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Why are high arched orthoses stiffer than low arched orthoses?

Discussion in 'Biomechanics, Sports and Foot orthoses' started by Simon Spooner, Mar 19, 2011.

  1. So, how does the compression of each area of the wedge per unit load (in other words the stiffness characteristics) relate to reaction forces at the foot-wedge interface at each area (1-5)?

    If we assume that the wedge is loaded at heel strike. The force at each area of the wedge (and at the foot-wedge interface) during this impact should be equal to the rate of change of momentum at each area, 1-5. Lets assume the mass of the body is equally distributed across the loading interface of the wedge and that equilibrium between each portion of the foot and wedge at the interface occurs when velocity equals zero and that the areas of the foot which come into contact with the areas of the wedge have the same initial velocity. Since each area of the wedge will compress more of less under loading due to it's length, then the time from initial contact to zero velocity should vary too, with the section which compresses the most under a given load (the least stiff) taking longer to reach zero velocity... hence the force of impact will be lower beneath the thickest portion of the wedge, since it is more compliant. This seems counter intuitive, since if we then calculated the centre of the forces (centre of pressure), the wedge should shift the centre of force (pressure) toward the thin end, compared to a flat piece of foam in which the centre of force would be in the middle. This is why my brain keeps turning to jam. Obviously we have assumed that the loading is equally distributed across the wedge and that the initial velocity of each portion of the foot is the same too, this probably isn't true. But it does make you think. If anyone can tell me why I'm wrong here, please, please, please chip in. This is so counter-intuitive its driving me mad (like you hadn't noticed), which is why I feel I must be making a mistake somewhere!
     
  2. David Smith

    David Smith Well-Known Member


    Simon

    Got a meeting to go to but here's a quick reply that might settle your mind

    Look at this page on Wikipedia I think it sums up and confirms my statement about strain and change of length due to a given force. http://en.wikipedia.org/wiki/Hooke's_law


    [​IMG]
     
  3. Dave, I've looked at that wikipedia page on numerous occasions. I'd rather you just state in blunt terms whether what I'm saying is right or wrong, in your opinion. I think we are in agreement and that what I am saying is correct. I'd actually think that as we move from the thin end of the wedge to the thick end, the deformation increase per unit load- is that what your diagram 1 and 2 shows? I think it is; I don't think diagram 3 is correct. I'll pull some springs out and try to show what I'm saying, it'll be in reverse though because they're tension springs. So what about the interface forces? Am I right on this? Looking at your diagram the impact should have been decelerated over a greater vertical distance at the thick end of the wedge, thus the impact force here should be lower than at the thin end of the wedge (Newtons 2nd law of motion). Thus, since Newton's 3rd law of motion is valid, should also mean that the orthosis reaction force is lower here.- Then, if we look at the shift in centre of pressure of the forces, it should be moved toward the thin end of the wedge, compared to a flat surface of the same material- right? (that's the counter intuitive bit in my head- since I learnt that wedges shift the centre of pressure toward the thick end- they may do, but not due to variation in deformation per unit load across their surface (if I'm right)), it must be due to the "splitting" of loads into vertical and horizontal components that occurs on inclined planes. Viz. the surface geometry at the foot-wedge interface. Or, frictional differences at the interface. So, in order to have a shift in the COP toward the "thick end of the wedge" there must be a balance between the angulation angle and stiffness of the wedge that allows this, thus we should be able to calculate this relationship, thus we should be able to use this relationship in orthosis prescription writing. "You may say that I'm a dreamer... but I'm not the only one"- Imagine: John Lennon. Enjoy your meeting.
     
  4. Moreover, a high arched shank dependent orthoses should have reduced stiffness characteristics compared to a low arched shank dependent orthoses manufactured from the same material for a point representing the apex of the medial longitudinal arch section of the device.:cool: C'mon Kevin, so you lost the initial moot point, which means you get the first beers in Manchester, I need some help here. Please.
     
  5. efuller

    efuller MVP


    Clinically, it is relevant whether an orthosis is shank dependent or independent. I can concede the mechanical part where it is "possible" to make a shank dependent device have the same surface deformation characteristics as a shank independent device. It would be very hard to make them the same. But, they would cease to be the same as soon as patient takes the device out and puts it in a different shoe. So, you change how you practice based on whether a device is shank dependent or not. You have to give the patient different instructions on what shoes they wear with the device.


