Solution of Nonlinear Equations: Graphical and Incremental Sea

Transcription

1 Outlines Solution of Nonlinear Equations: Graphical and Incremental Search Methods September 2, 2004

2 Outlines Part I: Review of Previous Lecture Part II: Sample Problems Solved with Numerical Methods Part III: Solution of Nonlinear Equations Review of Previous Lecture Overview of Numerical Methods Types and Sources of Error Significant Digits Rounding

3 Outlines Part I: Review of Previous Lecture Part II: Sample Problems Solved with Numerical Methods Part III: Solution of Nonlinear Equations Sample Problems Solved with Numerical Methods Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials

4 Outlines Part I: Review of Previous Lecture Part II: Sample Problems Solved with Numerical Methods Part III: Solution of Nonlinear Equations Solution of Nonlinear Equations Introduction General Form of the Problem Types of Nonlinear Equations Graphical Interpretation Example: Fluid Mechanics Incremental Search Method Homework

5 Overview of Numerical Methods Types and Sources of Error Significant Digits Rounding Part I Review of Previous Lecture

6 Overview of Numerical Methods Types and Sources of Error Significant Digits Rounding Overview of Numerical Methods 4 steps in engineering analysis Types of equations Analytical solution benefits and limitations Numerical solution benefits and limitations

7 Overview of Numerical Methods Types and Sources of Error Significant Digits Rounding Types and Sources of Error Mathematical modeling Blunders Input errors Machine errors Truncation

8 Overview of Numerical Methods Types and Sources of Error Significant Digits Rounding Significant Digits How to count them How they affect mathematical operations

9 Overview of Numerical Methods Types and Sources of Error Significant Digits Rounding Rounding Rounding up on 6 or higher Rounding down on 4 or lower Rounding to nearest even digit on 5

10 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Part II Types of Problems Solved with Numerical Methods

11 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Solution of Nonlinear Equations: Natural Frequencies of a Vibrating Bar

12 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Governing Equations Natural frequencies ω of axial vibration of a bar, fixed at one end and carrying a mass M at the other end, satisfy the equation ( ) ωl cot = M ωl E/ρ ρal E/ρ where l is the bar s length, E is the bar s elastic modulus, ρ is the bar s density, and A is the bar s cross-sectional area.

13 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Solution of Governing Equations Frequency Equation Solution Intersections Represent Solutions M/m=0.5 cot(ω L / c) Intersections of the red and blue ωl lines represent values of E/ρ that solve the previous equation: 4 3 ω 1 l E/ρ = ω 2 l E/ρ = ω L / c ω 3 l E/ρ = 6.579

14 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Solution of Simultaneous Linear Algebraic Equations: Static Analysis of a Scaffolding 3 bars supported by 6 cables form a simple scaffolding. Given the positions and magnitudes for 3 loads applied to the bars, find the tension in each cable.

15 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Governing Equations for Bar 1 Force equilibrium Fy = 0 T A + T B T C T D T F P 1 = 0 Moment equilibrium M = 0 9T B + T C + 4T D + 7T F + 5P 1 = 0

16 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Governing Equations for Bar 2 Force equilibrium Fy = 0 T C + T D T E P 2 = 0 Moment equilibrium M = 0 3T D + 2T E + P 2 = 0

17 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Governing Equations for Bar 3 Force equilibrium Fy = 0 T E + T F P 3 = 0 Moment equilibrium M = 0 4T F + P 3 = 0

18 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Assembling Equations At this point, we have six independent equations (two for each bar), and six unknowns (cable tensions). Reformat the six equilibrium equations to isolate the unknown tensions on the left-hand side of the equations. Make sure the tension variables are in the same order in each equation: T A +T B T C T D T F = P 1 9T B +T C +4T D +7T F = 5P 1 T C +T D T E = P 2 3T D +2T E = P 2 T E +T F = P 3 4T F = P 3

19 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Solution of Governing Equations T A T B T C T D T E T F = P 1 5P 1 P 2 P 2 P 3 P 3 If P 1 = 2000 lb, P 2 = 1000 lb, P 3 = 500 lb, various solution methods detailed in Chapter 3 can solve for T A T F : T A = lb T C = lb T E = 375 lb T B = lb T D = lb T F = 125 lb

20 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Eigenvalue Problems: Critical Loads for Buckling a Column A long column with elastic modulus E and cross-sectional moment of inertia I is subjected to an axial load P. If there is a small deformity in the column due to misalignment during construction or some other reason, its strength is considerably reduced. The deformity will cause the column to buckle long before a shorter column would have been crushed.

