Review Final Exam Fall 12

7/30/2019 Review Final Exam Fall 12

1/112

Review and Final ExamFormat

Fall 20121

Mechanical Response of

Engineering Materials:

EMch 315


2/112

Final Exam Format

When: December 18, Tuesday, 10:10AM -12:00

Where: 100 Thomas (all sections)

Conflict Exam: December 17, 1:00-4:00PM, 302A EES

6+1 Problems (16x4 + 18x2 +5(bonus) = 105)

All 10 chapters covered

2

Closed book,3-3x5 inch cards (or one sheet of the paper, hand

written only on one side) with equations are allowed

Bring a functioning scientific calculator.


3/112

3

Topics

One problem from: (1)

Analysis of Strain (Ch 1)

Analysis of Stress (Ch 2)

Elastic Behavior of Materials (Ch 3)

Tensile and Compressive Properties (Ch 4)

Yield Safe Design (Ch 5)

One problem from each: (5)

True Stress - True Strain (Ch 6)

Fracture (Ch 7)

Fatigue Failures (Ch 8) Viscoelasticity (Ch 9)

Creep (Ch 10)

Bonus problem : true/false, definitions, fill-in the gaps, etc.


4/112

How to prepare for exam?

Go over all the home-work solutions:

understand how they have been solved

Go over all In-class problems:

understand how they have been solved

Go over your quizzes with correctanswers

Go over your class notes.

4


5/112

How to raise your exam score? First start working on the problems you find easy to solve,

attempt the difficult ones towards the end

Sketch of the problem: What is known/given and what is tobe determined?

Identify basic equation(s) stated in general form to solve

problem: simplify/rearrange it to suit your problem

Necessary assumptions stated (if any)

Make sure proper Units used or converted

Show all your steps to get to the answer

Final answer clearly indicated

Garbage erased or crossed out

Before submitting the answer sheet, recheck your answers

and methodology if time permits.

If any doubt, please ask. 5


6/112

Review

Chapters 1 -10

6

Note: In this presentation the equations indicated by orange

arrow you should put on your equation cards.


7/112

The final exam will assess your ability to do the following :

Calculate strains from given strain-gage measurements.

Use the Mohrs circle approach to determine principal strains

from a 3-strain gage arrangement.

Use Hookes law to calculate strains given stresses and vice-versa

7

Chapters 1 - 4


8/112

8

X


9/112

9

(From Hibbeler)

True stress-strain Curve

Both curves are same until ______y


10/112

10

Sometimes the transition between elastic and plastic behavior is

gradual. How do we deal with this?__________________________________________________

______________________________________________________

______________________________________________________

__________________________

Tensile Response

0.002

EE

UTS = Ultimate

Tensile Strength

We define a term called the yield strength (YS, y). y is the

stress level when you have a very small amount of permanent

deformation.

pl

y

TS or UTS

failure

f= strain at failure

This small amount has been set by ASTM at

0.2% (.002) of strain

0.2% Offset Method

Governed by ASTM E8

= Pmax

/ A0


11/112

11

Ductility is expressed as:

% elongation = __________________ x 100

or

% reduction in area = ___________________ x 100

Ductility

0.002

EE

Lf- Lo

Lo

Ao - Af

Ao

= f

Easy to measure

than area/radius

The ductility of a material, orthe extent to which it can be

permanently deformed, is the

strain at fracture.


12/112

12

Sometimes you may not have the stress-strain curve. An estimate

of the materials toughness can be made by computing themodulus of toughness, T.

T = _________________

The maximum elastic strain energy that can be stored, is a

response parameter called the modulus of resilience, Ur.

Toughness

E

U rp l

2

2

s

StrainS

tress

X

f

y TS

s

2

splTS

Strain

Stress

Strain

Stress

y

f

TS

Toughness

Modulus of

Toughness


13/112

13

Transverse strains are induced by axial loads

yy zz d d

Poissons ratio ( ) = - Transverse strain/Longitudinal (axial) strain

= - yy or zz / xxis +ve number and its values range __________ (see table)

for metals it ranges from __________

Transverse strain is opposite (-ve) in a sense to axial strain,

however in some materials it is +ve and is -ve

Poissons Ratio

Rubber is 0.5, Foam varies 0.1 to 0.4, Clay is 0.3 to 0.45

i C i i i


14/112

14

These equations imply that ______________ are the same in allthree directions -- this is only true for isotropic materials. Not

all materials are isotropic.

Elastic Constitutive Equations

z zy yx xx x E sss -1

z zx xy yy yE

sss -1

y yx xz zz zE

sss -1

and


15/112

15

Elastic Constitutive Equations

Just as E relates normal stress to strain, there is a constant that

relates shear stress to shear strain. This constant, G, is called the

_______________ or ___________________.

There is no Poissons effect for shear (no transverse strain effect).

The constitutive equations for normal strains and shear strains

together constitute Hookes Law for isotropic materials.

and G =

xyxyG

1

yzyzG

1

z xz xG

1

In general G < E

E

2(1 + )

Shear Modulus Modulus of Rigidity

Shear strain

in x-y plane

Shear stress

in x-y plane

G


16/112

16

Elastic Constitutive EquationsBy adding thermal strain to our Hookes Law equations we get

xyxy G

1

yzyzG

1

z xz x G

1

TE

z zy yx xx xD- sss

1

T

E z zx xy yy yD- sss

1

TE

y yx xz zz zD- sss

1

Mechanical load

Thermal stress

Shear load


17/112

1. All mechanical forces due to thermal component and

external forces must be balanced in order to

maintain the mechanical and dimensional integrity

of the component.

