Four Point Bending From Theory to Practice and StandardsFour Point Bending From Theory to Practice and Standards Ad Pronk University of Delft; The Netherlands 2 nd European 4PB Workshop

Four Point Bending

From Theory to

Practice and Standards

Ad Pronk

University of Delft; The Netherlands

2nd European 4PB Workshop

University of Minho, Guimarães

24-25 September 2009

2nd European 4PB Workshop, 24/25-09-2009, Univ. of Minho

Ad Pronk ; Delft University of Technology

CONTENTS• “”QUIZ””• 4PB Theory• CEN Standard

(Shear - Calibration - Damping - Device Stiffness)

• 4PB Practice (Calibration - Stiffness Modulus - Fatigue)

• Fatigue Damage Model – Partial Healing• Healing – Endurance Limit ?• Temperature



The following slides of a table are an idea of my colleague and friend Rien Huurman.

He presented these slides already on the 1st

4PB workshop which was held in 2007 at the Delft University of Technology.

In my view this example is an excellent “tool” to visualize the “mistakes” we can make.

I have extended his example a bit.

LOAD, GEOMETRY & MATERIALThis is a top view of a table.What is the force in each leg?This is not a joke!

Rien Huurman1st 4PB Workshop



LOAD, GEOMETRY & MATERIAL

This is the 20 cm x 20 cm load area. The load is 100 kg.




There are

4 legs.




Indication of thedeformations and the stress within the material.




The materials!

Steel

Rubber




“HOLY TRINITY”: LOAD GEOMETRY (structure) MATERIAL Including Boundary/Time Conditions





1) Load 2) Geometry

Response3) Material+ Boundary conditions

⎫⎪⎪⎪ ⇒⎬⎪⎪⎪⎭



???

1) Load 2) Geometry

Correct Material Model+

Response

⎫⎪⎪→⎬⎪⎪⎭

QUESTION



-1

-0,5

0

0,5

1

Uni-axial Push-Pull (UPP) Test

Force

Deflection



Huet-Sayegh

Model

This model covers

already a wide

frequency range



00,20,40,60,8

1 Force

Response

Permanent



Modified

Huet-Sayegh

Model

Linear Dashpot in series



Another Example: 4PB Fatigue testing with Haversine Deflection Signals



0

0.5

1

1.5

2Haversine

Force

Deflection

You want to have this loading



0

0.5

1

1.5

2

Time

Def

lect

ion

-1

-0.5

0

0.5

1

Forc

eDeflection

Force

But this is what you get !!!!



Why don’t you get what you want?

REASON

Asphalt is a viscous-elastic-plastic material

There are two causes why you can’t remain the desired wave shape for deflection & force



1) Any static component in the applied loadwill lead to permanent deformation.

The Result: Within a few cycles you have a a bended beam and the force becomes a sine.In order to get back to a zero deflection you need to pull/push back the bended beam.



2) When a constant deflection is applied

(as in a Haversine signal), the Force (or better the stress in the material)to obtain this deflection level willdisappear (relaxation)

The Result: A STRESS FREE BENDED BEAM



ASTM “Constant Strain Haversine”

Using the total strain (peak-peak value) an endurance limit of around 100 μm/m or more is calculated !!!!.

But, according to me, this is not correct and the endurance limit is half this value: 50 μm/m



Four Point Bending: THEORY



A prismatic beam resting on 4 supports

The inner supports can move up and down

When the top of the beam is in tension the bottom will be in compression and vice versa

The centre line of the beam will keep itsoriginal length



Boundary Conditions for Beam Theory

• Free rotation at supports

• Free horizontal translation at supports

• No vertical deflection at outer supports

• Neutral “fiber” halfway the beam



Boundary Conditions

•Deflection V = 0 at x = 0 & x = L

•Moment M = 0 at x = - Δ & x = L + Δ

•Shear Force D = 0 at x = - Δ & x = L + Δ

•1st derative of V = 0 at x = L/2

•Only Steady State Solution



Applied Load

Sinusoidal Point Load at the two inner supports: F = Fo .Sin(ω t)

Response is a delayed deflection:

V.Sin(ω t + ϕ )



At this point my first thoughts were to deal with the background of the differential equations for the bending of a slender beam: Thimoshenko.

