BIOMECHANICS OF WORK

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 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

Muscles

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: 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 (åF=0, å 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 åF can be calculated

Problem in Text 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! 20kg

Problem in Text 1. Convert mass to Force 20kg*9.8 m/s2 = 196 N 2. Divide by # of hands. 196N/2 hands = 98N/hand 98 N

Problem in Text åF=0 1. Convert mass to Force 20kg*9.8 m/s2 = 196 N 2. Divide by # of hands. 196N/2 hands = 98N/hand 3. Calculate F elbow. åF=0 Felbow – 16N – 98N = 0 Felbow= 114N [up] Felbow 16 N 98 N

Problem in Text Felbow 16 N 98 N 1. Convert mass to Force 20kg*9.8 m/s2 = 196 N 2. Divide by # of hands. 196N/2 hands = 98N/hand 3. Calculate F elbow. åF=0 Felbow – 16N – 98N = 0 Felbow= 114N [up] 4. Calculate M elbow. åM=0 Melbow-16N*.18m +(-98N)*.36m=0 Melbow=38.16N*m Felbow .36m .18m 16 N 98 N

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 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 (single 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

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 Biomechanical criteria Epidemiological criteria Physiological criteria

Lifting Equation RWL=LCxHMxVMxDMxAMxFMxCM LC: load constant, maximum recommended weight 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

Lifting Equation Multipliers can all be obtained from tables (11.1, 10.2, 10.3, 11.2, 11.3) Multipliers are unitless Multipliers are always less than or equal to 1 (they reduce the maximum load or load constant)

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.

Information h=16” v=44” d=18” A=80degrees F=3 lifts/minute C=good job duration = 8 hours/day weight = 15lbs

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.

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