Presentation on theme: "BIOMECHANICS OF WORK Chapter 11 in your text. The Musculoskeletal System Bones, muscle and connective tissue supports and protects body parts maintains."— Presentation transcript:
The Musculoskeletal System Bones, muscle and connective tissue supports and protects body parts maintains posture allows movement generates heat and maintains body temperature
Bones 206 bones Body “framework” Protective: rib cage and skull Provide for action: arms, legs linked at joints by tendons and ligaments Tendons: connect bone to muscle Ligaments: connect bone to bone
Joints Connection of two or more bones Movement no mobility joints (e.g. in skull) hinge joints (elbow) pivot joints (wrist) ball and socket joints (hip and shoulder) 3DOF
Muscles 400 muscles 40-50% of your body weight half of your body’s energy needs
Muscle Composition bundles of muscle fibres, connective tissue and nerves fibres are made of long cylindrical cells cells contain contractile elements (myofibrils) both sensory and motor nerves motor nerves control contractions of groups of fibres (motor unit)
Muscle Contraction Concentric: (also called isotonic) muscle contracts and shortens Eccentric: muscle contracts and lengthens (overload) Isometric: muscle contracts and stays the same length
Muscle Strength proportional to muscle cross-section usually measured as torque force applied against a moment arm (bone) to an axis of rotation (joint) Static strength: measured during isometric contraction Dynamic strength: measured during movement
Basic Biomechanics Statics model ( F=0, Moments=0), isometric contraction Force at the point of application of the load Weight of the limb is also a force at the center of gravity of the limb F can be calculated
Problem in Text 20kg Person holding a 20kg weight in both hands. What are the force and moment at the elbow? Given: Mass =20kg Force of segment = 16N Length of segment =.36m Assume: COG of segment is at the midpoint!
Problem in Text 1. Convert mass to Force 20kg*9.8 m/s 2 = 196 N 2. Divide by # of hands. 196N/2 hands = 98N/hand 98 N
Problem in Text 1. Convert mass to Force 20kg*9.8 m/s 2 = 196 N 2. Divide by # of hands. 196N/2 hands = 98N/hand 3. Calculate F elbow. F=0 F elbow – 16N – 98N = 0 F elbow = 114N [up] 98 N 16 N F elbow
Problem in Text 1. Convert mass to Force 20kg*9.8 m/s 2 = 196 N 2. Divide by # of hands. 196N/2 hands = 98N/hand 3. Calculate F elbow. F=0 F elbow – 16N – 98N = 0 F elbow = 114N [up] 4. Calculate M elbow. elbow m +(-98N)*.36m=0 elbow=38.16N*m 98 N 16 N F elbow.18m.36m
Multi-segment models Repeat for each segment, working the forces and moments back How would you work out the Force and Moment in the shoulder? What information would you need?
Lower Back Pain estimated at 1/3 of worker’s compensation payments may affect 50-70% of the population in general Both in high lifting jobs and jobs with prolonged sitting
Biomechanics of Lower Back Pain Calculation in text 300N load to 5458N back compressive force Back must support many times the lifted load, largely due to the moment arms involved Calculation of compressive forces vs. muscle strength can identify problems
NIOSH Lifting Guide Sets numbers that are associated with risk of back injury Two limits (for simple lifts) Action limit (AL): small proportion of the population may experience increased risk of injury Maximum permissible limit (MPL): Most people would experience a high risk of injury. 3xAL AL Injuries rare Injuries inevitable MPL Weight
NIOSH Lifting Guide Recommended Weight Limit (RWL): a load value that most healthy people could lift for a substantial period of time without an increased risk of low back pain Covers more complex lifts Biomechanical criteria 3.4kN at L5/S1 Epidemiological criteria show damage at 4.4kN Physiological criteria to set repetition rate at 2.2- 4.7kcal.min
Lifting Equation RWL=LC*HM*VM*DM*AM*FM*CM General form RWL = max possible load * modifiers Modifiers reduce the RWL so that RWL<=LC (all modifiers <1)
The Modifiers LC: load constant, maximum recommended weight for a simple lift HM: horizontal multiplier, decreases weight with distance from spine VM: vertical multiplier, lifting from near floor harder DM: distance multiplier, accommodates for vertical distance that must be lifted AM: assymetric multiplier, reductions for torso twisting CM: coupling modifier, depends on whether loads have handles for lifting FM: frequency modifier, how frequently is the load lifted
Modifiers (diagrammatically) HM VM DM Originating height AMCMFM
Lifting Equation Multipliers can all be obtained from tables (11.1, 11.2, 11.3) Multipliers are unitless Multipliers are always less than or equal to 1 … why?
Example in the Text A worker must move boxes from 1 conveyor to another at a rate of 3 boxes/minute. Each box weighs 15lbs and the worker works for 8 hours a day. The box can be grasped quite comfortably. The horizontal distance is 16 inches, the vertical is 44 inches to start and 62 inches to finish. The worker must twist at the torso 80 degrees.
Example in the Text A worker must move boxes from 1 conveyor to another at a rate of 3 boxes/minute. Each box weighs 15lbs and the worker works for 8 hours a day. The box can be grasped quite comfortably. The horizontal distance is 16 inches, the vertical is 44 inches to start and 62 inches to finish. The worker must twist at the torso 80 degrees. FM Weight CM duration HM VM DM AM
Multipliers HM (T11.1): 10/h=10/16=.625 VM (T11.1):(1-.0075|v-30|)=.895 DM (T11.1): (0.82+1.8/d)=0.82+1/8/18=.92 AM (T11.1): 1-.00032a=1-.00032x80=.744 FM(T11.2): 0.55 (v<75, work 8hrs, 3lifts) CM (T11.3): 1 (good, v<75cm)
Calculation of RWL RWL=LCxHMxVMxDMxAMxFMxCM RWL=51lbx.625x.895x.92x.744x.55x1 RWL= 10.74lbs The load is greater than the RWL so there is a risk of back injury Lifting Index = RWL/Load IF >1 then the load is too high LI= 10.74/15 = 1.4
Designing to avoid back pain More importantly, NIOSH equation gives ways to reduce injury reduce horizontal distance keep load at waist height reduce distance to be travelled reduce twisting add handles reduce frequency of lifts