ME 130 Applied Engineering Analysis Chapter 2 Mathematical Modeling

1. ME 130 Applied Engineering Analysis Chapter 2 Mathematical Modeling Professor Tai-Ran Hsu, Ph.D. Department of Mechanical and Aerospace Engineering San Jose State University San Jose, California, USA (2017 edition)

2. Chapter Learning Objectives ● Mathematical modeling in engineering analysis ● Major components of mathematical modeling ● Functions and variables: The roles of functions and variables in engineering analysis Continuous and discrete-valued functions Curve fitting for discrete-valued functions using polynomial functions ● Differentiation and derivatives in engineering analysis ● Integration and application of integration in engineering analysis Plane areas Volume of solids of revolution Centroid of plane areas ● Differential equations –types and derivations

3. Mathematical modeling involves: Translating physical situations into mathematical Expressions, and vice versa It is a similar action of writing MUSIC from the melodies in the minds of great composers, e.g., Beethoven, Mozart, etc. into the notes. The musicians may play the melodies by the written music notes-a reversed action! Mathematical modeling in engineering analysis = Physical phenomena Mathematical Formulations

4. We recognize Engineers’ duties include: CREATION, DECISION MAKING, and PROBLEM SOLVING Performing each of these duties involves a process in reaching solutions FUNCTIONS AND VARIABLES: The SUBJECTS in the math modeling processes are represented by FUNCTIONS, and the factors that deterfmine the values of these subjects are VARIABLES. VARIABLES include: ● Spatial variables: represented by coordinate systems with set reference points. Commonly used coordinate systems are: (x, y, z) in rectangular coordinates, or (r, θ, z) in cylindrical polar coordinates ● Temporal variable: time (t) ● x,y,z and t are INDEPENDENT variables The process for solutions is to include the FUNCTIONS with VARIABLE in APPROPRIATE MATH MODELS, and reach math solutions

5. Frequently Used Functions in ME Engineering Analyses ● Mass (m), weight (W), Length (L), Area (A), Volume (V) of solids ● Forces (F) ● Stress (σ), Strain (ε) in deformed solids ● Distance traveled by a moving rigid body (S) ● Temperature in solids and fluids (T) ● Velocity of a rigid body or fluid (V) Properties of Functions ● Functions may change their values with the change of “independent variables” (spatial and temporal) - So, functions are “dependent variables” ● The value of a function is a CONSTANT-depending on the values of the associated independent variables.

6. Continuous functions and Discrete-valued functions The value of functions are determined by the values of the associated variables The variation of the value of the function may be a CONTINUOUS process = CONTINUOUS FUNCTION, or such variation may occur at distinct values of the variable = DISCRETE-VALUED FUNCTION Example of CONTINUOUS FUCTION - The daily temperature variation at SJ airport: Example of Function with discrete values - Load on a beam by a family of four:

7. Math Expressions of Continuous Functions and Functions of Discrete-Values Example: Math expressions for the “bending moment of a cantilever beam induced by: (a) the weight of a roller, and (b) a concentrated force: • A continuous varying bending • moment M(x) induced by a • rolling roller – continuous change • of loading locations • continuous loading function (b) Bending moment M(x) = FL1 – a distinct value induced by a concentrated force F

8. The Derivatives – Accounts for theRATE of the change value of functions with respect to the changing values of the associated variables ● Functions represent physical quantities in engineering analysis ● These physical quantities change their values with the change of associated independent variables, e.g., (x,y,z,t) ● Change of the value of physical quantities (i.e., the functions) can be ‘CONTINUOUS,” or “INCREMENTAL”: Example: The value of a continuous varying function can be approximated by “piece-wise” linearly varying functions (liner function = function change its value following straight lines) = X(t) Position of the car at time t X 0 x Continuous variation: Real Incremental variation: Unreal (Approx.) X(t) X(t) 0 t 0 t

9. Definition of Derivatives Mathematical expression representing the RATE of CONTINUOUS VARIATION of functions Rate of a continuous variation can be viewed as variation of a function with INFINITESIMALLY SMALL increments of the associated independent variables: ∆x→0 and/or ∆y→0 and/or ∆z→0 and/or ∆t→0 Continuous variation: Real Incremental variation: Unreal P(t) P (t) IV II III I 0 0 t t t t ∆t→0 ∆t >>0 The rate of change of P(t): The rates of change of P(t): (2.1) is the DERIVATIVE of function P(t) No single derivative for all stages!