17

3. Constitutive equations for strains.

Three Steps to solve a problem with thermal

strains/stresses

2. Compatibility of deformation. (Series and Parallel)


18/112

Constrained Thermal Stress and Strains Series/Parallel

Approach to Problem Solving

18

1. Series

deformations

add together

(like components

in series)

or = 0


19/112

Constrained Thermal Stress and Strains Series/Parallel

Approach to Problem Solving

19

2. Parallel

Deformations are

tied together ( like

components in

parallel)


20/112

Mohrs Circle

20

Amethod to transform stresses/strains from

one coordinate system to another system.

Real space: x, y, z

Mohr space: I, II, III

Please note, there is

no shear stress

3 Di i l St T f ti


21/112

21

3-Dimensional Stress Transformations

zz

yyyx

xyxx

ss

s

000

0

yy, CWA:

B

A

xxxy

yx

yyyx

yy

xx zz

xx, CCW

B: xy

1. Draw stress or pseudo-stress element: mark normal and shear

stresses

2. Identify coordinates of A and B

xx> yy> zz

3 Di i l St T f ti


22/112

22


zz

yyyx

xyxx

ss

s

000

0

yy, CWA:

B

A

xxxy

yx

yyyx

yy

xx zz

xx, CCW

B: xy


stresses

2. Identify coordinates of A and B3. Draw axes for Mohrs space and x, y plane stresses

4. Locate A and B using the coordinates.

5. Draw first circle in I-II plane using AB as the diameter

Assume: xx> yy> zz


23/112

23

(CW)

zz= III

(CCW)

R

C

II I

State of stress

on x - plane

State of stress on y - plane

A

B

xxyy

2

y yx xC

ss 2

2

2xy

yyxxR

ss

-

2

tan

yyxx

xywheress

-

= C + RI

= C - RII

yx

xy

3 Di i l St T f ti


24/112

24


zz

yyyx

xyxx

ss

s

000

0

yy, CWA:

B

A

xxxy

yx

yy

yx

yy

xx zz

xx, CCW

B: xy


stresses

2. Identify coordinates of A and B

3. Draw axes for Mohrs circle and x, y plane stresses4. Locate A and B using the coordinates.

5. Draw first circle in I-II plane using AB as the diameter

6. Mark and draw z = III axis using

7. Draw 2nd circle in III-II plane

8. Finally draw 3rd circle in III-I plane

zz = III


25/112

25

(CW)

zz= III

(CCW)

R

C

max

II I

State of stress

on x - plane

State of stress on y - plane

III

State of stress

on z/III - plane I - III

2max =A

B

xxyy

2

y yx xC

ss 2

2

2xy

yyxxR

ss

-

2

tan

yyxx

xywheress

-

n

= C + RI

= C - RII

= I+ III

2n

yx

xy

xy

yx

0 > >


26/112

26

(CW)

(CCW)

R

C

max

II IIII

State of stress

on z - plane

zz

yyyx

xyxx

s

s

s

00

0

0

I > II > III

xx> yy > zz

D

Cx

y

/2

x-axis CBy-axis CA

Principal axis II CF

Principal axis I CE

GE

I

II

I

II

Mohrs space

Real space

F

Principal StressElement


27/112

27

x

y

z

Principal Stress

Element

Showing of principal stresses and their orientation in real space

Using principal stress element

0

Sh i l


28/112

28

(CW)

(CCW)

R

C

max

II IIII

State of stress

on z - plane

zz

yyyx

xyxx

s

s

s

00

0

0

I > II > III

xx> yy > zz

D

xxyyzz

FE

I

III

45

IIIII

max

I

n

max

x

The arrows of

maxmust pointtowards

the largest

principal stress

Principal axis III DF

Principal axis I DE

Show in real space

using principal axes.max


29/112

29

IIII II

xx

yy

zz

max

xy

yx

III II

I

xy

yx

2

tan

yyxx

xywhere

ss

-

2

2

2xy

yyxxR

ss

-

= C + R

= C - R

2

y yxxC

ss

2ma x

I I II ss

-

3 Dimensional Strain Transformations


30/112

30

3-Dimensional Strain TransformationsExample: A state of strain at a point is shown here.

/2

IIII II

zz

yy

yx

xy

xx

ij

00

0

2

02

(CW)

(CCW)

Determine the principal strains and the orientation of the

principal planes with respect to the x and y axes.

Where xx yy zz

C + R

C - R

zz

zz

yy xx

Max shear strain

yx

2

xy

2

R2= {( xxyy)/2}2 + ( xy/2)

2

R

C

C xx + yy

xx

yy

yx

xy

I > II > III

zz

max/2

O


31/112

31

Strain Measurement

Since in a 45-45 rossette, gages a and c are 90 apart in real space,

they must be 180apart on Mohrs circle.

b must lie midway between a and c.

This is equivalent to fitting a T (of equal arm lengths) on the circle.

a

b

c

a

b

c

Rossette technique requires that one principaldirection be perpendicular to the plane of the

rossette, which is true on a free surface. Other two

principal axes lie in the plane of the rossette but not

necessarily aligning with the gages. And hence

Mohrs circle.

(c) (b) (a)

45

45

How to determine principal strains and


32/112

32

1. Mark off

2. Center of circle

C =

3. R = square root of the sumof the squares of the triangle

sides

R = ( a- C)2 + (C - b)

2

4. From geometry

How to determine principal strains and

shear strain?

ab

c

/2

ab

c

/2

R

R

R CR

a, b, c

a + c

C + R

C - R

RR

RC

y x

xxyy

Cos = ( a - C) / R

S i M


33/112

33

Strain Measurement

ab

c

/2

RR

R C

R

ab

c

/2

R

R

R CR

In real space the angle is /2

xy = 2 b( c+ a)

There is only one correctorientation for the T when you fit it

into Mohrs circle

(abc ccw)

y x

xxyy

ccw

cw

S i M


34/112

34

Strain Measurement

ab

c

/2

RR

R C

R

ab

c

/2

R

R

R CR

In real space the angle is /2

xy = 2 b( c+ a)

y x

xxyy

y

xI

II

/2

For the principal strains, we

will not order them using the

convention that I >II > III.

