Beam Theory Explained - How Spars Work

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  1. Dec 29, 2017 #1

    wsimpso1

    wsimpso1

    wsimpso1

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    Introduction to Beam Theory

    I have wished we had an explanation of this basic and important structural topic for a while. So here it is. I hope that you find it useful.

    We use beams a bunch in structures. Given the recurring discussions on various types of beams on Homebuiltairplanes.com, I thought that a discussion of how beams work is in order. I will follow with a discussion on composite beam theory, and then a third note discussing some of the “new” theories that we see.

    Beams are generally designed to carry external loads perpendicular to the long axis ofthe beam. These loads then resolve into shear and bending and sometimes torsion in the beam. The theory for it was developed over two centuries ago. Beam deflection and stresses under load are described by conventional beam theory to a very high degree ofaccuracy. We really can trust this stuff…

    Usually when this is taught, the professors get right into the math of how we do this without spending much (if any) time on why it works. I like to get into the physical reality, then add math… It gives you a basis for how it works, then why the math relates to it. Before we go any further, we need some ground rules and a couple assumptions. These make it easier to work through:


    • The beam has a long axis that is straight when the beam is unloaded and is coincident with the centroid of the beam cross section;
    • The neutral axis of the beam is coincident with the long axis when unloaded, and follows the centroid of the beam cross section when the beam is deflected;
    • We shall limit external loads to the vertical plane that goes through the long axis, no horizontal loads and no torsion. These can be added back in easily later on if you want;
    • The material we are using follows Hooke’s law, that is load and deflection are related by Young’s Modulas – strain and stress in the material then varies linearly with load. This also means that we are working within the yield strength of the material, not in any of the failure regimes;
    • Material strains (deflection divided by lengths) are small;
    • When the beam is deflected in pure bending, plane surfaces remain plane – Various pieces that make up the beam are firmly connected together and move (deflect) together when the beam is under load;
    • X axis is the long axis of the beam and is horizontal, Y axis is the narrow axis of the beam that is also horizontal and perpendicular to the X axis, and Z axis is the narrow axis of the beam that is vertical and perpendicular to the X axis;
    • The entire beam is made out of the same stuff. Same Young’s Modulus and strength everywhere makes this simpler.

    So, how do beamswork? We support the beam and add external loads. The loads result in stresses in the beam and get reacted off to the support.

    The easiest beam is a teeter-totter, it sits on one support in the middle of a beam and a weight on each end. We know how levers work, so if we know the lengths and the load at one end, we can find the load at the other end. The force picked up at the single support is equal to both weights.

    Shear is abbreviated as V. This is the vertical force carried by the beam from one place to another. From the first weight to the support, the shear is equal to the first weight. From the second weight on the other side to the support, the shear is equal to the second weight.

    Moment is abbreviated M. This is a couple that tries to bend the beam. It hasunits of force and length, like a torque, but when bending a beam, itis called a moment and it comes from the shear and distance betweenparts of our lever. We know how levers work with moments too – themoment at the support is equal to the weight at one end times timesthe length from that weight to the support. And we know from levertheory that if it is balanced, the moment from one side is equal tothe moment from the other side.

    What is the moment elsewhere along our teeter-totter? Starting at the first weight, the weight times the distance from the weight is the moment at any one spot right up to the support, where the moment is weight times the length on that side. Go to the other weight and work towards the support, and the moment is the weight on that end times the distance from that weight. Get to the support and the moment is again equal to the other moment… Other beams get more complicated.

    So now we know the shear and moment along the beam. What good is that? Well, the moment tries to deform the beam by bending it, and shear tries to deform the beam by shearing it. One thing at a time.

    Bending – if we take a little section of the beam and put a bending moment on each end of the section, the little section of beam becomes curved (a tinyamount) and stresses build up across the section in relation to how far we are from the neutral axis. And because plane cross sections remain plane, even in the deflected beam, the stresses are linear with distance from the neutral axis. And so the resistance to bending goes with E (we know that strain and stress are related linearly byYoung’s Modulus, E) and with I, the Second Area Moment of Inertia of the section. So now we have: Bending stiffness at any point along a beam is EI.

