ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER 2 May 2013 COURSE NAME: CODE: Mechanical Engineering Science [8 CHARACTER COURSE CODE] GROUP: AD-ENG 1 DATE: TIME: DURATION: "[EXAM DATE]" "[TIME OF PAPER]" 2 hours 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. DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Instructions: Answer any FOUR (4) questions. 1
[Question 1] (a) A uniform beam AB has a length of 4.0 m and a mass of 30 kg. The beam rests on two supports at its ends. A 200 N load acts at a distance 1 m from end A and a 300 N load acts at a distance 1.5 m from end B. (i) Draw a free-body diagram showing all the forces acting on the beam. (ii) Determine the reactions at the supports A and B. [3 marks] [10 marks] (b) The diagram in Figure 1 shows a rectangular steel lamina of length 2.2 m and width 1.3 m. A square segment of side 0.6 m is removed from the lamina. y 2.2 m 0.6 m 0.6 m Figure 1 0.4 m 1.3 m x (i) Find the position of the centroid of the lamina from the y-axis. (ii) Find the mass of the weight of the lamina if its thickness is 0.8 cm. Density of steel = 7800 kg/m 3. [8 marks] [4 marks] [Question 2] (a) Define the terms: (i) mechanical advantage, (ii) velocity ratio, (iii) efficiency. [3 marks] (b) With the aid of a suitable labelled diagram, explain the operation of one (1) of the following simple machines. Screw jack, Differential wheel and axle 2
Wheel and axle Block and tackle pulley system [8 marks] (c) The table below gives values of load and effort from an experiment on a machine. Load W (kn) 1.0 2.0 3.0 4.0 5.0 6.0 Effort E (kn) 1.35 1.70 2.05 2.30 2.65 2.95 i. Use the data in the table above to draw an effort versus load graph. [6 marks] ii. State the law of the machine described by the data. [4 marks] iii. If the ideal mechanical advantage is 5. What is the effort required to overcome friction at a load of 4.5 kn? [4 marks] [Question 3] (a) Define the following: (b) (i) plane stress (ii) plane strain. [2 marks] (b) Sketch the stress-strain graph for a ferrous material and show the following points 1. the stress at the limit of proportionality ( lim ). 2. the ultimate tensile stress (UTS). 3. the yield stress ( yield ). [7 marks] (c) A steel reinforcing bar has a diameter of 32 mm and a gauge length of 6.0 m. Calculate tension in the bar if it extends by 0.5 mm. [8 marks] (d) (i) Define the terms: 1. shear stress, 2. shear strain. [2 marks] (ii) It is required to cut a cylindrical steel bar of diameter 32 mm using a shearing force of 300 kn. Determine the ultimate shear stress for the bar. [6 marks] 3
(Question 4) (a) Three point masses A, B and C of magnitudes 4 kg, 5 kg and 3 kg respectively are mounted on a shaft as shown in Figure 2. The distance of masses A, B and C, from the axis of the shaft are 10 cm, 15 cm and 20 cm respectively. If the shaft turns at 300 rev/min., determine B 5 kg 15 cm A 45 0 10 cm 4 kg 3 kg C 20 cm Figure 2 i. the magnitude and direction of the out of balance force. [11 marks] ii. the kinetic energy of the masses. [3 marks] (b) The speed of a rotating wheel decreases uniformly from 300 rev /min to rest in 10 seconds. Find: i. the angular deceleration of the wheel, [3 marks] ii. the number of revolution turned during the deceleration. [3 marks] (c) A spanner of length 35 cm was used to turn a bolt. A force of 140 N was applied perpendicular to the handle of the spanner. Find: i. the torque applied, [2 marks] ii. the work done if the spanner makes 6 revolutions. [3 marks] 4
