biomechanical modelling of musculoskeletal...

The University of Sydney Slide 1

Biomechanical Modelling of Musculoskeletal Systems

Presented by

Phillip Tran

AMME4981/9981

Semester 1, 2016

Lecture 6


The Musculoskeletal System


The Musculoskeletal System

Skeletal System

Provides support,

structure, and

protection

Made up of:

– Bones

– Ligaments

– Cartilage

– Joints

Muscular System

Provides movement

Made up of:

– Muscles

– Tendons


Bones

Cortical/Compact bone

– Hard, dense bone

Cancellous/Spongy bone

– Trabeculae align along lines of

stress

– Contains red marrow in spaces

Medullary Cavity

– Contains yellow marrow


Synovial Joints

Articular Cartilage

– Act as spongy cushions to absorb

compressive forces

Joint Cavity

– Space that contains synovial fluid

Articular Capsule

– Synovial membrane secretes

synovial fluid

– Fibrous layer holds the joint together

Synovial Fluid

– Reduces friction between cartilage

Ligaments

– Reinforce the entire structure


Tendons and Ligaments

Tendons

– Dense connective tissue that attach muscles to bones

– Made up of collagen fibres that are aligned along the length of the tendon

– Transfers pulling force of the muscle to the attached bone

Ligaments

– Connect one bone to another

– Provides stabilisation


Skeletal Muscles

Muscle Contraction

– Concentric: muscle shortens

– Isometric: no change in length

– Eccentric: muscle extends

Force of Muscle Contraction

– Number of muscle fibres recruited

– Size of fibres

– Frequency of stimulation

– Degree of muscle stretch


Anatomical Position and Planes

Anatomical Position

– Arms at side with palms facing forward

– Legs straight and together with feet flat on the ground

– Movements of the body are described in relation to this position

Anatomical Planes

– Coronal (front/back)

– Sagittal (left/right)

– Transverse (top/bottom)


Anatomical Directions


Movement

Flexion/Extension Abduction/Adduction


Movement

Rotation Supination/Pronation Dorsi/Plantarflexion


Biomechanical Modelling


Solving a Mechanical Problem

– Forces are applied to a body

– Geometry is known

– Finding the internal stresses

– Finding the resultant motion

F=1000N

F=1000N

0.1m

1m

5 m

E=210GPa, v=0.33 x

y 1mA


Solving a Mechanical Problem

• Known ForcesF

• Equations of Motion∑𝐹 = 𝑚𝑎 = 𝑚 𝑟

• Double Integration∬

• Displacementr

Forces(known)

Motion(unknown)


Solving a Biomechanical Problem

Internal Forces

– Active muscles

– Reactions at joints

– Reactions at ligaments

External Forces

– Inertial forces due to acceleration

of a segment

– Load applied directly to a body

segment

External force

Internal force


Solving a Biomechanical Problem

• Known Displacementr

• Double Differentiation𝑑2

𝑑𝑡2

• Equations of Motion∑𝐹 = 𝑚𝑎 = 𝑚 𝑟

• ForcesF

Internal Forces(unknown)

Motion(known)

External Forces(known)


Direct/Inverse Problems

Direct Problems (Mechanical)

Using known forces to determine

movement

– Requires accurate measurements

of the geometry

– Requires knowledge of external

forces

Inverse Problems (Biomechanical)

Using known movements to determine

the internal forces:

– Requires full description of the

movement (displacement, velocity,

acceleration)

– Requires accurate measurements

of anthropometry (measurement

of the human body)

