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Thought Experiment #1: Tie Tensile Force in Loaded Arch

Discussion in 'Biomechanics, Sports and Foot orthoses' started by Kevin Kirby, Feb 12, 2006.

  1. Members do not see these Ads. Sign Up.
    When an arch stands up by itself and resists external loading forces, there must be internal forces acting within the arch in order to resist deformation by the external loading forces.

    In this first thought experiment for Podiatry Arena, there is given that there are two arched structures of the exact same dimensions with 10.0 cm beams hinged at top of arch. These arches are built to rest under load at different arch angles (30 degrees on left, 10 degrees on right). A tie is attached toward the end of the arch to prevent arch collapse, 1.0 cm from the end of the beam. The tie, therefore, is longer on the right arch than the left arch since the right arch is flatter than the left arch.

    Please solve for the tensile force that must be present in the tie in each arch structure to cause rotational equibrium (rotational velocity = 0) of each arch structure. Assume no frictional force in arch hinge or where beam rests on the ground.

    1. What is the tension in the tie for the 30 degree arch structure?

    2. What is the tension in the tie for the 10 degree arch structure?

    3. Why are the tensile forces that are present in the tie in the arches of different magnitudes even though the ground reaction force (GRF) and the vertical loading force acting at the hinge the same in both arch structures?

    4. How is this arch example similar or dissimilar to the longitudinal arch of the foot during relaxed bipedal stance?

    5. What might this thought experiment imply regarding the usefulness of pressure mapping of the plantar foot during gait in regards to using that information to predict the internal loading forces acting on the structural components of the foot?
  2. Kevin,

    I won't spoil everyone's fun by solving the math. For those unfamiliar with this kind of analysis, there are some nice sites on the physics of trusses, but you may need to brush up on your trig first.

    Taking your other questions Kevin:
    4. How is this arch example similar or dissimilar to the longitudinal arch of the foot during relaxed bipedal stance?

    While the model provides a nice introduction and helps provide a basic understanding of this kind of modelling. I believe it is somewhat limited for the following reasons:

    The longitudinal arch of the foot has more than one joint.

    The foot system is not friction-less.

    The bone segments of the arch are not straight as in the model.

    Bending and deflection of the compressional members occur in the foot.

    The model is only a 2D analysis.

    The model only incorporates minimal components and ignores the real anatomy.

    etc., etc. ;)

    5. What might this thought experiment imply regarding the usefulness of pressure mapping of the plantar foot during gait in regards to using that information to predict the internal loading forces acting on the structural components of the foot?

    In isolation pressure mapping tells us little about the internal loading. Equally, looking at structure in isolation tells us little about internal loading.

    Best wishes,
  3. By not answering questions 1-3, Dr. Spooner, you only get a 40% on your exam score. ;)

    Maybe someone else will try to solve the little problem in statics that I have presented above. Come on now, all of you who are interested in foot biomechanics, understanding this first in a series of thought experiments that I will be presenting, will help you better understand the importance of such things as longitudinal arch height on internal loading forces within the foot, external versus internal loading forces, function of plantar fascia and plantar ligements, etc.
  4. Thought Experiment #1: Tie Tensile Force in Loaded Arc

    Hello Kevin and all interested,

    Kevin, instead of giving me the opportunity to end my weekend in tranquility, you are shaking my grey cells for a last mental effort.
    It is already late, but, if you apply free-body-analysis on either of the beams, you have also to introduce 2 horizontal forces, one at the hinge, being F1h, and an equal one at the tie F2h. The respective vertical forces are F1v, being 100 N and the other the equal front GRF F2v.
    Rotational equilibrium yields the equation for force moments

    F1v .10 . sin60° + F2h . 1 . sin30° = F1h . 10 . sin30°

    where F2h=F1h and F1v=100N

    The solution for F2h= 192 N for the 30° case and F2h=629 N for the 10° case, the latter tension in the tie being about 3.3 x larger than the former one. This is because the lower the beam, the smaller the lever arm of the tension force and therefore the higher the tensile force to create an equal moment of force.
    But another question to you Kevin. Why choosing a GRF as low as 200N, resulting in a body weight of about 40 kg ? Are you referring to kids or are they midgets? Anyway, for a normal man of about 80kg, one has to double the tension in the tie.
    So far for question 1 to 3

    As for question 4, I agree with Simon. More, the fascia plantaris is assisted by other passive tissues and also intrinsic muscles, implying that the tension forces above will be much lower than calculated, fortunately.

