SEMESTER TEST 1 MEMORANDUM 7 to 10 September 2020 COURSE: SUBJECT: MECHANICAL ENGINEERING THEORY OF MACHINES MKE3B21/MKEMCB3 INTERNAL MODERATOR: INTERNAL ASSESSOR: • • • • • • • • • TIME: MARKS: 3 days 60 Prof AL Nel Dr CR Bester This questionnaire comprises 3 questions on 3 pages. Number your answers clearly and in agreement with the questions. Open-book test without a formula sheet. Answer all the questions in English. Type your report in MSWord. Only .doc or .docx files will be accepted. No pdf-files, photocopies, cellphone photos or scans of handwritten documents will be marked. Use a legible text font e.g. Arial or Times New Roman, and a dark colour, preferably black – NOT RED. You may type your equations in the default MSWord font for equations, i.e. Cambria Math. Don’t waste your time converting it to text font. Note the use of the decimal comma. ANY PLAGIARISM THAT IS DETECTED WILL BE SEVERELY PUNISHED. OFFENDERS WILL BE PROSECUTED AND MAY FACE EXPULSION FROM UJ. Question 1 (24 marks) Examine an 8-cylinder 4-stroke inline engine with a 1-3-5-7-8-6-4-2 firing order for primaryand secondary force- and moment balance. A schematic of the crankshaft is shown in Figure 1. Take the centre of the crankshaft, i.e. midway between cranks 4 and 5 as the reference plane for moments. The cranks are angularly spaced as shown in Figure 2. Determine if the primary and secondary forces and moments are balanced. Use R as the symbol for the total reciprocating mass per cylinder, r for crank radius and l for the spacing between the centerlines of adjacent cylinders. Do not draw the force- and moment polygons. Figure 1: Crankshaft of four-stroke 8 cylinder inline engine with 1-3-5-7-8-6-4-2 firing order 1,8 90° 3,6 7,2 90° 90° 5,4 Figure 2: Crank spacing of four-stroke 8 cylinder inline engine with 1-3-5-7-8-6-4-2 firing order 2 Solution (Balancing – similar to class examples / homework / tutorial problem) The primary- and secondary cranks are shown in Figures 3 and 4 respectively 1,8 90° 3,6 270° 7,2 180° 5,4 Figure 3: Primary cranks 1,5,8,4 180° 3,7,6,2 Figure 4: Secondary cranks 3 Primary forces Plane 1 3 5 7 8 6 4 2 Totals R i ri (kg m) Rr Rr Rr Rr Rr Rr Rr Rr i (°) 0 90 180 270 360 450 540 630 Riricosi (kg m) Rr 0 -Rr 0 Rr 0 -Rr 0 0 Ririsini (kg m) 0 Rr 0 -Rr 0 Rr 0 -Rr 0 (4 marks for each of columns 3 to 5) Primary forces are balanced (1 mark) Primary moments Plane 1 3 5 7 8 6 4 2 Totals xi (m) -7l/2 -3l/2 l/2 5l/2 7l/2 3l/2 -l/2 -5l/2 i (°) 0 90 180 270 360 450 540 630 Ririxicosi (kg m2) -7Rrl/2 0 -Rrl/2 0 7Rrl/2 0 Rrl/2 0 0 Ririxisini (kg m2) 0 -3Rrl/2 0 -5Rrl/2 0 3Rrl/2 0 5Rrl/2 0 (4 marks for each of columns 2, 4 & 5) Primary moments are balanced (1 mark) 4 Secondary forces Plane 1 3 5 7 8 6 4 2 Totals R i ri (kg m) Rr Rr Rr Rr Rr Rr Rr Rr 2i (°) 0 180 360 540 720 900 1080 1260 Riricos2i (kg m) Rr -Rr Rr -Rr Rr -Rr Rr -Rr 0 Ririsin2i (kg m) 0 0 0 0 0 0 0 0 0 (4 marks for each of columns 3 to 5) Secondary forces are balanced (1 mark) Secondary moments Plane 1 3 5 7 8 6 4 2 Totals xi (m) -7l/2 -3l/2 l/2 5l/2 7l/2 3l/2 -l/2 -5l/2 2i (°) 0 180 360 540 720 900 1080 1260 Ririxicos2i (kg m2) -7Rrl/2 3Rrl/2 Rrl/2 -5Rrl/2 7Rrl/2 -3Rrl/2 -Rrl/2 5Rrl/2 0 Ririxisin2i (kg m2) 0 0 0 0 0 0 0 0 0 (2 marks for each of columns 4 & 5) Secondary moments are balanced (1 mark) Total 48 marks; ½ marks → 24 5 Question 2 (20 marks) The driving torque TD of an engine is given as follows in terms of angle of rotation : TD = 5 000 + 2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ Nm The engine drives a machine with a constant resistance torque TR at a mean speed of 180 RPM. The flywheel mass and radius of gyration are 640 kg and 1,25 m respectively. Determine the maximum angular velocity of the crankshaft and the crank angle where it occurs, the minimum angular velocity of the crankshaft and the crank angle where it occurs, the coefficient of fluctuation of speed and the power of the engine. 