– Requires knowledge of external

forces


Movement: Trajectories of Motion


Movement: Motion Tracking


Movement: Motion Tracking in Movies

– Andy Serkis in movies

https://www.youtube.com/watch?v=XM9Pvfq1KhE&nohtml5=False

https://www.youtube.com/watch?v=XM9Pvfq1KhE&nohtml5=False


Measuring Movement

Angles

– Calculated using the position of two markers placed along the long axis of

the body segment

𝜙𝑖𝑗 = 𝑡𝑎𝑛−1𝑦𝑗 − 𝑦𝑖

𝑥𝑗 − 𝑥𝑖

Linear velocities and accelerations

– Calculated using the position of one marker on two frames

𝑣𝑥𝑖 =𝑥𝑖+1−𝑥𝑖−1

2Δ𝑡, 𝑎𝑥𝑖 =

𝑣𝑥(𝑖+1)−𝑣𝑥(𝑖−1)

2Δ𝑡

Angular velocities and accelerations

– Calculated using the angle of a body segment on two frames

𝜔𝑖 =𝜙(𝑖+1)−𝜙(𝑖−1)

2Δt, 𝛼𝑖 =

𝜔(𝑖+1)−𝜔(𝑖−1)

2Δt


Types of Motion

Translation

– All particles of the

body move in

parallel trajectories

Σ𝐹𝑥 = 𝑚 𝑎𝑥Σ𝐹𝑦 = 𝑚 𝑎𝑦Σ𝑀𝐺 = 0

General Motion

– The body performs

translation and

rotation

Σ𝐹𝑥 = 𝑚 𝑎𝑥Σ𝐹𝑦 = 𝑚 𝑎𝑦

Σ𝑀𝐺 = 𝐼𝐺

Rotation

– All particles of the

body move about a

point

Σ𝐹𝑛 = 𝑚 𝑎𝑛 = 𝑚𝜔2𝑟𝐺Σ𝐹𝑡 = 𝑚 𝑎𝑡 = 𝑚𝛼𝑟𝐺

Σ𝑀𝑂 = 𝐼𝑂𝛼


Anthropometry

Measurement of the human body

– Segment length

– Segment mass

– Position of centre of gravity

– Density


Anthropometry

Body Segment Length (% of

height)

Distance of centre

of mass from distal

joint (% of limb)

Mass (% of body

mass)

Head 9.4 50.0 5.7

Neck 4.5 1.3

Thorax+Abdomen 25.0 30.3

Upper Arm 18.0 43.6 2.6

Forearm 26.0 43.0 1.9

Hand 50.6 0.7

Pelvis 9.4 14.0

Thigh 31.5 43.3 12.8

Shank 23.0 43.3 5.1

Foot 16.0 50.0 1.3


Anthropometry

Body Segment Density (g/cm3) Mass moment at centre of

mass per segment length

(km·m2/m)

Head 1.11

Neck 1.11

Thorax+Abdomen

Upper Arm 1.07 0.322

Forearm 1.13 0.303

Hand 1.16 0.297

Pelvis

Thigh 1.05 0.323

Shank 1.09 0.302

Foot 1.10 0.475


External Forces

Gravitational Forces

– Acting downward through the centre of mass of each segment

Ground Reaction Forces

– Distributed over an area

– Assumed to be acting as a single force at the centre of pressure

Externally Applied Forces

– Restraining or accelerating force that acts outside the body

– Mass being lifted


Biomechanical Modelling: Body Segments

– Body segments can be modelled as rigid bodies

– Free body diagrams can be drawn for each segment

– Forces and moments acting at joint centres

– Gravitational forces acting at the centres of mass

– Accurate measurements are needed of:

– Segment masses (m)

– Location of centres of mass

– Location of joint centres

– Mass moment of inertia (I)


Biomechanical Modelling: Assumptions

– Rigid body motion (deformation is small relative to overall motion)

– Body segments interconnected at joints

– Length of each body segment remains constant

– Each body segment has a fixed mass located at its centre of mass

– The location of each body segment’s centre of mass is fixed

– Joints are considered to be hinge (2D) or ball and socket (3D)

