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T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supination

Discussion in 'Biomechanics, Sports and Foot orthoses' started by Kevin Kirby, Apr 8, 2006.


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    In Thought Experiment #6, the same two feet are analyzed as in Thought Experiment #5. A foot with a medially deviated subtalar joint (STJ) is illustrated (left) along with the same foot, now with a foot orthosis underneath its plantar arch (right). Both feet have a medially deviated STJ axis so that the medial weightbearing surface of the foot is 3 cm medial to the STJ axis and the lateral weightbearing surface is 5 cm lateral to the STJ axis.

    As opposed to Thought Experiment #5, however, in Thought Experiment #6, the posterior tibial (PT) muscle has been added to the model so that the foot can actively supinate at the STJ. STJ supination will occur just after the interosseous compression force within the sinus tarsi has dropped to zero. [I have kept the reaction forces under the plantar foot the same in Thought Experiment #6 as in Thought Experiment #5 even though, in the non-rigid human foot, with STJ supination, the reaction forces would be shifted laterally with STJ supination.]

    In both the medially deviated STJ axis foot without the foot orthosis (left) and the medially deviated STJ axis foot with the foot orthosis (right), it is given that the contractile force within the PT muscle is producing just enough tensile force within its tendon so that the interosseous compression force within the sinus tarsi has been reduced to 0. However, the STJ is given to still be in its maximally pronated position. The distance from the PT tendon to the STJ axis is given as 4 cm in each foot (i.e. the supination moment arm of the PT tendon is 4 cm). The PT tendon cross-sectional area is given as 25 square mm.

    Both the medial and lateral weightbearing surfaces of the foot have a ground reaction force of 200 N in the foot without an orthosis (left). In the foot with a foot orthosis (right), the orthosis has been designed to shift the reaction forces on the plantar foot to a more medial location. The orthosis reaction force (ORF) can be thought to act with a magnitude of 200 N at a point 0.9 cm medial to the STJ axis (i.e. the center or pressure of the orthosis is located 0.9 cm medial to the STJ axis). Note that since the foot orthosis is now bearing half of the plantar forces from the foot, the medial and lateral weightbearing surfaces of the foot now each have 100 N of GRF (right).

    The PT tendon force is F1 and the vertically-directed compression force acting through the STJ axis is F2 in the foot without an orthosis (left). In the foot with an orthosis acting on its plantar aspect (right), the PT tendon force is F3 and the vertically-directed compression force acting through the STJ axis is F4. Both feet are given to be in translational and rotational equilibrium.

    Here are my questions for Thought Experiment #6:

    1. What is the magnitude of PT tendon tensile force (F1) required in the foot without an orthosis (left) to bring the sinus tarsi compression force to equal zero? What will be the STJ axis compression force (F2) as a result?

    2. When a foot orthosis is added to the plantar aspect of the foot with a medially deviated STJ axis (right) so that it shifts the reaction force on the plantar foot in a medial direction, what is now the magnitude of PT tendon tensile force (F3) required to bring the sinus tarsi compression force to equal zero? What will be the STJ axis compression force (F4) as a result?

    3. Why is there a greater magnitude of PT tendon tensile force required to reduce the sinus tarsi compression force to zero and initiate STJ supination in the medially deviated STJ axis foot without an orthosis than in the medially deviated STJ axis foot with the orthosis?

    4. By what factor (i.e. 2x, 3x, 4x, etc) has the foot orthosis reduced the magnitude of PT tendon tension required to initiate supination when comparing the medially deviated STJ axis foot without an orthosis to the same foot with an orthosis? How has the orthosis done this even though it, by itself, did not cause STJ supination motion?

    5. Calculate the stress within the PT tendon in both the foot without the orthosis and in the foot with the orthosis in Newtons/meters squared. [1 N/meter squared = 1 Pascal (Pa)]

    6. If the PT tendon had a tear in it that reduced its cross-sectional area to half of its previous value (i.e. after the tear, the PT tendon cross-sectional area is now 12.5 square mm), how would that affect the values for F1 and F2? How would this tendon tear affect the values for the tensile stress within the PT tendon in the foot without an orthosis and in the foot with the orthosis?

    7. Which foot would be more likely to develop posterior tibial tendinitis/dysfunction during weightbearing activities? Why?

    8. How might one design the foot orthosis to further reduce the stress within the PT tendon in the foot on the right? Why will these design modifications work to relieve the PT tendon tensile stress?

    Here's one of many good internet resources for those interested in learning more about the mechanical characteristics of tendons and ligaments.Link on Ligament and Tendon Viscoelastic Mechanical Characteristics
     
  2. Donna

    Donna Active Member

    Hi Kevin :D

    Here are my answers for TE #6...some I'm not too sure about, so please let me know where I am wrong here... :eek:

    1. F1 = 100N, F2 = 500N

    2. F3 = 5N, F4 = 405N

    3. There is a greater pronation moment about the STJ axis in the foot without orthoses (4Nm compared with 0.2Nm in the foot with orthoses), so the PT nees to generate more tensile force to overcome this and produce enough supination moment to make sinus tarsi compression force zero.

