Download Biomechanics 2012

Biomechanics, LTH, 2012
Biomechanics 2012
Responsible teachers: Ingrid Svensson, Div. of Solid Mechanics, (Hållfasthetslära), Mhouse, 5:th floor, south part. Telephone: 070 234 10 29, [email protected]. Hanna
Isaksson, Div. of Solid Mechanics, [email protected] .
Assistants: Lorenzo Grassi, [email protected] , Neashan Mathavan,
[email protected]
Course literature:
Basic Biomechanics of the Musculoskeletal System, 3-rd or 4-rd edition, M. Nordin
and V.H. Frankel, Price: Approx. 500 skr. (Note: You can find it at internet
bookstores!)
• Introduction to Biomechanics for technical university students, Ingrid Svensson, 2012
• Illustrative exercises for the lectures, 2012
• Exercises for self-studies, 2012
• Assignment 1, 2 and IBL
The last four items are found on the homepage:
http://www.solid.lth.se/education/courses_swedish/biomekanik_fhl110/.
•
Times and places: Monday 10-12 first week in Sparta C and D, weeks 2-4 in M:Q, week 5
in KC:F, week 6 KC:B, week 7 M:E, Wednesday 15-17 in M:B, Thursday 13-15 in M:E.
You find a preliminary schedule in detail on the next page.
Types of teaching: Lectures where traditional lecturing is mixed with supervised problem
solving and exercises for independent problem solving with access to assistants for
discussions of the exercises, quizzes and assignments.
Examination: The examination of the course consists of three compulsory assignments. In
these assignments skills in modelling and computations as well as understanding of the
medical concepts presented during the course are tested. Assignment 1 and 2 are performed in
groups of 2-3 students. Assignment IBL is performed bigger groups (8-9 students) in
collaboration with students at the physiotherapist education program. The course is finished
by an individually written home exam that is orally defended. The home exam starts with a
short quiz on paper that is filled in individually without access to any books. The quiz gives
maximum 15 p, home exam with defence gives maximum 45 p and that means that the total
maximum is 60 p. Grades: < 30 gives Fail, 30-39.5 gives 3, 40-49.5 gives 4, 50-60 gives 5.
When solving the assignments three shorter reports are written and handed in and the result of
the third assignment is also presented orally at a seminar. The written reports are marked
Need Supplement or Passed. Please observe that all three assignments and three quizzes have
to be approved in order to get final mark on the course.
Good to know: The instructions for the three compulsory assignments are in the stack of
paper that you can buy when the course starts. Matlab is recommended for solving the
assignments. Magnus Tägil, surgeon at the Lund University Hospital, Eva Horneij,
physiotherapist, Lund University, are going to contribute during the course.
Check on the next page the themes written in italic and make sure you can attend at one of
them!
Biomechanics, LTH, 2012
Preliminary schedule (updated version!)
Week
1
2
3
4
5
Date
3/9
5/9
6/9
Monday
Wednesday
Thursday
7/9
Friday
10/9
12/9
Monday
Wednesday
13/9
Thursday
14/9
Friday
17/9
19/9
20/9
Activity and theme
L: Biomechanics, what is that?
L: Anatomy and physiology
L: Building stones of the human
body, bone and cartilage
Tasks
Quiz 1. Anatomy
Mail description
IBL (1 per group)
E: Statics and dynamics
L: Building stones of the human
body, tendons and ligaments,
muscles
L: Locomotion, the knee and the hip
(ankle, shoulder)
Example 1-5
Quiz 2. Mechanics
Monday
Wednesday
Thursday
E: Statics and dynamics
L: Mechanics of the spine
Collecting things up
Example 6-8
24/9
Monday
E: Deformable bodies
25/9
Tuesday
26/9
27/9
28/9
Wednesday
Thursday
Friday
1/10
Monday
3/10
4/10
6
Day
8/10
10/10
11/10
L: Gait and Sport Mechanics
Collecting things up
L: Biomaterial and
Tissue Engineering
Wednesday
Seminar* in E:B
Note! Try to book from 13.15-17.00!
Thursday
Seminar* in M:E
Note! Try to book from 13.15-17.00!
Monday
Wednesday
Thursday
L: Mechanobiology
L: Implants and prosthesis
L: The story of a surgeon.
Consult
PT-stud
by mail
Get answers from
PT-students
Hand
in
assignment 1.
of
Example 9-12
Get question from
PT-students
Answer questions
from PT-students
Consult PT-stud
Quiz 3. Mixed
Hand in of report
IBL
Presentation
of
IBL assignment
Presentation
of
IBL assignment
Hand
in
assignment 2.
of
Biomechanics, LTH, 2012
7
15/10
Monday
17/10
Wednesday
22/10
Monday
24/10 26/10
General feedback on the
assignments. Presentation of master
projects in biomechanics
Short Quiz exam 15.15-16.15
Book an appointment for defence!
L – lecture, E - Exercise
*) Note! You are supposed to attend one of these seminars.
Home exam is
handed out
Home exam is
handed in before
12.00
Biomechanics, LTH, 2012
Reading instructions (updated version!)
All chapters refer to: Biomechanics of the Musculoskeletal System, M. Nordin
and H. Frankel, forth edition.
Week 1. Anatomy and physiology, Building stones of the human body
Introduction to Biomechanics for engineering students
Biomechanics of Bone, chapter 2 (first part)
Biomechanics of Articular Cartilage, chapter 3
Week 2. Building stones of the human body, Locomotion
Biomechanics of Tendons and Ligaments, chapter 4
Biomechanics of Skeletal Muscle, chapter 6
Biomechanics of the Knee, chapter 7
Biomechanics of the Hip, chapter 8
Week 3. Locomotion
Biomechanics of the Lumbar Spine, chapter 10
Week 4. Gait and sport mechanics
Biomechanics of Gait, chapter 17
Hand-outs
Week 5. Biomaterial, Tissue Engineering
Hand-outs
Biomechanics of Bone, chapter 2 (last part)
Seminars with presentations of assignment 3
Week 6. Mechanobiology, implants and prosthesis
Hand-outs
Biomechanics of Fracture Fixation, chapter 15
Biomechanics of Arthroplasty, chapter 16
Guest lecture, hand-outs
Week 7.