Wh i M h i l ti d


35/112

35

When using Mohrs circle equations, I andII are ALWAYSdefined as

I

= C + RII = C R

When considering Mohrs equations, the

convention I > II > III is only valid for I >

II > III > 0, where III 0. Otherwise,

disregard the convention and do notreorder/rename the principal stresses.

Stress Concentration


36/112

Calculations based on uniform distribution of stress members having stress

concentrators have led to disastrous failures for highly stressed components.

The error is that the stress at the edge of the hole is not the nominal stress it isa much higher stress. This higher stress is obtained with one of the following

equations and knowledge of the Elastic stress concentration factor K .


Where w is the far field stress and nom is the nominal stress.

The stress concentration factor is a function of the geometry of the

stress concentrator.

)(s t r e s sno m ina l

s t r e s slo c a le dco nce ntra t

n o m

Ks

s

)(s t r e s sfieldfa r

s t resslo c a le dco nce ntra t

w

Ks

s

or(uses net cross-section)

(uses gross cross-section)

Specific values of K for various geometries are generally

reported in handbooks related to Stress analyses.

St C t ti


37/112

The tangential stress at the hole is given by


As x , tends to ______.

At the edge of the hole, or x = r, = ____. In this case, the stress

concentration factor, K = ___.

2r

3 W

Wx

23

2

12

2

4

4

x

r

x

rwss

w

3

1 1

3 W


38/112

A more typical example of a stress concentration factor would be

a flat plate with an elliptical flaw in it. (more realistic situation)

Defects in materials more typically have elliptical

geometry (Flaws, inclusions, voids etc.)


w

2b

2a

= radius of curvature

of the ellipse edge

(crack tip)


39/112

The geometry of the ellipse is defined by its major and minor axislengths 2a and 2b or by one axis length and the radius of the

curvature (measured at the sharp tip). For this case, the stressconcentration factor is given by

K =

and when


40/112

The final exam will assess your ability to do the following:

Utilize your understanding of the simple parabolic law

that was used to model the plastic strain-hardening

response

Understand the mathematical relationships betweentrue and nominal/engineering stress-strain values and

the underlying assumptions for the relationships to be

valid

Determine a measure of true stress and true strain at the

load instability, which corresponds to the maximum

load point (at the UTS) in the engineering stress-strain

response. 40

Chapter 6: True Stress and True Strain


41/112

41

(From Hibbeler)

True stress-strain Curve

Both curves are same until ______y

T St d T St i


42/112

42

A. Instantaneous or true stress

true =

only need when you are beyond y

B. Instantaneous or true strain

True Stress and True Strain

PAi

i

o

L

Lt

L

dL _____________ = _________ = _______

ln Li - ln LoLi

Loln

D

o

o

L

LLln

t = ln ( nom + 1)


43/112

43

True strain can also be computed from changes in the cross-

sectional area.

Experimentally it has been observed that plastic straining

generates no volume change. Therefore, in plastic region at any

instantaneous value of length (Li) and cross sectional area (Ai):


Volume = LoAo = LiAiSubstitute into our t equation.

so LiLo

AoAi

=LiLo

lnt = =

AoAi

ln

During plastic deformation, the instantaneous stress is same

as the true stress: i = t = P/Ai



44/112

44

What governs plastic portion of the Stress-Strain curve?

A relationship between true stress and true strain exists in theregion of plastic deformation. Known as Parabolic law for ductile metals.


t

n

= ot

strain hardening

exponent

strength coefficient

Valid only in plastic region

0 looks similar to E, but not exactly and much smaller than E

n represents rate of rise of parabolic stress-strain curve.

Elastic portion of Stress-Strain curve is governed by Hookes Law


45/112

45


If we take the log of both sides of this equation we get

tn= ot

strain hardening

exponent

strength coefficient

n log t=log t log o +

= +y b mxn is the slope of response line in log-log space

0 is true stress at an arbitrary t = 1.0 b

m

x

y

nlog

t

log tlog o

=log t log o

=t o when t = 1.0

n log t+



46/112

46

a b c d

X

max

1 2 3 4

Necking

Strainhardening

X


t = P/Ai nom = P/A0

t = nom A0/Ai

Recall t =A

o

Ailn

t = ln ( nom + 1)

Ao

Ai= ( nom + 1)

( nom + 1)= nom

t ( nom + 1)= nom

nomA0 = P

t = ln ( nom + 1)

Load instability in Tension


47/112

Load instability in Tension

n

n

t

t

Material still

exhibits instability

We know that there is a maximum

load that the material can carry

from engineering stress-strain

curve, however, we want to knowwhat is it using true stress-strain

curve

Want _____ which

corresponds to Pmax.

We will call this

stress:_____

t

____= onn

t*

t*

tn=

o

t

Engineering/nominalStress-strain curve

Instability begins

Pmax

P*

n P P*


48/112

P = P* = Pmax

dP = 0

n

n

Xwant

at maximum load, slope is zero and

We want a quantitativemeasure of stress when

load becomes unstable.

Pmax = P*

P = tAi

t * = n

True strain at maximum load or strain corresponding to load

instability = strain hardening coefficient Onset of necking

t *= o t *n

t *= ln ( nom* + 1) = n

t* = n

Valid only at the max. load

Onset of necking

= onn

t = ln ( nom + 1)

True tensile strength


49/112

49

No volume change during plasticity!

