Examples in Mechanics of Solids

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Real Life

Examples in

Solids

Lesson plans and solutions

Eann A Patterson, Edito



Real Life Examples in Mechanics of Solids




First edition produced September 2006

Second Edition, 2008Third [Electronic] Edition, 2011



Copyright © 2006, 2008, 2011 Eann A Patterson (editor)

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that any content on such websites is, or will remain, accurate or appropriate.

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This edition is distributed free of charge by the ENGAGE project(www.EngageEngineering.org), which is supported by the National Science Foundationunder Grant No. 0833076.



Real LifeExamples in

Mechanics of Solids


Suggested exemplars within lesson plans for Sophomore Solids Courses. Prepared as part

of the NSF-supported project (#0431756) entitled: “Enhancing Diversity in the

Undergraduate Mechanical Engineering Population through Curriculum Change”

Eann A Patterson, Editor

The University of Liverpool, England

[email protected]



This work was developed during the NSF-supported project (#0431756) entitled:“Enhancing Diversity in the Undergraduate Mechanical Engineering Population throughCurriculum Change”. Any opinions, findings, and conclusions or recommendationsexpressed in this material are those of the authors and do not necessarily reflect the viewsof the National Science Foundation. The following were members of the project andcontributed to the many discussions which led to this production of this volume:

Ilene Busch-Vishniac Johns Hopkins University (Project leader )Suzanne Gage Brainard University of WashingtonPatricia B. Campbell Campbell-Kibler Associates IncConstantin Chassapis Stevens Institute of Technology

Glenn Ellis Smith CollegeAshley Emery University of WashingtonDarrell Guillaume California State University, Los AngelesJeff Jarosz Johns Hopkins UniversitySusan Staffin Metz Stevens Institute of TechnologyEann Patterson Michigan State UniversityPradosh K. Ray Tuskegee UniversityHelen Ryaciotaki-Boussalis California State University, Los AngelesKuei-Wu Tsai California State University, Los AngelesHorace Whitworth Howard University




3

CONTENTS

page no.

Introduction and acknowledgements 5

ELEMENTARY STRESS SYSTEMS

1. Stress and strain in unixial solid and hollow bars 6

iPod, suspension bridge, femur, cello

2. Displacement plus deformation in control cables 9

Derailleur gears, yacht rudder

3. Stress in pressure vessel wall 11

Bike pump

STATICALLY INDETERMINATE PROBLEMS

4. Combined use of principles of compatibility and equilibrium 13

iPod, dinosaur display

TORSION

5. Stress and strain due to applied torque 17

Bottle closures

STRAIN ENERGY

6. Conservation of Energy/Energy Methods 19

Slingshot, bungee jumper

7. Helical springs 21

Bicycle suspension, pogo stick

BEAM BENDING

8. Bending moments and shear stress diagrams 24

Skateboarder, unicyclist on plank

METHOD OF SUPERPOSITION

9. Eccentric loading 29

Basketball goal

10. Thermal stress and statically indeterminate loads 31

Rail tracks, jewellery pendant

TWO-DIMENSIONAL STRESS SYSTEMS

11. Mohr’s circle of stress 34

Sausages

12. Combined bending and torsion 36

Wind-up clock, motor




5

INTRODUCTION

These real life examples and supporting material are designed to enhance the teaching ofa sophomore course in mechanics of solids, increase the accessibility of the principlesand raise the appeal of the subject to students from a diverse background. The exampleshave been embedded in skeletal lesson plans using the principle of the 5Es: Engage,Explore, Explain, Elaborate and Evaluate. Such lesson plans were developed by theBiological Sciences Curriculum Study1 in the 1980s from work by Atkin and Karplus2 in1962. Today they are considered to form part of the constructivist learning theory and anumber of websites provide easy to follow explanations of them3.

This booklet is intended to be used by instructors and is written in a style that addressesthe instructor, however this is not intended to exclude students who should find theexamples interesting, stimulating and hopefully illuminating, particularly when their

instructor is not utilizing them. In the interest of brevity and clarity of presentation,standard derivations and definitions are not included since these are readily available intextbooks which this booklet is not intended to replace but rather to supplement.Similarly, it is anticipated that these lessons plans can be used to generate lectures/lessonsthat supplement those covering the fundamentals of each topic.

Acknowledgements

Many of these examples have arisen through lively discussion in the consortiumsupported by the NSF grant (#0431756) on “Enhancing Diversity in the UndergraduateMechanical Engineering Population through Curriculum Change” and the input of thesecolleagues is cheerfully acknowledged as is the support of NSF. The influence of theeditor’s mentors and peers at the University of Sheffield is substantial and is gratefullyacknowledged since many of the ideas for these examples originate from tutorialquestions developed and used in the Department of Mechanical Engineering in Sheffieldover many years.

1 http://www.bscs.org/library/BSCS_5E_Instructional_Approach_July_06.pdf2 Atkin, J. M. and Karplus, R. (1962). Discovery of invention? Science Teacher 29(5): 45.3 e.g. http://www.science.org.au/primaryconnections/constructivist.htm




6


1. Principle: Stress and strain in uni-axial solid and hollow bars

Engage:

Take your iPod into class and dangle it by the earphonecable. Cut open the cable on an old set of earphones toexpose cable and insulation.