    And for question 5, I do not think this model is in any way useful to analyse plantar foot pressure or vice versa, as the model only represent two pressure points where the normal foot leave a complete pressure map on the pressure plate.

    Sorry guys, but I am now putting my cells to rest before they shut me down.
    Have a nice start of a fresh week.
    Regards to all
  5. You win the prize, Bart. Maybe when I get a chance to do some more drawings, I will show a little simpler way to calculate the same results. However, Bart is correct that the force in the tie is about 3.3 times greater and also the force at the hinge is 3.3. times greater in the arch with a 10 degree angle than a 30 degree angle. Therefore, one cannot assume that you will always know the internal forces within the joints and ligaments of the foot just because you know the locations and magnitudes of the plantar forces acting on the foot.

    The arches were not meant to be feet, just arches with convenient numbers for doing the calculations.

    I would say it is similar to the arch of the foot in that it capable of supporting weight from above by using compression bearing elements along with "plantarly" located tension bearing elements, to form a truss type structure. It is also similar to the arch of the foot in that the external forces acting on the foot are not the only determinants of the internal forces. The geometry of the compression and tension bearing elements of the structure in both the arch example and foot are also main determinants of the magnitudes of internal forces. Otherwise, Bart and Simon's comments on why it is dissimilar, I would agree with.

    I agree more with Simon, here. The thought experiment that I have presented here is a very simple way to demonstrate to the non-engineer, non-biomechanics PhD podiatrist that internal forces can be largely affected by arch geometry in the foot and that plantar pressures, by themselves, tell us only about plantar pressure, not about internal loading forces in the foot.
  6. David Smith

    David Smith Well-Known Member


    Didn't have time to reply earlier but I think that this is simmilar to your rotational equilibrium method or joint method, which is best for this problem as in this case we have a statically determinate truss which is determined by a rule of thumb which => M-((J*2)-3)=0 where M = members and J = Joints. (And, as stated, where the joints are assumed to be pin joints that transmit no moments. Also that there are no frictional forces at the ground contact since this would act as a second tie and the analysis would become statically indeteminate)
    Therefore can the equation be like this (F1*L1)-(F2*L2)=0 where F1= GRF at one member and L1 = lever arm of F1 about apex pin joint and F2 = tension in Tie and L2=lever arm of F2 about apex joint. Which gives values of tension which, as you have already calculated is 8.66N/.045m=192N and 9.8475N/.01562m=630N

    Here is a question for you Bart is there any simple analysis method which can be employed to determine simultaneous forces in the Plantar and spring ligament, the Plantar facia, and ground reaction frictional forces which, in effect gives 3 ties to the truss members that also in effect have varying modulus of elasticity or resistance to stretching of the distance between members.
    Or is it only possible with an FEM analysis?

    Regarding Kevins
    Q3 Shorter lever arms for F2 with more acute angle.

    Q4 There are many more forces to consider in a real foot and they are 3D plus some might say that, if modeled as arches, some are not loaded at the apex (but when does an arch become an truss?) which then introduces substantial bending forces.

    Q5 I would think that if one could determine the change in plantar foot length and also measure the relative/comparative change of loadings on the foot one could make clinical assumptions about the relative change in tensional forces within each band of the plantar facia.
    Also perhaps assumptions could be made about the tensional force impulses within the PF. I.E. more time on the f/foot = increased time of tension in the PF.

    Cheers Dave
  7. David,

    Congratulations too.

    As for your question about calculation of additional forces (spring ligament tension and frictional GRF's), the problem is that under this condition there are too much unknowns for the number of equations (3) available. So the system of equations is underdetermined and, therefore cannot be solved, unless for some of the unknown variables values are introduced or additional equations are created though realistic assumptions or optimalisation criteria.
    Finite element may do the job but this is a completely different approach and not my cup of tea. But then there you have to make a whole lot of stress measurements.