But, at

instability:

Hence:

t *Ai* = TSnomAo

AoAi*

ln = t* = n

Ai*Ao

= e-n

Ai*Ao

TSnom = onn = on

n

t *= onn

P* = Pmax

e-n

Nominal/engineering tensile strength

Ch t 5 7 8


50/112

The final exam will assess your ability to do the following :

Determine whether failure occurs based on failure theories that

predict yielding for multi-axial stress states

Use the fracture-free criteria based on the concept that a crack

becomes unstable when the stress intensity factor reaches a

critical value, referred to as the fracture toughness.

Determine the critical flaw size, given the fracture toughness

Assess whether fracture is the result of plane stress or plane strain

Interpret the stress amplitude, referred to as the endurance (orfatigue) limit and designated e, given an S-N curve

Calculate the equivalent completely reversed stress amplitude

that can be superimposed when m 0 given the S-N data

How to use Miners Rule to find out fatigue life.50

Chapters 5, 7, 8


51/112

Chapter 5

51

Safe Design for YieldingMSST: Maximum Shear Stress Theory

DET: Distortion Energy Theory

Yielding for Multiaxial Stress States

i i f i i S S


52/112

52

Predicting safe design limits for a material subjected to

combined (and perhaps complex) stresses requires the use of a

failure (yield) criterion.

We will discusstwo failure criteria (methods for determining upper

limits for elastic designfor ductile materials only, and only under

uniaxial stress), there are other failure criteria also in the book:

1) Maximum Shear Stress Theory (MSST): Plastic flow starts

when the maximum shear stress (in a complicated state of

deformation) reaches a value equal to the maximum shear stress

at the onset of flow in uni-axial tension (or compression).

2) Distortion Energy Theory (DET): Plastic flow arises from a

complicated state of stress (in multiaxial situation) when the

distortional or shear deformation energy is equal to the

distortional (or shear) deformation energy in uni-axial stress.


I - II = y


53/112

53

I

II

y

y

y

y

Locus of yield points as predicted by MSST

Pure tension

Pure compression

Twisting (Torsion)

Bending

Bending

No hydrostatic load

Because does not

cause yield.

y

II - III = y

III - I = y

2. Distortion Energy Theory (DET)


54/112

54

2. Distortion Energy Theory (DET)

aka: Von Mises Theory or Octahedral Shear Stress Theory

Von Mises derived an equation to describe yielding using energy criterion.

- He grouped the principal stresses and squared them (energy terms).

( I - II )2 + ( II - III )

2 + ( III - I )2 = 2 y

2 = K

DET is another yield criterion that is used for ductile metals.

DET is an alternate to MSST.

Plastic flow arises from a complicated state of stress (in multiaxial situation)

when the distortional or shear deformation energy is equal to the distortional

(or shear) deformation energy in uni-axial stress.

( x - y )2 + ( x - z )

2 + ( z - y )2 + 6( xy

2 + xz2 + zy

2) =

= 2 y2 = K

In terms of applied normal and shear stresses in a Cartesian coordinate

system, the DET can also be expressed as:


55/112

55

Distortion Energy Theory

Again we get K from a tension test, where:

I =

II =

III =

In plane stress the DET equation:

0

0

P

A

( I - II )2 + ( II - III )2 + ( III - I )2 =2 y2 = K

I2 + II

2 - I

II= y

2

simplifies to for :

DET Equations

III = 0

Yi ldi f M lti i l St St t


56/112

56


For our other yield criteria (DET) the boundary is an ellipse.

I

II

MSST

DET

Data from experiments

falls between the two

boundaries.

More conservative design

approach

Also tells you the planes o

which yielding occurs

Use if you need to design

close to the tolerance limits

of the material

(load path)

Safe

I2 + II

2 - I

II= y

2


57/112

Factor of Safety (FOS)

57

FOS =Stress causing failure (UTS, y, f)

Stress in service / Predicted yield stress (MSST, DET)

A safety factor of 2 implies that the failure/yielding will occur

at twice the stress value in service or predicted by yield theory.


58/112

58

Catastrophic Failure of Engineering Materials

Fracture and Fatigue

X

FractureX

Ductile

Brittle

P ti l P i t


59/112

Practical Points

We use components that contain cracks and defects.In Chapter 7 we seek to answer two questions that are important

in terms of safety and design of engineering components.

Given a certain applied design stress, what is the largest defect (crack)

that can be tolerated in a member without failure of the member. Given a certain size defect in a member, what is the largest stress that

can be tolerated without failure of the member.

The answers to these two questions will require knowledge ofa material property called Fracture Toughness.

59

Fracture (Brittle materials)


60/112

Knowing this ___________________________________

____________________

Griffiths analysis of the fracture of glass was based on an energy balance

between competing processes occurring at the tip of the flaw.

dV

daa

Stress at

crack tipApply energy

balance at thecrack tip.

very high strength materials can be made

in the form of fine fibers.

Griffth observed: strength increases inversely proportional to the

size (area/diameter, V) of the material.

( )

We need to analyze the crack propagation

b


61/112

The elastic strain

energy stored in dV is

a dadV

If a crack extends an amount da, the volume element cannot support the

stress (it is cracked), and the energy of the structure is reduced by

dV. However, the material also possesses a resistance to crack extension.

This resistance is denoted as ___ and is the reversible, adiabatic energy

necessary to create a new surface. As the crack lengthens by da, new surfaceis created and the energy of the solid is increased by:

2

2E

dU

dU

dV

E2

2s

dA

Area= (oa.ab)

= (s/E).s

o a

Fracture


62/112

62

The total change in energy of the system is then:

You would solve this equation for any structure that is cracked and

load bearing to determine if the crack will lead to catastrophic

failure.

actu e

dAdVE

dUtotal s 2

2

driving force for crack extension

materials resistanceto crack

02

2

dAdVE

s

Instability criterion for the

structure to allow the

crack to propagate and

lead to catastrophic failure

dUtotal = 0 (Critical)

dUtotal < 0 (Unstable)

dUtotal > 0 (Stable)

Determination of critical point

involves geometry, loading and

nature of the material.