Explore:

Pass around class lengths of copper wire and lengths ofempty hollow insulation and invite students to stretch them.

Discuss relative extensions and stiffness. Someone will probably snap one so talk about ultimate tensile stress. Besure have to enough lengths that every student has at leastone to play with while you are talking.

Explain:

Work through the example below:

An iPod, with a mass of 30 grams is dangled from its earplugcable.

a)

Assuming that the copper wire of diameter 0.40mminside the cable carries the entire load, evaluate thestress in the wire due to the weight of the iPod.

b) If the wire in (a) is 1.50m long, by how much will itstretch?

c) Assuming that the plastic insulation, which fits snuglyover the wire and has an external diameter of 1mm,carries the entire load, evaluate the stress in theinsulation due to the weight of iPoD.

d) If the insulation is made from uPVC calculate the

extension of the insulation in the circumstancesdescribed in (c).

Solution:

a) Stress,

41040.0

81.91030

4 23

3

2

d

mg

A

F 2.34×10

6 N/m

2

where F , is force applied, A is cross-section area, m is mass, g is gravitationalacceleration, and d is the diameter of the cross-section.




7

b) Extension,

41040.010110

50.181.91030

4 239

3

2

d E

mgL

AE

FL

E

L L 31.9×10-6m

where L is the length of the wire and E is the Young’s modulus of copperobtained from a data book .

c) Stress,

4004.0001.0

81.91030

4 22

3

22

io d d

mg

A

F 0.29N/m

2

where d o and d i are the outer and inner diameters of the insulation.

d)

Extension,

4104.01102

5.181.91030

4 6229

3

22

io d d E

mgL

AE

FL

=3.35×10-4m

Elaborate:

In practice the load will be borne by the wire and insulation together, discuss how thiswill influence the extension of both of them. The copper and plastic are bound together

and must extend by the same amount, i.e. wire = insulation. Consequently the wire willextend less and the insulation extend more; causing more tension in the insulation thancalculated and less tension in the wire.

Evaluate:

Invite the students to undertake the following examples

Example 1.1

At the center of a suspension bridge, the 20mm diameter suspension cables supportingthe deck are 30m long. Calculate the extension of the steel cables when a 44tonne truck passes along the deck if this load is shared between twelve cables on each side of thedeck.

Solution:

Asked for extension due to truck and hence can ignore the loading due to the weight ofthe deck:

Change in extension,

41020102124

3081.91044

4 2310

3

2

d nE

mgL

nAE

FL0.0082m

where F is the applied load, L the length of the cable, n number of cables carrying load, A is a cable cross-section area, E is the Young’s modulus of steel from a databook, and d isthe cable diameter.




8

Example 1.2

a) Estimate the stress in your femur when standing still and upright with your weightdistributed evenly on both feet.

b) Repeat the exercise in (a) for an adult African elephant and for an adult mouse.Only rough estimates of the bone dimensions and mass are necessary.

c) Assuming that the strength of bone in humans, elephants and mice isapproximately equal, discuss the relative susceptibility to fractures.

Solution

a) Stress,

000,800

410252

81.980

4 232

d n

mg

A

F N/m2

where F , is force applied, A is cross-section area of the bone, m is mass, g isgravitational acceleration, d is the diameter respectively of the cross-section and n

is the number of legs. b) Repeat for an elephant with for example: n=4, m=6 tonnes and d = 200mm, giving

= 360,000 N/m2; and for a mouse with for example n=4, m=25 grams and

d =1mm giving = 78,600 N/m2

c) The stress is lower in the legs of the mouse so that they are much less likely to break their legs.

Example 1.3

On a cello the 0.68m long steel strings are tuned by winding one end around a peg or fret.

For peg diameter of 15mm, calculate how many turns will be necessary to achieve atension in the 1.36mm diameter string of 84 N (approx middle G)

(see http://gamutstrings.com/tensions/cloten.htm for more information of cello strings).

Solution

Extension,

923 1021041036.1

68.084

AE

FL

E

L L 0.00019m

And this wrapped is around the circumference of the peg,

003.0

1015

00019.000019.03

d

n times or about 1.4 degrees.

The peg needs to be turned 1.4 degrees.

Example 1.4

Ask students to look for two other examples in their everyday life and explain how theabove principles apply to each example.




9


2. Principle: Displacement plus deformation in control cables

Engage:

Ride a bike into class changing gear in front of the students.

Explore:

Upturn the bike on the bench or floor. Invite a student to helpyou by turning pedals. Change gear to demonstrate mechanismfor those who have never thought about it. Release the tensionin the derailleur cable and try to change gear.

Explain:

Explain that with the tension released the motion of the gear lever is being used to take upslack resulting in no motion at derailleur. Tension is required in the cable to transmit themotion of the gear lever to the derailleur arm. The cable acts against the spring in thederailleur system, so force in the cable is higher in hill-climbing gears.

Elaborate:


Assume the cable to be 1.2mm in diameter and of length 1.8m. If the spring forceexerted when shifting one gear is 100N then estimate the linear motion required at thehandle bars when the derailleur needs to move 4mm.

