Link Segment Analysis

8/15/2019 Link Segment Analysis

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Link Segment AnalysisApplied forces

Joint moments

Net joint force

Muscle forces, and

Joint compression and shear forces


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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.


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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.


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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.


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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.


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Free Body Diagrams

The FBD opposite is for

a combined forearm

hand system.The problem is

Statically

Indeterminate.

Fm3

Fm2

Fm1

mg

F j


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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


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Static Equilibrium Equations

Σ Fy = 0

Σ Fx = 0

ΣM = 0

Ø This system can now be solved toobtain the joint moment


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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


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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)


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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.


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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).


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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


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Two Segments

m1g

Rx1

Ry1

M1

m2g

-Ry1

Rx2

-Rx1

Ry2

-M1

M2


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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


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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.


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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


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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


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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


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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


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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


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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


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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”.


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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).


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L5

L4

5-6 cm

S1


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Back to a Link Segment Model

Fm

mg

F jThe problem is

reduced to a single

equivalent muscle.Now we can solve it!

load


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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


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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


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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


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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.


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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


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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


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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

Documents

Link Segment Analysis