Feedback and evaluation
Presentation of master projects in biomechanics
Biomechanics, LTH, 2012
Introduction to Biomechanics for engineering
students
by
Ingrid Svensson
2012
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Biomechanics, LTH, 2012
Hello and welcome to the course in Biomechanics!
During the next seven weeks you are going to get a glimpse in how the human body works
and get an understanding and training in how this can be expressed in mechanical terms. To
get you faster into the medical sphere of concepts I have put together some pages with “An
Intense Course in Anatomy and Physiology”. You can consider these pages as a framework
for the lectures in anatomy and physiology.
The text is aimed to suit students with different native languages so some key expressions are
given both in English, Latin and Swedish.
I hope that you will find the pages interesting and that you will enjoy the course!
Sources:
Människans anatomi och fysiologi, Bertil Sonesson och Gun Sonesson, Liber 1993.
Människans fysiologi, Olav Sand, Øystein V. Sjaastad, Egil Haug, 2002.
MEDICINE Engelsk-Svensk-Engelsk Fackordbok, P. H. Collin,Norstedts, 2002.
Lund in July 2012
Ingrid
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Biomechanics, LTH, 2012
An Intense Course in Anatomy and Physiology
Physiology is in general the science of the normal function of living creatures. In this course
we will limit our self’s and concentrate on the human body. In order to try to get an overview
of how all the complicated processes interact in the human body it is very useful to divide the
processes into some main groups. The groups often used are the digestion, the respiration, the
circulation, the locomotion, the neural system, the endocrine system, the lymphatic system,
the excretion system and the sexual organs. A principal engineering sketch of how all these
groups work together is presented in figure 1.
air food
matspjälkningsapparaten –
apparatus of digestive
andningsapparaten –
apparatus of respiration
cirkulationsapparaten –
apparatus of circulation
intercellulärt vätskerum –
intercellular room for fluid
kroppscell – body cell
blod plasma – blood plasma
intracellulärt vätskerum –
intracellular room for fluid
urinvägar – paths for urine
Fig. 1. Principal sketch of the functional structure of the human body
The cell is the smallest functional unit in living structures. A human being starts his life as a
unicellular organism in the moment of fertilization when the egg and the sperm melt together.
The fertilized egg carries predisposition to all organ systems that are going to be developed in
the becoming individual in the embryonic stem cells. The life functions of the cell go on in
the cell organs and in the human body the different organ systems take care of this. So, both
the single cell as well as the multi cellular organism, the human being, represents life on earth.
The benefit in nature in creating multi-cellular organisms of high complexity is the possibility
to get stable environments for individual cells, homeostasis. A human being is more capable
of adaptation to changes in the environment than the individual cells are.
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As 65% of the human body consist of water a lot of the cells in the body is surrounded by
water. But, the water appears both within the cells, intracellular, and out of the cells,
extracellular, see figure 1. The extracellular fluid is found in the intercellular room and in the
blood. The composition of this fluid (e.g. the salt content, the pH factor) and its temperature
constitute the inner environment of the body and has a great importance for the possibilities of
surviving for the cell.
Some of the main organ groups will briefly be described here and the intention is to give an
overview of the subject in order to make the mechanical modelling of the biological tissues
easier further on.
The circulatory system
vener – veins
artärer – arteries
kapillärer – capillaries
hjärta - heart
Fig. 2. The circulatory system
The circulatory system is built up by the heart (cor) and a system of arteries and veins and
makes the blood circulate in the body, see figure 2. The blood vessels that carry blood from
the heart are called arteries and the ones carry blood back to the heart are called veins. The
heart works as the pump in the system and makes the blood flow through contractions that
give pressure differences. The primary issue of the circulatory system is to distribute nutritive
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Biomechanics, LTH, 2012
substances and construction elements to the cells in the body and to carry away waste. But the
circulatory system is also responsible for heat control, it transfers pressure changes and it
distributes different signal substances that control and coordinates the function of the different
organ systems in the body.
The blood flow transports oxygen from the lungs and nutrients from the liver through the
elastic parallel-connected arteries and capillaries to the tissues. The exchange of the oxygen
for waste matter, such as carbon dioxide, takes place in the capillaries. The waste is taken
back to the lungs to be expelled (the big circulation). At the same time the blood obtains more
oxygen in the lungs to be taken out to the tissues (the small circulation, pulmonary
circulation).
So, actually, the heart works as two separate pumps and the circulation pattern is as follows:
blood returns through the veins to the right atrium of the heart; from there it is pumped by one
of the pumping functions trough the right ventricle into the pulmonary artery, and then into
the lungs. From the lungs it returns through the pulmonary veins to the left atrium of the heart
and it is pumped, by the other pump mechanism, from here through the left ventricle into the
aorta and from the aorta further on into the other arteries.
In rest, each half of the heart pumps about 5 litres of blood per minute. This corresponds to
the total volume of blood in the system. The importance of the circulatory system as a
transport system can be illustrated by the fact that unconsciousness will set in already 30-40
seconds after the heart stops to beat and there is a risk of permanent damage in the brain
tissues if the heart beat is stopped for more than about 3-4 minutes. But, the circulation of
blood to the brain has the highest priority and in a crisis, the blood flow is redirected through
changed flow resistance in other organs.
In the blood there are white corpuscles (leucocytes, leuko-white) that take part in the
immunodeficiency. Leucocytes are also present in the bone marrow, i.e. the marrow acts as a
storage space for them. The leucocytes defend the body against microorganisms through
antibodies. When the body is threatened by a serious infection, a big number of leucocytes are
released from the bone marrow and they invade the infected area. The red corpuscles
(erythrocytes) contain haemoglobin that carries oxygen to the tissues in one way and carbon
dioxide on the return. The transport capacity will of course decrease if the number of red
corpuscles is reduced from the normal value but if there are too many of them, the viscosity of
the blood increase and the heart has to work harder. Another type of blood cell is the platelet
(thrombocyte). The thrombocytes are involved in the haemostasis, i.e. the mechanisms
stopping the bleeding when blood vessels are ruptured. The thrombocytes form plugs that
mechanically stop the bleeding. But this process can also be destructive. A heart attack or a
stroke might occur if a vein or an artery is blocked by a blood clot (thrombosis). Fibrinogen
is a substance in blood plasma that produces fibrin, a protein, which helps the blood
coagulate.
