a
central canal. The small central channel, called a haversian canal, contains blood
vessels and nerve fibres. Between the lamellae there are small cavities, called lacunae,
that contain osteocytes. Each osteocyte is linked to the haversian canal and other
lacunae by minute channels, called canaliculi, along which nutrients are carried from
the blood vessels. These canaliculi are about 0.2 m in diameter. Collagen fibres
interconnect the layers of lamellae within the haversian system. Each haversian system
is surrounded by a cement-like ground substance, calcified mucopolysaccharide, from
which collagen seems to be absent. This is the weakest part of the bone’s microstructure, probably due to the absence of collagen fibres.
HAVERSIAN SYSTEM
LAMELLAE
HAVERSIAN CANAL
LACUNAE
CANALICULI
haversian canal
haversian system
(A)
cancellous bone
compact bone
haversian canal
lacuna containing
a bone cell
(B)
lamellae
canaliculi
200 m
FIGURE 2. (A) A SECTION THROUGH THE SHAFT OF A LONG BONE (B) AN HAVERSIAN SYSTEM.
In cancellous bone, the basic structural unit is the trabecula. Trabeculae are arranged in
a latticework of branching sheets and columns (Figure 3A). Trabeculae are similar in
structure to haversian systems consisting of layers of lamellae with lacunae containing
osteocytes connected by canaliculi. The main difference is that trabeculae do not
Unit 1 - Bone 2
TRABECULAE
contain haversian canals (compare Figure 2B with Figure 3B). Haversian canals are not
needed in cancellous bone as blood vessels pass though the marrow filled spaces
between the latticework of trabeculae, supplying nutrients to the osteocytes through the
canaliculi. The latticework structure of cancellous bone allows it to vary greatly from
site to site according to the mechanical requirements. As a consequence of this the
density of cancellous bone may vary from as little as 0.1 g cm-3 to as much as 1.0 g
cm-3. However, even at its most dense, the density of cancellous bone does not approach
the density of compact bone which is typically around 1.8 g cm-3. It is also interesting to
note that cancellous bone with a density of 0.2 g cm-3 has a porosity of around 90%, i.e.
it is mainly space!
spaces containing
marrow and blood
vessels
(A)
trabeculae
canaliculi
lamellae
(B)
lacuna containing bone cell
FIGURE 3. (A) TRABECULAR STRUCTURE OF CANCELLOUS BONE (B) SECTION THROUGH A TRABECULA.
The overall structure of both compact and cancellous bone is non-uniform - they both
differ along different directions and planes. As a consequence of this their mechanical
properties vary accordingly (see Section 2.4). Bone tissue is therefore often described as
being anisotropic. An anisotropic material being one that does not exhibit uniform
mechanical properties in all directions. Conversely a material which does exhibit
uniform behaviour is said to be isotropic.
ANISOTROPIC
SAQ 2 - Which type of bone tissue contains haversian canals and why
are they needed?
2. BASIC MECHANICAL PROPERTIES
The behaviour of a material, such as compact bone, can be described and to some extent
predicted using its mechanical properties. The mechanical properties describe the way
the material reacts when it is loaded.
The simplest way in which a material can be loaded is in tension or in compression:
In tension the load is acting to stretch the material like in a rope.
TENSION
Unit 1 - Bone 3
In compression the load is acting to compress the material like in the supporting
column of a building.
COMPRESSION
Two examples of the tensile and compressive loadings that can occur in bones are
shown in Figure 4. The radius and ulna are both put under a tensile load when an object
is lifted as shown in Figure 4A and the vertebrae are under a compressive load due to
the upper body weight as shown in Figure 4B.
F
(B)
(A)
F
F
F
tensile load
compressive load
FIGURE 4. (A) TENSILE LOAD ON THE RADIUS AND ULNA (B) COMPRESSIVE LOAD ON THE VERTEBRAE.
To help describe the behaviour of a material under tensile and compressive loads two
parameters, called stress and strain, are used.
2.1 Stress
Stress is defined as the force per cross-sectional area.
STRESS
Stress is a measure of the concentration of the load being supported by a material. As
the magnitude of the load increases so does the stress in the material supporting the
load. For example, if the load on a supporting column is doubled then the stress will
also be doubled. If the area over which a load is distributed is increased then the stress
will decrease. For example, if the cross-sectional area of a supporting column is doubled
then the stress in the column will be halved.
The definition of stress can be expressed in the form of an equation as:
Stress =
Force
Area
Stress has the SI units newtons per metre squared (N m-2). Remember that the unit for
force is the newton (N) and the unit for length is the metre (m).
2.2 Strain
Strain is defined as the change in length divided by the original length.
STRAIN
Unit 1 - Bone 4
Strain is a measure of the amount of deformation a material has undergone. It is the
ratio of the change in length of the material to the original length of the material. Thus,
the larger the deformation the larger the strain. For example, if the length of a rubber
band is doubled, the change in length of the rubber band is equal to its original length
and therefore the strain is equal to one.
The definition of strain can be expressed in the form of an equation as:
Strain =
change in length
original length
Strain has no units, since it is a ratio of two lengths.
Worked Example
A specimen of human cortical bone is subjected to a tensile load of 360 N (Figure 5).
The specimen is found to stretch by 0.05 mm. Given that the specimen had an original
length of 10.0 mm and a cross-sectional area of 4 mm2, calculate the stress and strain in
the bone.
F = 360 N
2 mm
2 mm
10 mm
10.05 mm
F = 360 N
FIGURE 5. CORTICAL BONE SPECIMEN.
The cross-sectional area of the specimen is expressed in millimetres and not metres.
There are 1000 mm in 1 m, thus to convert the area from mm 2 into m2 we need to divide
by 1000 × 1000:
Area =
4
= 4 × 10-6 m2
1000  1000
The definition of stress can be expressed in the form of an equation as:
Stress =
Force
Area
Unit 1 - Bone 5
Thus substituting in the values for the applied force and cross-sectional area:
Stress =
Force
360
= 90 MN m-2