    On why high arched devices are stiffer:
    I liked the explanation that someone gave on effective cross sectional stiffness. The load is mostly vertical and in the high arched device will have more shell thickness in the vertical direction because of the steeper slope. Or it might be expained that more of the force is compressive because in the high arched device the vertical force is more in line with the material. In testing the observation it does seem true. I'm having a hard time thinking of the experiment that could separate the two above hypotheses.

    Eric

    Eric
     
  6. David Smith

    David Smith Well-Known Member

    AAhh! I see your confusion now. The acceleration or deceleration is angular not linear. So as the angular velocity increases from the centre to the perimeter (thin end to thick end) the change in angular velocity is also greater at the perimeter. The time from initial velocity to terminal velocity is constant at any point from the centre to the perimeter.

    In terms of the dynamic evaluation - The force applied along any part of the interface a initial contact is zero and the force applied at the terminal contact i.e. max compression is maximum and even along the interface. This must be so since this is proved by Hookes law. E.G. if you take each one of your foam columns of varying length and apply an equal weight or force to each say 10kg / 98.1N then they will all compress to the maximum and all will have the same force applied at max compression. As the foam compresses the force will rise from zero to max in the same amount of time but therefore the accelerations will be different.

    So in terms of the static evaluation - at any point in time between initial contact and maximum compression the contact force must be even along the interface.

    You will notice then that if force is constant across the interface and displacement from initial contact to full compression is increasing proportionally across the interface from centre to perimeter then the work done must increase proportionally from centre to perimeter.

    Force over distance/displacement = work and work divided by time = power
    Therefore there is more power at the thick end.

    Its late and I can't think how to prove that mathematically at the moment but I think the concept is good.

    regards dave
     
  7. Here is my contribution to the discussion based on Dave's suggestion that the low arched orthosis exerts more horizontal force than does the high arched orthosis and that this is the reason the high arched orthosis will be more stiff than the low arched orthossi.

    In the lower illustration, a low arched orthosis has a 260 N force directed inferiorly onto the apex of its arch and will have 130 N forces directed superiorly at the supporting ends of the orthosis. However, due to the low arch angle, the laterally directed thrust from the ends of the orthosis will be 715 N or 2.75 times the vertically directed force on the apex of the arch.

    In the upper illustration, a high arched orthosis has a 260 N force directed inferiorly onto the apex of its arch and will have 130 N forces directed superiorly at the supporting ends of the orthosis. Due to the higher arched orthosis having an arch that is two times the height of the low arched orthosis, the laterally directed thrust from the ends of the orthosis will only be 350 N or only 1.35 times the vertically directed force on the apex of the arch.

    Therefore, the lower arched orthosis has more laterally directed force at its ends than does the higher arched orthosis of the same base dimensions when they are subjected to equal vertical loading forces. As a result, the lower arched orthosis wil be more likely to deform or, said another way, the lower arched orthosis will be less stiff than the higher arched orthosis. Of course, if the ends of the orthosis are prevented from moving away from each other by being fixed onto the supporting surface when the 260 N load is applied, then the orthoses should have equal stiffness since they each will undergo minimal deformation during vertical loading.

    Dave, does the illustration and explanation make sense to you? Your idea seems like the best explanation so far.
     

    Attached Files:

  8. One more thing, I did some research on the area moment of inertia and its relation to stiffness and how orthosis thickness may affect orthosis stiffness in a shank independent orthosis.

    Here is the formula, for area moment of inertia in a rectangular beam: I = bh^3/12

    I = area moment of inertia

    b= base width of beam

    h = height of beam

    http://www.engineeringtoolbox.com/area-moment-inertia-d_1328.html

    So as the height (i.e. thickness) of the beam is doubled, the stiffness will increase by 8x. If the thickness of the beam is increased by 1.5x, then the stiffness will increase by 3.375. In order to double the stiffness of a beam, the thickness of the beam must be increased only by a factor of 1.26. In other words, an orthosis that is 3 mm thick is approximately half the stiffness of an orthosis that is 3.8 mm thick.

    However, if the width of the beam is doubled, then the stiffness is doubled. Therefore, an orthosis that is twice the width of another orthosis of equal anterior to posterior span, then the wider orthosis will be twice as stiff.

    Here are some nice calculators that allow you to plug in values for beams supported at both ends.

    http://www.engineersedge.com/beam_bending/calculators_protected/beam_deflection_1.htm

    http://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.html
     
  9. This and Roberts argument regarding moving foot orthosis from one pair of shoes to another seems to make the assumption that only the shank dependent device will be influenced by the shoe. This simply isn't true.
     
