mm1dm1 2008-2009 exam with solutions

MM1DM1-E1 Solution 2008

SECTION A

1. (a) What is the main objective of shaft connection in the context of machine element?

The main objective of

power between machine elements.

(b) Outline three factors that should be considered in selecting a method for shaft

hub connection. Any of the three factors below:

- The magnitude and direction of the forces(E.g. torque, radial forces due to belt tension etc. and axial forces due to

reactions from helical gears or a fan).- The need for axial restraint

shaft or not) and/or

a particular place- The need for angular location

never be allowed to rotate on other shaft where a timing action is required)

- The need for (relatively more or less) accurate

and the shaft.

(c) List and briefly describe (using sketchestypes of positive drives for shaft

Pin: A hole may be drilled radially through the hub and shaft and a pin inserted

connecting the two. The pin may be straight, tapered or a 'roll' or 'spring' pin. The load is

by shear in the pin. Pins provide positive

light loads.

Key: Keys make a positive shaft coupling and can be used to transmit loads of all

type. - Plain keys do not provide any axial restraint to the hub. The key may be held

in position either by fitting in a profiled keyway or by set screws.

Solution 2008-2009 Design and Manufacture 1

1

SECTION A – Machine Elements

What is the main objective of shaft connection in the context of machine

The main objective of shafts connection is to transmit between machine elements.

Outline three factors that should be considered in selecting a method for shaft

Any of the three factors below:

magnitude and direction of the forces between the hub and the shaft. . torque, radial forces due to belt tension etc. and axial forces due to

reactions from helical gears or a fan). axial restraint (should the hub be able to move freely along the

shaft or not) and/or axial location (does the hub need to be located axially at

a particular place?) angular location of the hub on the shaft (e

never be allowed to rotate on the hub – e.g. camshaft on a car engine or any other shaft where a timing action is required) The need for (relatively more or less) accurate alignment

and briefly describe (using sketches if necessary) the three of positive drives for shaft-hub connection.

A hole may be drilled radially through the hub and shaft and a pin inserted

. The pin may be straight, tapered or a 'roll' or 'spring' pin. The load is

Pins provide positive axial and rotational restraint but are only used for relatively

Keys make a positive shaft coupling and can be used to transmit loads of all

keys do not provide any axial restraint to the hub. The key may be held


Design and Manufacture 1

What is the main objective of shaft connection in the context of machine [1 marks]

is to transmit rotational motion and

Outline three factors that should be considered in selecting a method for shaft-

[3 marks]

between the hub and the shaft. . torque, radial forces due to belt tension etc. and axial forces due to

(should the hub be able to move freely along the (does the hub need to be located axially at

of the hub on the shaft (e.g. where hub must

g. camshaft on a car engine or any

alignment between the hub

the three most common [6 marks]

A hole may be drilled radially through the hub and shaft and a pin inserted

The pin may be straight, tapered or a 'roll' or 'spring' pin. The load is transmitted

restraint but are only used for relatively

Keys make a positive shaft coupling and can be used to transmit loads of all

keys do not provide any axial restraint to the hub. The key may be held



- Taper keys provide some axial restraint due to friction; the radial force between the shaft and hub allows som

friction between shaft and hub.- A GIB head taper key may be used to assist removal.- Woodruff keys are easy to make and locate but can only be used for light

loads. - Frictional connectors (

wherever possible.

Splines: A spline is a series of axial ribs machined onto a shaft with corresponding

grooves machined into the hub. As the ribs are integral with the shaft they do not weaken the shaft in Also the load is shared between a number of ribs rather than being taken on one

key. Splines are often used for heavy loads.The other main application of splines is where

between the hub and the shaft.

(d) An 80mm diameter shaft is made from a material with

A 20mm square key is to be used, as

a material with yield strength of 300MPa. Assuming the shear yield stress is half the uniaxial yield stress and the factor of safety required is 2. Consider both

shear stresses and bearing stresses on the shaft and key, determine the required length (to the nearest mm) shaft.


2

Taper keys provide some axial restraint due to friction; the radial force between the shaft and hub allows some of the torque to be transmitted by

friction between shaft and hub. taper key may be used to assist removal.

keys are easy to make and locate but can only be used for light

Frictional connectors (e.g. taper locking devices) are preferred to keys

wherever possible.

