Nombre d’exemplairesContraintes en MPa

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Lecture 1 - Static resistance

Recalls: internal forces, stresses, criteria etc.

CDIM Team @GM

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2

Intro – Why do we need to compute mechanics?

• Elements are subjected to mechanical loadings

• Axial load ( tension/compression),

• Radial load ( bending),

• Excentered (non-radial) load ( torsion)

• Elements have non-homogeneous shapes  stress concentration

• Shoulder (with rounds),

• Holes (screw/pins),

• Grooves (for keys),

• Threads

 The elements should ensure two criteria :

• The material should not degrade under the load (static/fluctuating)

• The deflection should not be too large to remain functional axial

radial excentered

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3

•

Forces / moments

• Joints / Wrenches

• Methodology

 Example : crank

•

Internal forces / moments

• Cohesion wrench

•

Stresses / strains

• From Forces/Moments

• Critical section

•

Materials

• Mechanical properties

• Stress criterion

•

Exercice 1a

Lecture 1 – agenda

A C B

Forces / Moments

A C B

Critical section

Design : choice of material, geometry… depending on

variables

0D : Load

@points 0D : Load

@points

1D : Load along axle 1D : Load along axle

2D : Stresses in a section 2D : Stresses

in a section

Maximal stress in section Maximal stress

in section

A C B

? ?

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4

Forces / Moments

• A joint :

• link k between part i and j

• n_c mobilities, n_s forces/moments

 in 3D, n_c+ n_s= 6

• Kinematics :

• Complete joint : NO degree of freedom (DoF) between part i and j : n_c = 0

• Welded joint, glued joint, interference fit etc.

• Partial joint : at least 1 DoF between two parts : n_c ≥ 1

• Revolute joint, sliding joint, cylinder joint etc.

• Forces :

• Perfectly rigid joint : no deformation

• Flexible joint : small deformations, elastic or visco-elastic behavior

liaison k pièce j

pièce i

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5

• Wrenches

• Kinematic wrench (translation / rotation)

• Associated Force wrench (force / moment)

• Associated? where X_ior Ω_i = 0  F_i or M_i ≠ 0, respectively…

… and inversely : where X_ior Ω_i ≠ 0  F_i or M_i = 0

• Joint is assumed perfect (no friction) : 1 DoF  no force/moment along that DoF

{ }

₁ ₂ ₃

1 2 3

1 1

2 2

, , ,

3 3 , , ,

G x x x

N M

R M T M

T M

 

 

=  

 

 

r r r

r r

{ }

₁ ₂ ₃

1 2 3

1 1

2 2

, , ,

3 3 , , ,

G x x x

X

X X

X

Ω

 

 

Ω =  Ω 

 Ω 

 

r r r

r r

1

2 3

⃗ ⃗

⃗

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6

• Loading types

• Quasi-static computation

• If translation / rotation speeds are constant

• If symmetrical parts / no unbalance

 if vibrations or fluctuations occur : need to account for it

• What about gravity?

• It can be neglected if other loads are much larger

• Need to check before neglecting it!

Constant Loads Fluctuating loads

System at rest Case 1 Case 2

System in movement Case 3 Case 4

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7

Forces / Moments : example crank

0. input :

geometry (complete),

external load q

A B

C

D

l l l

4l

2l q

input : l=0.1m, d=20mm, q=5000 N/m

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8

Forces / Moments : example crank

0. input :

external load q

1. unknowns :

reaction wrench in A :

2. equations of equilibrium :

A B

C

D

l l l

4l

2l q

,

input : l=0.1m, d=20mm, q=5000 N/m

0 ! ". $ 0 0

! ". 0

$ 0 0

%&' 0 $ &( ∧ ". * $ " ∗ 2 0

$ 0 0

! " ∗ 11 2 0⁄

3. solution :

0

". 500

0

* !2" !100 . 0

0

" ² ∗ 11 2 275 . 0⁄

0D : Load

@points 0D : Load

@points

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9

• Methodology

Objective: Determine the mechanical actions on each solid of the system

Isolate the solid from the system; represent graphically the solid in the same position as in the system

