L3 Deflection of a Horizontal Beam: Boundary Value Problems Rev 1/12/2016 she was quite surprised to find that she remained the same size: to be sure, this generally happens when one eats cake, but Alice had got so much into the way of expecting nothing but out-of-the-way things to happen, that it seemed quite dull and stupid for life to go on in the common way. In this lab, we examine of the role of boundary conditions (BCs) in forming the solution to DEs with constraints. As an example, it makes a huge difference whether the ends of an object are free to move or are fixed in place. And that brings up the tried and true construction method of horizontal beams attached to vertical supports. Complex structures: houses, office buildings, bridges, schools, etc, are built from intricate systems of this simple structural element. For the purposes of structural engineering, we define a beam to be any solid object that is much longer than it is wide or thick (width and thickness form the cross-sectional area, perpendicular to the long axis of the beam). The material of the beam must be strong enough so that the beam is capable of supporting a load. Width, or b for base Thickness, often h for height Length, usually L Cross-sectional area A = b h There are two distinct classes of loads: Dead load -- The beam must support its own weight and live load -- the weight of any individual objects on the beam. These may occur in the form of point loads or distributed loads. Loads are normally specified in units of force per unit length. What s a beam do when loaded? Resist the forces that make it bend. Unfortunately, gravity usually prevails and the beam bends downwards; we call the vertical distance from the horizontal at any point along the beam s deflection.
Construction companies want the cheapest possible solution and that usually requires long beams, supported by as few vertical posts as possible. Unfortunately, the longer the beam, the more likely it is to bend when loaded and that s where the engineering design comes in. The structure is thus a compromise between strength and cost of materials. Why is this important? When beams don t bend, they break. And that can be fatal. http://www.ideaphotos.com/bridge- Construction/PhotosMinneapolisBridgeCollapse.html What follows is a very elementary discussion of the vertical deflection of a horizontal beam under static conditions: The beam is supported at one or both ends in some manner and is loaded. The loads do not move nor does the beam vibrate -- we ll save these time-variant cases for another occasion. Note that we are also carefully ignoring the complicated issue of torsion, when beams twist around an axis parallel to their length. The solution to any static beam problem is a deflection function, y(x). When the beam bends, it has to change shape y(x), slope dy 2 dx and curvature d y 2 dx. If the curvature is as illustrated in the grossly exaggerated diagram below, the top surface must be compressed and the bottom surface stretched and that gives rise to internal shear forces 3 d y proportional to. The grossly exaggerated diagram also implies that there exists a neutral 3 dx surface where there is neither compression nor tension -- inside the beam! 2
y Upper surface shorter -- in compression Lower surface longer -- in tension x The static deflection of a one-dimensional beam under load is described by the Euler- Bernoulli Equation x d y ( EI ) w( x), which may also be written as the fourth-order DE dx 2 2 2 2 EI u xxxx = - w(x), or just u xxxx = - w(x) / EI where w(x) is the linear load density: The weight of the beam and whatever is on it per unit length. Euler is on lots of stamps. Daniel Bernoulli s father (Johann) is on a stamp, but not Daniel; apparently they didn t get along. The constants E and I are determined by the material the beam is made of and its shape in crosssection (we ll stick with rectangular cross-section beams for now). E is the elastic modulus (also called Young s modulus), a measured value of the strength of a solid material. I is the beam s second moment of area, a measure of its resistance to bending forces, ie, the stresses set up along the length of the beam by the 3
load. For a solid rectangular cross-section of width b and height (see the illustration on 3 bh the first page of this lab), I. Unlike our old friend rotational inertia, this I has 12 nothing to do with mass. There are published tables for the very different I formulae of a wide variety of shapes of beams. Of course, different shapes use differing amounts of material, which translates to vastly different costs and greatest strength for lowest cost is what structural engineering is all about. Engineering is also peculiar in its use of units: E is measured in force / unit area (in English, pounds per square inch, abbreviated psi); the units of I must therefore be inches 4. The product EI comes out in pound-inch 2. If our beam length is measured in feet and our load density w in pounds per foot (lb/ft), we must convert EI from pound-inch 2 to pound-foot 2, so that 4 d y 2 3 EI ( pound ft )( ft ) matches the units of load density (pounds per foot). 