Area 4  106
The definition of strain can be expressed in the form of an equation as:
Strain =
change in length
original length
Thus substituting in the values for the change in length and the original length:
Strain =
change in length 005
= 0.005

original length
100
The stress in the cortical bone sample is 90 MN m-2 and the strain is 0.005.
SAQ 3
(a) What happens to a material under a tensile load?
(b) What is the definition of strain?
(c) What are the SI units of stress?
2.3 Stress-strain Curves
As you may already suspect, stress and strain are not independent. If a force is applied
to a material then there will be a resulting deformation. The relationship between stress
and strain is not always a simple one but for many materials we can make some
simplifications.
A useful tool in conveying the mechanical properties and behaviour of a material is a
stress-strain curve. It consists of a graph of stress plotted against strain, showing how
the material deforms as it is loaded. Different materials exhibit different stress-strain
relationships according to their characteristic behaviour. Stress-strain curves therefore
also provide a useful way of comparing different materials to see which is relatively
more or less stiff, tough, ductile and/or brittle. Although some materials may
predominantly exhibit one of these behaviours, they will all exhibit all of these
behaviours to a certain degree depending upon the magnitude of the load to which they
are subjected.
STRESS-STRAIN CURVE
A typical stress-strain curve for cortical bone is shown in Figure 6. You will notice that
the stress increases with increasing strain - as the bone is increasingly deformed it
becomes increasingly harder to deform it further. The stress-strain curve is divided into
two regions: the elastic region and the plastic region. The division between the two
regions is marked by the yield point A. The amount of strain at the yield point is called
the yield strain and the amount of stress is called the yield stress.
In the elastic region, the curve is linear - the stress is directly proportional to the strain.
Thus if the strain is doubled in the elastic region the stress will also double. Provided
that a bone specimen is not deformed beyond its yield point by a load then it will return
to its original size and shape once the load is removed. This is termed elastic
behaviour.
ELASTIC REGION
In the plastic region, the curve is not linear. The bone yields to the applied load - for a
small increase in stress the bone deforms by a large amount. When a bone specimen is
deformed beyond its yield point, it will not completely recover its original size and
shape when the load is removed - it is permanently deformed. This is termed plastic
behaviour.
PLASTIC REGION
Unit 1 - Bone 6
At the point B on the curve the bone fractures. The strain at this point is called the
ultimate strain and the stress is called the ultimate strength or ultimate stress.
ULTIMATE STRENGTH
stress
B
ultimate strength
fracture
A
yield stress
strain
elastic
region
yield strain
plastic
region
ultimate strain
FIGURE 6. STRESS-STRAIN CURVE FOR CORTICAL BONE.
2.4 Young’s Modulus
As you already know, the initial part of the stress-strain curve is linear - the stress and
strain are directly proportional. This means the stress is equal to the strain multiplied by
a constant:
stress = strain × constant
This constant is known as Young’s modulus. Rearranging the equation gives:
Young’s modulus =
stress
strain
Young’s modulus is the ratio of the stress to strain. It has the same SI units as stress, N
m-2, since strain has no units. It describes how flexible or stiff a material is. A material
with a small Young’s modulus requires only a small amount of stress to produce a large
strain; i.e. it is flexible. For example, rubber has a Young’s modulus of approximately
0.01 GN m-2. A material with a large Young’s modulus requires a large amount of stress
to produce a small strain; i.e. the material is stiff. For example, diamond has a Young’s
modulus of approximately 1200 GN m-2. On a stress-strain curve a stiff material will
have a larger gradient than a flexible material, since for the initial linear region the
gradient is equal to the Young’s modulus.
SAQ 4
(a) How many times greater is the Young’s modulus for diamond
than that for rubber?
(b) Calculate the Young’s modulus for the test sample in the Worked
Example in Section 2.2.
(c) Figure 7 shows the initial part of the stress-strain curves for
several materials along with the corresponding Young’s modulus.
How stiff or flexible is cortical bone compared to the other
materials?
Unit 1 - Bone 7
YOUNG’S MODULUS
stress
steel (210 GN m-2)
aluminium (70 GN m-2)
glass (50 GN m-2)
cortical bone (17 GN m-2)
wood (11 GN m-2)
strain
FIGURE 7. STRESS-STRAIN CURVES AND YOUNG'S MODULI FOR CORTICAL BONE, WOOD, GLASS, ALUMINIUM AND
STEEL.
As you already know bone tissue is anisotropic. This is particularly marked in compact
bone and as a consequence of this its mechanical properties vary according to the
direction in which it is loaded. This can be illustrated very nicely using stress-strain
curves, as in Figure 8 which shows the curves for four test samples of cortical bone.
Each test sample has been taken from the shaft of a long bone with a different
orientation to the vertical aligned haversian systems and loaded under tension. The
curves clearly demonstrate that compact bone is much weaker, having a lower Young’s
modulus and ultimate strength, when loaded transversely (T), across the haversian
systems, as compared to when it is loaded longitudinally (L), along the alignment of the
haversian systems.
test sample
stress
L
30°
60°
T
shaft of
long bone
strain
FIGURE 8. TENSILE STRESS-STRAIN CURVE FOR CORTICAL BONE SAMPLES TAKEN FROM THE SHAFT OF A LONG BONE
AT DIFFERENT ORIENTATIONS.
2.5 Typical Values
All the parameters mentioned so far can be determined for bone tissue by testing
samples of bone. Some typical values for these are given in Tables 1 and 2. It is
however, important to remember that the actual values for any particular sample of bone
tissue will be dependent on a number of factors. These will include:
 the age of the subject from which the sample was taken (see Section 7)
 the bone from which the sample was taken - each bone has to withstand different
loading patterns and will adapt accordingly (see Section 5)
Unit 1 - Bone 8
 the site from which sample was taken - within a bone the loading stresses will vary
from site to site and the bone tissue will adapt accordingly (see Section 5)
 the direction in which the bone is loaded - bone tissue is anisotropic (see Section
2.4)
 the density of the bone - this is particularly true for cancellous bone (see Section 1)
 the wetness of the sample - dry bone and wet bone have very different mechanical
properties
Compact bone
Young’s Modulus
(GPa)
Ultimate strength
(MPa)
Tension
17
130
Compression
18
200
Tension
13
50
Compression
12
130
Loading direction
Longitudinal
Transverse
Loading Mode
TABLE 1. TYPICAL MECHANICAL PROPERTIES OF COMPACT BONE.
Cancellous bone
Compressive
Compressive
Young’s Modulus (MPa)
Ultimate strength (MPa)
0.2
25
3
0.5
350
15
Density (g cm-3)
TABLE 2. TYPICAL MECHANICAL PROPERTIES OF CANCELLOUS BONE.
SAQ 5
(a) Why is cortical bone so weak when loaded transversely?
(b) Is cortical bone more or less flexible in compression than
cancellous bone? Justify your answer.
3. TYPES OF LOADING
So far we have only considered how bones are loaded under compressive and tensile
loads. However bones are subjected to many other types of loading (Figure 9). These
include shear, bending and torsion which will be examined in the following sections.
unloaded
tension
compression
bending
shear
torsion
FIGURE 9. TYPES OF LOADING.
3.1 Shear Loading
In shear loading two forces acting in opposite directions tend to cause layers within the
material to slip or shear. Two common examples of shearing that occur in orthopaedics
Unit 1 - Bone 9
SHEAR LOADING
are shown in Figure 10. The screw is being sheared by the fracture fixation plate and
bone, and the bone cement is being sheared by the hip prosthesis and bone.
femur
fracture fixation plate
hip prosthesis
sheared surface
shaft
of long
bone
screw
sheared surface
(A)
bone cement
(B)
FIGURE 10. (A) SHEARING OF A SCREW (B) SHEARING OF BONE CEMENT.
Any shear loading will give rise to shear stress and shear strain (Figure 11). Shear
stress (denoted by the character  [tau]) is defined as being equal to the magnitude of
the shearing force, F, divided by the sheared area, A:
shear stress,  =
SHEAR STRESS
shearing force F