6 Solution (Crank effort and flywheel design – new problem, from H & S p. 126, probl. 8, altered) Given (no marks): TD = 5 000 + 2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ Nm TR = 5 000 Nm m = 640 kg k = 1,25 m The maximum- and minimum angular velocities occur where the angular acceleration is zero. The angular acceleration is given by: θ̈ = 1 Tnet (TD – TR ) = I I (in homework assignment 4, no marks) where TD = 5 000 + 2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ Nm TR = 5 000 Nm (given, no marks) (given, no marks) Net torque is given by: Tnet = TD – TR = 5 000 + 2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ − 5 000 = 2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ Nm (1 mark) The moment of inertia of the flywheel is calculated as follows: I = mk2 = (640)(1,25)2 = 1 000 kg m2 (2 marks) The following expression for angular acceleration of the flywheel is obtained using Newton’s second law of motion: θ̈ = Tnet 2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ = = 2 sin θ − sin 2θ + 0,4 sin 3θ rad/s2 I 1 000 (2 marks) 7 At minimum- and maximum angular velocities the angular acceleration is zero: ∴ 2 sin θ − sin 2θ + 0,4 sin 3θ = 0 (1 mark) where sin 2θ = 2 sin θ cos θ (1 mark) sin 3θ = sin (2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = 2 sin θ cos2 θ + (cos2 θ − sin2 θ)sin θ = 2 sin θ (1 − sin2 θ ) + (1 − sin2 θ − sin2 θ )sin θ = 2 sin θ − 2 sin3 θ + sin θ − 2 sin3 θ = 3 sin θ − 4 sin3 θ (1 mark only because it is in Maths books) ∴ 2 sin θ − 2 sin θ cos θ + (0,4)(3 sin θ − 4 sin3 θ) = 0 ∴ 3,2 sin θ − 2 sin θ cos θ − 1,6 sin3 θ = 0 ∴ (3,2 − 2 cos θ − 1,6 sin2 θ) sin θ = 0 ∴ (3,2 − 2 cos θ − 1,6 (1 − cos2 θ)) sin θ = 0 ∴ (1,6 − 2 cos θ + 1,6 cos2 θ) sin θ = 0 ∴ 1,6 cos2 θ − 2 cos θ + 1,6 = 0 or sin θ = 0 (7 marks) The first equation is quadratic in cos . Its roots are obtained using completion of squares: cos θ = 2 ± √(2)2 − 4(1,6)(1,6) 2 ± √− 6,24 = ; 2(1,6 ) 3,2 Complex roots: Reject (3 marks) The roots of the second equation, i.e. sin = 0, are: θ = 0 or π rad (2 marks) The largest fluctuation in energy is calculated as follows: E = ∫ Tnet dθ = ∫(2 000 sin θ − 1 000 sin 2θ + 400 sin 3θ) dθ 0 0 400 = [− 2 000 cos θ + 500 cos 2θ − cos 3θ]| 3 0 = [− 2 000 cos π + 500 cos 2π − 400 400 cos 3π] − [− 2 000 cos 0 + 500 cos 0 − cos 0] 3 3 = [2 000 + 500 + 133,33] − [− 2 000 + 500 − 133,33] = 4 266,7 Nm (4 marks) 8 The mean angular velocity is calculated as follows from the rotational speed: ω ̅= 2πN = 18,850 rad/s 60 (1 mark) The fluctuation in angular velocity is calculated as follows from the largest energy fluctuation: E = 1 1 1 I (ω21 − ω22 ) = I (ω1 + ω2 )(ω1 − ω2 ) = I (2ω ̅ )(ω1 − ω2 ) = Iω ̅ (ω1 − ω2 ) 2 2 2 ∴ ω1 − ω2 = E 4 266,7 = = 0,22635 rad/s (1 000)(18,850) Iω ̅ (3 marks) The maximum angular velocity is the sum of the mean velocity and half the velocity fluctuation: ω1 = ω ̅+ ω1 − ω2 0,22635 = 18,850 + = 18,963 rad/s 2 2 (2 marks) The minimum angular velocity is the difference between