MECH-325-2016-Midterm-Solutions1

pdf

School

University of British Columbia *

*We aren’t endorsed by this school

Course

325

Subject

Mechanical Engineering

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by ColonelHyenaMaster902

MECH 325 - Midterm Book 1 Tuesday, October 25, 2016 Name: ] O l\)L!D'\ ' Student Number: Circle your section Section 101: TIA (11:00am tutorial) T1B (3:30pm tutorial) TIE (9:00am tutorial) Section 102: TI1C (2:00pm tutorial) TID (5:00pm tutorial) TIF (9:30am tutorial) Signature: Part 1 MC Mark /14 Part | SA Mark /16 Part 2 LA Mark /30 Total ____ /60 Instructions There are two parts to this exam with different instructions for each. Please read carefully. Part 1 - Closed-Book (30 marks) Multiple choice (14 marks: 1 marks per question). Complete all 14 questions by marking your response in pencil in the computer score card. Write your name and student number on the computer card and mark your student number in the “ID Field”. Short answer (16 marks: 2 or 4 marks per question). Complete all 6 questions by marking your response in this exam booklet. After you have completed Part 1, hand in your booklet and Scantron scorecard and you will receive a handout for Part 2 and a separate exam booklet for your answers. You may not return to Part | after you hand it in. Part 2 — Open-Book (30 marks) Leng answer (30 marks): Complete ail parts to the questions for Part 2 by marking your responses in the separate exam booklet. This portion of the exam is open-book and open-notes. Only non-programmable calculators may be used (or programmable calculators with purged memories). Page |

Part 1A: Multiple Choice Answer all 14 questions. Each is worth 1 mark. Choose the single best response — a given question may have more than one choice that is correct but marks will only be given for the bess answer. 1. What is the best description of the type of motion between the tecth of two properly meshed spur gears? a) Purely rolling Some sliding when teeth engage and disengage, but predominantly rolling c) Approximately equal parts sliding and rolling d) Some rolling when teeth engage and disengage, but predominantly sliding e) Purely sliding 2. When creating involute profile gear teeth, which of the following diameters is used? a) Pitch Circle b) Addendum Circle ¢) Dedendum Circle d) Clearance Circle @ Base Circle 3. Which of the following gears allows the shafts to be offset and non-intersecting? a) Spur gear b) Planetary geartrain c) Helical gear d) Bevel gear e) )Hypoid Gear 4. A gear train has gears with pitch diameters of 20 cm (pinion) and 40 cm (gear). The pinion is turned at a speed of n = 300 rev/min by a motor supplying 500 W of power. What is the torque delivered to the gear? a) 5.6Nm b) 23.9Nm ¢) 25.0Nm . @) 38Nm Typo 1m E¥ES e) 53.7Nm 5. A spur gear with a pressure angle of 14.5° has 17 teeth and a pitch diameter of 2.5 in. The gear transmits 1 hp at 1800 rpm. If the gear is replaced by a spur gear with 24° pressure angle, with all other quantities remaining the same, which one of the following will increase? a) Diametral pitch b) Torque transmitted c) Transmitted force Radial force e) Axial force Page 2

6. On the following gear set, gear 5 is the driving pinion. Gear 5 has 19 teeth, Gear 4 has 38 teeth, Gear 3 has 16 teeth and gear 2 has 64 teeth. A 1600 rpm motor is driving the gear box in the direction shown. What are the speed and direction of the output shaft, relative to the drawing on the left? a) 300 rpm clockwise b) 300 rpm counter-clockwise c) 200 rpm clockwise 200 rpm counter-clockwise e) 600 rpm clockwise 7. For the planetary gear train shown, the ring gear rotates at 100 rpm counter-clockwise and the arm is fixed. How fast does the sun gear rotate and in which direction? a) 500 rev/min (counter-clockwise) b) 500 rev/min (clockwise) c) 1000 rev/min (counter-clockwise) @ 1,000 rev/min (clockwise) e) 1,500 rev/min (counter-clockwise) 8. A power screw engages with a nut with 10 threads. Approximately how much of the load is carried by the first 7 engaged threads, combined, in the power screw? a) 10% b) 20% ¢) 50% 9. Rank the following flexible drives from lowest to highest power capacity. Assume approximately equal dimensions and operating conditions. a) V-Belt Flat belt Roller Chain b) V-Belt Roller Chain Flat belt @ Flat belt V-Belt Roller Chain d) Flat belt Roller Chain V-Belt e) Roller Chain Flat belt V-Belt Page 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