sheared area
A
Shear strain (denoted by the character phi]) is defined as the angle sheared:
shear strain, = angle sheared
The SI unit of shear stress is the same as that for axial stress, i.e. the newtons per metre
squared (N m-2). The SI unit for shear strain is the radian (rad).
force
f
A
force
FIGURE 11. A CUBE SUBJECT TO SHEARING STRESS.
The shear modulus, also known as the modulus of rigidity, is equal to the gradient of
a shear-stress/shear-strain curve up to a limiting stress and is defined as:
modulus of rigidity =
shear stress
shear strain
Unit 1 - Bone 10
SHEAR STRAIN
The modulus of rigidity is measured in newtons per metre squared (N m-2) or pascals
(Pa) and is usually represented by the character G.
Human cortical bone is more flexible and weaker in shear than in tension or
compression. The shear modulus is only around 3 G N m-2 and the maximum shear
stress it can withstand (its ultimate shear strength) is only around 70 M N m-2 (compare
these with those given in Section 2.5). To illustrate the relative strengths of cortical
bone under different loadings, the bar chart in Figure 12 shows the strength for
compression, tension and shear.
200
200
150
ultimate
strength
(MN m-2)
130
100
70
50
compression
tension
shear
FIGURE 12. ULTIMATE STRENGTH OF CORTICAL BONE UNDER COMPRESSIVE, TENSILE AND SHEAR LOADINGS.
Despite the relatively low strength of bone in shear, fractures caused by shearing alone
are quite rare. One of these rare examples is an intra-articular shearing fracture of the
femoral condyles shown in Figure 13.
femur
fracture
patella
tibia
fibula
FIGURE 13. INTRA-ARTICULAR SHEARING FRACTURE OF THE FEMORAL CONDYLES.
SAQ 6
(a) What is the definition for shear modulus?
(b) Is cortical bone stronger in shear or tension?
Unit 1 - Bone 11
3.2 Bending Loading
In bending loading, loads are applied to a structure that tend to cause the structure to
bend. Bending can be achieved in number of ways. Two common types of bending are
cantilever and three point bending. In cantilever bending one end of the object is fixed
and a load is applied to the other end causing the object to bend. A diving board is an
everyday example of this type of bending. In three point bending three forces are
applied to the object (Figure 14A). A seesaw is an everyday example of three point
bending.
Force
BENDING LOADING
Force
neutral
axis
tensile
stress
(A)
compressive
stress
fracture
site
Force
Force
(B)
(C)
FIGURE 14. (A) NEUTRAL AXIS IN A SYMMETRICAL BEAM (B) BENDING OF THE FEMUR (C) FRACTURE OF THE FEMUR
CAUSED BY BENDING.
When a structure is bent, one side of it is elongated and the other is compressed.
Between the two sides of the structure there is a neutral axis along which no
deformation occurs. On one side of the neutral axis the material is elongated and on the
other side the material is compressed. In a symmetrical structure, such as the beam
shown in Figure 14A, the neutral axis is along the structure’s geometric centre.
However, in non-symmetrical structures the path of the neutral axis can be quite
complex. The shape of the femur causes it to be bent when it is loaded vertically, for
example during standing, with the medial side being compressed and the lateral side
elongated. In this case, the neutral axis in the bone shaft runs approximately along the
centre of the femur, as shown in Figure 14B.
When considering bending it is important to remember that bone is stronger under
compression than under tension (see Figure 12). Therefore when a bone is subjected to a
large bending load it will tend to fracture on the elongated surface of the bone which is
under tension (Figure 14C). Also the bone tissue on the outer surface will be deformed
most and therefore under most stress.
The maximum bending stress that a bone can withstand, its bending strength, is
dependent not only on the type and strength of the bone tissue, but also the bone’s
cross-sectional area and its cross-sectional shape. As you would expect, the stronger the
bone tissue and the larger the cross-sectional area, the stronger the bone is. The crosssectional shape is important because most of the load is carried by the bone tissue
furthest from the neutral axis. As a consequence of this the bone tissue closer to the
neutral axis carries very little load. It is therefore much more efficient to have the bone
tissue distributed as far as possible from the neutral axis. You can easily test this
yourself by bending a ruler holding it in the two ways shown in Figure 15. You can see
that there is more ruler material further away from the neutral axis in Figure 15B which
makes it more difficult to bend.
Unit 1 - Bone 12
NEUTRAL AXIS
(A)
(B)
cross sectional view
(full size)
FIGURE 15. BENDING A RULER IN THE HORIZONTAL PLANE.
The second moment of area is used to quantify how a material, in this case bone tissue,
is distributed. It is dependent upon the cross-sectional shape - the further the material is
concentrated away from its neutral axis the larger its second moment of area. For
example, Figure 16 shows the cross-section of four imaginary bone shafts. In each case
the same amount of bone tissue has been used - the cross-sectional areas are equal. The
second moment of area has been calculated twice - once for a bending in the horizontal
plane and one for a bending in the vertical plane. For the first two cases, these values
are different, but for the last two they are the same due to their symmetrical shape. The
imaginary bone with the largest second moment of area for bending in a vertical plane,
and therefore the largest bending strength, the I-shaped section bone, (B), has a similar
cross-sectional shape to the standard steel joists used for holding up floors and ceilings
in buildings. However, it is comparatively weak when bent in a horizontal plane. It is
therefore of little use as a bone since bones are generally loaded in many directions. The
bone with the hollow circular cross-section is therefore the best shape as it offers a high
bending strength in all directions. Not surprisingly, this design is similar to that found in
the shafts of all long bones which have a strong outer compact bone layer.
