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pastpaper0809 - Foundations of Engineering Mechanics...

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Unformatted text preview: Foundations of Engineering Mechanics (ENGGiO i 0) SECTION A f the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN and have a centre of gravity at H. Arm ABCD has a weight of 2 kN and a centre of gravity at B; arm DEFG has a weight of i kN and a centre of gravity at E. The weight of the hydraulic cyiinders can be ignored. A}. Figure A1 shows a sirnpiified sketch 0 cc, determine the force (a) Draw a free body diagram for the complete mechanism. Hen (5 marks) in the horizontal cylinder CJ, ermine the (’9) Draw a free body diagram for the bucket and its contents. Hence, det (4 marks) force in the horizontal cylinder E1. gram for arm DEFG. Determine ail the forces acting on this (c) Draw a free body dia (11 marks) arm for the position shown. ‘5 9‘ 1'0 _//-‘~\ 5 Figure Ai (13:110.) Foundations of Engineering Mechanics (ENGGmiG) Page 3 A2. (a) Two wooden pianks, each 12 mm thick and 225 mm wide, are connected by the dry mortise joint, as shown in Figure A2(a). Assume that the wood used shears off aiong its grain when the average shear stress exceeds 8 MPa. (i) Sketch the failure mode of the joint. (2 marks) (ii) Determine the magnitude of P at which the joint fails. (5 marks) ._..! 16 mm 3 (b) The structure shown in Figure A203) consists of two rods AB and BC, which have cross-sectional areas 400 and 800 mmz, respectively. (i) Assume that one of the rigid walls is removed; determine the totai elongation of the rods when the temperature is increased by 50 OC. (2 marks) (ii) Hence or otherwise, determine the axial forces in the rods when the 3 temperature is increased by 50 DC. (7 marks) (iii) Further to part “ii”, determine the displacement of iocation B. Assume that the coefficient of thermal expansion and elastic moduius of the two rods are lO'S / 0C and 200 GPa, respectively. (4 marks) C A B 400 mm 400 mm Figure A2(h) (PTO) ENGGIOIO — Examination in December 2009 Solution (Prof. A.K. Soh) A1(a} With reference to the free body diagram for the complete mechanism, by taking EMA=0, we obtain P(1.8 cos 30°) — 2(0.9 sin 30”) — 1(33 sin 30°)w10(0.3 + 4.5 sin 30°) 3 O [/‘D P = 18.0 kN (b) E 1.2m. 30° 6.3m. _ 63‘ 30m With reference to the free-body diagram for the bucket and its contents, by taking 2M3 : 0, we obtain '17 EU .2 cos 30°)—10(o.3)= 0 a E = 2.89 kN (e) 2g=oflcrze9=m oxzzgsm EFyzozt‘Gy—1050 Gy=1om .6.' 0° 4): tan" = £9.10?” 3.0 c0360° 2MB 1: 0, therefore, F{§.2 cos(30° - 19.:07°)1— 2.8902 cos 30°) w~ 1(06 sin 30°)-10(1.8 sin 30") x 0 i-e-, F = 10.44 kN EFX n 0, :fi Dx + 2.89 +1044 C0519.107° ~ 2.89 = 0, DK 1 -9.86 W EFY‘JO,:>Dy+1B.44Sin19.107°—§—10=0, ' Dyx7.58 kN D = flux)? +(Dy)2 = J(— 9.36)2 +(7.58)2 =12.44 kN Solution for A2(a) shear area Total shear area at failure, A = 6x12x16 : 1 152 mm2 " .'.P=8><A:8><1152N=92E6N ‘ , Solution for A2(b) (g) Total esongation = ZaLT z 2 X 30-50(00) 50 x 0.4 mm ~ FL , (i1) gig-F (111): 0 4 00 F 1 1 ) x W0 '4 jw’flfl «5'1"WY 200x10 400x10 800x10 :>1I«‘(3750)=—200><106 :F = —53.33><103N : —53.33kN _ ’1 3 (iii) 58 = E+aLT : 5333*” X400 H04 x50><400 AE ~400><200><103 = "0.06667 mm Location B is displaced by 0.0667 mm towards the left. {Lt/’WJ'I ,/' PageZ Fcundations of Engineering Mechanics {'ENGG 1 0 i 0) SECTION A A1. A hydrauiic-lift table is designed f0 lifting is controlled by a hydrauiic actuatora supported by two iinkagcs at B and D. Assume that the weight of the automobiie is 18 kN. (a) Draw a free-body diagram for the 31ft tabie.1-Ience, determine the forces at B and D. (12 marks) (b) Draw a free-body