Condition for crack propagation


63/112

6310

Condition for crack propagation

Values ofKI for some standard loads & geometries:

aKI s aKI s1.1

KI KIcStress Intensity Factor (SIF): Fracture Toughness: measure of a

materials resistance to brittle fracture

inksior

mMPa

Kofunits I :

Adapted from Fig. 8.8,

Callister 6e.

--Depends on load &

geometry. --Depends on the material,temperature, environment, &

rate of loading.

s

aa

s

2a2a

Y= correction

factor

Ys a

Ductile Materials: Plastic deformation at the crack tip,


64/112

64

If we looked at a

specimen that was thick

(no longer plane stress),

wed need to consider zz

in the DET approach.

2

2

2

16.0

y

IC

p

K

r sPlastic Deformation at the

Crack Tip and PlasticZone Sizes

Plane Stress (thin plate)

Plane Strain (thick plate)

Plane strain is the

conservative approach

needed for design, zone is

about 1/6th of plane stress

case

2

2

2 y

IC

p

K

r s

Using a DET

approach we

can determine the size

of the plastic zone, rp,

ahead of the crack tip.

If we looked at a

specimen that was

thin, wed only need

to consider stresses

the x and y direction.

z z

No plastic zone in brittle materials

F t


65/112

65

Fracture

How do we know if we have plane strain?

Start out with

2

2

2 y

ICp

Kr

s

and compare this to thickness of the specimen

if

if

rpc

t < 1/5 or 0.20

rpc

t> 1.0

Plane strain deformation

Plane stress deformation

rpc =

If it is in between: mixed situation


66/112

66

Chapter 8

Failure due to Fatigue

What is Fatigue? Cyclic Loading


67/112

67

Automobile scrap yards are filled with broken parts - the

majority of these components failed at stresses below the yield

strength. These failures were not a result of imperfections inthe material, but of the phenomenon called fatigue.

Fatigue is the name given to the:

______________________________________________

_____________________________________.

Eventual loss of the load-bearing capacity of a structural member subjected

tofluctuating loads repeatedly. The failure is inevitable.

Loads are much less than the tensile strength or even yield strength, but the

component still fails due to cyclic loading for numerous times. Many

electronic items we buy may fail after the warranty expires?

Time

yTS

Low cycle fatiguerelatively few cycles to

failure, below 103~104

High cycle fatiguemany cycles to failure

~107 or greater

Fatigue


68/112

68

g

Stress versus Time behavior:

Time

a= ao

The majority of engineering

failures involve cyclic (or

fluctuating) loading of oneform or another.

Examples of this kind of loading include:- Alternating stresses associated with a rotating shaft,

- Pressurizing and de-pressurizing in an aircraft fuselage

at takeoff and landing

- Load fluctuations affecting the wings during flight.

a = _______________

ao =

if m = 0

+

0stress amplitude

completely reversed

stress amplitude (amplitude on

tension side equals that on the

compression side)

One cycle

Mean stress, ma

Tension

Compression

Endurance, N = Number of cycles to failure, 103 to 1010 or infinite


69/112

69

du a ce, N Nu be o cyc es o a u e, 0 o 0 o e

ao = Reversed amplitude

S-N curve

e = Endurance limit, fatigue limit/strength

Endurance limit for some steels,

Mo, W (bcc metals)

No endurance limit for

Al, Cu, Au, Ag (fcc metals)

107

= No failure at 107

cycles

e

e

SAFE, no failure, N is

Statistical Variation


70/112

70

From this type of data we would find:

Log N

95% Survive

80% Survive

50% Surviveao

S-N plot implies an average. Unless otherwise noted, the line on

an S-N Curve implies that 50% of the specimens will

fail/survive at the stresses and N that the line represents.

Statistical Variation

Fail

Effect of Mean Stress


71/112

71

a

0 Stress T

en

sion+

-Co

mpression

a

0Stress Te

nsion+

-C

ompression

mean

m

min

max

Fully reversed stress

Cyclic stress superimposed on a mean stress

AB

Under identical stress histories, sample B would fail in fewer cycles than sample A

Mean tensile stresses tend to reduce the fatigue endurance while compressive

stresses will have the opposite effect. Compressive stresses are good to reduce

fatigue failures

a0

mean= 0

= ( max- min) a = ( max- min)/2 m = ( max+ min)/2R = min/ max

Fatigue ratio

Effect of Mean Stress


72/112

72

Mean stress is an important variable in the evaluation of a materials fatigue

response. Therefore, we need fatigue life data as a function of both ao and m.This information can be obtained from a simple linear relation. Many over the

years had proposed a correlation, but Goodmans correlation is widely accepted:

m

uts

The Goodman relation is strictly an empirical correlation.

m= __________=___________

a

ao

Goodman

equation

unsafe

safe

a = a0 [1- m/TS]

mean stress

tensile strengthuts

o

oo o o

o

m 0

Constant-Life Diagram

When a is 0, m= TS

a0 = a / [1- m/TS]m = ( max+ min)/2

Stress Fluctuation and Cumulative Damage Concepts


73/112

73

In addition to constant mean stress, stress conditions may also

vary with time. Combination of cyclic stresses.

Many structures are subjected to a range of _________________

____________________________.

How do these different cyclic histories contribute to

eventual failure?

