ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /15 PAPER A

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1 SSOCITE DEGREE IN ENGINEERING EXMINTIONS SEMESTER /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 consists of SIX questions. 2. Candidates must attempt NY FOUR questions on this paper. 3. ll working MUST be CLERLY shown. 4. Keep all parts of the same question together. 5. The use of programmable calculators is permitted. DO NOT TURN THIS PGE UNTIL YOU RE TOLD TO DO SO

2 [Question 1] (a) Figure 1.1 below shows four forces acting on a particle at the origin. Determine the magnitude of P and the orientation of the 800 N force so that the particle is in equilibrium. [10 marks] 800 N y 700 N 60 0 P Figure N (b) Epress the force F in Figure 1.2 in Cartesian vector form. [7 marks] z F y 100 N Figure 1.2 (c) Determine the moment of the couple formed by the two forces F and F acting on the bar B. z [8 marks] F = {-6 i + 3 j + 3k } N -F = {6 i - 3 j - 3k } N B y Figure 1.3

3 [Total = 25 Marks] [Question 2] (a) The loading on the beam B in Figure 2.1 consists of three point loads and a couple moment. Replace the loading on the beam by a single resultant force. Specify where the force acts measured from. [15 marks] 450 N 300 N 700 N 60 0 B m 4 m 3 m 1500 Nm Figure 2.1 (b) Find the magnitude of the projected component of the force F along the cable C in Figure 2.2. [10 marks] z C 4 m 2 m B 6 m F = 500 N y 4 m Figure 2.2 [Total = 25 marks]

4 [Question 3] (a) girder B shown in Figure 3.1 has four concentrated loads acting on it. 7 kn 7 kn 7 kn 7 kn 2 m 2 m 2 m 2 m 1 m B Figure 3.1 Neglecting the weight of the girder, determine: (i) the reactions at and B. [5 marks] (ii) Draw the shear force diagram. [8 marks] (iii) Draw the bending moment diagram. Indicate the maimum value of the bending moment. [12 marks] [Total = 25 marks] [Question 4] Figure 4.1 shows a plane frame which is simply supported. 12 kn 6 kn 6 kn H G F E 2 m B C D 2 m 2m 2m Figure 4.1 (a) Determine the reactions at and D. [6 marks] (b) Identify the zero-force member(s) and justify your answer. [3 marks] (c) Using the method of sections determine the forces in the members BC, GC and GF. [16 marks]

5 [Total = 25 marks] 5. (a) uniform crate has a mass of 40 kg and rests on a floor for which s = The crate is pushed with a force F at an angle of 30 0 above the horizontal as shown in Figure 5.1. F 30 0 crate Figure 5.1 floor Determine the smallest force F required to move the crate. [13 marks] (b) crate of weight 200 lb is supported by a cord which wraps over a pipe as shown in Figure 5.2. F 200 lb Figure 5.2 (i) Determine the minimum force F which is required to prevent the crate from slipping downwards if the cord passes once over the pipe ( = ). [8 marks]

6 (ii) What tension is required to allow the crate to move downwards at constant velocity when = [4 marks] s = 0.25 and k = 0.2. [Total = 25 marks] [Question 6] (a) Eplain the meaning of the term moment of inertia. [2 marks] (b) With the aid of a diagram eplain the parallel ais theorem. [4 marks] (c) The cross-section of a beam is shown in Figure 6.1. y 120 mm 20 mm 50 mm 125 mm 20 mm Figure 6.1 (i) Determine the location of the centroid of the beam from the -ais. [8 marks] (ii) Compute the moment of inertia of the area about the -ais. [11 marks]

7 [Total = 25 marks] END OF EXM SSOCITE DEGREE IN ENGINEERING SOLUTIONS SEMESTER MY COURSE NME: ENGINEERING STTICS CODE: GROUP: D-ENG 2 DTE: MY 2015 TIME: DURTION: 2 HOURS SOLUTIONS: 1. (a) F = 0; 700 cos cos = 0... (1) [2] F y = 0; 700 sin sin - P = 0...(2) [2] From (1): 800 cos = 700 cos = 650 cos = 650/800 = = cos -1 (0.8125) = [3] Substituting into ((2):

8 P = 700 sin sin = 700 sin sin P = = 1093 N [3] (b) F = 100 N F cos 40 0 = F = 100 F = 100/cos40 0 = 131 N [2] F = 100 cos 60 0 = 50 N [1] F y = 100 cos 30 0 = 86.6 N [1] F z = 131 cos 50 0 = 84.2 N [2] F = { 50 i j k} N [1] (c) M = r F = (0.6 k j) (- 6 i + 3 j + 3 k) = [3] i j k [1] = i j k [2] = i ( ) - j (0-3.6) + k ( ) M = {4.2 i j k} Nm [2] 2. (a) 450 N 300 N 700 N 60 0 B m 4 m 3 m 1500 Nm F = 450 cos cos 60 0 = = -125 N [3] F y = -450 sin sin N = N F = N R [3] F R = [( ) + (-1296) 2 ] = 1302 N

9 = tan -1 (1296/125) = [2] M = sin sin = M = Nm [3] d N M = d sin = 9535 d = 9535/(1302 sin ) d = 7.36 m [4] (b) C 2 m B F = 500 N 4 m 6 m 4 m r B = {- 6 i + 2 k} m [1] r C = {- 4 j + 4 k} m [1] u B = (- 6 i + 2 k)/ ( ) = (- 6 i + 2 k)/6.32 [2] u C = (- 4 j + 4 k)/ ( ) = (- 4 j + 4 k)/5.66 [2] F = (500) (- 6 i + 2 k)/6.32 [1] The projected component of F along C is given by F. u C.