Solution:

Displacement at handle-bars, d h where d is the displacement required at the

derailleur and is the extension of the cable.

Deflection of the cable, 4

92 106.71021040012.0

8.1100

AE

FL

E

L

L m

So the movement required at the handle-bars is 76.476.04 h mm.




10

Evaluate:

Ask the students to complete the following examples:

Example 2.1

A cable of diameter 3mm and length 3.5m connects the rudder to the wheel on a yacht.Assuming the force exerted on the cable by the rudder is approximately 20kN find thedisplacement that must be applied at the wheel to achieve a displacement of 10cm at therudder.

Solution:

The extension, of the cable is the difference in the displacements of the two ends, so for

the unknown displacement at the wheel, w

r w where r is the displacement at the rudder,

now the longitudinal strain,

L L

e r w

where L is the length of the cable.

Also we can define longitudinal strain in terms of stress, ; i.e.

42d E

P

AE

P

E

where E is the Young’s modulus, A is the cable cross-section area and d is the diameter.Hence equating these two definitions gives:

42d E

P

L

r w

thus

147.01.0

410310210

5.31020

4 239

3

2

r wd E

PLm

The required displacement at the wheel is 14.7cm.

Example 2.2





11


3. Principle: Stress in pressure vessel wall

Engage:

Take the front wheel from your bike into class with a large bike pump. Its good to have a flat tire, but no puncture.Inflate the tire in front of the class. Ask them to what pressure you should inflate the tire. Answer is written onthe wall of the tire.

Explore:

Compare with the pressure for your car tires. Ask studentsin pairs to discuss

(a) why it is difficult to pump up car tires with a bike pump (displaced volume of pump);

(b) the force you need to exert to pump against

maximum pressure ( A pF max where A is the

cross-section area of piston in pump) and

(c) the work done on each stroke ( Fd W where d is the length of the stroke, i.e.distance moved).

Then invite different pairs to give an explanation to each problem.

Explain:

The tire and inner tube is a complex stress system. First, due to the toroidal geometry,and second, because the tire constrains the free expansion of the inner tube, reducingtensile stresses in tube. However, the pump can be considered as a simple cylindrical pressure vessel and wall stresses can be evaluated.

Elaborate:


Assume pumping against maximum pressure allowed in the tire as a worse case scenario,

let 6max p bar with a pump of external diameter 40mm and wall thickness 1.25mm.

Evaluate the circumferential and longitudinal stress in the wall of pump. Compare thesestresses to the stress in the rod connecting the piston to the handle if it is made of 10mmdiameter tube with a wall thickness of 2mm.




12

Solution

Circumferential stress, 6

3

5

1 106.91025.1

02.0106

t

pr N/m2

where r is the radius of the cylinder and t is the wall thickness

Longitudinal stress,

6

3

5

1 108.41025.12

02.0106

2

t

pr N/m2

Rod stress,

6

622

235

109.1310610

105.38106

rod

piston

A

pA

A

F N/m

2

Note that the compressive stress in the rod is higher, so compare with yield stresses formaterial of your pump. Discuss the probability of failure, actually the rod is more likelyto fail in buckling.

Evaluate:

Ask students to design an emergency foot-pump for pumping up car tires. Ask them todesign it in aluminum or plastic so that it is lightweight and easy to handle.




13

STATICALLY INDETERMINATE PROBLEMS

4. Principle: Combined use of principles of equilibrium and compatibility

Engage:

Bring your iPod into class dangling it from the earplugcable and also the front wheel of your bike again. Askabout the connection between them. They are bothstatically indeterminate systems – you need to consider both forces and displacements to find the stresses. Remindthem how in the earplug cable, the wire and insulationshare the load and must have the same extension ordisplacements. Similarly in the front wheel, the tire

constrains the inner tube and they have the same radialdisplacement.

You might want to consider breaking the class into pairs, and asking them to draw thefree-body diagrams for the iPod cable and insulation and then construct the equations ofequilibrium and attempt to solve them. This will illustrate that the problem isindeterminate when only considering the principle of equilibrium.

Explore:

Highlight that redundancy is a feature of statically indeterminate systems. From astructure viewpoint we don’t need the wire and the insulation. Of course we do for signaltransmission. Similarly in the front wheel we don’t need the tube and tire from astructural perspective, but the tire would leak and the tube would puncture so we need both for a safe and comfortable ride. This problem has been solved in cars…

Explain:

Redundancy results in there being insufficient information available to solve for internalstress using equilibrium of forces alone. Need to also consider compatibility ofdisplacements for various elements in the structure.

For an iPod dangling from earplug cable:

Equilibrium of forces implies,

Weight of iPod, insulationwire F F mg (i)

Compatibility of displacements requires extension of wire and insulation to be

equal, i.e. insulationwire (ii)




14

Elaborate

So use the expressions above to solve for the same example as previously, i.e. a 30 gramiPod dangling from its earplug cable consisting of a copper wire of diameter 0.4mm withsnugly fitting plastic insulation of external diameter 1mm:

AE

FL

E

L L

so from (ii)

insulationio

insulation

wire

wire

E d d

LF

E d

LF 222 44

And

wire

insulation

wire

ioinsulationio

insulationwirewire F mg

E

E

d d

d

E d d

F E d F

22

2

22

2

Substituting for the forces from (i) and takingi

d d

wireinsulationio

wirewire

E d E d d

mg E d F

222

2

268.0101104.01024.01

81.903.0101104.092922

92

wireF N

thus N F mgF wireinsulation 026.0268.081.903.0

and now

13.24.04

268.0

4 22

d

F wirewire N/mm

2

039.0

4.014

026.0

4 2222

io

insulationinsulation

d d

F N/mm2

compare these values with those obtained when we considered the whole load to be borne by the wire or insulation, i.e. 2.34 N/mm

2 and 0.29 N/mm

2 respectively.