    In the foot with orthoses, the ORF helps to decrease the total pronation moment about the STJ axis. From TE #5, sinus tarsi compression force is less in the foot with orthoses, therefore less PT tension required to supinate and reduce sinus tarsi compression to zero.

    4. 20x... the GRF forces have decreased secondary to the addition of the orthosis (ORF), so with the orthotic, the supination moment medial to the STJ axis is 4.8Nm, and the pronation moment lateral to the STJ axis is 5Nm, so 0.2Nm of torque or 5N of tension needs to be produced by PT to reduce the sinus tarsi compression to zero.

    5. Stress in PT tendon without orthoses = 4 x 10^6 N/m squared (Pa),
    with orthoses = 2 x 10^5 N/m squared (Pa).

    6. Wouldn't change F1 or F2 as the model is in equilibrium :confused:
    Stress in torn PT tendon without orthoses = 8 x 10^6 N/m squared (Pa),
    with orthoses = 4 x 10^5 N/m squared (Pa).

    7. The foot without orthoses would be more likely to develop PT dysfunction because there is significantly greater tensile force required to produce supination moment and therefore also significantly more stress within the tendon as it produces this tension. So over time, the PT tendon in the foot without orthoses is more likely to develop degenerative changes, eg. tearing, due to the large amounts of tensile force and stress it is subjected to.

    8. An orthosis that increases the length of the supination moment arm will reduce the PT tensile stress, eg. Kirby medial skive technique, deep heel cup, lateral flange.

    Regards

    Donna :)
     
  3. Donna:

    You are simply amazing. All of these answers are right...100%!! I couldn't even get you with my trick question in #6 since you understood that a tendon tear wouldn't lessen the magnitude of tensile force within the PT tendon that is necessary to achieve rotational equilibrium.

    Send me your e-mail address privately since I would like to send you a little something for all the hard work you have put into answering these Thought Experiments on Podiatry Arena. Excellent work!
     
  4. Donna

    Donna Active Member

    Aww shucks... :eek: I'm just grateful for having the opportunity to learn something new! :cool: So thanks for making these Thought Experiments! I'm sure there are others who are secretly solving the problems themselves and learning from the experience! ;)

    Regards

    Donna :)
     
  5. Posterior Tibial Tear Causes Increased Tensile Stress

    In question #6, I gave the following example:

    6. If the PT tendon had a tear in it that reduced its cross-sectional area to half of its previous value (i.e. after the tear, the PT tendon cross-sectional area is now 12.5 square mm), how would that affect the values for F1 and F2? How would this tendon tear affect the values for the tensile stress within the PT tendon in the foot without an orthosis and in the foot with the orthosis?

    If the PT tendon has 100N tensile force within it, as it does in the foot without an orthosis (F1 = 100N), then it will have 4,000,000 N/m squared tensile stress within the tendon. But when the tendon loses one half its cross-sectional area to a tendon tear from 25 square mm to 12.5 square mm (as illustrated below), then the PT tendon stress in the area of the tendon with the tear will increase to 8,000,000 N/m squared. I hope the drawing below helps to better illustrate this mechanical concept so that it is more clear as to the difference between tendon tensile force and tendon tensile stress. Remember, just like pressure, stress = force/surface area, so that as the surface area of the tendon decreases, the stress also increases.

    Using these concepts, we now have a basis for understanding the etiology of PT dysfunction. I have described these mechanical principles in detail in the following publication (Kirby KA: Foot and Lower Extremity Biomechanics II: Precision Intricast Newsletters, 1997-2002. Precision Intricast, Inc., Payson, AZ, 2002; Kirby KA: Conservative treatment of posterior tibial dysfunction. Podiatry Management, 19:73-82, 2000).

    Craig....can we do superscript notation on Podiatry Arena?...that would be much more convenient for these problems.
     
    Last edited: Apr 9, 2006
  6. Louise B

    Louise B Member

    Kevin,

    Thanks for the 6 'Thought Experiments', I have enjoyed completing them! I am a relative newcomer to the Podiatry Arena (only in the last week) and have worked my way through them from the first to this latest one. I think they are so useful that my team of musculoskeletal Podiatrists are all going to complete them at our team meeting this month (they just don't know it yet!!)

    Thanks once again,

    Louise
     
  7. Louise,

    Thanks for the kind comments. These Thought Experiments are meant to introduce the non-engineer clinician to the concept of free-body diagram analysis and in a way that is clinically relevant enough to warrant their continued interest in the subject. For myself, I know that once I am forced to sit down and solve a problem, I find that I have learned something new that makes the next similar problem much easier to solve.