But I’m afraid I will loose your attention.

Nevertheless you will see quite a lot of slides with equations. But I will go through them quickly and only point out some details.



2 2 2 4*

2 2 2 2 2{ , } { , } { , }b b bE I V x t B H V x t I V x tx x t x t

ρ ρ⎧ ⎫∂ ∂ ∂ ∂

+ − +⎨ ⎬∂ ∂ ∂ ∂ ∂⎩ ⎭

Ψ

4 2 2*

* 4 2 2{ , } { , } { , }{ , } b bB H I V x t E I V x t Q x t

G B H t t xρ

ρ⎡ ⎤⎧ ⎫∂ ∂ ∂

+ − =⎨ ⎬⎢ ⎥∂ ∂ ∂⎩ ⎭⎣ ⎦

Ψ2 * 2 2 2*

2 2 2 2

{ , } { , }sG B HI E V x t

B Ht x t xρ

ρ⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂

− − +⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦

Ψ2 2 2

* *2 2 2{ , } { , } { , }s

IG B H V x t E Q x tB Ht t x

ρρ

⎡ ⎤∂ ∂ ∂+ = −⎢ ⎥∂ ∂ ∂⎣ ⎦

General Differential Equations



b bE I V x t bh V x t Q x tx x t

2 2 2*

2 2 2( , ) ( , ) ( , )ρ⎡ ⎤∂ ∂ ∂

+ =⎢ ⎥∂ ∂ ∂⎣ ⎦

* * iE E .e ϕ=

BENDING DEFLECTION Vb

{ } { }2

2mass total massF t Mass V x ,tt

∂=

∂



( ) { }

Ψ

Ψ

sG b h V x t Q x tx

EG b h b h

2*

2

**

{ , } ( , ) ( , )

& ,2 1

αμ

∂= −

∂

= =+

SHEAR DEFLECTION Vs

{ } { }2

2mass total massF t Mass V x ,tt

∂=

∂



( ) ( )

( )

( ) ( )

( )

Δ Δ

Δ Δ

t t

nt

t

t t

nt

t

ASin n Sin nL LF

Q x t Sin tL xSin n

L

ACos n Cos nL LF

Q x t SinL xCos n

L

0

1

0

1

2 1 2 12

( , ) ( )

2 1

2 22

( , ) (

2

π π

ω

π

π π

ω

π

∞

=

∞

=

⎡ ⎤⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪− − − ×⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭= ⎢ ⎥

⎛ ⎞⎢ ⎥× −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪− ×⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭= ⎢ ⎥

⎛ ⎞⎢ ⎥× ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∑

∑ t)



b bE I V x t bh V x t Q x tx x t

2 2 2*

2 2 2( , ) ( , ) ( , )ρ⎡ ⎤∂ ∂ ∂

+ =⎢ ⎥∂ ∂ ∂⎣ ⎦

* * iE E .e ϕ=

BENDING DEFLECTION Vb



{ } { } { } { }( )= + +i t i tb a c dV x e V x V x V x eω ω

Solutions Homogenous dif. Equation

Particular Solution satisfying the load condition



( ) ( )

( ) ( )

( )

( )

( )

Δ Δ

n

t t

n nt

an

Sini Arctg

Cos

t

ASin n Sin nL L

n Cos

F LV x

E I

xSin n eL

2

22 1

2 44

2 42 1 2 13

04 *

1

2 1 2 1

2 1 1 2

2{ }

2 1

ϕ

ωϕξ

π π

ω ωϕ

ξ ξ

π

π −

− −∞

= ⎛ ⎞⎜ ⎟⎜ ⎟

− ⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞+− − −⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥×⎢ ⎥⎢ ⎥− − +⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

⎛ ⎞⎢ ⎥× − ×⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥

⎣ ⎦

∑

( )4 422 1 4

2 1−

−=

*

nt

E In.