Evaluate

Ask students to attempt the following examples:

Example 4.1

In the lobby of a natural history museum, a square horizontal plate made of polycarbonateis suspended at the corners by means of cables 6m long and 3mm diameter attached to theceiling. The weight of the plate and the dinosaur skeleton displayed on it is 250N and thecables are adjusted so that the weight is evenly distributed between them. It is decided toadd the skeleton of the dinosaur’s prey attached directly below one corner. If theadditional skeleton weighs 160N find the stress in each wire for the complete exhibit.




15

Solution

160N

250N

F A

FB FC

FD

A

B C

D

2a

2a

160N

250N

F A

FB FC

FD

A

B C

D

2a

2a

Using equilibrium of forces

Moments about BC: D A F F aaa 21602250

Thus D A F F 285 (i)

Moments about DA: C B F F aa 2250

Thus C B F F 125 (ii)

Moments about CD: C A F F aaa 21602250

Thus A B F F 285 (iii)

From (i) and (ii) D B F F (iv)

this can be deduced from the symmetry of the problemUsing compatibility of displacements

By similar triangles: C B B A (v)

All the wires are identical in geometry and material so F

And C B B A F F F F (vi)

Solving simultaneously

Substituting (iv) in (vi) Bc A F F F 2 (vii)

Substitute in (iii) 5703 c A F F (viii)

Substitute (ii) in (viii) 2503 c A F F (ix)

Hence 5.182 AF N, 5.22C F N, and 5.102 BF N

Now the cross-section area of a cable is 6

23

1007.74

103

A m2




16

Thus, 6

6108.25

1007.7

5.182

A N/m2

6

6105.14

1007.7

5.102

B N/m2

6

61018.3

1007.7

5.22

A N/m2

The stresses in wires are 25.8, 14.5, 14.5 and 3.18MPa.

Example 4.2





17

TORSION

5. Principle: Stress and strain due to applied torque

Engage:

Enjoy a drink the evening before class, providing it has a screwtop (similar to those on individual bottles airlines give you oninternational flights) and take the empty bottle and its top intoclass. Some non-alcoholic drinks have the Stelvin closure(opposite) and would allow you to offer drinks to the wholeclass! Other screw tops would work but the aluminum top isthe simplest for analysis. Discuss the stress and strain system produced just before you break the seal when opening the

bottle.

Explore:

Discuss the forces induced when opening a bottle and how the torque is transmitted fromone hand to the other along the bottle as shear stress. Discuss the mode of failure in theclosure. Noting that aluminum is a ductile material and thus weaker in shear thantension, thus ensuring closure remains sealed until twisted.

Explain:


Ultimate strength in shear for aluminum alloy is 240MN/m2. So, to release the cap weneed to achieve this stress level in the closure.

Shear stress due to torsion, J

Tr

where T is applied torque, r is the radius at which the shear stress occurs and J is the

second polar moment of area, “ 24 R J ” and for a thin-walled tube is approximated by

t R J 32 , where R is wall radius and t the wall thickness. Thus the torque required to

open the bottle when R =1.2cm and t =0.1mm:

22102400001.0012.022 622 t RT Nm

This is about three times the average hand-grip torque strength of an adult4. Perforationsimply that the load bearing area is reduced by about a quarter thus reducing the torquerequired to =22/4=5.5Nm.

4 Imrhan, S.N., Farahmand, K., ‘Male torque strength in simulated oil rig tasks: the effects of grease-smeared gloves and handle length, diameter and orientation, Applied Ergonomics, 30(1999)455-462.




18

Elaborate

Consider the effect of the torque on the glass neck of the bottle, if the thickness of theglass is 3mm:

0.332018.0024.0

012.05.544

J

Tr MPa

Since the mean strength of soda glass is about 65MPa there is no danger of failure evenallowing for a stress concentration of three in the threads of the bottle, i.e.

9max SCF MPa. Now if the cap is damaged by a dent then it may jam and a

strong person could exert three times the typical torque for an adult, i.e. about 22N then

36 MPa (including the stress concentration) – still no problem.

However, if the thread on the bottle is mis-formed so that the wall thickness is reduced by1mm and the cap is damaged and a strong adult attempts to open the bottle, then

4.62332018.0022.0

012.02244max

SCF J

Tr MPa

Failure is likely! This simplistic analysis ignores the presence of cracks etc. and insteadfocuses on using the torsion stress-strain relationships.

Evaluate

Example 5.1

Ask students to repeat the analysis for a half size bottle of wine.