    Once the clinician starts thinking about their patient's musculoskeletal pathologies in regards to the forces and moments that not only cause the injuries but also in regards to the forces and moments that will need to be applied to the body to heal the injuries, then that clinician will start to develop a deeper understanding of the mechanical nature of musculoskeletal pathologies and their treatment.
     
  8. David Smith

    David Smith Well-Known Member

    Dear Kevin

    Like the thought experiments, each one getting a bit tougher and introducing new concepts of stress and strain as applied to the musculo skeltal system of the lower limb.
    Regarding the post tib torn tendon. If, as you have done, you are simplifying the example for the purpose of analysis then the tendon being reduced in cross sectional area, due to damage, would increase the internal stress as you indicate.

    However in steel, for instance, stress/strain relationships can be saftely calculated assuming no change in cross sectional area due to stress. so stress is expressed as
    Stress = force/area
    Strain can be expressed as a dimensionless term defined by the simple equation
    Strain = (L2-L1)/L1 where L=length,
    (1001-1000)/1000 = 0.001,
    and so stress/strain = young's modulus(E)/modulus of elasticity
    i.e.500N/mm^2/0.001 = 500kN/mm^2.
    Calculations for more elastic materials though must also take an account of the relative change in cross sectional area, I.E. Poissons ratio = lateral strain/longitudinal strain.
    The upshot of this is that as the tendon is subject to greater stress it will become proportionaly longer and therfore, because of it elastic nature, significantly thinner (poissons ratio). So the change/increase in tensile stress will be much higher.(the Poisson effect)

    The concept here is that not only is the tendon thinner because it is damaged but, under the same load, it will become thinner and therefore more greatly stressed because it will stretch significantly. So there may be an exponentially increasing stress load to a damaged tendon. Perhaps this would lead to more sudden failure.

    Just thought this might be an interesting concept to consider.

    Cheers Dave
     
  9. Good one Dave. How does the Poisson's ratio for tendon compare to steel and other material?
     
  10. Donna

    Donna Active Member

    Very interesting... the article describes "shortening" of the flexor tendons in horses due to the horse taking the weight off the affected ("stump") foot, or preferential weightbearing...

    So this sounds almost like the opposite of what is happening with the PT tendon in the TE model... I am finding it hard to get a proper grasp on the concept of equine biomechanics though ;)
     
  11. David Smith

    David Smith Well-Known Member

    Kevin

    I can't say exactly but the poissons ratio for rigid materials eg steel and other metals for construction is 0.25-0.33. The poissons ratio for elastic materials is around 0.50.
    But the elastic material, such tendons, will stretch much more for a given load so therefore the cross sectional area will decrease much more for that same load.

    So then if the tendon of 1.00cm^2 and 1metre long (its only pretend) has a load of 500kg and a poisson's ratio of 0.5 then as it stretches 10mm the cross sectional area transverse dimensions will decrease by 5mm.
    csa original = 100mm^2 csa new = 25mm^2 very significant.

    Steel E (young's modulus) = 200GPa (200kN/mm^2) of dimension 1metre X 1cm^2 and axial load of 500kg will increase in length by 0.0025cm. so its transverse dimension will only decrease by 0.0025 x 0.25 = 0.000625cm which is almost insignificant.
    original cross sectional area 10 x 10 = 100mm^2, new csa= 9.999375 x 9.999375 = 99.9875mm^2.

    1 pascal P = 1N/m^2

    For the same parameters rubber, E=1.7MPa, would stretch by 2.9metres (obviously in reality it would break before this extension was reached)


    Cheers Dave
     
    Last edited: Apr 12, 2006
  12. David Smith

    David Smith Well-Known Member

    Hi all interested

    Looked thru my files to find E of tendon.

    Young's modulus (E) Collagen = 1 X 10^9 N/m^2 (Alexander,1975, p.13)
    (0r 1GPa/1kN/mm^2, About 200 times less than steel)
    "Although the Young's modulus for collagen is more or less constant, we can of course make ligaments stiffer, simply by increasing the cross sectional area. In fact it seems reasonable that the stiffness of ligament is adjusted so that manageable deformations are produced by normal loads on the joints. There's no point in having an extremely stiff ligament any more than a really lax ligament would be useful.
    We could do a sample calculation where we found how much load it would take to elongatea ligament 5mm x 5 mm by 2%:"

    E = stress / strain
    E= 1 x 10^9
    1 x 10^9 =( xN /25 mm^2)/ 2%
    =(xN x 106 x100)/ 2 x 25
    x= 50 x 10^9/10^8 N
    = 500 N
    = 50 Kg weight

    Ref http://medicalsciences.med.unsw.edu.au/somsweb.nsf/resources/ANAT313105/$file/FA1-21-TENDONS.pdf
     
    Last edited: Apr 12, 2006
  13. David Smith

    David Smith Well-Known Member

    Kevin

    I wasn't convinced by that example from medicalsciences.med etc.
    So I did some calcullations of my own.

    lets look at steel first.