21 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Governing Equations for Discretized Column The continuous differential equation of deflection d 2 y dx 2 + P EI y = 0 can be discretized with the following substitution: d 2 y dx 2 y i+1 2y i + y i 1 ( x) 2

22 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Solution of Discretized Equations At any given point i, the governing equation evaluates to y i+1 2y i + y i+1 ( x) 2 + λy i = 0 where λ = P/(EI ). Dividing the column into 4 segments (a total of 5 points), evaluating the equation at points 2, 3, and 4 yields: y 1 (2 λl2 16 y 2 (2 λl2 16 y 3 (2 λl2 16 ) y 2 + y 3 =0 ) y 3 + y 4 =0 ) y 4 + y 5 =0

23 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Solution of Discretized Equations Since the column is pinned on both ends, we assume that the deflections y 1 = y 5 = 0. We can then convert the previous three equations into the matrix form ( ) ( 1 ) λl2 16 ( λl λl2 16 ) y 2 y 3 y 4 = 0 0 0

24 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Statistics: Realistic Design Properties of Materials A shipment of AISI 1020 hot-rolled steel your company bought has a textbook yield strength of psi. Upon testing 50 samples of the material, you notice that almost no samples measured a yield strength of psi. Assuming these samples are typical, what strength should your designers assume as a minimum, so that 95% of the time, the material they use will meet or exceed that minimum?

25 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Yield Strength Data (in ksi) # S y # S y # S y # S y # S y

26 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Statistical Characteristics of Samples One basic statistical characteristic is the mean or average, indicating the central tendency of the data. Add up all the yield strength measurements and divide by the number of samples to calculate it: X = 1 n x i = 30.1 n i=1 Another characteristic is the sample standard deviation, indicating the predictability of the data. Small standard deviations come from data that is predominantly clustered around the mean; large standard deviations come from data that is more scattered. s = 1 n ( xi X ) 2 = 1.36 n 1 i=1

27 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Predictions From a Normal Distribution We assume that the processes controlling the steel s yield strength are random, even if they re well controlled. Many random phenomena in engineering follow a normal or Gaussian probability distribution. Standard tables exist that allow us to predict probabilities of finding a particular range of results from a set of random measurements, or to take a particular probability and convert it back to a range of results.

28 Natural Frequencies of a Vibrating Bar Static Analysis of a Scaffolding Critical Loads for Buckling a Column Realistic Design Properties of Materials Using Statistical Tables The usual set of statistical tables show relationships between a variable z and a definite integral Φ(z). Without going into too much detail, z is a function of the measured variable (x in the general case, S y in ours) and the variable s mean and standard deviation. Our 95% success requirement corresponds to a Φ(z) value of 0.05, which is attached to a z value of If z = 1.64, then the S y corresponding to that z is 27.9 ksi. If your designers work off an expected yield strength of 27.9 ksi, your supplier will be able to meet that requirement 95% of the time.

29 Introduction Example: Fluid Mechanics Incremental Search Method Homework Part III Solution of Nonlinear Equations

30 Introduction Example: Fluid Mechanics Incremental Search Method Homework General Form of the Problem Types of Nonlinear Equations Graphical Interpretation General Form of the Problem Many engineering problems involve finding one or more values of x that satisfy one of the following forms of equations: 1 Form 1: 2 Form 2: 3 Form 3: f (x) = 0 g(x) = C f (x) = g(x) C = 0 f (x) = g(x) h(x) = f (x) g(x) = 0

31 Introduction Example: Fluid Mechanics Incremental Search Method Homework General Form of the Problem Types of Nonlinear Equations Graphical Interpretation Types of Nonlinear Equations Polynomial equations Transcendental equations Exponential equations Logarithmic equations Trigonometric equations Hyperbolic equations

32 Introduction Example: Fluid Mechanics Incremental Search Method Homework General Form of the Problem Types of Nonlinear Equations Graphical Interpretation Graphical Interpretation Solutions to equations of the form f (x) = 0 can be seen as places where the graph of f (x) crosses or touches the x axis.

