FMDF cours 181025 - lem3.univ-lorraine.fr

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A. Zeghloul Fracture mechanics, damage and fatigue – Stress concentration 1

The stress gradient at the root of the hole ( ), which will be noted to simplify

the writing ( ) , is obtained by differentiating the following relationships :y x a

x a

d dxσ =

=

)0 ( )

1( )

Elliptical hole

y y N x

D x

σ

σ=

∞ = +

) 2 4

0 1 31

2 2

Circular hole

y y a a

x x

σ

σ=

∞ = + +

2 4

3 5

6 7

Circular hole

y

x ax a

d a a

dx x x a

σ σσ∞

∞

==

= − − = −

( ) ( )( ) ( )

2 2 2 2 2

2 2 2 2 2

( ) ( 2 ) ( )With

( )

N x a a b x x c x c ab a b x

D x a b x c x c

= − − − − + −

= − − −

))

))

2 2 2 3

2 2

( ) 2 ( ) ( ) ( )From where and

'( ) ( ) (2 3 ) '( ) 3 ( )x a x a

x a x a

N x ab a b D x a b b

N x a a b a b D x ab a b= =

= =

= − = −= − − = −

3

(4 3 )

Elliptical hole

y

x a

d a a b

dx b

σσ ∞

=

+= −


The two previous relationships, can be put in a unique form

maxy

x a

d

dx

σ σγρ=

= −

The coefficient of proportionality γ has the same

expression for the two hole geometries

12

tK

γ = +

max

Circular hole

3aρ σ σ ∞= =2

max

Elliptical hole

1 2b a

a bρ σ σ ∞ = = +

2 3γ< <

7

Circular hole

y

x a

d

dx a

σ σ ∞

=

= −

3

(4 3 )

Elliptical hole

y

x a

d a a b

dx b

σσ ∞

=

+= −

max max7 17 2

3

Circular hole

ta K

σ σ σρ ρ

∞ − = − = − +

1/

max max4 3 4 32

2 2

Elliptical hole

tK

a b a b b

b a b a b

σ σ σρ ρ ρ

∞ + + − = − = − + + +

��

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Figure 7 gives the variations of stress σy along the axis of the notch,

for an elliptical hole (a/b=3, Kt=7) and for a circular hole (Kt=3).

This figure shows that the decrease of the stress is faster in the case

of the elliptical hole2.

Figure 7 : Changes in the stress

along the axis for an elliptical

hole and a circular hole2

Elliptical hole

a/b=3

Circular hole


Tutorial 3 : Stress gradient

A notched plate is loaded in tension. The stress gradient at the root

of the notch with stress concentration factor Kt and with notch

bottom radius ρ, is given by :max

maxis the maximum stress reached1

2 where at the root of the notch

y nom

t

t

dK

dx K

σ σ σ σρ

= − + =

Determine by linearizing this relationship

a- Distances δ along the axis of a 5mm diameter circular hole, where

the stress drops are respectively 2,5% and 10%.

b- The average stress in all grains adjacent to the root of a notch

where the stress drop does not exceed 10%. The plate loaded at σnom

= 20MPa, is made of low-alloy manganese steel with ferrite - pearlite

structure whose average grain size is dg=12µm. The notch is

elliptical in shape with a=2,5cm and b=0,5cm.

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c- the notch root radius ρ with factor Kt=5, knowing that the stress

drop is 5% at a point away than 50µm from the root of the notch.

Same question for a distant point 100µm from the root of the notch.

d- The stress concentration factor Kt of a notch root radius ρ=2mm,

knowing that the stress drop is 2,5% at a distance of 20µm.

e- The minimum and maximum distances where the stress drop is

5% for a notch whose root radius is ρ=1mm.


a- For a circular hole in a plate loaded in tension, the stresses

concentration factor is equal to 3. The notch root radius being

ρ=2,5mm (5mm diameter), the distance δ where the stress drop is

2,5%, is given by :

max max0,025 12

3 2,5

d

dx

σ σ σδ

− = = − +

The distance δ where the stress drop is 10%, is given by :

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b- The stress concentration factor is Kt=1+2*2,5/0,5=11. The

maximum stress is σmax=11*20=220MPa and the notch root radius is

ρ=0,52/2,5=1mm. The stress σ g1 reached at the end of the first grain,

is given by :

