S Series lectures – Introducing: Structural Distillation...

S Series lectures – Introducing:Structural Distillation, Modelling and Synthesisfor Structural Integrity

based on

Introduction to

Engineering Design— Modelling, Synthesis and Problem Solving Strategies

Andrew Samuel – John Weir

Oxford: Butterworth Heinemann, 1999 ISBN 07506 42823

This lecture series S01 to S12 introduces the notion of structuraldistillation, and guides the student through a range of modelling experiencesfocussed on basic structural components in engineering. The overriding aim isto develop confidence and wisdom in design synthesis for structural integrity.

We regard this aspect of the design experience as foundational. Not onlydoes it equip the novice designer with skills in structural integrity design thatare a core component of their professional repertoire, but it also provides anaccessible and very practical proving ground for exploring the full range ofconcepts and problem solving strategies introduced under the broader agendaof engineering design. Design for structural integrity gives tangible substance tootherwise abstract notions in engineering design. Moreover, due to itsquantitative and physically-based modelling approach, its content isrecognisable to students who are already becoming versed in allied engineeringscience subjects. For these and other reasons explained in the preface to ourbook, we generally present these S-series lectures before the D-series, howeverthe lectures have been successfully presented in either order.

In lectures S01 to S04, some essential groundwork is covered inpreparation for later work in structural modelling and design to resist failure.Lectures S05 to S12 then explore the behaviour, failure phenomena and designmethods associated with eight categories of engineering structural component.

While these lecture slides represent substantial elements of the lectureseries, they need to be seen as complementing and supporting the presentationbased on our textbook. Much of the material in the slides needs “unpacking”for the student, either by the lecturer or by reference to the textbook.

Caveat

what is failure

• what are the distinguishing features of engineering failure;• is there some classification system to identify the type of failure we are

dealing with;• the nature of engineering risk;• seriousness : economic, environmental and social impact of failure;

how to best avoid failure;

• tools for evaluating the performance of engineering structures;• designing for structural integrity;• factors of ignorance; (we call these factors of safety)• maximum credible accident;

some examples of failure;

Lecture S01 — Reflections of failurebased on

Introduction to




Lecture outline

S01–1

Lecture S01 — Structural integrityand the nature of failure

reference:Andrew Samuel and John Weir

Introduction to Engineering Design — Modelling, Synthesis andProblem Solving Strategies, Oxford: Butterworth-Heinemann, 1999

Chapter 1.1

Engineers use materials,

whose properties they do not properlyunderstand,

to form them into shapes, whosegeometries they cannot properly analyse,

to resist forces they cannot properlyassess,

in such a way that the public at large hasno reason to suspect the extent of theirignorance.

John Ure, Engineer, 1998

S01–2

Proposition 1:

Engineers are inherently concerned withfailure and our vision of success is todevelop modelling tools to avoid it

Proposition 2:

Engineering failures may be categorisedas:

— technical;

— operational; and

— unpredictable;

S01–3(refer: Samuel and Weir, section 1.1)

— technical failures are mostcommonly due to insufficientinformation about the nature ofthe structure, its material, itsloading or its operatingconditions;

Six oil storage tanks (two are shown)15.5 m diameter x 16.5 m height on thecoast of West Africa collapsed in highwinds; Loss >$US 1 million


— operational failures are mostcommonly due to improperoperating practices or conditions;

Broken Liberty ship (circa 1940)


— unpredictable failures are mostcommonly the result of specialcircumstance or “acts of God”.

Collapsed Kobe–Osaka highway 1995

The quake (Richter 7.2) lasted 40seconds. Experts claim that the highwayfailure was initiated by shear failures ofthe piers.

The total loss is estimated in billions ofdollars.


Proposition 3:

Incentives to avoid engineering failuresare related to “failure intensity” – thedegree of seriousness of the failure;

— degree of seriousness of a failure,or “failure intensity” is measuredby the economic, environmental orsocial impact caused by thefailure.

Proposition 4:

Failure is essentially related to risk;

— given extreme conditions, allstructures or systems can fail;


On 24 April 1993 an IRA car bomb - thelargest used to date- destroyed parts ofBishopsgate in London. One person waskilled, 40 injured . Loss > $US 1 billion.The major loss was due to fire damage.


The engineers task is to:

— generate design specifications thatbest meet the required operatingconditions of the structure orsystem within acceptable levels ofrisk;

— identify the limits of “know–how”associated with the structure orsystem and assign “factors ofignorance” (commonly referred toas factors of safety) to cope withthese limits - also withinacceptable levels of risk;

— this brings with it the notion of“worst credible accident”;


Some examples of failures andtheir causes:

1. Structural failures:

These failures are often spectacular, buttheir impact is most commonlyeconomical;

Hoarding around construction site destroyed byhigh winds in Dubai (approximately 62 km/hrwinds). Loss $US 400,000 Cause : although thedesign was specified to withstand 115 km/hrwinds, review of the design indicated that therewas a serious error in calculations.


Shortly after a routine maintenance inspectionthe turbo-generator “exploded” causing 85%destruction of the turbine and generator. Theloss was estimated at $US 40 million.

Cause: The insurer’s experts speculated on thecause, but they could not find any substantivereason for the failure. The popular wisdomsuggested that a bearing or rotor instabilitycaused the failure.

Failure of 600 MVA turbo-generator


High pressure preheater explosion

During startup of a 300 MVA unit in aconventional steam power plant, a highpressure preheater exploded duringemergency shutdown of the turbine.Estimated loss was $US 3 million.

Cause: An undetected crack in a weld(examined by X-ray inspection prior tocommissioning) coupled with a suddenrise in pressure during the emergencyshutdown.


Collapsed pile driver

A 30 m high track mounted pile driveroverbalanced and fell into a river duringoperation. The loss is estimated at more than$US 1 million.

Cause: The supportstructure for the hammerwas hydraulically controlledduring operation. On oneside of the controller thehydraulic valving becameclogged, causing thestructure to overbalanceand collapse.


Tools used for failureavoidance

— Engineering modelling –generation of mathematicalmodels amenable to analysis forpredicting behaviour;

Cross-sectionarea A;

Gaugelength

l

P

P

SU

Stress

εStrain = δ l/l

Sy

σ = P/A


— Decomposition (structuraldistillation) – abstract reasoningabout structures, which allow usto identify elements or sub-systems of engineering structuresand engineering systems, whichmay be open to engineeringmodelling;

Notch and hole stress raisers


Photo-elastic model

Fir-tree design of blade attachment toturbine disc


Failed turbine discfailed turbine rotor where failure wasinitiated by unacceptably high localisedstress intensities;


air screw where failure in the hub regionwas initiated by high local stressintensities near a screw hole.


Engineering components subjected to timedependent stresses can fail at a lower stressintensity than the ultimate failure stress. Underthese time varying loads the material appearsto become “tired” and we refer to this a timedependent failure mechanism as fatigue failure;

Fatigue failure in a rotating shaft. Lateralimposed bending load is fixed in direction whilethe shaft rotates. Fibres of the shaft away fromthe neutral axis will experience tension andcompression maxima as the shaft rotatesthrough a full turn.

Once a failure crack is initiated, it willpropagate through the shaft section, leavingmarks similar to the ones seen in the photo.These are called “beach marks”.


Fatigue failure of propeller blading

Engineering components subjected totime dependent stresses can fail at alower stress intensity than the ultimatefailure stress. Under these time varyingloads the material appears to become“tired” and we refer to this a timedependent failure mechanism as fatiguefailure;


Broken crank shaft – fatigue failure


Tools used for failureavoidance (concluded)

Engineering specifications –

carefully drafted specificationswhichidentify the predictive models usedin deriving the design andthe specific conditions of operationunder which the design is“safe” within specified,acceptable,levels of risk.

design for structural integrity of engineering systems• ‘design for structural integrity’ defined• ‘structural integrity’ in the wider context of engineering design

simple engineering components• LINK; COLUMN; BEAM; SHAFT; FATIGUE ELEMENT• VESSEL; BOLTED JOINTS; BEARING; SPRINGS

more difficult ‘simple’ engineering components• BEAM-COLUMN; PLATES; VACUUM VESSEL

some basic results you should know• beam & cantilever with concentrated end load• beam & cantilever with uniformly distributed load

decomposition of complex structures into simple components• looking at objects in a new way

• a pair of pliers• how are multigrips different from pliers?• fold-back clip• coat hanger• diaphragm pump spring• tow bar• (objects suggested by the class)

• challenge to view artefacts with the engineer’s eye

case examples

Lecture S02 — Structural Distillationbased on

Introduction to




Lecture outline

S02–1

— engineering design for thestructural integrity of systems

— engineers “distil the essence” ofproblems

— decomposition of complexstructures into simplecomponents:

• some basic components

• some simple results

• examples

Lecture S02 — Structural Distillationreference:Andrew Samuel and John Weir


Chapter 1.3


Axialload

Bending Moment Diagram (BMD)

Shear Force Diagram(SFD)

Bending+

A simple example —The Director’s chair

1. Identify critical components instructure

2. Develop simple mathematicalmodel for component

3. Estimate loads imposed onstructure

4. Assign comonent sizes to meetStructural requirements


BMD

BMD

BMD

Idealisedframe

F

Basketframe

a

W

b

Mo

c

M1

Mo

M1

A

Critically loaded location

A less simple example —Supermarket trolley

Free body representation

Physical object

Structural distillation


‘design for structural integrity’defined:

Need to:— specify/predict the actual

LOADS— specify likely mode(s) of

FAILURE— and hence predict STRENGTH

(i.e. loads which would causefailure)

“The designer’s responsibility is tochoose the material and dimensionsof an element to ensure that it hassufficient STRENGTH not to FAILunder the LOADSwhich it experiences in service …

… (or to reduce the probability ofsuch failure to an acceptably lowlevel)


manufacturing &forming processes

engineering sciencemathematics

marketing

economics

industrial designlife sciences

environment

aesthetics

ergonomics

design formanufacture

design foreffectivefunction

designfor the

human user

sociallyresponsible

design

economicallyresponsible

design

Engineering Design

includes design forstructural integrity

✻✻

‘structural integrity’ in thewider context ofengineering design:


axial tension:LINK(tie, cable, strut)

axial compression:COLUMN(strut, brace)

transverse bending:BEAM(cantilever)

torsion (and bending):SHAFT(axle)

?

stresses vary with time

various shapes

σ = σ(t)fluctuating loads:FATIGUE ELEMENT(various)

Simple engineering components:


internal pressure:VESSEL(tank, pipe, drum,manifold, …)

sealing forces:BOLTED JOINTS(seal, gasket)

localised surfacecompression:BEARING(roller, ball, cup & cone)

torsion & bending(special case):SPRINGS(various)

More simple engineering components:


“Not so simple” engineeringcomponents:

… and others …

(these are beyond our scope)

combined bendingand compression:BEAM–COLUMN

2D bending & shear:PLATES(web, shell, hull, …)

external pressure:VACUUM VESSEL(tank)


Things we expect you to know:

(a) simple beam behaviour

M

EI

d y

dxzz

=2

2 M EI

d y

dxzz=

2

2

(b) conventions for SF and BM

R

B

y

δ y

AM M

x

δθ

σb

− σb

Bending moment

Shear force

Positive Negative

NegativePositive



(c) simple beam formulae

Point loading

BEAM

W(N )

x

y

=MmaxWL4

W 2

W 2W 2 L

− W 2

Loads

SFD

BMD

Loads

SFD

BMD

CANTILEVER

− MmaxF= W 2

x

y

L 2

F

− W 2

Mmax = F = WL4

L /2


BEAM

CANTILEVER

BMD

SFD

Loads

Loads

SFD

BMD

Distributed loading

q(N/m)

x

y

L

Mmax =qL2

8

F = q= qL 2

l

−Mmax

Mmax = −

= −qL2

8

2

q 2 l

=L 2l

qL 2qL 2−

qL 2

qL 2


(c) simple beam formulae (cont’d.)