Example 5.2





19

STRAIN ENERGY

6. Principle: Conservation of Energy/Energy Methods

Engage:

Bring a slingshot and a handful of rubber balls intoclass, pull the elastic band back and release a fewrubber balls into the class.

Explore:

Ask students to work in pairs and to identify the conservation of energy during theloading, firing and trajectory of the balls. Invite some pairs to talk through to the classtheir understanding of the energy conversions. Tell them about elastic strain energystored in the elastic band. Discuss how strain energy is stored in the material and isavailable for instantaneous release. Ask them in their pairs to reconsider conservation ofenergy during loading, firing and flight of projectile.

Explain:

Strain energy is defined as the energy stored in a material when work has been done onthe material, assuming the material remains elastic and no permanent deformation occurs.

Equate strain energy to work done and derive an expression relating strain energy, U todeflection, and applied force, F i.e. F U 2

1 . Consider the stress-strain curve for

rubber and hence estimate the strain energy stored per unit volume, u upto the elasticlimit (u

≈3.6MJm-3

).

Elaborate

Estimate release velocity of projectile, i.e.

For an elastic band of length, 150mm and cross-section 16mm2 the volume, V is

2400mm3, so the energy stored when pulled to yield:

6.81024106.3 96 uV U J.

Kinetic energy supplied 2

2

1 Mv ,

so equating strain energy stored with kinetic energy supplied to a yellow dot squash ball:

271024

6.8223

M

U v ms

-1




20

Evaluate


Example 6.1

Consider a bungee jumper with a mass of 50kg leaping from a bridge who breaks theirfall with a long elastic cord having an axial rigidity, EA=2.1kN. If the bridge parapet is60m above the water, and if it is desired to maintain a clearance of 10m between the jumper and the water, calculate the length of cord that should be used.

Solution

Potential energy lost by jumper = work done on cord or strain energy gained

So,2

F mgh (i)

where m is the mass of the jumper, g is acceleration due to gravity, h is the distance fallen

by jumper, F is the maximum force in the cord and is the maximum extension of thecord. Now, by definition:

AE

FL

E

L L

, so L

AE F

(ii)

Substituting in (i) gives:

L

EAmgh

2

2

(iii)

Also know that Lh , so substituting in (iii) leads to:

L LhLh EA

L Lh EAmgh

22

2

222

So, 22 22 EAL EAhL EAhmghL

Thus, 02 22 EAhhLmg EA EAL

Solution is

EA

EAhmg EAhmg EAh L

2

442222

3

232323

101.22

50101.2481.950101.250481.950101.2502

L

6.254200

151680259050

L m

Required length of cord is 25.6m.

Example 6.2





21

STRAIN ENERGY

7. Principle: Helical Springs

Engage:

Ride you bike into class again and while braking bounceon the front suspension.

Explore:

Discuss why a suspension system is necessary. Discusswhat components are needed in a suspension system.Discuss the energy transfers and dissipation.

Explain:

Ask the students to consider what would happen if only a spring was employed and henceexplain why a damper is necessary. There is a nice graphic and explanation athttp://travel.howstuffworks.com/mountain-bike4.htm . This discussion could be extendedto cars and handling characteristics http://auto.howstuffworks.com/car-suspension.htm

Elaborate

Consider the design of the front suspension springs. These are hidden inside the frontshock absorbers, but physically they must have a coil diameter of about 45mm and it isdesirable that they have a deflection of about 10cm. Let us assume that the suspension isdesigned to absorb the vertical impact of rider plus bike of 120kg from 0.5m, then

Potential Energy lost in fall = Strain Energy absorbed by suspension system

2

W mgh

where W is load applied to springs and is the spring deflection, so

11772

1.0

5.081.912022

mghW N (for both springs)

From formula book, shear stress in spring:3

8

d

WD

where D is the coil diameter and d is the wire diameter, for Nickel-Chrome steel

yield 650 MPa, then

6

3

10650

025.0588688

WDd and d = 8.3mm




22

Also, deflection of spring,4

38

Gd

nWD

where n is the number of coils and G is shear modulus respectively.

hence

nn 3

49

3

1087.10083.01082025.058868

(i)

5.531087.1

1.0

1087.1 33

n coils!

Conclusion: springs alone cannot absorb this amount of energy with this distance oftravel. Need energy dissipation in the damper to contribute to the total energy absorption.Actually the front suspension often uses compressed air springs rather than mechanicalsprings.

Evaluate

Ask students to attempt the followingexamples:

Example 7.1

A pogo stick contains a Nickel-Chromesteel spring of free length 600mm andmean coil diameter 50mm made fromround stock. If the spring is designed to be fully compressed so that the coils just

touch each other under a mass of 100kg,then calculate the wire diameter andnumber of coils required if a factor ofsafety of 2 is employed.

Will this spring work for a pogo stick?

Solution: Extract from US Patent #2,793, 036 by G. B. Hansberg

From formula book, shear stress in spring:3

8

d

WD

where W is the applied load, D is the coil diameter and d is the wire diameter, sofor a factor of safety of 2 and yield 650 MPa, then

6

3

10325

05.081.910088

WDd and d = 7.27 mm

Also, deflection of spring,4

38

Gd

nWD

where n is the number of coils and G is shear modulus respectively.