    What else? Well, if we know the moment bending the beam, we can also calculate the axial stresses any place up and down along the cross section. Axial stress sigma is M*z/I where the moment is M, z is how far up or down we are from the neutral plane, and I is the Second Area Moment of Inertia ofthe section. Highest stress is at the farthest point on the cross section from neutral axis.

    Has anyone noticed that I have not discussed the shape of the beam yet? That is because if you somehow get the same I and z and M, the stress is the same, no matter how you get I and z and M. Also, if you have the same EI, you get the same bending stiffness, no matter how you get EI. So what difference does shape make? Well, some shapes get enough I and EI with less weight than others. Looking at the formula for I, material farther from the neutral plane contributes more to I than material closer to the neutral plane. It would seem that putting as much material as far from the neutral plane as possible would be wise. Wide-flange beams do this and for exactly the reasons discussed.

    Let’s finish up with bending moments. When you go to calculate I, each little piece of area is multiplied by the distance to the neutral plane squared. Why? Well, if plane surfaces stay plane, the stresses go with the distance from the neutral axis and the distance the section moves also goes with the distance from the neutral axis. Area is in there once, distance from the neutral plane is in there twice. Stress times area is force, and force times length is a moment. It all fits together...

    Shear – If we look at any little element of the beam with the support off to the right and the weight off to the left, the left side of the beam has a downward force across it, and the right side of the beam has an upward force across it. It turns out that these forces are balanced by forces on the top and bottom of the element. Now the little element of the beam has forces trying to distort it from a squareshape into a diamond shape. That is how shear shows up in parts…

    Now we get to where shape of the beam matters. In all beams, the top and bottom surface have no shear stress but go to maximum axial stress.

    In a rectangular beam, shear is distributed parabolically, with max shear stress at the neutral axis at 1.5*V/A, where V is shear, and A is area of thesection.

    Go to a beam with flanges and a shear web or a hollow box, and the shear is puny in the flanges, but nearly constant through the web and at about V/Aw where Aw is the area of the shear web only.

    Now this assumption of plane surfaces staying plane? Well, it gets violated when we add in shear on the beam, but an interesting thing comes out in the end. The shear deformation changes the shapes of those cross sections, but the deformation from any one place to the next one results in the axial stresses across the sections following the stresses seen from simply bending the beam. Several important things result from that:


    • Axial stresses in the beam calculated from bending alone do not change when shear is added;
    • Deflection of the end of the beam from bending deformation happens on the scale of the length of the beam, as it accumulates along the length of the beam, and so can be pretty big;
    • The deflection of the end of the beam from shear deformation only happens on the scale of the depth of the beam, and so is quite small;
    • Beam deflection calculations based upon bending moment on the beam, with shear neglected, is a VERY accurate predictor of total deflection. Effect of shear is commonly neglected.

    Now let’s talk about each of the calculations, why they are important, and how they are done.

    Centroid – this is where the neutral axis is, and all of the dimensions for the sections derive from the centroid. What is it? It is the “center forgravity” of the section. For the centroid to work and stay put, the assumptions of small strains and linear behavior must remain in place. You can calculate it by finding the First Area Moment of Inertia divided by the area. Your z axis can be measured from any datum you want, and you establish a new datum at zbar that everything will be re-dimensioned to:

    zbar = Sum(deltaA*z)/Sum(deltaA) = Integral (z*dA)/Integral(dA)

    Where A is the Area of the cross section and y is how far each little piece of the cross section is from the datum. deltaA and dA are small pieces of A, useful in calculus. You can repeat this in any axis, but we are sticking to the z (vertical) axis.