The average stress in this first grain, is :


The same calculation for the three other grains, gives :

The distance δ at which the stress droop reached 10%, is :

=4dg i.e. four grains

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c- Calculation of the notch root radius ρ for Kt=5, when the stress

drop is equal 5% at a distance δ=50µm, from the root of the notch

along the x-axis :

If δ=100mm, then the notch root radius ρ is :


d- Calculation of the stress concentration factor Kt, for a notch with

root radius ρ=2m, knowing that the stress drop is 2,5% at a distance

δ=20µm from the notch root along the axis.

e- Calculation of the minimum and maximum distances where the

stress drop is 5% for a notch with root radius ρ=1mm.

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Stress gradient along the edge of a notch

When structures are designed with notches, fatigue crack initiation usually occurs

at the root of the notch.

The surface state (surface roughness, presence or absence of scratches which

amplify the concentration of the stress ...) has a strong influence on this initiation.

Also, it is interesting to consider how the stresses vary along the edge of the notch.

Changes in stress at the edge of a circular hole are illustrated in figure 8, where the

main isostresses lines (lines of equal main stress) are determined by finite element

calculations : these lines are at levels of 90%, 80% and 50% of the maximum

stress3.

Figure 8 : (a) Variation of the stresses

at the edges of a circular hole

determined by finite element

calculations (b) Finite element mesh



These lines of equal main stresses, obtained thanks to the

development of computing resources at the beginning of the 80's,

confirm qualitatively measurements of stresses by the method of

photo elasticity ten years ago4.

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Large notches have a larger area of material along their edge and

therefore the risk of fatigue crack initiation is higher.

This relationship shows that the stress gradient is inversely

proportional to the notch root radius ρ.

It is therefore necessary to examine the scale effects on the stresses

concentration in the vicinity of a notch.

maxmax

is the maximum stress reached12 where

at the root of the notch

y nom

t

t

dK

dx K

σ σ σ σρ

= − + =


Influence of geometry and loading

on the stress concentration factor Kt

The theoretical determination of the stress concentration factor Kt (as

developed in the preceding paragraph) considers that the size of the

notch are very small compared to the dimensions of the structure.

Only the notch dimensions (a and b for an elliptical hole) involved

in the theoretical approach.

Let us now consider a tensile-

stressed plate, of width W and

length L, with a circular hole of

diameter D in its centre.

The figure against shows two

plates with this loading

configuration, of different sizes,

but having the same D/W and

D/L ratios.

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The SCF Kt is a dimensionless parameter. So, it depends only on

geometric ratios.

However, the plate 2 has a greater notch area heavily loaded and

therefore the probability of fatigue crack initiation is higher.

This observation makes it possible to apprehend the effects of

scale.

If all the dimensions of the

plate 2 are double those of

the plate 1 (figure against),

Kt factor and the maximum

stress reached at the root of

the notch are the same.


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Several specialized manuals provide the Kt values for a large number

of geometries and loading configurations4,5.

The figure against shows the

change in Kt factor for a plaque

of finite dimensions, edge

notched or with a circular hole in

its centre.

The variation of Kt factor for the

plate pierced in its centre, is given

by Heywood6 relation.


* Practical formulas for calculating Kt factor

Neuber7 formula are widely used to calculate the stress concentration

factor in structural elements such as plates or shafts with grooves or

shoulders.

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2 2

11

1 1

t

p q

K

aK bK

= +

+

where 1 12

p

dK

r= + −

1and

2qK

r

D d

=

−

Neuber formulas

The coefficients a and b in the Neuber formulas have different values

according to the geometry of the structural element and the type of

loading.

In shaft with semi circular groove,

loaded in tension a=1,197 b=1,871

for bending load a=0,715 b=2

loaded in torsion a=0,365 b=1


In shaft with a single shoulder,


for bending load a=0,541 b=0,843

loaded in torsion a=0,263 b=0,843

In shaft with a double shoulder,

If L > 2d, a and b have the same values

that in shaft with single shoulder

Else (L < 2d) must take Dequ=d+0,3L

and keep the same values a and b.