    Sorry for not being of more help.

  8. David Smith

    David Smith Well-Known Member

    Dear Bart and Kevin

    Thanks for your reply, I have no experience of FEM but I am interested in finding simple methods to analyse the foot structure. I have looked at methods of analysing statically indeterminate truss structures but these are not similar to foot structure in many ways.

    These are just my thoughts if you are interested.
    One model I have considered is the medial arch as a truss, with the apex at the talus, and the lateral arch as an arch with the apex at the calcaneo-cuboid joint (CCJ). This is simmilar to a two arches model, e.g. Kapanji Physiology of the joints, except that;
    If you consider the medial arch as an arch then if the apex is at the Talo-navicular joint (TNJ) then, as the loading is not over the apex (like the truss model), there are increased bending moments which makes the 1st ray more difficult to stabilise and requires forces external to the arch structure, eg muscles and ligaments, to do this.
    With the 1st ray element in the correct position the foot can be stabilised, against pronatory moments, mostly by compression forces in the ray.

    With regard to tensional forces stretching of the plantar fascia and forces on the foot mesured by a pressure mat.
    If one was to measure the change in length of the PF then if one made the assumtion that the stifness ratio of the plantar fascia was 200N/mm, Gefen, Foot and ankle, 2003, Kitaoka et al 1994, (also Youngs Mod about 30KPa I think) then could one make the assumption about the maximum tension in the PF and subtract that value from the calculated total tension in terms of the change in geometry of the Truss and GRF and this might be the tensional and friction force applied by other structures. Also if one knew the insertion locations of the plantar ligaments then one could, from the geometric changes, calculate changes in length and, as with the PF, make assumptions about tension in these tissues.
    Would this be clinicaly useful or are the 'known values' of stiffness to variable to be valid.
    Just exploring,

    Cheers Dave
  9. As promised is an alternative method by which to determine the tie tension in the loaded arch. Note that I have assumed that the left hand side beam has a fixed axis at the apex of the arch with the counterclockwise (CCW) moments being equal to the clockwise (CW) moments (rotational equilibrium = static equilibrium). Next, the tie force (TT) is resolved into its rotational component (TTR) that pulls at right angles to the beam at a 9 cm moment arm in a CCW direction. And the ground reaction force of 100 N is resolved into its rotational component (GRFR) that pushes at righ angles to the end of the beam with a 10 cm moment arm in a CW direction.

    The rest of the calculations are simply math with the tie tensile force, TT, being equal to 192.4 N in the 30 degree arch example. The exact same calculations could be carried out for the 10 degree arch example using this method.

    For me, this is an easier way to calculate these values, but Bart's method is probably a more commonly used method of calculating for the unknown forces in my examples.

    Hope this encourages those of you who were confused by Bart's and David's postings to try to see how these calculations are performed so that you can start to get a "feel" about what we are talking about in regards to arch geometry and the variation of internal forces. And you thought you would never need to use trigonometry again!
  10. David Smith

    David Smith Well-Known Member


    How about this for a simple graphic method without using trig or calculating moments.

    Draw the truss to scale of the GRF IE the hieght = 100 in this case, consider only one side of the truss (NB sometimes the truss may not be symmetrical) ie draw a line bisecting thru the apex to the base to make a right angle triangle, measure the length of the base on that side = (Lb), Draw in the tie to scale, ie using the 30dg example the length of the member (hypoteneuse) would be 200(Lm), so the tie would go in at 20(Ly) from the base of the member. Find the length from tie to apex along the member =180(Lr), divide 180(Lr) by 20(Ly) =9, divide 100 by 9 = 11.11% multiply (Lb) by 1.1111 which in the 30dg case= 173.2(Lb) x 1.1111 = 192.4(Lt) => the tension in the tie. Ans the 10dg example => 567.128 x 1.1111 = 630.12
    If the tie was at 8mm instead then this would => 567.128x1.25=708.9.