    I – Second Area Moment of Inertia - This one relates how much the beam is curved to how much strain and thus stress exists due the curve in the beam. The assumptions of linear behavior and plane cross section surfaces remaining plane have to be here for this to work. It is calculated by:

    I = Sum(deltaA*z^2) = Integral (z^2*dA). You can repeat this in any axis, but for now we are sticking to the z (vertical axis);

    There are formulas for most of the shapes you can imagine. A very common one is for an area offset from the neutral axis – I = A*z^2, where A is the area of the part, and z is how far from the neutral plane in the zdirection the centroid of the area is. This does not include the effect of the internal stiffness of this region, but it still has very little error for a wide thin rectangle, like a spar cap.

    You can build a beam out of smaller sections, as long as they are well attached to each other. Let’s look briefly at what I mean.

    If you try to make a beam out of a stack of veneer, each of the veneer plies are very flexible. Just stack up the veneer, put load on it, and they slide on each other. This is the leaf spring on a truck. The bending stiffness of the stack is simply the bending stiffness of one ply times the number of plies. But, put glue between each piece and the next, clamp it, and let the glue set. Now it is MUCH stiffer. This is the difference between having plane surfaces stay plane and allowing plane surfaces to separate… If this attachment between the pieces fails, you are back to having several separate flexible beams. So,the various parts must be solidly connected to each other for stiffness and strength.

    Now lets look at shear. In a solid rectangular spar, max shear stress (tau) is at the neutral axis, where Axial stress (sigma) is zero. And tau is zero where sigma is maximum. Where is the worst place for total stress? Let’s look at the math a bit.

    Shear stress tau =V/(2*I)*(h^2/4-z^2)) where V is shear force, I is second area moment, h is height of spar section, and z is how far you are from neutral axis. Sigma is Mz/I. If you know Tau and Sigma, you can combine them using Mohr’s circle and get the Principle Stresses for that spot, then compute von Mises stress and compare it to yield strength. Automate this a little and search out the highest stress spot.

    Go to a beam with flanges and webs, and the shear picture changes. The story on flange strength is comparing axial strength to yield strength of the material. Webs get a combination of axial stress and shear stress in the web. When the whole spar is made of one material, axial stress at the top and bottom of the web plus V/Aw in the web is combined using Mohrs Circle, Principle Stresses and von Mises stress comes out, and you can check that against yield strength.

    Want to do more complicated systems with more complicated supports? I like Timoshenkoand Gere for all this, but there are other excellent Mechanics of Materials texts. Roark’s is good too. They explain in more detail how to compute shear and moment diagrams for various types of support and loading.

    Cantilever beams with elliptical lift distribution are particularly interesting, as that is what happens in airplane wings and tails. The shear and moment at the tips is zero. The shear at any spot along the span is equal to the lift outboard of that spot, and the bending moment at the same spot is the total reacted lift outboard of that spot times the distance of each little piece of lift to the spot. Most of us do this numerically. Gets more complicated when we put in joints and stuff.

    But once you have the worst case for shear and moment defined along the span, you can design and analyze the spar, beef it up where you must, slim it down when you can.

    Design of Beams

    We see this a lot – someone tells you that you size the spar caps based upon forces inthe caps. You take the bending moment and divide it by the distance between the caps, and that is the load in the caps. Then you divide the load in the caps by the yield strength of the cap material to find the area you need in each cap. Then the same folks tell you tosize the web area by dividing the shear force by the shear strength. Is this actually OK? Well, if you read what I wrote carefully, you will know it is NOT OK! Yet we see it repeated over and over inarticles and on forums.

    The web may carry the shear nicely distributed, but the shear web also carries bending stress, and near the webs that bending stress is nearly as high as the bending stress seen in the caps. Do the von Mises stress thing and you will find that the shear only sizing of the web will result in failed parts at much lower load than intended. This may drive youto increase web thickness – after all, that is the thing that is overloaded, right? If you set up your calcs to let you adjust bothweb thickness and cap thickness, you can find the lightest combination that makes strength. Turns out that the lightest is always beefing the caps some and the web some, as opposed to beefing up just the web or just the caps.