decomposingcomplex structures into

simple components:

LOOKING AT OBJECTSIN A NEW WAY …

— Identify applied forces, moments(i.e. “loads”)

— Simplify the geometry(maybe subdivide the system)

— Sketch the free-body diagram

— Show internal shear forces,bending moments, torques


Some case examples

(a) Pliers

F

P

q

(b) Multigrips


F

R

a b c

q

x

y

F

R

q

(a) Pliers

Mathematical model


F

R

a b c

q

x

y

F

R

q

Mmax= -Fc

SFD

BMD

(a) Pliers

-qa= - 2Fc(a+2b)

F

Complete structuraldistillation


(b) Multigrips

M

- qa

- Fc

x

- F

y

M

a b

c

F

F

P

q

- qa b = - Fc

P= F 1+c2

b2

Px

Py

Fq

y

x

q

SFD

BMD


Some further exercises instructural distillation

This region in bending and torsionPaper loads applied here

Hinge

Cutting load

(c) Paper clip

(d) Scissors

Things We Expect You To Know …— a general knowledge quiz

estimation — a basic engineering skill— some estimation exercises

• time trial• calculation quiz• mental arithmetic

— the importance of estimation in engineering— exercises in rational approximation

• the birth rate in Victoria• barbers in California• some observations

units & dimensional reasoning— dimensional manipulation exercises

• confirmation of some dimensionless groups• checking a wind drag formula• conversion of thermal conductivity from Imperial to SI

— methodical manipulation of units (conversion)

professional engineering calculations— a calculation exercise

• chess board, wheat and constipated sparrow• sample calculation layout

— key elements of clear engineering calculations— checking: a cautionary tale

Lecture S03 — Estimation, Units & Calculationsbased on

Introduction to




Lecture outline

S03–1

Lecture S03 — Estimation, Units &Calculations

— Things We Expect You To Know …

— estimation as a basic engineeringskill

— clear calculations and professionalengineering competence

— units and the skillful handling ofdiverse physical phenomena

reference:Andrew Samuel and John WeirIntroduction to Engineering Design — Modelling, Synthesis andProblem Solving Strategies, Oxford:Butterworth-Heinemann, 1999

Chapter 1.2

S03–2

General Knowledge quiz:

1. What is the speed of light in a vacuum (in ms-1) ?

2. What is the 2nd moment of area of a rectangularsection of width b and height d about a horizontalcentroidal axis ?

3. What is the 2nd moment of area of a circularsection of diameter D about a centroidal axis ?

4a. What is the density of steel (in kg.m-3) ?

4b. … of aluminium ?

5. What is the speed of sound in air (in ms-1) ?

6a. What is the value of Young’s Modulus for steel ?

6b. … for aluminium ?

7a. What does the unit prefix ‘giga’ indicate ?

7b. … ‘pico’ ?

8. What is the diameter of the earth ?

(refer: Samuel and Weir, section 1.2.3)

These are all things we would expect a youngengineer to know. But wait! There’s more …

S03–3

General Knowledge quiz(continued):

9a. What is the area of a sphere, as a function of itsradius ?

9b. … the volume of a sphere ?

10a. What is the maximum bending moment in a simplysupported beam of length L when loaded by acentral concentrated force P ?

10b. … by a total force P distributed uniformly along itslength?

11. What is the shape of the BMD for a cantilever witha uniformly distributed load ? (Sketch it.)

12. What is the expression for the bending stress in abeam having a 2nd moment of area IZZ which issubject to a bending moment of M ?

13. What is the rotational analogue of the relationship:

F m d x dt= ( )2 2 ?

14. What is the yield stress of mild steel ?

15. What is the current human population of the earth(to one significant figure) ?


S03–4

answers to General Knowledge quiz:


1. c = × −3 0 108. m.s 1

2. I

b dZZ =

3

12

3. I

DZZ = π 4

64

4a. ρst = −7 800, kg.m 3

4b. ρal = −2 700, kg.m 3

5. 300 m.s 1−

6a. Est = 210 GPa

6b. Eal = 70 GPa

7a. 109

7b. 10-12

8.

de = 12 756, km

(equatorial)

9a. A R= 4 2π

9b. V R= 43

3π

10a. M

PLconcmax( ) =

4

10b. M

PLdistmax( ) =

8

11.

12. σB

ZZ

M yI

=

13. T J d dt= ( )2 2θ

14. Sy = 240 MPa

15. 6 000 000 000, , ,

S03–5

They also need to be comfortable with …

ESTIMATION

… a basic engineering skill.

Some exercises to test your estimating skill:

— a time trial

— a calculation quiz

— a mental arithmetic brain-buster

Engineers are people who work in the physicalworld with physical quantities and concepts.They need to “just know” certain things.


S03–6

ESTIMATION (1) … a time trial

(a) What mass can the average person lift andcarry comfortably over 100 metres ?

(b) What is the height of a bull elephant ?

(c) What is the mass of a bullant ?

(d) What is the weight of a moderate-sizedapple ?

(e) What is the maximum squeezing forcewhich could be delivered to a pair of pliersfrom the average hand ?

(f) How many locusts in a locust swarm ?

Rules: You must answer all six questions in 60 seconds.


Approximate answers:

(a) 30 kg (b) 3 m

(c) 0.1 g (d) 1 N

(e) 200 N (f) 109

S03–7

ESTIMATION (2)… guessing with calculators

(a) What is the contact pressure under astiletto heel ?

(b) How many cars cross the Golden GateBridge each day ?

(c) What is the average number of cars on theGolden Gate Bridge at one time ?

(d) What is California’s average rate ofelectrical power consumption ?

(e) How much water is flushed down London’stoilets in one day ?

Rules: You can use your calculator;Accuracy required (rather than many answers).



(a) 10 MP a (b) 50,000 cars/day (?)

(c) 50,000 / (24 × 60 min/day) × 4 min/crossing

(d) 2.5 GW (e) 20 × 106 litres

S03–8

ESTIMATION (3)… mental arithmetic

(a) How many seconds in a year ?

(b) Evaluate 800.

(c) The Empire State Building has 102 floorsand a TV mast above ground. What is itsoverall height ?

(d) What volume of water is contained in anOlympic swimming pool ?

(e) How many Monday afternoons have therebeen since 1 A .D. ?

Rules: You cannot use your calculator;Try not to make a sound.



(a) 32 million seconds (b) 28.3

(c) 449 metres

(d) 2,000 cubic metres (e) 104,000

S03–9

the importance of estimation in engineering …

• realism in approximation

• confidence checks during calculation

• used when devising routines forapproaching the unknown


S03–10

some rational estimation exercises …

(1) the national birthrate

Estimate the number of babies born eachyear in the country in which you live

(2) opportunities for barbers in California

Calculate the number of barbers whichcould be fully employed within the state ofCalifornia (population = 30 million).

(refer: Samuel and Weir, sections 1.2.2 & 1.2.3)

S03–11

some observations on rational estimation …

• requires you to focus on the essentialissues behind an unfamiliar problem

• helps you to gain insight into the way keycontrolling parameters influence a processor outcome

• helps to concentrate on primary ratherthan second-order effects(“see the wood for the trees”)

• helps identify major sources of uncertainty

(refer: Samuel and Weir, section 1.2.3 (d) )

S03–12

Units and dimensional reasoning

DIMENSIONAL MANIPULATION EXERCISE (1):

• derive the units of β in the GrashofNumber (a dimensionless group used innatural convection studies)

Gr

g T T xw=−( )∞β

ν

3

2

• note that [ν] = m2/swhere “[…]” means “the units of …”

(refer: Samuel and Weir, exercises 1.7 & 1.8)


• check the “wind drag” formula …

q C

VDD o=

ρ 2

2

… for dimensional consistency

• note that [q] = N/mand drag coefficient CD is dimensionless

S03–13

Units and dimensional reasoning


“Conversion between sets of units”

• convert the thermal conductivity ofcopper from Imperial units(i.e. 225 Btu/hr-ft-°F) to SI units.

• note: 1 Btu = 1.055 kJ

Note on the methodical conversion of units:

— arrange everything to be clearly in thenumerator or the denominator;

— place units (in parentheses) alongsideeach simple quantity;

— convert each simple quantityone-at-a-time;

— cancel & calculate.

(refer: Samuel and Weir, section 1.2.3 (e) )

S03–14

Professional engineering calculations

An engineer’s calculations …

— are like a journalist’s grammar

— are like a doctor’s stitches

— are like a solicitor’s briefs

(refer: Samuel and Weir, section 1.2.3 (f) )

Key elements of clear engineering calculations:

— well spaced

— assumptions & simplifications are stated

— argument unfolds in a logical sequence

— balance between algebra and calculation

— useful intermediate values are calculated

— important intermediate results are highlighted

— related results are summarised (maybe tabulated)

— you know when you’ve got a ridiculous answer

— awareness of ‘meaningful accuracy’ (sig. figs)

S03–15

CALCULATION EXERCISE (1):

“Chessboard, wheat, and a constipated sparrow”

According to legend, the inventor of the gameof chess asked from his grateful mogul lord onlyone thing in payment — all the wheat whichwould be needed on a 8x8 chess board if onewere to place 1 grain on the first square, 2grains on the second, 4 on the third, and so on.

Imagine that all this wheat is eaten by asparrow. Calculate the diameter of such asparrow, assuming the bird does not excreteany wastes until it finishes its meal.

(refer: Samuel and Weir, exercise 1.9)

PS: the inventor was executed for trying to be a bit too clever

S03–16(refer: Samuel and Weir, exercise 1.9)

Engineering Essentials Pty Ltd

S03–17

CALCULATION EXERCISE (2):

“Energy from a cuppa”

Calculate the height to which a cup of tea couldbe lifted above sea level if all the thermalenergy in the hot fluid were converted intomechanical energy.

(refer: Samuel and Weir, exercise 1.10)

S03–18(refer: Samuel and Weir, exercise 1.10)

Engineering Investigations Pty Ltd page 1 of 1

date 14/2/99

name aes

S03–19

Checking: a cautionary tale …

Checking designs is equally important asdesigning itself

(refer: Samuel and Weir, case example 1.1)

by James FerryENGINEER5 AUSTRALIA, February 1995, p.52–53

“… The case concerned the design andconstruction of a relatively simple andinexpensive dolphin enclosure.”

“… A crucial calculation, using anequation known as Merion’s equation,was made at conceptual designstage.”

“… The calculation contained a simpleand easily identifiable arithmeticalerror. However, it was neverchecked.”

“… almost immediately after the vinylenclosure was constructed, [it] failedas a result of the presence of longwaves in the marina.”

S03–20

a cautionary tale … (continued)

(refer: Samuel and Weir, case example 1.1)

“… [the client] ultimately claimed …the losses it had suffered as a resultof the defective design. These losses,inclusive of interest, by the end ofthe hearing totalled $602,845.68,together with legal and expertwitness costs of some $150,000.00.”

The original firm of designers and thecompany responsible for checking wereboth sued. The judge held them both tobe equally liable for the failure.

“… The case illustrates the importanceof risk management systems where alldesigns are reviewed and checked.”

RECAP: the designer’s responsibility in design for structural integrity— key concepts (strength, failure, loads) & relevant issues

modes of failure— summary of some key modes of structural failure— failure modes exercise (bicycle wheel & bicycle spoke)

failure prediction— why do engineers need failure predictors (“theories of failure”)?

• example of a spoke (uni-axial stress)• example of a shaft (bi-axial stress)• example of a pressure vessel (tri-axial stress)• problem: at what stress will each component yield?• material tests in the laboratory: uniaxial tension• the need for failure predictors

— examples of failure predictors• generic• maximum direct stress• maximum shear stress• maximum shear strain energy

factors of safety— what is a safety factor, and why do we use it?— rational estimation of Fd

design method for structural integrity— general form of strength-predicting equations— material selection (this is covered elsewhere in the course)

Lecture S04 — Failure Predictors and Safety Factorsbased on

Introduction to




Lecture outline

S04–1

Lecture S04 — Failure Predictorsand Safety Factors

— modes of failure

— failure prediction (sometimescalled “theories of failure”)

• why we need failure predictors

• major failure predictors

— factors of safety

• how are they used ?