    Steel tube with dia1 of 350mm and dia2 of 300mm = cross sectional area of
    Pi,r^2(1) - Pi,r^2(2) = 25,529mm^2 csa.

    Length 5metres. Axial load 2000kN tensile. E = 200kN/mm^2 (young's mod)

    Find stress, strain, poisson effect.

    Hooke's law e = S/E
    Strain (e) = change in length L1/L1

    Stress (S)= F/A (2x10^6)/25529 => S= 78.34N/mm^2

    Strain ratio (hooke's law) = e=S/E => 78.34/(200 x 10^3) = 0.000392

    Strain (e)= 5000 x 0.000392 = 1.95mm extension.

    Poisson effect. Poisson ratio = 0.3 1.95x0.3 = 0.579mm insignificant decrease in csa.


    Do the same for the Post Tib tendon no damage

    Length 100mm dia E = 1kN/mm^2 load = 100N csa = 25mm^2

    Stress = F/A 100/25 = 4N/mm^2

    Strain ratio (Hooke's law) e=S/E e= 4/1000 => 0.004

    Strain (e) 100 x 0.004 = 0.04mm extension

    Poisson effect = Poisson ratio = 0.5 => 0.5 x 0.04 = 0.02mm probably insignificant reduction in csa of 25mm^2

    Now PT with damage in normal walking ie dynamic single leg load = force of 250N in PT tendon.

    Load = 250N E=1kN/mm^2 lenght = 100mm csa = 12.5mm^2

    Stress = F/A 250/12.5 = 20N/mm^2

    Strain ratio (Hooke's Law) e=S/E e = 20/1000 = 0.02

    Strain (e) = 100 x 0.02 = 2mm extension

    Poisson effect = Poisson ratio = 0.5 0.5 x 2 = 1mm

    Reduction of csa by 1mm is this significant?

    Reduce diameter by 1mm = reduce radius by 0.5mm
    original csa =12.5mm^2 = radius 1.99 reduced radius = 1.99 - 0.5 = 1.49mm

    Pi x 1.49^2 = 6.976mm2

    New stress value = 250/6.976 = 35.84N/mm^2 This is a very significant increase in the PT tendon stress load.



    Kevin I have assumed a tendon length of 100mm is the reasonable.
    In walking normal peak GRF is 1.25 x body weight so I thought 250N PT force was reasonable.


    Cheers Dave
     
  14. Donna

    Donna Active Member

    Wow Dave! That is very impressive! Where did you learn all of this? I am intrigued! :eek:

    Regards

    Donna
     
  15. martinharvey

    martinharvey Active Member

    Is there something fishy here?

    I'm concerned that the poisson effect translates as 'fishy' effect. Is there somethin' that an ol' country chiropodist is missin' here. Hope my betters will enlighten me. Regards, Martin
     
  16. Donna

    Donna Active Member

    What do you mean by "fishy"? :confused:
     
  17. Dave:

    Thanks for doing those calculations for me since I like seeing how the math is done. For the average clinician, however, I think it helps for them to see the stress-strain curve for a tendon/ligament as a better way to visualize how the posterior tibial tendon may respond to stresses/forces at its elastic limit (and beyond) and how it behaves when it starts to experience plastic deformation/tearing/failure.

    Eric Fuller and I should finally have our chapter published on Subtalar Joint Equilibrium and Tissue Stress Approach to Biomechanical Therapy of the Foot and Lower Extremity (that we finished writing about a year ago) in a new book on lower extremity biomechanics within the next six months. In it, we have a section on the load-deformation characteristics of tendon/ligament with a nice graph that describes the major points along the curve. I think this is a great way to educate clinicians on the mechanical characteristics of ligaments and tendons. An example of a load-deformation curve for tendon/ligament is demonstrated on the following webpage: http://www.engin.umich.edu/class/bme456/ligten/ligten.htm
     
  18. David Smith

    David Smith Well-Known Member

    Kevin and Donna

    It gives me good practice and just by chance I am writing an assignment on 'Errorrs associated with stress strain measurements'. The experiment was to calibrate a Caribina using strain gauges to measure the strain caused by the force applied to it by isolated wrist and shoulder actions. Then use the voltage output data to calculate force and the moments about those joints.
    One indication of error would be to calculate the strain differential in the caribina and compare it to the stress calculated from the voltage potential from the strain gauges. So this fits in very well with this discussion.

    Martin
    Fishy? Poisson is french for Fish ok!!? but if you are thinking that the calculations don't fit with your 'real world' intuition then you are probably correct. These examples are only that. They give indications of the stress levels but extremely simplified. Structural diagrams, for instance, indicate only the PT tendon as the supinating force but there would be many other structures involved , which would recuce the stress in PT. .
    But it enables one to get the idea or the concept of stress actions on a certain material or structure of interest. I.E. Small increases in lever arms 0r material/tissue properties can have a massive effect on forces, moments and stress within a structue.
    Is this what you intended to say?