L bhπ

ξρ



{ }

( ) ( )

( )

( )

{ } ( ) ( ) ( )Δ Δ

nia n n

n

tn

n n

n

n

nt t t

V x A T x e

F LA

E I n Cos

SinArctg

Cos

A xT x Sin n Sin n Sin nL L L

*2 1

2 1 2 11

30

2 1 2 444 *

2 42 1 2 1

*2 1 2

22 1

2 1

{ }

2

2 1 1 2

2 1 2 1 . 2 1

ϕ

ω ωπ ϕ

ξ ξ

ϕϕ

ωϕξ

π π π

−

∞−

− −=

−

− −

−

−

−

⎡ ⎤= ⎣ ⎦

=

− − +

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟

−⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+= − − − −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

∑



( ) i

c n tn

V x C Cos x L e 42 1 0

1{ } 2

ϕ

β∞ −

−=

⎛ ⎞= −⎜ ⎟

⎝ ⎠∑

( ) i

d n tn

V x D Cosh x L e 42 1 0

1{ } 2

ϕ

β∞ −

−=

⎛ ⎞= −⎜ ⎟

⎝ ⎠∑



( ) ( )i i

n t n tC Cos L e D Cosh L e4 42 1 0 2 1 02 2

ϕ ϕ

β β− −

− −

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

{ } ( )

( )

Δ Δ

Δ

nii

n n n t

i

n t

A T e C Cos L e

D Cosh L e

*2 1 4

2 1 2 1 2 1 0

42 1 0

2

2 0

ϕϕ

ϕ

β

β

−−−

− − −

−

−

⎛ ⎞+ −⎜ ⎟

⎝ ⎠⎛ ⎞

+ − =⎜ ⎟⎝ ⎠

Boundary Conditions



{ } { }

( )

( )

nib a n n

n

i

c n tn

i

d n tn

V x V x A T x e

V x C Cos x L e

V x D Cosh x L e

*2 1

2 1 2 11

42 1 0

1

42 1 0

1

{ }

{ } 2

{ } 2

ϕ

ϕ

ϕ

β

β

−

∞−

− −=

∞ −

−=

∞ −

−=

⎡ ⎤= = +⎣ ⎦

⎛ ⎞+ = −⎜ ⎟

⎝ ⎠

⎛ ⎞+ = −⎜ ⎟

⎝ ⎠

∑

∑

∑

Complete solution without extra mass forces



( ) { }

Ψ

Ψ

sG b h V x t Q x tx

EG b h b h

2*

2

**

{ , } ( , ) ( , )

& ,2 1

αμ

∂= −

∂

= =+

SHEAR DEFLECTION Vs



( ) ( )

( )

( ) ( )

( )

Δ Δ

Δ Δ

t t

nt

t

t t

nt

t

ASin n Sin nL LF

Q x t Sin tL xSin n

L

ACos n Cos nL LF

Q x t SinL xCos n

L

0

1

0

1

2 1 2 12

( , ) ( )

2 1

2 22

( , ) (

2

π π

ω

π

π π

ω

π

∞

=

∞

=

⎡ ⎤⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪− − − ×⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭= ⎢ ⎥

⎛ ⎞⎢ ⎥× −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎡ ⎤⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪− ×⎢ ⎥⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭= ⎢ ⎥

⎛ ⎞⎢ ⎥× ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∑

∑ t)



{ }

( )( )

{ } ( ) ( ) ( )Δ Δ

*2

*2

2 21

2*0

2 222 *

2

{ }

2 12

2

2 2 . 2

n

n

ig n n

n

tin ni

t

nt t t

V x H U x e

LFH e

L E e b h n

A xU x Cos n Cos n Cos nL L L

ϕ

ϕ

ϕ

μϕ ϕ

π α

π π π

∞−

=

−

+

⎡ ⎤= ⎣ ⎦

+= ⇒ =

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+= −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