• how are they estimated ?


Chapter 2.3

S04–2

(RECAP …)

(refer: Samuel and Weir, section 1.3)

‘design for structural integrity’ defined:

“The designer’s responsibility is tochoose the material and dimensionsof an element to ensure that it hassufficient STRENGTH not to FAILunder the LOADSwhich it experiences in service …

… (or to reduce the probability of suchfailure to an acceptably low level)”

Need to:

— specify/predict the actual LOADS

— specify likelymode(s) of FAILURE

— and hence predictSTRENGTH (i.e. loadsthat would cause failure)

12345671234567123456712345671234567123456712345671234567123456712345671234567

TODAY’SLECTURE

12345671234567123456712345671234567123456712345671234567123456712345671234567

… ANDTHIS …

S04–3

(RECAP …)

RELEVANT ISSUESin design for structural integrity:

— Load

— Material

— Environment

— Geometry

— Stability (e.g. sensitivity to smallchanges in geometry or load)

— Duty Life

— Failure Mode

(refer: Samuel and Weir, section #.#.n)

S04–4

(RECAP …)

FRACTURE

— static loading • brittle

• ductile

— dynamic loading

— impact loading

— stress corrosion

— creep

EXCESSIVE DEFLECTION

— elastic (linear)

— plastic (non-linear)

INSTABILITY

WEAR & SURFACE DAMAGE


Modes of Failure

S04–5

FAILURE MODES EXERCISE (1):

how can it fail?


BICYCLEWHEEL

S04–6

FAILURE MODES EXERCISE (2):

how can it fail?


BICYCLESPOKE

S04–7

When listing possible Failure Modes, we areconcerned with the question:

“how will it fail?”

But for a given mode of failure, we still need toanswer the question:

“when will it fail?”

This requires the designer to predict thecircumstances under which a particularmode of failure will occur.

To make this prediction, designers sometimesneed to use so-called

Failure Predictors

— these are mathematical models of howstresses combine to cause failure byyielding or fracture

— they are often called “theories of failure”

Question: Why do engineers needfailure predictors? …


S04–8

Problem: PREDICTION OF FAILURE BY YIELDING

Case (1) — BICYCLE SPOKE


essentiallyuniaxial tension

Question: when will it fail ?

uniaxial stressσ1

S04–9

MATERIAL TEST IN LABORATORY

— used to determine materialstrength properties

— performed under conditions ofUNIAXIAL TENSION


experimental results

SU

Sy

Stre

ssσ

= P

/A

Strainε = δl / l

testconditions

l

P

P

cross-section area, A

gauge length

S04–10

Problem: PREDICTION OF FAILURE BY YIELDING(continued)

Case (2) — SHAFT with bending



biaxial stressσ1, σ2

(a shaft is a rotating element that transmits torque)

σx

x

y

τxy

τyx

σx

σy

σy

τxyτyx

highly stressed

element of surface

L

T

W

T

S04–11

Problem: PREDICTION OF FAILURE BY YIELDING(continued)

Case (3) — PRESSURE VESSEL



triaxial stressσ1, σ2, σ3

(a vessel is an element that contains fluid pressure)

σ1

highly stressedelement of surface

P

σ1

σ2

σ2

σ3

Section

σ3

S04–12

DIFFICULTY:

The designer faces a difficulty in gettingfrom here … to here …

Q : How do we predict failure whenmore than one principal stress ispresent ?”

A : Use a “FAILURE PREDICTOR”(note — this is why engineersneed failure predictors!)


UNIAXIAL TEST DATA

MULTIAXIALLY STRESSED COMPONENTS

?

σ3

σ2

σ1

σ1 σ

ε

M AT E R I A L L A B S

R E A L W O R L D

Sy

when does it fail ?

S04–13

FAILURE PREDICTOR“X”:

Failure Predictors

FAILURE OF A COMPONENT WILLOCCUR WHEN THE VALUE OF …

… ANYWHERE IN THE COMPONENTREACHES THAT VALUE OF …


… ATTAINED IN A UNIAXIAL TESTPIECE WHICH IS AT THE POINT OFFAILURE …

some function of stresses,

X σ σ σ1 2 3, ,( )

X σ σ σ1 2 3, ,( )

namely: X X Syfail = ( ), ,0 0

S04–14

MAXIMUM PRINCIPAL STRESSFAILURE PREDICTOR :

Failure Predictors (continued)





maximum principal stress, σ MAX

σ MAX

namely: σ MAX US=

[ note that σ σMAX = 1 ]

S04–15

MAXIMUM SHEAR STRESSFAILURE PREDICTOR :






maximum shear stress, τ MAX

τ MAX

namely: τ MAX yield yS, = 2

[ note that τ σ σ

MAX = −1 3

2 ]

S04–16

DISTORTION ENERGYFAILURE PREDICTOR :






distortion energy per unit volume, udistortion

udistortion

namely: u

ESdistortion yield y, = + ( )1

32µ

uEdistortion = + −( ) + −( ) + −( )

13 2

1 2

2

2 3

2

3 1

2

µ σ σ σ σ σ σ

S04–17

Failure Predictors (summary)

1. Maximum Principal Stress

— for brittle materials only

σ1 = SU

We use three main failure predictors:


2. Maximum Shear Stress (Tresca)

— for ductile materials only

— conservative failure estimate

σ σ σ σ σ1 3

1 2 32 2− = > >( )Sy

3. Distortion Energy (von Mises)

— for ductile materials only

— more accurate failure estimate

13 2

13

1 2

2

2 3

2

3 1

2

2+ −( ) + −( ) + −( )

= + ( )µ σ σ σ σ σ σ µE E

Sy

S04–18

FACTORS OF SAFETY

The “design factor of safety”, Fd …

• for handling UNCERTAINTY systematically

• a number greater than 1

• mostly used as a divisor to reduce theassumed strength of a component,for example: σ allowable = S Fy d

• a product of several factors:

— a factor reflecting the consequences of failure;

— and factors for uncertainty due to “LOAD”;

— and factors for uncertainty due to “MATERIAL”

• Evaluation of Fd can be:


“implicit” — based on experience,sometimes dictated by standards

“explicit” — determined by arational estimation of uncertainty

S04–19

FACTORS OF SAFETY (continued)

• Formally, Fd is the ratio of:

— the “strength” of a component, Sd(i.e. the component’s ability to resist a giventype of failure, usually measured as the loadwhich would cause that failure), to

— the actual “load” Ld exerted on thecomponent

• so that F S Ld d d= .

• Note that Sd and Ld have the sameunits (usually stress), for example:

S Sd y =250MPa (yield stress of the material= )

Ld MAX =50MPa (maximum stress applied= )σ

• In general, there will be a different Fdapplying to each different mode of failurefor the component (because consequencesand uncertainties may vary from onefailure mode to another).


S04–20

Overall,

F F s s s s sd = × × × × × × × ×0 1 2 3 1 2 3 4 5l l l

FACTORS OF SAFETY (continued)


FACTO R Related to … Usual Range

CONSEQUENCES:

F0… seriousness of failure 1.0 — 1.4

LOAD:

l1… load magnitude ? 1.0 — 1.6

l2… rate of load application (shock) ? 1.2 — 3.0

l3… load-sharing assumptions ? 1.0 — 1.6

MATERIALS:

s1… variation in material properties 1.0 — 1.6

s2… manufacturing process 1.0 — 1.6

s3… environment (temp, corrosion) 1.0 — 1.6

s4… effect of stress concentrations analytical

(can be high)

s5… reliability of mathematical model,

assumptions made.1.0 — 1.6

S04–21

SELECTION OF MATERIALS

In choosing a material, ask:

— will it work ?

— will it last ?

— can it be made ?

— can it be done within cost and weightlimits ?


ENGINEERINGMATERIALS

METALS COMPOSITES NON-METALS

NON-FERROUSFERROUS

(We will restrict our scope mainly to ferrous materials)

what is fatigue?— a description of the fatigue phenomenon— factors affecting fatigue— a microstructural explanation of fatigue crack propagation

time-varying stresses— categories of stress cycle:

• reversed, repeated and fluctuating stress cycles• general (regular) and multi-level stress cycles

— mean stress and stress amplitude defined— examples of time-varying stresses

fatigue strength (reversed stress cycles, zero mean stress)— types of fatigue exp’t (reversed axial & reversed bending)— experimental results (S/N diagram) for ferrous materials— concept of Se' (for ferrous metals)— approximate finding (Se' = 0.5 Su) for common steels— probabilistic S/N curves (& no Se') for non-ferrous materials

fatigue strength (fluctuating stress cycles, non-zero mean stress)— experimental results (A/M diagram) for ferrous materials— significance of limiting curves on A/M diagram— basis of the modified Goodman line

design against fatigue— construction of ‘Design Line’ on the A/M diagram— significance of ‘load lines’— factors affecting fatigue: KS and KF— example fatigue calculation (a notched link)

Lecture S05 — Fatiguebased on

Introduction to




Lecture outline

S05–1

Lecture S05 — Fatigue(design for dynamic loading)

— what is fatigue?

— time-varying stresses

— fatigue strength

• reversed stress cycle(σm = 0)

• fluctuating stress cycle(σm ≠ 0)

— design against fatigue


Chapter 2.5

S05–2

the fatigue phenomenon

“ compared to behaviour underconstant loads, things break atlower stress levels when thestress varies cyclically with time ”


— microstructural explanation(stepwise crack propagation)

— basis of understanding is experimental(accumulated experimental data)

— probabilistic nature of fatigue

S05–3

fatigue stress cycles

A. simple sinusoidal cycles

(i) fluctuating stress cycle

(ii) repeated stress cycle

(iii) reversed stress cycle


Stress

Time

σ

σ a

σ a

StressamplitudeMean stress

σmax

σmean

σmin

σ m

σσ a

σ a

σmax

σmean

0

σmin

σ

σ a

σ a

σmax

0

S05–4

fatigue stress cycles (continued)

B. sinusoidal, multi-level


C. non-sinusoidal cycles

σ

0

σmin

σ

σ a

σ a

σmax

0

σmean

Time

S05–5

load cycle examples


Time

σ

0

rotating axle

σ

0Time

TimeCycle time

Pressure

σ

Time0

0

(a) Railway axle; sinusoidal load cycle

(b) Fluttering flag; sinusoidal, variable amplitude load

(c) Truck front axle; random load

(d) Internal combustion engine; repeated load cycle

S05–6

factors influencing fatigue

Fatigue is affected by factors such as:

— Load characteristics(mean & amplitude of variation)

— Geometry(esp. stress concentrations)

— Surface finish(micro-cracks are important)

— Size of section(larger sections more likely to fail)

— Level of reliability required

— …


Standard laboratory fatigue tests aim to eliminate allbut the first of these effects (i.e. they examine only theeffects of load mean and amplitude) by using carefullyprepared test specimens.