10. Math Expression of Derivative of Function y(x): One would readily see that the value of the function Y depends on the value of the associated variable x in the diagram The derivative of function = the rate of change of y with respect to the change of the value of variable x But the variation of y is continuous with the change of the value of x But if we look the variation of y within a segment of variable Δx for the variation of x Between Pt. A and Pt.B: The RATE of change of the y-values can be shown to be: Δy/Δx if we look at the “cord” between A and B. The cord AB obviously cannot represent the curve AB as shown. However, the two will become very closely resembled if the gap Δx becomes small. So, for a CONTINUOUS FUNCTION y, the rate of value change of y can be evaluated with infinitesimally small increment of x, i.e. Δx → 0, or mathematically in the form: = Derivative of function y with variable x with abbreviation:

11. Orders of Derivatives = the first (1st) order derivative = the second (2nd) order derivative, y”(x) = the third (3rd) order derivative, y’’’(x) = the fourth (4th) order derivative, yiv(x) ME analyses almost never involve derivatives with orders higher than 4

12. Physical Representations of Higher Order Derivatives Deflection Curve of a Bent Beam: = The slope of the deflection curve of the bent beam evaluated at location x. = Bending moment at x with C being a constant. = Shear force at x, with α being a constant.

13. The Curve Fitting Techniques Curve fitting allows engineers to derive CONTINUOUS FUNCTIONS to describe an existing curve, for example the curved edge of a plate: OR to derive a CONTINUOUS FUNCTIONS that will pass (fit) the discrete point values (e.g., the values of sample points) of an engineering event, e.g., the function T(t) in the following diagram: Temperature, T Time t The derived continuous function for the above case can be used to interpolate or extrapolate in data that are not shown or recorded in the above diagram There are several curve-fitting techniques available to engineers. The simplest one is called: “Curve fitting by Polynomial Function” method.

14. Applications of Curve-fittings in Engineering Analysis 1. To fit a limited number of discrete data points from a field testing or laboratory experiment. Example: To derive a CONTINUOUS FUNCTION to fit the following four (4) field test data. 2. To interpret the data between given sample data, e.g., the temperature at 3.7 minutes, or at 6.5 minutes ORextrapolate the sample data range. e.g., the temperature at 2 minutes or at 15 minutes. Measured sample points ● ● ● 3. To find a continuous function to fit an existing shape, such as the area of archway doors 4. Determining the maximum or minimum values of the available sample data

15. The Polynomial Curve Fitting Technique Y(x) Let: Yi = Given value at discrete sample point xi with (i = 1, 2, 3, ……….,n+1) n = the highest order of an assumed polynomial function: The sample points: (xi, Yi) (x4,Y4) Y2 Y4 Yi Y1 Y3 Y5 x x1 x4 0 xi x2 x3 x5 (a) Y(x) = Ao + A1x + A2 x2 + ………………….+ Anxn where Ao, A1 , A2 , ……………., An in Equation (a) = constant coefficients to be determined from given sample points of Yi at xi (i = 1,2,3,…., n+1) We assume that the assumed function Y(x) in Equation (a) PASSING all the sample points with specified (given) coordinates: (x1,Y1), (x2,Y2),…., which lead to the following expressions for the 5 sample points: Solve for the 5 unknowns A1, A2, A3, A4 and A5 in the above 5 simultaneous equations will lead to the function in Equation (a) passing the 5 sample points

16. Example 2.1 on generating a function fitting specified data points. Derive a polynomial function that will pass the following three (3) data (sample) points: (1, 1.943); (2.75, 7.886); and (5, 1.738) We first realize that the total number of sample point is 3, i.e. (n+1) = 3, from which we have n = 2 to be the highest order of the polynomial function for the case Y(x) = Ao + A1x + A2 x2 (a) Fitting the three given data (sample) points would given us the following conditions: x1 = 1 and Y1 = 1.943 x2 = 2.75 and Y2 = 7.886 x3 = 5 and Y3 = 1.738 Substituting the above sample points into Equation (a), we will have the following three simultaneous equations: For i = 1: Ao + A1 (1) + A2 (1)2 = 1.943 For i = 2: Ao + A1(2.75) + A2 (2.75)2 = 7.886 For i = 3: Ao + A1(5) + A2 (5)2 = 1.738 Solve for: Ao = -5.6663, A1 = 9.1414 and A2 = -1.5321 And the function that passes all these 3 data points to be: Y(x) = -1.5321x2 + 9.1414x – 5.6663 We may use the above function for interpolation and Extrapolation of data values other than these 3 data points

