Analysis of a Simple Structure BRIEF ABSTRACT In this multivariate calculus application, we determine the position of a truss joint by minimizing potential energy. Physics preliminaries are provided. GENERAL INFORMATION FileName: STRUCT Full title: Analysis of a Simple Structure Last Update: 6/5/96 Developer: Jerry Fine Department of Mechanical Engineering Rose-Hulman Institute of Technology Terre Haute, IN 47803 Phone: 812-877-8353. Email: fine@nextwork.rose-hulman.edu FAX: 812-877-3198 Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Klebanoff@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM A simple structure is shown in the figure below. It consists of 3 bars which share a common pin joint. The bars are also joined to rigid supports using pin joints. In this type of arrangement, called a truss, the bars act as linear springs. They remain straight as a load is applied, stretching and compressing, but not bending. In the problem shown a force load is applied at the common joint. Your task is to find the motion of the joint under that loading. How far does it move horizontally and vertically? Your answer may be given with two numbers: the horizontal and vertical components of the joint's change in position. (Call the former "u" and the latter "v.") Useful background information: read this to learn about spring behavior. A linear spring is a mechanical device which transmits a force, F, that is directly proportional to the stretch, or length change, d, (from its standing equilibrium) that it experiences. We write F = k d, where the constant of proportionality, k, is called the spring stiffness. The units of stiffness are force per length. The length change, d, is calculated as the difference between final length, Lf and initial, unloaded length Li. d = Lf - Li. The truss bars in this problem act as though they were springs. Their stiffnesses and initial lengths are given in the figure. As a linear spring changes length, it stores elastic potential energy, U, a quadratic function of d. In fact, U = 1/2 k d2. Note that the same energy will be stored if the spring stretches one inch or compresses one inch. Useful background information: read to learn what static "equilibrium" means. The structure is assumed to be able to resist the load without failing in any way. The common joint will move (we hope only slightly) and the members will share the load, transmitting it to the grounded supports. There will then be no further motion of any kind. This situation is called static equilibrium. Solving this problem means coming up with a mathematical description of static equilibrium. There are two alternatives. The first has to do with forces. One might write down equations which state that the sum of the forces acting on each bar is zero. A good approach is to just consider the equilibrium of the common joint. The forces acting on the joint must have a zero vector sum. Therefore, the forces in the bars and the load must also sum to zero. The second approach has to do with energy. As the joint moves, each of the bars strains and stores elastic potential energy. At the same time the load also moves, reducing the potential energy of the whole structure. Equilibrium is reached when the total potential energy (elastic minus potential energy of loads) is at a minimum. THE KEY TO THE PROBLEM IS TO MINIMIZE THE POTENTIAL ENERGY OF THE STRUCTURE. This is a problem in the calculus of several variables. KEYWORDS Structural analysis, elastic potential energy, static equilibrium, minimization of function TEACHER NOTES ISSUES RELATED TO THE PROBLEM This is not commonly known outside of the world of structural engineering, but describing the static equilibrium of structures mathematically amounts to looking for stationary points for a function of multiple variables. These variables are often the displacements (position changes) of key points in the structure. Therefore, the structures problem is strongly linked to multi-variate calculus. This problem is intended to motivate the study of multivariate calculus by providing a "real world" example. The difficulty with the example lies in its complexity. Even for a two-variable problem, the function to be minimized is complex both to write down and to deal with. It would be hoped that the student has access to technology which would allow the student to deal effectively with this complexity. An important issue with almost any structure is "linearity." If the structure behaves linearly: (1) the members have linear force displacement relationships, F = k d; AND (2) the "stretch" or length change is a linear function of u and v, the displacement components. In the first solution given below, the structure is only linear in the first of these two senses. However, it makes very great sense to linearize the "stretch" displacement relationship for most structures, because the displacements are so small compared to the lengths of the bars. This is an approach that might be considered to help the analysis if there is not a CAS available. It is illustrated below. The energy functional becomes quadratic, and the equilibrium equation, obtained through partial differentiation, are linear. This approach is illustrated in the solution section. Prerequisites The student may never have thought about structural analysis. He or she might not be very interested in it from an engineering point of view, but the problem does relate the minimization of a function of two variables to an important class of problems in technology. The student needs to have some familiarity with simple geometry -- such as the distance formula -- in order to find the new lengths and hence the stretches, in terms of the displacement components, and the data. The student needs to know how to calculate the work done by a constant force acting in a two dimensional plane. This is needed to find the potential energy contribution of the load -- it is just the work of this force with the sign changed. If the external load force (gravity for example) does positive work on the structure, then the potential energy decreases, and visa-versa. The student needs to know about energy storage in springs. This is included in the explanation of the problem statement given above. The student