7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

1/9

CONCORDIA UNIVERSITYFACULTY OF ENGINEERING AND COMPUTER SCIENCEDEPARTMENT OF MECHANICAL AND INDUSTRIAL ENGINEERING

MIDTERM EXAMINATION MECH 31l-MANUFACTURING PROCESSESFALL 2009

Date: October 15 ,2009 Duration: I hourMaterials allowed: NO Calculators: YESClass instructors: sections T and X: Dr. M. Packirisamy and Mr. A.ChangiziInstructions: Answer the multiple choice type questions on the first page of the first answersheet by darkening the correct answer on the corresponding line indicated by the number ofthequestion. Select the most appropriate answer. For an example, if a question would ask: "Whatis the benefit of hardening the surface of the tool?" and it would offer option an swers as: a)Higher cutting speed b) Extended life of the tool, c) Reduce thermal effect on the tool or d)Lower lifespan cost of the cutting tool. You should select b) as your answer because d) is aconsequence of the extended life ofthe tool.The objective answer questions will be solved in your answer book and only the finalrequested answers will be written in the separate answer tables. No answer in the answer boxwill receive zero marks. Make sure that your name is written on the exam book and the answersheets. You can do the calculation in your exam book itself.At the end of the exam, gather the question book and the answer sheets in the answer book andsubmit them all. Please, do not forget to sign out.

Student's name: ............................... .Student's I.D. #: . . . . . . . . . . . . . . .

Student's Signature: . . . . . . . . . . . . . . . . .MARKS' Total 40

I 1 I I I

I

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

2/9

Select the MOST appropriate answer. (Total Marks 20)Ql. What are the two most significant parameters that influence the tool life as perTaylor's model

A: I. The material of the tool and the material ofthe workQ) The material of the tool and the cutting speed3. The material of the tool and the cutting feed4. The material of the work and the chip thickness

Q2. During metal cutting, a significant amount of heat is generated. The heat isdistributed among the chip, tool, work and the cutting fluid. What is the type ofregime at which the cutting is recommended?A: I. Slow speeds to enable heat removal2. Optimal speed, such that the heat is optimally distributed

(DHigh speed, for better distribution of heat among the four bodies4. The optimization of heat is carried out through the optimal feed rate.

Q3) The effectiveness of the cutting expressed through the shear angle can be adjustedthrough a judicious sharpening of the cutting tool. Which geometric feature of the cuttingtools would you recommend to improve the efficiency of the cutting, according to thetheory of the orthogonal cutting?A: I. Relief angle

2. Tool wedge angle3. Tool width@l Back rake angle

Q4) A cutting edge of the following machining tool has varying rake angle:I. Shaping2. Planing3. Boring(9 Drilling

2

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

3/9

Q5) What is the most appropriate sequence of hot hardness from the best to the worst interms of their cutting speed.a) Ceramicsb) Diamondc) HSSd) Carbon and Low/medium alloy steelsA:(D b-a-c-d2. c-a-b-d

3. d-b-a-c4. d-c-a-b

Q6) In a finishing orthogonal metal cutting operation with the cutting speed of 20m/minand 0 back rake angle, what would'be the chip velocity if the material shears at 45?

A : d) 20mlmin2. 10mlmin3. 28 .28m1min ~ ~ . \" Vl ~ If J4. 14.142 mlmin V 1: '2A:> 'IIJ&II:'"

Q7) Assume you are production in-charge to establish a manufacturing process formaking a part with a tolerance range of O.lmm. What will be the maximum standarddeviation you would aim in order to qualify the process for production?

A: 1. 0.1252mm(J) 0.01252mm3. 0.0250mm4. O.IOOOmm

Q8) In a finishing orthogonal metal cutting operation with the cutting speed of 100m/minand 0 back rake angle, what would be the chip thickness if the material shears at 45and the uncut chip thickness is 2mm? 1>:: K

sin ,/, r, cosa 2cosa r l : ; 0 /- - - ' - - - - , tan I" = , & = ) " t ~cos(-a) I - r , sina I+sina ~ v - i : { ~ ~c.( . ~ ~ 'c'-~

t(Note: r =- =, t,A: 0 2mm

2. Imm3. 1.414mm4. 2.828mm

3

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

4/9

Q9) Identify a method that will be used for surface crack detection.A: ( [ ) FPI

2. Eddy current3. X-Ray4. Ultrasonic

QI0) Identify the critical parameter that influences the thrust force in drilling.A: 1. Helical angle

2. Point angle0':l Chisel edgeV Lip . 'Qll) Which is the right sequence for machining a hole in order to get very closepositional tolerance and size tolerance:,

A : ~ Centering drill-drilling ,reamingCentering drill-drilling,boring. Drilling-centering drill, boring4. Centering drill-drilling-boring-reaming

Q12) Identify the angle maked below for the cutting tool .