(A)
(B)
5.86 mm
(C)
(D)
30.0 mm
17.3 mm
15.0 mm
30.0 mm
26.0 mm
30.0 mm
3.00 mm
IHORIZONTAL
IVERTICAL
503
2640
2490
17300
13200
22500
2490
17300
FIGURE 16. THE CROSS-SECTIONS OF IMAGINARY BONE SHAFTS AND THEIR HORIZONTAL AND VERTICAL SECOND
MOMENT OF AREAS.
SAQ 7 - Check that the area of each of the cross-sections in Figure 16
are the same.
Unit 1 - Bone 13
A typical example of a bending fracture is the so called ‘boot top’ fracture sustained by
skiers. The boot top fracture is a result of three point bending. As the skier falls forward
over the top of the ski boot a force is exerted on the proximal end of the tibia. As the
distal end of the tibia is fixed in the boot, the tibia is bent over the top of the rigid ski
boot as shown in Figure 17. If the bending load is large enough, the tibia will fracture.
Hence development of quick release ski bindings!
Force
Force
Force
FIGURE 17. BENDING OF THE TIBIA IN A SKI BOOT.
SAQ 8
(a) Name two types of bending.
(b) What is the neutral axis?
(c) On which surface will a fracture most likely occur when an
excessive bending load is applied to a long bone?
3.3 Torsional Loading
Torsional loads occur when a bone is twisted about its longitudinal axis. Torsional
loads often occur when one end of the bone is fixed and the other end is twisted. If the
torsional load is excessive then torsional fractures can result which have a characteristic
spiral appearance as shown in Figure 18A. Torsional fractures of the tibia are quite
common in many sports such as football, rugby and skiing, occurring when the foot is
held in a fixed position and the rest of the body is twisted.
neutral axis
no distortion
(A)
(B)
most
distortion
FIGURE 18. (A) SPIRAL FRACTURE OF THE TIBIA CAUSED BY A TORSIONAL LOAD (B) TWISTING OF A SOLID BAR.
Unit 1 - Bone 14
TORSIONAL LOADS
When a structure is subjected to a torsional load the stress and strain within the structure
are not evenly distributed. If you examine the diagram in Figure 18B you will see that
the centre of solid bar is not distorted by the applied torsional load - there is a neutral
axis extending through the centre of the bar - and the outer surface of the bar is the most
distorted. In fact the stress and strain are greatest on the outer surface of the bar - just
like during bending. The material which forms the core of the bar is carrying only a
small proportion of the torsional load whilst the outer portion of the bar is carrying the
vast majority of the torsional load. This means that if the torsional load on the bar is
increased, the resulting fracture will start at the outer surface where the stress and strain
are the greatest.
Long bones are designed to resist torsional loads efficiently - they are hollow with
strong cortical bone forming the outer layer. This is the most efficient way of
distributing the bone tissue to resist torsional loads. If the same quantity of bone tissue
were used to construct a completely solid bone rather than a hollow one the bone would
be smaller in diameter and would be less able to resist torsional loads, i.e. it would
break easily. If the bone was the same size but completely solid then the weight of the
bone would increase dramatically with only a small increase in its capacity to resist
torsional loads. The hollow structure of bones maximises their strength-to-weight ratio
(Figure 19). The frames of mountain bikes are also constructed from large hollow bars
to maximise the strength-to-weight ratio.
strength
weight
strength-to-weight ratio
(A)
(B)
(C)
FIGURE 19. COMPARISON OF STRENGTH-TO-WEIGHT RATIOS IN THREE DIFFERENT BARS. THE SOLID BAR (B) HAS THE
SAME OUTSIDE DIAMETER AS THE HOLLOW BAR (A) GIVING IT A GREATER STRENGTH AND LARGER MASS BUT
OVERALL A LOWER STRENGTH-TO-WEIGHT RATIO. THE SOLID BAR (C) HAS THE SAME MASS AS BAR (A) GIVING IT A
LOWER STRENGTH AND THUS A LOWER STRENGTH-TO-WEIGHT RATIO.
Fractures of the tibia are often caused by torsional loads. Most of these fractures are
found to occur distally. The reason for this is quite simple - the distal cross-sectional
area of the tibia is smaller than the proximal cross-sectional area and although the
amount of bone tissue is the same, the distal part is less able to resist torsional loads and
therefore is most liable to fracture (Figure 20).
Unit 1 - Bone 15
STRENGTH-TO-WEIGHT
RATIO
neutral axis
proximal
tibia
site of
torsional
fracture
distal
FIGURE 20. DISTAL TORSIONAL FRACTURE SITE OF THE TIBIA.
SAQ 9 - When considering torsional loads, do hollow or solid bars
have the best strength-to-weight ratio?
3.4 Combined Loading
In reality, bones are rarely subjected to only one type of loading; they are usually
subjected to a combination of two or more types of loading. The presence of more than
one type of loading is known as combined loading. Combined loadings result from
irregular geometry of bones, and the combined actions of gravitational forces, muscle
forces and ligament forces. For example, the irregular geometry of the femur means that
when a compressive load is applied to the head of the femur a combined loading results,
consisting of compressive and bending loadings. For example, examine the loading of
the femur in Figure 14B. The two forces applied to the femur produce both bending and
compression. It is important to note that fractures usually result from combined
loadings. Rarely can a fracture be attributed to one type of loading.
SAQ 10
(a) Name five types of loadings.
(b) What is combined loading?
4. INFLUENCE OF MUSCLE ACTIVITY
So far the way we have considered bones to be loaded has been quite unrealistic. You
will know from your knowledge of anatomy that the tendons of muscles are attached to
bones at many different points and that muscles contract to produce movements. When
a muscle contracts it will also load the bone in addition to any external loadings such as
those acting at the joints. The load applied by the muscle alters the distribution of stress
within the bone.
Muscles will often contract, not to cause movement, but to alter the stress distribution
within a bone. If a muscle contracts, producing a compressive load on a bone, it can