∑ { } 0,eV x t H=



Applied Solution for Back Calculation

• First order approximation for Vb

• Neglect overhanging beam ends (Δ=0)

• Neglect in first instance the deflection Vs

→

Solution is similar to the solution of a viscous- elastic spring mass system with complex spring constant K and an equivalent mass Mequiv



Applied Solution for Back Calculation

{ }{ }

( )

( )

{ }

{ }{ } { }

{ }

*0 13

0

4

* 2 2 40 0

*

3

2

,1 2

i t

b

equiv

R xA xSin

F L eV x tR x E I Cos

R x

Sin

E IM xK x

L

LK x

L

ω ϕ

ϕ ζ ω

π

π

ζ ω

ζ

π

−

=⎛

=− +

= =

⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



Replacing R{x} by R*{x} of the pseudo-static

solution leads to smaller deviations.

{ }*2 2

12 1

3 3

LR xA x x A

L L L

=⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



Back Calculation

{ } { } { }( ) { } { }( )

{ }( ){ }( ) { }

{ } { } { }

{ } { }{ } { }

{ }( )

32* *0

*

*

*

2. 0

0

*,*

. 24

,

1 2b

bequiv

i mass ibeamequiv

mass i

F LE Cos x Z x Z x

V x R x I

Sin x

Cos x Z x

V xZ x M x

F

R A M xMM x R x

R x

ϕγ

ϕϕ

ϕ

ω

π

= + +

=+

=

⎛ ⎞⎡ ⎤⎜ ⎟⎢ ⎥= +⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

∑



CEN Standards

“Don’t shoot the pianist, he is doing his best”

I wasn’t allowed to finish the ‘job’

• Deflection Vs due to shear

•

Calibration - Reference beams - Damping Term

• 4PB Device Stiffness



In the EN standards of CEN only the deflection Vb

is used for the determination of Smix . The deflection

due to shear forces is not taken into account.

The effect of extra mass inertia forces is “dealt” with

but the coupling with the shear deflection is ignored.

How to deal with this ?



Using the complete analytical solution

is not practical for routine back calculations.

So, another solution (patch) is needed.

Proposal: Use the pseudo-static solution as a

starting point.



Online correction for deflection Vs due to shear

( )( )

( )

b t

LVs HL L AVb

L HA ProblemL

L LV V

2

2 2

2

4 123 4

2

36 1???

3 23

0.962 2

μα

υ αα

⎧ ⎫⎨ ⎬ +⎩ ⎭ =⎧ ⎫ −⎨ ⎬⎩ ⎭

+ ⎛ ⎞= → =⎜ ⎟⎝ ⎠

⎛ ⎞ ⎧ ⎫≈ ⎨ ⎬⎜ ⎟⎝ ⎠ ⎩ ⎭



Procedure for the estimation of α

Compare the analytical solution with 3D

finite element calculations.

Problem: How to “load” a 3D beam



Constant Deflection Amplitude

The constraints are due to ‘touching’ the beam



Is it possible to bend a 3D beam without

physically touching it?

Answer: Yes, by applying shear forces in

the interfaces at the locations of the

supports (“fooling” ABACUS) and by

using the principle of St. Venant

2nd European Workshop on 4PB, 24-25 September 2009, University of Minho, Guimarães, Portugal

1st Step 2nd Step

Deviations


-25.0

-16.7

-8.3

0.0

8.3

16.7

25.0

0.00 0.05 0.10 0.15Shear stress [MPa]