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Biomechanics, LTH, 2012
The respiratory system
näshålan - the nasal cavity
struphuvudet - the voice box
(larynx)
luftstrupe - the wind pipe
(trachea)
lungorna - the lungs
(pulmo)
Fig. 3. The respiratory system
There has to be a constant flow of oxygen into the body to supply the process of oxidation of
nutrition so energy could be released. In human the exchange between oxygen and carbon
dioxide takes place in thin tissues in the lungs. The total area of these tissues is very big,
about 75-80 m2 for a grown up. The tissues are well protected in the moist environment. It has
to be moist because the gases must be resolved in liquid if they should be able to diffuse from
blood to air in the lungs.
The respiration process takes place in the organs of respiration: the nasal cavity, the throat
(pharynx), the voice box (larynx - struphuvud), the wind pipe (trachea – luftstrupe), the
bronchi (luftrör) and the lungs (pulmo - lunga). The two lungs are situated in the chest cavity,
protected by the rib cage and supported by the diaphragm. The heart is situated between the
lungs. The right lung has three lobes, the left only two. Air goes down into the lungs through
the trachea and bronchi. The cilia (flimmerhår) at the walls filtrates and moistures the air in
order to protect the easily damaged mucous membranes (slemhinnor) of the alveoli. The air
passes to the alveolar sacs. The alveolar sacs are made up of clusters of alveoli like grapes in a
bunch. The gas exchange, oxygen-carbon dioxide, occurs in the blood vessels that are
wrapped around each individual alveolus. The oxygen is needed for the metabolism of the
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Biomechanics, LTH, 2012
cells and through the respiration carbon dioxide is ventilated and balance of the pH factor is
obtained in the blood and other body fluids.
The respiratory system is also the base of the speech. Air is pressed from down under through
the narrow slit, the glottis between the cords (plica vocalis - stämband) in the voice box. A
pulsating air stream builds up in the glottis and propagates through longitudinal oscillations
into the resonance areas, i.e. the throat, the oral cavity, the nasal cavity, the sinus and the
chest, where the oscillations are filtered and amplified and the result is speech or song.
The difference in the voice of different people is controlled by the activity of several muscle
groups during speaking. Varying the activity in the respiratory muscles will influence the
volume of the sound. The frequency of the longitudinal oscillations is given by the stresses in
the cords that come from contractions of the muscles in the voice box. The airflow through
glottis makes the cords vibrate almost perpendicular to the airflow and thus the cords open
and close the slit. The slit is closed for a longer period than it is open when sound of low
frequency is created and the opposite goes for high frequencies. On top of that, the cords are
thinner for high frequencies than for low. The spectrum for the human voice goes from about
40 to 2000 Hz.
The apparatus of digestive
The food is processed mechanically and
chemically in the apparatus of digestive in
order to split the food so the nutrients, salts
and water can be assimilated. The process
starts with chewing in the oral cavity
where the food is divided into pieces and
mixed with saliva. The salvia consists of
more than 99% water but also mucin, a
compound of sugars and protein
facilitating the chewing. Enzymes, like
amylase that converts starch into maltose
and lysozyme that in combination with
antibodies reduces the amount of bacteria
in the oral cavity.
The process continues in the throat, in the gullet (oesophagus-matstrupe), in the stomach and
in the guts (intestine – tarmar). The partly digested food comes first to the small intestine
where nutrients are absorbed and continues then to the large intestine where most of the water
is absorbed, (9.9 litres of 10). The digestive tube (from mouth to anus) is about 7 m long.
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Biomechanics, LTH, 2012
In addition there are glands (körtlar) as the liver and the pancreas (bukspottskörtel)
involved in the chemical part of the digestion. Secretion containing different types of enzymes
necessary for the process is produced in the glands. The gall, secreted by the liver, is used to
digest fatty substances and to naturalize acids. The pancreas produces enzymes that convert
fat, carbohydrate and proteins into components that can be absorbed in the guts. The result is
energy released to the body and also distribution of components for heating, growing and
reconstruction. Leftover leaves the body as excrement and urine.
The apparatus of locomotion
The organ system in focus in this course is the apparatus of locomotion and we will go into
detail with the different parts further on in the course. However, in order to give a short
overview, some facts in general of the apparatus of locomotion are given here.
The organs for motions, i.e. the skeleton, joints and muscles, are responsible for more than
half of the body mass, i.e. the skeleton makes about 15 % and the muscles 45 % of the body
mass of an adult human. The skeleton gives stability for the construction of an upright
position and it also provides fixation sites for the soft tissues. The skeleton bones function as
shelter for inner organs, e.g. the brain and the heart, as well as a prerequisite of locomotion,
talk and the performance of outer work. More than 200 individual bones interconnected by
joints build up the skeleton. The bones also serve as depots of minerals (e.g. calcium and
phosphor) and further, the red blood cells are produced and stored in the red bone marrow
within the long skeletal bones.
To provide locomotion the skeletal bones are joined together through the joints that have a
very low friction due to the cartilage covered contact layer and a lubrication system. The
joints are stabilized through systems of supporting ligaments and tendons and the joints have
different amount of freedom of movement. In mechanical terms we can consider them as
friction free joints with a varied number of degrees of freedom.
The muscles are responsible for the posture, the balance and the motion of the body. They
also contribute to keeping the body warm through generating heat when contracting. The
skeletal system and the muscle system are strongly coupled both structurally and functionally.
The muscle tissue continues through the tendons into the bone tissue and ensures secure
fixations. Physiologically, the muscles and the bones are also coupled through the content of
calcium. The contractions in a muscle can only take place if the concentration of calcium is
within quite narrow limitations. Since the most of the calcium depot is in the skeleton,
diseases in the skeleton may also influence the muscle function. On the other hand, the
muscles also influence the skeleton. During hard muscle work, like intensive sport practicing,
the muscles are growing and the skeleton becomes denser, it gets stronger. On the contrary, if
you stay in bed, you soon will notice both smaller muscles and a weaker skeleton.
Some skeletal bones and muscles that we are going to mention often during the course are
given in the next pictures and the related table.
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Biomechanics, LTH, 2012
1.
2.
3.
4.
