Stress - Strain

Mechanical properties

tensile, compression, torsion test

3

1. Initial 2. Small load 3. Unload

Elastic = reversible

The volume is not

constant during elastic deformation

F

ΔL

Bonds became

longerReturn to the initial

shape

F

L

Linear elastic

Non-linearelastic

Elastic deformation

4

Plastic = remaining

The volume is constant

during plastic deformation

Plastic deformation

F

Lelastic+plasticLplastic

F

ΔL

Lplastic

Linear

elastic

ΔLelastic

1. Initial 2. load 3. Unload

Bonds became

longer,

planes slide

Planes remain

in the new

position

5

In elastic state

= E

(Hooke’s law)

0

0

l

ll

0S

F

S

F

Stress

Strain

l0

Δl/2

F

F

l l0 l

F

F

S0

S S0S

Δl/2

Δl/2

Δl/2

tension compression

Tensile and compression

6

0S

F

S

F r

I

M

p

S0 F

F

Simple shear Torsion

M

r

G

In elastic state

Shear

7

F, N

II.I.

Fu

Fm

III.

FeL

L, mm

FeH

tensile diagram

I. Elastic deformationII. Uniform plastic deformationIII. Necking (non-uniform plastic deformation)

L=L-L0

Tensile test

8

0

2,0

2,0

00

0

,

S

FR

S

FR

S

FR

S

FR

p

p

eLeL

eHeH

ee

%100

S

SSZ

0

u0

0S

FR m

m

Stress Deformation

Yield stress (MPa)

Tensile strength (MPa) %100L

LLA

0

0u

Contraction

Elongation(engineering strain at fracture)

Mechanical quantities

9

Engineering stress, strain True stress strain

10

0

0

S

S

l

ll

S

S

l

l

0

0

ln

ln

Strain

StressF

S

energy per unit

volume (J/cm3)

u

M

c

0

W d

u

c

0

W d

M

0

F

S

Mechanical quantities

10

z

r

z

zzrr

g

zz

d

d

d

d

Rr

rr

0

min

min

0

min

22

min

ln

ln2

21ln1

strain equivalent

,e

stressquivalent

Stress state at contraction

11

u

u

m

e

M M

0F S S 1

ln 1

Strain

Str

ess

M

Stress-strain curves

true stress-strain

engineering stress-strain

12

Elastic modulus:

E (Young-modulus)

Hooke’s law

σ = E ε

Poisson coefficient, ν:

metals: ν ~ 0,33ceramics: ν ~ 0,25

polymers: ν ~ 0,40

units:

E: (GPa) or (MPa)ν: no dimension

σ

Linear elastic

1

E

ε

ε r

ε

1-ν

F

FUniaxial load

r

r - radial strain

Eceramics > Emetal >> Epolymer

Linear elastic properties

13

Shear modulus, G

τ = G γ

Bulk modulus, K

G

E

2(1 )

)21(3

EK

Torsion

test

under hydrostatic pressure:

initial volume: Vo

volume change: ΔV

p = -KΔVVo

τ

1

G

γ

p

ΔV

1-K

Vo

p

p p

M

M

p

Hooke’s law

Linear elastic properties

14

Engineering strain

En

gin

ee

ring

str

ess

brittle - if the remaining (plastic) deformation ≈ 0ductile - if the remaining (plastic) deformation is significant

brittle

ductile

Ductile – brittle behavior

Deformation - strain

continuum mechanical description

16

u

X

3

0

1

X

1,x

,x

X ,x

3

P

Q

22

X

x

,x(x ,x )21 3(X

P,X ),X

Q

1 2 30

0

dsdS

The motion of the body is described in coordinate system. Thepoints, line and volume elements of the body are described in thissystem during the deformation

Motion of a body

17

Lagrangian description: quantities (temperature, velocity, stress,etc.) are bind to material points.

Euler description: Quantities are described in a specific point ofcoordinate space

t,xXXt,x,x,xXX

3,2,1i,t,Xxx,t,X,X,Xxx

321ii

321ii

Mapping function

Inverse mapping function

Displacement

Xt,Xxt,Xu

Motion of a body

18

Strain tensors characterize the deformation of the material:

E - Lagrangian strain tensor e - Eulerian strain tensor

Strain tensors

𝑭 =

𝜕𝑥1𝜕𝑋1

𝜕𝑥1𝜕𝑋2

𝜕𝑥1𝜕𝑋3

𝜕𝑥2𝜕𝑋1

𝜕𝑥2𝜕𝑋2

𝜕𝑥2𝜕𝑋3

𝜕𝑥3𝜕𝑋1

𝜕𝑥3𝜕𝑋2

𝜕𝑥3𝜕𝑋3

Deformation gradient F

𝑰 =1 0 00 1 00 0 1

𝑬 =1

2(𝑭𝑇𝑭 − 𝑰) 𝒆 =

1

2(𝑰 − 𝑭−𝑇𝑭−1)

19

dS ds

Stretch ratio, engineering strain

dS

ds

Stretch ratio

Engineering strain

11dS

ds

dS

dSds

20An (small) sphere in the environment of point P0 at t=0 transforms to an ellipsoid during the deformation.

