Download - BIOMECHANICS OF WORK
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BIOMECHANICS OF WORK
Chapter 11 in your text
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The Musculoskeletal System
Bones, muscle and connective tissue supports and protects body parts maintains posture allows movement generates heat and maintains body
temperature
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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
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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
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Muscles
400 muscles 40-50% of your body weight half of your body’s energy needs
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Muscles
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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)
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Muscle Contraction
Concentric: (also called isotonic) muscle contracts and shortens
Eccentric: muscle contracts and lengthens (overload)
Isometric: muscle contracts and stays the same length
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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
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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
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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!
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Problem in Text
1. Convert mass to Force
20kg*9.8 m/s2 = 196 N
2. Divide by # of hands.
196N/2 hands = 98N/hand98 N
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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
3. Calculate F elbow.
F=0
Felbow – 16N – 98N = 0
Felbow= 114N [up]
98 N
16 N
Felbow
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Problem in Text1. 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.
F=0
Felbow – 16N – 98N = 0
Felbow= 114N [up]
4. Calculate M elbow.
elbowm +(-98N)*.36m=0
elbow=38.16N*m
98 N
16 N
Felbow
.18m.36m
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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?
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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
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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
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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
ALInjuries rare Injuries inevitable
MPL
Weight
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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
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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)
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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
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Modifiers (diagrammatically)
HM VMDMOriginating height
AM CMFM
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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?
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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.
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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.
FMWeight
CM
duration
HMVM
DM
AM
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Information
h=16” v=44” d=18” A=80degrees F=3 lifts/minute C=good job duration = 8 hours/day weight = 15lbs
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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)
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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
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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