  10. the difference between heel height and forefoot height for 1 will make a huge difference.

    In all the pictures of Super Dave and Kevin the base of the arc of the device begins and ends on the on the same surface, but an orthotic in a shoe will not, the heel will be higher that the forefoot - so the mechanics maybe different and will in my way of thinking effect stiffness.

    After the success of my last picture :D Ive drawn another, will the heel height effect the calculations of force ?
     

    Attached Files:

  11. David Smith

    David Smith Well-Known Member

    Yes that's it Kevin, if I had one of your spacetime manipulating machines I would do some really good diagrams too.

    But, as I mentioned earlier this is an intuitive or conceptual illustration. The reason is this is a static analysis.

    In your diagram the force opposing the horizontal forces would be friction the friction is what stops the ends moving apart. Imagine that the supporting ends were on frictionless rollers and the same force was applied. The only forces would be vertical and they cause a bending moment about the apex of the arch, therefore the arch would lower until its stiffness to resist bending matched the applied forces and moments.

    If you were applying a given weight to the arch e.g 80kg then that is a potential 785N but as the arch deflects it cannot have an opposite reaction force of 785N until it reaches a point of equilibrium, i.e. stiffness to bending = applied forces and moments. So until that point there will be a force applied somewhere between 0 and 785N and so there also will be a proportional increase in the horizontal forces.

    Now getting back to friction, if the frictional reaction force available is above the applied horizontal forces then there is a stabilising effect, i.e. the frictional reaction forces apply a moment about the apex that is in the opposite direction to the moments from reaction vertical forces. Once frictional forces are overcome the model becomes one like the frictionless one described above and the arch deformation occurs until resistance to bending stiffness and frictional forces are in equilibrium with the maximum applied force i.e. weight times gravity.


    As a side note, which might be helpful to Simon, if the arch has a significant thickness and the reaction force vector stays within the constraints of the material thickness, i.e. where the vector points from base support to apex, then there are no bending and shear forces applied to the arch walls between the apex and the supports. This is more readily achieved with a low arch than with a high arch for the same wall thickness. So if the arch walls were segmented then the compression effect of the reaction vector would make the low arch more stable than the high arch.

    Dave
     
    Last edited: Mar 23, 2011
  12. Thanks Dave.

    Let me try to summarise our foam wedge example so that i may check my understanding:
    1) the wedge will deform more under a given load at the thick end than the thin end.
    2) since stiffness = force / deformation, the wedge is less stiff at the thick end than the thin end.
    3) If we impact the wedge with a given linear/ rotational velocity say 10m/s then the intiai impact force will be greater at the thin end of the wedge because the rate of deceleration will be greater due to the stiffer material (walking into a lamp-post versus walking into a mattress) - shorter braking distance = faster rate of deceleration. f=ma mass is unchanged, acceleration (or deceleration) varies across the wedge due to stiffness differences
    4) Once equilibrium is reached the reaction force will be the same across the wedge.
    5) energy transfer will be greater to the thick end of the wedge since greater deformation occurs here.
    6) More work is done on the thick end of the wedge.
    7) the rate of work will be higher at the thick end of the wedge.
     
  13. Phil Wells

    Phil Wells Active Member

    Simon

    Nice summary!
    Just something to add to the pot is the material properties of a Shank dependent foam e.g closed cell EVA.
    Its loading characteristics are non-linear, visco-elastics and temperature dependent. Also wear and tear changes its load attenuation properties.

    Lets not let this become too easy!!!!

    Cheers

    Phil
     
  14. Thanks Phil, although I'm not sure about point 3)
    This seems to make sense to me but Dave said: "- at any point in time between initial contact and maximum compression the contact force must be even along the interface."

    Which seems to contradict my point 3)??
     
  15. Visco-elastic material are strain rate dependent then the faster the rate of loading the "stiffer" they become. So in our wedge, the loading rate is fastest at thin end of the wedge and slowest at thick end....?
     
  16. All attached is an image of two sections across the heel cup area of two devices. The device on the left has a "standard" form whereas the device on the right has a "heel-skive" the arrows indicate the point where each of the devices should be stiffest (deform the least amount for a given vertically directed load applied to their superior surfaces). Note that in the heel skive the thickness of the material immediately to the left of the arrow (lets call this medial) is increased compared to the same area in the standard heel cup. It should therefore be less stiff. The deeper the heel skive is made the thicker this section of the orthosis should be, therefore the more compliant it should be too.
     