(Question 5) (a) Explain the meaning of the following terms: (i) coefficient of limiting friction, (ii) angle of friction. [4 marks] (b) A 50 kg crate is pushed up an inclined plane at constant velocity by a 320 N force acting parallel to the plane. The angle of the inclined plane is 30 0, g = 9.80 m/s 2. (i) Draw a diagram and show all the forces acting on the crate. [3 marks] (ii) Determine: 1. the angle of friction. [2 marks] 2. the coefficient of friction between the crate and the inclined plane [4 marks] 3. the force acting parallel to the inclined plane which is required to make the crate slide down the plane with constant velocity. [4 marks] (c) Figure 3 shows how the torque acting on the shaft of a rotating machine varies over one complete revolution. Find the work done in one revolution. [8 marks] /(Nm) 150 100 (Question 6) 0 90 180 360 / 0 Figure 3 (a) (i) Differentiate between heat capacity and specific heat capacity. [2 marks] (ii) A piece of metal of mass 5.0 kg is heated to 200 0 C and immersed into 4.0 kg of water at 25 0 C contained in a brass container. The mass of the brass container is 20 kg. 5
Determine the final temperature of the system. [8 marks] Specific heat capacity of water = 4200 J/(kg K). Specific heat capacity of brass = 350 4200 J/(kg K). (b) (i) Distinguish between gauge pressure and absolute pressure. [2 marks] (ii) A U-tube water manometer is used to measure the pressure of a gas in a container. If the gauge pressure of the gas is 12 kn/m 2 and atmospheric pressure is 101.3 kn/m 2. 1. What is the absolute pressure of the gas in bars? [2 marks] 2. What will be the difference in the levels of the water in the two arms of the manometer? [5 marks] (c) A compressed air tank has a volume of 1.5 m 3. A pressure gauge fitted to this tank reads 1600 kn/m 2 when the barometric pressure was 1020 millibars. If the temperature of the air in the tank is 45 0 C, find the volume the air would occupy at s.t.p. Standard atmospheric pressure is 101.3 kn/m 2. [6 marks] ****END OF PAPER***** 6
ASSOCIATE DEGREE IN ENGINEERING SOLUTIONS SEMESTER 1 2009 DECEMBER COURSE NAME: CODE: Mechanical Engineering Sciencce [8 CHARACTER COURSE CODE] GROUP: "[AD-ENG 1 OR 2]" DATE: TIME: DURATION: "[EXAM DATE]" "[TIME OF PAPER]" 2 hr Solutions [Question 1] (a) (i) 200 N 300 N 300 N 1 m 1 m 0.75 m 1.25 m A R A 4.0 m B R B [3] (ii) M A = 0; - 200 x 1-300 x 2-300 x 2.75 + R B x 4 = 0-200 - 600-825 + R B x 4 = 0 7
(b) (i) 4 R B = 1625 R B = 406 N [5] M B = 0; 200 x 3 + 300 x 2 + 300 x 1.25 R A x 4 = 0 600 + 600 + 375 - R A x 4 = 0 4 R A = 1575 R A = 394 N [5] Check: F y = 0; 406 + 394-200 - 300-300 = 0 0 = 0 y 2.2 m A 0.6 m 0.6 m B 1.3 m 0.4 m x Consider the lamina is composed of a full rectangular lamina A and the cut-off part B of negative mass. A = 2.2 x 1.3 x A = 1.1 m = 2.86 m 2 B = - 0.6 x 0.6 x B = 2.2-0.4-0.3 = - 0.36 m 2 = 1.5 m [4] (A + B) x = A x A + B x B (2.86-0.36)x = 2.86 x 1.1-0.36 x 1.5 2.5 x = 3.146-0.54 = x = 1.04 m [4] (ii) V = A x w = 2.5 x 0.008 V = 0.02 m 3 m = V = 800 x 0.02 m = 16.0 kg [4] 8
[Question 2] (a) (i) Mechanical advantage is the ratio of load to effort for a machine. [1] (ii) Velocity ratio is the radio of the distance moved by the effort to the distance moved by the load. [1] (iii) Efficiency is the ratio of the work output to the work input. [1] (b) Labelled diagram and operation of a simple machine. [8] (c) (i) 3 Effort (kn) 2 1 0 0 1 2 3 4 5 6 4.5 kn Load(kN) [6] (ii) slope = (2.95-1.1)/(6-0) = 1.85/6 = 0.308 a = 0.308 b = 1.1 kn Law of the machine: E = 0.308 L + 1.1 kn [4] 9
(iii) From the graph, when the load is 4.5 kn the effort is 2.5 kn Ideal effort = 4.5/5 = 0.9 kn Friction effort = 2.5-0.9 = 1.6 kn [4] [Question 3] (a) (i) Plane stress is the ratio of the force per cross-sectional area of the material. [1] (ii) Plane strain is the ratio of the extension to the original length of the material. [1] (b) Stress UTS yield lim [7] Strain (c) A = π d 2 /4 = π (32) 2 /4 = 804 mm 2 A = 8.04 x 10-4 m 2 [2] L = 6.0 m e = 0.5 x 10-3 mm E = 2.03 x 10 11 Pa. E = F L/(A e) F = E A e/l [1] = (2.03 x 10 11 x 8.04 x 10-4 x 0.5 x 10-3 )/6 = (2.03 x 8.04 x 0.5 x 10 11 4-3 )/6 F = 13.6 x 10 3 N (13.6 kn) [5] (d) (i) 1. Shear stress is the ratio of the force to area, where the force acts parallel to the area. [1] 2. Shear strain is the ratio of the deformation of the material parallel to the force to the length perpendicular to the force. [1] (ii) A = 8.04 x 10-4 m 2 (from (c) above) Shear stress = force /area = 300 000 N/8.04 x 10-4 m 2 Shear stress = 37.3 x 10 7 N/m 2 [6] 10