needs to be comfortable with the mathematical description of static equilibrium as the minimal potential energy state of the structure. The student needs to know how to minimize a function of several variables. Time allotment - time management This problem really needs the technological boost of a computer algebra system. It would be possible to work it by hand if the student were: (1) very patient and accurate (2) was good at shortcuts -- which in this case amounts to a linearization of the elastic strain energy function. Still, it would take 2 or 3 hours to produce a carefully worked solution. With the use of a computer algebra system, and assuming that the student understands what he or she is about, the problem could be completed and explored in about an hour. The background material might take 15 minutes to assimilate. Expectations Some students, perhaps those with some physics or beginning statics under their belt, may try to write down force equilibrium equations for the common joint. This is a very messy way to approach the problem, as they will find out. Hopefully, they will follow the hint and take the energy point of view. An early source of trouble might be writing the "d" for the spring based on the displacement components, and the data. A very common problem would be to get the sign of the load term wrong. Students who are adept at using the CAS will like this problem. Future payoffs This problem makes an excellent introduction to a more advanced study of structures. It also has relevance to explanations of the finite element method. Extensions Instead of having two degrees of freedom, one could have as many as 8. After all, there are 4 joints in the problem with a u and a v at each one. (The symbol u stands for horizontal displacement and v stands for vertical displacement.) So one could write the total potential energy in terms of these eight unknowns. The problem is that there are obvious constrains on 6 of the variables. These could be imposed with Lagrange multipliers, and the problem could be solved with the same answer as above. But, there would now be more information. The Lagrange multipliers have a physical interpretation: they are the reactions at the supports. In other words, the Lagrange multipliers are the x and y force components at the fixed joints which hold the structure in static equilibrium. Clearly, structures with much greater complexity could be considered if one had a CAS. References and Sources A First Course in the Finite Element Method by Daryl L. Logan. PWS Kent, 1986, ISBN 0-534-05394-7. See Chapter 3. POSSIBLE SOLUTION(S) Input the Data ki = stiffness coefficient in lb/in of bar i Li = length of bar i fi = angle from the horizontal of bar i. F = external load fF = orientation angle of external load with respect to + x direction. data = { k1 -> 150000, L1 -> 100, f1 -> -30 Degree, k2 -> 200000, L2 -> 75, f2 -> 55 Degree, k3 -> 200000, L3 -> 75, f3 -> 125 Degree, F -> 1500, fF -> 30 Degree }//N; Set up the elastic potential energy contributions of each bar Bar length changes u -- horizontal displacement of common joint v -- vertical displacement of common joint. d1, d2, d3 --- net length changes in bars 1, 2, and 3 respectively. d1 = Sqrt[ (L1 Cos[f1]+u)^2 + (L1 Sin[f1]+v)^2 ] - L1 /. data; d2 = Sqrt[ (L2 Cos[f2]+u)^2 + (L2 Sin[f2]+v)^2 ] - L2 /. data; d3 = Sqrt[ (L3 Cos[f3]+u)^2 + (L3 Sin[f3]+v)^2 ] - L3 /. data; Elastic Potential Energy Remember that the elastic energy stored in a single bar or spring is 1/2 k d^2. U[u_,v_] = 1/2 k1 d1^2 + 1/2 k2 d2^2 + 1/2 k3 d3^2 /. data 2 2 75000. (-100. + Sqrt[(86.6025 + u) + (-50. + v) ]) 2 + 100000. (-75. + 2 2 2 Sqrt[(-43.0182 + u) + (61.4364 + v) ]) + 2 100000. (-75. + Sqrt[(43.0182 + u) + 2 2 (61.4364 + v) ]) Potential Energy of the Load P[u_,v_] = F Cos[fF] u + F Sin[fF] v /. data 1299.04 u + 750. v Total Potential Energy of the Structure T[u_,v_] = U[u,v] - P[u,v]; Here's a picture of total potential energy -- looks like there is a minimum in there somewhere. So we go after it with the Findminimum function. Plot3D[ T[u,v], {u,-.03,.03},{v,-.03,.03}] -SurfaceGraphics- Easiest Way to Get the Answer --Nonlinear prob. We seek the values of u and v which will minimize this function of two variables T(u, v). We use the power of Mathematica and apply the FindMinimum command with a reasonable starting estimate for u = 0 and v = 0. sol1=FindMinimum[ T[u,v], {u,0}, {v,0}] {-5.53616, {u -> 0.00633173, v -> 0.003796}} And now let us explain just what we have actually found here and interpret it in our problem, e.g., with our determination of u and v we have found "exactly" how far the load has moved. Sqrt[ u^2 + v^2] /. sol1[[2]] 0.00738244 A More Explicit Way to Get the Answer -- Nonlinear Problem Derivatives of the total potential energy. e1 = D[ T[u,v], u ] == 0; e2 = D[ T[u,v], v] == 0; Solution of the resulting non-linear system FindRoot[ {e1,e2}, {u,0}, {v,0} ] {u -> 0.00633173, v -> 0.003796} Linear Solution -- linearize the expressions for the displacements by expanding in a first degree Taylor series truncating higher order terms. d1l =Series[ d1, {u,0,1}, {v,0,1}] //Normal //Expand 0.866025 u - 0.5 v + 0.00433013 u v d1l = d1l - d1l[[3]] //Chop 0.866025 u - 0.5 v d2l =Series[ d2, {u,0,1}, {v,0,1}] //Normal //Expand; d2l = d2l - d2l[[3]] //Chop; d3l = Series[ d3, {u,0,1}, {v,0,1}] //Normal //Expand; d3l = d3l - d3l[[3]] //Chop; Write an elastic potential energy function -- a quadratic Ul[u_,v_] = 1/2 k1 d1l^2 + 1/2 k2 d2l^2 + 1/2 k3 d3l^2 /. data 2 75000. (0.866025 u - 0.5 v) + 2 100000. (-0.573576 u + 0.819152 v) + 2 100000. (0.573576 u + 0.819152 v) Linearized Total Potential Energy -- also a quadratic function Tl[u_,v_] = Ul[u,v] - P[u,v]; Equilibrium equations are linear! e1l = Expand[D[Tl[u,v],u]] == 0 -1299.04 + 244096. u - 64951.9 v == 0 e2l = Expand[D[Tl[u,v],v]] == 0 -750. - 64951.9 u + 305904. v == 0 Solution of linear equations Solve[ {e1l,e2l}, {u,v}] {{u -> 0.00633197, v -> 0.0037962}} Comparison with nonlinear solution -- good to around the 4th significant figure. ISSUES IN SOLUTION An important issue involves the ease with which the student is allowed to find the minimum. For some CAS experts, the use of FindMinimum is obvious. But do they know what is happening?