Hei

ght [

mm

]

mid clamp response

1st response at clamp

2nd response at clamp


Cowper 1966 (10+10μ)/(12+11μ) 0,8517

Timoshenko 1974 5/6 0,8333

Timoshenko 1922 (5+5μ)/(6+5μ) 0,8710

Olson 1935 (20+20μ)/(24+15μ) 0,9231

Pickett 1945 24,612(1+μ)/(29,538+5,942μ+64,077μ2) 0,8419

Tanji 1972 (6+12μ+6μ2)/(7+12μ+4μ2) 0,9354

Pai 1999 5/(6+(μ/(1+μ))2(H/B)4[1-

(90/π4)Σ{Tanh(nπB/H)/(n5(π(B/H))}])

0,8269

Hutchinson 2001 5(1+μ)/(6+5μ-(μ/(1+μ))2(H/B)4[1-

(90/π4)Σ{Tanh(nπB/H)/(n5(π(B/H))}])

0,8763

ABACUSABACUS 20092009 Comparison analytical solutions and finite element Comparison analytical solutions and finite element calculationscalculations

0,8590,859



In the former slides I have dealt with most

of the theoretical aspects of the 4PB test.

Is that enough for the practical application

of the 4PB test?

In theory YES but in practice NO.

Calibration of the 4PB device is required !!



At the moment an European calibration

4PB test is “running” and it turns out to

be very complex and time consuming.

For calibration you need a reference material with a KNOWN stiffness modulus

E.g. Aluminum: 71.3 GPa



• The product of E.I is the relevant parameter.

• Calibrate over a wide range of EI values.

If calibration fails due to not traceable errors

• Try to model your device with FEM

•

Introduce an extra term: Damping correction coefficient.



Lesson I learned:

Never underestimate calibration efforts

Sometimes calibration seems to be overdone in view of the natural variation in material properties of asphalt mixes.

But it is, at least in my view, necessarily to carry out high quality calibrations in order to exchange results between laboratories and to investigate (new) material models.



In a glance I showed you some theoretical aspects of the 4PB device and how to deal with these aspects.

When I was writing the EN standards I tried to ‘help’ 4PB device owners by introducing a so called damping term. Sometimes you have checked “everything” but still a deviation is present in the system.



The damping term is meant for these cases.

E.g. You back calculate too high E values in your calibration tests. In spite of all efforts the ‘error’ is not traceable but you have the idea that there is too much friction at the supports.

For such a case I introduced the damping term which, in my view, is allowed after extensive testing (different frequencies etc.)



Too low back calculated E values

In principle you can use a ‘negative’ damping term, but I advise you to check the clamping forces and “device stiffness”.

Commonly a default clamping force is used in all cases. However, I think you should verify this using reference beams with different beam stiffness's E.I.



4PB Device Stiffness

The bending beam theory is based on the assumption that you have an infinitely stiff bending device. But of course this is not true.

Using a relative deflection measurement (supports resting on the beam) this wouldn’t matter. But in case of an absolute deflection measurement you should correct for the device stiffness by using a very, very stiff beam.



“PRACTICE”

• Calibration Protocol

• Stiffness Modulus Measurements

• Presentation of Fatigue Results



Calibration Protocol

•

Perform at least once a year a complete calibration test with at least three reference beams (representing E values for asphalt ranging from 6 to 12 GPa).

•

Perform in front of each 4PB project a routine stiffness test with a “DELRIN” beam and start a data base of the obtained figures.



Calibration Protocol (Extra)

Even the E value of a reference beam is not known ‘exactly’. Therefore, given the small applied strains, a very good indication (or verification) can be obtained by tests like:

• Pulse velocity measurement

• Resonance frequency measurement



Stiffness Modulus - EN Standards

• Frequency sweep

• Limit number of cycles & strain amplitude

• Practice Experience: ε < 50 μm/m

• No indication of fatigue damage



Frequency Sweep Test

•

If possible perform a frequency sweep test in front of each 4PB (fatigue) test.

•Test the beam in controlled deflection mode.

• Limit the strain amplitude to 40 – 50 μm/m.