10. A V-belt drive has a driving pulley operating at 1800 rpm, a drivetrain value of e = 0.33, a nominal power of H,,., = 0.75hp, a service factor of K, = 2.2, and a design factor of ny = L.5. TWO belts are to be used in parallel. The following belts are available in the catalogue. Which is the smallest belt that will satisfy the design requirements? a) Allowable power H,= 1.0 hp Allowable power H, = 1.4 hp c) Allowable power H, = 1.5 hp d) Allowable power H, = 2.5 hp e) Allowable power H, = 5.0 hp 11. Which of the following is the best description of how a hydrodynamic bearing works? a) The shaft moves by sliding against the bearing material b) Righ pressure fluid is injected into the bearing to support the shaft @ The rotating shaft pulls fluid to form a layer between it and the bearing d) The shaft is supported by high pressure air fed through a porous material €) Metal-on-metal contact is avoided by packing the bearing with a thick grease 12. Which of the following design or operating parameters should be increased in order to be able to reduce the required bushing length of a boundary-lubricated bearing in service? (assume all other parameters remain fixed) a) Operating rotation speed, N b) Applied force, F c) Coefficient of friction, £, (PV) rating of the bushing e) Wear factor, K 13. Which of the following is NOT a possible or recommended consideration when sigunificantly increasing the power carried by a roller chain drive? (assume the linear speed is moderate) a) Increasing chain pitch size @ Upgrading the lubrication system c) Increasing both driving and driven sprocket diameters d) Increasing the number of teeth on the driving sprocket only e) Using a higher strand chain (e.g., duplex or triplex instead of simplex) 14, What type of bearing is depicted in the image to the right? a) Deep-groove ball bearing Angular contact bearing c) Tapered roller bearing d) Tapered ball bearing €) Self-aligning bearing N“"J TAR\ WL n‘ B Page 4

Part 1B — Short-Answer Questions (16 marks) Answer all questions in the spaces provided. Each question is worth 2 marks. 15. a) In gear design, what is meant by interference? How can it be avoided (2 marks)? I ‘ltv{v it H[ kr "’V [ "’DO’{\-. (ﬂ\k—ng" .\,\\A o :\]0{- {CSV(WGL t* 3&\);(1 j{a» Sfﬂbll\ )| \;\M‘ Rh«] L Cavt MY e, } Todordonee 15 G007 &4 by hewg N 3uv edl ¢ b{—\uuv W 5(@»\ W on Xf\/ \)c,h rdy fovmd \,1 He b\‘uw re ow\l«. One cen X!.,l( ! aeas will qm.la nombir ol et b) Two parallel shafts with 6 inch center distances are to be connected by a pair of gears having a diametrical pitch of 16 teeth/inch. The spur gears have a contact pressure angle of 24° and provide a gear reduction of 3.0. Determine the pitch diameter and number of teeth in the pinion gear (2 marks). o= 67 PR al e= 30125 e r@/b"?:5'3 e qrpié” rp: LS7 qnd Ve~ Ys™ So JF'\},I qucl cJC,'— 7” A7l = 48 feedt 9 x16 = |y fuhh ( h-\‘( Page 6