5.
6.
7.
8.
9.
20.
21.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
Fig. 4 Skeletal bones in Swedish, English and “Latin”
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
kranium, skalle
ansiktsskelett
nyckelben
skulderblad
överarmsben
bröstkorg
kotpelare
strålben
armbågsben
bäcken
handrotsben
mellanhandsben
fingerben
lårben
knäskål
vadben
skenben
fotrotsben
tåben
skull
face skeleton
collarbone
shoulder blade
bröstben
breastbone
chest
column
carpal bones
finger bones
kneecap
toe bones
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cranium
viscerocranium
clavicula
scapula
humerus
thorax
columna vertebralis
radius
ulna
pelvis
carpus
metacarpus
phalanges
femur
patella
fibula
tibia
tarsal
phalanges
manubrium
sternum
Biomechanics, LTH, 2012
Anterior view
Posterior view
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Biomechanics, LTH, 2012
Useful expressions in Latin
Bumps in the skeleton:
Capitulum, litet huvud, small round end of a bone
Caput, huvud, the head (in plural: capita)
Facies, yta, surface
Condylus, ledhuvud, rounded end of a bone that articulates with another
Spina, bentagg, bone prickle (spine)
Trochanter, benknöl, bony lumps (e.g. the two lumps on either side of the top
end of the femur where muscles are attached)
“Inverted bumps” in the skeleton:
Alveol, urholkning, hålighet, hollow
Cavum, håla, cavity
Fossa, grop, pit
Sinus, hålighet, hollow
Openings in the skelton
Apertura, öppning, opening
Fissura, springa, spricka, crack
Porus, öppning, opening
In vivo, levande, alive, an experiment that takes place in the living body
In vitro, död, dead, an experiment that takes place in the lab on isolated tissue
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Biomechanics, LTH, 2012
Planes of reference, motions, and directions
Sagittal plane – Divides body into right and left halves
Frontal plane – Divides body into front and back halves
Transverse plane – Divides body into upper and lower
halves
Terms
Posterior (Dorsal) – in frontal plane, towards the backside
Anterior (Ventral) – in frontal plane, towards the front
Superior (Caudal) – in transverse plane, above
Inferior – in transverse plane, below
Medial – in sagittal plane, towards the midline
Lateral – in sagittal plane, away from the midline
Superficial – towards the surface
Deep – away from the surface
Proximal – towards the trunk
Distal – away from the trunk
Palmar – palm of hand
Plantar – bottom of foot
**Supine (Anatomical) Position – standing erect, facing forforward, hands by side, palms facing forward
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Biomechanics, LTH, 2012
Directions
•
If a motion takes place in the transversal plane along the transversal axis and in
directed inwards the centre of the body it is called medial. If it has the opposite
direction, it is called lateral.
•
A motion in the frontal plane, along the vertical axis, directed upwards is called
superiort and if it is directed downwards inferiort
•
It is possible to move in the sagittal plane, along the sagittal axis, forwards is the
motion called anteriort and backwards it is called posteriort.
•
The part of a skeletal bone that is closest to the heart is called the proximal part while
the part that is most far from the hart is called the distal part.
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Biomechanics, LTH, 2012
Test your skills
Fill in the blank or provide a short answer
1. Groups of cells that have a common function are termed ______________.
2. The larynx is an organ of the _________________ system.
3. The system that has the ability to store minerals, such as calcium, is called the
________________ system.
4. The breakdown of ingested foods into simple molecules that can then be absorbed into
the bloodstream is called _______________.
5. The ____________________ refers to all the chemical reactions in the body.
6. The body’s ability to maintain stable internal conditions is referred to as
____________.
7. The navel is ____________ to the spine.
8. A cut that is made along the midline is called a ______________ section.
9. The function of the ____________system is to control the body functions via
hormones.
Answers:
1. Tissues 2. Respiratory 3. Skeletal 4.Digestion 5. Metabolism 6. Homeostasis
7. Ventral or anterior 8. Mid sagittal or median 9. Endocrine
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Biomechanics, LTH, 2012
Biomechanics
Illustrative exercises for the
lectures
Ingrid Svensson
2012
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Biomechanics, LTH, 2012
1. To practise the use of free-body diagram, consider the problem of analyzing the stress
in man’s back muscles when he does hard work. The figure below shows a man
shovelling snow. If the snow and the shovel weigh 10 kg and have a centre of gravity
located at a distance of 1 m from the lumbar vertebra of his backbone, what is the
moment about that vertebra?
The construction of a man’s backbone is sketched in the figure above. The disks
between the vertebrae serve as the pivots of rotation. It may be assumed that the
intervertebral disks cannot resist rotation. Hence the weight of the snow and shovel
has to be resisted by the vertebral column and the back extensor muscle. Estimate the
loads in his back muscle, vertebrae, and disks.
Lower back pain is such a common affliction that the loads acting on the disks of
patients were measured with strain gauges in some cases. It was found that no
agreement can be obtained if we do not take into account the fact that when one lifts a
heavy weight, one tense up the abdominal muscles so that the pressure in the abdomen
is increased. A free-body diagram of the upper body of a man is shown in the figure.
Show that it helps to have a large abdomen and strong abdominal muscles.
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Biomechanics, LTH, 2012
2. Compare the bending moment acting on the spinal column at the level of a lumbar
vertebra for the following cases:
a) A student at LTH when buying beer, lifting the case with (i) the knees straight
and (ii) the knees bent.
b) A water skier skis with (i) the arms straight and (ii) with elbows hugging the
sides.
Discuss these cases quantitatively with proper free-body diagrams.
3. A tree is to be considered as a uniform cylinder with a specific gravity of 0.9 and a
diameter of 0.3 m. Young’s modulus is 1.379 GPa and the mass is 100 kg determine
and the mass of the trunk can be ignored.
How long can the tree be without loosing its straight posture as equilibrium?
4. Many biological materials are built up of more than one constituent, e.g. bone can be
considered as a mixture of collagen and mineral. The individual stiffness and
structural arrangement of the constituents will give the elastic modulus of the mixture.
Since the structural arrangement is quite complex, two different mathematical models
known as the Voigt and Reuss models, are often used to derive the upper and lower
limit for the mixture. The Voigt model assumes that the strain is equal in the two
constituents whereas the Reuss model assumes equal stresses.