diagram for the linkage AFB. (4 marks) (6 marks} (0) Determine the force on the piston rod EF. diameter of ressure in the hydraulic actuator, which has an internal {3 marks) (d) Determine the p 100 mm. Figure A1 (Firm l’age 3 Foundations of Engineering Mechanics (ENGGlOlO) ith a 70 mm x llO mm rectangular cross section will he (450 :3 es 90°) at its mid-section, as shown in for the glue are 5.0 MP3 in tension and A2. (a) A wood tension member w fabricated with an inclined glue joint Figure A2{a). If the allowable stresses 2.9 We in shear, determine (5 marks) (i) the optimum angle gfi for thejoint; and (4 marks) (ii) the maximum safe load P for the member. Figure A2(a) sectional area A, coefficient of thermal expansion (13) Show that when a truss of cross E is subjected to a tensile force F and temperature at, length L and eiastic modulus change 1’“, its elongation is: 5 = fl— ‘ etLT A E (4 marks) Figure A2(b) shows a bi-metallic temperature sensor, which is designed to have a short steel tube of outer diameter 65 mm and inner diameter 60 mm surrounding a solid copper rod of 50 mm diameter. At 20°C, the rod and the tube have the same length and are unstressed. If a load of 80 RN is placed on top of the rod and cylinder, determine (i) the stresses in the two components when the whole assembly is subject to a (9 marks) temperature change of 45°C; and e at which the copper rod would take up all the loading. (ii) the temperatur (3 marks) are 20? GPa and HM GPa, Assume that the moduli of elasticity of steel and copper f thermal expansion are respectively, and the corresponding coefficients 0 l2x10‘6/°C and 18.5x10‘6 1°C. Steel tube WED) (38 (Prof. Sch) ENGGlGlfl w- Examination in December 20 Solution for A} For the table, 21MB 2 0 gives 2.5 R. 008 400 =18000(1_13) '1 R =10620.8 N ZMD z: 0 gives 13342.5) = 18000 (2.5 — 1.13) P}. z 9864 N 2F; 3 0 gives 1% 3 R sin 49° :- 68269 N For link AFB) 0.25 + AFSinSO” mm? = w-w , 2.7M AFCOSSO‘ With Ki“: 1, 8: tan" 0.4939 2 26538” and ziAFE 12’ 1800 — 58” — 26.38“ = 103.7” ZIWA : 0 gives Q sin(ang16 APE) PX (1.5 sin 50”) P}. {LS COS 50°) Q = (6826.9 x 1.5 Sin 50“ + 9864 81.5 (:05 5005/5511 103.7” ' := 17863.5 N Pressure in the actuator = - 2.2? MP5! @ 7r 5. (100)? 5'4 W Solution for A2(a) {i} N11111:: I 6,. A“ = P sin :9 Vmax 1 in An = P cos Q) (ii) A : 70(110) = 77100 mm2 A 7709 An 2 W : ‘W—‘F sir;ng Sin 59.89” 3.0 x 8901.08 sin 59.89” fl 1 55 Solution for A200) Total strain X mechanicai strain +therma1 strain 5/12 3: 0:15 + aT 0 FL gr 5 ~m + {IL-T (i) Let F: and F5 be the compressive forces on the copper rod and steel tube, rcspcctiveh-z ’ FCL Elongation of copper rod} 56 I a}. LA!" — =— ACEC fl _ “ FSL Elongatlon 05: steel tube, 65 I aSLAT ~ A F p , S Compatibility: ac -— és :> (1553?“ ._ “mew” : @551? M _m#s__ (g A555 ATM”, AC x 34035)? ==1.963><1O’3m2 AX = fig—(0.0652 — 0.062) x 23.90% 10"1 :72?" Thus, Eqn. (i) becomes: m.-wa 5+ 5.30:2.925x10'4 101.6x10 204.2x10 136/2042 w F’s/101.6 = 292.5 (ii) 20 = 0 :> F; + x 80000 01;} From Eqns. (ii) & (11;), 12 : 73.27 kN , F; r: 6.73 RN Compressive stress in copper rod = ETC/Ar : 73.27 x 1030.963 x 10'}: 37.33 x 10610; O = FEMS Compressive stress in Staci tube = 6.73 x 103/4909 >< 104:1311 x :06 Pa (ii) Substitute Fe E 80000 & F; I 0 in Eqn, (i), it becomes mm :— 05 x1046“ ” 20) 2;» T = 80.27% {3) 204.2 X 106 ...
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This note was uploaded on 09/24/2011 for the course ENGG engg1010 taught by Professor Wong during the Spring '11 term at HKU.

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pastpaper0809 - Foundations of Engineering Mechanics...

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