    Just to be boring, Here's one area of 'error'.

    Recalculate the tension in PT damaged tendon but only regarding the damaged area which is say 10mm long.

    original csa = 12.5
    Stress original = 20N/mm^2
    Strain ratio = 0.02
    Strain Extension = 0.2
    poisson ratio = 0.5 = change in Diameter of 0.1mm
    New csa = 11.82 = New Stress of 21.15N/mm^2

    See how small input changes = large outcome changes.



    Cheers Dave
     
  19. Donna

    Donna Active Member

    Hi Dave,

    Thanks for showing those extra calculations, I am gradually understanding more of the stress-strain relationship and how different diameter-lengths of tendon are affected... :cool:

    and thanks also for translating that poisson = fish, as you can tell I speak no french. :rolleyes:

    Regards

    Donna
     
  20. David Smith

    David Smith Well-Known Member

    Ne pas problem Donna. Actuellment c'est le meilluer pour la poisson avec pomme de fritte, dejeurne!! Bonsoir.

    Now that's just showing off.

    Bye Dave
     
  21. David Smith

    David Smith Well-Known Member

    I made an error in those calculations so I thought I had better put the record straight

    OOOPS!!

    Error in calculations (see changes in red)

    Steel tube with dia1 of 350mm and dia2 of 300mm = cross sectional area of
    Pi,r^2(1) - Pi,r^2(2) = 25,529mm^2 csa.

    Length 5metres. Axial load 2000kN tensile. E = 200kN/mm^2 (young's mod)

    Find stress, strain, poisson effect.

    Hooke's law e = S/E
    Strain (e) = change in length L1/L1

    Stress (S)= F/A (2x10^6)/25529 => S= 78.34N/mm^2

    Strain ratio (hooke's law) = e=S/E => 78.34/(200 x 10^3) = 0.000392

    Strain (e)= 5000 x 0.000392 = 1.95mm extension.

    Poisson effect. Poisson ratio = 0.3 x 0.000392 = 0.0001176 x 50mm (pipe thickness) = 0.00588mm reduction which = insignificant decrease in csa.

    Do the same for the Post Tib tendon no damage

    Length 100mm dia E = 1kN/mm^2 load = 100N csa = 25mm^2

    Stress = F/A 100/25 = 4N/mm^2

    Strain ratio (Hooke's law) e=S/E e= 4/1000 => 0.004

    Strain (e) 100 x 0.004 = 0.04mm extension

    Poisson effect = Poisson ratio = 0.5 => 0.5 x 0.004 = 0.002 or 0.2% diameter reduction probably insignificant reduction in csa of 25mm^2
    Now PT with damage in normal walking ie dynamic single leg load = force of 250N in PT tendon.

    Load = 250N E=1kN/mm^2 length = 100mm csa = 12.5mm^2

    Stress = F/A 250/12.5 = 20N/mm^2

    Strain ratio (Hooke's Law) e=S/E => 20/1000 = 0.02

    Strain (e) = 100 x 0.02 = 2mm extension

    Poisson effect = Poisson ratio = 0.5 0.5 x 0.02 = 0.01

    Reduction of radius by 1% is this significant?

    Reduce diameter by 1% = reduce radius by
    original csa =12.5mm^2 = radius 1.99 reduced radius = 1.99 – 1.99 x 0.01 = 1.9701

    Pi x 1.9701= 12.196 =

    New stress value = 250/12.196 = 20.5N/mm^2 This is a significant 2.5% increase in the PT tendon stress load.
    (But not so drastic!!)


    I multiplied the poisson ratio by the actual change in length instead of ratio of change in length. Which gives more reasonable results.

    Dave
     
  22. Donna

    Donna Active Member

    Still very impressive Dave ;)
     
  23. David Smith

    David Smith Well-Known Member

    Donna

    It might look impressive but really its not.
    This because these calculations are made using a form of stress /strain / poissons ratio relationship that is called engineering stress analysis (e). This is really only reliable for when the material of interest has a linear or near linear stress/strain curve within the limit of proportionality (similar to the elastic limit) and the ratio of linear to transverse strain is very low and has a positive value.
    This means that strain (stretch) has a linear and proportional function of stress (load/mm^2) to a certain limit of total stress. Whereas elastic limit simply means that the material will return to its original length after it is stretched by a certain load, but not necessarily in a linear way. (Oh blimey! how do I make this more understandable, help me out Kevin). Rigid Non bio materials usually have an elastic limit which exceeeds their limit of proportionality.

    Unfortunately biological connective tissue deformation is not linear in relation to stress. It is a Visco elastic material that has many properties that make calculation of deformation difficult. The form of calculation for this is called True stress/strain analysis the maths is a lot more involved and I'm working on it now.