– The moment of inertia of each body segment about any point is constant

during any movement


Case Study

Cycling


Cycling


Modelling Process

– Mathematical model based on

trigonometry

– Position coordinates for the joints

– Differentiation to compute linear

velocities and accelerations

– Forces and moments calculated at

the joints

– Compute power

Aim: To determine if ankling will

improve performance


Angle of the Ankle

-40

-30

-20

-10

0

10

20

30

40

0 90 180 270 360Angle

Crank Angle

Normal

Ankling


Comparison

– Ankling– Normal Cycling


Free Body Diagrams


Force on the Pedal (Tangential)

-100

-50

0

50

100

150

200

250

300

0 90 180 270 360

Forc

e (

N)

Crank Angle

Normal

Ankling


Validation: KAvideo to Calculate Hip Forces


Validation: KAvideo to Calculate Hip Forces

0

50

100

150

200

250

300

350

400

0 20 40 60 80 100

Resu

ltan

t Fo

rce (

N)

Percentage of Cycle (%)

Calculationfrom exsitingdata

Model Results


Validation: Torque at the Ankle

– Literature – Model

Torq

ue (

Nm

)

Crank Angle in degrees


Examples


Arm Analysis: Part 1

A flexed arm is holding a ball of Wb=20 N with a distance of 35 cm to the

elbow centre. What is the force required in the biceps (B) if the forearm weighs

Wa=15 N and the centre of mass for the forearm is 15 cm from the elbow

centre of rotation? Also find the reaction force at the elbow joint. Assume the

forearm is in the horizontal position and the angle between the forearm and

upper arm at the elbow is 100 degrees. The biceps tendon is inserted 3 cm

from the elbow centre of the forearm, and at the proximal end of the upper

arm, which is 30 cm in length.



Free Body Diagram

Using trig formulae:

θ = 74.5°



Arm is in static equilibrium

Σ𝐹 = 0 and ΣM = 0

Scalar equations

ΣFx = 0 → 𝑅𝑥 − 𝐵 cos 74.5° = 0

Σ𝐹𝑦 = 0 → 𝑅𝑦 + 𝐵 sin 74.5° − 15 − 20 = 0

Σ𝑀 = 0 → 0.35 × −20 + 0.15 × −15 + 0.03 × (𝐵 sin 74.5°) = 0

Solve the equations

𝐵 = 319.97𝑁𝑅𝑦 = −273.33𝑁

𝑅𝑥 = 85.51𝑁



The ball is lifted from the horizontal forearm position with an angular

acceleration of α=2rad/s2. Determine the additional force required by the

bicep to provide this movement. The radius of the forearm is 4cm. Assume that

the upper arm remains stationary.



Arm is rotating about the elbow

Σ𝐹𝑛 = 𝑚𝜔2𝑟𝐺 , Σ𝐹𝑡 = 𝑚𝛼𝑟𝐺 and ΣM = IO𝛼

Mass moment of inertia

𝐼𝑂 = 𝐼𝐺 +𝑚𝑑2 =1

12𝑚 3𝑟2 + 𝐿2 +𝑚𝑑2 = 0.05

Scalar equation

ΣM = IO𝛼 → 0.35 × −20 + 0.15 × −15 + 0.03 × 𝐵 sin 74.5°= 0.05 × 2

Solve the equation

𝐵 = 354.56𝑁Δ𝐵 = 34.59𝑁


Summary

– The skeletal and muscular systems work together to provide movement for

the human body

– The body can be modelled biomechanically

– Inverse method to derive the internal muscle forces and joint reactions

– Movement

– Anthropometry

– External forces


Joint Reaction Analysis

A person stands statically on one foot. The ground reaction force R acts 4cm

anterior to the ankle centre of rotation. The body mass is 60kg and the foot

mass is 0.9kg. The centre of mass of the foot is 6cm from the centre of rotation.

Determine the forces and moment in the ankle.