16. The diagram shown below is the tension profile for a flat belt. On this diagram, sketch the tension profile for a V-belt belt that is transmitting the same torque, but otherwise has the same linear speed, initial tension, pulley diameters, and belt weight per unit length. (2 marks) 17. The relationship for flat belt tension versus initial tension is shown in the diagram below. Complete the diagram by giving the name or equation for each of the four quantities identified. Put your answers in the boxes provided (2 marks) (Tisht Side T, Fl Df'»‘"") TCh 1\ mlx‘ M’g‘»% E “ Tension 2 Dehvwesd |ovw CQ'\J(V;s“)" Fc ~.7 Initial Tension L’ '/@ WU GOy U Page 7

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

18. The following figure illustrates a rear differential of a rear wheel drive vehicle. a) Explain why a differential gearbox is used on this vehicle. (2 mark) A S lredial i ued o ede Wb LK Chr e Y Tw @ ard, T "mv'\c\u %vefux boeakh okl el epedey e A‘n&&vtxk Jv{{\r L b VAR Xr\l'v‘-\-] pors N b) I you happen to park the car with one of the wheels on ice and the other on dry pavement, will you be zb,lemijave the parking spot when it is time to leave? Explain. (2 mark) | mbe MO\/ ‘\)\) X’W:\,JL e M Yvans fored L \'\\ \N\\\-L\ (I8 A‘-, ‘)cd(,k.(\*-, ‘*\‘N$(/ <\,L wheel om e wil r"'\\- whlt My ot R I)r«\-\$‘wwl : | Page 8

19. A power screw is self-locking when the screw will not turn once a lifting torque has been removed. The equations for the raising torque, Tk, and the lowering torque, T; for a power screw are given below: ) ) What is the condition for self-locking? Derive the thread efficiency equation for the power screw for the raising condition. (2 marks) SCH" lD(.,O"\j whew ‘IT£J~ -170 oot 7/ +> L/l Thect off- = Tolae frhe) u|u~ :[n (}1'\ . ﬂ /7 L Te 20. A) Why is lubrication important in a gear train system? (1 mark) TR= B) What is meant by “galling?” (1 mark) A~ R\)o'\é YV$SM/\ '\(‘7 ‘N\L"‘a\ (a\-&s\.“‘ ’ - r((\_ugx Srv'. Lk:\ n 2 - \ww\ét Luv\\"\j B [\ 4' L\h$\«-\ qlimnivn \N/\\ vVH Q ‘w:\ktkl Eb\:& WMLtKJS(\)-¥ ¢k 3(\, l(owkm\-)\h v syt Tl o VJ/\\ okélv w s har a4 S'G\\ uwh'\) ,.'Q'\m\ bonds 4 \ ‘m LY D oM. Page 9

Lona Answer Problem (30 marks) By attaching an idler gear between the pinion and gear, one can create a 3-gear geartrain where the output shaft turns in the same direction as the input shaft (the pinion meshes with the idler while the idler meshes with the gear). Specifications for the gear box are as follows: You can work in either Imperial or SI units. However, do not mix units as the conversions are not exact. Input Power to pinion Driving Motor Driven Load Speed of pinion Pinion Diametrical Pitch Face width of teeth ‘Number of Teeth on Pinion Number of Teeth on Idler Gear Gear Reduction Ratio Pressure Angle Gear Material (all 3 gears) Brinell Hardness Pinion (through hardened) Brunel Hardness ldler (through hardening) Brinell Hardness Gear (through hardened) Gear Quality (all 3 gears) Design Lifetime (all 3 gears) Reliability (all 3 gears) Size Factor, K, (all 3 gears) Load Distribution Factor K,, (all 3 gears) Rim Thickness Factor Kz (all 3 gears) 20 hp (15kW) 2 cylinder gasoline engine. moderate shock 2500 rpm 6 teeth/inch (module = 4 mm.) 2 inches (50 mm.) 17 teeth 20 teeth 3.5:1 20° Grade 2 Steel 325 Bhn 325 Bhn 275 Bhn 6 10 cycles 0.90 1.0 1.6 1.0 21. Determine a) the number of teeth on the gear, b) the centre line distance from the pinion to the gear, c) the force transmitted from the pinion to the idler and from the idler to the gear, and d) the radial load on the pinion shaft and the idler shaft. (9 marks). 22, Assuming that the transmitted load on the pinion gear is 450 Ibf (2000N), determine all the AGMA factors for the gear tooth bending stress equation for the pinion, and the idler. Estimate the bending stress number for the pinion and idler gears. Justify all assumptions. (10 marks) 23. Assume that the gear bending stress number for the pinion is 10000 psi (70MPa), and the idler is 13000 psi (90Mpa). Estimate the minimum safety factor for bending stresses for the pinion and idler gears. (8 marks) 24, This geartrain is to operate 8 hours a day during regular business days (Monday through Friday). Given the calculations you have performed, are you confident the geartrain will operate for at least five years? Explain. Which gear would you expect to fail first? Why? Are there any recommendations you can make to the client to improve the lifetime of the gearbox? (3 marks). Page 2