4 dx Example: a 1 thick (height) by 6 wide (base) structural steel bar has E = 28.6x10 6 psi 3 bh (http://www.engineeringtoolbox.com/young-modulus-d_417.html) and I = 0.5 in 4. 12 The product EI = 14.3x10 6 pound in 2, which is reasonably close to 10 5 pound-ft 2. Compare this value to the EI of the same beam turned 90 o so that h = 6 and b = 1. (BTW, the square inch is sometimes known as the squinch, but square feet are never squeet. As an example of the typical engineering sense of humor, if you have square feet, you should see a doctor). The various derivatives of y with respect to x have physical significance which are given meaning by the boundary conditions at the ends of the beam: y -- the slope of a tangent line to the beam at any point, which is nonzero only when an end is free to rotate y -- the quantity EI y xx is known as bending moment, describing how internal forces of tension and compression combine; this is 0 at beam ends when they are free to change slope. As a second derivative, this goes through 0 wherever the displacement of the beam changes concavity. Where the magnitude of the bending moment reaches a maximum is often the point where the beam will fracture. - EI y is the shear force on the beam - EI y in the Euler-Bernoulli equation, this is equal to load density w(x). 4
In its simplest state, the Euler-Bernoulli equation is just a 4 th order ODE. The homogenous case (w = 0) just requires four integrations of the left hand side: 2 3 y C0 C1x C2x C3x, but this assumes the beam itself has no mass but this is engineering, folks; beams have weight. No ideal beams for you! We obtain the load function w(x) by combining the weight/foot of the beam itself with any point loads and/or distributed loads at specific locations. Boundary conditions define the values of the various derivatives at the beam ends (and that is how we find the four constants): Simply supported ends both ends rest on a support and are free to tilt, but maintain 0 curvature and experience no shear: y (0) = y (L) = 0, y (0) = 0 = y (L) We won t use this very often, because although nothing is holding the ends down, y(0) = y(l) = 0 as well. Pinned ends the ends cannot move up or down, but they are free to change slope. If the pin (usually driven through a hole in the beam) is frictionless, there are no bending moments: y(0) = y(l) = 0; y (0) = y (L) = 0 as in the simply supported case. Fixed (clamped) ends ends can neither move nor change slope because they are rigidly attached to their supports: y(0) = y(l) = 0, y (0) = 0 = y (L); y will be nonzero. Cantilever one end is clamped; the other has no support whatsoever: y(0) = y (0) = 0, y (L)= y (L) = 0 (this end is free ) The last case suggests that both ends do not necessarily have the same boundary condition. An excellent reference on these conditions: http://www.geom.uiuc.edu/education/calcinit/static-beam/support.html A very detailed explanation of bending moments, complete with an excellent on-screen deflection calculator: http://www.doitpoms.ac.uk/tlplib/beam_bending/bend_moments.php 5
Examples and Problems to work Note that all of these cases exclude failure conditions: We want to keep our beams from breaking. Nor do we address (yet) outside limitations like how much slope and/or curvature our customers are willing to accept. Set up both of the following Mathematica statements; one or the other will usually work: beam[x_, y_, w_, BC1_, BC2_]:= DSolve[ {EI D[y[x],{x,4}] w, BC1, BC2}, y[x], x]//chop beamn[x_, y_, len_, w_, BC1_, BC2_]:= NDSolve[ {EI D[y[x],{x,4}] w, BC1, BC2}, {y[x],y [x]}, {x,0,len}]//chop The length len is the distance between adjacent support points; it is known as the span. Where have you heard that word before? We will use the first when we can, but DSolve will inevitably fail with more complicated loads functions; we will need NDSolve to rescue us (and reveal more of Mathematica s quirky behavior with differential equations). Define some expressions for the possible boundary conditions: clampend[x_, y_, pos_]:= {y[pos] 0, y'[pos] 0} pinend[x_, y_, pos_]:= {y[pos] 0, y''[pos] 0} freeend[x_, y_, pos_]:= {y''[pos] 0, y'''[pos] 0} The value in pos is a position along the length of the beam. With NDSolve, you must specify len explicitly; with DSolve, you ll specify the length as the pos of the particular end of the beam. Define some simple load functions: A uniform load is just a constant weight per unit length. The same steel bar we used in the example above (1 thick x 6 wide) weighs 20.5 pounds per foot of length (see http://www.kaelinsteel.com/chequa.htm, also http://www.onealsteel.com/calculators.html). So w = -20.5 (recall that this bar s EI = 10 5 ). A point load is almost what it says: there s something heavy and relatively short (compared to the length of the beam) at a specific point on the beam. Something like this: pointload[x_, pos_, width_, w0_]:= Evaluate[If [Abs[x - pos] < width/2, w0, 0]] pointload2[x_, pos_, width_, w0_]:= Which[x < pos, 0, pos-width/2 < x < pos+width/2, w0, x > pos, 0] Plot these point loads; they re not all that realistic and our solvers will usually have trouble with sharp edges. Remember to make w always point down (ie, negative). 6