A spline is a series of axial ribs machined onto a shaft with corresponding

grooves machined into the hub. As the ribs are integral with the shaft they do not weaken the shaft in the same way as a key. Also the load is shared between a number of ribs rather than being taken on one

key. Splines are often used for heavy loads. The other main application of splines is where axial movement

between the hub and the shaft.

n 80mm diameter shaft is made from a material with yieldA 20mm square key is to be used, as shown in figure 1(d). The key is made from

a material with yield strength of 300MPa. Assuming the shear yield stress is half l yield stress and the factor of safety required is 2. Consider both

shear stresses and bearing stresses on the shaft and key, determine the required (to the nearest mm) of the key based on the torque value of the gross

Figure 1(d)


Taper keys provide some axial restraint due to friction; the radial force e of the torque to be transmitted by

keys are easy to make and locate but can only be used for light

taper locking devices) are preferred to keys

A spline is a series of axial ribs machined onto a shaft with corresponding

grooves machined into the hub. As the ribs are integral with the shaft they do

Also the load is shared between a number of ribs rather than being taken on one

axial movement is required

yield strength of 400MPa. shown in figure 1(d). The key is made from

a material with yield strength of 300MPa. Assuming the shear yield stress is half l yield stress and the factor of safety required is 2. Consider both

shear stresses and bearing stresses on the shaft and key, determine the required of the key based on the torque value of the gross

[15 marks]

MM1DM1-E1 Solution 2008-2009 Design and Manufacture 1

3

Given: Factor of safety, Fs = 2 Shear yield stress = ½ × uniaxial yield stress

Shaft:

• σy,s = 400 MPa

• D = 80 mm

Key:

• σy,k = 300 MPa

• Height = Width = 20 mm For the shaft:

Working normal stress for the shaft based on the factor of safety of 2

MPa 200

2

MPa 400,

==sw

σ

Working shear stress for the shaft is thus

2

since MPa 1002

MPa 200,

y

ysw

σττ ===

[1 marks]

For the key: Working normal stress for the key based on the factor of safety of 2

MPa 1502

MPa 030, ==kwσ

Working shear stress for the key is thus

2

sinceagain MPa 752

MPa 150,

y

ykw

σττ ===

[1 marks]

From the equation of torsion, rJ

T τ= ⇒

r

JT

⋅=

τ

For a circular solid shaft, the polar second moment of area,

32

4dJ

⋅=

π

⇒ ( ) ( ) 46

664

10021238597.432

106796351.128

32

1096.40

32

80mmJ ×=

×=

×⋅=

⋅=

ππ

[2 marks]

Hence the maximum torque that the shaft can carry, based on the gross shaft section is

( )( )mNmmN

mm

mmmm

N

r

JT

⋅=⋅=×

=

⋅=

05309649.1049.1005309640

10021238597.4100 462

τ

[2 marks]


4

From torque equation, FrT ⋅= The force at the shaft surface (which is the shear force acting on the key) is

Nmm

mmN

r

TF 4123.251327

40

49.10053096=

⋅==

[2 marks] Hence the required length of the key (based on the bearing stress on the

shaft) can be calculated from

( ) ( )swkeykey

swHeight

Fl

lHeight

F

,

,

22σ

σ

=⇒

=

⇒

( ) mm

mmNmm

Nl 6637062.125

2002

20

4123.251327

2

=

=

Note: This is essentially the compressive stress acting on the shaft surface due to the contact with the key.

[2 marks] The required length of the key (based on the bearing stress on the key) can

be calculated from

( ) ( )kwkeykey

kwHeight

Fl

lHeight

F

,

,

22σ

σ

=⇒

=

⇒

( ) mm

mmNmm

Nl 5516082.167

1502

20

4123.251327

2

=

=

Note: This is essentially the compressive stress acting on the key surface due to the contact with the shaft.

[2 marks]

The required length of the key (based on the shear stress on the key) can be calculated from

( )( ) ( )( )kwkeykey

kwwidth

Fl

lwidth

F

,

, ττ =⇒=

⇒ ( )( ) mm

mmNmm

Nl 5516082.167

7520

4123.251327

2

==

[2 marks]

Therefore, the required length of the key is 168 mm. [1 marks]


5

2. (a) List and describe (using sketches if necessary) four different types of lubrication used in plain bearings. [8 marks]

Any four from the following:

- Boundary lubrication occurs when the film of lubricant is not thick enough to

prevent contact between the surfaces. The lubricant carries only part of the load and helps to reduce the coefficient of friction and wear. This type of

lubrication is used where the loads are not heavy and speeds not too high. Examples are small electric motors, slowly rotating wheels (children's bicycles), and linkages in car steering mechanism.

