BIOMECHANICS OF WORK
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Transcript BIOMECHANICS OF WORK
BIOMECHANICS OF WORK
Chapter 11 in your text
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
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: (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
Person holding a 20kg
weight in both hands.
What are the force and
20kg 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/s2 = 196 N
2. Divide by # of hands.
98 N 196N/2 hands = 98N/hand
Problem in Text
1. Convert mass to Force
Felbow
20kg*9.8 m/s2 = 196 N
2. Divide by # of hands.
16 N
196N/2 hands = 98N/hand
98 N
3. Calculate F elbow.
F=0
Felbow – 16N – 98N = 0
Felbow= 114N [up]
Problem in Text
1. Convert mass to Force
Felbow
20kg*9.8 m/s2 = 196 N
.36m
.18m
2. Divide by # of hands.
196N/2 hands = 98N/hand
16 N
3. Calculate F elbow.
F=0
98 N
Felbow – 16N – 98N = 0
Felbow= 114N [up]
4. Calculate M elbow.
elbow-16N*.18m +(-98N)*.36m=
elbow=38.16N*m
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
Weight
Injuries inevitable
Injuries rare
AL
MPL
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 2.24.7kcal.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
Originating height
AM
FM
DM
CM
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 FM
from 1 conveyor
Weight
duration
to another at
a rate of 3 boxes/minute. Each
box weighs 15lbs and the worker works for 8
CM
hours a day. The box can be grasped quite
VM
comfortably. The horizontal
distanceHM
is 16
DM
inches, the vertical is 44 inches to start and
62 inches to finish. The worker must twist at
the torso 80 degrees.
AM
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
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