S05–7

fatigue experiments

reversed axial tests (reversed tension)

reversed cantilever tests (reversed bending)(developed by Wöhler ~1870)


P

both tests produce fully reversed axial loading

0 Time

σ

σ σmin = − a

σ σmax = a

S05–8

material behaviour under cyclic loads

ferrous materials(these exhibit an “endurance limit”, ′Se )


non-ferrous materials(note the lack of endurance limit)

0 t

for reversed testsσS'e

SU

finite lifeinfinite life

low cycle high cycle

100 106102 104 108

Number of cycles to failure

number of stress cycles to failure N

alte

rnat

ing

stre

ss (

MPa

)

104 105 106 107 108

probability of failure

150

200

250

300

350

0.010.10

0.500.90

0.99

(S-N

dia

gram

)(P

-S-N

dia

gram

)

S05–9

material endurance limit

— ′Se is the stress amplitude below which amaterial will never fail in a reversed axialfatigue test

— applies to ferrous metals only

— is closely correlated with the material’sultimate tensile strength, SU


Hence can approximate: ′ ≈S Se U0 5.

for STEELS with SU < 1400 MPa

400 800 1200 1600 U Tensile strength S (MPa)

hardness (BHN)100 200 300 400 500

700 MPa14

00 M

Pa

wrought steel

S’e = 0.5 SU

cast iron andcast steel

S´

Endu

ranc

e lim

it e

(MPa

) 1000

800

600

400

200

00

S05–10

design to resist fluctuating stresses

— so far, we have considered only thesimplest case, namely “fully reversed”stresses (with σm = 0)

— we now consider the more general case of“fluctuating” stresses (with σm ≠ 0)

— in this case, an A-M diagram is used …… “A” for stress amplitude, σa

… and “M” for mean stress, σm


Stress

Time

σ

σ a

σ a

σ a

Stress amplitude

Stre

ss am

plitu

de

Mean stress

Mean stress

σmax

σmean

σmin

σ m

σ m

(σa, σm )

This stress cycle …

… corresponds to this point on the A-M diagram

S05–11

fluctuating stresses (continued)

— experimental data for a large number offluctuating stress tests are summarised onthe following A-M diagram

— here, axes have neen normalised using SU


A/M Diagram for steels, unnotched

curves ofconstant

fatigue life

1.0

0 0.2 0.4 0.6 0.8 1.0– 0.2– 0.4– 0.6– 0.8–1.00.0

103

104

106∞

σ

tσmin

σmax

σ

t

σmax

σ = 0min

"fluctuating":

"repeated":

"reversed":

corresponds to R≈1.0

corresponds to R=0.0

corresponds to R=–1.0σ

tσmin

σmax

0.2

0.4

0.6

0.8

R=–1

.0

R=–0

.5

R=0.0

R=0.5

R=1.0 σm

US

σa

USσ σσ σ

σσ

m a

m a

−+

=

min

max

R = stress ratio =

S05–12


— two extreme curves on the previous A-Mdiagram correspond to …

(a) infinite fatigue life (N=∞ )

(b) static failure (N=1)

— aim to stay “below the infinite-life line”


simplified A-M Diagram for unnotched steel

infinite fatigue life (N= ∞ )

0SU

σa

σm

SU

′Se

σm +σ a =

static failure(N=1)

SU

S05–13


— the “Modified Goodman Line” is astraight-line approximation to thecurve of infinite fatigue life

— it is drawn from ′Se on the σa axis,to SU on the σm axis

— the “Soderberg Line” is an even moreconservative design approach


Modified Goodman diagram for unnotched steel

Modified Goodman line

curve of infinite fatigue life

0

σa

σm

′Se

SUSy

Soderberg line

S05–14



σm + σ a = Sy

approximate experimental curve for infinite fatigue life

0

σm + σ a = SU

σ a

σm

SU

Sy

Sy

SU

′Se

Se=′Se

Kf KsSe

modified Goodman line (test-piece)

modified Goodman line(component)

(yield failure line)

(fatigue failure line)

Sy

Se

Se /Fd

Sy /Fd

yield design line

fatigue design line

safe design space

σ a

σm

Sy SU

modified Goodman line(component)

U dS /FSy /Fd

0

(a) Construction of A-M diagram for a steel component

(b) Construction of A-M diagram (including Fd )

S05–15

other factors influencing fatigue

We consider explicitly the effects of …

• Surface finish(define a surface finish factor, Ks )

• Stress concentrations(define a stress concentration factor, Kf )

Other effects such as …

— size of section

— environment (temperature? corrosion?)

— manufacturing process(residual stresses? surface hardening? plating?)

— metallurgical condition (grain size? inclusions?)

— reliability, experimental scatter

… may also be considered, but here we willsimply include them in the safety factor, Fd


S05–16

effect of surface finish

Ks is the ratio by which fatigue strength isreduced due to the effect of surface finish.

It depends on manufacturing process and SU .


1.0

4.0

5.0

0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8tensile strength SU (x103 MPa)

2.0

3.0

surfa

ce fi

nish

fact

or

K S

machined or cold drawn

polished

ground

as forge

d

hot rolled

S05–17

effect of stress concentrations

Kf is the ratio by which geometrical stressconcentrations reduce fatigue strength.

Kf depends on:

— component geometry

— loading configuration

— material properties q and SU


K q Kf t= + −( )1 1

Kf is calculated from …

where:

Kt = theoretical stress concentration factor

q = material’s notch sensitivity factor

stress on section XXσ

σave = P/Ar

X

X

P P

S05–18

theoretical stress concentration factor,Kt — example chart


notch sensitivity factor, q

1

1.4

1.8

2.2

2.6

0.05 0.10 0.15 0.20 0.25 0.30

Kt

r/d

dD

r

D/d = 1.50

1.101.05 1.02

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Not

ch s

ensit

ivity

inde

x, q

0 .5 1.5 2.01.0 2.5 3.0 3.5 4.0

Notch radius mm

steels, SU = 1.40 GPa (400 Bhn)steels, SU = 0.98 GPa (280 Bhn)steels, SU = 0.70 GPa (200 Bhn)steels, SU = 0.56 GPa (160 Bhn)steels, SU = 0.42 GPa (120 Bhn)steels, SU = 0.35 GPa (100 Bhn)

elements in bending or axial loading:

wrought aluminium alloys 2024-T6

S05–19

design endurance limit

— factors Ks and Kf both serve to reduce ′Se ,the material endurance limit for a smooth,un-notched test specimen

— along with the factor of safety (Fd), Ks andKf are used to define a practically usefuldesign endurance limit:

S

SK K Fe

e

s f d

= ′

— Se may be regarded as the endurance limitfor an actual component, whereas ′Se isthe “ideal” endurance limit of the material


Example: SU =600MPa (so ′Se ≈ 300MPa) &machined finish. Find Ks , Kt , q, Kf , and Se

R3 (=‘r ’)

30(=‘D’) P

P

20(=‘d’)

factors affecting shaft design— (example 1) electric motor shaft:

• bending moment (overlay)• torsion (overlay)• stress concentrations (overlay)

— (example 2) hot air fan shaft:• applied forces (axial & transverse) and torque• peak bending stresses• peak torsional stresses

excessive deflection as a mode of failure— torsional (twist) deflection— lateral (beam) deflection

strength-based failure prediction (recap)— max shear stress & shear strain energy for uniaxial test piece— summary of Tresca & Von Mises failure predictions— as above, but modified for shafts (Se replaces Sy) (overlay)

principal stresses on the surface of a shaft— summary of the Mohr's circle construction— Mohr's circle for an element on the shaft surface— derivation of principal stresses (from bending & torsional)— evaluation of max shear strain energy— basis of shaft design formulae

Lecture S06 — Shaftsbased on

Introduction to




Lecture outline

S06–1

Lecture S06 — Shafts: Bending andTorsion

reference:Andrew Samuel and John Weir


Chapter 3.1

A very expensive shaft failure in a 600 MW turbo generatorrepair costs amounted to $US40 million

S06–2(refer: Samuel and Weir,Chapter 3.1)

Multistage gearbox

Close coupled gearmotor


Output shaftassembly

Output shaft

Electric motor Gearbox

Example 1: Geared motor


Radial loadsforcing gears apart

Tangential loadsdue to appliedtorque

π

Plane ofresultant loads

Load diagram


Transverse reactionsfrom bearings

Resultant transverseforce - combined radialand tangential force

Mmax

Bending momentdiagram

Torsional momentsfrom pinion and couplings

Torque diagram

T

D2D1 D3 D4 D5

keywayssteps


PulleyImpeller Shaft

Inlet

Outlet

belt load F

out of balance

load (shaft)

out of

balance

load (pulley)

out of balance

load (impeller)

torque = F x r

r

weight

(pulley)

weight (impeller)

fluid

load

k

I1

I2

angular twist;

torsional vibration

Example 2: Sludge pump


Mmax

BENDING (derived from force equilibrium)

fanload

pulleyload

BMD

SFD

Ttorquediagram

TORSION (derived from power transmission)

x

x

x

stresses arising from BENDING:

stresses arising from TORSION:

σ B,max = Mmax ymax

Izz

= 32Mmax

πd 3

τ T,max = Trmax

IP

= 16 Tπd 3

Bending and torsion of pump shaft


Failure by yielding (static stress)

• MSFP (Maximum Shear Stress FailurePredictor)

• DEFP (Distortion Enenrgy Failure Predictor)

σ σ σ σ σ σ1 2

2

2 3

2

3 1

2

2

− + − + −=

( ) ( ) ( )Sy

Failure by fracture (static stress)



2

2 3

2

3 1

2

2

−( ) + −( ) + −( )= SU


σ σ1 3

2 2− =

Sy

σ σ1 3

2 2− = SU


Failure by fatigue fracture(dynamic loading)



σ σ1 3

2 2 2− = =

S Sy e


2

2 3

2

3 1

2

2

− + − + −

= =

( ) ( ) ( )

S Sy e


τ T

−τ T σ B

σ B x-directionτ T−τ T

y-di

recti

onM

T

M

T

x-direction

y-direction

element on the surface of a shaf t(at extreme bending fibres)

σ B = 32 M

πd 3

τ T =16T

πd 3

Analysis of most highly stressedelement of shaft


(a)

σ

σ x

τ xy

τ yx

τ σ y

τ max

σσ

+x

y(

)/2

φ2

0

radi

us o

fM

ohr's

circ

le =

τ m

axτ

σσ

xyx

y2

2

2+

−

=

σ1

σ2

radi

us =

σ

τ σ B/2

0

τ Tσ B

,(

)

0, −

τ T)

(

στ

BT

22

2(

)+

σ 3σ 1

(b)

Moh

r’s c

ircle

ana

lysis


Failure prediction for dynamic loading

• DEFP design rule:

SF d

M Te

d

≥ +323

2 2

π

SF d

MTe

d

≥ +32 343

22

π

d

FS

MTd

e

3 2232 3

4≥ +

π

S

dM Te ≤ +32

32 2

π

S

dM

Te ≤ +32 3

432

2

π

• MSFP predicts failure when

• DEFP predicts failure when

• MSFP design rule:


Shaft Design Equations from AS1403

Case 1 — (general) fully reversed torque:

d QK M

Tsf

32 23

4≥ ( ) +∗

Case 2 — pulsating torque:

d

Q K MT K

sfsf

3

2 22

34

12

≥

( )[ ] + +

∗

d Q K M

Tsf

32 23

4≥ ( )[ ] +∗

Case 3 — steady torque:

Q

FS

K K K M MFdd

esf s f=

′= = +∗10

8; ;


δx

y

P

L/2

Izz=bd3/12z z d

b

beam sectionL

Failure by excessive lateral deflection

δ = PL

EIzz

3

48

δ δ< allowable


Failure by excessive torsional deflection(excessive twist)

θγL

T

T

φ θ=

L φ γ= 2

d γ τ= T

G

τT

P

T d

I=

( )2

φ = T

GIP

GE=+( )2 1 µ

θ φ= = ≈L

LTGI

LT

EdP

26 54

.

