View
218
Download
0
Category
Preview:
Citation preview
8/15/2019 Link Segment Analysis
1/33
Link Segment AnalysisApplied forces
Joint moments
Net joint force
Muscle forces, and
Joint compression and shear forces
8/15/2019 Link Segment Analysis
2/33
Back Moments
Ø So far we have focused on backmoments with simple models that
assumed we knew the location of the
upper body centre of mass.
Ø These models are not very accurate,because the true centre of mass for
the upper body depends on the
position of each segment.
Ø Dealing with the model segment bysegment allows you to calculate
moments about each joint.
8/15/2019 Link Segment Analysis
3/33
Forces Acting on the LinkSegment Model
Ø Gravitational forces.
Ø Ground reaction forces andother external forces.
Ø Muscle and ligament forces
Ø Joint reaction forces.
8/15/2019 Link Segment Analysis
4/33
Link Segment Models
Ø By looking at each segment insequence we can work our waythrough the body and calculate
the joint moment due to externalforces and the NET joint forcesat each joint.
Ø However, unless we cancalculate the muscle and
ligament forces acting acrossthe joint we cannot calculate thetrue articular (bone-on-bone)compression and shear forces.
8/15/2019 Link Segment Analysis
5/33
Joint Strength
Ø In some cases, analysis of external forces
and joint moment is sufficient. Joint
moments and positions (angles) can thenbe compared to data on joint strengths.
Ø In these circumstances analysis of
internal forces acting on each segment
(muscle, ligament, and joint articularforce) is not necessary.
8/15/2019 Link Segment Analysis
6/33
Free Body Diagrams
The FBD opposite is for
a combined forearm
hand system.The problem is
Statically
Indeterminate.
Fm3
Fm2
Fm1
mg
F j
8/15/2019 Link Segment Analysis
7/33
However this system can be reduced
to a generalised (net) joint moment
and a net joint force
mg
M1 is the net moment due
to all internal (muscle)forces for segment 1.
Rx1
Ry1
M1
Ry1 and Rx1 are the net
joint force components
due to both muscleand articular forces
This diagram is mechanically equivalent to the
indeterminate problem in the previous slide
8/15/2019 Link Segment Analysis
8/33
Static Equilibrium Equations
Σ Fy = 0
Σ Fx = 0
ΣM = 0
Ø This system can now be solved toobtain the joint moment
8/15/2019 Link Segment Analysis
9/33
Solution
mg
M1 is the internal or reaction joint
moment that balances the moment
due to external forces (mg) .Rx1
Ry1
M1 Solution for Static
Equilibrium
mg + Ry1 = 0
Rx1 = 0
M1 + Mmg = 0
8/15/2019 Link Segment Analysis
10/33
Solution (cont.)
mg
Rx1
Ry1
M1
Rx1 = 0
Ry1 = -mg
Hence, Rnet = - mg
r è
M1 + mg * r cos è = 0, or
M1 = - mg * r cos è
where
r = distance from joint
center to segment centerof mass, and
è = angle of segmant to
horizontalNote: M1 is +ve (anti-clockwise)
8/15/2019 Link Segment Analysis
11/33
Limitations of this Method
Ø The net force which is calculated at a jointwill be the vector sum of the muscleforces and the true joint reaction force
(bone-on-bone articular force).Ø It is sometimes referred to as the joint
“reactive” force in textbooks, butremember that it is not the real joint force.
Ø Unfortunately this net joint force cannot beseparated into its muscular force and jointreaction force components.
8/15/2019 Link Segment Analysis
12/33
Ø Note that we can reach this point without any
knowledge of internal muscle and joint forces at all.
Ø HOWEVER, the muscle moment (M1) must always
be considered to act in concert with net joint force.
Ø Simply put, the net joint force (due internal forces) isequal and opposite to the external forces
(gravitational and applied load).
Ø Similarly the net joint moment (due to internal forces)
is equal and opposite to the external moments (dueto gravitational and applied loads).
8/15/2019 Link Segment Analysis
13/33
Multiple linked segments
Ø The above analysis can be extended to multiplelinked segments
Ø The net joint force and joint moment acting at
the proximal end of segment 1 must bebalanced by an equal and opposite force andmoment acting on the distal end of segment 2
Ø Thus we can link together a series of segments
(hand – forearm – upper arm etc.) a shown inthe next slide
8/15/2019 Link Segment Analysis
14/33
Two Segments
m1g
Rx1
Ry1
M1
m2g
-Ry1
Rx2
-Rx1
Ry2
-M1
M2
8/15/2019 Link Segment Analysis
15/33
Static Equilibrium Equations
Segment 1
m1g + Ry1 = 0
Rx1 = 0
M1 + Mm1g = 0
Segment 2
Ry2 + m2g - Ry1 = 0Rx2 - Rx1 = 0
M2 - M1 + Mm2g + M-Rx1 + M-Ry1 = 0
8/15/2019 Link Segment Analysis
16/33
Advantages
Ø The advantage of this method is that net joint
moments can be calculated for all the joints
(with the HAT model shown in the previous
lecture on lifting, we are calculating the moment
only about one joint).
Ø If you calculate a moment across a joint system
where you can accurately model the muscles asa single equivalent muscle, then you can
determine muscle force and joint reaction force.
8/15/2019 Link Segment Analysis
17/33
Analysis of a single joint using the
link segment model
Ø In some work situations you may only be interested in the loading
at a single joint - for example the L5/S1 lumbar joint.
Ø Using a link segment model you can calculate joint force and
moment directly without considering the intervening links.
Ø Note that the net joint forces and moments of adjacent segments
are equal and opposite. Therefore they cancel out in the general
equation.