Express the total elastic modulus for the mixture if the elastic moduli for the
constituents are E1 and E2 and the volume fractions are V1 and V2.
Assume the volume fractions of apatite in bone to be 45 % (elastic modulus 114 GPa)
and consequently the volume fraction of collagen to 55 % (elastic modulus 1.3 GPa).
Derive the upper and lower limits for the stiffness in bone and compare these values
with experimental data.
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Biomechanics, LTH, 2012
5. Consider the question of the strength of human tissues and their margin of safety when
we exercise. Take the example of the tension in the Achilles tendon in our foot when
we walk and when we jump. The bone structure of the foot is shown the figure. To
calculate the tension in the Achilles tendon we may consider the equilibrium of forces
that act on the foot. The joint between the tibia and the talus bones may be considered
as a pivot.
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Biomechanics, LTH, 2012
6. The head in the picture below is considered as a rigid body for the present purpose. It
rocks on the occipital condyles C where an axial force FA, a shear force, V, and a neck
torque To, resist motion. An uppercut (treated as an instantaneous load B at an angle of
63o to the horizontal) is applied to the chin, and the initial linear acceleration, a, of the
mass centre, G, is photographically determined to be 140g. For a head mass of 3.5 kg,
a moment of inertia about an axis perpendicular to the sagittal plane of 0.0356 kg m2,
and the dimensions shown, what are the reactions at the occipital condyles if torque
T0, which requires muscle activation, is temporarily neglected?
7. Resilin is a protein found in e.g. insects where it functions as a very effective energy
storage. Studies presented in literature show that it is able to store about 2Ÿ106 J/m3
when loaded near the failure point. Compare this ability with steel with the stress of
425 MPa and elastic modulus 200 GPa. (Assume that the steel is loaded to the yield
point and has approximate linear elastic behaviour.)
Consider a flea with the mass 0.45Ÿ10-6 kg having dual resilin springs containing a
total of 3Ÿ10-13 m3 resilin. How high can it jump?
5
Biomechanics, LTH, 2012
8. The cervical disc can be considered as a Kelvin-Voigt solid. In the cervical region, the
average section of the fist six discs is 150 mm2 and the average thickness is 5 mm.
The Young’s modulus (spring constant) of the disk is 30 MPa and the viscous constant
is 25 Ns/m. A mass of 20 kg is suddenly placed and maintained for some time on the
head. Assume that the entire load is carried by the discs and determine:
a) the strain in the discs after 10 s
b) the total change in length of the neck (cervical region) after 10 s
9. A serious dental problem is bruxism, i.e. the unconscious sideways gnashing of teeth
that produces major material damage by flattening out sharp teeth as well as producing
erosion to the root. In order to quantify this problem, consider the digestion of food by
biting involving (i) vertical forces only, fig. 1, and (ii) sideways chewing involving
horizontal loads which act cyclically on the tooth creating this horizontal motion, fig.
2. Assume that the two force components required to chew the food are the same, 50
N. The root of the idealized tooth has a width of 4 mm and a depth of 3 mm. The root
is located 10 mm below the surface where the load is applied.
a) What is the direct stress acting on the root of the tooth for case (i)?
b) What is the maximum bending stress acting on the root of the tooth for case
(ii)? The area is the same as in (a).
c) What is the factor of safety with sideways chewing after 1 year and 50 years if
200 such motions occur per day ?
Figure 1. Simple model of a tooth.
Figure 2. Fatigue curve.
6
Biomechanics, LTH, 2012
10. The safety factor for a total hip replacement is to be determined.
(inches)
The problem is to estimate the actual load on the prosthesis. For this purpose, a study
of an ordinary hip is performed. To get a “worst static case” the situation of standing
on one leg is considered. The free-body diagram is presented below.
Make an estimation of the size and direction of the joint reaction force using the freebody diagram in the hip in order to get the loading to use in the calculation of the
safety factor.
7
Biomechanics, LTH, 2012
Sources:
• Biomechanics, Mechanical Properties of Living Tissues, Y.C. Fung, 1993
• Introduction to Bioengineering, S.A. Berger, W.Goldsmith, E.R. Lewis, 2000
• Fundamentals of Machine Elements, B.J. Hamrock, B. Jacobson, S.R. Schmid, 1999
8
Biomechanics, LTH, 2012 Biomechanics Exercises for self-­‐study Ingrid Svensson 2012 1 Biomechanics, LTH, 2012 1.
Two students are trying to move a heavy stone
block. Assume that the students are equally
strong and apply forces with the same magnitude
of 200 N. One of the students is pushing the
block toward north and the other is pushing to the
east. Determine the magnitude and net force
applied by the students on the block.
2.
The students decide to change strategy and find a
rope that they tie in a ring in the corner of the
block. One of the students takes the rope and pull
with a force of 200 N in the northeast direction
and the other is now pushing with the same
magnitude and direction. Determine the
magnitude and net force applied by the students
on the block.
3.
The course book rests on a table. Calculate the
average pressure applied by the book on the table
top.
Hint: The mass of the book is 1.1 kg.
4. A student, Tom, with one hand in plaster after a wrist fracture, wants to move the
course book on the table by pushing it. Assume that the static coefficient of
friction between the book and the table top is 0.3. What is the minimum force
required to move the book over the table?
2 Biomechanics, LTH, 2012 5.
Tom works on rehabilitating his wrist and uses the exercise apparatus that you see
in the figure. He holds a handle that is attached to a cable. The cable is wrapped
around a pulley and attached to a weight pan. The weight in the weight pan
stretches the cable and produces a tensile force ! in the cable. This force is
transmitted to Tom´s hand through the handle. The force makes an angle ! with
the horizontal and is applied to the hand at the point B in the figure. Point A
represents the centre of gravity of Tom´s lower arm and the point O represents the
centre of rotation of the elbow joint.
Assume that the points O, A and B and the force ! all lie in the same plane and
the magnitudes of the forces ! and ! are 50 N and 20 N in respectively.
Determine the net moment generated about O by ! and !.
Dimensions: ! = 15 cm, ! = 35 cm, ! = 30!
6.