17. The derived polynomial function allows the engineers to do the following: • Interpolation between data points: for instance: the Y-value at x=2.25 (between x = 1 and 2.75) Y(2.25) = -1.5321 (2.25)2 +9.1414(2.25) -5.6663 = -7.7563 + 20.5682 -5.6663 = 7.1456 2) Determine the maximum value of the function: Step 1: To determine WHERE ym = ymin or ymax happen at xm by: from which, we solve for x =xm=2.983 Step 2: Check the sign of the value of β: from A +ve sign for minimum value β for Ymin, whereas a –ve sign β for Ymax at x=xm So, Y(2.983) =7.9694 = Ymax at xm = 2.983 3) To determine the AVERAGE value of this discrete data set in the whole range of x = 1 and 5, or in the selected range of for example x=1.5 and 4:: Ym=7.9694 ● xm =2.983 Given dataset: (1, 1.943); (2.75, 7.886); and (5, 1.738) HOW WILL YOU DO THESE????

18. The Integrals ● Integration is a reverse process of differentiation ● Differentiations evaluate the rate of change of function values with infinitesimal increments of the associate variables, e.g., the rate of change function P(t) between t = t and t = t+∆t is: P(t) 0 t t ∆t→0 t = t+∆t ● Integration SUMS UP the function values obtained in all infinitesimal increments of variables associated with the function, e.g., P(t) P(t) 0 = ∑ Element areas with Δt Function value at t 0 P(ta) t t 0 t t ∆t→0 ∆t→0 Function Range t = ta

19. Integration of Function P(t) between ta and tb Element Area, Ak P(t) P(t) Function value at ta Function value at tb P(tb) Pk P(ta) 0 t 0 t t t ∆t→0 t = tb t = ta t = ta t = tb ∆tk → 0 ● Mathematical Formulation for Area Under the Curve Represented by Function y(x): y(x) Because the increment ∆x is so small, the area of Element k under the curve can be made to equal: Consequently the total area under the curve between X = a and x = b is: yk Element k yk-1 (2.2) 0 NOTE: You need to formulate the function y(x) to obtain the area by Eq. (2.2) x ∆x x = b x = a εk-1 εk

20. Example 2.1: Determine the area of a triangle: C 4 units A B A 2 units Step 1: To set the plane in a coordinate system with reference at O: y y C (0,4) y(x) y(x) (0,2) OR A B 0 x 0 x (2,0) (4,0) Step 2: Determine the function y(x): y(x) = -2x + 4 Step 3: Use Equation (2.2) to determine the area under function y(x): unit square ● Read: Examples 2.4 (p. 22) and 2.5 (p. 23)

21. Average of a CONTINUOUS varying physical quantity represented by a function) Function y(x) represents the variation of a physical phenomenon: y(x) y(x) y(x) Area A yav A 0 x x x = b x = a 0 x = b x = a b - a The average value of function y(x) between x=a and x=b is obtained by:

22. Example 1: A function with discrete values at: (1, 1.943); (2.75, 7.886); and (5, 1.738) Find the average value of the dataset between x=1 and x=5 Solution: We may find the equation of a function that involves all these data points from the previous example to be: x=1 x=5 Y(x) = -1.5321x2 + 9.1414x – 5.6663 Example 2 Determine the average temperature of a summer day in San Jose with measured temperature as:

23. Plane Area Bonded by Two Curves y(x) Function y1(x) A Function y2(x) 0 x a b The area defined by the two functions between limits x = a and x = b is: (2.3)

24. Example (Not in the printed lecture notes) Determine the area of a plate with a geometry defined by two curves of ellipse and arc with dimensions shown below: y Half-ellipse Half-circle A a = 4m 0 x b =1m y We may determine the area bonded by these two curves in an (x,y) coordinate system as shown→ y1(x) for ellipse ∆A y2(x) for circle (using the geometric symmetry about the y-coordinate) x 0