the mean velocity and half the velocity fluctuation: ω2 = ω ̅− ω1 − ω2 0,22635 = 18,850 − = 18,736 rad/s 2 2 (2 marks) The angles where the maximum- and minimum angular velocities occur are determined as follows: E > 0 ∴ ω21 − ω22 > 0 The crankshaft accelerates from 0 to rad. The maximum velocity 1 occurs at rad and the minimum velocity 2 at zero rad. (4 marks) The coefficient of fluctuation in angular velocity is calculated as follows: α= ω1 − ω2 0,22635 = 100% = 1,2008 % ~ 1,2 % ω ̅ 18,850 (2 marks) Engine power is the product of mean engine toque and mean angular velocity: ̅ω Ẇ = T ̅ = TR ω ̅ = (5 000)(18,850) = 94 248 W (2 marks) Total 40 marks; ½ marks → 20 9 Question 3 (16 marks) Figure 5 shows a schematic of a simple pendulum, consisting of a concentrated mass m suspended on a weightless rod with a length l, which rotates about a pivot point P. The angular displacement of the pendulum is denoted by . The angular velocity θ̇ is positive in the direction shown in Figure 5. The pendulum motion is undamped. (i) Derive an equation for the force T in the rod in terms of mass m, gravity acceleration g, angular displacement , displacement amplitude A, natural frequency n and time t (ii) Simplify the equation for T derived in (i) above for small angular displacements P l θ̇ m Figure 5: Simple pendulum 10 Solution (Vibrations: Force in rod of simple pendulum – theory) (i) Rod force T Figure 6 shows a vector diagramme of the forces acting on the pendulum mass. T denotes the tensile force in the rod, mg the weight and Fc the centrifugal force due to the rotation of the pendulum about P. The term mg cos denotes the radial component of the gravity force. Figure 6: Forces acting on mass The radial component of the gravity force is given by: Fgr = mg cos θ Equation 1 (1 mark) If = 0, then Fgr = mg; If = /2, then Fgr = 0 (no marks) In the radial direction the mass is in static equilibrium. The tensile force T in the rod balances the combined centrifugal force and radial component of the weight: T = Fc + Fgr Equation 3 (1 mark) 11 The angular displacement of the pendulum is expressed as follows as a harmonic function of time: θ = A sin ωn t Equation 4 (1 mark) where n denotes the angular natural frequency given by: ωn = √ g l Equation 5 (1 mark) Differentiation of Equation (4) with respect to time gives the following equation for angular velocity: θ̇ = ωn A cos ωn t Equation 6 (1 mark) The centrifugal force in the rod is given by: 2 Fc = mlθ̇ Equation 7 (1 mark) Substitution of Equation (6) into Equation (7) gives: Fc = mlω2n A2 cos2 ωn t Equation 8 (1 mark) Substitution of Equation (5) into Equation (8) gives: g Fc = ml ( ) A2 cos2 ωn t = mgA2 cos2 ωn t l Equation 9 (2 marks) Substitution of Equation (1) and Equation (9) into Equation (3) gives: T = mgA2 cos2 ωn t + mg cos θ Equation 10 (1 mark) Equation 11 (1 mark) Collecting terms in Equation (10) gives: T = mg (cos θ + A2 cos2 ωn t) 12 (ii) T for small angular displacements For small angular displacements cos θ ≈ 1 (1 mark), so that T in Equation (11) may be approximated by: T ≈ mg (1 + A2 cos2 ωn t) Equation 12 (1 mark) Significance of the terms on the right hand side of the above approximate equation is considered next. For small oscillations A2 is significantly smaller than 1: A2 ≪ 1 Equation 13 (1 mark) Furthermore, cos2 ωn t is bounded by ±1: cos2 ωn t ≤ 1 Equation 14 (1 mark) Combination of Equations (12) to (14) gives: T ≈ mg Equation 15 (1 mark) 13