23

hence

n

n 3

49

3

1028.400727.01082

05.081.91008

(i)

also 6.0 nd free length (ii)

substitute (i) in (ii) to give:

6.01028.41027.7 33 nn and n=51.9 coils

So, =4.28×10-3

×51.9=0.22m and equating potential energy with strain energy stored

then 11.02 h m so the spring will be fully compressed for a jump of only 11cm!

Example 7.2





24

BEAM BENDING

8. Principle: Bending moments and shear stress diagrams

Engage:

Ride a skateboard into class.

Explore:

Discuss the shear forces and bending moments set-up inthe skateboard when you stand on it sideways balancedon your heels, i.e. approximating a point load. Whenyou stand of the board more normally, how do the shearforces and bending moments change? Discuss whereyou need to stand to induce a zero bending moment.

You might want to ask students to work in pairs to draw schematics of these loadingschemes.

Explain:

Plot the shear force and bending moment diagrams for the case where you were rockingon your heels.

Considering the complete beam

Resolve vertically: P R R B A

Moments about A: 0 L RPa B

Thus: L

Pb R A

L

Pa R B

Considering the cut section (0< x<a)

Resolve vertically: L

Pb RF A

Moments: L

Pbx x R M A

Considering the cut section (a< x<l)

Resolve vertically:

L

PaP

L

PbP RF A

Moments:

x L L

Paa xP x R M A

A B

R A RB

L

P

a b

A

R A

x

P

a

F

M

A

R A

xF

M

F Pb

L

-Pa

L

x

Pab

L

M

x

A B

R A RB

L

P

a b

A

R A

x

P

a

F

M

A

R A

xF

M

A B

R A RB

L

P

a b

A

R A

x

P

a

F

M

A

R A

x

P

a

F

M

A

R A

xF

M

A

R A

xF

M

F Pb

L

-Pa

L

x

F Pb

L

-Pa

L

x

Pab

L

M

x

Pab

L

M

x




26

For maximum M C :

0

x

M C and a x L L

Px x R M AC 22

2

So, 042

a x L x

M C and42a L x

Thus, 22

16424222

2ˆ a L

L

Pa La

a L L

L

P M C

So for a 1.8m plank and typical skate board (a=65cm) carrying a 65kg person,

11665.06.3316

81.965ˆ 2

C M Nm

If the plank is 13cm wide and 1.8cm thick, then the maximum bending stress is

5.16018.013.0

1166

12

2ˆˆ23

bh

h M

I

y M C C MPa

This compares to compressive ultimate strengths for common woods in the range 35 to55 MPa parallel to the grain and 4 to 10MPa perpendicular to the grain.




27

Evaluate

Example 8.1

Ask students to repeat the analysis above but for unicyclist crossing the plank.

Solution:

Considering the complete beam

Resolve vertically: P R R B A

Moments about A: 0 L RPs B

Thus: L

s LP R A

L

Ps R B

Considering the cut section (0< x<s)

Resolve vertically: L

s LP RF A

Moments:

L

xs LP x R M A

Considering the cut section (s< x< L)

Resolve vertically:

L

PsP

L

s LPP RF A

Moments:

x L L

Pss xP x R M A

For maximum bending moment:

0

s

M C and 2

s Ls L

P

L

ss LP x R M AC

So, 02

s L

s

M C and2

Ls

Thus, 2874

8.181.965

442ˆ

22

PL L L

L

P M C Nm

A B

R A RB

L

P

s

A

R A

x

P

s

F

M

A

R A

xF

M

F P(L-s)

L

-Ps

L

x

MCM

x

A B

R A RB

L

P

s

A

R A

x

P

s

F

M

A

R A

x

P

s

F

M

A

R A

xF

M

A

R A

xF

M

F P(L-s)

L

-Ps

L

x

F P(L-s)

L

-Ps

L

x

MCM

x




28

41018.013.0

2876

12

2ˆˆ

23max

bh

h M

I

y M C C MPa

Maximum stress of 41MPa induced when unicyclist at the middle. The position could

have been deduced without analysis.

Example 8.2





29


9. Principle: Eccentric loading

Engage:

Bounce a basketball into class. There are some free(preview) clips of basketball coaching athttp://www.magicfundamentals.com/clips.htm

Explore:

Discuss the loading on the basketball pole duringdifferent types of play, e.g.

o

Static compression with low level bending due tooffset of backboard and goal;

o Additional low level bending during a goal;

o Dynamic bending when the ball bounces off the backboard from a long shot plus torsion if theshot is wide; and

o High level compression and bending during aslam dunk.

Explain:

Ask the students, working in pairs and sketching, toidentify forces and moments acting about the centerof the cross-section of the pole that are equivalent tothe weight of a player hanging on the rim (solid

arrow). The solution is a compressive force and amoment (dashed arrows).

Explain that if these two forces only produce linearelastic deformation then their effects can be addedtogether, or superimposed. Discuss principle of

superposition.

