mm1dm1 exam 2012-13 with solutions

8/12/2019 MM1DM1 Exam 2012-13 With Solutions

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MM1DM1-E1

MM1DM1-E1

The University of Nottingham

DEPARTMENT OF MECHANICAL, MATERIALS AND MANUFACTURING ENGINEERING

A LEVEL 1 MODULE, SPRING SEMESTER 2012-2013

DESIGN AND MANUFACTURE 1

Time allowed ONE Hour THIRTY Minutes

Candidates may complete the front cover of their answer book and sign their desk card butmust NOT write anything else until the start of the examination period is announced

Answer ONE question from each Section

Only silent, self contained calculators with a Single-Line Display or Dual-Line Display arepermitted in this examination.

Dictionaries are not allowed with one exception. Those whose first language is not Englishmay use a standard translation dictionary to translate between that language and English

provided that neither language is the subject of this examination. Subject specific translationdictionaries are not permitted.

No electronic devices capable of storing and retrieving text, including electronic dictionaries,may be used.

DO NOT turn examination paper over until instructed to do so

ADDITIONAL MATERIAL: None

INFORMATION FOR INVIGILATORS: None

Turn Over


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Section A

1.(a) The following types of lubrication types are available:

Boundary

Dry Hydrodynamic Hydrostatic NoneFor each application below (i-v), choose the most suitable lubrication type.

i) A car wheel[2 marks]

AnswerBoundary

ii) A car crankshaft[2 marks]

Answer

Hydrodynamic

iii) A barrel lock in a door[2 marks]

Answer

Dry

iv) A toddlers bike wheel (A steel axle running in a polymeric bearing)[2 marks]

Answer

None

v) A slowly rotating shaft which experiences a high radial load[2 marks]

AnswerHydrostatic

(b) Explain why a cylindrical roller bearing can support a greater radial load than aball bearing.

[5 marks]

Answer

The cylindrical roller distributes the load across a line contact rather than a pointcontact.

(c) Sketch a deep groove ball bearing mounted on a shaft with a fixed inner raceand a floating outer race.[10 marks]


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Answer

(d) A shaft of20mm running in a plain bearing rotates at 100 RPM in a journal thatis 30mm long. A radial load of 500N is carried by the bearing. Calculate thelimiting PV value for the bearing.

[25 marks]Answer

The Pressure is calculated as

=

= 833,333 (10 marks)The operating Velocity is calculated as

2

2

(10 marks)

The limiting PV is calculated as

(5 marks if units are correct else 2 marks )

(e) A pulley (75 mm) on a crankshaft is used to drive an alternator pulley (30mm) using a vee belt. The distance between the pulley centres is 500 mm. Thecrankshaft rotates at 3000 RPM. The vee belt has a cross sectional area of1.44x10-4 m2 and the maximum permissible strength of the belt is 12 MPa. Thecoefficient of friction between the belt and the pulleys is 0.8.

You may find the following equations useful: Belt length

+) 12

) ( Angle of contact on smaller pulley

Angle of contact on larger pulley

Eytleweins Formula

eF

F

2

1


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i.) Determine the speed of the alternator pulley in RPM.[5 marks]

Answer

=

= 3000

75/2

30/2= 7500

ii.) Determine the maximum power that can be transmitted from the crankshaft tothe alternator assuming no belt slippage

[35 marks]Answer

The tight side load in the belt is limited by the maximum permissible strength of thebelt. = 1012 = 101.44 = 1728

(5 marks)

To transfer the power, no slippage can occur, so Eytleweins formula must be satisfied

eF

F

2

1

Where slippage is likely to occur on the smallest pulley as it has the least contact area,so

= () = 2 sin 2 = 2 sin

0.075 0.030

2 0.500

180

3.052 =(10 marks)

The Eytleweins formula can be solved for the slack side force, F2

= = 1728.. = 150(10 marks)

The belt speed is calculated as

= 2

= 3000 160

21

0.075

2 = 11.781 /

(5 marks)

So the maximum power transmitted is

)= ( =(1728 150) 11.781 = 18,590(5 marks)

iii.) Determine the torque produced at the crankshaft[10 marks]

Answer

)= ( =(1728 150) 0.0752

= 592.(a) If you wanted to prevent axial and rotational movement of a gear on a shaft

would you make the connection with a roll pin or a plain key?