For a steel shaft of diameter d andlength L, the total deflection is:

The nature of mechanical springs— Linear elastic phenomena— Common types of mechanical spring— Some special types of springs and their applications

Linear compression springs, static loading— Nomenclature— Modelling stress in compression springs (static loading)— Modelling deflection in compression springs (static loading)

Stresses in linear compression springs, dynamic loading

Other practical issues for compression springs— Buckling— Bottoming— Binding

Tension springs

Torsion springs

Properties of materials used for springs

Vibrations and natural frequency

Lecture S07 — Mechanical springsbased on

Introduction to




Lecture outline

S07–1

Lecture S07 — Mechanical springs:a special case of torsion and bending

The nature of mechanical springs

Linear compression springs understatic loading

Dynamic loading

Other practical issues forcompression and tension springs

Spring materials

Vibrations and natural frequency

Torsion springs


Chapter 3.3

S07–2

Springs have been around for ages …

Some early beam and torsion springs


sapling-poweredtrap

recurve bow

— provided a dramaticmilitary advantage inhorseback warfare

Roman ballista

S07–3

The nature of mechanical springs

Linear elastic phenomena

— all materials exhibit a linear elastic region,in which the deflection, x, is linearlyproportional to the imposed load, F

— it took Rober Hooke (1660) to formulatethis inta a “law” …

F = k.x

… where k is the constant ofproportionality, called the “stiffness”


force

deflection

∆x

∆F slope, k = ∆F / ∆x

stored energy,

S07–4

(helical compression spring, beam springs,torsion spring)

Types of mechanical springs


(b) Leaf spring

(c) Clock type torsion spring

(d) Clothes peg

(a) Simple helical compression spring

S07–5

some special spring applications

wooden clothes peg


negator spring(can have kT=0… or even kT<0 )

Belleville spring

unfurled cross-section shape

h

t

F

forc

e, F

deflection, x

0

h/t=0 0.71.4

2.12.8

3.3

x

S07–6

Springs in parallel and series

example: Belleville springs


k2

k1

k1 k2

series parallel

k

k kTOTAL = +

−1 1

1 2

1

k k kTOTAL = +1 2

S07–7

spring modelling examples …

example: diaphragm spring


example: railway buffer

S07–8

more spring modelling examples …

example:plastic moulding


example: torsion bar

16

300

250

S07–9

Summary: the purpose of springs

The essential purpose of a springis to provide:

— a predictable deflection

— or rate of deflection

… when a load is applied to it.


Springs also store energy …

… in the form of elastic potential energy.

Mechanical springs have evolved from thelinear elastic behaviour of all materials(at small deflections)

S07–10

Analysis of linear compression springs,static loading

— D = coil pitch diameter

— d = wire diameter

— n = number of coils

— p = linear pitch

— l0= free (undisturbed) length of spring


Fn coils

d

p

FD

l0

maindesignvariables

S07–11

modelling of static stresses

— free-body diagram reveals …… direct shear force, F… torsional moment, T

— hence two shear stresses, τF and τT

— these are parallel to each other along thehorizontal diameter of the wire (only)

— overall shear stress by superposition


T

F

F

Torsionalsheardistribution

Super-position

Directsheardistribution

coil

cent

relin

e

d/2 D/2

τ πF F d= ( )2 4

τ πT T r d= ( ). 4 32

S07–12

modelling of static stresses (continued)

— moment equilibrium gives T

FD=2

— so at inner and outer extremes ofhorizontal diameter of wire, superpositionof shear stresses gives:

τ

π= ± +Td

IF

dP24

2

— worst case is at the inner surface of wire,where the upwards shear stresses add:

τπ π π

π

max.= + = +

= ⋅

8 4 81

0 5

8

3 2 3

3

FD

d

F

d

FD

d C

FD

dKshear


— where C is the coiling ratio, or“spring index”: C D d= ( )

— and K

Cshear = +

10 5.

S07–13

modelling of static deflections

— shear strain, γ, and rotation,dα, ofelemental length ds due to torque T

— for small γ, must have γ α⋅ = ⋅d ds d 2

— Hooke’s Law for circular bars:

γ τ

π= = ⋅ =

GTdI G

FD

GdP21 8

3

where G

E=+( )2 1 µ

— G is the shear modulus of elasticity,(approximately 90GPa for most steels)


d

d s

γ

T

αdQ

Q

p

S07–14

static deflections (continued)

— total rotation α by integration:

α γπ

π π= =

=

∫ ∫2 16

16

0 40

2

4

ds

FD

d Gs

FD n

d G

n D n Dd d

— since wire is distance D/2 from coil axis,any rotation dα of wire leads to axialcompression of d dx D= ⋅α 2 in the coil

— therefore overall compression of spring is

x

D FD n

d G= ⋅ =α

28 3

4

— the spring “rate”, k, (or its stiffness) is theforce F required to cause defelection x ofthe force F at the centre of the coil

k

Fx

d G

D n=

=4

38

— can get same result by energy methods


S07–15

modelling of dynamic stressesin linear compression springs

— torsional and direct shear stresses may besuperposed to obtain maximum shearstress in coil wire, as before

— this occurs at the inner surface of coil

— this is also the location of a stressconcentration effect for curved bars, firstreported by A.M. Wahl (1929)


τ max = Td2 IP

T T

τ max > Td2I P

TT

straight bar:

curved bar:

S07–16

modelling of dynamic stresses(continued)

— Wahl found …

τ

πpeak = KFD

dWahl

83

where K

CC CWahl = −+

+4 14 1

0 615.

— this includes effects of both direct shearand torsional shear

— this peak stress disappears due to localyielding when the spring is “set” …

… so this stress concentration effect canbe ignored for static loading

— but it must be considered if the load isimposed repeatedly (i.e. dynamic loading)

… (why?)


S07–17

resisting dynamic stresses

— springs that experience dynamic loadingmust be designed to have stresses belowthe shear endurance limit

τ peak < ′Sse

— must also keep τ peak < Ssy

— there is very little data on the values of

′Sseand the best is that of Zimmerli …

′Sse = 310 MPa for unpeened springs

′Sse = 465 MPa for peened springs

(F.P. Zimmerli — Human failures in springapplications, “The Mainspring”,No.17, August/September 1957)

Peening is a process of bombarding the surfacewith fine steel particles. This tends to removesurface defects and work-hardens the surface.The residual compressive stresses it leaves inthe surface also impede fatigue crack growth.


S07–18

practical issues

(a) compression springs

Need to avoid:

— buckling

• not a serious problem if l0 4D( ) <

• can be avoided by guiding the spring

— binding

• coil diameter D increases withcompression (& decreases withtension), and so can bind on guides orother coil enclosures

— bottoming

• compression springs can “bottom out”(i.e. become solid when adjacent coilsare fully in contact with each other)

• this can be helpful since it limits stressin the section


S07–19(refer: Samuel and Weir, section 3.3.6)

practical issues (continued)

(b) tension springs

— in tension springs the stresses in the endsupports may be more significant than inthe coils, due to stress concentrations

— here are some typical stress-reducing endconnections for tension springs …

Standard ends Ends withreduced stressconcentration

S07–20

properties of spring materials

— SU values are generally fairly high(refer Table 3.1)

— tensile strength varies with heattreatment and with wire diameter(higher at small diameters)

— temperature and shock loading arecritical factors in material choice

— since failure is by shear, we need a valuefor yielding in shear: Ssy

— but Ssy is not always well reported.

— If tabulated values not available, useas an approximation:


S S Ssy y U= ≈0 577 0 433. .

S07–21

vibration of springs

— for springs supported between two plates,natural frequency is

f

km

= 12π

— mass, m, of spring may be found from

m

d Dn= π ρ2 2

4

— if spring is very rapidly compressed,adjacent coils can bump against oneanother, producing axial wave motioncalled “surging”. This can occur atfrequencies up to 13th harmonic of spring

example:

engine at 5000rpm; valve springs operatingat 2500cpm. The 13th harmonic of this is32,500cpm, or 542Hz. Spring must bedesigned to have natural frequency wellabove this value.


S07–22

torsion springs

— the wire in torsion springs is in BENDING(not in TORSION !)

— bending stress in torsion springs given by

σB B

c

KMZ,max = where

Z

Ic

dc

zz= = π 3

32

— KB is a correction factor, K

CCB = −+

4 14 1

— total deflection: θ = 64

4

FRDn

d E (caution!)

— torsional spring constant: k

TT =

θ


Fθ

Ra

M

tensile loading— simple illustration of a general design method

compressive loading: buckling of “long” columns— mathematical model (due to Euler)— failure is elastic, no yielding— critical buckling load— column stress, slenderness ratio, graph of these

factors affecting column strength (critical buckling load)— effective length, l (comment on end fixity)— radius of gyration, r (comment on cross-sectional shape)— plane of buckling (i.e. XX and YY may have different l or r )

buckling of “short” columns— Johnson parabola (curve-fitting to experimental results)— application of Fd for safe design line (overlay)— significance of the transitional slenderness ratio (overlay)

design algorithm— algorithm to find column geometry (A, r )— algorithm to find column strength (Pcr )

eccentrically loaded columns— “secant formula” used in Mechanics of Solids— not used here (too complex), but is basis of Structures Code

concept map for columns

Lecture S08 — Columns (design for axial loading)based on

Introduction to




Lecture outline

S08–1

Lecture S08 — Columns(design for axial loading)

— tensile loading

— compressive loading (‘columns’)

• “long” columns

• factors affecting columnstrength

• “short” columns

• design algorithm

— concept map for columns


Chapter 3.4


design against failure in tension

— the simplest possible case! (trivial)

— included here because it illustrates aGENERAL DESIGN METHODapplicable to all structural integrity designproblems

σaxial = PA

σaxial

Modes of Failure

Failure Predictor(theory of failure)

Key Quantitiesto model ?

MathematicalModel

Design Inequality

Limiting Value

Geometry Found

yield, fracture,

excessive deflection

in general: in this case:

:

:

:

:

:

:

:

τ MAX ,fail = Sy 2

⇒ σ axial, fail = Sy

σaxial ≤ σallowable

σallowable =Sy

Fd

A ≥P Fd

Sy


design against failure in compression

Modes of Failure for columns

— elements loaded axially in compression arecalled, in general, COLUMNS (also struts,braces, props, …)

— columns which are long and slender tendto fail by BUCKLING …

…characterised by large lateral deflections

— columns which are short and stumpy tendto fail by YIELDING …

— in between these extremes, columns ofintermediate slenderness exhibit acombined failure mode involving bothyielding and large lateral deflections

S08–4

Buckling of "Long" Columns

— failure is elastic, no yielding

— a form of elastic instability

— mathematical model for failure prediction(due to the Swiss mathematicianLeonhardt Euler, 1707–1783) …

P P

y

x

l

— for this arrangement, the critical bucklingload of the column is found to be:

P

E Icr

zz= π 2

2l

(Note: Pcr is the load at which thedeflection, y, first becomes indeterminate.)

— Pcr is thus a measure of the column’s“strength” in resisting failure by buckling.


S08–5

Some terminology …

Since P

E Icr = π 2

2l and, by definition, I Ar≡ 2

it is useful to define

PA

E

r

cr

=( )π 2

2l

where:

PAcr

= “column stress”

(since it has the units of stress)

l

r

= “slenderness ratio”


(an inverse-squarerelationship)

r( )

Pcr

A

Pcr

A

= π 2E

r( )2

l

l

S08–6

Effective Length of a column …

— … depends on the nature of its endsupports

— distinguish effective length (l) from actuallength (L )

— define the ratio m L≡ ( )l


(Note: Shigley recommends a more conservative valueof m ≈ 1.1 for both cases C. and D., since fully “fixed”end supports are difficult to achieve in practice.)

l = 0.5L

l ≈ 0.7L

A. Fixed–Free

B. Pinned–Pinned

C. Fixed–Pinned

D. Fixed–Fixed

Pl = 2L

L

l = L

l= 0.6992L( )

theoreticalm = L l( )

m = 0.5

m = 1

m ≈1

0.7= 1.43

m = 2

S08–7

Radius of Gyration, r


— As defined earlier, we have

I Ar≡ 2 or alternatively r

IA

≡

— r applies to a particular axis throughcross-section,

for example: I A rZZ ZZ≡ ( )2

where rZZ is the radius of gyration “aboutthe Z-axis”

— Compare this with I y AZZ ≡ ∫ 2d

— In general, I IZZ YY≠

— A physical interpretation of r is:

rzz

equivalent sectionsequal areas, equal Izz

dA

z z z z

y }} very thin,

very long

S08–8

Designers’ note …

… IT IS POSSIBLE TO HAVE DIFFERENTEFFECTIVE LENGTHS “l” APPLYINGTO DIFFERENT AXES OF BUCKLING


PP

x

y

z

Question: How to determine the direction inwhich this column will buckle first?