I m a g e : x e d o s 4 / F r e e D i g i t a l P h o t o s . n

e t

M=Pe

PP

e

M=Pe

PP

e

M=Pe

b

h

M.h/2

I P

b

h

P

bh

M=Pe

b

h

M.h/2

I P

b

h

P

bh




30

Elaborate

For a pole 10cm square manufactured from aluminum with a 60cm offset when a playerhangs from the front of the ring at an effective distance from the backboard of 50cm, themaximum tensile stress in the pole occurs on the back of the pole:

Maximum tensile stress = Maximum bending stress – compressive stress

bh

P

bh

Pe

bh

P

bh

hPe

bh

P

I

h M

23max

6

12

22

So, for a 90kg player, ignoring the mass of the backboard etc.

9.58829058271401010

81.990

1010

10506081.99062232

2

max

MPa

EvaluateAsk students to attempt the following examples:

Example 9.1

Calculate the tensile maximum stresses when a 90kg basketball player hangs from theside of the ring for a goal mounted on a 12cm square section pole with wall thickness of3mm with an offset of 1m from the pole center to ring center. The ring diameter is 42cm.

Solution:

Hanging from the side will induce two bending moments about the front-back axis( M fb)and about the side-side axis ( M ss) of the pole, thus:

bh

P

I

br P

I

heP

bh

P

I

h M

I

h M

fbss fb

fb

ss

ss

2222

max

Since the pole is square fbss I I and 64242 1021.3104.11101212

1 I m4

226

2

6

2

max

1012

81.990

1021.3

10642.081.990

1021.3

106181.990

4.2361312693117816502803max MPa

Maximum tensile stress induced in the pole is 23.4MPa.

Example 9.2





31


10. Principle: Thermal stress & statically indeterminate loads

Engage:

Use BBC News report at http://news.bbc.co.uk/1/hi/wales/north_west/3125813.stm andhttp://news.bbc.co.uk/1/hi/uk/3126441.stm to illustrate very real problem of thermalexpansion in train tracks.

Right picture from http://www.volpe.dot.gov/sdd/buckling.html (public information from RITA)

Explore:

Long slender components subject to compressive loads usually show a preference to fail buckling rather than by material failure. A plastic ruler is an easy example. This can alsohappen in railway tracks. The rail is stretched before laying to reduce expansion in hotweather; but at very high temperatures rails can expand so much that they buckle. The

heat alone can cause buckling but so can the passage of a train over the rail. To reducethe risk of buckling speed restrictions are imposed on high speed lines from aboutlunchtime until early evening.

Explain:

Consider the stress and load induced by the temperature changes mentioned above. Forexample in a rail laid at 20ºC so that it is stress-free the compressive stress induced by a30ºC temperature rise is given by:

E T E

Where is the strain induced, E is the Young’s modulus of the rail (≈

206GN/m2), T the

temperature rise and the coefficient of thermal expansion (for rail steel ≈ 12×10-6

perºC). So,

7410206301012 96 MPa

If the cross-section area of the rail is 14.5cm2 then the buckling load is about

108105.141074 46 AP kN




32

Elaborate

The free expansion of the rail if it was 25m long would be:

930251012 6 T L mm

So, if a 4mm expansion gap is present then the stress will be reduced:

2.4110206

27

1049 93

E MPa

A reduction of 44%. However modern welded tracks have very few expansion gaps inorder to give a smooth ride and hence are more susceptible to buckling at hightemperatures. If rails were pre-tensioned to avoid buckling in summer then they would be more susceptible to cracking in winter under high tensile loads.

Evaluate


Example 10.1

A jewellery pendent is to be made by fitting a gold ring 12mm wide over a platinum ring12mm wide at 1020ºC. The discs fit snugly inside one another at the elevatedtemperature with a common diameter of 50mm and so will lock together on cooling toroom temperature. Both rings are 12mm thick. Calculate the radial pressure at theinterface and the circumferential stress set-up in both rings at 20ºC.

Material Properties (http://www.goodfellow.com/csp/active/gfPeriodic.csp?form=All )

(/ºC) E (GPa) y(MPa)

Gold 14×10-6 78.5 205

Platinum 9×10-6

170 185

Solution:

On cooling from 1020ºC

Change in circumference in gold = Change in circumference in platinum

Since gold and platinum have different coefficients of thermal expansion an interferencefit will be set-up with each exerting an equal and opposite force on one another:

Pt Au F F (i)

These circumferential forces together with the thermal contractions will cause thechanges in circumferences, i.e.

Pt Pt

Pt Pt

Pt Pt

Au Au

Au Au

Au Au E A

LF LT

E A

LF LT (ii)

Substitute (i) in (ii) and:




33

9.19

1017010144

1

105.7810144

1

10914500

119696

6

Pt Pt Au Au

Pt Au

Au

E A E A

T F

kN

Circumferential stresses are

13410144

109.196

3

Au

Au

Pt Au A

F MPa

Also Circumferential stress,t

pr circ from thin-walled pressure vessel theory so

610134

r

t p

32

1050

1012101346

66

r

t p

MPa

Could argue that thin-walled pressure vessel assumption does not apply because r/t =

4.166 which is not greater than 10.

Example 10.2





34

TWO DIMENSIONAL STRESS SYSTEMS

11. Principle: Mohr’s circle of stress

Engage:

Use a hot plate in the classroom to fry somesausages and share with the class. Find a brand of high quality sausages that will notcome undone at the ends and do not piercethem so that they burst longitudinally whilecooking.