Identify all the contact zones

Draw a calculation scheme (represent vectors) on the system skeleton

Balance of the formulation (Unknowns / Equations)

Possible resolution ? Isolate other solids

to get more equations

NO Problem solved :

all mechanical actions are determined YES

0D : Load

@points 0D : Load

@points

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10

•

Forces / moments

•

Internal forces / moments

• Cohesion wrench

•

Stresses / strains

•

Materials

•

Exercice 1a

Lecture 1 – plan

A C B

Critical section

variables

0D : Load

@points 0D : Load

@points

in a section

in section

? ?

Forces / Moments

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11

• Cohesion wrench

Internal Forces / Moments

3_⊕→⊖ 7

7 _8,9_:_,9_;_,9_<

⊖ ⊕

⃗

Virtual section cut

= ⊖

⃗_⊕→⊖ ! ⃗_⊖→⊕

⊕→⊖ ! _⊖→⊕

>⃗ >⃗ >⃗

: normal force (tensile force) 7 _, : shear forces

: torsion moment

, : bending moments

?

Global basis

Local basis

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12

Internal Forces / Moments : example crank

0. input :

external load q

reaction wrench in A

A B

C

D

l l l

4l

2l q

input : l=0.1m, d=20mm, q=5000 N/m G

&? ⁸⁸

8

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13

0. input :

external load q

1. unknowns :

reaction wrench in G∈[AB] :

2. equations of equilibrium :

A B

C

D q

input : l=0.1m, d=20mm, q=5000 N/m

3. solution :

G

&? ⁸⁸

8

A/C 7

7 _8,D9_:_,9_;_,9_<_E

/C 0 $ 0

$ 7 0

/C%?' 0 $ ?& ∧ $ A/C%?'

/C%?' * !2"

0

" ² ∗ 11 2⁄ $ ! 8

! 8

∧ 0

". 500 0 $

A/C

0

7 !". !500

7 0

A/C%?' * 2" 100 . 0

0

!" ∗ 11 2 $⁄ ₈. ".

Calculation details :

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14

0. input :

external load q

A B

C

D q

input : l=0.1m, d=20mm, q=5000 N/m G

&? ⁸⁸

8

A/C%?' * 2" 100 . 0

0

!" ∗ 11 2 $⁄ ₈. ".

A/C

0

7 !". !500

7 0

3. solution :

0

7

0 7

0

0 0 0

4l

4l -ql

2ql²

-11/2.ql²

-3/2.ql²

A B

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15

•

Forces / moments

•

Internal forces / moments

•

Stresses / strains

•

Materials

•

Exercice 1a

A C B

Critical section

variables

0D : Load

@points 0D : Load

@points

in a section

Max stresses in section Max stresses

in section

? ?

Forces / Moments

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16

Stresses from Force Wrench

7 7

_8,9_:_,9_;_,9_<

F , F , F

GF HI

J

7 GF HI

J

7 GF HI

J

G F ! F HI

J

G F HI

J

! G F HI

J

F I $ K ! K F 7

I ! K F 7

I $ K

2D : Stresses in a section 2D : Stresses in

a section

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17

Stresses in critical section : example crank

A/C

0 100 . 0

7 !500 0

7 0 ! 275 . 0 _,9

:,9_;,9_<

A

F I $ K ! K 275

7,854. 10^NO

F 7

I ! K !500

3,14. 10^NQ$ 100 1,571. 10^NR

F 7

I $ K !100

1,571. 10^NR I SH

4 31400 3,14. 10^NQ0

K SH^Q

32 1,571. 10^Q00^Q 1,571. 10^NR0^Q K , SH^Q

64 7,854. 10 00^Q 7,854. 10^NO0^Q

M

F F

F

&

F 275

7,854. 10^NO0,01 3,5. 10^RUV

F !500

3,14. 10^NQ 1,59. 10^XUV

F 100

1,571. 10^NR0.01 6,36. 10^YUV Input :

& 0 H/20

Solution for the critical point M in the critical section (in A):

Solution for the critical section (in A):

Calculations:

2D : Stresses in a section 2D : Stresses in

a section

in section

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18

•

Forces / moments

•

Internal forces / moments

•

Stresses / strains

•

Materials

•

Exercice 1a

Lecture 1 – agenda

A C B

Critical section

variables

0D : Load

@points 0D : Load

@points

in a section

in section

? ?