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Peobla_2 ? < a) l\)uhhw JL 'I’CULL o Lok T\I QN Piddh= Cl‘“%& h'\Oéu\L 2 Lf b ® 3 g I’J) =59, = 0) Ov GO iggﬂ;\, (‘0 ‘-L G(((P}QGL) L) A$L ()L por = /3«,\.\ % -E g}l"k}\dl dia, of Priwes = pr\ S EEL NS cgnu. Lo of idlr = 2% = 333 ) dic of bl = 200 - gD dia. OC Gear = COA’ 10n, i S of Year = GOXY = E‘fOA'-x (é ¢ ot gesr ¥ = 9. 83\\,) ‘ (é' of o > 59 = 2 30hm, ° Cth’(’v lM qif‘lfik(rt = rf ik é J‘(v L r?e.“p "‘k ‘(:ovmu\ﬂ‘}i (A,\Lv \M x\.;\m(c B 1,(?. 13 ) <337 *‘()O ) (enler ‘\-L dilame 9. P mdw Cor 906 C) ‘ inu-_ Toh-g. v X 5"“5 P e T-.fi . 70415& " Torgu < Fort <o .OT\nn W (3 T/d/ 504/(2 $3 /fi) \l\,t 375(’ alL’( T_Mfi\l u\r.) Ul \N-[ }3000“] —— o — . ohp (550 fly (i) (2500 ] " ®Ycp ) £ Y (€33 8043 (2yDn. = 23U mm, o 232 ne @l form b \$O)f(- Sotmt Jklw\f will m| Y J(\a\ Cavee o5 bramenittcl load SR \l\_/.t__'_,__... Seme '(: orcd is ""'ﬂ'ﬂslc“cl ‘E"ﬂ"" lc)‘z ‘J G- - lf |S m(u ‘u‘ W, ~ 600MH - (0000 (I5kW) | & F4 39 IE ( .&7 TQE’GX\“‘ )(asowr») o l.(,%m— @ T, \ \ = W, = 356, all,c Ve 0"\ DL ,moL P/\)rn -~ )) qé \IJU WR W Jar 4 (w‘:(l (,;meﬂ. - (3561 ] 4am ) = 6l - N~ o — \Rfig\ﬂ‘ laq ok |“rb = W |29fz\L 643‘ e d uy’pfxi" Iu)ra;) Cj