    Attached Files:

  17. David Smith

    David Smith Well-Known Member

    Simon

    you wrote

    Now your making me think and go all counter intuitive like, so I did some (two) experiments (damn you I'm missing lunch now:craig:!!)

    First I took two identical elastic bands both 6 inches long. I then hung an 18oz mallet from one band and the displacement (stretch) was 6.5inches.
    I then hooked the two bands together in series and again hung the 18oz mallet from them. The measured displacement was 13.5 inches.

    I then made 12 x 25mm * 10mm blue EVA blocks and stuck them on top of each other with double sided tape in a tower arrangement.
    I then put the tower of eva (120mm high) and placed them inside a small cardboard box on top of a set of electronic bathroom weighing scales. I marked the unloaded initial height on the inside of the box. I then, using my thumb, vertically applied 11 kg weight to the tower and marked the displacement (compression under load) on the side of the box. I repeated 3 times.

    I then removed 6 blocks cut down the box and did the same again vertically applying 11 kg to the 60mm tower and marking the displacement (compression) and repeating 3 times.

    120mm tower displacement = 10mm +/- 1mm
    60mm tower displacement = 5mm +/- 1mm

    Obviously this confirms the reasoning made earlier, which is just as well since I don't like to look too dumb.

    Obviously the EVA has a material stiffness rating which does not change just because it is stacked higher. That stiffness is defined by Hookes law. E (young's modulus)= stress/strain or strain (change in length) = E*stress

    Force/displacement is not strictly the same as stress/strain

    Got a customer now - More later

    Regards Dave
     
  18. Dave,
    Happy to make you think. I did pretty much the same experiments yesterday in my materials testing jig rather than a cardboard box- cool idea though. I even did the eva wedging strip experiment too. The fact that the materials act roughly like springs didn't surprise me. The counterintuitive bit is the effect on the impact forces- at impact the wedge should shift the cop pressure toward its thin end if we ignore the geometrical effects on the force vectors as velocity should be decelerated more rapidly here. Moreover if the velocity of the foot is decelerated to zero faster under the thin end of the wedge than it is under the thick end of the wedge this should tend to cause the foot "to roll" towards the thick end of the wedge?
     
  19. David Smith

    David Smith Well-Known Member

    Simon

    This may be where you thinking is stuck, the time to zero is the same at the thin end as the thick end. The change in velocity is greater at the thick end since the displacement is greater for the same time.

    UUrggh now I'm doubting what I say :bang: stop doing that:mad:

    Regards Dave
     
  20. so it is the rate of change of velocity that is different
    The force is still related to this since force equals mass times acceleration? And rate of change of velocity is acceleration. So we still have higher forces at the thin end of the wedge compared to the thick end- right? So, this should tend to shift the cop toward the thin end. Counter-intuitive- right? Cool.
     
  21. Centre of force (pressure) shift in relation to relative deformation under load across the surface of a wedge of material.... is there a relationship between the angulation of the wedging and the stiffness of the material?
     
  22. Perhaps orthotic wedging beneath the rearfoot "works" by transferring more energy to the thick end of the wedge as the foot impacts upon it, and then returning this energy to help to resupinate the rearfoot as the heel unloads? Thus, the energy returned doesn't have to push the centre of mass of the whole body forward, just the centre of mass of each of the foots segments in the right direction, at the right time, which might ultimately result in more efficient forward progression of the centre of mass of the whole body:cool: Maybe orthoses work differently to how conventional thought dictates and maybe we need to think outside of the box..... or perhaps its a bit of both. Perhaps, I'm wrong in my analyses. Please tell me if you think that I am wrong and why you believe this.
     
  23. You lost me hear David.... a "spacetime manipulating machine".....what does that mean??

    Is my CorelDraw X4, the software program I do all my illustrations with, a "spacetime manipulating machine"?

    This is exactly right, David. Depending on the frictional force at the supporting surfaces of the orthoses, the stiffness of the device will vary. The orthosis, when loaded, is always trying to resist deformation or, in other words, is always applying an internal force that tends to bring the two ends of the orthoses closer to each other. The force from the foot moves the two ends of the orthosis further away from each other. When the two forces are in equilibrium, then the ends remain at constant distance to each other and the arch height of the orthosis is stabilized.

    Thanks for the help with this. Maybe an idea for a new newsletter!
     