ØTo solve for net joint force and moment, you need only considerthe external forces acting on each segment and their respective
moment arms about the joint in question (e.g. L5/S1).
Ø Moment arm is the perpendicular distance from the line of action
of each external force to joint centre
8/15/2019 Link Segment Analysis
18/33
Two Segments
m1g
m2g
Geometrical MethodNeed to calculate the moment
arms for each centre of mass.
While this is conceptually easierto understand the resultant
geometry can be tricky when we
get into multiple segments and
external force vectors.
M2
Rx2Ry2
8/15/2019 Link Segment Analysis
19/33
Two Segments
m1g
m2g
Joint moment:
M2 + Mm2g + Mm1g= 0
M2 + m2g * x2 + m1g * x1= 0
Joint force:
Rx2 = 0
Ry2 + m2g +m1g = 0
M2
Ry2 Rx2
8/15/2019 Link Segment Analysis
20/33
Muscle and Joint Forces
The FBD opposite is for
a combined forearm
hand system.The problem is
Statically
Indeterminate.
Fm3
Fm2
Fm1
mg
F j
load
8/15/2019 Link Segment Analysis
21/33
Multiple Muscle Moments
Fm3
mg
F j Even if you knoweach muscle’s line of
action and insertionpoint you still can’t
establish what force
to attribute to each
muscle.
Fm2
Fm1
load
8/15/2019 Link Segment Analysis
22/33
Multiple Muscle Moments
Fm3
mg
F jØCan solve this problem
by using “optimisation”
techniques.
ØFor example: minimizemuscle or joint force;
or minimize muscle
stress (Fm/Am).
Ø Alternatively, can use
EMG data to assign
muscle forces.
Fm2
Fm1
load
8/15/2019 Link Segment Analysis
23/33
Simple Solution:
A Single Equivalent Muscle
Ø There are not many joints that we can claim
that one muscle (or group with a common
line of action and insertion) produces the
moment (e.g. quadriceps in knee extension).
Ø For forearm flexion three prime movers.
Ø For lumbar spine we have bilateral erector
spinae and latissmus dorsi .Ø We often lump such muscle groups together
and term them a “single equivalent muscle”.
8/15/2019 Link Segment Analysis
24/33
Bone-on-Bone Forces
Ø Assumptions:
Ø there is a single equivalent muscle,
Ø the line of force action of that muscle isknown, and
Ø The moment arm is known.
Ø This allows us to calculate equivalent
muscle force and the true (approximate) jointreaction force (the bone-on-bone
compression and shear forces).
8/15/2019 Link Segment Analysis
25/33
L5
L4
5-6 cm
S1
8/15/2019 Link Segment Analysis
26/33
Back to a Link Segment Model
Fm
mg
F jThe problem is
reduced to a single
equivalent muscle.Now we can solve it!
load
8/15/2019 Link Segment Analysis
27/33
In this diagram joint reaction moment M1represents the moment due to muscle
force Fm : or, M1 = (Fm x d).
Fm
mg
F jd
load
Solution:
M1 + Mmg + Mload = 0
M1 = - Mmg - Mload
Fm = M1 / d
8/15/2019 Link Segment Analysis
28/33
Sample Problem
What flexor muscle moment is needed to hold theforearm/hand segments in the position shown?
Use 50% male anthropometry from Kin 201
Taking moments about the elbow. Hence the system inquestion is the forearm and hand. Draw a diagram.
To calculate the answer the first step is to calculate themoment arms from the elbow.
Forearm com = 10.9 cm
Hand com = 25.3 + 9.2 = 34.5 cm
Moments: 0.109 x 1.2 x -9.81 = -1.294 Nm
0.345 x 0.4 x -9.81 = -1.354 Nm
Total = -2.65 Nm
Therefore the elbow flexor muscle moment is +2.65 Nm
8/15/2019 Link Segment Analysis
29/33
Additional Question
If the “forearm flexors” insert 3 cm from the axisof rotation of the elbow, what is the muscleforce and bone-on-bone force?
Moment = Force x ⊥ Distance
2.65 = F x 0.03 ∴ F = 88.333 N
Looking at the free-bodydiagram again.
ΣF=0
88.33 -11.77 - 3.92 + FR=0
FR = + 72.64 N
8/15/2019 Link Segment Analysis
30/33
Link Segment Models Assumptions
Ø Each segment has a fixed mass located asa point mass at its centre of gravity.
Ø The location of the centre of gravity remainsfixed during movement.
Ø The joints are considered to be hinge (pin) joints (2 dimensional models).
Ø The moment of inertia of each segmentabout its mass centre (or distal and proximal joint centres) is constant during movement.
8/15/2019 Link Segment Analysis
31/33
Problem
What is the muscle moment atthe wrist, elbow and shoulder
for our 50th percentile male if
he is carrying a load of 300 N?
Assume the load acts at the
hand centre of gravity.
m2g
Ry2
M2
m3g
-Ry2
Ry3
-M2
M3
Ry1
m1g -300-Ry1
M1
-30o
-M1-30o
-80o
8/15/2019 Link Segment Analysis
32/33
Problem as before
Remember this can be donegeometrically. All four vertical
forces will contribute to the
moment M3.
m2g
m3g
M3
m1g 300 N
-30o
-30o
-80o
8/15/2019 Link Segment Analysis
33/33
Predicted Strength
Ø There are tables that suggest safe limits formuscle moments for various joints.
Ø Other tables provide equations that predict
joint strength. These generally factor in jointangles.
Ø With the use of link-segment models, thesetables can be consulted to compare demandsof the job with worker capabilities.
Ø It is for the ergonomist, designer, etc, todecide if task is suitable.
Ø This is discussed in the next lecture
Recommended