Cyclic stress history for one aircraft flight

load fluctuations,

mean levels, and frequencies

IdleTake-off

Cruising

Landing

Stress Fluctuation and Cumulative Damage


74/112

74

Miners Rule

The task is to predict, based on constant amplitude test data,

the life of a component subjected to a variable load history. The

component has been subjected to many stress states, their

combined effect eventually contributed to the fatigue failure.

How to predict it?

We need a concept of damage to describe the contribution ofeach stress state i.

di = ni / Ni where: di = damage from stress state i

ni

= number of cycles applied ati

Ni = fatigue life at i(# of cycles tofailure if only condition i were applied)

i = i th stress level

Obviously the failure will occur when ni = Ni or di = 1

Miners Rule


75/112

75

Miner s Rule

Add fatigue damages

from all the stressconditions and set = 1

to predict the fatigue failure. 1 i

ii

Nnd

i = 1,2,3,. . . i = 1,2,3,. . .

D(d) = 1# of

Days/flights to

failure

Cumulative damage from

all cyclic stress conditions in

one day

Fatigue Crack Propagation


76/112

76

Two ways to predict fatigue endurance of a structural member: 1. Macroscopic

approach to test samples for finding out N w/o concerning the underlying defects, (2)

Microscopic approach to predict the number of cycles required to propagate a crack to a

critical size.

For this, information on fatigue crack growth rates at constant amplitude cyclic stress

with a0 size of a pre-existing crack, is necessary.

a

aj

ai

2 1

N

jadn

da

iadn

da

12 ss

growth is slow at first

ao

ao =size of pre-existing crack

It is reasonable to believe that the fatigue crack growth rate, da/dN, will be

proportional in some way to KI.

Dowling,Norman E. Mechanical Behavior of Materials Chapter 11: Fatigue Crack Growth. p. 545, 2007.

Fatigue Crack Growth/propagation, Growth rate curves


77/112

77

Crack growth in length = a by application of a number of

cycles by N,

Fatigue crack growth rate = da /dN

Stress Intensity Factor range: K

Dowling, Norman E. Mechanical Behavior of Materials Chapter 11: Fatigue Crack

Growth. Figure 11.5 p. 542, 2007.

P=

Depends

Upon geometry



78/112

78

So, da/dN _____ KI (stress intensity factor range)

If we plot log da/dN vs. log KI it will be as shown: There are

several interesting things emerging from this response:I. No propagating cracks, II. Linear relationship, III. Rapid crack

growth. IV. Threshold value of

log KI

logda/dN

I II III

Kth

Below Kth no crack propagation analogous to endurance

strength (limit).

Nopropagatin

gcracks

Powerlawrelationship

permitsfatig

uelife

calculations

Rapidunstable

fatiguecra

ckgrowth

YaK

YaK

I

I

s

s

DD

With cyclic loading,

use instead of :

= max - min

and

KI = ( a)1/2 Y

Kth

Paris Crack Propagation Law


79/112

79

The fact that we have a straight line suggests a power law relates da/dN

and KI.

The endurance, or number of cycles to failure, Nc , is what we are

concerned about. The Paris law can be utilized by solving for Nc

.

D-

D

c

o

c c

o

a

a nI

oc

N

N

a

a n

I

KA

da

NN

KA

dadN

)(

)(00

Need coefficients n and A for a material (see book)

No =

Nc =

da/dN = A( KI)n Paris crack propagation law (after P.C. Paris)

A and n are material dependent constants

actual # of cycles accumulated at

initial crack length a0, can be set to zer

critical # of cycles for failure

(# of additional cycles above No

needed for failure)



80/112

80

Upper limit

-- ac is found from the crack instability criteria.

KI,max = max( ac)1/2 Y = KIC

Lower limit

-- ao

set it equal to the minimum detectable limit at this point in

time 0.001inch

Set KI, max = KIC and solve

for ac

How to find upper and lower limits of cracks?: Biggest and smallest



81/112

81

For a small edge or surface crack of depth a in a tensile

stress field that ranges through , the stress intensity range is

approximately: KI = ( a)1/2

-sD-

-

-

2

2n

c

2

2n

o

2/nnc

a

1

a

1

A)2n(

2N

where:ao = initial crack size

ac = critical crack size

Nc = critical number of cycles ( # of cycles to failure)

Using Paris equation, prove that the number of cycles to failure, Nc is:

Chapter 9 and 10


82/112

The final exam will assess your ability to do the following:

Demonstrate your understanding of the viscoelastic response andwhich response corresponds to a given imposed condition

Interpret and utilize the Maxwell model for a viscoelastic material

response

Utilize the Larson-Miller approach to evaluate the time-to-rupture

Define the Larson-Miller parameter when given the material

parameter t0 or (C)

Ability to utilize the Larson-Miller approach when given amaster-life curve

Understand the steady-state creep approach to find out material

parameters.82

Viscoelasticity


83/112

83

1) Stress Relaxation - loss of gripping strength with time

2) Creep - deformation under constant load at a constant

temperature increases with time.

Eyeglasses with plastic frames lose gripping power with time Plastic bolts lose their gripping power

Nylon earring posts lose their gripping power

rubber band

turbine blades

Definition: Viscoelasticity is the relationship among time,

strain/stress, and temperature that can lead to permanent

deformation of a material (which we call creep) or a loss in

holding force (which we call stress relaxation) with time.

Undergo length

changes with time

Two types of manifestations of viscoelasticity:

Viscoelastic Model Elements


84/112

84

To understand and predict the mechanical response of viscoelastic

materials, we have to devise some mechanical models based on some

mechanical analogues that are consistent with these two properties:elasticity (instantaneous) and viscosity (time dependent).The linear elastic behavior of a material can be described by an ideal spring.

When a force is applied, the deformation is proportional to the force. The spring

responds immediately.