25. y Finding the functions for the elliptical part: ●The equation for an ellipse shown in the figure at left is: y1(x) for ellipse ∆A y2(x) for circle from which, we get: x 0 Thus, we have: ● The equation for a circle is: Thus, the function ● The area ∆A is by using Equation (2.3): The above integral can be evaluated either using the CRC Math Tables, or by calculator to be: ∆A = 4.71 m2. This leads to the total area defined by the half-ellipse and circle To be: A = 2∆A = 9.42 m2

26. Volume of Solids of Revolution Solids of revolution: Solids with their geometry symmetrical to an axis of revolution. They are commonly used in machine component design. Examples; Cylinders; cones, wine and coke bottles Mathematically, they are defined as: The volume of revolution about the x-axis can be obtained by: (2.4)

27. The volume of revolution about the y-axis can also be obtained by: (2.5) Example 2.6: Determine the volume of a right-cone by using the integration method. y(x) Function y(x) = 0.5 x y(x) x – Axis of revolution 0 x = 8 Use Equation (2.4) to determine the volume of revolution:

28. Example 2.8 (p. 27) Determine the volume of wine that can be contained in a bottle shown below Geometry: from “curve fitting” Section 1 Section 3 Section 2 ● Wine bottle is a “solid of revolution,” with the coordinate z being the axis of revolution ● Because the axis of revolution coincides with a “vertical” coordinate, we will use Eq. (2.5) to get the volume of the 3 designated sections ● The volume of Section 1 and 2 are right cylinders. There is no need to use integration method Volume of Section 1: Volume of Section 2:

29. 15 z = -4r2 + 274 Approximated by curve fitting using polynomial function using 3 sample points Volume of Section 3 – the curved section: By using Equation (2.5): The total volume inside the wine bottle is: V = V1 + V2 + V3 = 31.4 + 703.36 + 106.76 = 841.52 cm3

30. Centroid of Plane Areas ● “Centroid” is the location in a plane solid, e.g., plates, at which situates the “center of gravity” ● It is an important parameter in “rigid-body dynamic analysis” and “computer-aided-design” Centroid location: = element area Define: ● Area moment about the x-axis: ● Area moment about the y-axis: Then, the centroid location is: and Expressions for area moments and centroid location for convenient cases (with one edge of the plane area coincides with x-coordinate):

31. Example 2.9 Determine the location of the centroid in a plate of semi-circular geometry Function y(x) a 2a The function y(x) can be derived from the equation of circle: x2 + y2 = a2 in the forms: or By using Equation (2.7), we have: leading to: The location is at x = 0 because of the symmetry of geometry about y-axis

32. Centroid of Plane Areas Enclosed by Multiple Functions – Not available in the printed notes y(x) Areas of individual elements: y2(x) y3(x) y1(x) B A2 A A3 A1 0 x a b c d Calculate centroids of individual elements using Equations (2.7a) and (2.7b): for Element 1 for Element 2 for Element 3 The centroid for the plane defined by Abad can be obtained by the following expressions: and

33. Centroid of Plane Areas Enclosed by Multiple Functions – Not available in the printed notes Examples: Determine the locations of the centroid in the following plates: A triangular plate A joint in a large mechanism Half-ellipse Half-circle A special plate A robotic arm

34. Differential Equations in Mathematical Modeling What are differential equations? Equations involving derivatives (of different orders) How differential equations are derived? They are derived from the laws of physics What are the laws of physics relevant to engineering applications? ●Fundamental laws of Physics: ● Conservation of energy – The first law of thermodynamics ● Conservation of momentum ● Conservation of mass ● Application laws of physics in ME: ● Newton’s laws for solid mechanics (static and dynamic) ● Fourier law for heat conduction in solids ● Newton’s cooling law for convective heat transfer in fluids ● Bernoullis law for fluid dynamics

35. Bonus Quizzes for Chapter 2 for voluntary submissions Quiz 1: Use the integration method to compute the opening area of a window in a REAL church, temple or some architectures, such as shown below, or the opening of the Caesar Charvez Arch on the SJSU campus. Include the name of the particular place where the opening is located with pictures and the measured dimensions and your calculation of the area of the opening in your report. OR OR Quiz 2: Use the integration method to compute the content volume of a REAL wine bottle. Show its picture and measured dimensions of the bottle and your computation of the volume of the bottle and check your computed volume with what is posted on the bottle. NOTE: No more than two students in each of these quizzes