10 F. u C = (500/6.23) (- 6 i + 2 k).(- 4 j + 4 k)/5.66 = = 113 N [3] 3. 7 kn 7 kn 7 kn 7 kn 2 m 2 m 2 m 2 m 1 m B (a) M = 0; R B 9 = R B = R B = 0 R B = 140/9 = 15.6 kn M B = 0; -R = 0-9 R = 0 R = 112/9 = 12.4 kn [5] (b)

11 7 kn 7 kn 7 kn 7 kn 2 m 2 m 2 m 2 m 1 m B V(kN) 12.4 kn 24.8 knm 5.4 kn 10.8 knm -1.6 kn -8.6 kn -3.2 knm knm knm kn M(kNm) 24.8 knm 15.2 knm 35.6 knm 32.4 knm

12 M ma = 35.6 knm 4. (a) R 12 kn 6 kn 6 kn 2 m 2 m 2m D R D M = 0; R B 6 = R B = 0 R B = 84/6 = 14 kn [3] M D = 0; -R = 0 6 R = = 60 R = 60/6 = 10 kn [3] (b) Zero force members are : B and FE. [3 ] (c) 12 kn H G 45 0 F GF 2 m F GC 10 kn 2 m B F BC [4] Using the section shown above:

13 F y = 0; F GC sin 45 0 = 0 F GC = (10-12)/ sin 45 0 = kn (S) [4] M G = 0; F BC = 0 F BC = 20/2 = 10 kn (T) [4] F = 0; F GF + F BC + F GC cos45 0 = 0 F GF (-2.83) cos45 0 = 0 F GF = cos45 0 = -8.0 kn (S) [4] 5. (a) F 30 0 crate [3] f floor N mg Equation of equilibrium: F = 0; F cos N = 0 F cos N = (1) [3] F y = 0; -Fsin g + N = 0... (2) [2] From (1): N = Fcos 30 0 / (3) Substituting equation (3) into (2) gives: -Fsin (9.8) + Fcos 30 0 /0.25 = F F = 0 or 2.96 F = 392 F = 132 N [5] (b) (i) T 2 = 200 lb = and = 0.25

14 T 2 = T 1 e 200 = T 1 ep( 0.25 ) T 1 = 200/ ep( 0.25 ) T 1 = 91.2 lb [6] (ii) T 1 = 200/ ep( 0.2 ) = 107 lb [3] 6. (a) Moment of inertia is the second moment of area of a cross-section about a given ais. [2] (b) rea = G d Consider the figure shown above with and centroid at point G. The centroidal ais is -. The moment of inertia about the centroidal ais is I. The moment of inertia about the new ais - which is at a perpendicular distance d from the centroidal ais is given by I = I + d 2 [4] (c) y 120 mm 2 20 mm 50 mm mm

15 Divide the cross-section in parts 1 and 2 as shown in the diagram: (i) 1 = = 2500 mm 2 ; y 1 = 125/2 = 62.5 mm [2] 2 = = 2400 mm 2 ; y 1 = /2 = 135 mm [2] ( ) ŷ = 1 y y 2 ( ) ŷ = [2] 4900 ŷ = = ŷ = /4900 = mm [2] (ii) I 1 = (1/12) (125/2) 2 = I 1 = mm 4 [5] I 2 = (1/12) (135) 2 = I 2 = mm 4 [5] I = I 1 + I 2 = ( ) 10 6 mm 4 I = mm 4 [1]

16 Eamination Paper nalysis ssociate of pplied Science Degree Module: Engineering Mechanics (Statics) Eaminer: Noel Brown pril 2, 2015 Syllabus Objectives Question B C D E F G H I J 1 X X X 2 X 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.

17 The paper need to be formatted properly. I have included some of the formatting. Please ensure that the paper is proofread before it is sent to the second eaminer. Overall both papers are provides good coverage of the syllabus objectives. The time allotted is also adequate for the student to safely complete four questions. ny of the papers could be used as the final.

ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER /13

ASSOCIATE DEGREE IN ENGINEERING EXAMINATIONS SEMESTER 2 2012/13 COURSE NAME: ENGINEERING MECHANICS - STATICS CODE: ENG 2008 GROUP: AD ENG II DATE: May 2013 TIME: DURATION: 2 HOURS INSTRUCTIONS: 1. This