So point loads aren t really points; they re spread over a short distance. A better point load function doesn t have sharp corners: betterptload[x_, pos_, w0_, k_]:= we 0 Plot this and experiment with various k values. 2 k ( x pos) In these load functions, the value of w0 is the maximum magnitude of load density; the total load is the integral of the load function over the distance it covers. In some cases, we specify load as weight per unit length, in others we specify the total weight of the load and the distance it covers (spread distance). A point load near the end of a beam need not be symmetric. If there is an edge, we can try edgeload[x_, pos_,w0_,k_] := Which[x pos, w0 e k(x - pos)^2, x > pos, 0] Many other load configurations are possible; most often you will add together several different types of loads make sure they are all negative values! Here is a trivial case, illustrating the syntax of what we will be doing: EI= 100000; trivial= beam[x, y, 0, clampend[x,y,0], clampend[x,y,10]] This is a fixed beam of length 10 feet, clamped at both ends; however, it must be ideal because the beam itself has no weight. What is the solution y(x) for this case? Did you really need to use Mathematica for that? Now let the beam have some constant weight per unit length and plot the solution: EI=100000; len=10; soln = beam[x, y, -20.5, clampend[x,y,0],clampend[x,y,len]] Look at the function that results; despite ugly constants, it is comfortingly straightforward. y inches 0.01 0.02 0.03 0.04 0.05 0.06 2 4 6 8 10 x feet When you plot this y[x] /. soln, you can adjust the plot option AspectRatio, which controls the ratio between the plot s horizontal to vertical scales. The default is AspectRatio Automatic. Note that the plot above is cleverly rescaled so the vertical axis is inches rather than feet. 7
Not bad: This 10 foot long, 1 thick by 6 wide steel beam sags by no more than ~0.06 inch under just its own weight (a mere 205 pounds). Of course, that was with the 6 side horizontal what happens if it s vertical? Verify with a comparison plot that beamn gives an extremely similar result. Try it with two pinned ends; then try one end clamped and the other pinned, plotting the results each time. What happens when we put a point load (something like a big bucket of cement or a refrigerator or a parked car) in the middle of the beam? You should find that DSolve doesn t like the way we set up a point load. Try the NDSolve version with the sharp-cornered point load. If NDSolve doesn t like that either, try the betterpointload shown above. Note that if you change the constant k in the betterpointload (which is actually a Gaussian or normal distribution), you have to adjust w0 so that the integral of the load density over its length is the desired amount. It is also useful to compare the total point load to the total weight of the entire beam. In this comparison plot (blue is just the point load (with w 0 = -50), red is the uniform load computed above): y inches 0.02 2 4 6 8 10 x feet 0.04 0.06 0.08 0.10 With these values, the point load has a much smaller impact than the weight of the beam itself. Plot the graph showing the combination of these two loads. Examine the resulting deflection function: What is the maximum deflection? Where does it occur? What is the maximum slope? Where is the bending moment (second derivative) changing sign? Where does it reach a min or max? Where is the beam most likely to break? Of course, when we use NDSolve, we don t actually have a solution function (NDSolve s so-called InterpolatingFunction is just a set of points), so you ll have to do some work to get this information. 8
Experiment with various point loads in other positions (continue including the weight of the beam itself). Try multiple point loads on the same beam and try different boundary conditions. Now try the cantilever beam (one free end, supporting only its own weight). DSolve handles this case: See what happens when you add a point load at the end of the beam: Experiment with other load configurations on the cantilever; consider an exponential load. What might be the criteria for physical limitations on the size of the load? Maximum deflection or maximum slope? Makes a big difference if you are walking the plank! For some unusual cantilever behavior, see http://www.youtube.com/watch?v=le7xar7j4nw and http://www.youtube.com/watch?v=om5e-stzeoe. This one s also fun: https://www.youtube.com/watch?v=0wn8rp7bz6q Here is a plot obtained with a single point load at the center of a beam. From the shape of the deflection function, how can you determine the boundary conditions? Galileo s 1638 drawing of a cantilever, supported by a not-sogreat looking wall. http://newtonexcelbach.files.wordpress.com/2008/02/galileo-beam.jpg 9
http://people.virginia.edu/~pjm8f/engr162/projects_two.htm http://people.virginia.edu/~pjm8f/engr162/projects_three.htm 10