ENB110-T1-S1-2010

ENB110-T1-S1-2010 - GUT I II Surname Given Name/s III-III-...

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Unformatted text preview: GUT I II Surname Given Name/s III-III- Examination Paper SEMESTER: FIRST SEMESTER EXAMINATIONS 2010 UNIT: ENB110 ENGINEERING STATICS AND MATERIALS — THEORY 1 DURATION OF EXAMINATION: PERUSAL: 10 MINUTES WORKING: 2 HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS FORMULA SHEET - ONE (1) PAGE ATTACHED EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS CALCULATORS - ANY TYPE INSTRUCTIONS TO STUDENTS: Students are prohibited from having mobile phones or any other device capable OI communicating information (either verbal or written) in their possession during the examination NOTES MAY BE MADE ONLY ON THE EXAMINATION PAPER DURING PERUSAL TIME ALL SIX (6) QUESTIONS ARE TO BE ATTEMPTED ALL QUESTIONS ARE OF EQUAL VALUE GUT Gardens Point 1 All 6 Questions are to be Attempted All Questions have Equal Weighting QUESTION 1 A first year Engineering student had performed ate carbon steel to understand the behaviour of the mate a. Sketch a typical stress-strain curve of the steel parameters in your graph. b. Discuss briefly five important properties using QUESTION 2 Crystal structures Stress-Strain Curve of Engineering Materials njile testing on plain- a1. and label all important your curve. Determine the crystallographic plane and direction within the cubic unit cells shown in Figure 1.Write down all the necess ary steps you followed to arrive at your results. The origin of the unit cell in Figure 1b is indicated by the letter “O”. (b) Figure 1 Crystallographic (a) plane and (b) direction in cubic unit cell. QUESTION 3 You have been asked to design a cable car to be use top of a building. The cable is made of steel of di Design of Lift Car for lifting people to the eter 12 mm and has a length of 24 m (6 floors). The elastic modulus and yield strength of the steel are 210 GPa and 250 MPa, respectively. If the lift car weighs 24,500 N, compute, a. The extension in the cable due to the weight of the car. b. The maximum number of people that can get into the lift Without plastic deformation of the cable. Assume one person weighs 65 kg. Explain the assumption you have made to achieve your final result. QUESTION 4 Alloying of Metals a. Define point defects and dislocations in solid materials and explain why they are important in metals. b. Which one of the following two alloying (substitutional solid solution vs interstitial solid solution) is a dominant strengthening mechanism in steel? Explain why. ENBllOTl.lOl cont/ . . . 2 QUESTION 5 Phase Diagram of Steel The carbon contents of ferrite (0t), pearlite and cementite (FeC3) are 0.02wt%, 0.77 wt% and 6.70 wt%, respectively. a. Calculate the proportion of pearlite in the microstructure of 0.4 wt% C steel just below the eutectoid temperature (7270C). b. Compute the percentage of primary phase (proeutectoid phase) of the steel (0.4 wt% C) just above the eutectoid temperature. Show all the necessary steps to arrive at your answers. QUESTION 6 Heat Treatments of Steel With the aid of the TTT- curve (Figure 2), determine the microstructure of a 0.77%C (eutectoid) steel that has the following heat treatments: a. Quenching of the steel from 800°C (austenite phase) to room temperature. b. Cooling of the steel from 800°C to 600°C and held for 10 seconds before quenching to room temperature. 