33 Introduction Example: Fluid Mechanics Incremental Search Method Homework General Form of the Problem Types of Nonlinear Equations Graphical Interpretation Graphical Interpretation Solutions to equations of the form f (x) = g(x) can be seen as places where the graphs of f (x) and g(x) intersect.

34 Introduction Example: Fluid Mechanics Incremental Search Method Homework Mathematical Model Water is discharged from a reservoir through a long pipe as shown. By neglecting the change in the level of the reservoir, the transient velocity of the water flowing from the pipe, v(t), can be expressed as v(t) ( t ) = tanh 2gh, 2gh 2L where h is the height of the fluid in the reservoir, L is the length of the pipe, g is the acceleration due to gravity, and t is the time elapsed from the beginning of the flow.

35 Introduction Example: Fluid Mechanics Incremental Search Method Homework Governing Equations v(t) ( t ) = tanh 2gh 2gh 2L Find the value of h necessary for achieving a velocity of v = 5 m/s at time t = 3 s when L = 5 m and g = 9.81 m/s 2.

36 Introduction Example: Fluid Mechanics Incremental Search Method Homework Solution of Equation Substitute the values for v, t, L, and g into the previous equation on the left side v(t) 2gh = 5 = (9.81)h h and the right side ( t ) ( ) 3 ( tanh 2gh = tanh 2(9.81)h = tanh ) h 2L 2(5)

37 Introduction Example: Fluid Mechanics Incremental Search Method Homework Solution of Equation Plot the two sides of the equation as separate functions of h, then find their intersections. In this case, the two graphs intersect at h = 1.45 m, so the original equation is satisfied with h = 1.45 m.

38 Introduction Example: Fluid Mechanics Incremental Search Method Homework Incremental Search Method Incremental search is the most basic automated numerical method for solving nonlinear equations. The method: 1 Pick a starting point x 0 and a step size x. Use a positive x if you want to search to the right, and a negative x if you want to search to the left. 2 Let x 1 = x 0 + x and calculate f (x 0 ) and f (x 1 ). 3 If the sign of f (x) changes between x 0 and x 1, it is assumed that a root of f (x) exists on the interval (x 0, x 1 ). 4 If the sign of f (x) does not change between x 0 and x 1, let x 2 = x 1 + x and repeat the process.

39 Introduction Example: Fluid Mechanics Incremental Search Method Homework Incremental Search Example Find the root of the equation f (x) = 1.5x 0.65 (1 + x 2 tan 1 2 ) ( ) x x 1 + x 2 = 0 using the incremental search method with x 1 = 0.0 and x = 0.1. Evaluate the function f (x) at x = 0.0, 0.1, 0.2, : x 1 = 0.0 f (x 1 ) = x 2 = 0.1 f (x 2 ) = x 3 = 0.2 f (x 3 ) = x 4 = 0.3 f (x 4 ) = x 5 = 0.4 f (x 5 ) = x 6 = 0.5 f (x 6 ) =

40 Introduction Example: Fluid Mechanics Incremental Search Method Homework Incremental Search Example Since the sign of f (x) changed between x = 0.4 and x = 0.5, we assume there is a root of f (x) between 0.4 and 0.5. Repeating this method with x 0 = 0.4 and x = 0.01 would allow us to make a more accurate estimate of the root.

41 Introduction Example: Fluid Mechanics Incremental Search Method Homework Incremental Search Limitations Only finds real-valued roots of f (x). It cannot find complex roots of polynomials. Only finds roots where f (x) crosses the x axis. It cannot find roots where f (x) is tangent to the x axis. May be fooled by singularities in f (x), such as in the tangent and cotangent functions. If the step size x is too large, you may miss closely-spaced roots by skipping over them.

42 Introduction Example: Fluid Mechanics Incremental Search Method Homework Example of Singularities

43 Introduction Example: Fluid Mechanics Incremental Search Method Homework Homework Find a root of f (x) = x 3 3 starting at x 0 = 1 and x = 0.1. Then, using the first root estimate, use a step size of x = 0.01 to find the root more precisely. How does each estimate differ from the analytical solution?