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Chapter 6 Occupational Biomechanics Models Why Model? • Models • simple representation of complex systems • improve understanding • even with gross simplifications and assumptions • rigid links vs complex anatomy of segment • SEM vs multiple muscles of the group • If error in the model is too large • improve our model parameters • % TBM in segment, CofM location • Increase complexity • individual muscles Why Model? • Biomechanical model allows to predict potentially hazardous loading conditions on NMS components without actual subject risk • manipulate parameters (loading, geometry) • Measure response • compare model prediction to real world • refine the model (limit interpretation) Why Model? • Biomechanical model allows to predict potentially hazardous loading conditions on NMS components without actual subject risk • Provide understanding of Guidelines to improve efficiency and safety in the workplace Our application: • estimate forces acting on different components of the body. • F ==> stress ==> injury • understanding risk • compare force (stress) predicted from model to force (stress) known to cause fatigue and/or injury • Identify dangerous situations and success of interventions. Planar Static Biomechanical Models Single-body segment static model • used when no movement no linear or angular acceleration is present. • standing with a load in the hands. • moving at constant velocity (isokinetic) • Planar analysis is limited to a 2D analysis. Planar Static Biomechanical Models • KEY: identify the magnitude of the external forces acting on the stationary mass • Always ==> gravity W = mg • where: • weight measured in Newtons; • mass measured in kg; • g is the gravitational acceleration (-9.8 m.s-2) Forces is a vector quantity, and has 4 characteristics: • 1. Magnitude. Vector • 2. Direction (-) • 3. Line of action. • 4. Point of application. Planar Static Biomechanical Models • Consider 10kg load in one-handed lift • Since a=0, forces on the body = ??? • Some other external force must be acting to counter the weight • Obviously 2nd force is provided by the hand pulling up the load Planar Static Biomechanical Models • Since the object is not moving, it is defined to be in static equilibrium. • This means that the additive effect of all external forces acting on a mass is zero F = m a dir dir Planar Static Biomechanical Models • Since the object is not moving, it is defined to be in static equilibrium. • This means that the additive effect of all external forces acting on a mass is zero F=0 dir Planar Static Biomechanical Models • Since the object is not moving, it is defined to be in static equilibrium. • This means that the additive effect of all external torques acting on a mass is zeroα T = I α axis axis axis Planar Static Biomechanical Models • Since the object is not moving, it is defined to be in static equilibrium. • This means that the additive effect of all external torques acting on a mass is zero T =0 axis Free body diagrams: • Force vectors are scaled in the drawing to indicate magnitude • Vectors are orientated in the direction of the force (tip) • Vectors are aligned on the body to indicate point of application and line of action. Solve for force in EACH hand F vert =0 Identify the forces to be summed: Weight + 2 hands = 0 2 hands = - Weight Each hand = - Weight / 2 Each hand = - (-98) / 2 Each hand = + 49 N Planar Static Biomechanical Models • Determine force on each hand to hold a 10-kg mass in static equilibrium • Extend the planar static analysis to estimate the elbow forces and moments (torques) with forearm horizontal • Assumptions • Average Anthro & Inertia • Load applied at CofM of hand • Forearm/hand is a single segment Calculate Force at Elbow: 1st condition of equilibrium F=0 vert W load + W f&h + R elbow = 0 R elbow = -W load - W f&h R elbow = - (-49) - (-15.8) R elbow = + 64.8 N Joint Reaction Force R elbow = + 64.8 N The NET tensile force created by ligaments and muscles holding the joint together. MUST be present to give a = 0 (no translational acceleration) Forces acting on the system will cause torque •If eccentric to an axis •W f&h & W load •Not if centric to an axis • R elbow Calculate torque at elbow: 2nd condition of equilibrium T=0 axis T load + T f&h + T elbow = 0 T elbow = - (T load) - (T f&h) T elbow = - (-49 x .355) - ( -15.8 x .172) T elbow = - (-17.4) - (-2.7) = 20.1 Nm (CCW, flexion) Net Joint Torque (net moment of force) T elbow = 20.1 Nm (CCW, flexion) Present at EACH elbow The NET torque created at the elbow joint by muscles. Which muscles? Ignores co-contraction MUST be present to give α = 0 (no angular acceleration) On own: • Calculate Rf at elbow and torque at elbow for the segment held at the horizontal