\Ii\

A: 1. Back rake angle

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

5/9

Q13) Identify the type of machining process depicted in the following figure.I. Deep Hole GrindingG> Ejector Drilling3. Lapping4. Forging

Q14) Assume that you have performed a process capability (PC) study and ended up with aprocess capability index C = 1.60 and a mismatch value D which is close to zero. Find thestandard deviation of the process if the upper and lower specification limits are 40.5 mm and40.05 mm, respectively. 1'-0 ') _"-'oJV S L ~ L S L : : - T r

12.(!>4.

0.0938mm0.0234mm0.0469mm0.1876mm

~ (). 't-S-tt-\ oO, ~ ~ - ; : ;.6')(h

QlS) A 4" diameter rod is cut by turning at a speed around 60 m/min. The approximate NRPM of the spindle ofthe lathe must be: , 1\0 J S"1I' of.V... 1- 00l-

@189rpm 2) 300 rpm 3) 412 rpm 4) 590 rpm ~ O ::. ~ 'f. 4-Q16) Which production process has less roughness?

I. Drilling(p Lapping3. Girding4. Sawing

Q17) For which case we cannot use MPI1. Detecting surface cracks2. Detecting sub-surface cracks(j) Detecting crack in aluminum4. Cracks perpendicular to flaws direction

Q18) How to make a square hole?I. Drillingq> BroachingMilling4. Shaping

~ ~

5

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

6/9

Q19) What are the name of these reamers and their application

1. Expansion reamers, to fit the hole size after regrindingo Chucking reamers, for soft materials3. Adjustable reamers, to fit the hole size after regrinding4. Taper reamer, for finishing hole to produce exact taper

Q20) The identified part of the drill is called

CD Web2. Flute3. Shank4. Point

6

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

7/9

PART B: TECHNOLOGY DESIGN PROBLEM

I. ) Consider the following component to be machined in a slab milling machine. Themachining has to be done only on the three surfaces marked. Other surfaces are left as cast.Assume that an allowance of Imm on the machined surfaces to be removed in single passon each side. The slab milling cutter of 70mm diameter and 160mm long with 8 teeth isused. The maximum allowed cutting speed is 20m/min. The maximum feed per tooth isO.lmm. The specific power, mechanical efficiency and correction factor can be assumed as1.3 Ws/mm" 75 % and 1.15. Neglect the cutting allowance. You can neglect themachining stock for calculation. (20 Marks)

Find out the following:a) The maximum rpm ofthe cutterb) The maximum possible width ofthe cut at any time, in mmc) The maximum feed per minute in mmlmind) The total cutting time to machine the three faces in mine) The maximum possible MRR in mm3/mint) The actual MRR with no allowance in mm 3/ming) The maximum motor power required in Wattsh) The total machining cost( Assume $2751hr) in $

P specijicpower *MRR *CFower = = - ~ ' - - - ' ' - - - - - - - - -77

. - - - - - - , - - - - - , - - ~ -

2 marks2 marks2 marks3 marks3 marks3 marks3 marks2 marks

100mm

1 0 0 ~ : : L : L-__________ ' - - ] 5 150mmI. 200mm I

7

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

8/9

#- ']f) = -=to M MtZ.- -::. I6 0 /'Y\ N'I .71:::

7/22/2019 Mech311 Midterm 2009 ,2011 and 2012

9/9

\rr/:. : ~ x 'ole 'V + I x. be tt:o"", \J,-f . tv-(:::::: c::l-[ ~ O C X 1 0 0 + i 2oo 'SOJX1 + 200 . / 00 - 1= dl. [ 2 ~ o o o + 5;oooJ + 2ooCC'

:=. ";fo000 '(Y\ IV\S

7Co..""",'" M ~ 1 1 - '"' iii ';Is)div,'o\e ~ 60

= 2'1- ~ X g'2S- :::: 6p