Explore:

Before they eat; ask students to examine thesplits in the sausages. Discuss whatdirection the maximum stress must be actingto cause the damage seen in sausage skins.

Explain:

Discuss the derivation performed earlier inthe course to obtain expressions forlongitudinal and circumferential stresses in a

cylindrical pressure vessel:

t

pr circ and

t

pr long

2

And hence explain longitudinal split due tocircumferential stresses.

Elaborate

Discuss that there are two stresses acting in

perpendicular directions so that an elementaligned to these directions experiences thestresses as shown. Then construct theMohr’s circle of stresses and explain how itis the loci of points representing planes at allorientations in the wall of the sausage.

long

long

circ circ

plane Aplane B

A

B

long

circ

long

long

circ circ

plane Aplane B

A

B

long

circ




35

Evaluate


Example 11.1:

How does the Mohr’s circle change for the sausage if there is some residual stress fromtwisting the sausage during manufacture? Where would you expect the sausage to splitnow?

Solution:

i.e. at an angle 2 to the longitudinal axis.

Example 11.2


long

long

circ circ

plane Aplane B

A

B

circ

T

long

T

T

Tp

C

plane A

plane C

long

long

circ circ

plane Aplane B

A

B

circ

T

long

T

T

Tp

C

plane A

plane C




36

TWO-DIMENSIONAL STRESS SYSTEMS

12. Principle: Combined bending and torsion

Engage:

Take a few old or cheap wind-up alarm clocks into class.Set one to go off at the start of the lecture! Dismantlethem to various stages and pass around.

Explore:

Discuss the forces, moments and torques acting insidethe mechanism. Ask them in pairs to figure out how the

clock mechanism works.

See http://home.howstuffworks.com/inside-clock.htm

Explain:

Explain that gears are often overhung and hence their shafts are subject to both torsionand bending.

Elaborate


The clock drives the minute hand through a spur reduction gear. The 4mm diameterdrive wheel transmits 2mW at 1 r.p.m. and has a pressure angle of 20º. The driven wheelhas a diameter of 24mm and is overhung 6mm from its bearing.

Power, NT t

T P

2

where T is torque, is angle turned, t is time taken, and N is rotational speed (revs.s-1).

For the driven (smaller) gear:

F

pressure angle

l

1.3W 1rpm

1/6 rpm

F

pressure angle

l

1.3W 1rpm

1/6 rpm




37

Torque, 019.0)601(2

002.0

2

N

PT Nm

Circumferential force,

55.9

2104

0019.03

r

T F circ N

And from the geometry of the pressure angle:

20cosF F circ , so 1020cos

circF F N

For the larger gear:

Torque in shaft, 12.080sin012.01180sin Fr T Nm

Moment of shaft, 06.0006.011 Fl M Nm

And, shear stress due to torque,

34

16

32

2

d

T

d

d T

J

Tr

xy

stress due to bending moment,

34

32

64

2

d

M

d

d M

I

My x

Using a Mohr’s circle of stress:

22

3

2

2

max

16

22T M M

d xy

x x

22

3

2

2

max

16

2 T M d xy

x

Therefore 322

max

16T M M d

or 3

22

max

16T M d

whichever is greater.

If the gears are made from polythene with a tensile strength of 30MPa, then it would bereasonable that the maximum tensile stress should not exceed 15MPa and the maximumshear should not exceed 7MPa, hence the diameter of the driven gear shaft must be atleast 4.6 mm based on shear stress and 4.0mm based on tensile stress, so the minimumshaft diameter is 4.6mm.

Evaluate


Example 12.1

A machine is driven by an 8kW motor through a spur reduction gear, the motor runs at1450 r.p.m. and its driving pinion has 21 teeth. The driven wheel on the machine has 100teeth and is overhung 120mm from a main bearing. The teeth of the gear are of 3mmmodule and the pressure angle is 20º. Find a suitable diameter for the shaft at the main




38

bearing of the machine if the tensile stress is limited to 80MPa and the shear stress to50MPa when the motor is developing its rated power.

Solution:

As above, for the small gear

Torque,

69.526014502

108

2

3

N

PT Nm

Now, Module = pitch diameter/number of teeth

So Pitch diameter = 21 × 3 = 63mm

Circumferential force,

3

310672.1

21063

69.52

r

T F circ N

And from the geometry of the pressure angle:

20cosF F circ , so 310779.120cos

circF F N

For the larger gear:

Pitch diameter = number of teeth × module = 100 × 3 = 300mm

Torque in shaft, 8.26280sin150.0177980sin Fr T Nm

Moment of shaft, 5.213120.01779 Fl M Nm

Thus,

75.328.2625.2135.2131080

16163

22

63

22

max T M M d mm

And

55.328.2625.213

1050

16163

22

63

22

max

T M d mm

So the diameter must be at least 32.75mm.

Example 12.2




39

NOTES



Junior So lid s C o urse : Sug g e ste d e xem p la rs w ith in le sso n p la ns

NOTES FOR INSTRUCTORS ON EXAMPLE APPLICATIONS

Prepared as part of the NSF-supported project ( #0431756) entitled:“

Edited by Eann A Patterson, University of Liverpool

Population through Curriculum Change”

Examples in Mechanics of Solids

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