•

End the frequency sweep test with the start frequency (e.g. 1-2-4-6-8-10-1 Hz)



Fatigue Measurements – EN Standards

• Why only a ε6 strain value?

•

Why are fatigue lives Nf,50 different when measured with different devices?

→ Need for Fatigue Damage Models



For Type Testing only the strain value ε6 is required in the EN standard.

This is strain amplitude for which the life time Nf,50 is 106 cycles.

So, the whole Fatigue characteristics are ‘combined in a single data point !!!



Personally I felt this a big omission.

At least the slope of the Wohler curve should be given as well.

Furthermore 2 temperatures should be used.

A confidence interval should be included.



In the EN standards different devices are allowed for the determination of the stiffness

and fatigue characteristics of asphalt.

Stiffness is not a problem but Fatigue is !

Nf,50 is different for different devices.

How to solve the problem?



Fatigue Damage Models

Starting point has to be a MATERIALMATERIALMATERIAL model

But in practice you measure the response of a SPECIMENSPECIMENSPECIMEN

Conclusion: The geometry of the specimen plays a large role



Uni-Axial Push-Pull Test

UPP

Herve DiBenedetto

In theory the response of the specimen is equal to that of the material



Force

Maximum

Strain

Two Point Bending Test

2PBLCPC, France

OCW, Belgium

Specimen Response

is not equal to

Material Response



The same is true for 4PB testing

•

Fatigue damage is less in outer sections (smaller strains and stresses).

•

Fatigue damage decreases from top and bottom inwards the beam and is theoretically nil at the ‘neutral’ fiber.

⇒ MATERIAL DAMAGE MODEL



Partial Healing (PH) Model•

It is a MATERIALMATERIAL model. After integration over the stress/strain field in the specimen the response for the SPECIMENSPECIMEN is obtained

•

Equations for the SPECIMEN response are not the same as the PH equations and can only be determined in a numerical way

•

Equations of PH model are used for a fit on the SPECIMEN response in order to get SEED values for the numerical determination



However,

The shape for the evolutions of the phase lag and weighed stiffness modulus of the specimen using the PH material model parameters in combination with a numerical integration over the specimen is nearly the same as the one obtained by a regression using the PH model equations directly for the specimen response.

t dQ tF t F e do d{ } .( ){ } . .1 10

τ β τα γ ττ

− −⎡ ⎤= − +∫ ⎢ ⎥⎣ ⎦

t dQ tG t G e do d{ } .( ){ } . .2 20

τ β τα γ ττ

− −⎡ ⎤= − +∫ ⎢ ⎥⎣ ⎦

Storage Modulus G{t}=Smix .Cos(ϕ)

Loss Modulus F{t}=Smix .Sin(ϕ)

disdis

Wd dQ t W t f F tdt dt T

π εΔ 2

0{ } { } { }=⎡ ⎤⎣ ⎦2nd European 4PB Workshop, 24/25-09-2009, Univ. of Minho




Healing – Low Endurance Limit

Can the PH model explains Healing?

Answer: Only Partially!



Beam 26-02

2000

3000

4000

5000

6000

7000

8000

0 40000 80000 120000 160000Number of Load cycles

Stiff

ness

mod

ulus

[MPa

]

Measured "Predicted"

Fitting on 1st

Load period only



The PH model underestimate the amount of Healing.

Given the “Strength/Power” of the PH equations what will be the evolution if the first 2 load periods are used for the

fitting?



3000

4000

5000

6000

7000

8000

0 40000 80000 120000 160000Number of Load cycles

Stiff

ness

mod

ulus

[MPa

]

Measured "Predicted"

Fitting on 1st & 2nd

Load period



A Healing project is submitted to the Centre of Transport and Shipping [DVS] (former DWW) of the Ministry for Transport, Public Works and Water Management.

The project includes 5 asphalt mixes which will be tested at 20 oC and 30 Hz using a rest- load ratio of 10 and different load periods.