    What do you think?
    If I could get the attachment thingy to work a picture would be so much easier.

    Cheers Dave
  11. I think I know what you are getting at, Dave. However, without a diagram I am lost. Make sure your JPG attachment is below 90 KB when doing an attachment for this forum. I use CorelDraw and then export the image as a JPG file to post the drawings that I made for this topic.
  12. markjohconley

    markjohconley Well-Known Member

    Re: Thought Experiment #1: Tie Tensile Force in Loaded Arc

    I thought i'd revisit the TE's from #1. Good intentions, I never got past

    I should have hung out the washing, watered the new shrubs, put on the crock pot for dinner, had a shower, check myself in the mirror for those little embarrassments, paid attention at the wound care meeting, prepared my presentation on Charcot's jts but noooo...been trying to work out what Bart's doing all day.

    Please will someone explain this approach to this problem, hopefully with a DIAGRAM.

    In BIG trouble with the missus, the laundry basket has develped splits, it's off to the nursery for new"er" shrubs this weekend, people did pull funny faces if I got too close, luckily no zits (phew), wound care (too many new products I never remember anyway), I've got plenty of sickies up (someone else can give a talk)
  13. Re: Thought Experiment #1: Tie Tensile Force in Loaded Arc


    There is more than one way to solve the problem. Did you look at the diagrams that I did in solving the problems? Do you understand those diagrams?
  14. David Smith

    David Smith Well-Known Member

    Re: Thought Experiment #1: Tie Tensile Force in Loaded Arc


    Bart has used the principle of rotational equilibrium IE the sum of all moments (and forces) = Zero. We must assume that all nodes are frictionless pin joint that can only transmit translational forces and not moments. Also the truss is supported only at the bottom nodes. In this example the force is vertical and the truss is symmetrical therefore the forces at each support must be 1/2 the total force applied. (Force-Force= 0 or equilibrium) This can be proved with rotational equilibrium. IE The downward force at the top node (c) * (horizontal) lever arm to node (a) = moment X. (Force c* lever ac=Moment X)Therefore the upward force at node (b) must be that that causes equal moments about node (a) for equilibrium. As this truss is symmetrical the lever ab is twice that of ac so the force for equilibrium must be half that applied at (c) therefore Fb * L ab = Mx and therefore (Fb*Lab)-(Fc*Lac)= 0. Equilibrium of Moments for Vertical forces. Now as the nodes are pinned the Force c will tend to collapse the Truss therefore there must also be horizontal forces required to resist this tendency. For Horizontal forces one must balance the moments about node c by forces at nodes a and b. This force must be applied by the Tie. IE (Fa * Lac) - (Ftie * Lh) = 0 (where h = height of truss and Lac = length from node a to bisection of base at c)
    This is the best way to resolve forces and moments in a truss or pinned arch where the forces are not perpendicular and the structure is not symmetrical.

    Cheers Dave
  15. markjohconley

    markjohconley Well-Known Member

    Messrs. Kirby and Smith, thankyou for replying.
    Kevin, my omission, I had no trouble with your posts nor explanation, excellent diagram.
    David, I will have to sit down and draw a diagram, shame about your "attachment thingy".

    With Bart van Gheluwe's equations, how does F1h = F2h, as wouldn't there be translation of the whole figure (truss), doesn't F1h = - F2h?

    Also in the "F1v .10 . sin60° + F2h . 1 . sin30° = F1h . 10 . sin30° " isn't that a CW + CCW = CCW, shouldn't it be the sum of all CW moments = sum of all CCW moments

    If this is getting tedious I understand. Thanks , mark c
  16. markjohconley

    markjohconley Well-Known Member

    David I can't get my thingy working either, but here is my attempt at describing my diagram from your instructions

    right-angle triangle,
    base AB (apex to right-angle) of ~173 units
    hypotenuese AC of 200 units
    other side BC of 100 units
    point D on hypotenuese 20 units from apex A
    point E on side BC ~11.1 units from right-angle corner B
    line joining points D to E, of 155.7 (173 x 0.9) units, which ain't 192

    David, please, for the love of *&!#@, where have I erred, thanks, mark c
  17. David Smith

    David Smith Well-Known Member


    Although I have been able to put attachments o this forum in the past, at preset each time I try I get an error message that I have exceeded my upload limit.
    My PDF upload is only 11kb and I only have 149kb saved from previous uploads and the PDF limit appears to be 4mb so I don't see why my limit is reached and I can't find how to delete previous uploaded image attachments. Perhaps the system should be modified to make it more user friendly as I am sure that both of us can't be as daft as brushes. Anyway I'll send you a PM if I have your email or you have entered it in your member details.

