ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /13

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1 ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /13 COURSE NAME: ENGINEERING MECHANICS - STATICS CODE: ENG 2008 GROUP: AD ENG II DATE: May 2013 TIME: DURATION: 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 programmable calculators is permitted. DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO

[Question 1] (a) i) What is the chief requirement for a body to be classified as rigid? [1] ii) Identify three (3) conditions under which a distributed force may be concentrated.

2 [Question 1] (a) i) What is the chief requirement for a body to be classified as rigid? [1] ii) Identify three (3) conditions under which a distributed force may be concentrated. [3] b) For the hook shown in Figure 1, determine the magnitude of the resultant force and its direction measured counter-clockwise from the positive x-axis. [10] Figure 1 c) Determine the angle θ between the edges of the sheet metal bracket shown in [11] Figure 2. Figure 2 [Total =25 Marks]

3 [Question 2] Use Figure 3 to answer i) through to iii). Figure 3 i) Why does this system of active loads qualify to be replaced with a single resultant? [2] ii) iii) Which of the active effects is free to move throughout the system without consequence? [1] Replace the load acting on the beam by a single resultant force in Cartesian (i and j format). Specify where the force acts, measured from end A. [10] b) Using Cartesian vector analysis, determine the resultant moment of the three forces about the base of the column at A in Figure 4. Take F 1 = { 400i + 300j + 120k } N. Leave your answer in terms of i, j and k. [12] Figure 4 [Total = 25 marks]

4 [Question 3] (a) Figure 5 shows a beam ABC which is supported by a pin-joint at C and a roller support at A. 10 kn 0.6 m 0.6 m A B 0.8 m 6 kn 0.4m 30 0 C Figure 5 If the beam is in equilibrium, determine: (i) the reaction at A. [5] (ii) the magnitude and direction of the reaction at C. [10] (b) The pipe assembly shown in Figure 6 is rigidly attached at the origin of the coordinate system. A force F acts at the end B of the pipe assembly and a force F acts on the collar at A. Determine the moment of the couple formed by the two forces acting on the pipe assembly. [10] z y 0.3 m x A 0.4 m collar F = (-6 i + 2 j + 2 k) N -F = (6 i- 2 j- 2k) N 0.6 m B Figure 6

[Total = 25 marks] [Question 4] Use Figure 7 to answer this question. Figure 7 (a) Determine the support reactions. [3] b) Starting at end A, determine the shear forces as you progress towards end B.

5 [Total = 25 marks] [Question 4] Use Figure 7 to answer this question. Figure 7 (a) Determine the support reactions. [3] b) Starting at end A, determine the shear forces as you progress towards end B. [3] c) Starting at end A, determine the bending moments as you progress towards end B. [8] d) On one full page of your answer booklet, draw the free-body diagram at the top of the page and the shear and bending moment diagrams aligned beneath it. [11] [Question 5] (a) Explain the meaning of the following terms: (i) dry friction (ii) kinetic friction [Total = 25 marks]

6 (iii) limiting frictional force (iv) angle of fiction. [6] (b) An object is placed on an inclined plane whose angle can be varied. Show that when the object starts to slip, the angle of inclination of the inclined plane is given by tan = s.. [7] (c) The crate shown in Figure 8 rests on a horizontal floor and is pulled with a force P using a tow rope. The weight of the crate is 900 N and its centre of gravity is at the point G. 1.0 m crate tow rope P G 1.0 m h floor 0.6 m Figure m (i) Draw a free-body diagram to show all the forces acting on the crate. [2 ] (ii) What is the horizontal force P required to move the crate? [5] (iii) Determine the height h of the tow rope so that the crate slips and tips at the same time. s = 0.4 [5] [Total = 25 marks]

7 (Question 6) Figure 9 shows the cross-section of a beam which has a line of symmetry X-X. y 25 mm 100 mm 45 mm 150 mm 25 mm 80 mm X X 150 mm x Figure 9 (a) Determine the position of the centroid of the cross-section relative to the x-axis and y-axis.. [10] (b) Determine the second moment of area of the cross-section about the centroidal axis X-X. [15] [Total = 25 marks]