- Hydrodynamic lubrication occurs where the two surfaces are moving fast enough for a wedge of lubricant to form between the surfaces holding them

apart so that there is no contact and low friction. The pressure within the fluid film is generated solely by the relative motion of the surfaces. This type of lubrication is often used for heavy loads operating continuously at high

speeds, e.g. crankshaft and camshaft bearings on a car engine, steam turbine shafts (wear is minimised and shock is taken by the lubricant film so there is

no damage to the bearing). Boundary lubrication may occur during start up. - In hydrostatic lubrication a film of lubricant also holds the surfaces apart but

the pressure is generated by some external means (e.g. a pump). It is applicable for low speed applications where the speed is insufficient to generate a hydrodynamic film. Examples include machine tools, measuring

instruments, any large load which needs precise positioning. - Dry lubrication in the form of a graphite powder or other dry lubricant is

sometimes used. It has the advantage of being easily retained in the bearing but as coefficients of friction are higher than for fluid lubrication it is restricted to low load/speed applications.

- No lubricant may be used in certain light duty applications where plastic materials are used on one of the bearing surfaces. Examples include domestic

appliances. (b) List and describe (using sketches if necessary) three different types of frictional

belt drives and their application. [6 marks]

Vee belts: These are the most common form of belt. - vee shape enhances friction - several belts side by side for heavy loads

- flexing of belt around pulley and wedging of belt into vee generates heat - hence losses - low efficiency.

- Poly-vee and cogged vee belts have higher efficiency. Flat belts: These were once very common and made from leather or woven

cotton. They are now making a comeback made from reinforced polymeric materials.

- quieter than vee belts - higher efficiency than vee belts ( up to 98%) - thinner cross section is more

flexible hence lower losses.

- barrelled pulleys prevent belt slipping off - require higher belt tension than vee belts


Linked belts: These are made of small sections of vee belt joined together by pins, which enable the belt to be split.

to be dismantled to fit a continuous belt. (c) Describe (in a few sentences) about chain drives and its advantages over

belt drive system.

- Chain drives are most commonly used in low speed high torque applications, especially where space is limited. The most common type of chain is chain (as on a bicycle).

- For high load applications two or more rollers can be placed side by side. - An alternative chain form is the

chain and may be run at higher speed. - Chain drives have the advantage of having longer lives than belts but they

need good lubrication.

(d) There are two general types of the functionality for each of

Rigid Couplings

- Rigid couplings are designed to connect two shafts together so that no

relative motion occurs between them.- Rigid couplings are suitable

- If significant radial or axial misalignment occurs, high stresses may result which can lead to early failure.

Flexible Couplings

- Flexible couplings are designed to transmit torqaxial, radial and angular misalignment.

- There are two sorts of flexibility couplings: o Kinematic flexibility that allows for misalignment but does not dampen

out shock, and

o Resilient flexibility that can absorb shock and vibratiofor misalignment.

- Generally flexible couplings are able to tolerate up to misalignment and up to 0.75mm parallel misalignment depending on their design.

(e) Name two main objectives of applying pre

- Pre-tensioning is essential to prevent the fastened components from moving

apart when under load.- Pre-tensioning reduces the magnitude of stress fluctuations in the bolt when

the loads on the component fluctuate and e.g. in the cylinder head bolts of an internal combustion engine.


6

These are made of small sections of vee belt joined together by pins, which enable the belt to be split. They are of use where a shaft would have

to be dismantled to fit a continuous belt.

Describe (in a few sentences) about chain drives and its advantages over

Chain drives are most commonly used in low speed high torque applications, especially where space is limited. The most common type of chain is

(as on a bicycle).

pplications two or more rollers can be placed side by side. lternative chain form is the inverted tooth. This is quieter than the roller

chain and may be run at higher speed. Chain drives have the advantage of having longer lives than belts but they need good lubrication.

There are two general types of shaft couplings, namely rigid and flexible.the functionality for each of these shaft couplings techniques

Rigid couplings are designed to connect two shafts together so that no

relative motion occurs between them. Rigid couplings are suitable when precise alignment of two shafts is required.

If significant radial or axial misalignment occurs, high stresses may result which can lead to early failure.

Flexible couplings are designed to transmit torque, whilst permitting some axial, radial and angular misalignment.

There are two sorts of flexibility couplings: Kinematic flexibility that allows for misalignment but does not dampen out shock, and

Resilient flexibility that can absorb shock and vibratiofor misalignment.