F = ___________________________________

For a force applied over a unit area this equation would become:

= ____________________

k * x

E *

deformation

spring constant

strain

Youngs Modulus

spring

Response is

Time Independent

Instantaneous response

Mechanical Analogue

for elasticity



85/112

85


How can we predict deformation with time? We will use a

different mechanical modeling element -- ______________.

Unlike the spring, it takes time to move a piston because ofthe gummy (viscous) fluid in it. Examples of dashpots:

______________________________________________.shock absorber and storm door closing mechanism

the dashpot

dashpot Piston-in-cylinder with aviscous fluid or air in it.

No- instantaneous response

Delayed response

Response is

Time Dependent



86/112

86

The simplest form of viscous fluid response is the Newtonian

fluid wherein:

= viscosity which is a measure of the materials resistance to

deformation with time (units of stress-time). Viscosity is temperature

dependent: decreases with increasing temperature and vice versa.

kT/H1

o

1 e D---

Activation potential energy

for sliding of molecules

dt

d

k = Boltzman constant =1.38x10-23J/oK or

k = 6.79x10-23 in-lb/oR

Viscoelasticity calculations depend heavily on

the viscosity variable, . The inverse of is

also known as fluidity, . The value of either

can be derived as a function of temperature.

Arrhenius type expression governs viscosity behavior vs T



87/112

87

The behavior of many materials lies between the behavior of a

spring or a dashpot and is described by a combination of theHookean (spring) and Newtonian (dashpot) elements.

A couple of possibilities:

time independentresponse

time dependent

response



88/112

88

(Maxwell Model)

The simplest viscoelastic model is a series combination of ourtwo elements. This viscoelastic model is called the

_______________.

To come up with our design equations well use the same steps

that we have used in the past in thermal problems (Chapter 3).

1) Force equilibrium

Maxwell Model

s = spring d = dashpot

2) deformation/compatibility

3) constitutive equations

R

Boundary Conditions


89/112

89

o

t

o

t

o

t

o

t

Stress relaxation

Creep: Time dependent viscous flowTime dependent strain

ResponseImposed Condition

1. Constant strain

2. Constant stress

Sun glasses,

Earring posts,

Nylon/plasticscrews

Silly putty

Rubber bands

Nylon strings

Wooden beams

Viscoelastic Model Elements(Maxwell Model)


90/112

90

(Maxwell Model)

Substituting the constitutive equations into the compatibility

step we get our governing equation for Maxwell material:

d s

d t1E

d

d t1

= + d

The total rate of change

in strain with time

equals the rate of change

of elastic deformationplus the rate of flow of

the material.

We now need to know whether our constitutive equation predicts real life

behavior. For this we have to model two specific behaviors (boundary

conditions) : stress relaxation and creep

d

d t1E

d

d t1

= +

Elastic response

Viscous flow



91/112

91

V

Governing differential equation for Maxwell solid

s

s 11

dt

d

Edt

d

Under stress relaxation conditions:

=______________=

dt

d

Separate variables and solve by integration

This equation indicates that the stress decays exponentially.

constant O ; = 0d

d t

0 s

s 11

dt

d

E

t = AeBt t



92/112

The equation governing stress relaxation of a Maxwell material is:

(Maxwell Model - Stress Relaxation)

e

t

This equation could be rewritten as:

et time constant or relaxation time =

t

The lnstress versus time relationship should be a straight line if

you have a Maxwell material (homework problem #1).

Where:

Viscoelastic Model Elements(Maxwell Model - Creep)


93/112

93

(Maxwell Model Creep)

Go back to our differential equation

s

s 1

dt

d

E

1

dt

d

o,0

dt

dss

s

t

dt1

d

1

dt

d

s

s

s

o

o

o

o

o

t

0

Creep is mechanical response where at constant load

or nominal stress strains accumulate with time.

o

t

Constant stress

xx

failure

Elastic

Solving by

Integration




94/112

94

(Maxwell Model - Creep)

Imposed condition

o

t

d

d t= 0

Response

t

s

o

o

Initially, only the spring responds. With time, the dashpot responds.

o

t

xfailurex

Elastic

contributionViscous flow

contribution

(t) = e + p

Viscoelastic Response Equation for a

M ll M t i l (S i t f t


95/112

Maxwell Material (Series arrangement of two

mechanical analogue elements)

Elastic,

Spring

Viscous flow,

Dash-pot

The total rate of change in

strain with time equals the

rate of change of elastic

deformation plus the rate of

flow of the material.

d 1 dsdt E dt= + 1

Governing Differential

Equation

Now apply boundary conditions:

1. Constant strain: Stress relaxation

2. Constant stress: Creep

Viscoelastic Model Elements(M ll M d l S R l i )


96/112

96

(Maxwell Model - Stress Relaxation)


o

t

Imposed condition -

strain is constant with time o

Actual behavior

Response predicted by

equation

Et

oe

s s

-




97/112

97

(Maxwell Model - Creep)

Imposed condition

o

t

d

d t= 0

Response

t

s

o

o


o

t

xfailurex

Elastic

contributionViscous flow

contribution

Alternate situation when the two imposed


98/112

98

conditions are not available

Elastic,

Spring

Viscous flow

Dash-pot

Take integration


99/112

99

Chapter 10Failure due to Creep considering

temperature and time effect

CreepD fi iti P t t i th t i f ti f


100/112

100

Definition: Permanent strain that increases as a function of

time and temperature under constant stress.

Creep is especially important in the design of chemicalprocessing equipment, electrical power generation equipment,

and aircraft gas turbine engines

Creep strains are important when the service temperature exceeds 30

to 40 % of the melting temperature (absolute degrees: Rankine orKelvin).