    If you look at the previous calculations you will notice that the diameter of the material becomes less as the stress and length increases. This is a very noticable effect with rubber. Get a bungie cord and stretch it.
    For steel this ratio is very small so the diameter of the loaded member would be very similar to the unloaded state, provided we are well within the limit of proportionality. Therefore it is reasonable to use engineers analysis for stress.

    If however the change of diameter is great (and non linear) compared to the change in length then the loaded diameter will be much smaller than the unloaded diameter so there will be a much more significant (and unpredictable) change in the rate of change of stress.
    True stress analysis looks at the rate of change of length to diameter change in small segments of time so that each segment give an instantaneous stress/strain value. A graph plotted in this way would not be linear but curved in some way. Tendon for instance starts off a relatively low stiffness then becomes more stiff and then less stiff again. Tendon also suffers, initially, from hysterisis which means its loading curve is different to its unloading curve. Some papers I have read recently hinted that at some stages biomaterial may be auxetic which means it has a negative poisson ratio IE it becomes thicker as the stress becomes greater.

    Anyway when (or if) I know more and if you havn't lost the will to live (yawn) I will post calculations of true stress analysis.

    Kevin's approach is better by the fact that it gets over the principles and concepts without getting lost in maths which, since many parameters are unmeasurable in the clinic , are not very useful for the clinician.
    Approximation and guesstimation become useful tools.

    Cheers Dave
     
  24. Donna

    Donna Active Member

    Hi Dave

    My little brain is learning quite a lot from Podiatry Arena, and I really appreciate you taking the time to explain these principles in such great detail... At uni the subtalar joint neutral approach was taught, so the tissue stress model is quite new and interesting to me, plus it actually makes a lot more sense. :cool: The stuff I learnt in high school physics is slowly coming back to me, and when applying these principles it does reinforce and improve my understanding of a lot of the concepts... so please continue to post your calculations as I find them fascinating ;)

    Regards

    Donna :D
     
  25. David Smith

    David Smith Well-Known Member

    Donna

    I think thank the tissue stress model makes a lot of sense but I also believe that positional models like STN are useful.

    This is because the joints of the foot are not a stable structure but a mechanism. A stable structure is stable when forces are applied from any direction. A mechanism is stable from forces in some directions but not in others. These forces can cause it to move. Also if the mechanism is not in a position of stability any force can cause it to move.

    Look at a square truss (2D) fixed on the ground and with pin joints, it is stable to vertical forces but not horizontal forces. It is only stable to vertical forces if it remains in its square position once it becomes a parallelagram it is a mechanism and unstable in both directions. The position would give an indication of instability. Just like you wouldn't stand on a stool with crooked legs because intuitively you could see that it may be unstable.

    In the foot tho the position tells you nothing about the forces acting about it. But if it were true that STN position is the position of intrinsic stability due to gravity thru the closed kinetic chain then moving out of that position would require extrinsic forces to maintain stability. How and when those forces become pathological is open to much debate and requires knowledge of tissue stress and mechanics.

    Thats only my take on positional versus tissue stress what do you think?

    Cheers Dave
     
  26. Donna

    Donna Active Member

    Hi Dave,

    I agree with what you have said, the tissue stress model, to me, makes more sense since you can see (with quantified forces) how the stresses and strains have damaged the tissues. I use the SJN approach as that was how I was taught at uni, and I think I'd be a bit lost if I abandoned using SJN theory for biomechanical assessments. It is also a lot easier to explain the basics of SJN to a patient when using bisection lines, as they can physically look and see the "misalignment" of the posterior heel and leg, and how an orthotic can straighten these lines. :cool:

    At present I am using the Tissue Stress Theory just in my own head, and am thinking things through a bit differently when I assess a patient and prescribe orthoses. :eek: I am very new to this, so I know it's going to take a while to become proficient, but the Thought Experiments have started me thinking, which is a good thing (I hope).

    I am currently reading Kevin Kirby's "Foot and Lower Extremity Biomechanics II" which explains a lot of the principles in great detail, and my understanding is gradually getting better. ;)

    Regards

    Donna :)
     
  27. efuller

    efuller MVP

    Hi Donna,

    The thing to do is figure out what parts of the STJ neutral theory you can discard. For example the nuetral part. However, there are parts that you have to keep. An uncompensated varus foot that has callus along the lateral plantar foot exits and is explained well by the STJ neutral theory. It is however, overcomplicated by dividing it into forefoot or rearfoot varus. The subtalar joint can't tell whether it is maximally pronated because of a rearfoot or forefoot deformity. The stress is still in the floor of the sinus tarsi.

    When you start using tissue stress explanations with your patients you will be amazed how they understand it just as well as the allignment explanation. It is amazing what people will believe when they hear someone with a few letters after their name explain it.