S08–9

Behaviour of “Short” and intermediate Columns

— Euler buckling theory clearly fails todescribe short columns, since it predictsinfinite resistance to compressive loads for

l r( ) → 0

— We know that very short columns fail(by yielding) when the axial compressivestress reaches Sy

— Many empirical and semi-empiricalmethods have been proposed for matchingthe experimental data

— The Johnson Parabola is one of thesecurve-fitting methods

— It is an inverted parabola, symmetricabout point 0,Sy( ) and tangent to theEuler curve

— These constraints dictate the point oftangency joining the Euler and Johnsoncurves to be at the coordinates

l

rPA

ES

S

y

y, ,

=

22

2π


S08–10

failure prediction graph

this value of l r( ) is called the

“transitional slenderness ratio”


2 π 2E Sy

Sy

2

Sy

Pcr

A

Euler:PA

π 2 E

Fd r( )2l

r( )l

=

Johnson:

PA

Sy

Fd1 −

Sy

4π 2E r

2

l

= Sy − constant × r( )2l

=

S08–11

Design Method

— introduce a Factor of Safety, Fd

—

PA F

PAd

=

design

cr1

— require

PA

PA

≤

design


Euler:

Johnson:

r( )

Sy

2

Sy

Sy

Fd

r( )∗ =2π 2 E

Sy

PA

≤ π 2 E

Fd r( )2

PA

≤

Sy

Fd1 −

Sy

4π 2E r

2

PA

l

l

l

l

S08–12

Design Algorithm for columnscase (a)

when finding geometry …


Pcr

A

= Sy 1−Sy

4π 2 EL

mr

2

no

Lm r

≥ r( )∗ ?

1Fd

Pcr

A

≥

PA

?

choose geometryA, r

guessdimensions

I = Ar2 =PL2 Fd

π2 Em2

assumelong column

r( )∗ = 2π2 ESy

transitionalslendernessratio

GIVEN: E, Sy , L, m, P, Fd

FIND: A, r (i.e. geometry)

solution OK(long column)

solution OK(short column)

increase A or r or both modifydimensions

yes

yes

no

S08–13

when finding allowable load …


Design Algorithm for columnscase (b)

Pcr

A

= Sy 1−

Sy

4π 2 E

Lmr

2

Pcr

A

=

π 2 E

L

mr

2Lm r

≥ r( )∗ ?

answer

is it a longcolumn?

r( )∗=

2π2 ESy

transitionalslendernessratio

GIVEN: E, Sy , Fd , L, m, A, rFIND: Pc r (i.e. rating)

yes

no

Pallow-able

=Pcr

Fd

Pcr = Pcr

A

⋅ A

long

short

S08–14

— such eccentric loading is beyond our scopehere, but is covered elsewhere in yourcourse (in Mechanics of Solids)

— the analysis gives rise to the“secant formula” …

— … which is rather complex and awkwardto use for design purposes. Nevertheless,the secant formula is the basis of manymodern structures codes.

What about eccentric loading?

— due to off-centre loads …


P P

e

e

— … or due to out-of-straightness …

e increases

~1

~0

P cr

A

Sy

l r( )

S08–15

Concept Maps

— a useful, flexible tool for organising ideas

— a network of nodes (concepts) & links(associations)

— individualistic, there is no “right” map

example: a Concept Map for columns …

GRAPH OF

FAILURE PREDICTION,

DESIGNINEQUALITY

TRANSITIONALSLENDERNESSRATIO,

(Euler ↔ Johnson)

EXPERIMENTALRESULTS

EULER JOHNSONBUCKLING COMBINED YIELDING

MATHEMATICALMODELLING

SECANT

ECCENTRICLOADING

MODES OFFAILURE

LENGTHSECTIONGEOMETRY

SLENDERNESSRATIO,

2nd momentof area

radius ofgyration

endfixity

effectivelength

(long) (intermediate) (short)(long) (short)

(eccentric)

dualprincipal

axes

lr( )

lr( )∗

PCRA

vs. l

r( )

PCR

P ≤ PCRFd

COLUMNSELEMENTS SUBJECT

TO AXIAL LOAD

— examples of pressure vessels

thin-walled cylindrical pressure vessels— design parameters— issues for the designer— the need for a failure prediction theory

stresses in pressure vessels— recap: maximum shear stress theory of failure— principal stresses in thin-walled cylinders— application of max. shear stress theory (basic design inequality)

other practical issues— welded joints and inspection— corrosion— design equations

stresses due to external bending— superposition of axial bending stresses (compressive, tensile)— effect on the Mohr’s stress circle— effect on the maximum shear stress

rarities and complexities (beyond our scope)— thick-walled pressure vessels— non-cylindrical pressure vessel shells— vessels subject to external pressure— thermal stresses; heat exchangers; fired pressure vessels

Lecture S09 — Vessels (design for internal pressure)based on

Introduction to




Lecture outline

S09–1

Lecture S09 — Vessels(design for internal pressure)

— elements subject to internalpressure: pressure vessels

— pressure stresses

— design equations

— stresses due to external bending


Chapter 3.6

S09–2

examples of pressure vessels …


S09–3

Thin-walled cylindrical pressure vessels

(refer: Samuel and Weir, section 3.6.3.1)

σR

Highly stressedelement of theshell

contains fluidpressure

Longitudinal weld

Circumferential weld

σA

σH

σR

σA

σH

= hoop stress= axial stress= radial stressσR

σA

σH

cross-sectionthrough vessel

po = 0 Pa (gaugepressure)

p

Do

tD

S09–4

Design of cylindrical vessels


The designer generally knows:

— volume of fluid, V

— operating pressure, p

— operating temperature, T

— nature of fluid

— external loads

— material properties ( S Su y, , …)

We restrict our scope to …

— steel vessels

— welded construction

… as these are the most commonly used.

S09–5

Design of cylindrical vessels (cont.)

What is the designer trying to DETERMINE?

— geometry, or more specifically,

— D and t

What is the designer trying to AVOID?

— failure, or more specifically,

— yielding

How does the designer PREDICT this failure?

— stresses too high

What sort of stresses act on the shell of thepressure vessel?

— tri-axial, therefore …


need a FAILURE PREDICTOR(i.e. a “theory of failure”)

S09–6

Design of cylindrical vessels (cont.)

• the material is generally DUCTILE …

• therefore use the Maximum Shear StressFailure Predictor (MSFP): τ MAX yS,fail = 2

• need to limit τ MAX within the vessel:

τ τ

τ

MAX allowable

MAX

≤

⇒ ≤S

Fy

d

2

• define “design stress intensity”, f

SF

y

d≡

(values of f are tabulated directly in mostnational design codes for pressure vessels)

• recall that τ σ σ

MAX = −1 3

2

• from the above we can write σ σ1 3

2 2− ≤ f

or alternatively, σ σ1 3−( ) ≤ f


S09–7

Stresses in Pressure Vessels(i) Hoop Stress, σH

(usually the largest of the three principal stresses)

pressure force on mid-plane,F 2

F 2reaction force in shell,

F

(ii) Axial Stress (due to pressure), σ A

pressure forceon end

Fend =

(iii) Radial Stress, σR

(usually the smallest of the three principal stresses)

po = 0p

inner surface of shell

outer surface of shell


S09–8

Evaluation of Pressure Stresses(thin-walled vessels)

(refer: Samuel and Weir, section 3.6.3.3

(i) Hoop Stress, σH

(ii) Axial Stress (due to pressure), σ A

(iii) Radial Stress, σR

t D ⇒

(pressure) × (fluid area) =(stress) × (shell section area)

equilibrium:

1

t tD

(pressure) × (fluid area) =(stress) × (shell section area)

equilibrium:

⇒

radial stress variesfrom 0 (outer wall)to -p

(inner wall)

⇒

p D tH× × = × ×( )( )1 2 1σ

σH

pDt

=2

p

DDtA×

= × ( )π σ π

2

4

σ A

pDt

=4

σR

p≈ −2

S09–9

Application of Maximum Shear StressFailure-Predictor (MSFP):


Summarising stress results:

• σ σ σ σ1 32 2

= = = = −H RpD

tp

;

(note: σ σ2 = A is not used in τ MAX)

• required “design inequality”

σ σ1 3−( ) ≤ f

• substitution leads to the following simpledesign rule:

pDt

pf

tpDf p

2 2

2

+ ≤

⇒ ≥−

S09–10

Other issues …

WELDED JOINTS AND INSPECTION

— use “welded joint efficiency”, η

— varies from 1.0 to ≈ 0.5, depending uponthe extent of weld inspection and thequality of the manufacturing process.

CORROSION

— use “corrosion allowance”, c

— around 1mm for each exposed surface(inner and outer), depending on the natureof the fluid.

MODIFIED DESIGN EQUATION

(refer: Samuel and Weir, section 3.6.3.5 to 3.6.3.7)

t

pDf p

c≥−

+2 η

S09–11

Stresses due to external bending

The shell can act as a beam as well as apressure container …

— the resulting bending stresses σB( ) arecombined with the axial stresses due topressure σ A( ) by superposition.

— (Why? What is so special about σ A ?)

— total longitudinal stress, σ σ σL A B= +


neutralplane

concentrated loading, W

distributed loading, q

exaggerated deflection

S09–12

Stresses due to bending (continued)

— use Mohr’s circle to represent 3D stresses

— pressure stresses only (no σB )

— (case 0): σB very small


τ

σσ 2 σ 1σ 3

σR = − p2

σ A = pD4t

σH = pD2t

τmax

τ

σσ 2 σ 1σ 3

σB

σ L

σ H

σ A

σ R

S09–13

Stresses due to bending (continued)

— (case 1): σ σA B+ can be higher than σH


τ

σσ 1σ 3

σB

σ L

σ H

σ A

σ 2

σ R

τ

σσ 1σ 3

σB

σ L

σ H

σ A

σ R

σ 2

— (case 2): σ σA B− can be less than − p 2

S09–14

Complexities and Rarities …

— thick-walled PVs(see text, section 3.6.4)

Φ = ( ) = ( )D D b ao

— pressure-bearingflat plates and ends

— non-cylindrical PVs

— vessels subject toexternal pressure

— thermal stresses

(refer: Samuel and Weir, sections 3.6.4 and 3.6.7

r

b

σR

σA

σH

σ

O

a

= 2.5Φ

The nature of contact phenomena— categories of contact— contact loading applications— failure modes

Spheres in contact— geometry— deformations— contact pressure distribution

Cylinders in contact

Stress distributions for bodies in contact

Rolling resistance and heat generation

Lecture S10 — Design for contact loadingbased on

Introduction to




Lecture outline

S10–1

Lecture S10 — Design for contactloading

— Nature of contact phenomenabetween elastic bodies

— Spheres in contact

— Cylinders in contact

— Stress distributions for bodies incontact

— Rolling resistance andheat generation


Chapter 2.7

S10–2

The nature of contact phenomenabetween elastic bodies

— consider two elastic bodies in contact

— a complex stress state is set up betweenthem (tri-axial stresses)

— stresses and deformations extend beyondthe region of direct contact


have you noticed this at the beach?

dry (tension)

direct contact(compression)

wet

S10–3

The nature of contact phenomenabetween elastic bodies (continued)

— high local stresses develop

— these attenuate rapidly in both the radialand depth directions

— both bodies deform

— the deformation leads to rolling resistanceand heating


deformationof flat surface

depthattenuation

radialattenuation

S10–4

typical contact loading applications

— gears

— wheels on rails

— bearings


S10–5

categories of contact

— contact area is elliptical in 8 and 4 …… and in case 3 if R R1 2≠

— otherwise contact area is circular(cases 1, 2, 7 — and case 3 if R R1 2= )