When reshuffling, Tom is using a rope and pulls a removal enclosure over a rough
floor. Assume that the mass of the block is ! = 50 kg, the magnitude of the force
is ! = 150 N and the angle between the rope and the horizontal is ! = 30! . The
coefficient of kinetic friction between the block and the floor is ! = 0.2 .
Determine the acceleration of the block if the bottom surface remains in contact
with the floor throughout the motion.
3 Biomechanics, LTH, 2012 7.
The figure shows Tom at the gym, doing shoulder abduction in the frontal plane. O
represents the axis of rotation of the shoulder joint in the frontal plane, line OA
represents the position of the arm when it is stretched out parallel to the ground
(horizontal), line OB represents the position of the arm when the hand is at its
highest elevation and line OC represents the position of the arm when the arm is
closest to the body. Assume that the angle between OA and OB, !! , is equal to the
angle between OA and AC so the motion is symmetric with respect to the line OA.
Also assume that the time it takes for the arm to cover the angles between OA and
OB, OB and OA, OA and OC, and OC and OA are approximately equal.
Derive expressions for the angular displacement, velocity, and acceleration of the
arm. Draw simple graphs of the expressions as functions of time. Take the time
period of angular motion of the arm to be 3 s and the angle !! to be 80! .
8.
Lisa is on winter vacation and gets the opportunity to try ski
jumping. A simplified sketch of the circular region of the ski
jump track is shown in the figure. The radius of curvature of the
track at this region is ! = 50 m. At the end of the track, the
direction normal to the track coincides with the vertical and the
direction tangential to the track coincides with the horizontal.
Consider Lisa (! = 70 kg) at the position in the figure and
decelerating at a rate of 1.5 m/s2 due to air resistance. If the
friction of the track is negligible and Lisa reaches the end of the
track with a horizontal velocity of ! = 20 m/s, determine the
forces applied on Lisa by the air resistance and the track.
4 Biomechanics, LTH, 2012 9.
Lisa is working with mechanical testing during the summer.
She is testing a circular cylindrical rod with radius ! =
1.26 cm in the lab in a uniaxial tension test. Before applying a
tensile force of ! = 1000 N, Lisa marks two points A and B
on the rod, separated a distance !! = 30 cm. After the force is
applied, she measures the distance between A and B to
!! = 31.5 cm.
Determine the tensile strain and stress generated in the rod.
Assume that the loading is in the elastic region, yielding has
not occurred, and give an estimation of the Young’s modulus
for the material of the rod.
10.
In the lab, they also test a fixation device consisting of a plate
and two screws that is used for stabilization of fractures.
Estimate the shear stresses in the screws due to the body
weight when standing well in balance on two feet, dividing the
load equal on both feet. The diameter of the screws is ! = 5
mm and the weight of the patient is ! = 700 N.
What happens with the shear stresses in the screws if a fourscrew fixation device is used?
11.
When working on the design of a new hip prosthesis, a bench
test is set up in the lab. The distal end of a human femur is
clamed to the bench and a horizontal force with magnitude 500
N is applied to the head of the femur at point P.
Determine the maximum normal and shear stresses generated
at a vertical distance ℎ = 16 cm measured from point P.
Compare the magnitude of the maximum of the normal and
shear stresses.
Determine also the directions of the normal stresses, tensile or
compressive, over the cross section.
Assume that the geometry of the femur at the section ! − ! is
circular with an outer radius of !! = 13 mm and an inner
radius of !! = 6 mm.
5 Biomechanics, LTH, 2012 12.
The human femur is now mounted in the grips of the torsion testing machine in
order to obtain the material properties of the bone in torsion. The length of the
bone at sections between the rotating (D) and stationary (E) grips is measured as
! = 37 cm. The femur is subjected to a torsional loading until fracture and the
applied torque versus angular displacement (deflection) graph is shown in the
figure. The femur is fractured at section ! − ! located ! = 25 cm from the
stationary grip, see the figure. The geometry of the bone at the fractured section is
approximated to a circular ring with outer radius !! = 13 mm and inner radius
!! = 7 mm.
Calculate the maximum shear strain and shear stress at the fractured section of the
femur and determine the shear modulus of elasticity of the femur.
6 Biomechanics, LTH, 2012 Answers
1.
2.
3.
4.
5.
6.
7.
283 N, northeast
400 N, northeast
The pressure is about 190 Pa (Note that Pa is a quite small unit!)
The friction force is about 3 N (Quite small force! Test!)
!! = 5.75 Nm (ccw)
The acceleration is ! = 0.94 m/s2 →
! = 1.4 sin 2.09 !
! = 2.93 cos 2.09 !
! = −6.12 sin 2.09 !
8. The force from the air resistance is 105 N to retard the skier and the track is
applying a force of 1246 on the skis. Note that that this force includes both the
effects of the weight and the rotational inertia !!! of the ski jumper.
9. ! = 0.05 or 5% and ! = 2 MPa
10. ! = 17.85 MPa, for a four-screw fixation device the shear stress in each screw
will be halved (! = 8.92 MPa)
11. !!"# = 48.6 MPa and !!"# = 2.4 MPa. The normal stress is positive, tensile on
the medial side and negative, compressive, on the lateral side.
12. The maximum shear strain is ! = 0.0123 rad and the maximum shear stress is
! = 56.9 MPa which, assuming the deformation being elastic, gives the shear
modulus of the bone ! = 4.6 GPa.
Source
•
Fundamentals of Biomechanics, Equilibrium, Motion, and Deformation, N.
Özkaya, M. Nordin, 1998.
7 Biomechanics, LTH, 2012
Assignment IBL
Biomechanics in practice
Background
You have probably been thinking in biomechanical terms of some situation in your daily life
before or during this course. It can be anything from making your bed, practicing some sport,
carrying food from the store or watching someone else working. Discuss together with your
colleagues and decide one special situation of interest for you all, 8-9 group members.
To your help, there are a group of physiotherapy students to use as consult in physiological
questions and questions about anatomy. They also take a course in Biomechanics at the
moment and have a similar assignment and you are supposed to be there as consults in
computational questions related to their assignment in exchange.
Schedule:
• Send a mail to [email protected] with a very short description of the chosen
situation (just one or two sentences) and the names of the students in the group before
12.00 on the 7:th of September
• Start to work with analysing the problem. Formulate a short description and questions
to the PT-students and send it to them before 12.00 the 13:th of September. Also send
a copy of this mail to [email protected].