TIME

Response

t1

2

1

I II III

t2 t3

0Linear

Elastic

X

rupture

t

Imposed condition

TIME

0

Creep is the slow deformation of a material when it is under the influence ofCreep


101/112

101

stresses. It comes about as a result of long term exposure to levels of stress

that are below the yield strength of the material.

There are three stages of creep:I. Primarywhere plastic deformation occurs at decreasing strain rate. In

this stage the strain rate is relatively high, however it slows down with

increasing strain. This is due to work hardening.

II. Secondarywhere the strain rate eventually reaches a minimum and

becomes near constant (steady state). This is because of the equilibriumbetween work hardening and recovery or annealing due to T.

III. Tertiarywhere the strain rate exponentially increases with strain

eventually leading to fracture (rupture).

t1

2

1

I II III

t2 t3

0

I. Transientdecreasing --initial working

II. Steady stateconstant --(most important)

III. Tertiaryincreasing --failure instability

Creep rate = d /dt

At constant T


102/112

102

I

II

III

At constant T

0 Isolated voidsMovement of

dislocations

Voids making

clusters

Clusters of voids coalasce

making microcracks

Macrocracks

leading to

catastrophic

failure

Creep


103/112

103

Two types of design situations occur in engineering practice

1. Design is limited by the total amount of strain that can accumulateor be tolerated by component.

(Steady State Creep)

2. Strains are not much important but lifetime of the component prio

to failure is critical. (Creep Rupture)

What we want

= f

Example: Turbine blades in a

jet engine, will elongate at elevatedservice T until hit the casing/housing

Example: High pressure steam lines, saggingdue to strain not a worry, but rupture is.

(material, , T, t)

Lead steam pipe

1. Steady State Creep

Many predictive equations have been proposed, for our analysis, let us start


104/112

104

tt ssn

o

Transient terms (stage I) are often insignificant in

comparison to steady state (ss) term

total=

ss = = steady state or linear creep rate

total=

t1

2

1

I II III

t2 trtime

tsso

12

12

tt -

-

ignoreinstantaneous transient

Steady state

0

ss

t= time, T=temperature,

ss = steady state strain rate

with strain accumulated as the sum of the following 3 contributions:

and n depend upon stresstemperature and the material.

For most materials 0


105/112

105

An Arrhenius type rate equation of the

form:

Where:

A = coefficient, material and stress dependent.

H = activation energy to derive creep.

k = Boltzmans constant (k = 1.38x10-23 J/K ork = 6.79x10-23 in-lb/R)

T = temperature (absolute degrees, oK or oR)

Kelvin = 273 + C (metric units)

Rankine = 460 + F (English units)

Ae - H/kTss =

describes steady state creep.

2. Creep Rupture


106/112

106

I II III

t

I

II

III

At constant T

0

ss= ( / 0)m e- H/kT Dorn-Miller creep

rate equation.

ss= B m

Ae

- H/kT

ss=

log ss= logB + m log

These curves imply that creep rate has power law dependence on

Steady State Creep


107/112

107

These curves imply that creep rate has power law dependence on

stress, i.e.

ss

ss= ( )m e- H/kT

, m, and H are material parameters.

In dimensionless form: Dorn-Miller creep

rate equation.

A

ss= ( / 0)m e- H/kT

m

log

log

s

s

10

10

Slope = m

= (1/ 0)m e- H/kTss= B m Where B

logss

= logB + m log Problem #1 of Home Work

Ae- H/kT

ss =

Simplified version of D-M eqn.

Creep Rupture (Catastrophic Failure)


108/112

108

Some design situations will be constrained by complete

failure due to creep. This is called creep rupture. (failure)

Rupture rate of a temperature dependent process = r

Arrhenius type equation

r = Ae - Hr/kT

r 1

tr

A = material parameter = t0-1

May be stress dependent

Activation energy here is for

rupture process and is different

from the activation energy needed

for steady state in earlier equation

tr-1 = to

-1e- Hr/kT

For creep rupture to occur, strains must

accumulate to the point where strain induced

voids grow and form microcracks which lead

to rupture. (the material on right is unstable)

Taken from

www3.toshiba.co.jp/ddc/eng/

materials/e_tan_3.htm

Grain boundary

Voids, holesInter-grainular failure

At grain boundaries

tr = time to rupture

Creep Rupture

L Mill E ti


109/112

109

Taking logs of both sides

rearranging terms

look at a collection of data (tr vs. T, and .)

There are two possibilities:Case 1 - different slope and same y intercept

Case 2 - same slope and different y intercept

-logtr = -logto - log eHr

kT log e = .434

.434 Hrk T

= - log to + log tr

Activation energy forcreep rupture

b yxm

Hr dependent

to dependent

tr-1 = t0

-1 e - H/kT

Larson-Miller Constant,

C = -logto

Larson-Miller Equation

For creep rupture

Going back to our equation

H


110/112

110

P = T( absolute)[C + log tr]

Move T over

Larson Miller

Master Life Curve

log

HrkT

= - log to + log tr0.434

Larson Miller Constant C (material dependent)

Hrk

= T (absolute) (C + log tr) = P0.434

P, function of T, t, materialfunction of stress

P = Larson-Miller

ParameterActual shape of the curve

depends on the material

and is the result of

numerous tests.

Average value is 20

(if not given use 20)

12-

Steady state

Summary and Equations


111/112

111

tsso

12

12

tt -ss=

total=

ss= ( / 0)m e- H/kT

ss= Bm

Ae - H/kTss =

log ss= logB + m log

P = T( absolute)[C + log tr]

HrkT

= - log to + log tr0.434

tr-1

=

to-1

e-H/kT

C = - log10t0 Slope = 0.434 Hr/k

Steady state

Elastic

Larson-Miller Parameter

Larson Miller Constant

Steady state creep

Creep Rupture


112/112

Documents

Review Final Exam Fall 12