without the hand held load. Arm not at horizontal (20º below) • Reaction force at elbow is the same because ... • Muscle torque at elbow is decreased because ..... Trig? SOH CAH TOA Two-body segment static model: • Start at segment with known external force (or only one unknown force) • FH: W f&h, R elbow, T f&h, T elbow • Horizontal position Two-body segment static model: • Start at segment with known external force (or only one unknown force) • FH: W f&h, R elbow, T f&h, T elbow • Horizontal position • Non-horizontal position • Note elbow load from task & posture Extend model to nonparallel forces • Preceding egs. have considered gravity as the only source of external forces (parallel force systems) • What if person is pushing or pulling on a load??? • Resolve force to orthogonal components • horizontal • vertical Planar static analysis of internal forces: • Extend the model technique to estimate the force on various musculoskeletal tissues • tension within a muscle (SEM) that creates the observed moment of force • bone on bone force (not just JRF) that accounts for the tension in the muscle Planar static analysis of internal forces: • Needed: the point of application and the line of action of muscle(s) tendons within the musculoskeletal structure • Our simplification: • concept of Single Equivalent Muscle (SEM) • only Biceps Brachii acting at the elbow joint, inserting 0.05 m from the axis Solve for Biceps Muscle Force Earlier: T load + T f&h + T elbow = 0 Becomes: T load + T f&h + T biceps = 0 Isolate: T biceps = - (T load) - (T f&h) Expand to: F bi x MA bi = - (T load) - (T f&h) F bi = (- (T load) - (T f&h)) / MA bi Substitute: F bi = (- (-49 x .355) - ( -15.8 x .172)) / 0.05 Solve: F bi = 20.1 Nm / 0.05 m F bi = 402 N Solve for Biceps Muscle Force Earlier: T load + T f&h + T elbow = 0 Becomes: T load + T f&h + T biceps = 0 Isolate: T biceps = - (T load) - (T f&h) Expand to: F bi x MA bi = - (T load) - (T f&h) F bi = (- (T load) - (T f&h)) / MA bi Substitute: F bi = (- (-49 x .355) - ( -15.8 x .172)) / 0.05 Solve: F bi = 20.1 Nm / 0.05 m F bi = 402 N ========> 8x > HHW Solve for Joint Reaction Force Earlier: W load + W f&h + R elbow = 0 Becomes: W load + W f&h + F bi + R elbow = 0 R elbow = -W load - W f&h - F bi R elbow = - (-49) - (-15.8) - (402) R elbow = + 64.8 - 402 N R elbow = - 337.2 N Joint Reaction Force Previous: R elbow = + 64.8 N (no Biceps) Now: R elbow = - 337.2 N The NET compressive force pushing DOWN on the forearm from the humerus (created by muscle squeezing the joint together) Minimum MUST be present to give a = 0 (no translational acceleration) ignores potential co-contraction Planar static analysis of internal forces • Technique more complicated if • consider > 1 muscle Planar static analysis of internal forces • Technique more complicated if • consider > 1 muscle • Overhead Scott & Winter, MSSE, 1991 • determine each muscle % contribution • move through the ROM (L/T) Models used to calculate Forces Results Planar static analysis of internal forces: • Compare relative lengths of MAs •Hand held load > muscle •To generate equal but opposite torque, force in muscle must be greater • mechanical disadvantage Multiple-link coplanar static modeling • Posture has no effect on calculated JRF, but has a very large effect on calculated JMF • • • • • note JRF constant at 549 N while moment increase is approximately 10x • Note: has not considered the increase in JRF from muscle tension to provide the moment (very complex musculature) What muscle group active in a), b), c)?? What happens to alignment of vertebrae? What happens if load added to hands? What happens if arm/ab used for support? Qualitative Low Back Load Importance: • Since skeletal muscle responds to the load moments (creates the calculated net JMF)... • simple static models give insight into what postures require • specific muscle groups to be active • to what relative magnitude each specific muscle group must be active Dynamic Biomechanical Models HPR 482: Advanced Biomechanics • The introduction of motion into biomechanical models introduces two types of complexity • kinematics must be quantified • position, velocity, acceleration • linear and angular • must account for inertial force and inertial torque in calculations • F = ma when a <> 0 • T = I (alpha) when (alpha) <> 0 Dynamic biomechanical model: • Analysis indicates the increased hazard of performing dynamic movements. • greater force for linear acceleration • speed up or slow down • greater torque for angular acceleration • speed up or slow down • Musculoskeletal load increases as speed of movement increases • greater accelerations • Add additional mass ??? • Additional segments???? Summary of dynamic biomechanical models: •Prudent to encourage workers to develop smooth