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[2 marks]

Answer

Roll pin

(b) Name 2 advantages and 2 disadvantages of keys for attaching elements to

shafts.[8 marks]

Answer

Advantages1. Inexpensive2. Provide positive driveDisadvantages1. Shafts requires a keyway to be machined2. The keyway gives rise to a stress concentration in the shaft

(2 marks/answer)

(c) Name 5 ways of achieving an interference fit between the hub of a gear and ashaft

[10 marks]

Answer

1. Cool the shaft prior to assembly2. Heat the hub prior to assembly3. Use a combination of 1 and 2 above4. Press the hub onto the shaft5. Expand the hub using high pressure oil

(2 marks/answer)

(d) A hub is secured onto a shaft using an interference fit. Pressure, P, is developedbetween the hub and shaft which keeps the hub from slipping on the shaft whena torque is applied. Show that the expression for the maximum torque capacityof this hub to shaft connection is

=

2

Where = coefficient of friction between shaft and hub

l = hub length in contact with the shaftD = shaft diameter

[30 marks]

Answer

The interface pressure, P, sets up a frictional force, N, between the hub and shaft surfacewhich acts tangentially to the surface and along the contacting length of the shaft, l. Whenthis frictional force is overcome, the hub will slip on the shaft, therefore the maximum torqueat the point of slip can be defined as

=

2 (5 marks)

The normal force, Nacts everywhere normal to the interface surface, A which is related to the


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interface pressure as=

(5 marks)

Where the interface surface is defined as

= 2 2 (5 marks)

So

= = 2 2

(5 marks)

Substituting for the normal force and simplifying

= 2

= 2

(10 marks)

(e) Estimate the fastener pre-load, F, required in the tapered hub-shaft connection shownbelow to achieve a maximum torque capacity, Tmax, of 500 Nm, assuming a frictioncoefficient of 0.2 on all surfaces. An M24 thread is employed. The taper angle is 12,the shaft end diameter, D1 is 30 mm and the hub length is 45 mm. Include the torquecapacity associated with the hexagonal nut having a major thread diameter of 24 mmand a distance across its flats of 36 mm.

(f)

Recall that the maximum torque carrying capacity for the tapered region only isexpressed as

= 2

+ Where = coefficient of friction between the shaft and hub as well as between the nut

and hubl = hub lengthDt= mean shaft diameter along the taper

[50 marks]

Answer

As stated, the maximum torque capacity generated in the tapered region is given by the

=12

45mm

D1D2


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expression

= 2

+ The additional torque capacity resulting from the nut is

= 2

(15 marks)

The total torque capacity is

=+ = 2

+ +(10 marks)

Solving for the pre-load force

= 2 + + (5 marks)

Where

= + 2

(5 marks)And = 2 (0.045 12) + = 2 (0.045 12) + 0.030 = 0.049

=0.030

+ 0.049

2 = 0.040 (5 marks)Also the average of the major nut thread diameter and the distance across the flats is

= 0.024 + 0.0362

= 0.030(5 marks)

Finally

=2

+ + =2 500

0.2 0.0400.2 12 + 12 + 0.030 = 38,723(5 marks)


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Section B

3.(a) Using a detailed figure in each case, demonstrate how 3 forms of milling tool

wear occur. Your answer should also include some commentary on the defectsto the machined surface which may result from this.

[30 marks]

Answer

Any three figures and discussion with three comments relating tosurface properties of the workpiece.

(b) Modern machine tools which are used to mill complex geometries are


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capable of 5-axis machining. Using a fully labelled diagram discuss theprincipal features of such a machine tool.

[30 marks]

Answer

Discussion of principle features (6 features for 5 marks each).Candidates may also choose to discuss trunion type arrangements andhorizontal machine tools

(c) Discuss the typical stages involved in developing a CNC program for a machine toolusing generic CAD/CAM software. Your answer should go on to make mention of thestages involved in setting up a job on a modern machining centre.

[40 marks]

Answer Import of CAD data Perhaps some mention of CAD formats and method of importing into the CAD package Placing the part inside a stock block Selection of tool type and profile Defining cutting conditions (feed and spindle speeds) Post processing to translate to the machine dialect G-Code created in readable format Typical commands may include: spindle ON (M05), Spindle speed (S4000) for 4000rpm

or M00 for all stop. Accept any sensible suggestion with meaning. Set up of job (may include figure with datum acquisition, fixturing and tooling).

4.(a) With the aid of a diagram discuss briefly the process of laser cutting. Your

answer should include some comment about the interaction of the laser with theworkpiece.


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[30marks]

Answer

Answer should include some commentary on the conversion of incident radiation toheat within the substrate, formation of dross and evaporation.

(b) With the aid of a labelled diagram explain the process of electrochemicalmachining. You should also discuss a typical application for this process in your

answer.[30 marks]

Answer

Answers should give a sensible application for this process, most likely finishing ofaero/biomedical components.

(c) The purchase of any machine tool and its incorporation into a manufacturing process

presents a significant challenge to manufacturing companies. Discuss, with the aid ofdiagrams the typical stages and considerations involved when selecting andcommissioning a new machine tool.

[40 marks]


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Answer

Some discussion around key points such as (but not limited to):

CostCapability/performance

QualityEnvironmentalAvailabilityFlexibility

The outlay associated with machine tools is significant (10K- 1000sK)

Good decisions must be made, SME large multinational. To expand or replace legacyequipment

Development of business plan/case for investment

Technical justification

Consideration of product life time.