0.1 1 10 102 10‘ 10‘ 105 Timeseoonds Figure 2 TTT curve of Eutectoid (0.77wt%C) steel. END OF PAPER ENB110T1.101 (i) FORMULA AND C ONS TAN T S SHEET Crystallography E = 2G (1 + v) p = "A Alloying and lever rules VCNA Elastic and Plastic Properties C A = Lx 100 m A + m B 0' 2 E8 CB =1—100% 2' = Gy W _ CB —C A _ ._—!1’_ 8x 8v C B — C A V : —— : —-—' 8 8 z — C W3 2 Co A CB _ CA Values of Selected Physical Constants Avogadro’s number, N A = 6.023x1023 molecules per mole. Boltzmann’s constant, k= 1.38><10'23 J/atom-K. Gas constant, R = 8.31 J/mole-K. leV=1.6x10'19 J .‘ 0°C=273 K ENB110T1.101 III Surname III-I..- SEMESTER: FIRST SEMESTER EXAMINATIONS 2010 UNIT: DURATION OF EXAMINATION: PERUSAL: 10 MINUTES WORKING: 3 HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS RESOURCE SHEET - ONE (1) PAGE ATTACHED EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS CALCULATORS — ANY TYPE INSTRUCTIONS TO STUDENTS: Students are prohibited from having mobile phones or any other device capable of (either verbal or written) in their possession during the examination Examination Paper Given Name/s MABZZO COMPUTATIONAL MATHEMATICS 1 - THEORY 1 communicating information NOTES MAY BE MADE QNLX ON THE EXAMINATION PAPER DURING PERUSAL TIME STUDENTS SHOULD AIM TO COMPLETE FOUR (4) QUESTIONS, BUT MAY ATTEMPT ALL FIVE (5) ALL QUESTIONS ARE OF EQUAL VALUE Queensland University of Technology GUT Gardens Point QUESTION 1 (a) Let g be continuous on [a,b] and let g(:c) belong to [a,b] for all .73 in [a,b]. Prove that g has a fixed point in [a, b]. (ii) Furthermore, let 9 be differentiable on ((1,1)) and let stant k < I exist such that |g’(:t)| S k for all x in (a the fixed point in [(1, b] is unique. (b) Consider the nonlinear equation tan—1(x) = a: — 1, which is known to have a solution on the interval [2, 3]. Showing at least sin: decimal places of working, a positive con— ,b). Prove that (i) apply four iterations of the bisection method to approximate the solution on the interval [2, 3]; (ii) find two fixed point forms x = 9(33), and demonstrate that one form will converge on the interval [2, 3]; (iii) use fixed point iteration to find the solution on the interval [2, 3] correct to two decimal places. Use x0 = 2.5; (iv) use Newton’s method to find the solution on the interval [2,3] correct to three decimal places. Use (to = 2.5. MAB220T1.101 cont/... QUESTION 2 (a) Give the name of an efficient algorithm for solving an upper triangular linear system of equations. (b) Describe the three kinds of elementary row operations, and state which of them are used in the standard Gaussian elimination algorithm without pivoting. (0) Rank the following three algorithms in order from fewes t floating point operations required to most floating point operations required: forward substitution, LU factorisation, Cholesky factorisat (d) If the Cholesky factorisation of a symmetric matrix fails, imply about the matrix? ion. what does that (e) Describe a scenario where an iterative method would be preferred over a direct method for solving a linear system of equations (f) State a property of the coefficient matrix of a linear system that guaran- tees the Jacobi method will succeed when applied to the system. (g) Use Cholesky factorisation to solve the following augmented linear system: 9 15 —3 —6 12 15 34 4 —1 11 —3 4 19 2 23 —6 —1 2 31 —80 (h) Showing at least six decimal places of working, apply three iterations of the Gauss-Seidel method to the following augmented linear system, using x(0) = [0,0,0]T: 5 3 —1 —17 —3 8 —4 12 2 —1 4 13 MAB220T1.101 cont/... QUESTION 3 (a) Consider the Newton divided difference form of the interpolating polynomial for the function f : n k—l ‘ =Zfleaxla"'amk] ) k=0 i=0 4 in the case where the abscissas are equally spaced, with 4121- = :60 + ih for 2': 0,1, . ..n and with f(cci) = y,» (i) Prove that the divided difference notation and the forward dif- ference