    Once the caps and web won’t yield in any conventional way, what else might go wrong? Cracking, yielding, buckling. There is also failure of the attachments between parts that then lets buckling in any of several modes occur. Let’***** each of those.

    Cracking happens when the part sees enough load cycles at high enough load to fatigue the material. How much is enough? Well fatigue does not really start until you have 1000 full positive to full negative and back load cycles at very nearly yield level stress. Most of us are designing to FOS of 1.5 or even 2.0 at high g levels, so we won’t get near yield, and it will take a pretty large number of cycles at design limits. And most of us don’t fly more than a few cycles near design limits. The rest of our flying is at significantly lower g’s. That being said, there are a few Aerobatic Bonanza’s and T-34’s that have grown cracks in their spars. So it can happen. When in doubt, increase the FOS a little and put it out of your mind. Or inspect frequently and be prepared to replace spars and such.

    Yielding is what we have been avoiding through most of this article, and with FOS of 1.5 or more and good analysis and design, it is unlikely.

    Buckling. This can happen in large scale or in small scale. In large scale buckling,some structure or element is loaded in compression and when it reaches a critical load, and the structure tries to collapse. This is generally for long slender elements, like control push rods and longerons in the fuselage. In wing spars, we generally have a bunch of support for the spar and large scale buckling is unlikely. In smaller scale buckling, some call it crippling, others talk of elastic stability, but we can get local wrinkling when a critical load is reached, where one portion of the structure wrinkles and unloads. If we are lucky, another portion of the structure can carry the load. If we are not lucky, the structure folds up. If we do our analysis properly during design and all of the pieces of our spars stay attached together, this won’t happen. If we were somewhat deficient in our design, the pieces do not all stay attached to each other, they move separately, they can then buckle in large scale or in small scale (wrinkle and cripple), and we have a problem. Seriously, if you are building up a spar, it all needs to be solidly connected together. Use Shear Flow theory to figure out what loads are seen in the assembly joints and design conservatively. Yep, that means you need to look up Shear Flow and then learn about fasteners and bonding and welding… Geez, this guy asks for a lot. Make sure the parts will all move together. If they don’t, they get to be like the plies of veneer without the glue that I mentioned above.

    Shear, Bending Moments, Curvature, Beam Slope and Deflections are all Related

    Yeah, you might expect so, but it is both more direct and more difficult than you might expect. I won’t do the whole thing, but I will give you aquick tour.

    Shear load is integrated once over the length of the beam to get the bendingmoment. You use definite integrals to do this, choosing integrationconstants that make the boundary conditions work out to the valuesyou know they have to be, usually at the various supports.

    Then you divide the bending moment function by EI to get curvature over the length of the beam. You do have to establish integration constants that again allow boundary conditions to be correct.

    Next you integrate it again over the length of the beam to get beam deflection slope. Select integration constants again to meet boundary conditions.

    Then you integrate it over the length of the beam one more time to get how much the beam deflects. Yeah, fix the boundary conditions with the integration constants again.

    Sounds complicated, but it is a very direct application of indefinite integrals. And it is also covered pretty thoroughly in books on Mechanics of Materials and in Roark’s, with a whole bunch of them already worked out and shown in formulas for you, so you do not need to do all of the math and calculus. You do need to know the loading on the beam, and EI and lengths and type of support for the beams, but a lot of them are already known. Some I have not found. Like a cantilever beam with an elliptical load distribution. So, you may have to do the math for wings yourself. Or send me a PM with your email addy and a civil request for help, and I will send you the functions.

    Ok, that is the basics of beams made of one material. Some things should be apparent– Beams are more sophisticated than they seem. They have to be well built and thought through. If they are, they will do what you have in mind for them. Once you understand this fairly well, you can graduate to composite beams. More complicated to be sure. I look forward to writing it, you should look forward to reading it.

