PORTMORE COMMUNITY COLLEGE ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY
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1 PORTMORE COMMUNITY COLLEGE ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY RESIT EXAMINATIONS SEMESTER 2 JUNE 2011 COURSE NAME: Mechanical Engineering Science CODE: GROUP: ADET 1 DATE: JUNE 28 TIME: DURATION: 9:00 am 2 hrs INSTRUCTIONS: 1. This paper consists of SIX questions. 2. Candidates must attempt ANY FOUR questions on this paper. 3. All working MUST be CLEARLY shown. 4. Keep all parts of the same question together. 5. The use of non-programmable calculators is permitted. 6. Use g = 9.81 m/s 2 1
2 DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Instructions: Answer any FOUR (4) questions. [Question 1] (a) State the conditions for equilibrium for a set of co-planar forces. (b) A simply supported beam AB is shown in Figure 1 below. (i) Calculate the reactions R 1 and R 2. (ii) How far should the 3.5 kn load be positioned from A so that the reactions R 1 and R 2 are equal? 4.5 kn 3.5 kn 1 kn A R 1 3 m 5 m 5 m 4 m R 2 B Figure 1 (c) A uniform ladder AB, 4m long and having a mass of 25 kg, rests against a smooth vertical wall at A and is supported on rough horizontal ground at B (Figure 2). The end A of the ladder is inclined at 60º to the horizontal so that it is 2 m from the base of the wall. ladder A 4 m smooth wall B rough ground m Figure 2 (i) Draw a diagram to show the three forces acting on the ladder. 2
3 (ii) By taking moments about point B, or otherwise, find the reaction of the wall. [Hint: The force exerted by the wall on the ladder acts horizontally] (iii) Determine the magnitude and direction of the reaction of the ground. [6 marks] [Total 25 Marks] [Question 2] (a) An electric motor takes 20 s to accelerate from rest to reach a maximum speed of 1200 rev/min. After doing so the power is switched off and the motor makes a further 240 revolutions before coming to rest. (i) Sketch a graph of angular velocity against time for the complete motion of the electric motor. (ii) Convert 240 rev/min to rad/s. (iii) What is the initial acceleration. [1 mark] (iv) Determine the time for the motor to come to rest after the power was switched off. (b) Three point masses A, B and C of magnitudes 3 kg, 4 kg and 2.5 kg respectively, are mounted on a shaft as shown in Figure 3. The distances of the masses from the axis of the shaft O are 15 cm, 20 cm and 30 cm respectively. B 4 kg 20 cm 15 cm O A kg 30 cm 2.5 kg C Figure 2 If the shaft turns at 300 rev/min, determine 3
4 (i) the magnitude and direction of the out-of-balance force. (i) the kinetic energy of the masses. [9 marks] [Total 25 Marks] [Question 3] A simple machine has a velocity ratio of 150. An effort of 100 N is required to raise a load of 2.0 kn and an effort of 200 N is required to raise a load of 6.0 kn. (a) (i) Sketch a graph of effort against load and show the two given pairs of values of effort and load. (ii) Determine the law of the machine. [8 marks] (b) If the load is 12.0 kn, find (i) the effort, (ii) the mechanical advantage, (ii) the efficiency, (iii) the ideal effort, (iv) the friction effort. (c) (i) What is the limiting efficiency of the machine. (ii) Can this machine overhaul when the load is 12 kn? Explain. [9 marks] [Total 25 Marks] [Question 4] (a) Define the following: (i) tensile stress, (ii) modulus of elasticity (b) Sketch a graph of stress against strain for mild steel under tensional load. On your stress-strain graph above, show the following points using the letters given below: 1. Limit of proportionality P 2. Maximum yield stress Y 3. Ultimate tensile stress U 4. Breaking Stress B (c) On your graph in (b) above, label the following region 4