(Question 4) (a) i. = 300 x 2 /60 = 10 rad/sec. [1] F = m 2 r F A = 4 x (10 ) 2 x 0.1 = 395 N F B = 5 x (10 ) 2 x 0.15 = 740 N F C = 3 x (10 ) 2 x 0.2 = 592 N [3] Determine the resultant of the three forces: x-component (N) y-component (N) F A 395 0 F B 0 740 F C 592 cos 225 0 = -418.6 592 sin 225 0 = -418.6 Total -23.6 321.4 [5] 321.4 N F R 23.6 N F R = (23.6 2 + 321.4 2 ) = 322 N [1] = tan -1 (321.4/23.6) = 86 0 [1] ii. K.E. = 0.5 m 2 r 2 = 0.5(10 ) 2 [ 4(0.1) 2 + 5(0.15) 2 + 3(0.2) 2 ] = 493.5[0.04 + 0.1125 + 0.12] K.E. = 134 J [3] 11
(b) i = 300 rpm = 10 rad/sec. f = 0 t = 10 s i. f = i + t 0 = 10 + 10 = - 10 /10 = - 3.14 rad/s 2 [3] ii. 2 2 f = i + 2 0 = (10 ) 2 + 2 x ( - 3.14) = 157 rad = 25 revs. [3] (c) i. = 140 x 0.35 = 49 N m [2] ii. W = = 49 x (6 x 2 ) W = 1847 J [3] (Question 5) (a) (i) The ratio of the limiting frictional force to the normal reaction for impending motion. [2] (ii) The angle between the reaction of the ground (frictional force plus normal reaction) and the normal reaction. [2] (b) (i) F k N 500 cos 30 0 30 0 500 sin 30 0 320 N [3] - (ii) 1. F y = 0; N - 500 cos 30 0 = 0 N = 433 N [3] F X = 0; 320-500 sin 30 0 - F k = 0 F k = 320-500 sin 30 0 = 70 N [3] 12
µ k = F k /N = 70/433 µ k = 0.16 [2] 2. = tan -1 (0.16) = 9.2 0 [2] 3. F k N 500 sin 30 0 30 0 P 500 cos 30 0 F X = 0; 500 sin 30 0 - F k - P = 0 250-70 - P = 0 P = 180 N (c) Find the work done in one revolution. [8] /(Nm) 150 100 0 90 180 360 / 0 Work done = area under graph = [100 x /2 + 0.5(100 + 150) x /2 + 0.5 x x 150] = [50 + 625 + 75] = (750) Work done = 2356 J [8] (Question 6) (a) (i) Heat capacity is the amount of heat absorbed or given off by a body when its temperature changes by one degree, whereas specific heat capacity is the 13
amount of heat given off or absorbed by 1 kg of a substance when the temperature changes by one degree, [2] (ii) Heat lost by hot metal ingot = heat absorbed by water + heat absorbed by brass container 5 x c ingot x (200-55) = 4 x 4200 x (55-25) + 10 x 350 (55-25) 725 x c ingot = 504 000 + 105 000 = 609 000 c ingot = 609 000/725 = 840 c ingot = 840 J/(kg K) [8] (b) (i) Gauge pressure is the excess pressure above atmospheric pressure and absolute pressure is the sum of the gauge pressure and atmospheric pressure. [2] (c) (ii) 1. absolute pressure = 12 + 200 = 212 kn/m 2. = 2.12 bars [2] 2. P = g h h = 12000/(1000 x 9.8) h = 1.22 m [5] A compressed air tank has a volume of 1.5 m 3. A pressure gauge fitted to this tank reads 1600 kn/m 2 when the barometric pressure was 1020 millibars. If the temperature of the air in the tank is 45 0 C, find the volume the air would occupy at s.t.p. Standard atmospheric pressure is 101.3 kn/m 2. P 1 = 1600 + 101.3 = 1701.3 kpa V 1 = 1.5 m 3 T 1 = 45 + 273 = 318 K P 2 = 101.3 kpa T 2 = 273 K [2] P 1 V 1 /T 1 = P 2 V 2 /T 2 V 2 = P 1 V 1 T 2 / P 2 T 1 = (1701.3) x 1.5 x 273/(101.3 x 318) V 2 = 21.6 m 3 [4] \ 14
Examination Paper Analysis Associate of Applied Science Degree Module: Mechanical Engineering Science Date: March 30, 2013 Examiner: Noel Brown Syllabus Objectives Question A B C D E F G H I J 1 X 2 X 3 X 4 X 5 X X 6 X X Question 1 Question is adequate for this level. Check editing. Question 2 Question is adequate for this level. Check Formatting. Question 3 Question is adequate for this level. B1 and B2 seems to be asking to identify the same point. Question 4 Question is adequate for this level. Question 5 Question is adequate for this level. Question 6 Question is adequate for this level. Check the solution 6b, it does not correspond to the question asked. Overall the paper provides good coverage of the syllabus objectives. Only a few changes are recommended, I have included some of the changes. Please ensure that the paper is proofread and all formatting are removed. Six questions are adequate for this exam, the time allotted (2 hours) is also adequate. The students should be able to complete four questions in 2 hours. 15