8 ASSOCIATE DEGREE IN ENGINEERING SOLUTIONS SEMESTER MAY COURSE NAME: ENGINEERING STATICS CODE: ENG 2008 GROUP: AD-ENG 2 DATE: MAY 2013 TIME: DURATION: 2 HOURS SOLUTIONS:

14 Question 3 (a) B 0.8 m 6 kn R C R A H C V C (i) M C = 0; -R A x ( cos30 0 ) + 10 ( cos30 0 ) + 6 x 0.4 = 0 -R A (2.24) = 0 R A = 18.79/2.24 = 8.39kN [5] (ii) F x = 0; H C - 6 cos 30 0 = 0 orh C = 5.20 N [3] F y = 0; V C - 6 sin = 0 V C = 0 V C = 6.81kN [3] R C V C = 6.81 kn H C = 5.20 N R C = ( ) = 8.57kN [2] Ø =\ tan-1(6.81/5.2) = [2] (b) r A = -0.3 k r B = 0.6 j-0.7 k r AB = r B - r A = (0.6 j-0.7 k) - (-0.3 k) r AB = 0.6 j-0.4k [3]

15 r AB x F = ij k = i(1.2 -(-0.8)) - j(0 - (2.4)) + k (0 - (-3.6)) = i(2.0)+j(2.4)) + k ( 3.6 ) r AB x F= { 2i+2.4 j +3.6 k} Nm [6]

16 Question 4

19 (Question 5) (a) (i) Dry friction refers to the frictional force between two unlubricated surfaces. [1] (ii) Kinetic friction occurs when there is relative motion between two surfaces. [1] (iii) The limiting frictional force is the maximum frictional force which occurs between two surfaces, when an object an object is just about to slip. [2] (iv) The angle of friction is the angle between the normal reaction and the reaction of the ground. The reaction of the ground is the vector sum of the normal reaction and the frictional force. [3] (b) N F S mg sin mgcos [3] The diagram above shows the forces acting on an object when

20 the frictional force is limiting.. For equilibrium: N = mg cos (i) F S = mg sin (2) [2] Dividing (2) by (i) gives: F S /N = tan = tan [2] (c) (i) 1.0 m crate tow rope P G 1.0 m 900 N N S h floor F s 0.6 m 0.6 m O [2] (ii) F x = 0; P - F s = (1)

21 F y = 0 ; N = (2) [2] From (2):N = 900 N F S = S N at the point of slipping F S = 0.4 x 900 = 360 N From (1):P= F S = 360 N [4] M O = 0 ; 900 x P h = 0 h = (900 x 0.6)/360 h = 1.5 m [4]

22 Question 6 (a) A 1 = 7500 mm 2 A 2 = A 3 = 2500 mm 2 x 1 = 12.5 mm x 2 = x 3 = = 75 mm [4] (A 1 + A 2 + A 3 ) x = A 1 x 1 + A 2 x 2 + A 3 x x = 7500 x (2500 x 75) = = [6] (b) Rectangle 1 I 1 = (1/12) (25) (300) 3 = (10 6 ) mm 4 [4] Rectangle 2 or 3 I 2 = I x + A d 2 x

23 = (1/12)(100)(25) 3 + (2500)(92.5) 2 = (10 6 ) + (21.39)(10 6 ) = (21.52)(10 6 ) mm 4 [6] I x - x = I I 2 = (10 6 ) + 2(21.52)(10 6 ) I x - x = 99.29(10 6 ) mm 4 [5]

ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /15 PAPER A

ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /15 PAPER A SSOCITE DEGREE IN ENGINEERING EXMINTIONS SEMESTER 2 2014/15 PPER COURSE NME: ENGINEERING MECHNICS - STTICS CODE: ENG 2008 GROUP: D ENG II DTE: May 2015 TIME: DURTION: 2 HOURS INSTRUCTIONS: 1. This paper