Generally flexible couplings are able to tolerate up to misalignment and up to 0.75mm parallel misalignment depending on their

Name two main objectives of applying pre-tensioning to fastened components.

tensioning is essential to prevent the fastened components from moving

apart when under load. tensioning reduces the magnitude of stress fluctuations in the bolt when

the loads on the component fluctuate and this reduces the effects of failure, e.g. in the cylinder head bolts of an internal combustion engine.


These are made of small sections of vee belt joined together by They are of use where a shaft would have

Describe (in a few sentences) about chain drives and its advantages over

[4 mark]

Chain drives are most commonly used in low speed high torque applications, especially where space is limited. The most common type of chain is roller

pplications two or more rollers can be placed side by side. . This is quieter than the roller

Chain drives have the advantage of having longer lives than belts but they

, namely rigid and flexible. Outline these shaft couplings techniques. [4 marks]

Rigid couplings are designed to connect two shafts together so that no

when precise alignment of two shafts is required.

If significant radial or axial misalignment occurs, high stresses may result

ue, whilst permitting some

Kinematic flexibility that allows for misalignment but does not dampen

Resilient flexibility that can absorb shock and vibration as well as allow

Generally flexible couplings are able to tolerate up to ±3° of angular misalignment and up to 0.75mm parallel misalignment depending on their

fastened components.

[3 marks]

tensioning is essential to prevent the fastened components from moving

tensioning reduces the magnitude of stress fluctuations in the bolt when

this reduces the effects of failure, e.g. in the cylinder head bolts of an internal combustion engine.


3. (a) Explain the formation of builtthese chips during machining process.

a) Layers of material from the workpiece that are gradually deposited on the tool. Adversely affect surface finish in cutting. BUE hardness increases

significantly (work hardening or material layer deposition)undesirable. But a thin, stable BUE can reduce the tool wear.increases the size of the BUE decreases (or may not form at all)

BUE: Decrease the depth of cut; increase the rake angle; using sharp tool and effective cutting fluid.

(b) Why does the temperature in cutting depend on the cutting speed, feed and

depth of cut? Explain in terms of the relevant process variables.

When cutting speed increases, time required to dissipate heat reduces, therefore increase the temperature at the cutting zone. When feed rate is increased, a large amount of metal removed is subjected to higher temperature and plastic

deformation because of an increase in cutting force and normal force which result in temperature. When depth of c

and a large cutting force is required. This large force will cause dissipation of high heat and increase the temperature.

(c) For a turning operation using a ceramic cutting tool, if the speed is increased by 50%, by what factor must the feed rate be modified to obtain a constant

tool life? Use n = 0.5 and y = 0.6. Explain how thtool life.

nx

nn fdVCT ⋅⋅⋅=−−11

n = 0.5 y = 0.6

21

2.1

2

2.1

1

2.1

15.02.0

1

2

93.0

084.1

ff

ff

fdVCx

=

=

⋅⋅⋅−−

−−−

Feed rate reduce by 7%


7

SECTION B – Machining

Explain the formation of built-up edge (BUE) chips. Identify an advantage of these chips during machining process.


significantly (work hardening or material layer deposition)a thin, stable BUE can reduce the tool wear.

increases the size of the BUE decreases (or may not form at all)

BUE: Decrease the depth of cut; increase the rake angle; using sharp tool and effective cutting fluid.

does the temperature in cutting depend on the cutting speed, feed and

depth of cut? Explain in terms of the relevant process variables.

When cutting speed increases, time required to dissipate heat reduces, therefore e temperature at the cutting zone. When feed rate is increased, a

large amount of metal removed is subjected to higher temperature and plastic

deformation because of an increase in cutting force and normal force which result in temperature. When depth of cut increases, more material is removed

and a large cutting force is required. This large force will cause dissipation of high heat and increase the temperature.

For a turning operation using a ceramic cutting tool, if the speed is increased by 50%, by what factor must the feed rate be modified to obtain a constant

tool life? Use n = 0.5 and y = 0.6. Explain how this modification will maintain

ny

f

( ) 2.1

25.02.0

1

225.1 fdVC

x

⋅⋅⋅= −−−

Feed rate reduce by 7%


up edge (BUE) chips. Identify an advantage of [6 marks]


significantly (work hardening or material layer deposition) BUE is generally a thin, stable BUE can reduce the tool wear. Cutting speed

increases the size of the BUE decreases (or may not form at all) To reduce the

BUE: Decrease the depth of cut; increase the rake angle; using sharp tool and

does the temperature in cutting depend on the cutting speed, feed and

depth of cut? Explain in terms of the relevant process variables. [9 marks]

