Biomechanical Modelling of Musculoskeletal Systems

Transcription

1 Biomechanical Modelling of Musculoskeletal Systems Lecture 6 Presented by Phillip Tran AMME4981/9981 Semester 1, 2016 The University of Sydney Slide 1

2 The Musculoskeletal System The University of Sydney Slide 2

3 The Musculoskeletal System Skeletal System Provides support, structure, and protection Muscular System Provides movement Made up of: Bones Ligaments Cartilage Joints Made up of: Muscles Tendons The University of Sydney Slide 3

4 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 The University of Sydney Slide 4

5 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 The University of Sydney Slide 5

6 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 The University of Sydney Slide 6

7 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 The University of Sydney Slide 7

8 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) The University of Sydney Slide 8

9 Anatomical Directions The University of Sydney Slide 9

10 Movement Flexion/Extension Abduction/Adduction The University of Sydney Slide 10

11 Movement Rotation Supination/Pronation Dorsi/Plantarflexion The University of Sydney Slide 11

12 Biomechanical Modelling The University of Sydney Slide 12

13 Solving a Mechanical Problem Forces are applied to a body Geometry is known Finding the internal stresses y 1m A F=1000N Finding the resultant motion E=210GPa, v=0.33 x 1m 5 m F=1000N 0.1m The University of Sydney Slide 13

14 Solving a Mechanical Problem Forces (known) Motion (unknown) F F = ma = m r r Known Forces Equations of Motion Double Integration Displacement The University of Sydney Slide 14

15 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 Internal force External force The University of Sydney Slide 15

16 Solving a Biomechanical Problem External Forces (known) Motion (known) Internal Forces (unknown) r Known Displacement d 2 dt 2 F = ma = m r Double Differentiation Equations of Motion F Forces The University of Sydney Slide 16

17 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 The University of Sydney Slide 17

18 Movement: Trajectories of Motion The University of Sydney Slide 18

19 Movement: Motion Tracking The University of Sydney Slide 19

20 Movement: Motion Tracking The University of Sydney Slide 20

21 Movement: Motion Tracking in Movies Andy Serkis in movies The University of Sydney Slide 21

22 Measuring Movement Angles Calculated using the position of two markers placed along the long axis of the body segment φ ij = tan 1 y j y i x j x i Linear velocities and accelerations Calculated using the position of one marker on two frames Angular velocities and accelerations v xi = x i+1 x i 1, a 2Δt xi = v x(i+1) v x(i 1) 2Δt Calculated using the angle of a body segment on two frames ω i = φ (i+1) φ (i 1), α 2Δt i = ω (i+1) ω (i 1) 2Δt The University of Sydney Slide 22

23 Types of Motion Translation All particles of the body move in parallel trajectories Rotation All particles of the body move about a point General Motion The body performs translation and rotation ΣF x = m a x ΣF y = m a y ΣM G = 0 ΣF n = m a n = mω 2 r G ΣF t = m a t = mαr G ΣM O = I O α ΣF x = m a x ΣF y = m a y ΣM G = I G The University of Sydney Slide 23

24 Anthropometry Measurement of the human body Segment length Segment mass Position of centre of gravity Density The University of Sydney Slide 24

25 Anthropometry Body Segment Length (% of height) Distance of centre of mass from distal joint (% of limb) Head Neck Thorax+Abdomen Upper Arm Forearm Hand Pelvis Thigh Shank Foot Mass (% of body mass) The University of Sydney Slide 25

26 Anthropometry Body Segment Density (g/cm 3 ) Mass moment at centre of mass per segment length (km m 2 /m) Head 1.11 Neck 1.11 Thorax+Abdomen Upper Arm Forearm Hand Pelvis Thigh Shank Foot The University of Sydney Slide 26

27 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 The University of Sydney Slide 27

28 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) The University of Sydney Slide 28

29 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 The University of Sydney Slide 29

30 Case Study Cycling The University of Sydney Slide 30

31 Cycling The University of Sydney Slide 31

32 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 The University of Sydney Slide 32