    Cheers,

    Eric
     
  28. Donna

    Donna Active Member

    G'day Eric! :D

    True what you're saying there! I guess I just need to "turn the screws" :p a bit more in regard to how I am going to use the tissue stress and SJN theories together, without them clashing badly...and when I get my head around it all I will be more confident in explaining it to patients :cool:

    Regards

    Donna ;)
     
  29. It's interesting to discuss the different theories. Here's my opinion of each theory:

    STJ neutral theory is a theory that assumes that the STJ neutral position is the ideal rotational position of the STJ and that attempts to explain the function of the foot and how to design your patient's orthotics using the STJ neutral position. It has some good aspects to it, but falls short as a good-all inclusive theory for foot function and orthosis design since it is based on position of the calcaneus and STJ rotational position, and does not at all take into account the large interindividual variation in spatial location of the STJ axis in the human population.

    STJ axis location/rotational equilbrium (SALRE) theory is a theory that explains how external and internal forces and moments acting on and within the foot are determined largely by STJ axis spatial location. SALRE theory does not, however, tell you how exactly to design your patient's orthoses but may be used as a basis for designing orthoses.

    Tissue stress theory is a theory which does not necessarily explain how the foot works but gives the clinician a guide by which to design orthoses, shoes or braces to allow their patients to heal their injury. In other words, tissue stress is a theory that explains how tissues are injured and how treatment may be directed to optimum healing of those tissues.

    SALRE theory combined with tissue stress theory work like hand in glove since, when used in combination, they offer a concise theory of function and treatment of mechanically based pathologies of the foot and lower extremity. SALRE theory explains the mechanical behaviour and function of the foot and tissue stress theory explains how mechanical treatment may modify internal forces.

    I suggest keeping the good parts of STJ neutral theory but with the realization that STJ neutral theory does not adequately explain why many injuries occur. The strict adherence to STJ neutral theory, which is still commonly taught in podiatry schools around the world, will prevent the podiatry student and podiatrist from understanding the mechanical etiology and most effective treatments of a number of common mechanical pathologies of the foot and lower extremity.

    The Subtalar Joint Equilibrium and Tissue Stress Approach to Biomechanical Therapy of the Foot and Lower Extremity, in my opinion, is the most common sense and efficient way of approaching the mechanical treatment of the foot since it blends SALRE theory with tissue stress theory to allow the clinician to better understand not only how the foot works but how to make it better with foot orthoses and other mechanical treatments.
     
  30. Donna

    Donna Active Member

    Thanks Kevin! :) That's very useful info!

    Regarding the SAL/RE, I have been practicing finding the STJ axis location, and am finding a lot of very medially deviated STJ's! Is this a normal finding in the general population, or are my fingers too weak? :eek: I am finding the medially deviated STJ axis in heavy adult patients but also in children...is it a case of practice makes perfect?! :confused:

    I also wonder if someone who was bigger and stronger than me assessing the same patients would get the same results? Is there a trick to it? :eek:

    Regards

    Donna
     
  31. The STJ palpation technique (Kirby KA: Methods for determination of positional variations in the subtalar joint axis. JAPMA, 77: 228-234, 1987) requires only a normal amount of manual strength to be performed successfully. About a 10 pound pushing force from the distal tip of the examaner's thumb onto the relaxed patient's foot medial to the STJ axis should produce enough supination motion (and lateral to the STJ axis should produce enough pronation motion) to be detectable by the examiner's hand that is grasping the 5th metatarsal head (to sense STJ motions). Make sure to reposition the foot into the "plantar parallel position" after each plantar push onto the foot. Also make certain that the push from the examiner's thumb is directly parallel to the midsagittal plane of the patient. You will get better as time goes on and I would try to practice on a few feet a day to start to feel more comfortable with the technique. In the photo below, this lady's right foot has PT dysfunction and a severely medially deviated STJ axis.

    In the general population as a whole, I would estimate that about 50% of the population has a medially deviated STJ axis, with 40% having a fairly normal STJ axis location and 10% having a lateral STJ axis location. I have not done a formal study on this so the above percentages are simply educated guesses from doing the technique for over 20 years on thousands of patients.

    You will also find that a good clinical weightbearing estimate of the STJ axis location comes from marking a point on the posterior foot at the posterior-lateral-superior calcaneus (i.e. approximate posterior exit point of STJ axis) and then marking the bisection of the talar neck from anterior (i.e. approximate anterior exit point of STJ axis). Next, using a pen or pencil at the anterior mark on the talar neck and "pointing" the pen or pencil toward the mark on the posterior calcaneus, this will give a fairly good representation of the STJ axis in many feet, as long as they aren't too deformed. The subtalar joint axis locator is a device that Dr. Simon Spooner and I collaborated on that actually locates the STJ axis with fairly good precision and this paper should be published next month in JAPMA (Spooner SK, Kirby KA: The subtalar joint axis locator: A preliminary report. JAPMA, In Press, 2006).
     