  24. You mean like this ......another picture

    Where in the 1st wedge the angle of the wedge will have a negative acceleration on the path of COM and the after the Tip Over Point the Wedge will have a positive acceleration of COM distally and as F=m*a this may add the body in Dorsiflexion at the MTP joints and establishing the winding of the drums in windlass.

    Also re my last diagram re heel height the heel height will effect the agglutination of the wedge with should also effect the speed positive and negative acceleration.

    or have I not understood the discussion correctly.
     

    Attached Files:

  25. Not really. I'll come back to this.
     
  26. How does this fit with Olsen's contention that stiffness = 1/ span length cubed? In a filled arch device the span length is zero, so this would mean stiffness = 1???? What on earth does this mean and where did he get this equation from????????
     
  27. Spent an hour or more today building a model for finite element analysis consisting of five springs of identical properties except for different resting lengths, connected at their bases by a flat bar and at their tops by a flat bar, due to the different resting lengths of the springs the two bars form an angle with one another. I'll try to run the analysis tomorrow- if the computer can cope with it. Now that I've thought about it, I could have just modelled a wedge, which would have taken about two minutes... doh. I'll do that too. Never mind, the 3d model I built with the springs will come in useful to illustrate a paper or a lecture or both.
     
  28. Attached are a couple of fea models in which both the area under the heel and distal edge of the devices were restrained, so basically we've got massively high friction under these areas, if you like. The first one is low arch the second is high arch. I can't remember what the exact difference in arch height was for these models, they were pretty different- I'll try to find them tomorrow, but the difference in maximal deformation under a given load of 1000N was only 0.041mm
     

    Attached Files:

  29. Simon:

    I'll bet that if you took the same two FEA models and didn't restrain the contact surfaces with any frictional forces so that the orthoses could elongate freely under load, then the higher arched orthosis would deform significantly less than the lower arched orthosis.
     
  30. One step ahead, I already thought of this and will try varying the restraints in the models -if I get time today, but maybe next week. The key is in trying to emulate the restraints that the orthosis would experience in the shoe. Trinidad in her Masters thesis restrained just under the heel and around the periphery of the heel cup. Full text of her thesis here: http://scholarworks.umass.edu/theses/213/

    The software I use won't allow me to have no restraints at all and I cannot vary the friction- the selected regions are either restrained or not.
     
  31. Simon is it possible to take a prefab test then heat bend increase the arch height test again to illustrate the change in stiffness that we were discussing with Craig P earlier in the thread ?
     
  32. I can do this as a physical test using the materials testing jig. I need one of those time-warp machines really though as I have a full list of patients today and tomorrow...... You could do a crude version yourself using the method Dave employed with his bathroom scales and a cardboard box.
     
  33. Now I remember why I placed the restraints at both ends of the orthosis, because I only have a "lite" version of the software I can't use sliding restraints, so if you don't restrain under the distal edge, the model acts like a spring board since the system thinks the free edge is floating in space. I tried. If someone would like to buy me the deluxe version of the software, that'd be great. A snip at $7500.

    Maybe, I'll do it with some physical testing instead.

    Couldn't get the spring model to mesh, as it was too complex.

    Did model the wedge- it deforms more at the thick end- but then we already knew that it would. Still fascinated by the implications of this on the reaction impulses at the foot orthosis interface.... would love to talk more about this- anyone?

    But best of all, I discovered I could vector the loading!!! Just a shame about those restraints.

    All this and a full list, with one turning up with her "own notes", "fibromyalgia" and literally a bin bag full of shoes- yeah, great. "Has anyone seen my camel?"- Jasper Carrott. I would say I'm glad it's Friday, and get out on it, but I'm working tomorrow too.
     
  34. David Smith

    David Smith Well-Known Member


    CRUUDE!!:craig: Them speriments was precision improvisation engineering in the field.

    Hrumphhh:mad:
     
  35. David Smith

    David Smith Well-Known Member

    BTW I think I've cracked the solution to the loaded wedge counter intuitive problem, got to get it on paper asap.

    Dave
     
  36. Sanitary technician= toilet cleaner ;)
     
  37. I'm all ears. It's only been driving me mad for the last four weeks. Send it private if you like Dave, but send it soon, I'm wearing out the carpet pacing around the house with a confused look on my face, even Grace is sick of this now.
     
  38. So how the arch section of the orthosis deforms under load when it is not in contact with the supporting surface will depend on whether the limits of static friction are overcome or not during loading at it's points of support. What factors influence the friction between two surface?
     
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