1 12

p

dK

r= + −

1

2qK

r

D d

=

−

2 2

11

1 1

t

p q

K

aK bK

= +

+

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The figure above shows, for given ration d/D, the changes in SCF Kt

for a shaft with shoulder loaded in tension or in bending : Neuber

formulas are used to calculate Kt factor as a function of r/D ratio.

Tension

Bending


A large connecting radius decreases the value of Kt factor, but in

practice it is not always possible to increase this radius. Also, we use

technological solutions such as the one shown in the figure below.

Solution to reduce

the Kt factor

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Notch overlay - Suppose that within a Kt1 factor notch is a Kt2

factor notch ; this schematic below is to be avoided in practice.

The maximum stress at point A located at the bottom of the small

hole is given by :

Where Kt=Kt1Kt2


let's now examine the pin-jointing of a hook with a clevis

(this case is often encountered in practice)

In section BB , the SCF Kt is high at point A.

Perforated pate

Hook

Pressure

distribution

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A

It is necessary to lubricate at point B and not at point A

where the stress concentration is high


A shouldered shaft of large diameter D=100mm, small diameter

d=64mm and fillet radius r=5mm, is subjected to combined tensile

force (F) , bending moment (Mf) and torque (Mt).

The shaft is steel yield strength σE =300MPa.

1- Determine the SCF Kt for each loading, using the Neuber

formulas.

2- Calculate the equivalent stress according to the Von Mises

criterion, for F= 50kN, Mf=100m.daN et Mt=500m.daN.

Does the shaft becomes plastic?

Tutorial 4 :Stress concentration in shaft

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3- Tensile and torsional loads are fixed to the previous values

(F=50kN and Mt=500m.daN), what is the value of bending moment

(Mf) is not exceeded to avoid the yielding of the shaft?

4- The radius of fillet is reduced to r=2mm. Does the shaft yielded if

the same loads are applied as in the question 2 (F=50kN,

Mf =100m.daN and Mt=500m.daN)?

Tensile and bending loads are fixed, what is the value of torque is

not exceeded to avoid the yielding of the shaft?


The shouldered shaft is replaced by a grooved shaft of large diameter

D=100mm, small diameter d=64mm and fillet radius r=5mm.

This shaft is submitted to the same loads (F= 50kN, Mf=100m.daN

et Mt=500m.daN). Same questions 1, 2 and 3 as in the previous case.

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1- Calculation of stress concentration factors

2 2

11

1 1

t

p q

K

aK bK

= +

+

where 1 12

p

dK

r= + −

1and

2qK

r

D d

=

−

Neuber formulas

In shaft with a single shoulder,


for bending load a=0,541 b=0,843

for torsional loading a=0,263 b=0,843

Tensile 2,10 Bending 1,80 Torque 1,44F T

t t tK K K= = =


Tensile

Bending

Torque

2- The equivalent stress according to the Von Mises criterion is :

The shaft does not yield, it remains elastic

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3- Tensile and torsional loads are fixed to the previous values

(F=50kN and Mt=500m.daN), what is the value of bending

moment (Mf) is not exceeded to avoid the yielding of the shaft?


4- The radius of fillet is reduced to r=2mm. Does the shaft

yielded if the same loads are applied as in the question 2

(F=50kN,

Mf =100m.daN and Mt=500m.daN)?

Tensile and bending loads are fixed, what is the value of torque

is not exceeded to avoid the yielding of the shaft?

Tensile Bending Torque

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2 2

11

1 1

t

p q

K

aK bK

= +

+

where 1 12

p

dK

r= + −

1and

2qK

r

D d

=

−

Tensile

Bending

Torque

The shouldered shaft is replaced by a grooved shaft of large diameter

D=100mm, small diameter d=64mm and fillet radius r=5mm.

This shaft is submitted to the same loads (F= 50kN, Mf=100m.daN

et Mt=500m.daN). Same questions 1, 2 and 3 as in the previous case.

Tensile 2,78 Bending 2,17 Torque 1,60F T

t t tK K K= = =

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Tensile

Bending

Torque


Influence of residual stresses on the stress

concentration factor Kt

- Residual stresses (or internal stresses) refer to a distribution of stresses following

inhomogeneous plastic deformation of the material.

- Residual stresses balance each other necessarily.