    Attached Files:

  32. Donna

    Donna Active Member

    Thanks again, Kevin! I will keep practicing my technique... :D

    Regards

    Donna :cool:
     
  33. Asher

    Asher Well-Known Member

    Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    Hi Donna and Kevin,

    I'm up to TE#6 and have surprised myself that I can do some physics calculations (I got a D for yr 12 physics and an A in everything else so physics is a sore point for me).

    I'm going fine with the clockwise forces = anticlockwise forces and got the correct answer (20 times less). But I don't understand how you worked out the calculations above.

    Your help would be appreciated.

    Rebecca
     
  34. Donna

    Donna Active Member

    Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    Hi Rebecca,

    The way I calculated the answers as above is:

    (F1 x 0.04) + (200 x 0.03) = 200 x 0.05
    (F1 x 0.04) = (200 x 0.05) - (200 x 0.03) = 4 Nm
    F1 = 4Nm/0.04 = 100N

    F2 = F1 + 200N + 200N = 500N

    (F3 x 0.04) + (100 x 0.03) + (200 x 0.09) = 100 x 0.05 = 5Nm
    F3 x 0.04 = 5Nm - (100 x 0.03) - (200 x 0.09) = 5 - 3 - 1.8 = 0.2Nm
    F3 = 0.2/0.04 = 5N

    F4 = F3 + 100 + 200 + 100 = 405N

    So F3, or the PT tensile force in the foot with the orthosis is 1/20 the amount of F1, which is the PT tensile force in the foot without an orthosis (ie. F1/F3 = 100/5).

    Hope this answers your questions and makes sense...

    Regards

    Donna :)
     
  35. Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    Rebecca:

    It was probably the teacher, not you. I had a great physics course with about 30 students during my junior year of high school (age 16-17) and had a great learning experience due to the teacher, Mr. Limpach. He did lots of neat demonstrations and was excited about the subject. I loved it. Then I got to UC Davis for my college education to again take a year long physics course (age 18-19) and struggled more with the subject since it was taught by someone who was less interested and it was less personal, with about 200 students in the classroom.

    I'll bet if I taught your physics course you would have done better. Physics explains the mechanics of the real world!;):drinks
     
  36. Asher

    Asher Well-Known Member

    Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    Thanks Donna for your prompt response. I have just now had a chance to concentrate on your post.

    Basically, the calculations you provided are the bits that I get. What I have no idea of is how to answer question 5: Calculate the stress within the PT tendon in both the foot without the orthosis and the foot with the orthosis in Newtons / Metres squared (1Nmetre squared = 1 Pascal (PA). Thanks in advance for your help!

    Also, I do have some questions about the calculations you have provided:

    Firstly, for others that might be trying the calculations also (as if), where you have 0.09, it should actually read 0.009 (unless I am mistaken?).

    But importantly, my 2 questions are:

    1. I didn't convert cms into metres - does that matter because you get the same answer?

    2. I got the same first line of the calculation as you, but my second line is different, which instead of coming up with another calculation, I just carried the first calculation further to find F1:

    (4 x F1) + (3 x 200) = (5 x 200)
    (4 x F1) + 600 = 1000
    therefore (4 x F1) = 400
    so F1 = 100N
    ... and so on.

    Will this stand me in good stead or have I found a way of calculating which makes me lucky now but will trip me up later.

    Sorry for roping you back into these calculations Donna and Kevin! :eek:

    Regards

    Rebecca
     
  37. Donna

    Donna Active Member

    Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    Hi Rebecca...

    Yes you are right...where I have 0.09 should have been 0.009... typo... I think it has something to do with the fact I am currently working close to a 60 hour week and am operating on minimal sleep... hmmmm...

    Question 5:
    Post tib tensile force/cross sectional area = Stress

    Without orthosis:
    100N/0.000025 = 4 x 10^6 N/m

    With orthosis:
    5N/0.000025 = 2 x 10^5 N/m

    Regards

    Donna:)
     
  38. Asher

    Asher Well-Known Member

    Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    THANKS FOR THE EQUATION. BUT WHY ARE THERE SO MANY ZEROS IN FRONT OF THE 25 MM CROSS SECTIONAL AREA? IF WE NEED IT IN METRES, ISN'T 25 MM EQUAL TO 0.025 METRES.

    I'M SURE YOUR RIGHT DONNA, ITS JUST MY POOR UNDERSTANDING, SO I NEED TO KNOW WHY.

    THANKS

    REBECCA

    PS: MY CAPS LOCK BUTTON IS STUCK - SORRY
     
  39. Asher

    Asher Well-Known Member

    Re: T.E. #6: Effect of Foot Orthoses on PT Tendon Force Required to Cause STJ Supinat

    APOLOGIES FOR MY FRIVOLOUS POST ABOVE. I HAVE JUST SPOKEN TO A MATHS TEACHER FRIEND WHO HAS ENLIGHTENED ME.

    BTW DONNA, IT SOUNDS LIKE YOU NEED A COUPLE OF WEEKS OF OVER CHRISTMAS / NEW YEAR.

    REGARDS

    REBECCA
     
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