— or a rectangular strip (cases 5, 6, 9)


Spheres in contactR2 >R1

1 2 3

4 5 6

7 8 9

Spheres in contactR2 >>R1

Skew cylinders in contact

skew angle = 90 0

Skew cylinders in contactskew angle < 90o

Skew cylinders in contactskew angle = 0o

Cylinders in contactR2>>R1

Ball in cup Ball in groove Cylinder in groove

S10–6

analysis of contact phenomena

Historical notes:

— elastic deformation between two sphericalbodies was worked out by Heinrich Hertz,German physicist (1857–1894)

— the work was first published in Journalof Mathematics (v.92, 1881, Leipzig)

— makes use of following result fromTheory of Elasticity for deflection dwdue to point force dF acting atdistance r on a “semi-infinite solid”

(refer: Samuel and Weir, section 2.7 & 2.7.1)

deformationof flat surface

undisturbed datum

dF

rdw

dw

dF

E r=

−( )1 2µ

π

S10–7

analysis of spheres in contact

— consider two elastic spheres in contact

— when forced together with contact force Fthey form a circular contact area

— the diameter of this contact area isdenoted as 2a

— a varying contact pressure q acts overthis contact area, distibuted as shown

(refer: Samuel and Weir, section 2.7.1(a) )

y

F

F

q

Circular contactarea; radius 'a'

2a

R2

R1

S10–8

spheres in contact (continued)

Geometry

— spheres with radii R1 and R2

— zero contact force

— no deformation (yet)

— assuming r is small (i.e. r R R<< 1 2, ) …

… obtain z

rR

zrR1

2

12

2

22 2= =;


R1

R2

N

M

y

O

Z1

z2

z1

r

Z2

S10–9

Geometry (continued)

— initial separation of points M and N atradius r is therefore given by

z z

r R R

R Rr1 2

21 2

1 2

2

2+ =

+( )= Γ

where geometry factor Γ is defined as

Γ = +R R

R R1 2

1 22

— the same definition of Γ is used for thecase of sphere-on-plate (i.e. flat) loading,but since R2 = ∞, we can write

Γflat = 1

2 1R

— same definition of Γ is also used for thecase of ball-in-cup (i.e. concave) loading,but since R2 is negative, we can write

Γconcave =

−R R

R R2 1

1 22


S10–10


Deformation

— we begin with a “thought experiment” inwhich the two spheres are held fixed inspace, and supported far from origin ‘O’

— imagine that the distributed load q isimparted, equal and opposite, to thesurface of contact (radius = a )

— the surfaces will separate (by distance δ )


N’

N

M

M’

R1

R2

w1

w2

z1

z2O

2a

q

q

δ

r

Z1

Z2

deflected

deflected

S10–11

Deformation (continued)

— points like M and N will deflect by localdeformations w1 and w2 respectively

— their separation will increase from z z1 2+to z z w w1 2 1 2+( ) + +( )

— within the surface of contact (i.e. r a< ),clearly δ = +( ) + +( )z z w w1 2 1 2 , so

δ − +( ) = +( ) =w w z z r1 2 1 22Γ

— in reality the contact surface does notseparate at all, and the distance δ is justthe net approach of the two spheres

— recall dw dF

r= κ

π

… where material constant κ µ= −1 2

E

… and dF q dA= ⋅ where dA is anelement of the contact surface


S10–12

Deformation (continued)

— Hertz combined all these equations,integrated over the contact surface(radius a ), and showed that q must be a semi-spheroidal pressure distribution


q0

2a

semi-ellipse

δ

κ κ π=

+( )1 2 0

2

q a

a

q=

+( )κ κ π1 2 0

4Γ

— maximum pressure q0 is at the centre

— key results:

… where κ µ κ µ

112

12

22

2

1 1= − = −E E

… and Γ = +R R

R R1 2

1 22

S10–13


— the peak pressure q0 is found by equatingthe applied load F, to the integratedpressure distribution over the contactsurface

q

F

a0 2

3

2=

π (can you show this?)

— exercise: confirm the following relationsfor two spheres in contact …

a

F=

+( )κ κ1 233

8Γ

δ κ κ= +( )9

8 1 2

2 23Γ

F

— exercise: are they dimensionally correct?


S10–14

analysis of cylinders in contact

— cylinders are considered as a special caseof general ellipsoidal bodies in contact

— Hertz solved this in 1881

(refer: Samuel and Weir, section 2.7.1(b) )

b

F R R

R R=

+( )+( )

4 1 2 1 2

1 2

κ κ

π l

δ

κ κπc

F Rb

Rb

=+( )

+ +

1 2 1 223

4 4l

ln ln

q

Fb0

2=π l

F

R1

R2

l2

l1

2b

FF

F

— for cylinders, lengths l l1 2, → ∞

— obtain:

S10–15

stress fields due to contact

— stress distributions for spheres in contactfirst calculated by A.N. Dinnik (Russia,1909) and M.T. Huber (Germany, 1904)

— an excellent review of the above (and much morethan you ever dared to ask!) is available inTimoshenko and Goodier, Theory of Elasticity, 1983

— main points to note:

• the maximum shear stress, τmax ,is the critical stress for yielding ofelastic bodies

• τmax occurs on the central axis, just

below the contact surface, at a depthof about half the radius of contact

• for µ = 0 3. , τmax .= 0 31 0q

• for brittle materials. the maximumprincipal (tensile) stress occurs at theouter edge of the contact area:

σ

µt q=

−( )1 2

3 0 (in radial direction)

(refer: Samuel and Weir, section 2.7.1(c) )

S10–16

stress fields due to contact (continued)


(Note: above stress values apply along the Z-axis only)

0

0.20

0

z/b

σx

σz

τzy

σy

σ/q0

0.40

0.60

0.80

1.00

0.50 1.00 1.50 2.00 2.50 3.00

y

z

0

0.20

0.40

0.60

0.80

1.00

0 0.50 1.00 1.50 2.00 2.50 3.00

σ/q0

z/a

σzσx= σy

τmax

y

z

S10–17

rolling resistance and heat generation


— original work assumed frictionless contact

— later work by others (refer especially Palmgren,1959) included friction and tangential loading

— when a body rolls on a support, forcesdeveloped in the area of contact act toresist the motion …

y

z rotation

"bow wave"

resultant force does not pass through centre of roller

— resulting friction causes uneven rolling ofballs or rollers in contact with bearing

• for ball bearings, ball will rotate aboutits axis under action of tangentialforces generated

• for roller bearings, rollers will slip

— resulting friction represents greatestcomponent of rolling resistance of bearings

bolted joints which maintain a seal— the function of a bolted sealing joint— failure modes— applications— the canonical bolted joint

a simple mathematical model— free-body diagram (three forces in equilibrium)— assume elastic behaviour of bolt & joint— concept of stiffness (evaluate k from one-dimensional Hooke's Law)— kj and kb are springs in parallel— kj derivable from kf1, kg and kf2 (springs in series)— areas and lengths of bolt and joint— basis of bolted joint equations (derivation)

bolted joint design equations— summary of bolted joint equations for Fb and Fj— graphical representation of bolted joint equations:

• graph of Fb versus F (overlays)• graph of Fb and Fj versus δ (overlays)

— fatigue of bolted joints

further details regarding threaded connections— tensile strengths of SAE bolting— nuts, dilation— threads & axial force distribution— estimation of pre-tightening torque

Lecture S11 — Bolted Jointsbased on

Introduction to




Lecture outline

S11–1

Lecture S11 — Bolted Joints(design for sealing contact)

— bolted joints that maintain a seal

— a simple mathematical model

— bolted joint design equations

— further details regarding threadedconnections


Chapter 4.2

S11–2

Function of a bolted (clamped) joint

Consider cases where …

— we must maintain a

• fluid-containing seal, or• frictional contact

between the joined components

— the external force (acting parallel to thebolts axis) varies with time


generic elements of half a clamped joint

Failure modes:

— lose sealing/clamping contact

— bolts yield (or break) in tension

tensile element

contact surface

anchor point

external force

S11–3

Applications


(these are examples of fluid-sealing joints)

(a) flanged pipes

(b) hydraulicequipment (c) machine elements

(d) pressure vessels

S11–4

Applications (continued)


(examples of clamped/frictional non-slip joints)

(e) friction joints(structural steelwork)

(f) appliance cover

(g) rustic bracket[U-bolt and plate]

(h) scaffolding clamp

(i) wall bracket[wood screws]

S11–5

the “standard” bolted joint


(a) external force from

pressure:(b) external force from

traction:

FTOTAL

FTOTAL

FTOTAL = p ⋅ A

S11–6

Note that the two forces appear misaligned.We need not worry about this. Why ?

the “standard” bolted joint(continued)


F

F

F =F TOTAL

n

S11–7

A simple mathematical model

Based on the following free-body diagramof the cover (three forces in equilibrium):


NOTE:

— all forces are “per bolt”

— e.g. F

Fn

TOTAL= (etc …)

(where n is the number of bolts used)

F

externalforce bolt

force

joint force (gasket)

(resu

ltant)

Fb

Fj

S11–8

Deformation behaviour of bolt & joint

Bolt & joint are considered to be elastic,

i.e. … (force) = k × (deflection)

where the stiffness — k — is assumed to be constant

• also implies∆(force) = k × ∆(deflection)

• can derive k from σ ε= E

(this is a statement of Hooke’s Lawin one dimension)

⇒ =k

AEl

(refer: Samuel and Weir, section 4.2.3(d) )

FbFj

F

kbkj

(compressive) (tensile)

S11–9

Springs in parallel and series

Bolt and joint behave like two springsconnected in parallel, since they areconstrained to have the same deflection …

Note also that the joint ( & hence k j ) is madeup of several different stiffnesses in series …

(refer: Samuel and Weir, section 4.2.3(d) and (e) )

deflected position of cover

kb

δ b

anchor point

anchor point

kj

δ j

≡kj

(flange)

(gasket)

(flange)

kf 1

kf 2

kg

‘equivalent’ ‘compound’

k

k k kjf g f

= + +

−1 1 1

1 2

1

S11–10

Factors influencing stiffness, k

AE=l

— Areas of bolt and joint:

— Lengths of bolt and joint:


Aj ≈πD j w j

n

D j

w j

db

Ab ≈ db2π

4

lg lb

≈ l j

S11–11

Derivation of bolted joint equations

Equilibrium — F F Fb j= + (1)

Initial conditions — F Fb i j i= (2)

Compatibility — ∆ ∆δ δb j= (3)

Rearrange (3) using basic definitions —

∆ ∆

∆ ∆∆ ∆δ

δb b b

j j j

b

b

j

j

F k

F k

Fk

F

k

= ( )= −( )

⇒ = −

which can be written as

F F

k

F F

kb b i

b

j j i

j

−= −

−

(4)

Use (2) to eliminate Fj i from (4)

F F

k

F F

kb b i

b

b i j

j

−=

−(5)


S11–12

Derivation of bolted joint equations(continued)

Finally, using (1) to eliminate Fj from (5), weobtain:

F F

kk k

Fb b ib

b j

= ++

(6)

… and using (3) to eliminate Fb from (5), we

obtain:

F F

k

k kFj b i

j

b j

= −+

(7)

Notes:

— these are the“bolted joint performance equations”

— they show the linear variation of Fb and Fj

with external force, F

— F varies between 0 and Fmax


S11–13

Graphical representation of bolted joint equations:

(A) — graph of Fb versus F


45o

BoltForce

ExternalForce,

Fb

Fb = Fb i + kb

kb + k j

F

Fb iFj = Fb − F

line:

F =F b

OF

General bolt force(i)