movements that reduce accelerations and decelerations, especially if heavy loads are being manipulated. Coplanar Biomechanical Models of Foot Slip Potential While Pushing a Cart. • Common activity in workplace • mailroom, scrap, TVs, luggage • Load may approach max strength • Common time to slip • What causes slipping??? Coplanar Biomechanical Models of Foot Slip Potential While Pushing a Cart. • Common activity in workplace • mailroom, scrap • Load may approach max strength • Common time to slip • What causes slipping??? • Low Friction between sole and surface • What prevents slipping??? Coplanar Biomechanical Models of Foot Slip Potential While Pushing a Cart. • Common activity in workplace • mailroom, scrap • Load may approach max strength • Common time to slip • What causes slipping??? • Low Friction between sole and surface • What prevents slipping??? • Adequate friction between sole and surface Coplanar Biomechanical Models of Foot Slip Potential While Pushing a Cart. • What is friction?? • Force that tends to resist slipping • Reflects nature of TWO surfaces Friction = u N “mu” Coefficient of Friction u= Max Limiting Friction Normal Reaction Force property of two materials placed in contact. 0.2 smooth & wet 0.9 rough & dry Coefficient of Friction Peak Shear Force u= Max Limiting Friction Normal Reaction Force Peak Normal Force Materials in Contact from http://www.fearofphysics.com/Friction/frintro.html Typical Coefficients Sticky Slippery Extent of problem of falls Pedestrian-fall accidents have been the second largest generator of unintentional workplace fatalities, accounting for nearly 11% and 20%, respectively, of all fatal and non-fatal occupational injuries in the USA. Redfern et al (2001). Biomechanics of slips. Ergonomics, 44:13:1138-1166. Workplace falls Friction Manipulation Occupational Health & Safety E-News 03-17-03 Our Safety Tip of the Week is courtesy of Manuel (Mel) Rosas, a safety consultant for Carolinas Associated General Contractors. "I consistently find employers do not have procedures in place to inspect the soles of the shoes their employees wear to work. During walk-around inspections I ask employees to lift up and show me the bottom of the work shoe (boot), and I find many with worn or nearly slick soles. Employers should address this issue to reduce the risk of injury due to a worn shoe or boot." Friction Manipulation floor design Friction Manipulation PVC dots on palmar surface Coplanar Biomechanical Models of Foot Slip Potential While Pushing a Cart. • To prevent slipping: • What are the peak normal and shearing forces expected at the foot/floor contact point? • Design footwear and floor surfaces that will provide friction under these circumstances. • ie provide greater coefficient of friction How does employee alter pushing action as the load gets heavier? What does this do to friction requirements? Special Purpose Biomechanical Models of Occupational Tasks: • Model specific areas that are prone to overuse and/or traumatic injury • low back • wrist/hand • knee • shoulder Low-back biomechanical models: • NIOSH suggests using the load moment about the lumbosacral disc (L5/S1) as the basis for limits when • lifting • carrying loads • Why L5/S1: • 85% to 95% of all disc herniations • loads in hands have the largest moment arms relative to this axis Earlier, showed • Large increase in Net moment at low back with lifting • What will happen with load in hands? • What about anatomical moment?? • Muscle force: erector spinae • Moment arm: 0.05 m • Abdominal pressure • pushes torso into extension • Role of abdominal muscles?? Resultant effect on vertebral column • HUGH compressive forces • some from the load itself (posture) • greatest % from the muscle force • STRESS on vertebral disk = Force / area • effect of posture • increase force • decrease area Low-back biomechanical models: • Initial calcs: used simple back model • Cadavers: compression forces created micro-fractures on intervertebral disks • weak spot for potential herniation? • Recent models incorporate: • more muscle groups • corrected moment arms • relative contribution of each muscle group • effect on compressive force????? Low-back biomechanical models: • NIOSH (1981) • recommended that predicted L5/S1 compression values • above 3400 N be considered potentially hazardous for some workers. • above 6400 N be considered hazardous for most workers. • Basis: repeated, large compression force may increase risk of disc degeneration & chronic low-back symptoms. • NIOSH (1993): to be discussed Summary • Models vary in complexity • All based on Newton II • Require adeptness with Tables • Require logical thinking • CM locations, Moment Arm length • Provide insight to joint & muscle loading • Underly postural & load guidelines