    Billski
     
    Last edited: Dec 29, 2017
  2. Dec 29, 2017 #2

    plncraze

    plncraze

    plncraze

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    Thanks Billski. You are a great companion to J. E. Gordon!
     
  3. Dec 29, 2017 #3

    plncraze

    plncraze

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    Thanks Billski. You are a great companion to J. E. Gordon!
     
  4. Dec 30, 2017 #4

    TFF

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    Very nice work.
     
  5. Dec 30, 2017 #5

    flyboy2160

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    Ya gotta have pitchers with this...
     
  6. Dec 30, 2017 #6

    flyboy2160

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    Ya gotta have pitchers with this...
     
  7. Dec 30, 2017 #7

    BBerson

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    If V is "horizontal shear" ( Maximum at neutral axis)
    What is "vertical shear"? ( mentioned in Peery, chapter 6.2)
     
  8. Dec 30, 2017 #8

    wsimpso1

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    Critics... Beams were already worked out when the Wrights invented our beloved airplanes.

    Seriously, there are several ways. I could just say that it is straightforward and left to the student. But I am not like that...

    I explained with loads in the z direction (vertical loads) and all of the action in the x-z plane to make it simple enough so it could be absorbed. No need to work in other directions, axial loads, or torsion, when you are starting out. And the primary discussion applies nicely to wing spars. Now, some loads do exist in other directions. Watch your nomenclature, and I advise you to get careful with matrix notation (Vyi, Vzi, Myi, Mzi, Ti, and so on).

    Methods:

    For loads that are only in the y direction, you do EXACTLY the same thing but the computations are run the other way across the cross section - the I computed and y dimensions are most likely different;

    If loads occur in z and y direction simultaneously, one way is to run the loads and EI's in each direction, compute the tau and sigmas at critical places around the cross section from each, add them up at each location, then at each place compute von Mises stresses. Watch your notation so that you keep what is going on in each plane straight;

    Or if you have simultaneous loads in both y and z directions and then you can compute the simultaneous direction of each of the combined loads, compute the I's for the combined sections in the direction of each of the combined loads, and superimpose them all;

    You can also apply the spanwise distribution of pitching moment (it is elliptical, sort of like lift, and accumulates like it too), compute r's and GJ's via similar methods, compute the stresses from those and compute von Mises stresses. Just to let you in on a secret, Ixx + Iyy = J;

    You can superimpose torsion on the other two directions;

    Or, you can go whole hog, produce a load vector with loads in Direct Shear and Bending in orthogonal directions, axial loads, and torsion, a stiffness matrix for all of the EA's, EI's, GA, and GJ to sit in then solve for the strain vector, decompose the strain to get local strain, then compute total stress state (von Mises stress) at all of the critical locations for each section.

    Billski
     
  9. Dec 30, 2017 #9

    wsimpso1

    wsimpso1

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    What is it with everything going in twice and taking so long to run anything?
     
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  10. Dec 30, 2017 #10

    tspear

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    Billski,

    You lost me at the end when combining loads from multiple axis.
    Is it a simple sum of the load in each direction, or was it more complicated?
    I always had the impression it was more complex because of the possible combination of compression and tension in the load matrix.

    Tim
     
  11. Dec 30, 2017 #11

    tspear

    tspear

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    Billski,

    You lost me at the end when combining loads from multiple axis.
    Is it a simple sum of the load in each direction, or was it more complicated?
    I always had the impression it was more complex because of the possible combination of compression and tension in the load matrix.

    Tim
     
  12. Dec 30, 2017 #12

    BBerson

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    Unable to edit post 7.
    Forum not working for me.
     
  13. Dec 31, 2017 #13

    wsimpso1

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    Look up "principle of superposition". It is covered in all of the Mechanics of Materials texts. I am giving you the basics here, but I can not begin to teach you the entire topic on a forum. Let's try to do some of it.