eliminate any tensile loading and produce an overall compressive loading on the bone.
Thus, since bones are stronger in compression than in tension the likelihood of a
fracture will be reduced.
Unit 1 - Bone 16
COMBINED LOADING
Consider when three point bending is applied to the tibia as shown in Figure 21. One
side of the tibia is in compression and the other is in tension. Now if the soleus muscle
contracts it will produce a compressive load on the tibia by pulling downwards on the
proximal end of the tibia. This compressive load is superimposed on the bending load.
The overall effect is to reduce or completely eliminate the tensile load and produce an
overall compression load throughout the cross-section of the tibia. This will result in a
higher compressive stress on the anterior surface of the tibia. However, as the bone is
stronger in compression, a fracture is less likely.
soleus
compression
tension
compression
FIGURE 21. INFLUENCE OF THE CONTRACTION OF THE SOLEUS MUSCLE.
There are many other examples of muscle activity beneficially altering the stress
distribution in the bones. An important case occurs during reciprocal gait. The loading
on the hip joint during the stance phase produces a tensile stress on the superior aspect
and a compressive stress on the inferior aspect of the femoral neck (Figure 22A).
However, the contraction of the gluteus medius muscle, which lies superior to the
femoral neck, produces a compressive stress that effectively neutralises the tensile stress
(Figure 22B). The overall result of the muscle action is that there is no tensile stress in
the femoral neck which allows the femoral neck to withstand much higher loads than
would otherwise be possible.
F
gluteus
medius
muscle
F
tensile
stress
F
compressive
stress
(A)
(B)
FIGURE 22. (A) STRESSES IN THE FEMORAL NECK DURING STANCE PHASE OF GAIT (B) THE CONTRACTION OF THE
GLUTEUS MEDIUS MUSCLE PRODUCES A COMPRESSIVE FORCE WHICH REDUCES THE TENSILE STRESS ON THE
INFERIOR ASPECT OF THE FEMORAL NECK.
Unit 1 - Bone 17
Tired athletes are more likely to fracture a bone than when they are fresh because
their muscles are fatigued and they are therefore less able to control the distribution
of stress within their bones.
SAQ 11 - Why is it sometimes desirable for muscles to contract even
when it is not to produce movement?
5. REMODELLING
Bone is not a dead inorganic material. It is served by a rich blood supply and has the
ability to repair itself and also to remodel itself; altering its size, shape and structure in
response to changes in the mechanical demands placed on it. Compact and cancellous
bone tissue is continually gained or lost in response to the amount of stress placed on
the bone. This phenomenon is expressed in Wolff’s law:
WOLFF’S LAW
Bone is laid down where needed and resorbed where not needed.
There are a number of different circumstances that result in remodelling. During
physical exercise, such as jogging, the bones are subjected to increased levels of stress.
The bones respond to this increase by laying down more collagen fibres and mineral
salts to strengthen the bones. Conversely, inactivity and lack of exercise leads to the
resorption of bone tissue. This is called bone atrophy. This can particularly be a
problem for patients who are bedridden or wheelchair bound for long periods - once
they begin to reuse their legs they are liable to fracture their lower limb bones quite
easily.
In orthopaedics, bone remodelling can cause problems. In fracture fixation, a plate is
fixed to the broken bone to immobilise it during healing. During fracture healing the
plate will carry most of the load on the limb. If the plate is not removed soon after the
fracture has healed then the bone will weaken as unstressed bone tissue is resorbed. This
is because the plate is carrying most of the load rather than the bone itself. This is
termed stress shielding. At the points in the bone where the screws are inserted the
opposite will happen. The bone will strengthen as the bone tissue at these sites will be
carrying a greater load than normally. The increase in bone tissue is called bone
hypertrophy.
fixation plate
bone hypertrophy
bone atrophy
FIGURE 23. REDISTRIBUTION OF BONE TISSUE UNDER A FRACTURE FIXATION PLATE.
Unit 1 - Bone 18
BONE ATROPHY
STRESS SHIELDING
BONE HYPERTROPHY
SAQ 12
(a) Can you think of any problems astronauts may suffer from if they
spend long periods in the weightless environment of outer
space?
(b) What is the difference between bone hypertrophy and bone
atrophy?
The importance of bone remodelling in sport was highlighted in a study of professional
tennis players (Jones et al., 1977, Humeral hypertrophy in response to exercise, J. Bone
Joint Surg. [Am.] 59:204-208). In the study, the inner and outer humeral diameters were
measured (from radiographs) in both the players’ playing and non-playing arms. The
average results are summarised in Table 3.
Males
Females
Playing
Non-playing
Playing
Non-playing
Average inner diameter (cm)
0.98
1.10
0.87
0.96
Average outer diameter (cm)
2.45
2.20
2.07
1.90
Thickness (cm)
0.74
0.55
0.60
0.47
TABLE 3.
There is both an increase in the outer diameter and a decrease in the inner diameter for
the playing as compared to the non-playing humerus (Figure 24). These two factors
result overall in a much increased thickness of the cortex. In men, for example, the
thickness was increased by 34%. These results show that there is a highly significant
amount of humeral hypertrophy as a consequence of the chronic stimulus of
professional tennis playing.
non-playing
playing
2.20
2.45
1.10
0.98
0.96
0.87
1.90
2.07
FIGURE 24. AVERAGE HUMERAL CROSS-SECTIONS FOR NON-PLAYING AND PLAYING ARMS OF MALE AND FEMALE
PROFESSIONAL TENNIS PLAYERS.
Unit 1 - Bone 19
What is the structural consequence of the humeral hypertrophy? As you already know
the bending strength is dependent on the cross-sectional area and the second moment of
area (see Section 3.2). Assuming that the diameter is uniform these can be calculated as:
A = (rOU2 - rIN2)
where
A
is the cross-sectional area
rIN
is the inner radius, and
rOU is the outer radius.
For example, for the male playing humerus:
  245 2  098  2 
2