5 (i) which shows plastic deformation (ii) where Hooke s law can be applied (c) It is required to design a 4 m long tie bar for a roof truss to carry a load of 80 kn. Calculate the diameter and elongation of the tie bar if the allowable stress for the material is 200 MPa and its modulus of elasticity is Pa. [10 marks] [Question 5] [Total 25 Marks] (a) A piezometer is connected to a pipeline which carries oil. The level of the free surface of the oil rises to a height of 340 cm above the centre line of the pipe and the gauge pressure is measured to be 30 kn/m 2. Find the density of the oil. (b) Make a sketch of the indicator diagram for a gasoline or diesel engine and show the region which represents the net work done. (c) A hydraulic ram has a piston of diameter 5.0 cm. Oil at a steady pressure of 16 MN/m 2 gauge acts on one side of the piston during a working stroke of 25.0 cm, the other side of the piston being exposed to the atmosphere. Determine (i) The work done on the piston in one working stroke. (ii) the average power developed if stroke is completed in 3.5 s. (d) With the aid of a diagram explain the operation of the Bourdon gauge or a gauge which uses a diaphragm or bellows. (e) Name two methods for calibrating a pressure gauge. [7 marks] [Total 25 Marks] [Question 6] (a) Explain the meaning of the following terms: (i) coefficient of kinetic friction, (ii) angle of friction. (b) A 50 kg crate is pulled with a rope at constant velocity along a rough horizontal floor, with the rope exerting a pull of 140 N and making an angle of 45 0 with the horizontal. (i) Draw a diagram to show all the forces acting on the crate. (ii) Determine 1. the normal reaction, 5
6 2. the frictional force, 3. the coefficient of kinetic friction between the crate and the floor. (c) For the crate in (b) above, determine the minimum pull which will cause the crate to move with constant velocity and the angle of the rope to the horizontal. (d) The same 50 kg crate was placed on an inclined plane so that it slides with constant velocity down the plane. If the coefficient of kinetic friction is 0.25, find the angle of the incline to the horizontal. [Total 25 Marks] ****END OF PAPER***** 6
7 ASSOCIATE DEGREE IN ENGINEERING SOLUTIONS SEMESTER DECEMBER COURSE NAME: CODE: Mechanical Engineering Science [8 CHARACTER COURSE CODE] GROUP: AD-ENG 1 DATE: TIME: DURATION: "[EXAM DATE]" "[TIME OF PAPER]" 3 hrs Solutions [Question 1] (a) Conditions for equilibrium: 1. The resultant force is zero. 2. The resultant moment about any point is zero. (b) 4.5 kn 3.5 kn 1 kn A R 1 x 3 m 5 m 5 m R 2 4 m B 7
8 (i) Take moments about point A: A CM = A ACM 4.5 x x x 9 = R 2 x = 5 R 2 R 2 = 27.5/5 = 5.5 kn Take moments about point B: B CM = B ACM R 1 x x 4 = 3.5 x x 4 5R = R 1 = ( )/5 = 3.5 kn (ii) Let the distance of the 3.5 kn force from point A be x: R 1 = R 2 = ( )/2 = 4.5 kn Take moments about point A: A CM = A ACM 4.5 x x + 1 x 9 = 4.5 x 5 = 22.5 x = ( )/3.5 x = 2.57 m (c). (i) R A A R B W = 245 N 3.46 m B m C rough ground (ii) AC = 4 sin 60 0 = 3.46 Take moments about point B: A CM = A ACM 25 x 9.81 x 1 = R A x 3.46 R A = (25 x 9.81)/3.46 = 70.9 N (iii) Use the triangle of forces: R B 245 N 8
9 70.9 N R B = ( ) = 255 N = tan -1 (245/70.9) = 74 0 [6 marks] [25 Marks] [Question 2] (a) (i) /(rad/s) 20 Area = 480 rads (ii) max = 1200 x 2 /60 rad/s max = 40 rad/s = 40 /20 = 6.28 rad/s 2. t t/s [1 mark] (ii) 240 revs = 240 x 2 = 480 rads Area of shaded region = 480 ½ t x 20 = 480 t = 48 s t = 48 s (b) (i) Calculate the centrifugal forces acting on each mass: = 300 x (2 /60) = 10 rad/s F = m 2 r F A = 3 x (10 ) 2 x 0.15 = 444 N F B = 4 x (10 ) 2 x 0.2 = 790 N F C = 2.5 x (10 ) 2 x 0.3 = 740 N Find the components of the forces and the resultant: [1 mark] x-component y-component F A 444 N 0 F B N F C -740cos 45 0 = N -740 sin 45 0 = N Total N N 9
10 266.7 N The resultant force is 278 N acting at above OA to the left. (ii) K.E. = ½ I 2 = ½ ( 3 x x x ) x (10 ) 2 = ½ ( ) x (10 ) 2 K.E. = 223 J F R 79.3 N F R = ( ) = 278 N = tan -1 (266.7/79.3) = [1 mark] [4 mark] [Question 3] (a) (i) E/N 200 E = a L + b [25 Marks] (ii) Equation is E = a L + b slope, a = ( )/( ) = 100/4000 = E = L + b Substitute E = 100 N, L = 2000 N. 100 = (2000) + b or b = 50 N Law of the machine is: E = L + 50 N. (b) (i) E = x E = 350 N (ii) M.A. = L/E = 12000/350 M.A. = 34.3 (iii) = M.A./V.R. = 34.3/150 = 0.23 (23 %) (iv) For an ideal machine, M.A. = V.R. L/N [6 marks] 10