Forces / Moments

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19

Materials : classification (David Duresseix)

Métaux et Alliages

Polymères

Céramiques, Verres et

Produits naturels Composites

Composites avec renfort à base

de fils métalliques (pneumatiques, ..)

Composites avec renfort à base

de fibres et poudres (outils carbures, ..)

Polymères chargés avec des poudres ou

des fibres de verre, carbone ou autres (courtes ou longues)

Metals

&

Alloys

Polymers

Ceramic Glass Organic

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Materials : mechanical properties

• Elasticity, plasticity : Young modulus, yield and ultimate stress

0

σ = E . ε σ_E

σ_R( R_m)

Limiteσ_E^{en MPa} ε = ∆l

l

47,5 50 52,5

25 50

Nombre d’exemplaires

Déformation

Contraintes en MPa

1 E

Caractéristiques sur une éprouvette

Caractéristiques sur un lot de pièces

50

Characterisics from one test sample

Characterisics from several test samples

Stress (MPa) Numberof samples

Strain

Statistical spread is generally small on mechanical properties of metallic alloys

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Materials : mechanical properties

• Resistance for the bulk material

• Criteria on deviatoric stresses

• Von Mises Z_{[\_}_{^_`} Zâ$ bcâ d ^Z_fê

• TrescaZ_{[\_}_{^_`} Zâ $ gcâ d ^Z_fê

• Resistance of the surface material

• Criteria on hydrostatic stress p_MAX<p_ADM

 Needed material properties

• Yield stress (limit elastic stress): Z_h,e

• Ultimate stress (strength): Z_i,j

• Elasticity Modulus: E (or Shear modulus: G)

• Poisson coefficient: ν

• Admissible pressure : p_adm

 Plasticity should be avoided

 Plasticity is due to dislocation motion

 Dislocation are driven by shear stresses

 Contact lead to high pressures

 Damage is engendered

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Materials : example crank

A M

F F

F

F 275

7,854. 10^NO0,01 3,5. 10^RUV

F !500

3,14. 10^NQ 1,59. 10^XUV

F !100

1,571. 10^NR0.01 6,36. 10^YUV

F_kl F $ 3m

& 0 H/20

Solution for the critical point M in the critical section (in A):

m= F $ F

Unknown: F F

F_kl F $ 3%F $ F ' 3,66. 10^RUV 366 UV

Solution for von Mises (at the critical point in the critical section):

The design/ dimensioning aims at checking:

• If known dimensions  verification

• If not  calculation of the minimum dimension α Is a safety factor depending on:

- Operating conditions (intermittent, continuous), regularity - Industrial filed: automotive, aeronautics, etc…

Z_n^ d Z_e f

in section

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Influence of element shape : stress concentration factor