1 1,x X

2 2,x X

3 3,x X

Logarithmic (true) strain

21

dS – sphere diameter, dsi – axes of ellipsoid. ds1>ds2>ds3

3

2

1

333

222

111

00

00

00

lnln,lnln,lnln

dS

ds

dS

ds

dS

ds

lll Ee 21ln2

121ln

2

1

Where el and El are the eigenvalues of Eulerian and Lagrangian

strain respectively.

Logarithmic (true) strain

22

31 21 2 3

1 2 3

31 2

1 1 2 2 3 3

, , , 0 ha

1 1 1, , , 0 ha

i

k

i

k

x xx xa a a i k

X X X X

X XX Xi k

x a x a x a x

Task 1.

A square based prism with size of A x B x H is deformed to the size of a x b x h.

Introducing the ratios a1=a/A, a2=b/B, a3=h/H.

The mapping function: x1= a1X1, x2= a2X2, x3= a3X3

Calculate the Eulerian, Langrangian and logarithmic strain.

if

if

23

2 21 1

2

2 2

2

2

32

3

1

2

3

1 11 1 0 0

1 0 0 22

1 1 10 1 0 , 0 1 0

2 2

11 10 0 1

0 0 122

ln 0 0

0 ln 0

0 0 ln

ij ij

ij

a a

E a ea

a

a

a

a

a

Task 1.

24

Equivalent strain, stain rate

Tensor quantity characterized with a scalar value.

2

23

2

13

2

12

2

3322

2

3311

2

2211 63

2

Strain rate

i

j

j

iij

x

v

x

v

2

1

2

23

2

13

2

12

2

3322

2

3311

2

2211 63

2

t

t0

dt

From velocity field:

Equivalent strain rate

Equivalent strain:

25

dS dS dS ds ds ds1 2 3 1 2 3 1 2 3 0

,0vdiv

0

1E21E21E21

1e21e21e21

332211

321

321

Volume constancy

1 1,x X

2 2,x X

3 3,x X

Stress

continuum mechanical description

27

fF

1

0lim

V V

tF

lim

A A0

f

t

Volume force density

Surface force density

x1

x2

x3

ΔAΔV

External forces act on a body with V0

volume and A0 surface, therefor it

undergoes deformation; Volume and

surface changes to V and A

respectively. The external forces canbe volume and surface forces.

Volume and surface forces

28

V

VdA

t

nII

I

Cut the body into two and apply surface forces on the

cut surface to have the same deformation.

1 11 1 12 2 13 3

2 21 1 22 2 23 3

3 31 1 32 2 33 3

, T

i ij jt n

t n n n

t n n n

t n n n

t σ n

σ – Cauchy stress tensor

333231

232221

131211

σ

Stress tensor

29

333231

232221

131211

σ

Diagonal elements: normal stress

Off-diagonal elements : shear stress

Normal stress: positive if tensile

negative if compressive

1x

2x11

31

21

2212

32

33

13 23

Stress tensor

30

Equivalent stress – von Mises or Huber-Mises-Henky

Equivalent stress

2

23

2

13

2

12

2

3322

2

3311

2

2211 62

1

Tensor quantity characterized with a scalar value.

31

L1 L

2

S1 S

2

S

3

T1 T3T

2T3

3

2

1

00

00

00

Stress state

Constitutive law

Stress - strain relation

33

Elastic behavior of the material

Hooke’s law

ν – Poisson coefficient

34

Plastic behavior of the material

Condition of plastic flow: f < 0 elastic state

f = 0 plastic deformation - yield

von Mises yield criterion

𝑓 = 𝜎𝑚𝑖𝑠𝑒𝑠− 𝜎𝑦𝑖𝑒𝑙𝑑 = 0

𝜎𝑚𝑖𝑠𝑒𝑠 =1

2(𝜎11 − 𝜎22)2+(𝜎33 − 𝜎22)2+(𝜎11 − 𝜎33)2+6(𝜎12

2 + 𝜎122 + 𝜎12

2 )

35

Plastic behavior of the material

Hardening

𝜎𝑦𝑖𝑒𝑙𝑑 = 𝜎𝑦𝑖𝑒𝑙𝑑(𝜑𝑝, 𝜑𝑝 , 𝑇)

𝜎𝑦𝑖𝑒𝑙𝑑

𝜑𝑝

𝜎𝑦0

T ↑

36

Plastic behavior of the material

Flow curves

𝜎𝑦𝑖𝑒𝑙𝑑

𝜑𝑝

𝜎𝑦0

𝜎𝑦𝑖𝑒𝑙𝑑

𝜎𝑦𝑖𝑒𝑙𝑑

𝜎𝑦𝑖𝑒𝑙𝑑

𝜎𝑦𝑖𝑒𝑙𝑑

for cold forming

for hot forming

𝜎𝑦𝑖𝑒𝑙𝑑

𝜑𝑝

𝜎𝑦0

T ↑𝜎𝑦𝑖𝑒𝑙𝑑 𝜎𝑦0