• You will receive an answer on your mail on the 14:th of September.
• The PT-students will send a mail to you with questions on the 24:th of September.
• You should send an answer to the PT-students (with a copy to [email protected])
on the 25:th of September.
• Write a short report (maximum 2 pages) and a Power Point presentation including the
topics presented at the next page. The report should be sent to [email protected]
before 12.00 at the 28:th of September. The presentations are scheduled to the 3:th and
4:th of October.
Biomechanics, LTH, 2012
Issues to cover in the report and in the PowerPoint-presentation
•
A motivation for the chosen situation
•
A proper biomechanical model for the problem (like free-body diagrams)
•
At least three equations governing the situation and a numerical calculation of
something relevant for the situation
•
An analysis of some risks for injuries coupled to the situation and a strategy to avoid
those risks
The PP should include 5-6 slides and you should prepare a 20 minute presentation where all
group members are active.
Biomechanics, LTH, 2012
Assignment 1.
Analysis of a kick
Background
At kick-off a football player kicks the ball, which initially is placed in the center of the
football field. During the kick, the player stands on one leg and let the other swing forwards
from a position up and behind the ball. The hip neither translates nor rotates during the swing
and the swing can be considered as a movement in one plane.
The mass- and length data for the player is collected in the table below:
Body
Upper leg
Lower leg
Mass (kg)
80
8.0
4.9
Length (m)
1.80
0.44
0.45
You are supposed to create a simplified model of the kick built up by an upper and lower leg.
The rotations can be considered as the lower leg rotates with the knee as a centre and the
upper with the hip as the centre (both rotations are rotations in the plane).
Biomechanics, LTH, 2012
Issues to cover in the report
1. Put the origo of the coordinate system in the right hip and express the velocity and
acceleration for arbitrary points along the left leg in the x- and y-directions indicated
in the figure.
Hint: Use the angles θ1 and θ2 in the figure and a coordinate r1 running along the upper
leg and r2 running along the lower (two polar coordinate systems). Start with points
along the upper leg, connect in the knee and continue with the lower leg.
2. Use the expressions from issue 1 and plot of the velocities of the knee and the ankle as
functions of time, both x- and y-components and absolute values.
Hint: Pick the values of t, θ1, θ2,ω1, ω2, α1 and α2 from the file data.txt that is
available from the homepage of the course (www.solid.lth.se). It is smart to use either
Matlab or Excel to create the plots.
3. Establish the equations of motion for the upper and lower parts of the left leg.
Hint: Do free-body diagrams of the two parts of the left leg. Use Newton’s second law
and the expressions of the accelerations from issue 1.
4. Plot the reaction forces in the hip.
Hint: Use your results from issue 3 and use either Matlab or Excel.
5. Establish the equations of the contact forces for the right leg (i.e. the leg in contact
with the ground).
Hint: Expand the free-body diagram from issue 3 and regard the rest of the body
(except from the left leg).
6. Plot of the vertical contact force between the right foot and the ground.
Hint: Use your results from issue 5 and use either Matlab or Excel.
The report should also include discussions of the following:
• At what time do you think the hit between the foot and the ball occurs?
• Is there any possible explanation of why the upper leg is slowing down (the angular
velocity is negative and there is an angular retardation) under the period 0.17-0.30
seconds?
• Is there a horizontal force on the foot in contact with the ground during the kick? How
does this force come up?
Note!
If you want to check your results, you can do the kick by your self, standing on a scale.
Biomechanics, LTH, 2012
Assignment 2.
Analysis of muscle forces in the lower arm
Background
-
To be or not to be, that is the question!
Analyze the forces in the active muscles in the lower arm of Hamlet if he had hold the
skull in his hand with his arm bent when he made his famous statement.
Data
A sketch of the lower arm and some of the active muscles is presented in the figure and data
for the muscles is given in the table. The elbow joint can be regarded as friction less.
Muscle
Brachialis (bra)
Biceps brachii (bri)
Brachioradialis (brd)
PCSA (m2)
4.6 ⋅10−4
7.0 ⋅10−4
1.5 ⋅10−4
r (m)
0.034
0.046
0.075
θ (deg)
80.3
68.7
23.0
PCSA = physiological cross section area, muscle volume/fiber length
r = flexion-extension moment arm (Note! Right angle distance!)
θ = angle between muscle force vector and long ulnar axis sagittal plane
W = force corresponding to mass of the arm
P = force corresponding to the load in the hand and the mass of the hand
R = reaction forces in the elbow
If you need more information about the geometry in order to solve the problem you can take
your own arm and measure!
1
Biomechanics, LTH, 2012
Issues to cover in the report
1. What about the model, do you think it is sufficient for the problem? Should there be
any more muscles included? (You can find information about the anatomy of the arm
and what muscles that are active when on e.g. different webb-pages.)
2. Consider the weight of the arm and the outer load as given. Establish the equations of
equilibrium. (Note that r in the table is the moment arm perpendicular to the force.)
Which of the equations gives the most information if we are interested in the forces in
the muscles and not the reaction forces in the elbow? Is the problem statically
determined?
3. There are different hypothesis for how the weight of the arm and hand and the outer
load are divided on the muscles. We are going to try different possibilities. Derive the
muscle forces considering:
i)
ii)
The muscle forces are equal, Fbra = Fbic = Fbrd .
Two of the muscle forces are zero, the third taking the whole load. Make this
calculation for each of the muscles being the one carrying the load.
4. Derive the muscle forces under different minimization criterion:
a)
min( Fbra + Fbic + Fbrd )
b)
2
2
min( Fbra
+ Fbic2 + Fbrd
)
3
3
c) min( Fbra
+ Fbic3 + Fbrd
)
Construct a loop so that the calculations are done for P = 0, 5, 10… 100 N and present
the results, i.e. Fbra, Fbic and Fbrd as functions of P in proper diagrams, one diagram for
each minimization criteria.
Hint: Use the command fmincon in Matlab. To make the numerical solution of the
problem easier, use the argument lb to specify zero as the minimum value of the
muscle forces and ub to specify the maximal values in accordance with the
calculations done in ii) (i.e. one muscle carrying the whole load). The value calculated
in i) (the forces being equal) can be used as an initial guess, x0. Note that the equation
of equilibrium should be fulfilled, use it as a sub condition to the minimization. Use
the attached example for facilitating the understanding of how to use fmincon.