• Some typical cases

0,1 0,5 0,8

2,2 2,6 3,0

d / w

D d F

F

max ) 0

σ(

σMax =Kt

Facteur de concentration de contraintes Kt

0,1 0,5 0,8

2,2 2,6 3,0

d / w

D d F

F _D _d F

F

max ) 0

σ(

σMax =Kt

Facteur de concentration de contraintes Kt

1,0

0,1 0,2 0,3

1,4 1,8 2,2 2,6 3,0

Facteur de concentration de contraintes

K_t

D / d = 3

D / d = 1,5 D / d = 1,1

D / d = 1,05 D / d = 1,02

r / d r

D d

M_f M_f

max ) 0

σ(

σMax =Kt

1,0

0,1 0,2 0,3

1,4 1,8 2,2 2,6 3,0

Facteur de concentration de contraintes

K_t

D / d = 3

D / d = 1,5 D / d = 1,1

D / d = 1,05 D / d = 1,02

r / d r

D d

M_f M_f

r

D d

M_f M_f

max ) 0

σ(

σMax =Kt

1,0

0,1 0,2 0,3

1,4 1,8 2,2 2,6 3,0

Facteur de concentration de contraintes K_t

r / d r

D d

F F

max ) 0

σ(

σMax =Kt

D / d = 1,5 D / d = 1,1

D / d = 1,05 D / d = 1,02

1,0

0,1 0,2 0,3

1,4 1,8 2,2 2,6 3,0

r / d r

D d

F F

r

D d

F F

max ) 0

σ(

σMax =Kt

D / d = 1,5 D / d = 1,1

D / d = 1,05

D / d = 1,02 ^1,0

0,1 0,2 0,3

1,4 1,8 2,2 2,6 3,0

r / d r

D d

F F

max ) 0

σ(

σMax = Kt

D / d = 1,5 D / d = 1,1

D / d = 1,05 D / d = 1,02 1,0

0,1 0,2 0,3

1,4 1,8 2,2 2,6 3,0

r / d r

D d

F F

r

D d

F F

max ) 0

σ(

σMax = Kt

D / d = 1,5 D / d = 1,1

D / d = 1,05 D / d = 1,02

F

_{l o}

p . F d F

_q

r

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Example : Shaft of a starter/alternator

A C B

D

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25

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26

t T

x

y 53° 53°

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27

• Problem definition

• Centers of bearings

• Center of pulley

• Center of the wire winding

• Data

• Unknown : F

_B

, F

_D

• Derive all calculation as a function of the shaft diameter

L

x z

F_A d

e

C_m

A

D B

e/2

C y

-C_m

F_B F_D

L e d F_A C_m

150 mm 90mm 25mm -2000N 80 N.m

σr σD σE

700 MPa 300 MPa 600 MPa

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28

•

Forces / moments

•

Internal forces / moments

•

Stresses / strains

•

Materials

•

Exercice 1a

A C B

Critical section

variables

0D : Load

@points 0D : Load

@points

in a section

in section

? ?

Forces / Moments

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29

Second moment of area (J. Colmars, GM)

• General definition • Examples

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30

Example : Shaft of a starter / alternator

• Calculation of all external actions

• Internal forces/moment, cohesion wrench

0D : Load

@points 0D : Load

@points

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Example : Shaft of a starter / alternator

• Stresses in section B+

^F _{I $ K} _{! K}

F 7

I ! K

F 7

I $ K

B M

F F

F s 0

H/20

,D9_:,9_;,9_<E

F F F

Solution for the critical point in the critical section:

Solution for the critical section :

F_kl F $ 3m

F_kl

=

m= F $ F F F

Solution for von Mises :

in a section

in section

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32

Example : Shaft of an starter / alternator

• Resistance criteria as a function of shaft diameter:

• One neglects the shear force 

• Expressed as a condition on d

• Using the provided material, compute the condition on d for:

Fkl d

d

H t

α=1  d=

α=2 d=

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33

Example : Shaft of a starter / alternator (correction)

• Calculation of all external actions

• Internal forces/moment, cohesion wrench

F

_B

$F

_D

$F

_A

0 et F

_A

.%d$e'$F

_B

.e 0 d’où : F

_B

-F

_A

.%d$e'/e et F

_D

F

_A

.d/e

•‚ ƒ ^‚ „ƒ•. …

†a !aƒƒƒ• ^a ƒ

†_b ƒ ^_b !‡ƒ•. … _ˆ^‰_,Š

‚,Š_a,Š_b 0D : Load

@points 0D : Load

@points

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34

Example : Shaft of a starter / alternator (correction)

• Stresses in section B+

F I $ K ! K 50 ∗ 64 SH^Q

F 7

I ! K !2000

SH !80 ∗ 32 SH^Q

F 7

I $ K 80 ∗ 32 SH^Q

B M

F F

F

s

,D9_:,9_;,9_<E

F 50 ∗ 64 SH^Q H

2 50 ∗ 32 SH

F !2000

SH F 80 ∗ 16

SH

F_kl F $ 3m

m= F $ F F F

s 0

H/20

,D9_:,9_;,9_<E

Solution for the critical point in the critical section:

Solution for the critical section:

Solution for von Mises :

Fkl F $ 3%F $ F '

Fkl 50 ∗ 32

SH $ 3% !2000

SH $ 80 ∗ 16

SH '

in a section

in section

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35

Example : Shaft of an starter / alternator (correction)

• Resistance criteria

• One neglects the shear force 

• Genralization, case of a transmission shaft, in the critical section:

F_kl 50 ∗ 32

SH $ 3% !2000

SH $ 5 ∗ 16

SH ' dFq

f

Fkl 50 ∗ 32

SH $ 3% 5 ∗ 16

SH ' dFq

f

H t f Fq

50 ∗ 32

S $ 3 80 ∗ 16 S

H t f Fq

, ∗ 32

S $ 3 ∗ 16

S

, ‹

F

_kl

F $ 3m

α=1  d=11,3 mm α=2 d= 14,3 mm

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36

Exercise 1: Bar bending torsion (correction)

• External action 0. input :

geometry,

external load F

1. unknowns :

reaction wrench in 0 :

2. equations of equilibrium :

F 0

O’

C l

a

,

0 ! $ 0 0

! 0

$ 0 0

%0' 0 $ =Œ ∧ %! ' ^• $ V 0

• $ 0 0

•! 0

3. solution :

•

• 0

•

!V0

0D : Load

@points 0D : Load

@points

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37

F 0

O’

C l

a Exercise 1: Bar bending torsion (correction)

• Cohesion wrench 0. input :

external load

reaction load in O

1. unknowns :

reaction wrench in G :

2. equations of equilibrium : 3. solution :

G

=? ⁸8

8

A/C 7

7 _8,D9_:_,9_;_,9_<_E

/C 0 ^• $ 0

•$ 7 0

• $ 7 0

/C%?' 0 • $ ?= ∧ • $ A/C%?'

/C%?' ^• !V

• 0

• $ ! 8

! 8

! ₈∧ ^• 0

•

• 0$

A/C

0

7 !

7 0

A/C%?' V

0

% ! ₈'

Calculation details :

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38

• Stresses in section O+

F I $ K ! K K

F 7

I ! K !

I !V K

F 7

I $ K V

K

O M

F F

F

=

,D9_:,9_;,9_<E

I SH

4 1,26. 10^N 0

Solution for A and B in the critical section:

F K H

2

F !

I

F V

K H 2

=& 0 H/20

Solution for the critical section:

Calculation :

in a section

in section

K SH^Q

32 2,51. 10^NY K K 2

F 0

F ^NŽ_J -_•^•Ž

<<

‘

F

=s 0

H/20

F 72 UV F 1,2 UV F 24 UV

=& 0 H/20

F 0

F 49 UV

=s 0 F H/20

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39

• Design with plain tube

O A

F F

F

F_kl F $ 3m

m= F $ F F F

Von mises :

F_kl F $ 3%F $ F ' Fkl & 83 UV

I SH

4 1,26. 10^N 0 Stresses values :

K SH^Q

32 2,51. 10^NY K K 2 F 72 UV

F 1,2 UV F 24 UV

=& 0 H/20

F 0

F 49 UV

=s 0 F H/20

F_kl s 85 UV

 α=3.93

 α=3.83

Discussion :

B

F F

F

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40

• Design with hollow tube

O A

F F

F

Von mises : Fkl &

I’ S%H !H_“' 4

Stresses values as a function of d_i (neglectingT₂) : K’ S%H^Q!H_“^Q'

32 K’ K’

2

F_kl s ?

 d_i=

Discussion :

B

F F

F

F K’ H F 0 2

F V

K’ H 2

=& 0 H/20

F 0

F ^NŽ_J - ^•Ž_•’

<<

‘

F

=s 0

H/20

with α=1