w000 = C €0 =L i 423b SU APl Yy D) 1Y o i 09 BT b g e Tl ipwe) E_(‘, o ‘2’0 e “ ‘11 Ltt \\/t t, "1\101 \“"N_J u'l\". Ai"‘\l.(fi né } ) = b} \ 9‘}] B \d ) 0 L 0 .£ @ ( I a1l .-r\«l_ -L' 4)|? ] fi\“- nn‘ul»d (Y‘ K 109 0 I “y (Y ) faaen) BP0 P U ( 1 ) 40100:‘ L‘.}woaﬂ Tt[“‘“ll'f (\"‘-{\N}-@ ) . 10),6 ] ’ :ﬂ g - e _eebgm () <O , Q) b st (o BTT] T g e 7‘.‘-\4 l T—“‘/{-j 252\[ = :i)( :i"‘) _!_,2)(:-'\4/,\“ 0058) iJM :f}/\ L\T.V)q )gf\ M‘.l \]T\& (3 0 \ z . Y ( \i oi gj-s 7 /]: = /\2 ! = < -\M'a\q $- "(f\“ l ) ‘)5‘%0{ N .-'\>| "’)4\0 N/\ 0 - — STy 8 L Geg) 16 ) §1 P IS TR B Bl (141936 “5) O(:r>l ool e TSy IR o hd ) <0 @ poeu b bapig T | A0 M =\ gm0l vhav F1h =AM 22d

fiy( Z,'z.'('ski—\ W U.O/I W Ko Kok P K I F 1 i ‘-(%olt/)(:s)u.g)umé%%& (Leyiv § = E_Lgi ps Qo\,hu\a\‘\ok l\v\ll k) Pm\bx 3 =0.30 | =518, = 19280 ps @ -,\L).Iél(lf T: O"j' ov 0,3) . CLoth ’JW(V UG?U( - T: 01?, o= 5% < Jam e (181 Droblee. 23, Gul= 2+ QA s.f = sally Lo | ;\: : . S.‘F Ki Kr Sy= all oweble btu&('v} sror number 2 Frow Fig -2 0 Hoo= 330 Bl ‘ 0 ks Sypmen = Spidle 25T kpsi Say 34° 50 kg % Ffik («'ml'i‘ Gt _cq,oqkof (,Ffi_. |‘1~-?,)L,,S,,T= {02 “B +16Y 00 ps c ; ¥ S"’F‘:‘“:S"i‘l’k‘, e !{%}?(330) 4«7”,““)0 ‘ Sp = G0lps 345 M Slw! CY(_,(L FGL"O:" ‘ : Use Fig 14 14 € cydle = (D2 Mﬂi @ gKT: 10 Famune T <25 'F* ,‘ [Slalt MN““me] QS = 0,95 TH D O R= (.90 @

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

mp ’\hnd “1 / /waE Mlm.v)\ (e @ ‘”-rl./"\ a):}}-. 1w (T J’@aé /rHsn’b -13‘15_»‘1 (I ;»Hb‘m«»rad am-ld,\aql\ -—\~ J\ali.a?fi‘u\uo])zi 'Zrmo' My, RS Jtl_ fiu\o /"”AJ/IJPZ) 6()]\,()'6 R L (ﬂa ' wﬂ‘f\'f!) L\‘\V‘JC‘ A‘TJ y)\,v\‘,@ O A T I e | A U "N“? -—,:l\ 2ﬂ ‘\\— ?:)au felyp]")Y) J— -Pmmwa r;'.;p_frb r(1 ruf ‘;‘nau My To [)f‘\ A ‘TI fyf 1‘!5 hl»)/;)%(n T0 M"t)T!l \Qubya\) A4NQ JVanxa Bn]”[\ Il.“ll (7 2571 1aS1) PP (X ST —_— —— ——— - oy 0 ol Yy 1052 % 7\4\{, /;\A’\/\ L, Y B 05N ) ) o ) “”“l”fw‘_‘ﬂ peyn /wal,g e n(wlmg() m e~ — T e —~ (S'LT{'h 1(:HDA) hh = jl§ 1870 Yo' ) r &E':)l hS¢ :Lkv{w oc,;:ps ,.1\§1’ (85-3% ea ) GIEETIG (520 ) o) w},‘\(\, e P e —_— e ~h-2h 1‘-ka) 6"{7::{‘3 £e-gs dbwa) 94 ad 1) Aen0’y§ 2 (20 )0 T) ny .JW Maps = (‘.H’QO(]/(H)j'S | vm\.u\_d ( LT\-‘\") ‘_2 .d Sop L hse) T LARQLIAs wnG oy