S11–14


(A) — graph of Fb versus F (continued)


atfailureof joint

F when joint fails

Fb

kb

k b +kj

kb

kb + kj

LARGE

(i.e. flexible gasket)

SMALL

(i.e. stiff gasket)

Fj = 0

45o

BoltForce

ExternalForce,

Fb

Fb = Fb i + kb

kb + kj

F

Fb iFj = Fb − F

line:

F =F b

OF

Bolt force with varying joint stiffness(ii)

S11–15


(B) — graph of Fb and Fj versus δ


pre-loadpoint

Force

deform’n

initial extensionof bolt (on assembly)

initial compressionof gasket (on assembly)

Fbi = Fj i

δ

bolt

forc

e, F b

(slop

e = k b )

joint force, Fj

(slope = k

j )

(i) General joint deformation

S11–16


(B) — graph of Fb and Fj versus δ(continued)


∆ Fj

∆Fb

Fj

Fb

externalforce (bysubtraction)

F

Fj = 0

joint opens(fails) when

(increases)

(decreases)

Fb = F

Fb = F

∆δ

Force

deform’n

initial extensionof bolt (on assembly)

initial compressionof gasket (on assembly)

Fbi = Fj i

δ

bolt

forc

e, F b

(slope = kb )

joint force, Fj

(slope = kj )

(ii) Joint deformation at failure

S11–17

Strength of bolted joints

Friction joints

— need to keep Fj above some minimum

— F

Fn

Fj dminmin

= ⋅transverse

µwhere Fd

is a suitable safety factor

(refer: Samuel and Weir, sections 4.2.4(g) & (h) and 4.2.6(a) )

Tensile strengths of bolts

— need to keep Fb below some maximum

— F

S AFb

fail s

dmax =

′where ′Fd

is another suitable safety factor

— Sfail may be taken as SP

(“proof stress”),or Sy , or SU

as appropriate

— values are tabulated (e.g. Tables 4.2, 4.3)

— As is the ‘stress area’ of the bolt

S11–18

Fatigue of bolted joints

For many applications the force F istime dependent — we need to use theA–M diagram to design our bolted joint.

• the minimum load on the bolt is Fb i

• the load amplitude is

12

kk k

Fb

b j+

• more particularly,

amplitude of bolt forceamplitude of applied force

=+

k

k kb

b j

• If we can reduce kb and increase k j wecan reduce the fatigue effect of the appliedload. This is done in many applicationswhere fatigue is a great concern.

Question: What does it “cost” to use suchreduced shank bolts?


straight shankreduced shank

S11–19

Terminology of threaded connections


major diameter= shank diameter= thread diameter

minor diameter= root diameter

pitch diameter

pitch

rootcrest

(a) flats (b) socket (c) conical recess

(d) slot

head shank threads

nut

torqueto nut

fillet

45o chamfer

(a) the archetypal bolt

threads

long neck

torqueto head

head

standard neck

neckfillet

(b) the archetypal screw

(… bolts have shanks)

S11–20

Further threaded connection issues

Nuts, dilation

The nut is axially compressed due to the applied force F.Moreover, the nut will dilate radially as a result of thethreads attempting to “force the bolt and nut apart”.

Threads & axial force distribution

In general a large proportion of the load in the nut iscarried on the first thread (about 230% of the averagethread stress on the first thread in a 12mm bolt of finethreads — 180% for coarse threads).

• a softer nut will allow more even distribution ofstresses on threads than a hard nut.

• a coarse thread has better stress distribution than afine thread — also less stress concentrations.

• required pre-tensioning torque — largely frictiondependent, but within 20% accuracy is given by

… T

F db i≈5

(where d is the nominal thread diameter)


typical pinned joints— typical applications— the canonical pinned joint— key dimensions defined

modes of failure— illustrated planes of failure (overlays)— summary

simple design inequalities— sample derivation— tabular summary

types of welded joint

welded joints under parallel loading— free body— mathematical model— allowable parallel load per unit length— a note about effective length of weld

welded joints under transverse loading— free body— mathematical model— determination of max. shear stress plane— allowable parallel load per unit length

welded joints subject to oblique loads

examples of parallel & transverse welds

transverse welds subject to net bending moments

Lecture S12 — Pinned & Welded Jointsbased on

Introduction to




Lecture outline

PinnedJoints

WeldedJoints

S12–1

Lecture S12 — Pinned & WeldedJoints

Pinned Joints:

— typical pinned joints

— modes of failure

— simple design inequalities

Welded Joints:

— types of welded joint

— welded joints under parallel/transverse/oblique loads

— welds subject to net bending


Chapters 4.3 and 4.4

S12–2

pinned joints


(a) Pinned truss

(b) Household hinge

(c) Scissors

In pinned joints, connected elements are always“pierced” to form a hole or holes …

… and linked by a shear connector (“pin”)

S12–3

standard pinned joint

main features of such joints:

— shear forces act transversely to the pin,and parallel to the eye-plates

— no moments

— in “2-force” members, forces act alongmembers only

— in “3-force members”, resultant forcedirection must be found, but always actstransverse to shear pin


Clevis

S12–4

standard pinned joint (continued)

— various eye-plate arrangements

— key dimensions for standard joint

(refer: Samuel and Weir, section 4.3.1(b) & (c) )

(these may need adaptation to suit other joint shapes)

(a) (b)

(c) (d)

a

a

b

d

do

D

D

S12–5

modes of failure for pinned joints

we are concerned with the various “modes” offailure characteristic of pinned joints:

(refer: Samuel and Weir, section #.#.n)

Note that Lshear in case (d) is conservatively estimated by

L

d dshear

o≈ −2

. Why is this a conservative estimate?

(e)

(c)

d

(b)

D

Lshear

(d)

do(a) — tensile failure

of rod

— transverse shearingof pin

— tensile failure of aneye

— shear failure of aneye

— compressivebearing failurebetween pin & eye

S12–6

pinned joint failure modes (continued)


1 12

31

2

13

45

6

78

7

8

Tensile surfaces

Shear surfaces

Compressive bearing surfaces

S12–7

1. tensile failure of the rod;

2. tensile failure in the eye;

3. tensile failure in the net area of the clevis;

4. shear of the pin;

5. shear failure in the eye;

6. shear failure in the clevis due to "tear out";

7. compressive failure in the eye due toexcessive bearing pressure of the pin;

8. compressive failure in the clevis due toexcessive bearing pressure of the pin.

summary of failure modes & locations


S12–8

Simple design inequalities

— choose failure mode

— identify stressed area (“failure surfaces”)

— estimate stresses

— compare with allowable value

example: case 6. “tear-out” of clevis eye


Clevis

Lshear a

P

P

S12–9

Summary of design inequalities

assumptions:

— uniform pin-to-eye-plates contact force

— ignore stress concentrations

— ignore bending of pin


Mode Stress Type FailureLocation Expression Limiting Stress

1 tensile rod σt = 4p/pD2 σt,allowable= Sy /Fd

2 tensile clevis σt =P/b(d0-d) σt,allowable= Sy /Fd

3 tensile clevis σt =P/2a(d0-d) σt,allowable= Sy /Fd

4 shear pin τ =2P/pd2 τallowable= Sy /2Fd

5 shear eye τ =P/b(d0-d) τallowable= Sy /2Fd

6 shear clevis τ =P/2a(d0-d) τallowable= Sy /2Fd

7 compressive eye σ c =P/db σc,allowable= Syc /Fd

8 compressive clevis σc =P/2da σc,allowable= Syc /Fd

S12–10

pinned joints — effects of pin bending


more pronouncedbending of pin

highlynon-uniformcontact forcedistribution

contact forcedistribution

S12–11

welded joints


weldparent metal

HAZ

(b) LAP weld

weld parent metal

heat -affected zone (HAZ)

(a) BUTT weld

weldparentmetal

HAZ

(c) FILLET weld

S12–12

practical welding techniques

multi-pass welding


edgepreparation

grinding back

thickplate

(d) double V butt(c) bevel

(a) square butt (b) V butt

(e) U butt (f) double U butt

(g) J joint (h) T joint

S12–13

spot welding

robotic spot welding in automated manufacture


weld sheetmetal

HAZplatefuseddiameter

S12–14

design of fillet welds for strength

idealised fillet weld geometry


types of loading considered

θ

sectioning planethrough weld

t

h

45o

(a) parallel loads (b) transverse loads

S12–15

fillet welds subject to parallel loading

— max actual shear stress is

τ = ( )W

h LP

e2

— where Le is the effective length of the weld

— define w W LP P e≡ ( )

allowable as the

allowable parallel load per unit length

— then w

hP ≡

2τallowable

— if τallowable MPa≈ 140 , and h is in mm …

… this gives: w hP ≈ 100 kN / m)(

(refer: Samuel and Weir, section 4.4.3(a) & (b) )

L

h

t

45o

WP

WP

worst-caseshear stress ison 45° plane

S12–16

fillet welds subject to transverse loading


L

WT

hθ

x

WT

loads on weld

WT

θ

45o

x cosθ

x

h

componentshear

=WT sinθWT

θ

h - x sinθ

x sinθ Geometry and force analysis

Shear and normal components of loads

xh

θ

shearcomponentof WT

WT

L

normalcomponent

WTof

S12–17

transversely loaded fillets (continued)

Geometry: x h xcos sinθ θ= −

so the width of the sectioning plane is given by

x

h=+cos sinθ θ

Shear stress:

τ θ= =W

AREAW

x Lshear T

e

sin

τ θ θ θ= +( )W

h LT

e

sin cos sin2

A bit of calculus:

dd

when

τθ

θ θ θ θ

θ θ

= − +( )= +( ) =

Wh L

T

e

cos sin sin cos

cos sin

2 2 2

0 2 2 0

so worst shear stress is when θ = °67 5.

Worst case:

sin . cos . sin . .67 5 67 5 67 5 1 212° ° + °( ) =


S12–18

transversely loaded fillets (continued)

— at worst value of θ, τ MAX = ( )W

h LT

e

1 21.

— compare this with the earlier parallel-

loaded case, for which

τ = ( )W

h LP

e2

— define w W LT T e≡ ( )

allowable as the

allowable transverse load per unit length

— then w

hT ≡

1 21.τallowable

— if τallowable MPa≈ 140 , and h is in mm …

… this gives: w hT ≈ 115 kN / m)(

— comparing this with wP suggests thatfillet welds subject to transverse loadingare stronger by about 15%

— so use wP for oblique loading … (why?)


S12–19

examples of parallel and transverseloading of fillet welds


Parallel loading

Transverse loading

S12–20

fillet welds subject to bending

— complex loading case

— simplify by treating weld throat as if itwere in the plane of the joint

— replace fillet welds by narrow rectangularstub, same thickness as throat

t h=( )2

example:


d

t

L

P

M

(b) Sectional view

d d+2tt

(d) Idealised weld model

(a)

dL

P

Loaded structure

(c) Weld replacement model

S12–21

fillet welds under combined stresses

— fillet welds often occur in locations whereboth normal and shear stresses on theweld throat are appreciable

— consider the circular welded bar in theprevious example

Bending Stress:

— σB ZZM y I,max max= where y dmax = 2

— for circular fillet with t d<< , we can

approximate: σ

πB nPL

d tf,max = =( )4

2

— fn is the maximum normal stress acting onthe weld throat

Shear Stress:

— τ

πmax = = =( )PA

Pdt

fv

— fv is the maximum shear stress acting onthe weld throat


S12–22

combined stresses (continued)

— welding codes recommend combining theseas an equivalent stress given by

σcombined = +f fn v2 23

— … and limiting this to 3 0 33× ×. SU(based on the strength of the electrode)

— the design inequality is therefore:

f f S

Sn v UW

UW

2 23 3 0 33

0 57

+ ≤ × ×=( )

.

.

— for circular bar example, this becomes:

16 30 57

2 2

2 4 2

2

2 2 2

P L

d t

P

d tSUWπ π

+ ≤ .


d

t

L

P

Md

L

P

weld size = h

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