A = (rOU2 - rIN2) = 3.14 ×  
  = 3.96 cm
  2 


2


The second moment of area (I) of a hollow bar can be calculated using the formula:
I=

(rOU4 - rIN4)
4
For example, for the male playing humerus:

314
   245
 098 
(rOU4 - rIN4) =

 


 2 
4
4  2 
4
I=
4
 = 1.72 cm4


Similarly the values for the other cases can be calculated (Table 4). These values
demonstrate that due to bone remodelling the bending strength of the playing humerus
is much larger than the non-playing. Consequently, for any specific loading, the stresses
in the playing humerus will be much less than the stresses in the non-playing humerus
as it contains more bone tissue which is also better distributed to carry bending loads.
Males
2
Cross-sectional area (cm )
4
Second moment of area (cm )
Females
Playing
Non-playing
Playing
Non-playing
3.96
2.85
2.77
2.11
1.72
1.08
0.87
0.59
TABLE 4.
Note: In Britain and Europe the property of a structural cross section that quantifies
its resistance to bending is called 2nd moment of area - units mm4. Rather
confusingly American text books call this area moment of inertia, or sometimes just
moment of inertia. The quantity we use to describe a body’s resistance to changes in
angular motion (moment of inertia) the Americans call the mass moment of inertia.
SAQ 13
(a) What is the percentage increase in the thickness of humeral
cortex in the female tennis players?
(b) Check that the cross-sectional area and second moment of area
values given in Table 4 are correct.
Unit 1 - Bone 20
6. FATIGUE FRACTURES
Bone fractures can either be caused by a single large load that exceeds the ultimate
strength of the bone or by a smaller load that is applied repeatedly. A fracture resulting
from the repeated application of a load that is smaller than the ultimate strength of the
bone is called a fatigue fracture. Fatigue fractures are also commonly known as stress
fractures and march fractures. The latter name derives from the fatigue fracture of the
second metatarsal which is often suffered by young army recruits after long marches
(Figure 25).
fatigue fracture
FIGURE 25. FATIGUE FRACTURE OF THE SECOND METATARSAL.
If a graph is plotted of the applied load against the number of repetitions required for
that load to cause a fatigue fracture, then it will look something like the graph shown in
Figure 26. The graph illustrates that if the load is small, A, then a great number of
repetitions is required to cause a fatigue fracture, and if the load is large, B, then very
few repetitions are required to cause a fatigue fracture.
B
injury occurs
above the line
magnitude of
applied load
safe
region
A
number of repetitions
FIGURE 26. FATIGUE INJURY CURVE.
As you have already discovered in the previous subsection, bone has the ability to
remodel itself to match the required load. Because of this ability, the frequency of
repetition (the number of repetitions in a given time) is also important. If the repetitions
are well spaced then the bone will have time to remodel itself and repair any damage. A
fatigue fracture will therefore only result when the frequency of repetition is too fast for
the remodelling process. For this reason fatigue fractures are usually only sustained
during a continuous period of strenuous physical activity. For example, long-distance
runners who train excessively can suffer from fatigue fractures of the metatarsals, the
Unit 1 - Bone 21
FATIGUE FRACTURE
tibia, the femoral neck and the pubis, and gymnasts can suffer from fatigue fractures of
the vertebrae.
Muscle fatigue is also an important contributory factor to fatigue fractures arising
during sustained periods of activity. This is because during strenuous physical activity
the muscles become fatigued as the activity proceeds and they become less able to
neutralise the tensile stresses exerted on the bones.
Also, with increased muscle fatigue there is a general loss of shock absorbing capacity
and movements are often altered resulting in abnormal loadings. These effects, taken
together, produce an increased likelihood of a fatigue fracture.
SAQ 14
(a) What are the commonly used names for fatigue failure?
(b) How is a fatigue fracture caused?
7. AGE-RELATED CHANGES
Bone is actively changing throughout life. There is a constant process of bone formation
and bone resorption. In young adults this process is balanced so that the total amount of
bone tissue does not alter.
In children, there is more bone tissue formation than resorption as they grow and
develop. Children’s bones also differ from young adults’ bones in that they contain a
greater proportion of collagen than adults’ bones. This greater proportion of collagen
gives children’s bones greater flexibility. In other words, children’s bones are less brittle
than adults. It is for this reason that greenstick fractures are more common in children
than in adults. A greenstick fracture is an incomplete fracture whereby one side of the
bone is bent and the other side is buckled (Figure 27). They are usually caused by
excessive bending or torsional loads.
buckling
periosteum
bone
ruptured
periosteum
incomplete fracture
FIGURE 27. A GREENSTICK FRACTURE.
From the age of around 35 to 40 years, bone tissue begins to be lost as resorption
exceeds formation. There is some thinning of the compact bone tissue and a larger
reduction in the amount of cancellous bone tissue due to the thinning of the longitudinal
trabeculae and the resorption of some transverse trabeculae (Figure 28). This process
reduces the strength of the bone tissue. This reduction may be as much as 15% over the
60 years between ages 25 to 85 years.
Unit 1 - Bone 22
GREENSTICK
FRACTURES
transverse trabecula
transverse trabecula lost
(A)
(B)
longitudinal trabecula
thinning trabeculae
FIGURE 28. CANCELLOUS BONE TISSUE (A) TOP PICTURE SHOWS THE LONGITUDINAL AND TRANSVERSE TRABECULAE
IN A YOUNG ADULT (B) SHOWS THE THINNING OF TRABECULAE AND LOSS OF TRANSVERSE TRABECULAE IN AN