    The whole idea is that you can analyze a structure under one load, then another, and so, getting axial stresses (sigma) and taus (shear) and deflections. Then, if the loads can occur simultaneously, you add the stresses and deflections to get the combined effects. For instance, there are published formula for many load cases on beams, and so instead of building up the entire shear and moment diagram, integrating the moment twice to get deflection diagram, you can break the load down to some simple cases that are published, fill in the formula to get stress and deflection curves for each load, then add them together.

    It also allows us to put together loads in different directions. For instance, if a beam has loads and deflections in the z direction, you can thus get the sigma and tau in the cross section of the beam. Then if you have loads and deflections in the y direction, and get the sigma and tau in the cross section, and add them together. Then if you also have torsion in that beam, it produces tau in the cross section, and you can add it in. When you have all of the stresses combined across the cross section, at any one spot, you can then combine stresses using Mohr's Circle and calculate von Mises stresses.

    Billski
     
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  14. Dec 31, 2017 #14

    wsimpso1

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  15. Dec 31, 2017 #15

    tspear

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    So not easy....
    An hour later, and I keep finding more threads to follow.

    Tim
     
  16. Dec 31, 2017 #16

    tspear

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    So not easy....
    An hour later, and I keep finding more threads to follow.

    Tim
     
  17. Dec 31, 2017 #17

    wsimpso1

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    Dupe
     
    Last edited: Dec 31, 2017
  18. Dec 31, 2017 #18

    wsimpso1

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    Tim,

    Did you think that designing light and strong airplanes was child's play? Beam analysis and design is a substantial part of an undergraduate mechanical and/or aeronautical engineering class on structures. This particular course is preceded by one semester each of linear algebra, differential equations, statics, dynamics, and at least two semesters of integral calculus, and we needed much of it to do the class. And this is just one part of it. You still have to do shear flow, fastener system design, rib design, skin and bay sizing, and others. Then you have buckling and crippling checks, inspection port placement and sizing, mountings, and so on.

    Funny thing is that I do find metal wing spar design and validation straightforward and relatively easy. Determine lift distribution, integrate it once to make shear curve, again to get moment curve. These integrations can be done analytically or numerically. If you have more than a simple load scheme. Then you can break the wing into 10 to 20 steps and design the spar at each of those points by sizing the caps and web by the method I gave you. Now you get to stresses. If the spar loading is simple, you find von Mises stresses and check against yield strength. If the load schemes are complicated, you will be using superposition to determine full stress state at each cross section, then stress state, von Mises, and yield check. The web and cap sizes will need to be iterated. All of this up to iterating/optimizing the web and cap sizing should be run through with pencil and paper and a calculator, and then can be automated in Excel, and rather easily too. Then you can check your work...

    Get into wooden wing and composite wing, and you will find that it gets much more involved. That is coming.

    Billski
     
  19. Dec 31, 2017 #19

    tspear

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    Billski,

    I thought it was complicated, but your intro made it simple enough to follow (I hope that was the intent!). Then toward the end, you made a statement that made me wonder.... And I recall those engineering classes, I fell asleep in two before I gave up on engineering (bad teachers). I actually completed Diff E-Q (only non-engineer in the class), Mathematical Models, Calculus 1 through 3; I was computer science track, so I was math/physics heavy. However, this was all twenty five plus years ago, and I do not use any of it professionally/personally.

    Tim
     
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  20. Dec 31, 2017 #20

    tspear

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    Billski,

    I thought it was complicated, but your intro made it simple enough to follow (I hope that was the intent!). Then toward the end, you made a statement that made me wonder.... And I recall those engineering classes, I fell asleep in two before I gave up on engineering (bad teachers). I actually completed Diff E-Q (only non-engineer in the class), Mathematical Models, Calculus 1 through 3; I was computer science track, so I was math/physics heavy. However, this was all twenty five plus years ago, and I do not use any of it professionally/personally.

    Tim
     
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