Rotation

Centre

Mass Centre

Ry

Rx

mg

Ax

Ay

MA



Foot is in static equilibrium

Σ𝐹 = 0 and Σ𝑀 = 0

Reaction force

𝑅𝑥 = 0 and 𝑅𝑦 = 60 × 9.81 = 588𝑁

Solving

Σ𝐹𝑥 = 𝑚𝑎𝑥 → 𝑅𝑥 + 𝐴𝑥 = 0

∴ 𝐴𝑥 = 0

Σ𝐹𝑦 = 𝑚𝑎𝑦 → 𝐴𝑦 + 𝑅𝑦 −𝑚𝑔 = 0

∴ 𝐴𝑦 = 𝑚𝑔 − 𝑅𝑦 = 0.9 × 9.81 − 588 = −579.2𝑁

Σ𝑀 = 0 → 𝑀𝐴 + 𝑅𝑦 × 0.04 − 0.9 × 9.81 × 0.06 = 0

∴ 𝑀𝐴 = −23𝑁𝑚



A person exercises his left shoulder rotators. Calculate the forces and moments

exerted on his shoulder.

F = 200 N

a = 25 cm

b = 30 cm

y

x

z

Rj

Mj

FA

a

b

B

C

a

b



Consider a quasi-equilibrium

Equations of motion

Force

Σ𝐹𝑥 = 0 ∴ 𝑅𝑗𝑥 = 0

Σ𝐹𝑦 = 0 → 𝑅𝑗𝑦 + 𝐹 = 0 ∴ 𝑅𝑗𝑦 = −𝐹

Σ𝐹𝑧 = 0 ∴ 𝑅𝑗𝑧 = 0

∴ 𝑹𝑗 = −𝐹𝒋

Moment

Σ𝑀𝐴 = 0 → 𝑴𝑗 + 𝒓𝑐 × 𝑭 = 0

But: 𝒓𝑐 = 𝑎𝒊 + 𝑏𝒌 and 𝑭 = 𝐹𝒋

∴ 𝑴𝑗 = −𝒓𝑐 × 𝑭 = 𝑏𝐹𝒊 − 𝑎𝐹𝒌


Muscle Analysis

A weight lifter raises a barbell to his chest. Determine the torque developed by

the back and the hip extensor muscles (Mj) when the barbell is about knee

height.

Weight of barbell, Wb = 1003N

Mass of upper body, mu = 53.5kg

a = 38cm, b = 32cm, d = 64cm

IG = 7.43 kg·m2, α = 8.7 rad/s2

aGx = 0.2 m/s2, aGy = -0.1 m/s2

Mj

Fj

O

y

x60°

Wb

a

mug

b

G

d


Muscle Analysis

Dynamic equilibrium of forces

Σ𝐹𝑥 = 𝑚𝑎𝐺𝑥 → 𝐹𝑗𝑥 = 53.5 × 0.2 = 10.7𝑁

Σ𝐹𝑦 = 𝑚𝑎𝐺𝑦 → 𝐹𝑗𝑦 −𝑚𝑢𝑔 −𝑊𝑏 = 53.5 × −0.1

∴ 𝐹𝑗𝑦 = 53.5 × 9.81 + 1003 − 53.5 × 0.1 = 1522.49𝑁

𝐹𝑗 = 𝐹𝑗𝑥2 + 𝐹𝑗𝑦

2 = 1522.53𝑁

Mass moment of inertia about O

𝐼𝑂 = 𝐼𝐺 +𝑚𝑑2 = 7.43 + 53.5 × 0.642 = 29.34𝑘𝑔 ∙ 𝑚2

Dynamic equilibrium of moments

Σ𝑀𝑂 = 𝐼𝑂𝛼 → 𝑀𝑗 +𝑚𝑢𝑔 × 0.32 +𝑊𝑏 × 0.38 = 29.34 × 8.7

∴ 𝑀𝑗 = −53.5 × 9.81 × 0.32 − 1003 × 0.38 + 29.34 × 8.

= −293.83𝑁𝑚

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