ELDERLY ADULT.
The reduction in the strength of bone tissue should mean that an elderly person is more
likely to break their bones than a young adult when subjected to the same loading.
Figure 29 shows typical stress-strain curves produced using complete bones of elderly
and young adults. The first section of the curves are similar, however, the curve for
elderly bone terminates at about half the strain of the young bone, illustrating the brittle
nature of elderly bone. Nevertheless, there is only a slight reduction in the ultimate
strength of the bones. The reason for this is that, not only are there changes in bone
tissue with age, there are also changes in the overall structure of bones.
stress
elderly
young
adult
strain
FIGURE 29. STRESS-STRAIN CURVES FOR YOUNG ADULT AND ELDERLY ADULT BONE.
The loss of strength of bone tissue in older adults is compensated for by changes in the
shape of the bone with age. This change manifests itself as changes in the diameter of
the inner and outer bone cortex. Typical values for the femur are given in Table 5 for
males and females aged 25 and 85 years. The thickness of the cortical bone, its crosssectional area and second moment of area have also been calculated (in the same
manner as in Section 5).
Unit 1 - Bone 23
Males
Females
25 years
85 years
25 years
85 years
Average inner diameter (cm)
1.50
2.00
1.25
1.50
Average outer diameter (cm)
3.00
3.50
2.50
2.50
Thickness (cm)
0.75
0.75
0.62
0.50
1.32
1.62
0.91
0.79
0.23
0.41
0.11
0.10
Cross-sectional area (cm )
2
Second moment of inertia (cm )
4
TABLE 5.
In males, both the outer and inner diameter of the cortex increases with age, however,
overall there is no change in the thickness of the cortex. Consequently there is an
increase in cross-sectional area and second moment of inertia with age. Thus, in males,
the structural changes more than compensate for the loss of tissue strength, so that risk
of fracture is generally decreased with age. However, in females the outer diameter
remains fairly constant with age, whilst, the inner diameter increases, resulting in a
reduction in cross-sectional area and second moment of area. Thus, in females, the
structural changes do not compensate for the decreases in tissue strength, but in fact
they contribute to an increased risk of fracture with age.
SAQ 15
(a) How would you recognise a greenstick fracture?
(b) Why are children prone to greenstick fractures but not adults?
(c) How is cancellous bone tissue affected by ageing?
(d) By what percentage does bone tissue strength decrease with age
from 25 to 85?
Unit 1 - Bone 24
SUMMARY
In this unit you have been introduced to the tissue mechanics of bone.
You have discovered that bones are subjected to a number of different types of loading
including tensile, compressive, shearing, bending and torsional. However, these
loadings are rarely applied in isolation. Instead a combination of different types of
loadings are usually applied as a result of the structure of the bones and the action of
muscles.
You have seen that the action of muscles is important, as they can compensate for the
comparative weakness of bone tissue in tension. The repetitiveness with which a load is
applied and the age of the person are also important factors in determining whether or
not a load will result in fracture.
You have also discovered that although the quantity of bone tissue is important, the
cross-sectional shape can be more important for bending and torsional loads. This is
highlighted in the changes in bone structure with age, which in men compensates for the
loss in strength of the bone tissue itself.
Finally, it is important to remember that bone is very different from any inorganic
structural material, not only in its microscopic structure, but also in that it can remodel
itself and so adapt to changing loads.
Unit 1 - Bone 25
SAQ ANSWERS
SAQ 1
(a) Bone cells are called osteocytes.
(b) The inorganic component, which consists mainly of crystals of calcium
phosphate, gives bone its characteristic hardness and rigidity.
SAQ 2
Compact bone tissue contains haversian canals. These contain blood vessels
which are needed to supply the bone tissue with nutrients.
SAQ 3
(a) When a material is under a tensile load it will elongate.
(b) Strain is defined as the change in length divided by the original length.
-2
(c) The SI units of stress are newtons per metre squared (N m ).
SAQ 4
(a) The Young's modulus for diamond is 120,000 times greater than that of
rubber (from 1200/0.01).
(b) Using the definition for Young's modulus:
Young's modulus =
stress
strain
Inserting the values calculated in the Worked Example in Section 2.2:
Young's modulus =
90
-2
= 18000 MNm
0005
= 18 GNm
-2
-2
Thus, the Young's modulus for the cortical bone sample is 18 GNm .
(c) Cortical bone is stiffer than wood (it has a larger Young's modulus than
wood) but more flexible than glass, aluminium and steel (it has a smaller
Young's modulus than all of these).
SAQ 5
(a) The haversian systems are only weakly bound together by a cement-like
ground substance (Section 1). This is the weakest part of cortical bone's
microstructure so when it is loaded transversely, the haversian systems
can be pulled apart comparatively weakly.
(b) Cortical bone is less flexible in compression than cancellous bone as
indicated by its larger Young's modulus.
Unit 1 - Bone 26
SAQ 6
(a) Shear modulus is the ratio of shear stress to shear strain.
(b) Cortical bone is stronger in tension than in shear.
SAQ 7
Area of rectangular cross-section:
A = depth × breadth = 30.0 × 5.86 = 175.8 = 176 mm
2
Area of I-shaped cross-section:
AMIDDLE = depth × breadth = (30.0 - 6.00) × 3.00 = 72.0 mm
AEND = depth × breadth = 17.3 × 3.00 = 51.9 mm
2
2
A = AMIDDLE + 2AEND = 72.0 + 2 × 51.9 = 178.8 = 175.8 = 176 mm
2
Area of solid circle:
2
 150 
2
2
A =  r = 3.14 × 
 = 176.625 = 176 mm
 2 
Area of hollow circle:
  300 
 260 