11 L/E = 150 Ideal effort = 12000/150 Ideal effort = 80 N (v) Friction effort = Friction effort = 270 N (c) (i) The machine cannot overhaul because the efficiency is less than 50%. (ii) lim = 1/(a V.R.) = 1 /(0.025 x 150) lim = 0.27 (27 %) [1 mark] (Question 4) (a) (i) Tensile stress is the ratio of force to cross-sectional area. (ii) Modulus of elasticity is the ration of stress to strain if the material is within the elastic limit. (b) U Stress P Y B [25 Marks] Plastic region (c) Hooke s law applied L = 4.0 m F = 80 x 10 3 N all = 200 x 10 6 Pa E = 200 x 10 9 Pa all =F/Aor A = F/ all A = 80 x 10 3 /(200 x 10 6 ) A = 0.4 x 10-3 m 2 Strain A = d 2 /4 = 0.4 x 10-3 m 2 d 2 /4 = 0.4 x 10-3 d = (4 x 0.4 x 10-3 / ) 1/2 d = 2.26 x 10-2 m (22.6 mm) [6 marks] E = stress/strain = stress/(e/l) e = L x stress/e = 4 x 200 x 10 6 )/(200 x 10 9 ) 11
12 Extension, e = 4 x 10-3 m (4 mm) (Question 5) (a) h = 3.4 m p = Pa p = g h = x 9.81 x 3.4 = kg/m 3 [1 mark]) (b) Indicator diagram of a four stroke engine. P Net work done = shaded area (c) Cross-sectional area = d 2 /4 = 5 2 /4 = 19.6 cm 2 Swept volume, V = 19.6 x 25 = 490 cm 3 (i) W = p V = 16 x 10 6 x 490 x 10-6 J W = 7840 J (ii) P = W/t = 7840/3.5 P = 2240 W (d) Diagram of a Bourdon gauge: V scale pointer pivot pin-joint Bourdon tube free end of Bourdon tube with pin-joint fixed end of Bourdon tube 12
13 Fluid under pressure - The pressure sensitive element is a thin metal tub, fixed at one end and closed at the other end. - Fluid pressure causes the tube to straighten and the free end moves outwards. - This small movement is proportional to the pressure. - The movement is magnified by two levers and two gears. - The pressure range is determined by the thickness of the bourdon tube. - The Bourdon tube is not sensitive to small pressures. [1 mark for each point] (e) Methods for calibrating pressure gauge: 1. use a dead weight tester. 2. use a manometer. 3. use a calibrated pressure gauge. [25 Marks] (Question 6) (a) (i) The coefficient of kinetic friction k, is given by F = k N, where F is the frictional force when the object slides over a surface and N is the normal reaction. (b) (ii) The angle of friction is the angle between the normal to the plane (over which the object slides) and the reaction of the plane (the vector sum of the normal reaction and the frictional force) (i) Normal reaction, N Pull =140N Friction, f crate 30 0 (ii) floor Weight = 50g N 1. Consider the forces acting vertically: N sin 45 0 = 50 g N = 392 N 2. Consider the forces acting horizontally````: f = 140 cos 45 0 = 99.0 N 3. f = k N k = f/n k = 99.0/392 =
14 (c) Angle of friction = tan-1(0.25) = 14 0 R W/sin 90 0 = P/sin14 0 P = Wsin14 0 = 50g x sin14 0 W 14 0 P = 119 N The pulling force is 119 N acting at 14 0 to the 76 0 horizontal P 14 0 (d) N Pull, P mg sin30 0 mg cos 30 0 Consider the forces acting normal to the incline plane: N = mg cos 30 0 = 50 x g x cos 30 0 N = 425 N Consider the forces acting parallel to the plane: P = mg sin k N = 50 x g x x 425 P = 352 N 30 0 [25 Marks] 14
15 ****END OF PAPER***** Examination Paper Analysis Associate of Applied Science Degree Date April 2, 2010 Module: Mechanical Engineering Science Examiner: Noel Brown Syllabus Objectives Question A B C D E F G H I J 1 X 2 X 3 X X 4 X 5 X 6 X Question 1 Question is adequate for this level. Question 2 Question is adequate for this level. Question 3 Question is adequate for this level. Question 4 Question is adequate for this level. Question 5 Question is adequate for this level. Question 6 Question is adequate for this level. The paper provide good coverage of the syllabus, however, it need to be properly formatted. I have included some of the formatting and redistributed the marks. Please ensure that the paper is proofread and all track changes removed before it is sent for printing. 15
16 The time allotted for the paper is adequate as students are required to do four questions from six in two hours. All the questions can be completed in 30 minutes if students are prepared. 16
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