notation are related by Ale f[xo,x1,...,xk] = my; (ii) Hence derive the Newton forward difference form of the inter— polating polynomial PM) = X"; Nye, k=0 where (S) = and 3 =m (x — x0)/h. k; k! (b) Consider the following table of coordinates sampled from a certain func- tion y = f(x): 1- 1.5 10 Showing at least four decimal places of working, (i) find the Lagrange form of the interpolating polynomial through the pOintS (x01 yO): (x1: 91), ($2: (ii) use your polynomial from to estimate the value of f (1); (iii) given that f = sin(x), find an upper bound on the error in your approximation of f (1) from part (ii); (iv) find the Newton divided difference form of the interpolating polynomial through the points ($0,310), ($1,111), ($2,142); (v) use your polynomial from (iv) to estimate the value of f (1); (vi) find the Newton divided difference form of the interpolating polynomial through the points (3:0, yo), (501,341), ($2,flF), ($3,313), (x4, 3/4). part (iv); You may re—use or extend any of your working fro (vii) use your polynomial from (vi) to estimate the value of f MAB220T1.101 cont/... QUESTION 4 (a) Derive the second order central difference equation for f’ (are): f(-’Bo+h) - f($o - h) ’12 m0) = 2h — -6-f”’(c) where c is between :50 —— h and $0 + h. (b) Approximate the derivative of the function f = sin(x) a (i) second order forward difl’erence; (ii) second order backward difference; (iii) second order central difference; with step size h = 0.1. (0) Let f be twice continuously differentiable on [$80,151], whe Show that f” ma) doc — Ewe )+ m )1 — h—Bf”(c) $0 — 2 ° 1 12 where c is between $0 and x1. (d) Consider the integral 3 . I=/ s1n(:c) dx. 1 117 Showing at least six decimal places of working, at 3:0 = 1 using rex1=xo+h. (i) approximate I using the trapezoidal rule with n =1 8; (ii) approximate I using Simpson’s rule with n = 8; (iii) approximate I by completing the first four rows of table. the Romberg Note: the exact value of this integral to six decimal places is 0.902569. MAB220T1.101 c0nt/... QUESTION 5 (a) Consider the initial value problem dy —= < < y(a)=a. (i) Show that 31i+1 = yr + g lf(ti7 311‘) + f(ti+1,yi+1)l + 0013) Where 3/1 = 3/02), yi+1 = y(ti+1), ti+1 = ti + h and 0 (ii) Hence derive the modified Euler method: 100:0, 1 . wi+1=wi+—(k1+k2), z=0,1,...n 2 Where 191 = hf(ti: wi), [62 = + 11,107; + k1) . (b) Consider the initial value problem dy —— =2cos dt Showing at least six decimal places of working, (t) _ ya = 1 ‘ (i) approximate y(0.2) using Euler’s method with h 1: (b — a)/n. == 0.1; (ii) approximate y(0.2) using the second order Taylor method with h = 0.1; (iii) approximate y(0.2) using the modified Euler method With h = 0.1; (iv) approximate y(0.2) using the classical fourth order Runge—Kutta method (RK4) with h = 0.2, where RK4 is given w0=a, by l w¢+1=wi+-6(k1+2k2+2k3+k4), i=0,l,...n—1 with ’91 == hf(ti7 wi): fl 2 h k3=hf(t¢+—,w;~+£2— 2 2)’ k4 =hf(ti+h,wi+k3) . h k2=hf (ti+§,wi+ END OF PAPER MAB220T1.101 (i) RESOURCE SHEET I Identities and Rules Triggnometric Identities sin + cos (x ) = 1 sin(a :I: b) = sin(a) cos(b) i cos(a) si cos(a, :l: b) = cos(a) cos(b) 3,: sin(a) si ! sin(2m) = 2 sin(a:) cos(x) cos(2x) = cos — sin I togarithm Identities log(ab) = log a + log b 10g(a/b) = log a — 10gb log(a) = bloga Exponent Identities (ab)" = anb" (an)m ___ anm Differentiation Rules I f(u($)) = f’(u($))u’(x) (u(x>v(m>> = u'mvm + u<x>v'(> (“($)/v($)) = (u’($)v(x) — u($)v’($) . Table of Derivatives and Integrals MAB220T1.101 ...
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ENB110-T1-S1-2010 - GUT I II Surname Given Name/s III-III-...

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