Consider the following distribution of residual stresses in a plate of thickness t :

-

-

+

+

In the absence of external loading, the balance

of forces and moments, leads to :

/ 2

/ 2

0

t

x

t

y dyσ−

=/ 2

/ 2

0

t

x

t

dyσ−

=

ext

aσext

moyσ →

ext

a aσ σ=

ext

moy moy resσ σ σ= +

If an external cyclic

loading (σaext , σmoy

ext )

is applied, we have :

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The residual stresses follow a plastic deformation that can have various origins :

- inhomogeneous plastic deformation near a notch ;

- Manufacturing processes that induce residual stresses (forging, stamping, deep

drawing …) ;

- Heat treatment ;

- Assembly of the components of a structure ….

Consider for example, a plate with a circular hole on its centre, loaded in

tension at σmax

> σE

+

-+

Residual stresses

after unloading

max nomtKσ σ<

• During unloading, the elastic part of platewill compress the plastic zone formed atthe root of the hole

• The residual stresses thus generated willbalance out


Plastic zone

Plastic partsElastic zone

Residual stresses


Calculation of residual stresses – Neuber assumption

When the maximum stresses remain less

than yield stress, the strains are proportional

at these stresses and then we have :

max t nomKσ σ= maxmax

t nomt nom

KK

E E

σ σε ε= = =

If the maximum stresses exceed the yield

stress, we have as shown in the figure opposite:

max t nomKσ σ<max t nom

Kε ε>

maxt

nom

K Kσσσ

= < maxt

nom

K Kεεε

= >

• Neuber hypothesis

(or Neuber rule)

2

tK K Kσ ε = ( )2

max max

t nomK

E

σσ ε =

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( )2

max max

t nomK

E

σσ ε =

The right term in this relationship, is

known.

We then determine σmax, graphically from

the constitutive law σ = σ(ε) .

t nomK σ

maxσ

resσ

( )σ σ ε=

( )2

t nomK

E

σσ ε⋅ =

maxε ε

At the intersection point A of the two

curves, σ = σ(ε) and σ .ε , σmax and εmax

satisfy the constitutive law.

In absence of plasticity, the maximum

stress is reached at point B.

The difference between Kt σnom (point B)

and σmax (point A) gives the residual

stress, which is a compressive stress.

res A B A t nomKσ σ σ σ σ= − = −

Graphic determination procedure

of σmax and εmax.


Tutorial 7 : Calculation of residual stresses

A steel plate of Young’s modulus E=200GPa and yield stress σE = 300MPa,

has a bilinear elastic plastic behaviour with a plastic Young’s modulus

Ep=E//20

Ai

1- Calculate the residual stress, and the

maximum stress reached at the root of a

notch of SCF Kt=2.5 ; the plate is loaded

at σnom = 200MPa and then unloaded.

Deduce the factors Kσ and Kε.

2- Same question as that at 1- with Kt=2

and Kt=3.

Constituve law ( )E p EEσ σ ε ε− = −

300At point 300 ( )

20A A

EA

Eσ ε= + −

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( ) ( )2 2

300Neuber hypothesis 300 ( )

20

t nom t nom

A A A

A

K KE

E E E

σ σσ ε σ

σ⋅ = = + −

2 285 12500 0 323,62A A A

MPaσ σ σ− − = =

1,62A

nom

Kσσ

σ= =

2

3,86tKK

Kσε = =

176, 4res A t nom res

K MPaσ σ σ σ= − = −

300At point 300 ( )

20A A

EA

Eσ ε= + −

Ai


300At point 300 ( )

20A A

EA

Eσ ε= + −

( )2

300NH 300 ( )

20

t nom

A

A

KE

E E

σσ

σ = + −

2a) 2 and 200t nomK MPaσ= =1,55A

nom

Kσσ

σ= =

2

2,57tKK

Kσε = =

2 285 8000 0 310,74A A A


89, 26res A t nom resK MPaσ σ σ σ= − = −

2b) 3 et 200t nomK MPaσ= =2 285 18000 0 338, 22A A A


261,78res A t nom res

K MPaσ σ σ σ= − = −

1,69A

nom

Kσσ

σ= =

2

5,32tKK

Kσε = =

Ai

FMDF cours 181025 - lem3.univ-lorraine.fr

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Transcript of FMDF cours 181025 - lem3.univ-lorraine.fr