Scientific metrics must often be devised and applied to facilitate these decisions.


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Section C

5.(a) Describe the process of calibration.

[40 marks]Answer

[8 marks for each of the steps described below]

1. A second instrument or standard of known accuracy is required for calibration.2. The value of the quantity being measured is varied over the range of interest.3. All other parameters (e.g. temperature, pressure, noise) kept constant as far as possible.4. Readings of both instruments are recorded

a. X - output of the known instrument/standardb. M - output of the instrument being calibrated.

5. These are plotted on a Calibration Graph as follows, a regression analysis is performed tocreate an equation that give X in terms of M

(b) The Vernier callipers listed in Table 1 below are available to measure a steel shaft ofdiameter 65 mm 0.2 mm, to ensure the manufacturing process is well controlled themeasurements must be accurate to 10% of the tolerance for the shaft.

i) Select the most appropriate Vernier calliper from those available (A to D) as shown inTable 1.

[30 marks]

ii) Of the four Vernier callipers state which one appears to be making an unreasonableclaim with respect to accuracy.

[10 marks]

Table 1

AnswerFor part (i) there are three steps required as follows

[15 marks] step 1


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From the table first calculate the absolute accuracy of each Vernier:

Absolute accuracy = ( % of range accuracy / 100 ) * Range

[10 marks] step 2

Then calculate 10% of the tolerance for the shaft = 0.2 / 10 = 0.02

[5 marks] step 3

Based on this only Verniers C and D are candidates but Vernier C has a range that is toosmall.

Therefore select Vernier D.

For part (ii) [10 marks]

The absolute accuracy calculated in part (i) for each sensor must be compared to the statedresolution. As the resolution for Vernier A is greater than the stated accuracy this brings intoquestion the accuracy of the instrument as it would not be possible to resolve measurementsto the stated accuracy.

(c) Based on Taylors Gauge Theory state the difference between a Go Gauge and aNot-Go Gauge, then describe the benefits of using Go and Not-Go gauges ascompared to taking instrument measurements.

[20 marks]

Answer[5 marks for each of the following] The GO gauge checks the maximum material condition and should check as many

dimensions as possible. The NOT-GO gauge checks the minimum material condition and should check one

dimension.

[5 marks for each of the following statements about the benefits of using Go and Not-Gogauges] Limit gauging is faster than direct measurement for giving conformance data Skill requirement and opportunity for human error is minimised

6.(a) Describe the meaning of each of the five symbols that may be used on an engineering

drawing as part of the geometric tolerance definition for orientation and location. Youmay support your description with sketches to illustrate the description of each symbol.

[40 marks]


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Answer

[4 marks each] for correctly naming the symbol

[4 marks each] for the description which may include an associated sketch as shown below.

(b) Describe the working principle of a stylus based surface texture instrument used tomeasure surface form and roughness; you may use a diagram to support your answer.

[30 marks]

Answer

A stylus based surface measurement instrument consists of:

[5 marks for each bullet points below, that are included as a description of the stylus

instrument] a skid that makes contact with the specimen to be measured protruding from the skid is a pointed stylus that rests on the surface of the specimen

a.

b.

c.

d.

e.


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It is connected to a pivot point and is allowed to move up and down in response to thelocal height of the specimen.

The vertical position is measured using a position transducer which converts displacementinto a voltage signal

The voltage is converted to a digital signal using and A/D converter The digital signal is passed to a computer for analysis and display

(c) A cylindrical shaft must be manufactured with a diameter of 8.5mm 0.03mm. Large

volumes of components are to be manufactured so for speed it is to be inspected usingGO and NOT-GO gauges.

i) Calculate the ideal gauge size required[10marks]

ii) Calculate the maximum and minimum gauge sizes if the wear limit of the gauge is0.002 mm, and the gauges can be manufactured with a tolerance of 0.005 mm

[20 marks]

Answer

i) Ideal size for GO gauge : shaft size + tolerance = 8.5 + 0.03 = 8.53 [5 marks]Ideal size for NOT-GO gauge: shaft size - tolerance = 8.5 - 0.03 = 8.47

[5marks]

ii) GO maximum: Ideal gauge size gauge wear limit = 8.53 0.002 = 8.528 mm

[5 marks]

GO minimum: Ideal gauge size gauge wear limit gauge tolerance = 8.53 0.002 0.005 = 8.523

[5 marks]

NOT-GO minimum: is the ideal not-go minimum size = 8.47 mm[5 marks]


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NOT-GO maximum: ideal minimum size + the gauge tolerance = 8.47 + 0.005 = 8.475mm

[5 marks]

END

mm1dm1 exam 2012-13 with solutions

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