    Cheers Dave
  18. markjohconley

    markjohconley Well-Known Member

    "Oh gad, won't this antipodean pedantic driveller just go away or shutup"
    Yes I'm back

    David Smith, thanks got your email and diagrams. I still don't get 192 for that interval as it can't be greater than the 173 base, it has to be 0.9 times the base's length. I appreciate the effort you have put into this.
    For sanity's sake and my marriage I am moving on and I'll stick to the approach Prof. Kirby uses.

    Haven't had an update on the little Aussie Paynes lately Admin
  19. markjohconley

    markjohconley Well-Known Member

    Re: Thought Experiment #1: Tie Tensile Force in Loaded Arc

    Working backwards, always a good way of checking, i get ...

    rearranging (1)

    F2h = ((F1h).(10).(sin 30 o) - (F1v).(10).(sin 60 o)) / ((1) . (sin 30 deg))


    192 = ((192).(10).(0.5) - (100).(10).(0.86)) / (1) . (0.5))

    = (960 - 860) / 0.5

    = 200

    Which although close doesn't fit. Bart did say he was about to "shut down". Any ideas which bit is wrong?
    After this I promise, no more, as long as somebody fixes it.
  20. David Smith

    David Smith Well-Known Member

    Re: Thought Experiment #1: Tie Tensile Force in Loaded Arc

    Yes sin 0.09m = 0.045 not 0.5

    So Resolving the Horizotal force is like this

    (Fv*Lever) - (Fh * Lever) = 0

    So (100 * 0.1cos30)- (Fh * 0.09sin30) = 0

    = (8.66) - (Fh * 0.045) = 0

    So 8.66/0.045=Fh

    Fh = 192.444N
  21. Mark:

    I have been reading your comments on this original Thought Experiment I did for Podiatry Arena over two years ago. I congratulate you on making the effort to try and understand the physics of this experiment. I totally understand your frustration with not being able to understand all the methods by which to solve this somewhat complicated model since I can also become very frustrated when I can't figure things out on my own.

    To give you some encouragement: you are making an effort to improve your knowledge of the inner workings of the foot that only a small percentage of podiatrists will ever understand. Understanding the physical realities of the foot through mathematics, which is much more of a pure science than our clinical world, to me is very comforting. I feel that by proving mechanical concepts of the foot with mathematics, I am standing on bedrock, rather than quicksand, ready to take on any challengers to my ideas.

    People like Bart Van Gheluwe and David Smith, because of their physics and engineering backgrounds, understand this very well since their early post-graduate educations were in fields that heavily rely on mathematics to prove the concepts that are used throughout their fields of specialization. However, for the average podiatrist, that has a limited mathematics, physics and engineering background, these concepts may appear foreign and completely incomprehensible.

    You have grown so much in knowledge over the years I have been reading your postings, Mark. Keep up the good work. Your continued striving toward increased knowledge will help you help your patients more than you can ever imagine.:drinks
  22. David Smith

    David Smith Well-Known Member


    Kevin wrote
    This is so right. We all use maths and mechanics intuitively everyday in our work, however not all principles are as intuitive as we think and basic misconseptions occur. When we understand and apply Newtonian mechanics to a problem the outcome is likely to be clearer and more precise. And, what I consider to be one of the most important aspects, much more clearly and precisely communicated.

    FYI Mark: If you first understand the concept that all mechanical systems, whether stationary or in motion, have equilibrium of forces and moments.