= 3.14 ×  

  2 
 2 

2
A =  (r
2
OUTER
-r
2
INNER)
= 3.14 × (225 - 169) = 175.84 = 176 mm
2



2
2
All the cross-sections have the same area of 176 mm calculated to three
significant figures.
SAQ 8
(a) Three point bending and cantilever bending are two types of bending
loading.
(b) The neutral axis is the line through a structure along which no distortion
occurs.
(c) When a bar is subjected to a bending load a fracture will most likely occur
at the outer surface.
SAQ 9
When considering torsional loads, a hollow bar has a better strength-to-weight
ratio than a solid bar.
SAQ 10
(a) Five types of loading are: tensile, compressive, bending, shearing and
torsional.
(b) Combined loading is when more than one type of loading is present.
Unit 1 - Bone 27
SAQ 11
It is sometimes desirable for muscles to contract even when it is not to produce
movement as they can alter the distribution of stress in bones to reduce the
amount of tensile stress and potentially prevent any damage to the bones.
SAQ 12
(a) If astronauts spent long periods in the weightless environment of outer
space then they may suffer from bone atrophy if they do not undertake
exercises that will help to stress their bones.
(b) Bone hypertrophy means an increase in bone tissue and bone atrophy
means a decrease in bone tissue.
SAQ 13
(a) Percentage increase =
=
change in thickness
× 100%
original thickness
 060  047
047
× 100%
= 28%
(b) For example, for the female non-playing arm:
4
2
  190
 
 096  
2
2

A = (rOU - rIN ) = 3.14  


 2 
 2  

= 2.11 cm
I
=
2
4
4

314   190
 
 098  
4
4

(rOU - rIN ) =



 2  
4
4   2 
4
= 0.59 cm .
SAQ 14
(a) Fatigue fractures are also commonly known as stress fractures and march
fractures.
(b) A fatigue fracture is caused by a load being applied repeatedly over a short
period of time.
SAQ 15
(a) A greenstick fracture is characterised by an incomplete fracture with one
side bent and the other buckled.
(b) Children are prone to greenstick fractures because their bones contain a
greater proportion of collagen than adult bones which gives their bones a
greater amount of flexibility.
(c) The amount of cancellous bone is reduced with ageing.
(d) Bone tissue strength decreases by around 15% from age 25 to 85 years.
Unit 1 - Bone 28
END OF UNIT EXERCISE
1.
(a) What percentage of the dry weight of bone is made up of:
(i)
collagen
(ii)
inorganic components?
What accounts for the rest of the dry weight?
(b) What are the mechanical characteristics of collagen?
2.
[6 marks]
(a) Describe the structure of cancellous bone tissue.
(b) How does the structure of cancellous bone tissue differ from cortical
bone tissue?
(c) How does the structure of cancellous bone tissue change with age?
[10 marks]
3.
(a) What do tensile and compressive loads do?
(b) For stress and strain write down the following:
(i)
Definition
(ii)
Equation
(iii)
SI units.
(c) If the compressive load on a bone is doubled, how will the stress and
strain be affected?
(d) What will happen to a bone when its ultimate strength is exceeded?
[12 marks]
4.
Compare and contrast the typical values for cortical and cancellous bone
given in Tables 1 and 2.
[6 marks]
5.
(a) State Wolff's law.
(b) Define the terms “bone atrophy” and “bone hypertrophy”.
(c) Compare and contrast the differences between male and female
tennis players using the data given in Tables 3 and 4.
[10 marks]
6.
Discuss the importance of the age-related changes to overall cortical bone
structure.
[6 marks]
Total = 50
Unit 1 - Bone 29