As you can see in the table, the three muscles have really different cross section areas.
It is therefore more physiological to use the stresses and not the forces as bounds in
minimization criterion. So to conclude, derive the muscle forces under a criterion
based on the stresses, i.e.:
⎛ ⎛ F ⎞2 ⎛ F ⎞2 ⎛ F ⎞ 2 ⎞
d) min ⎜ ⎜ bra ⎟ + ⎜ bic ⎟ + ⎜ brd ⎟ ⎟
⎜ ⎝ Abra ⎠ ⎝ Abic ⎠ ⎝ Abrd ⎠ ⎟
⎝
⎠
2
Biomechanics, LTH, 2012
Here Ai is the PSCA of each muscle. Introduce different maximal values of the
F
stresses σ i = i as upper bounds in case d) and compare the result when using the
Ai
maximal stress values and when not using them. Present also the result in case d) as
one plot of the three muscle forces as functions of P.
3
Biomechanics, LTH, 2012
Example for facilitating Assignment 2
We are going to use MATLAB in order to minimize a quadratic function of two variables, e.g.
2
2
⎛ ⎛
1 ⎞ ⎛
1 ⎞ ⎞
min ⎜ ⎜ x − ⎟ + ⎜ y − ⎟ ⎟
⎜ ⎝
2 ⎠ ⎝
2 ⎠ ⎟⎠
⎝
We shall find this minimum by using the function in MATLAB called fmincon. Write help
fmincon in your MATLAB-window to get familiar with this function. Note that this function
is a part of the “Optimization toolbox”. So if you do not have access to this toolbox you
would not find the function fmincon.
We start by putting the variables x and y in a vector:
⎡ x(1) ⎤ ⎡ x ⎤
x = ⎢
⎥ = ⎢ ⎥
⎣ x(2) ⎦ ⎣ y ⎦
so that x(1) = x and x(2) = y. We can then define the function that is going to be minimized in
a m-file in MATLAB with the following content:
%This is the file funk.m which defines the function
%that is going to be minimized. The function has been
%implemented so that the size of the vector x, and thereby the
%number of variables is read automatically by using size.
function out=funk(x)
[m,n]=size(x);
out=0.;
for i=1:m
out=out+(x(i)-0.5)*(x(i)-0.5);
end
We will then write a m-file that contains the call to fmincon and the definitions of the input
and how the output is handled.
%This is the file testcall.m which can be called by writing
%testcall at the MATLAB promt. fmincon will then
%minimize the function in the file funk.m
clear all
x0=[1;1];
lb=[0;0];
ub=[1;1];
a=[];
b=[];
aeq=[];
beq=[];
xut=fmincon(‘funk’,x0,a,b,aeq,beq,lb,ub);
xut
1
Biomechanics, LTH, 2012
The following arguments for fmincon have been used:
•
‘funk’ tells that the function that fmincon is going to minimize is defined in the file
called funk.m. This is the file that is going to be called from fmincon.
•
x0 is a first guess of the solution and it is used as an initial value for the numeric
solution procedure. Note that ; means new line, i.e. x0 is a column matrix:
⎡ x0 (1) ⎤ ⎡ x0 ⎤
x0 = [1;1] = ⎢
⎥ = ⎢ ⎥
⎣ x0 (2) ⎦ ⎣ y0 ⎦
•
a, b, aeq and beq are matrices used for defining the sub-conditions. We do not have
any sub-conditions yet, but these arguments can not be left out and they are therefore
defined here as empty matrices. Sub-conditions are included in the end of this example
by using aeq and beq.
•
lb and ub stand for lower bound and upper bound in respectively. They are used in
order to limit the area where the solution is searched from. Here the limits tell that the
solution is searched within the area:
⎡0⎤ ⎡ x(1) ⎤ ⎡1⎤
⎢0⎥ ≤ ⎢ x(2) ⎥ ≤ ⎢1⎥
⎣ ⎦ ⎣
⎦ ⎣ ⎦
Note that in this example is it not a part of the formulation of the problem that the
solution should be searched in this area; instead the area have been limited in order to
make it easier for fmincon to find the solution, i.e. of numerical reasons. It is possible
to think of other problems where similar conditions show up as a part of the physical
modelling. If one want to keep the different kinds of limitations apart, the convention
can be to let lb and ub denote conditions that are introduced of numerical reasons
while a and b stands for the limits which are a part of the formulation of the problem.
The minimum of the function can now be calculated by writing the name of the file
containing the call of fmincon by the MATLAB promt. We assume that the file is called
testcall:
>> testcall
Warning: Trust region method does not currently solve this type of problem,
switching to line search.
>In /sw/matlab/5.3/toolbox/optim/fmincon.m at line 190
In /your/matlab/directory/testcall.m at line 11
Optimization terminated successfully:
Search direction less than 2*options.TolX and
maximum constraint violation is less tha options. TolCon
No Active Constraints
xut =
0.500
0.500
>>
2
Biomechanics, LTH, 2012
Finally we are going to see how the sub-conditions can be introduced by using the
matrices a, b, aeq and beq. They are interpreted by fmincon as the solution has to fulfil
the following:
a•x ≤ b
aeq • x = beq
If we e.g. want to introduce the following sub-condition
x = 2 y ⇔ x(1) - 2x( 2) = 0
this is given by
⎡ x(1) ⎤
aeq = [1 −2] , beq = [0] ⇒ [1 −2] ⎢
⎥ = [0]
⎣ x(2) ⎦
If we change the m-file testcall.m to the following:
%This is the file testcall.m which can be called by writing
%testcall at the MATLAB promt. fmincon will then
%minimize the function in the file funk.m
clear all
x0=[1;1];
lb=[0;0];
ub=[1;1];
a=[];
b=[];
aeq=[1 -2];
beq=[0];
xut=fmincon(‘funk’,x0,a,b,aeq,beq,lb,ub);
xut
and write testcall by the MATLAB-promt, we will instead get
xut =
0.600
0.300
which is the minimum of the function fulfilling the sub-condition.
3