What might be a reason that the PH model underestimates the amount of healing?

1. There is at least a second reversible damage term with a small β value. It can’t be determined in continuous tests.

2. Irreversible Fatigue damage occurs above a certain strain/stress level → Low Endurance Limit



LOW ENDURANCE LIMIT

“Does it exists?”



Does the PH model includes

an endurance limit?

In the first version: NO !

Regardless the value for the applied strain

amplitude permanent damage will occur



In the (original) PH model the following assumption is made:

The reversible and irreversible fatigue damages per cycle can be written as the

products of CONSTANTS times the Dissipated Energy per cycle.

→ α1 , α2 , β, γ1 , and γ2 are constants



How can we investigate the validity of this assumption:

One possibility is carrying out UPP Tests.

E13D 180-17 α1 α2 β γ1 γ2Individual fit 83 1187 51440 44 127

E13D 80-14 α1 α2 β γ1 γ2

Individual fit 82 850 50120 3 17



0

20

40

60

80

100

120

140

0 50 100 150 200

Strain amplitude [μm/m]

Para

met

er v

alue

γ1 &

γ2

Endurance Limit ?



Consequences of an Endurance Limit

If an Endurance Limit exists it may explain the underestimation of the healing effect.

Also the damage will be more concentrated in the top and bottom of the mid span of the beam. It has to be investigate what the influence of this non-homogeneity will be on the assumption: constant deflection means constant strain !!!



TEMPERATURE

The viscous dissipated energy per cycle Wdis = π σ ε Sin(ϕ) in a fatigue test is ‘completely’ transformed into Heat.

Due to this the temperature will increase and as a consequence the Stiffness modulus will decrease. How serious is this?



E13D 80-14

0,00,10,20,30,40,50,60,7

0 1000000 2000000 3000000 4000000 5000000

N

Tem

pera

ture

incr

ease

[o

C]

Centre oCEdge oC

Temperature Increase During Fatigue

UPP Test ε = 82 μm/m



E13D 180-17γ =0.604 ; δ=− 1.43 Ε(−05)

0,0

0,5

1,0

1,5

2,0

2,5

0 20000 40000 60000 80000 100000

N

Incr

ease

in

Tem

pera

ture

[oC

]Centre oCEdge oC

Forced Convection

Heat Transfer Coefficient h = 40 W/m2/K

UPP Test ε = 172.5 μm/m



E13D 180-17

0200040006000

8000100001200014000

0 20000 40000 60000 80000 100000 120000

N

Stiff

ness

[Mpa

]

0,0

0,5

1,0

1,5

2,0

2,5

Tem

pera

ture

In

crea

se [o

C]

Decrease in Stiffness due to

Temperature



E 13D 180-17H eat T ran sfer h =4 kg /s3 /o K

0,0

1 ,0

2 ,0

3 ,0

4 ,0

5 ,0

6 ,0

7 ,0

0 50000 100000 150000

N

Tem

pera

ture

Incr

ease

[o

C]

Centre oC

Edge oC

Free Convection instead of Forced Convection




E 13D 180-17H eat T ransfer h =13 kg /s3/o K

0,0

0,5

1,0

1 ,5

2 ,0

2 ,5

3 ,0

3 ,5

4 ,0

0 50000 100000 150000

N

Tem

pera

ture

Incr

ease

[o

C]

Centre oC

Edge oC

“Bad” Forced Convection




4PB test, ε = 200 μm/m, T = 00C



Source: LCPC; Chantal de La Roche



Conclusion/Recommendation

Use Force Convection in the 4PB cabin

Check by temperature measurements the effectiveness of the cooling (using temperature

couples inside the specimen)

The heat ‘creation’ is related to: σ.ε or (ε 2), Temperature increases fast with higher strains





Documents

Four Point Bending From Theory to Practice and StandardsFour Point Bending From Theory to Practice and Standards Ad Pronk University of Delft; The Netherlands 2 nd European 4PB Workshop