33 Angle Angle of the Ankle Crank Angle Normal Ankling The University of Sydney Slide 33

34 Comparison Normal Cycling Ankling The University of Sydney Slide 34

35 Free Body Diagrams The University of Sydney Slide 35

36 Force (N) Force on the Pedal (Tangential) Normal Ankling Crank Angle The University of Sydney Slide 36

37 Validation: KAvideo to Calculate Hip Forces The University of Sydney Slide 37

38 Resultant Force (N) Validation: KAvideo to Calculate Hip Forces Percentage of Cycle (%) Calculation from exsiting data Model Results The University of Sydney Slide 38

39 Torque (Nm) Validation: Torque at the Ankle Literature Model Crank Angle in degrees The University of Sydney Slide 39

40 Examples The University of Sydney Slide 40

41 Arm Analysis: Part 1 A flexed arm is holding a ball of W b =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 W a =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. The University of Sydney Slide 41

42 Arm Analysis: Part 1 Free Body Diagram Using trig formulae: θ = 74.5 The University of Sydney Slide 42

43 Arm Analysis: Part 1 Arm is in static equilibrium ΣF = 0 and ΣM = 0 Scalar equations ΣF x = 0 R x B cos 74.5 = 0 ΣF y = 0 R y + B sin = 0 ΣM = (B sin 74.5 ) = 0 Solve the equations B = N R y = N R x = 85.51N The University of Sydney Slide 43

44 Arm Analysis: Part 2 The ball is lifted from the horizontal forearm position with an angular acceleration of α=2rad/s 2. 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. The University of Sydney Slide 44

45 Arm Analysis: Part 2 Arm is rotating about the elbow ΣF n = mω 2 r G, ΣF t = mαr G and ΣM = I O α Mass moment of inertia I O = I G + md 2 = 1 12 m 3r2 + L 2 + md 2 = 0.05 Scalar equation ΣM = I O α B sin 74.5 = Solve the equation B = N ΔB = 34.59N The University of Sydney Slide 45

46 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 The University of Sydney Slide 46

47 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 A y M A A x Mass Centre R x mg R y The University of Sydney Slide 47

48 Joint Reaction Analysis Foot is in static equilibrium ΣF = 0 and ΣM = 0 Reaction force R x = 0 and R y = = 588N Solving ΣF x = ma x R x + A x = 0 A x = 0 ΣF y = ma y A y + R y mg = 0 A y = mg R y = = 579.2N ΣM = 0 M A + R y = 0 M A = 23Nm The University of Sydney Slide 48

49 Joint Reaction Analysis 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 R j y a b M j z A F a C b B x The University of Sydney Slide 49

50 Joint Reaction Analysis Consider a quasi-equilibrium Equations of motion Force ΣF x = 0 R jx = 0 ΣF y = 0 R jy + F = 0 R jy = F ΣF z = 0 R jz = 0 R j = Fj Moment ΣM A = 0 M j + r c F = 0 But: r c = ai + bk and F = Fj M j = r c F = bfi afk The University of Sydney Slide 50

51 Muscle Analysis A weight lifter raises a barbell to his chest. Determine the torque developed by the back and the hip extensor muscles (M j ) when the barbell is about knee height. Weight of barbell, W b = 1003N Mass of upper body, m u = 53.5kg a = 38cm, b = 32cm, d = 64cm I G = 7.43 kg m 2, α = 8.7 rad/s 2 W b G d y F j a Gx = 0.2 m/s 2, a Gy = -0.1 m/s 2 m u g 60 O b a M j x The University of Sydney Slide 51

52 Muscle Analysis Dynamic equilibrium of forces ΣF x = ma Gx F jx = = 10.7N ΣF y = ma Gy F jy m u g W b = F jy = = N F j = F jx 2 + F jy 2 = N Mass moment of inertia about O I O = I G + md 2 = = 29.34kg m 2 Dynamic equilibrium of moments ΣM O = I O α M j + m u g W b 0.38 = M j = = Nm The University of Sydney Slide 52