    Then you can start all your equations with the following premise

    (This) - (that) = zero

    F - MA=0

    Upward force - downward force = 0

    Inward force - outward force = 0

    Clockwise moments - AntiClockwise moments = 0

    Find the knowns

    Then rearange to find the unkown.

    So in Kevins TE#9 the forces and anatomy are conveniently configured to resolve without bits left over. IE the Ach ten is perpendicular as are the forces. It is a convention for modeling that things (models) are simplified to convey the essential concept that is trying to be conveyed. Assumptions are made to allow the useful estimation of forces of interest.

    John A. Adam, in his book Mathematics in Nature: Modeling Patterns in the Natural World, Princeton Uni Press ISBN 0-691-11429-3

    Explains very well the principle of simplification, assumption and estimation in terms of modeling to produce useful information.

    Kevin's model simplifies the real anatomy and estimates the plantar fascia tension.
    In this case we can understand the limitation of such a model but at the same time it conveys useful concepts about the nature of how forces change in a truss or arch mechanism and what the relationship is between applied external forces and resisting internal forces.
    For me this is an essential concept to grasp. Because there is always equilibrium of forces and moments then we should understand how external forces do affect internal forces and how these internal forces resist the deformation or deflection of an joint or limb.

    Perhaps sometime in the future Kevin will add some obliqity into one of his Thought Experiments. In TE#9 for instance if the Ach Ten was oblique instead of perpendicular then it would be immpossible to balance the moments produced by GRF by that one muscle and another muscle would have to be chosen to balance the equation. In this way it would build up an understanding by the reader of how muscles are synergistic and antagonistic to each other.

    As you add more obliquity of muscles and forces then the maths becomes more involved but the same principle still applies.

    (This) - (that) = 0 and so always equlibrium.

    All the best Dave
  23. markjohconley

    markjohconley Well-Known Member

    Re: Thought Experiment #1: Tie Tensile Force in Loaded Arc

    i'm a goose
    If I had used the more precise figures of 192.4 and 0.866 the RHS of the equation would be 192 (3 sf)
    Apologies Bart van Gheluwe for ever doubting you
    I've finally worked out Bart's free-body-analysis onto a diagram and
    1) still can't see why F1h = F2h, surely it's F1h = - F2h, for F1h to exert a CCW moment the force has to be applied 180 deg opposite direction to F2h, and
    2) also why is the moment arm for the F2h the 1 unit to the base, why not the 9 units to the apex (hinge). Obviously I'm wrong on this one but can't understand why?
    The calculation of moments about one "fixed" point, a la Kirby, is far simpler to understand, but I'd still like to know why Bart's preferred method works, Mark c
  24. David Smith

    David Smith Well-Known Member


    Bart does say this doesn't he? But you are correct that if you use a convention that assigns positive and negative values to the force and moments vectors then you should use those terms in the equation.

    In Bart's original equation he has 2 unknowns but as F1h = F2h they can be considered the same. So what Bart has done is use a freebody diagram with only the left member for analysis and assigned equal force to resolve the rotational equilibrium. To resolve the horizontal force at either end simply divide the monents by vertical force by the lever arm available to the horizontal force and there is the answer. IE (100 * 0.0866)/0.09sin30 = 8.66/0.045 => 192.44

    Does that do it for you

  25. markjohconley

    markjohconley Well-Known Member

    Yes, thanks David, I'll definitely give it a rest now, all the best, mark c
  26. David Smith

    David Smith Well-Known Member


    Just to say I went back to Bart's first post and did the calculations his way. I have to admit that at first look it is a bit counter intuitive. But if you set up the free body diagram with only the left side then one can see that he adds up the clockwise moments about the left base node, which are by forces F1v and F2H and they must equal the anticlockwise moments by force F1h. IE the formula (this - that) = 0, but before you can resolve that equation tho, you must first find one of the horizontal forces, which equals both the horizontal forces, whereby you already have the answer to the tie tension problem and don't need to resolve the first equation anyway.:dizzy:

    Bart please make your equatios more intuitive in future as Mark has gone thru several packs of Anadin (aspirin) over the last week. :bash:

    Cheers all:drinks Dave

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