When cutting speed increases, time required to dissipate heat reduces, therefore e temperature at the cutting zone. When feed rate is increased, a

large amount of metal removed is subjected to higher temperature and plastic

deformation because of an increase in cutting force and normal force which ut increases, more material is removed

and a large cutting force is required. This large force will cause dissipation of

For a turning operation using a ceramic cutting tool, if the speed is increased by 50%, by what factor must the feed rate be modified to obtain a constant

is modification will maintain [10 marks]


8

4. (a) Explain the reason for the development of cutting inserts. [5 marks]

Cutting inserts were developed to make changing of cutting tool to be fast. They also avoid the entire cutting tool to be removed from the machine for grinding which would therefore require resetting. Cutting inserts also made the

manufacturing of the cutting tool to be cheaper because only the cutting tip has to make out of carbide while the shank can be made of HSS.

(b) Thermal conductivity and coefficient of thermal expansion are properties that can

be important to the life of a cutting tool. Explain the reasons. [10 marks]

Thermal conductivity is the ability of the heat to flow. Coefficient of thermal

expansion is the dimensional response to temperature change. Tool wear occurs due to the high temperature produced during cutting. A high thermal conductivity will make sure heat flows and dissipate faster from the cutting zone.

Coefficient of thermal expansion will also reduce thermal fatigue and thermal cracks due to expansion and compression of tool materials due to high heat.

(c) A machining operation is being carried out with a cutting fluid that is an effective

lubricant. Explain the changes in the mechanics of the cutting operation if the fluid is shut off. [10 marks]

Friction at tool-chip interface will increase Chip will become thicker

Built-up edges likely to occur Temperature in cutting zone increase Surface finish deteriorate

Dimensional tolerance difficult to control


9

SECTION C – Metrology & Tolerancing

5. (a) Explain the specifications that need to be considered when choosing a measuring instrument. [10 marks]

Range Accuracy

Resolution Repeatability

(b) Explain the need for automated inspection and give an example of an industrial automated inspection system. [10 marks]

Automated inspection is required to decrease the inspection time which will

decrease the overall production time. Automated inspection also will reduce

human error during inspection which will cause product quality problem. Automated inspection also will allow quality to be maintained at a tighter

tolerance band. Example of automated inspection is vision system to inspect part geometry and infra-red sensor to detect changes in liquid level in bottle.

(c) Describe how flatness of a surface can be measured. [5 marks]

Flatness of surface can be measured using a stylus in a surface measuring instrument. (Student expands on this).

6. (a) Explain with the aid of neat diagrams what is meant by clearance fit,

interference fit and transition fit. [9 marks]

(b) A shaft with nominal diameter of 50 mm has to be connected to a hole with the same diameter. Calculate the maximum and minimum required to produce:-

(i) interference fit [5 marks]

50H7p6 50H7s6

0042.0

0026.0

0025.0

0000.0 5050 0059.0

0043.0

0025.0

0000.0 5050


10

(ii) transition fit [5 marks]

50H7k6 50H7n6

0018.0

0002.0

0025.0

0000.0 5050 0033.0

0017.0

0025.0

0000.0 5050

(c) Explain the difference between dimensional tolerance and geometrical tolerance and their importance in manufacturing. [6 marks]

Necessity of Dimensional Tolerance It is almost impossible (and sometimes uneconomical) to maintain the strict

degree of accuracy as listed on a plan. To accommodate this, it is normal to display measurements with a plus or minus (+/-) tolerance which allows for

some margin of error. Care needs to be taken however when determining such +/- tolerance, particularly where there are mating parts. For example, a shaft which is machined to its maximum tolerance may not fit a gear center that has

been machined to it minimum tolerance or an unsatisfactory loose fit would result from the shaft being machined to its minimum tolerance with the gear

center machined to its maximum tolerance. Usually, the dimensional tolerance is decided at the design stage and a Machinist

must take care to apply the required dimensional tolerance and to ensure that

discrepancies are not introduced as a result of poor workmanship of measuring techniques.

Necessity of Geometrical Tolerance Geometric tolerances specify the maximum variation that is allowed in form or

position from true geometry. The geometric tolerance is, in essence, the width or diameter of tolerance zone within which a surface or axis of hole or cylinder

can lie which results in resulting feature being acceptable for proper function and interchangeability.

If a tolerance of form is not specified on a drawing for a feature, then the feature

as made will be acceptable regardless of form variation. The tolerances of form control straightness, flatness, parallelism, angular displacement etc. etc.

MM1DM1 End

mm1dm1 2008-2009 exam with solutions

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